Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Do photons experience every moment in time and position in space simultaneously? Would it be more correct to say that a photon, traveling at the speed of light, would experience all points in time simultaneously, and therefore be everywhere at once? It might be just our perception that seems to portray them as moving?
| A photon traveling at speed of light has a lightlike worldline. It has one place of emission and one place of absorption. The spacetime interval between both points is empty (=0), that means that no spacetime is between them. That means, if a photon would experience something, it would experience both points as simultaneous. But there is no reference frame of photons, photons don't experience anything.
Concerning the points of the worldline between the place of emission and the place of absorption, they correspond to the empty spacetime interval between the point of emission and the point of absorption. For this reason, from the hypothetical point of view of the photon only the point of emission and the point of absorption would exist.
This empty spacetime interval corresponds only to the hypothetical point of view of the photon. In contrast, all observers perceive the light wave as moving at c. By consequence, and as you are suggesting, the movement at c is not more and not less than an observation. This fact is also taken into account by the formulation of the second postulate of special relativity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What type of matrices do we mean by $I_{D2}$ and what is Chandrasekhar decomposition? Suppose that $A_{2\times 2}$ is a hermitian matrix, so it has real eigenvalues $\lambda_1$ and $\lambda_2$ and corresponding orthonormal eigenvectors $\underline u_1$ and $\underline u_2$ and we know it can be written as the sum of two $rank-1$ hermitian matrices: $\text{T is the symbol for transpose and * for complex conjugate}$
$$A=\lambda_1\underline u_1\otimes\underline u_1+\lambda_2\underline u_2\otimes\underline u_2=\lambda_1\underline u_1\underline u_1^{*T}+\lambda_2\underline u_2\underline u_2^{*T}\qquad $$
Both $\underline u_1\otimes \underline u_1$ and $\underline u_2\otimes \underline u_2$ are hermitian so $(\underline u_1\otimes \underline u_1+ \underline u_2\otimes \underline u_2)$ is hermitian. In a book on page 50, it says that:
As the two unit orthogonal eigenvectors verify $\underline u_1\underline u_1^{*T}+\underline u_2 \underline u_2^{*T}=I_{D2}$, it
follows the Chandrasekhar decomposition of the wave given by
$$J=(\lambda_1-\lambda_2)\underline u_1\underline u_1^{*T}+\lambda_2I_{D2}=J_{CP}+J_{CD}$$
My question is what do we mean by $I_{D2}$ matrix?
Also what is Chandrasekhar decomposition? Can you introduce me online resources to study?
Or can you guide me to the Chanrasekhar's original paper in which he introduced this decomposition?
| From the general properties of 2x2 matrices, if ${\underline u}_1$, ${\underline u}_2$ are the eigenvectors of $A_{2x2}$, then
$$
{\underline u}_1 {\underline u}^{*T}_1 + {\underline u}_2 {\underline u}^{*T}_2 = I_{2x2}
$$
As for the Chandrasekhar decomposition, perhaps the explanation on pgs.269-271 in "Direct and Inverse Methods in Radar Polarimetry" (Google Books link) can help. It seems it is a decomposition of the coherency matrix J into a pure state (perfectly polarized) component, $(\lambda_1 - \lambda_2){\underline u}_1 {\underline u}^{*T}_1$, and a randomly polarized, noise term, $\lambda_2 I_{2x2}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/200984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the time period of an oscillator with varying spring constant? It is well known that the time period of a harmonic oscillator when mass $m$ and spring constant $k$ are constant is $T=2\pi\sqrt{m/k}$.
However, I would be interested to know what the time period is if $k$ is not constant. I have searched hours after hours for right answers from Google and came up with nothing. I am looking for an analytical solution.
| To obtain some kind of practical answer, you have to determine how k varies. For example, if k varies with temperature, I would determine its value at -50, 0, and 50 degrees, then use those values and calculate T (which varies inversely as the square root of k). I would Use more points if a higher accuracy is required. If a formula is required, I would use the "best fit curve" through the points, to generate it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
A heavy rope is attached to one end of a lightweight rope
If one end of a heavy rope is attached to one end of a lightweight rope, a wave can move from the heavy rope into the lighter one.
(a) What happens to the speed of the wave?
(b) What happens to the frequency?
(c) What happens to the wavelength?
My instructor hasn't gone over any of this in class (it's for a reading assignment), so what I've guessed so far just off the equations the book gives.
(a)
$v=\sqrt{\frac{F}{\mu }}$
So, as the mass per unit length ($\mu$) goes down, the velocity will increase
(b) $v=\lambda f$
Now I am unsure. There is no way to tell (from this one equation) whether the wavelength ($\lambda$) will increase, decrease, stay constant. Is it determinable at all?
(c) Same problem as with (b).
|
here is an animated version of image
for more in formation refer to http://www.physicsclassroom.com/class/waves/Lesson-3/Boundary-Behavior
and a good tool for playing with
https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_en.html
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does a moving fan seem transparent? We all know when fan starts moving faster, we cannot see its blades. Why is this?
First I assumed persistence of vision may be the reason. But that can happen with blade also right? Image of blade can remain in our memory and moving fan can appears as a circular plane with blade color.
Why only image of rear side of fan is remaining in our memory?
Note: I tried with fan whose blade area is almost same as that of non blade area
| The eyes are measuring the number of photons of each color that are hitting a given point of the retina – that are coming from some direction.
This is a function of time, $f(t)$, for each point. However, when this function is changing too quickly, the eye can't see the changes. Effectively, the eye may also see the average of $f(t)$ in each period of time which is as short as 1/50 second or so. That's why 24 or 25 or 30 or 50 frames per second are usually enough for a TV screen.
If the fan frequency is at least 1 blade per 1/50 second, which is the same as 10 rotations per second for a 5-blade fan, for example, the following is true:
During 1/50 seconds, each point of the image where fan blade may either be or not be sees a full period, so the perception is no different from the perception in which the color is averaged over those 1/50 seconds. But the averaged color of each point is pretty much the same. It's a weighted average of the (RGB) color of the objects behind the fan at the given point; and the color of the fan blade. The weights in the weighted average are determined by the thickness of the fan blades (relatively to the circumference), and these weights may actually depend on the radial coordinate $r$.
So what we see is not "quite" transparent – the contrast is lower – but it's enough to see what's behind; the color of the things behind the fan is mixed with the color of the blades; and this mixing occurs pretty much independently of the location relatively to the axis of the fan (if the fan blades' color is uniform), and independently of time (because of the averaging over the 1/50 second time intervals).
Note that the 1/50 second resolution depends on the neurology – abilities of the eye, nerves, brain etc. However, even if the brain were perfect, there would exist certain limitations that couldn't be beaten. The number of photons coming to each retina cells per second is finite and the inverse of this number basically determines the best possible time resolution one can have for the given "pixel".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "39",
"answer_count": 4,
"answer_id": 0
} |
The Thomson scattering optical depth for a photon ar radius r I am looking to understand some more about the physics of gamma ray bursts. In particular I am looking at the origin of the "prompt emission". Some of the energy associated with this prompt emission is thought to come from the photosphere of the expanding fireball at the centre of the GRB, which expands at relativistic speeds.
The Thomson scattering optical depth for a photon at radius $r$ is give n by
$$\tau = \int \frac{dr}{c} (c-v) \sigma_T n_e $$
where $c$ is the speed of light, $n_e$ the electron number density and $\sigma_T$ the scattering cross section and $v$ the velocity of the expanding fireball.
Can anyone give me a physical explanation behind this equation? Also, at the photospheric radius, the optical depth becomes 1. Why should this be so? Thanks
| The origin of this equation is reasonably well explained in Abramowicz (1991).
If you take a relativistically expanding enevelope and only consider Thomson scattering, then as the electron scattering cross-section in the co-moving frame $\sigma_T$ is independent of frequency, then the mean free path of a photon in the co-moving frame is independent of the envelope velocity as seen in the "stationary" (observers) frame.
Abramowicz et al. shows that in the case of photons moving "downstream", i.e. with the flow
$$ d\tau = \gamma (1 - \beta) d\tau_0,$$
where $\beta = v/c$, $\gamma = (1-\beta^2)^{-1/2}$ is the Lorentz factor, $\tau_0$ is the optical depth in the co-moving frame and $\tau$ is the optical depth in the observer's frame.
Now $\tau_0(r) = \int_{r}^{\infty} n_{e,0} \sigma_T\ dr$, so
$$ \tau(r) = \int^{\infty}_{r} \gamma (1 - \beta) n_{e,0} \sigma_T\ dr,$$
where $n_{e,0}$ is the electron density in the co-moving frame.
But if $ v\ll c$ then $\gamma (1 - \beta) \simeq (1- \beta)$ and $n_{e,0} \simeq n_e$ thus
$$ \tau(r) \simeq \int_{r}^{\infty} (1 - \beta) n_e \sigma_T\ dr = \int_{r}^{\infty} \frac{(c-v)}{c} n_e \sigma_T\ dr$$
The integral is carried out from some physical depth in the wind out to infinity. The Thomson scattering photosphere can be defined as where this optical depth is approximately unity. This is because this will be roughly where the radiation decouples from the flow and photons can escape.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The pressure in a container of water is based on depth. So what happens if I remove the bottom of the container? So I understand that if we have a system that involves a container of water the pressure will equal atmospheric pressure at the top and as we go further down the container the pressure will increase with depth until it is at its maximum pressure at the bottom of the container. In terms of a 1-D fluid equation (neglecting viscosity) we have
$$\frac{\partial v}{\partial t} = \frac{1}{\rho} \frac{\partial p}{\partial z} - \frac{\partial v}{\partial z} + g$$
$$0 = \frac{1}{\rho} \frac{\partial p}{\partial z} - 0 + g$$
$$\frac{\partial p}{\partial z} = \rho g$$
$$p = \rho g z + const$$
$$p = \rho g z $$
where we have taking the constant to be $0$.
Now say you removed the bottom of this container. Obviously the water starts to flow out the bottom. So the velocity changes with time. As the fluid is incompressible we still have $\frac{\partial v}{\partial z} = 0$.
So
$$\frac{\partial v}{\partial t} = \frac{1}{\rho} \frac{\partial p}{\partial z} + g$$
But if pressure still balances with gravity we have no fluid flow! But the fluid does flow so the pressure must not balance with gravity anymore. So what is the pressure in the system now that the bottom of the container has been removed?
| Take a 1-cm square tube and place it vertically in the container from top to bottom, touching the bottom so that the bottom of the container is the bottom of the tube.
The pressure at the bottom of the tube is nothing but the weight of water it is supporting - the water in the tube.
Supporting means to keep from falling.
(Forget the air pressure - that's just confusing the issue.)
If you take away the bottom of the container, you are taking away the bottom of the tube, so it is no longer supporting the column of water, so the weight of water above goes to zero because it is not being supported. So the pressure goes to zero because it's nothing more than the weight of water being supported.
You don't need equations to understand this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Peskin eqn 7.2 contradiction They state $$\langle\Omega|\phi(x)|\lambda_{\bf p}\rangle=\langle\Omega|e^{iP\cdot x}\phi(0)e^{-iP\cdot x}|\lambda_{\bf p}\rangle \tag{7.4}$$
where $|\lambda_{\bf p}\rangle$ is a state of momentum ${\bf p}$. They then rewrite this as
$$\langle\Omega|\phi(0)|\lambda_{\bf p}\rangle e^{-ip\cdot x}\tag{7.4}$$ with $p^0=E_p$.
So they've basically said $$\langle\Omega|e^{iP\cdot x}=\langle\Omega|.$$ This would be fine if the interacting vacuum had zero energy and momentum. It does have zero momentum but on page 86 they've defined $E_0=\langle\Omega|
H|\Omega\rangle$ so really I'd expect
$$\langle\Omega|
e^{iP\cdot x}=\langle\Omega|
e^{iE_0t}.$$
I've tried seeing what happens if I just subtract the term $E_0$ from the Hamiltonian to give the interacting vacuum zero energy. Subtracting it from the interaction term basically gave me an infinite S-matrix, while subtracting it from the free part gives asymptotic states at infinity negative energy and seems to mess up their proof of the expression of the two point correlator.
What should I be doing?
| Perhaps I am overlooking the real problem here, but isn't it clear that by doing
$$
H\rightarrow H+E_0
$$
both the vacuum and the one particle state energies get shifted by the same amount
$$
H|\Omega\rangle = E_0\,,\qquad H|p\rangle =E_0+E_p
$$
so that $E_0$ Actually cancels out
$$
\langle\Omega|\phi(x)|p\rangle=e^{iE_0 t}\langle\Omega|\phi(0)|p\rangle e^{-i px}=\langle\Omega|\phi(0)|p\rangle e^{-i E_p t+i\vec{p}\vec{x}}
$$
that is you are free to set $E_0=0$ without affecting the result.
Moreover, subtracting a constant term from the Hamiltonian can't possibly give you a divergent S-matrix because it simply rescales all fields by the same identical phase, and in QM the states are defined defined up to a phase.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/201988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Does sound show wave-particle duality? We know that light and electrons both show wave-particle duality. Or in other words we can say that they can be both seen as a wave and a particle. Can a similar theory be applicable for sound? Can sound also be explained as a particle as well as a wave?
| The notion you should look up and learn about is the phonon. It is a quasiparticle that arises in the quantum description of acoustics in condensed matter. The description is simplest and clearest in regular lattices of atoms / quantum particles, so it doesn't work so well for sound in a gas. But phonons can be thought of as quantums of sound in solid lattices.
Basically, a lattice is modelled as a system of coupled quantum harmonic oscillators, whose Schrödinger equation is very like a classical model of point masses linked by ideal massless springs. The system has eigenmodes with natural frequencies $\omega_j$, and the energy level of $j^{th}$ eigenmode can change only by integer multiples of $\hbar\,\omega_j$, whilst its ground state has energy $\frac{1}{2}\,\hbar\,\omega_j$. The quantum of this energy change $\hbar\,\omega_j$ corresponds to the phonons of the acoustic eigenmode in question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Directional subwoofer? I was thinking. The subwoofers that I've seen are a circular parabolic surface section (or perhaps a circular circlic(?) surface section?) and are considered omni directional.
I would guess that this is because the longitudinal waves would have to move through the focus of the parabola/circular section, dispersing the wave in all directions in front of the speaker (and behind, depending on the acoustic shielding).
However, if a subwoofer was made from a circular triangular surface section, whose height is the same as the radius of the circular section:
Would this make the subwoofer directional? I.e. could I point at someone very far away and it would be heard but wouldn't be heard by those not in it's path?
| Audio speakers are, in general, not particularly directional. That is to say, the sound emitted from the speaker spreads out in all directions. With the exception of ultrasound frequencies, it is impossible to 'beam' sound energy along a narrow path with a speaker of practical dimensions.
This is due to the relative size of sound-waves and speakers. A normal stereo speaker will have a diaphragm 5-30 cm in diameter; a middle-C note has a wavelength just over a meter. So sound typically disperses from a speaker in all directions - similar to the way that light disperses in all directions when it passes through a pinhole.
If you wanted to have a directional sub-woofer, with sound wavelengths in the range 1-10m (~30-300 Hz), it would require an enormous active surface - bigger than the side of a barn - in order to focus or project sound in a significantly directional manner. So speaker shape will have only a very insignificant effect on sound-volume as a function of angle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Can we reconstruct 1D potentials in QM from the spectrum? Knowing the potential, we can find the spectrum of the Schrödinger operator. The converse question is: Knowing the spectrum, can we reconstruct the potential? As an example, a harmonic potential has an equally spaced spectrum. But is the converse true?
This is, of course, similar to the 'hearing the shape of the drum' problem, which has a negative answer. But we also should notice that in classical mechanics, if the potential is symmetric, we can recover it from the oscillation period as a function of the energy of the particle. This is due to ingenious work by Abel.
| The answer is no, I am afraid. As you may well know, the self-adjoint Laplace operator $-\Delta$ on $L^2(\mathbb{R})$ has purely absolutely continuous spectrum $\mathbb{R}^+$.
Now let $V\in L^{\infty}(\mathbb{R},\mathbb{R}^+)$ be an arbitrary bounded positive function. Then $-\Delta_x +V(x)$, where $V$ acts as a multiplicative operator is self-adjoint and has spectrum $\mathbb{R}^+$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Why array of telescope is used? To increase the resolution of an instrument, smaller wavelength and larger aperture is desirable. It is mentioned in some textbooks that the "effective" diameter of a telescope can be increased by using arrays of smaller telescopes. I just wonder why it is possible because every telescope is separated.
| Picture yourself looking into a large mirror on the wall. Now picture the mirror is made up of smaller, tiled mirrors. You will still see your reflection. If you begin to remove the tiles, so that there are only a few left, you can still use them to reconstruct the image of your face that was given by the original mirror. This is what is happening with an interferometer. Astronomers are constructing an image measured by the "full mirror" (the longest baseline) based on the information they get from a few tiles (individual antennas).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Creating nanosecond pulses from a CW laser source Experimental setup question.
If anyone aware of a technique to create pulses of several nanoseconds at a around 10Hz repetition rate from a CW laser source?
| You could use an electro-optic modulator. These don't need kV supplies, can have very fast rise/fall times, and can be fully programmable by using a digital delay generator (these can also be triggered optically for extremely good accuracy, preventing timing drift between the delay generator and whatever source you're using). You usually need the following:
1) Electro-optic modulator.
2) DC power supply (20 V or so).
3) RF amplifier.
4) Delay generator.
Using a delay generator (such as the Stanford Research Systems DG645) you can create modulated RF signals with a very wide range of waveforms which can then be used to open and close the electro-optic modulator to transfer the RF waveform to an amplitude modulation on the CW laser output.
This will not be cheap, though.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/202881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Nucleon-meson interaction Suppose interaction lagrangian between neutron-proton doublet and $\pi$-mesons:
$$
\tag 1 L_{\pi pn} = \bar{\Psi}\pi_{a}\tau_{a}(A\gamma_{5} + B)\Psi , \quad \Psi = \begin{pmatrix} p \\ n\end{pmatrix}
$$
Is it possible to derive it from the first principles? I realize that proton and neutron aren't pseudogoldstone bosons, like pions, and thus their lagrangian cannot be derived simply. But, maybe, it is possible to derive directly $(1)$.
| This interaction cannot be derived from QCD. It is also not quite correct.
1) QCD conserves parity (for $\theta=0$), and the pion field is a pseudoscalar, so $B$ must be zero.
2) The pion is an (approximate) Goldstone boson, so it is derivatively coupled
$$
{\cal L}=\frac{g}{f_\pi}\bar\psi\tau^a\gamma_5\gamma_\mu\partial^\mu\pi^a\psi
+ \ldots
$$
In chiral perturbation theory, this is the first term in an infinite tower of pion nucleon interactions, involving higher derivatives and higher powers of $\pi^a$. The value of $g$ has to be determined from experiment (or from lattice QCD). Chiral symmetry implies that $g$ is related to the axial-vector coupling of the nucleon, $g_A$, which can be measured using neutrino scattering or neutron beta decay.
3) The Skyrmion model is (as the name suggests) a model, not something that can be derived from QCD. First of all, the idea that the nucleon can be described by some kind of classical field can only be true in the large $N_c$ limit. Second, the nature of the classical field is not clear (why the chiral lagrangian?). Third, even if I accept the idea that the nucleon is a solitonic solution of the chiral lagrangian, its properties cannot be determined. For the soliton to be stable, the solution has to exist in the regime where all powers of the gradient expansion are of the same order, and there is no predicitive power.
Additional Remarks: In Witten's paper (http://inspirehep.net/record/140391?ln=en) the large $N_c$ limit is used to motivate a classical (mean field) picture of baryons, in which the nucleon might emerge as a soliton. If $N_c$ is not large, quantum corrections are $O(1)$ and the soliton picture makes no sense. Witten argues that the chiral lagrangian is a natural candidate for the mean field lagrangian (and that thanks to the WZ term the quantum numbers work out), but he does not claim to derive this. Finally, with or without vector mesons (why vectors? why not spin 5 mensons?) all terms in the gradient expansion are of the same order.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/203074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is the electrostatic field really static? Does thermal vibrations not affect it? We know that if a conductor has any net charge, the charges reside on the surface. The electric field immediately outside the surface is perpendicular to the surface. But the charged particles, say the conductor has net electrons, will be in thermal vibration and increase in temperature will increase the vibrations. So, the net electrons vibrating on the surface will lead to changing electric fields outside the surface, not perpendicular all the time. Doesn't that cause a magnetic field, however small?
| Yes, these thermally generated currents (Johnson noise) generate magnetic fields. This means that even non-magnetic materials generate a very-small magnetic noise if they are conductive. This actually places a limit on very-sensitive magnetic field measurements in shielded environments because the shields are usually conductive. The following Review of Scientific Instruments article (abstract) discusses this issue. Alternately, one can use high resistance materials such as ferrites for magnetic shielding as discussed in this
Applied Physics Letters article (PDF).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/203276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fringe width and spacing and number of slits in diffraction experiments In a single slit experiment, the fringes are not equally spaced and aren’t of equal widths—the central maximum is the widest, the secondary maxima grow narrower and narrower outward, and the minima grow wider and wider outward.
In a double slit interference pattern, the fringes are equally spaced and of equal widths.
With a diffraction grating (lots of slits), the fringes are highly focused, with small widths and unequal spacing.
What are the reasons for the differences in fringe spacing and widths as the number of slits increases, specifically in each of the three scenarios I’ve presented above?
| When you pay attention to the left or right area of your double slit experiment you will see at the end the typical intensity distribution of a single slit. So a multi-slit arrangement is nothing all as the sum of two single slits. Of course the intensity distribution depends from the distance between the two slits. Artfully created, one would see a clear distribution like in your case. Then bigger the distance, then more one see the single slit distributions.
Go a step back and reflect, that even behind an edge fringes appear. Move two edges closer and closer together one get the intensity distribution you describe. And multi-slits are the sum of all of the edges. To break it down, the question is, why there are intensity distributions behind edges.
The answer is simple. Photons are moving units with oscillating electric and magnetic fields. Then thinner the edges shape (for example eraser blade), then higher the electrostatic potential from the surface electrons of the edge. This electrons and the photons form a quantized field and the projection of this field are the intensity distributions behind the edge.
Such an explanation helps to understand why even single photons, emitted over time, form intensity patterns behind edges. The surface electrons are in motion and differ with their energy, the distance from the photons source to the edge differs and all this led to the intensity distribution.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/203367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
What's the difference between semiconductor and insulator (besides band gap)? The typical classification of electronic materials is metal-semiconductor-insulator. Is there any actual difference between a semiconductor and an insulator, besides the size of the bandgap?
| Basically they are the same from a physics point of view, if you only look at crystalline materials. A semiconductor is defined to be insulating at $ 0 K $, while conducting at room temperature, although I don't recall, what level of conductivity is required to count as semiconductor.
Technically, insulators are a more general group of materials, since they could also be amorphous, while semiconductor materials are in the best case monocrystalline with very few defects or at least polycrystalline with a certain grain size in order to show proper semiconducting behavior due to the bandstructure. If there are to many defects, grain boundaries, impurities, the real bandstructure can differ from the ideal one by a lot.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/203577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is potential energy frame dependent in special relativity? In newtonian mechanics, As far as I'm aware, only kinetic energy is dependent on frames of reference, since kinetic energy is a function of velocity(squared) and velocity is dependent on frames of reference, therefore kinetic energy is frame dependent.
In newtonian mechanics, assuming that all observers use the same reference for zero potential energy($U=0$), it's frame independent . Potential energy of point particles in force fields, like gravity and electric field are given by: $U=constant\dfrac{\xi_1\xi_2}{r}$, where $\xi$ is either the gravitational or the electric charge of a particle. Since both charges and the distance $r$ between them is invariant under all frames of reference, therefore potentential energy is frame independent.
In SR, one expects that the potential energy of an object in a force field should also be a function of the charges and the distance between them. However since the distance $r$ between the charges is relative to the choice of the frame of reference(it's $r$ in a rest frame and $\dfrac{r}{\gamma}$ in a frame that's moving relative to the two charges, owing to lorentz contraction), therefore it seems to me that potential energy becomes frame dependent in SR.
Is this the case?
| Yes potentials are frame dependent. Let us take the electric and magnetic fields as an example. The electric field can be written as:
$$\vec E=-\frac{1}{c} \frac{\partial \vec A}{\partial t}- \nabla \phi $$
Where $\vec A$ is a vector potential and $\phi$ a scalar potential. Like wise, the magnetic field can be written as:
$$\vec B=\nabla \times \vec A$$
Where $\vec A$ is the same potential as that that appears in the magnetic field. Associated with these potentials we have a four vector, called the electromagnetic four-potential and given by:
$$A^\mu=(\phi, A_x, A_y, A_z)^T$$
Like all four vectors this has to transform via the Lorenz transform matrix (assuming the relative motion between the two frames occurs in the $x$ direction):
$$L=\begin{pmatrix} \gamma & -\beta \gamma &0 &0\\ -\beta \gamma & \gamma & 0 &0\\ 0&0&1&0\\ 0&0&0&1\end{pmatrix}$$
Such that:
$$A^{\mu'}=LA^\mu $$
So potential does depend on frame.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/203671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find constant acceleration with only initial speed and distance Given the problem:
"A car moving initially at 50 mi/h begins decelerating at a constant rate 60 ft short of a stoplight. If the car comes to a stop right at the light, what is the magnitude of its acceleration?"
While this problem seems simple, I can't seem to find the correct formula to use. Most formulas I am finding require the use of time (t) which is not given in the problem statement. What formula(s) do I use to solve this problem? Am I supposed to use distance as the unit of time somehow? Or should I use some sort of derivation to get the number needed?
| The total distance traveled during the constant acceleration is 60 feet. Over the course of 60 feet the velocity has to go from 50mph to 0mph. That means the average velocity is 25mph. So, what amount of time going 25mph is equal to 60 feet? Then take that time and divide the change in velocity by it to arrive at the acceleration.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/204103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Rockets and distance I am trying to create an equation which allows for me to change the aspects of the rocket so i can calculate the distance traveled vertically. My idea is for a rocket that only moves vertically; with this i can calculate the amount of time it would take to make it past the first Lagrange point. So far I have made an equation which calculates the acceleration of the rocket, this being:
$$
\frac{32T}{W_0+F-Bt}-\frac{Gm_e(W_0+F-Bt)}{r^2}
$$
$T$=Thrust
$W_0$=Initial weight
$F$=initial weight of fuel
$B$=Burn rate(lbs of fuel per second)
$t$=time in seconds
$G$=gravity constant
$m_e$=mass of earth
$r$=radius from earths center of gravity
Thank You for taking the time to read this.
| You've almost got it!
The constant thrust comes from a mass rate $\mu$ of fuel being expelled at a velocity $v_f$ as opposed to the speed of the rocket itself $v$. Therefore the equation is instead:
$$ (m_0 - \mu t) \frac{dv}{dt} = \mu v_f - \alpha \frac {(m_0 - \mu t)}{r^2},$$
where $\alpha = G M.$ Hence the gravitational term you wrote as $G m_e (W_0 + F - Bt) / r^2$ is deeply incorrect as an acceleration (it is a force!) and your equation has a type error. The corrected equation is simply:
$$ a = \frac{dv}{dt} = \frac{\mu v_f}{m_0 - \mu t} - \frac{\alpha}{r^2}.$$
Of course if $\alpha = 0$, the solution to this is famously $$\int dv = v - v_0 = \int dt~\frac{\mu ~v_f}{m_0 - \mu t},$$ and defining $m = m_0 - \mu t;\;dm = -\mu~dt,$ one gets$$v - v_0 = -\int dm ~ \frac {v_f}{m} = - v_f~\ln\left(\frac{m}{m_0}\right).$$In other words, starting from rest, you have $v(t) = -v_f \ln\left( 1 - \frac{\mu t}{m_0}\right).$
A crude heuristic then equates $v_1$ at the end of rocket acceleration with the escape velocity of the planet you're exiting, $v_e = \sqrt{2\alpha/r_0},$ hence you need $m_0 / m_1 = \exp(v_f / v_e).$ Earth's escape velocity is 11.2 km/s, some rocket exhaust speeds can be in the 3-4 km/s range, so that suggests that you need $m_1/m_0$ to be something like 10-50ish, or in other words your payload tends to be somewhere between 2% to 10% of the needed weight.
If you're going to model this equation in more depth than that, you should probably add a friction force $a_f = -\gamma v^2 / (m_0 - \mu t)$ and simply solve the equations numerically. It's about a day's coding, maybe less, to get some reasonably fast explorations of the parameter space in NumPy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What's the difference between the work function and ionisation energy? In a particular textbook, the work function of a metal (in the context of the photoelectric effect) is defined as:
the minimum amount of energy necessary to remove a free electron from the surface of the metal
This sounds similar to ionisation energy, which is:
the amount of energy required to remove an electron from an atom or molecule in the gaseous state
These two energies are generally different. For instance, Copper has a work function of about 4.7eV but has a higher ionisation energy of about 746kJ mol-1 or 7.7eV.
I've sort of figured it's because the work function deals with free electrons whilst ionisation is done with a valence electron still bound within the atom. Is the difference due to the energy required to overcome the attraction of the positive nucleus?
| On first reading they do sound similiar, but they are entirely independent energies and concepts.
The work function of a metal refers to the minimum energy required to release an electron from the surface of a metal by a photon of light. The work function will vary from metal to metal. You might have a read of these: Compton Effect, this previous answer, Work function 1 and this Wikipedia article Work Function 2.
The Ionization energy is the energy needed to release electrons from their bound states around atoms, it will vary with each particular atom, with one outer electron around that atom needing less energy to release it than a lower, more closely bound electron, which requires greater energy because of the greater electrostatic force holding it closer to the nucleus.
Also, to complicate things, you need to allow for the Shielding Effect of Inner Electrons
Periodic trends for ionization energy (IE) vs. proton number: note that within each of the seven periods the IE (colored circles) of an element begins at a minimum for the first column of the Periodic table (the alkali metals), and progresses to a maximum for the last column (the noble gases) which are indicated by vertical lines and labelled with a noble gas element symbol, and which also serve as lines dividing the 7 periods. Note that the maximum ionization energy for each row diminishes as one progresses from row 1 to row 7 in a given column, due to the increasing distance of the outer electron shell from the nucleus as inner shells are added.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
When does the concept of electric field in classical electrodynamics fail, and QED is needed? It is really hard to find reference to when the traditional concept of electric wave, especially TEM wave, fails, and needs to be replaced by quantum electrodynamics.
So when does the concept fail? At high frequencies of electric field?
| It fails when the photon number is small. Since the electromagnetic field can never be zero because of the third law of thermodynamics this automatically couples the temperature, the effective volume and the photon number to each other. As a result it is experimentally impossible to do experiments with single photons at low frequencies because we can't produce the required ultra-low temperatures to make the photon number small.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Problem imagining how does a black hole merger look like from far away Suppose we have two black holes moving on a path of direct frontal collision. Is it correct that from far away, due to time dilation, we can never "detect" that the two black holes merging or colliding? Would we perceive the relative motion of the two black holes slow down before the two event horizons touch? if that is the case:
question: Will a planet orbiting the system
from far away feel a uniform distribution of mass located at the center or that of two blobs next to each other?
| Luckily black holes emit Hawking radiation, so in some sense we can see them.
So first there are two glowing balls approaching each other, after some time there will be one glowing ball that is twice as large as one of the initial glowing balls. (I'm considering black holes of same size)
And during the collision the balls distort to other shapes and shake violently emitting gravity waves.
All that happens quite fast when observed from far away. From closer observation point it happens even faster.
Just so that things don't become too absurd, all kinds of measuring instruments, like eyes and gravimeters, should agree what is happening during the merging.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Searching for a textbook on Imaging I'm hoping someone can recommend a book that will cover the physics involved in optical range image capture. I have a degree in physics, so I am looking for something on the technical end. To expand, I would like the book to cover the chemistry of how images are recorded on film and how we translate that data. From there, I would like the book to move into an in-depth treatment of CMOS and CCD sensor's and the data processing required to make images. Particularly, I'm hoping to gain a better understanding of the limits of digital sensors, but I am also looking to expand my knowledge in all areas of this topic. Can anyone make a recommendation?
| I think a good recommendation for a book that covers the physics involved in imaging is (regarding my Astrophysics background) 'Observational Astrophysics' by Pierre Lena, Daniel Rouan, Francois Lerund, Francois Mignard and Didier Pelat. Another recommendation, more basic, is 'Optics' by Eugen Hecht. Furhtermore, there are several Universities that have tons of lectures notes on imaging and such on their websites. For instance, here I share the website for the lecture notes on the course 'Image Science and Engineering', given by Dr. R. A. Schowengerdt, of the University of Arizona:
https://uweb.engr.arizona.edu/~dial/ece425/ece425.html
I don't know of any book that covers the chemistry behind the process of recording images on a film. However, I think any resource about the history of printing images would give at least a little of insight on that subject.
I hope this answer will help you, although almost 7 years have happened since you posted your question :(
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Help me understand Gauss law Suppose I have the following, the gaussian surface is the drawing in the middle. So charge enclosed is zero, and then eletric field must be zero since the area of the gaussian surface is not zero. But, clearly the eletric field is not zero in the middle, because if you put a charge there it will move. Why I'm getting the concept wrongly? Edit: you can consider the middle figure as a sphere.
| Gauss law says that the total flux going through a closed surface is equal to the charge inside the surface. You can think of flux as the number of field lines going in (or out) through the surface.
In your example there is no charge inside the sphere so the total flux through the surface of the sphere is zero. On one end (the left hand side) the field lines go in (negative flux) and on the other end (the right hand side) the field lines go out through the surface (positive flux). Gauss law in this case says that we have as many field lines going in as going out through the surface.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Question Regarding torricelli's theorem/Law I recently studied about bernoulli's equation/principle.
After the derivation of the said equation , my book gave some applications of the principle, which include torricelli's theorem/law.
In deriving torricelli's law from bernoulli's principle, the pressure at the opening of the tank in which the fluid is contained , is said to be equal to the same pressure which is applied at the the top surface of the applied fluid , namely the pressure of the atmosphere.
But my book also states that the pressure drops (according to bernoullis principle) when the fluid passes through a narrow pipe or opening and its velocity increases.
So why does the pressure remain the same in this situation ? Why doesn't it change?
Any help would be much appreciated , THANKS.
Could you please answer in simple and easy to learn terms , Thanks AGAIN.
| I agree with don_Gunner94's answer.
If the fluid come out from the constricted passage to atmosphere,it will experience atmospheric pressure,which is same as the pressure acting at top of the container.
Even according to Bernoulli's principle,
Static pressure + Dynamic Pressure = Constant
Therefore, the pressure acting on the fluid when it is inside the container would be static and when it will come out of the passage and open to atmosphere would be dynamic,all together total pressure remains constant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the difference between flow and expansion? Fluids (both liquids and gases) will move from one point in space to another due to a potential gradient. Some examples may be:
1) horizontal pipe flow, a fluid will move from a region of high pressure to a region of low pressure.
2) inclined open channel flow, a fluid will flow from high elevation to low elevation.
Both liquids and gases have a degree of compressibility, gases more so than liquids. Fluids expand or compress when subjected to a change in pressure, volume, and/or temperature.
In the first example concerning horizontal pipe flow, if the fluid was gas moving from a high pressure to low pressure, would it be said that the gas was flowing or expanding through the pipe? What features about the gas' movement is different than the liquid's movement described in the second example -- open channel movement of a liquid down an incline?
What are the definitions of flow and expansion and how does that distinguish the two when describing fluid movement from one point in space to another?
| *
*A gas can expand by filling more volume than before. Like a balloon in a pressure chamber where the pressure is suddenly lowered. No net motion (no flow) happens here.
*A water stream can flow continously without simultaneous expansion. Consider a circular stream that ends where it starts. As a bathtub where there is a big plastic bucket in the center. By twirling around in the water a flow can be established. No expansion happens.
The definition of flow is about a net motion of fluid particles. The definition of expansion is about the relative size or space taken up by the fluid particles. These concepts are fundamentally different.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/205947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why is the K shell electron preferred in the photo electric effect? I have read in many books and on Internet as well that photoelectric effect is only possible when an electron is emitted from the K shell of the metal. Why not other bonded electrons?
| The premise of your question is false. The photoelectric effect will occur whenever radiation interacts with the bounds electrons of an atom.
There is a sharp increase in absorption when the energy of the radiation exceeds the binding energy of a particular "shell". For Rb, for example, the NIST XCOM database shows clear K and L edges:
For atoms with lower Z (for example potassium) you don't always see such an edge except for the K shell electrons - it depends on the binding energies and available states. But the plot above shows it is possible to get photoelectric absorption from electrons in other shells.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/206263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
What happened here in this Taylor-Couette Flow experiment? I came across this video of Taylor-Couette Flow on YouTube. Originally I was looking for a visualization of the wavy Taylor vortices induced by the angular motion of the inner cylinder.
However, I found something strange (as you can see at the end of the video), the experimenter at the beginning injects three different dyes in a viscous liquid:
And then he begins to stir slowly in a specific direction until the three dyes are fully diffused into the main liquid:
Finally, he stopped the stirring and began to stir in the opposite direction returning the three dyes to its original spots (well sort of!):
So, how is this reversibility even possible? shouldn't the diffusion of dyes into the liquid be irreversible?
| This is what's happening in the video. I've drawn just a single drop, and for convenience I've ignored the curvature of the plates (it's harder to draw curves!):
It looks as if the (red in this example) ink drop is being mixed with the fluid, but actually it's just being stretched out into a thin sheet. When you turn the cylinder back again the sheet is pushed back into a drop.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/206579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Infinitesimally change a operator in QM Reading Balian, "From Microphysics to Macrophysics", I've found the following identity:
If we change the operator $\hat{{\mathbf{X}}}$ infinitesimally by $\hat{{\delta\mathbf{X}}}$, the trace of an operator function $f(\hat{{\mathbf{X}}})$ can be differentiated as if $\hat{{\mathbf{X}}}$ and $\delta\hat{{\mathbf{X}}}$ comutted:
$$\delta\operatorname{Tr}f(\hat{{\mathbf{X}}})=\operatorname{Tr}\left(\delta \hat{{\mathbf{X}}}f'(\hat{{\mathbf{X}}})\right).$$
What does "change an operator by $\delta \hat{{\mathbf{X}}}$" mean mathematically in this context? How I can prove that identity?
| $f(\hat{X})$ usually "means" $\sum a_n \hat{X}^n$. So, big hint: Let $y$ be your infinitesimal and let $\hat{Y}$ be some operator.
\begin{align*}
&\mathrm{Tr}f(\hat{X}+y\hat{Y})-\mathrm{Tr}f(\hat{X})=\\
&\int \langle q| \sum a_n (\hat{X}+y \hat{Y})^n|q\rangle \mathrm{d}q-\int \langle q| \sum a_n \hat{X}|q\rangle \mathrm{d}q
\end{align*}
expand ignoring higher powers of $y$ ($(\hat{X}+y \hat{Y})^n\approx \hat{X}^n+n y \hat{Y}\hat{X}^{n-1}$ isn't true, but you use cyclicity of the trace to do the same thing)
obviously there are huge mathematical problems here (with the integral of an infinite sum) but those should be ignored :)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/206687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Fluid speed and fluid density How does fluid density affect fluid speed?
Basically I am trying to figure out if, with all other quantities remaining constant, would an increase in fluid density cause the fluid speed to increase/decrease?
For example, would water and honey have different fluid speeds in a pipe, because their densities are very different? I know that:
$$Av = Av$$
and
$$P + ρgh + (1⁄2) ρv^2$$
But does an increase in density lead to an increase/decrease in fluid speed? How so?
| You could look at viscosity and its effects for yes, and at some never happened story of a guy dropping something from a tower and concluding something for no.
We're not allowed to give straight answers, so I hope this is vague enough.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/206867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Coffee Straw Physics When I put my little, cylindrical coffee straw into my coffee, the liquid immediately rises about half a centimeter up the straw without provocation. This is also the amount of coffee that the surface tension of the coffee will allow to stay in the straw when removed from the liquid in the cup.
Keep in mind that all the while, the top end of the straw is open.
Why does the level of the liquid in the straw insist on being higher than the level of all the liquid in the cup?
| Just to extend my comment about Capillary action which is the reason for the liquid rising through the capillary (straw in your case), I show this animation of how the diameter of the capillary (d) effects the height of the liquid (h) that rises above the contact surface.
The relation of $h$ and $d$ used to simulate this is taken from wiki page which is
$$ h=\frac{4\gamma cos\theta}{\rho g d}$$
It is for water and with the value of all constants substituted we get
$$h\approx \frac{2.96\times10^{-5}}{d}m$$
The diameter varies from $0.3mm$ to $2mm$
Also the answer to this question would make this true for your coffee as well.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/206971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Invariance of law of conservation of angular momentum under a Galilean transformation Given a reference frame O' moving at a constant speed $\vec{V}$ in relation to another reference frame O, I want to prove that
$\vec{r_{1B}} \times m_1\vec{v_{1B}} + \vec{r_{2B}} \times m_2\vec{v_{2B}} = \vec{r_{1F}} \times m_1\vec{v_{1F}} + \vec{r_{2F}} \times m_2\vec{v_{2F}}$
in O is equal to
$\vec{r'_{1B}} \times m_1\vec{v'_{1B}} + \vec{r'_{2B}} \times m_2\vec{v'_{2B}} = \vec{r'_{1F}} \times m_1\vec{v'_{1F}} + \vec{r'_{2F}} \times m_2\vec{v'_{2F}}$
in O'. The particles 1 and 2 are colliding (elastic collision). B stands for before the collision and F, after the collision.
Galilean transformation gives :
$\vec{r} = \vec{r'} + \vec{V}t$ and $\vec{v} = \vec{v'} + \vec{V}$
Substituting these expressions into the first equation and developing the cross products, I obtain, after having cancelled out 8 terms (because of the conservation of linear momentum),
$\vec{r'_{1B}} \times m_1\vec{v'_{1B}} + \vec{r'_{1B}} \times m_1\vec{V} + \vec{r'_{2B}} \times m_2\vec{v'_{2B}} + \vec{r'_{2B}} \times m_2\vec{V} = \vec{r'_{1F}} \times m_1\vec{v'_{1F}} + \vec{r'_{1F}} \times m_1\vec{V} + \vec{r'_{2F}} \times m_2\vec{v'_{2F}} + \vec{r'_{2F}} \times m_2\vec{V}$
My problem is that I do not know how to cancel out the $\vec{r'_{1B}} \times m_1\vec{V}$ terms... How do I do that?
| It is true that angular momentum is conserved in all frames, but the actual value of the angular momentum will, in general be different.
If you look at the extra terms you will find that they correspond to the change in the angular momentum of the centre of mass about your chosen origin.
\begin{align}
\mathbf{r}_{CM}\times M\mathbf{V} & = \frac{1}{M}(m_1\mathbf{r}_1 + m_2 \mathbf{r_2}) \times M\mathbf{V}\\
& = \mathbf{r}_1 \times m_1 \mathbf{V} + \mathbf{r}_2 \times m_2 \mathbf{V}
\end{align}
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/207185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to determine the acceleration and position as a function of time
An object is losing mass at a rate k in kg/s. The object is acted on by a force F. Determine the acceleration and position as a function of time.
I know the answer for the position function is
$$
x(t) =\left (\dfrac{Fm_0}{k^2}\right)\left[\left(\dfrac{-k}{m_0}\right)t - ln\left(1-\left(\dfrac{k}{m_0}\right)t\right)\right]
$$
but I'm not entirely sure how to get there. If anyone could just help me out by listing a few equations that would relate the rate of loss of mass to force I would much appreciate it.
| Recall Newton's 2nd law in the form $F=\frac{dp}{dt}$, where $p$ is momentum, namely $p(t)=m(t) v(t)$. Reformulating the equation gives you $F=\frac{dm}{dt}v+\frac{dv}{dt}m$. Using $m(t)={m}_{0}-kt$ yields the differential equation $$F=-kv+\stackrel{.}{v}\left({m}_{0}-kt\right)$$
Solve the equation to get $v(t)$, integrate to get $x(t)$ and differentiate to get $a(t)$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/207410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why can scalars have a sign? I wondered to myself why some scalars have a sign, if they do not have a direction. After all, the plus and minus indicate the direction of the scalar on a one-dimensional axis.
So, for example, why can temperature have a sign? Why can't mass?
| There is case where the definition is not crystal clear:
Chemical reaction rates.
The rate of a chemical reaction, $n R \rightarrow m P$ (where $n$ is number of reactants $R$ changing into $m$ Product $P$ molecules on every event), is given by:
$v = - \frac{1}{n} \frac{d R}{d t}$
Now in theory this should be a scalar, because there is no direction in space, yet there is direction in a $1$-dimensional "conceptual" frame of the reaction coordinate, frequently used to plot Energy vs. chemical change.
Reaction Coordinate diagrams, free textbook.
In this case, $v$ known as "reaction velocity", or "reaction rate", but never "reaction speed", and it is positive if within the system under study the net flow of molecular change if from $R$ into $P$, but it is negative for the opposing direction. The distinction between velocity and speed appears to deal with the scalar nature of the variable, however, it could (and perhaps it should) be viewed as a vector in the $1$-dimensional space of the reaction coordinate, where the sign is not part of the magnitude, but an indication of direction.
This has been a long standing, yet unsettled discussion in the Faculty where I teach. It does not change the state of the art for the fields of chemical kinetics, or enzymology and nevertheless, It would be nice to reach some agreement.
What do you make of it?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/207477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 5
} |
Velocity of an Electron as it Passes through a Uniformly Charged Ring I've been presented with a problem in which an electron is placed a certain distance x from the center of a positively charged ring and allowed to move freely. The ring has a known charge density λ. I am tasked with finding the velocity of the electron as it passes through the center of the ring.
First of all, I know that when the electron is initially positioned it will have a certain electric potential energy and that as it moves in towards the ring that potential energy will be converted into kinetic energy until it reaches the center, at which point all of the potential energy will be converted into kinetic energy.
The best way that I can think of to find the initial potential energy is to first find the electric potential at that point and then translate that into potential energy via the equation $V = \frac{U_e}{q}$. However, I am not certain what charges should be used in each equation. I have tried using the total charge of the ring (extrapolated from λ) in the electric potential calculation, and the dividing the charge of the electron out of that expression to produce the electric potential energy. Which produces the following equations:
$$
V = k\int\frac{dq}{r} = k\int\frac{dq}{\sqrt{R^2 + x^2}} = \frac{kQ}{\sqrt{R^2 + x^2}}
$$
$$
U_e = \frac{V}{q_e} = \frac{kQ}{q_e\sqrt{R^2 + x^2}}
$$
$$
U_e = \frac{1}{2}mv^2
$$
$$
velocity = \sqrt{\frac{2U_e}{m_e}} = \sqrt{\frac{2kQ}{m_eq_e\sqrt{R^2+x^2}}}
$$
I'm told that this is an incorrect solution. Could somebody please explain to me where I went wrong?
| The total energy at the initial point should equal the final energy.
i.e. $$V(x)=V(0)+KE$$ ($x,0$ are along the axis of the ring.)
The potential at a point on the axis of the ring is given by: $$V(x)=\frac{kQ}{\sqrt{R^2+x^2}}$$
From this, we get: $$-\frac{keQ}{\sqrt{R^2+x^2}}=-\frac{keQ}{R}+\frac{1}{2}mv^2$$ From this you can obtain the velocity. The P.E. has a negative sign because the electron and the ring have opposite charges and are attracted to each other. The point at the center of the ring has the least potential energy locally, and that is why the electron moves towards it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/207727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Do photons decay as they travel in free space From maxwell's equations, it occurred to me that photons are stable. Decrease in electric field creates magnetic field and vice versa and somehow there is a harmony that allows photon to exist as long as it travels with c. Therefore I wouldn't think photons would decay as they travel.
However, I know that radiation decays with 1/r from the source and a photon is a form of radiation. Does that mean photons can travel in space with c, but they also have to decay? I kind of imagined them like harmonic sine waves.
| No - assuming they don't hit anything they don't decay.
The distance dependant "decay" is the drop in the number of photons per volume as the volume gets bigger - it's not a decay of individual photons. It's the same as a crowd dispersing as it leaves a subway exit - nobody is disappearing.
Photons can lose energy as they collide with gas or dust in space but can also gain energy if they interact with a high energy particle - so the wavelength of light from a distant object can change.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/207919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How is $ \frac{dv}{ dt} = a $? I know how , in the physical sense -
$$\frac {dv}{dt} = a$$
But, even after thinking a lot, I am not able to see the fault in this -
$$\frac {dv}{dt} = \frac {d(st^{-1})}{dt}
= \frac {sd(t^{-1})}{dt}
= s*(-1)*t^{-2}
= \frac {-s}{t^2}
= \frac {-v}{t}
= -a$$
I know something is terribly wrong here but I'm just not able to figure out what or where.
Please keep in mind I'm just a curious 16 year old. Any help would be greatly appreciated.
|
I know how , in the physical sense - $$\frac {dv}{dt} = a$$
But, even after thinking a lot, I am not able to see the fault in this
- $$\frac {dv}{dt} = \frac {d(st^{-1})}{dt} = \frac {sd(t^{-1})}{dt} = s*(-1)*t^{-2} = \frac {-s}{t^2} = \frac
> {-v}{t} = -a$$
I know something is terribly wrong here but I'm just not able to
figure out what or where. Please keep in mind I'm just a curious 16
year old. Any help would be greatly appreciated.
Here's an example. Suppose an object is traveling at a constant velocity. Then
$$
s=v_0t
$$
where $v_0$ is constant in time. I.e., we know that $dv_0/dt=0$ and $ds/dt=v_0$.
So, for this example, you might get confused if you rearrange and write
$$
v_0 = \frac{s}{t}
$$
and then the RHS looks like it might not be constant... but it has to be, so what gives?
Well, in general:
$$
\frac{d}{dt} \left( \frac{s}{t} \right)
=-\frac{s}{t^2}+\frac{1}{t}\frac{ds}{dt}
$$
because I have to differentiate both terms: the $s$ term; the $\frac{1}{t}$ term. This is just an application of the product rule of differentiation. The above result holds in general.
And, for the example case of constant velocity, plugging in $s=v_0t$, the above equation becomes:
$$
-\frac{s}{t^2}+\frac{1}{t}\frac{ds}{dt}=-\frac{v_0t}{t^2}+\frac{1}{t}v_0=0
$$
So, the problem you are having is that you have only differentiated one of the terms, the $1/t$ term, and not the other, the $s$ term.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is the relation between energy levels of hydrogen atom in Bohr's solution to that of Dirac solution In Dirac solution for hydrogen atom, the energy levels are calculated as positive
\begin{equation}
E=\frac{mc^{2}}{R(t)\sqrt{1+\frac{z^{2}\alpha^{2}}{\left(n+\sqrt{\left(j+\frac{1}{2}\right)^{2}-z^{2}\alpha^{2}}\right)^{2}}}}
\end{equation}
, while in Bohr's model the energy levels are negative
\begin{equation}
E=\frac{-Ze^{2}}{8\pi\epsilon_{0}r}
\end{equation}
How are these two related to each other?
| The $E$ in your expression is the quantity calculated using the operator $i\hbar\frac{\partial}{\partial t}$, so it is the total energy. As $n \rightarrow \infty$ this energy $E$ goes to the rest energy of the electron $m_ec^2$ as we'd expect. For finite $n$ the energy is lower than $m_ec^2$ with the difference being the binding energy of the electron.
To compare the Dirac $E$ with the Bohr energies just subtract off $m_ec^2$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why does everything written on a page appear in concentric circles when we rotate that page? When i put a pencil in the middle of a paper and rotate it very fast, whatsoever is written on it, will appear in concentric circles. What is the reason behind this phenomenon and what is it called?
| Fast things appear blurry to your eyes, and the distance from a point on the paper to your pencil is fixed. The distance from an ink dot on the paper, to the tip of your pencil where you press down, cannot change. Call this distance $d$. The only positions the point is allowed to be at when you spin the page around, are positions that satisfy the equation $x^2+y^2=d^2$, with $d$ being constant. The set of solutions of that equation is a circle, so things appear as circles.
To add a lesson that has nothing to do with your question, $x^2+y^2$ has an intimate connection with euclidean geometry. $x^2-t^2$ ("$t$" for "time") has an intimate connection with special relativity! The analogous things to circles in euclidean geometry are hyperbolas in special relativity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles? How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles?
It would appear to a layman such as myself that these heavier unstable particles are just transient interplay of the stable forms.
|
How can the unstable particles of the standard model be considered particles in their own right if they immediately decay into stable particles?
Nobody has an issue calling the electron a particle. Ditto for a neutron. It's stable in a nucleus, and the fact that a free neutron decays in circa 15 minutes doesn't much matter. It's similar for a muon, which lasts for a couple of microseconds, because that's quite a long time in particle physics. Muons leave tracks, as do other non-stable particles. And because we can see those tracks, I don't think many people have an issue calling them particles.
Things start to get interesting when we move on to shortlived baryons where you hear the word "resonance". I'm afraid to say I always think of a twanged ruler when I hear that. Something that doesn't last, that's more of an action or an event than a thing. Particularly when it comes to the Z-boson which lasts for circa 3 x 10$^{-25}$ seconds. Nobody has actually seen a Z-boson, or a track. Its existence is inferred. It's similar for some other particles: "Experiments don’t detect the Higgs boson directly – instead, its existence is inferred by looking at the particles left behind when it decays".
It would appear to a layman such as myself that these heavier unstable particles are just transient interplay of the stable forms
I guess particle physicists call them particles, and everybody else goes along with that. But maybe somebody like anna v can give a better answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 7,
"answer_id": 3
} |
Are white noises always Markovian? Are white noises always Markovian? I am a bit confused about it. As white noise always has a constant power spectrum， its auto correlation function must contain a delta function of time. Thus the correlation time of the noise vanishes. But I don't know whether they can be called Markovian.
| Whether a noise process is truly memoryless is very hard to test. Strictly speaking it's impossible to say that it is memoryless because the memory could be longer than the longest time series we can analyze. In practice a lot of low-entropy pseudo-random generators can make synthetic signals that are physically indistinguishable from the "real thing", so it always depends on the application if one can make that assumption and get away with it, or not.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Temperature of a falling meteor I am reading "What if?" article https://what-if.xkcd.com/20/ and I'm interested in it's scientific background. Mr. Munroe writes:
As it [the meteor] falls, it compresses the air in front of it. When the air is compressed, it heats it up. (This is the same thing that heats up spacecraft and meteors—actual air friction has little to do with that.) By the time it reaches the ground, the lower surface will have heated to over 500℃, which is enough to glow visibly.
How can one make such estimation? I wanted to use PV = nRT, but I don't know the volume and the difference in pressure. I tried to sum up all the kinetic energy of all air molecules of the air, bumping into the meteor, but the answer is nowhere near. Does anyone have an idea? Such an interesting problem.
| The most significant contribution to the heat generation is the the exothermic reaction of iron in the meteorite with atmospheric oxygen and nitrogen. The oxides and nitrates that form at the leading entry surface are not uniform. They will distribute themselves so that there is the greatest possible entropy among them. This will be another ongoing process. So there will be heat generated from molecules avoiding any definable order among themselves. Furthermore as the oxygen and nitrogen are heated under pressure a great deal of nitric and nitrous acids will form with help from water vapor in the air....which will react further with the iron, contributing another exothermic reaction. Unless that meteorite is made of some highly unreactive substance(s) calculating the heat it will generate or temperature it will attain is going to be a really tough problem.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Clarification about Wave-particle duality Okay,so I am learning about the double slit experiment done with electrons. I saw this picture, which shows the interference pattern being built up slowly with increasing number of electrons:
I just wanted to confirm whether I have the correct understanding. The fact that the first image has a random distribution, shows that each electron interferes with itself and strikes a point on on the screen which would be dictated by the probability function.
The interference pattern is the result of the same interference of many electrons and is a statistical property of many electrons.
Also, does this mean the electron travels as a wave, but then it obviously must strike as a particle since it hits a well defined spot on the screen?
| From the theory of light waves we know that for similiar experiment an interference pattern occurs when light quanta interacts with the system. Now with electrons there is an unique "wave" for that particular experiment that guides those electrons that hit the screen. Initially electrons must have equal speed and direction for that clear pattern to emerge.
We can assosiate that "wave" $\psi$ with any electron in an abstract sense, and just like with light we have $|\psi|^2$ that stands for intensity of the hits on the screen (probability distribution). So for every electron there is a wave that depends on the system (e.g. double slit experiment, hydrogen atom, ...)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Why does wavelength determine the energy of a photon? The professor for my first-year university chemistry class remarked that the wavelength of a photon determines its energy. Why is it that the case?
I've only completed high-school physics so far, so please bear that in mind in answering this question.
Thank you.
| Why does the wavelength determine a photon's energy?
In the 19th century, it was thought that the energy of light was determined only by its intensity. Then, experiments, particularly the photoelectric effect, showed that this was not so: a low-intensity short-wavelength light can cause reactions that intense light of a longer wavelength cannot. Thus, from experiment, shorter wavelength must mean higher energy.
Next, there was the problem of black body radiation. In thermal equilibrium, bodies glow but the classical theory predicted the wrong spectrum. In 1905, Einstein created a theory that predicted the correct black body spectrum on the assumption that light is made of particles whose energy is inversely proportional to wavelength.
So, in other words, to get theory and experiment to match, we have to assume that the energy of a photon is inversely proportional to its wavelength.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/208942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trying to understand lowest configurations of carbon My study group is debating about which are the three lowest configurations of carbon. I've been arguing that the electron has to jump to the 3s level for the configuration to be different. Others have suggested that the two valence electrons just have to change their $m$ and $s$ numbers on the 2p level. We are using Morrison's Modern Physics and having trouble settling this issue within the text. We are aware of Hund's rule, so some of the problem is about exactly what is meant by "configuration." We want to understand this problem and do the work ourselves, but we are installing doubt in one another. Can someone clarify "configuration" and maybe suggest the general approach appropriate here?
| By 'electron configuration' can be understood the way an atom's electrons are arranged in atomic orbitals, in accordance with Pauli's Exclusion Principle, the Aufbau Principle and Hund's Rule, of the lowest possible total energy (known as the Ground State).
For carbon (Z=6), six electron have to be placed in the correct atomic orbitals.
The first 2 occupy the lowest energy atomic orbital possible, that is 1s, so we have $1s^2$ for the first term.
For the remaining four electrons, the next two lowest available atomic orbitals are 2s and 2p and following the above rules that gives us $2s^2$ and $2p^2$. Bearing in mind that to satisfy Hund's Rule the latter two 2p electrons are divided over one $p_x$ and one $p_y$ sub-orbital, each with one electron of the same spin quantum number ($m_s=-\frac{1}{2} \text{or} +\frac{1}{2}$).
Overall we can write the electron configuration of carbon as:
$1s^2 2s^2 2p^2$ or with some added detail $1s^22s^22p_x^12p_y^1$ and because $[He]=1s^2$, carbon's electron configuration (ground state) can be written as:
$[C]=[He]2s^22p^2$ = $[He]2s^22p_x^12p_y^1$.
The first excited state of carbon $C^*$, and the one that explains the existence of $C(+4)$ chemical compounds, is $[He]2s^12p_x^12p_y^12p_z^1$ where all three lone 2p electrons have the same $m_s$ value.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/209246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Gamma matrices and trace operator I'm trying to show that the trace of the product of the following three Gamma (Dirac) matrices is zero, i.e. $$\text{tr}(\gamma_{\mu} \gamma_{\nu} \gamma_{5})=0 \text{.}$$ I attempted to use the fact that the trace operator is invariant under cyclic permutations and linear, and that $$\gamma_{\mu} \gamma_{5}= -\gamma_{5} \gamma_{\mu}, \text{ } (\gamma_{5})^{2}= I_4 \text{ (4 $\times$ 4 identity matrix)} \text{,}$$ where $\gamma_{5} \equiv i\gamma_{0} \gamma_{1} \gamma_{2} \gamma_{3}$. But whenever I do that, it seems that I keep going in circles. Any idea on how I should proceed?
| Substitute two values for $\mu$ and $\nu$.
If $\mu=\nu$, then using $(\gamma^\mu)^2 =\pm1$ and $tr(\gamma ^5) =0$ you have finished.
If $\mu \neq\nu$, then use $tr(\gamma ^\mu\gamma ^\nu) =0$ with the two remaining $\gamma$ .
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/209445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Definition of a line charge with Dirac delta function Is the following statement correct for a line charge distribution $λ(x)$?
$$ρ(\mathbf r)=λ(x)δ(y)δ(z)$$
If yes - what does it say?
|
$$\rho(\mathbf{r})=\lambda(x)\delta(y)\delta(z)$$
describes a charge density in the form of a (possibly infinite, depends on what your allowed x values are in the system) line in 3D space, where $\lambda(x)$ is the linear charge density as a function of x. The delta functions indicates the charge density is concentrated at one point in the yz plane, but extended in the x axis
For a 1D description of the above, you will simply use
$$\rho(x)=\lambda(x)$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/209922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is it easier to drop on to a downslope? On a bicycle, why is it easier to land from a drop or jump on a slope going downwards than landing on a flat surface or on an upslope?
I've already heard answers such as "because that's how a bike can best keep going with all the momentum it's carrying from the drop" but I'm asking for a more elaborate answer that can give a good understanding of the physics involved.
| Maybe this will satisfy more:
When you imagine a body (or a bike) think of the velocity vector in space, that vector upon impact will cause a reaction in the direction of the normal vector of the surface you are falling uppon:
As you can see, the upwards reaction from the surface on the velocity vector will be multiplied by cos(Θ) where Θ is the angle between the surface normal and the velocity vector.
When the angle between the direction you are falling and the slope is nearly 0, your speed is barely absorbed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What has the potential energy: the spring or the body on the spring? Particles have gravitational potential energy due to its position in the gravitational field. We say the particle has potential energy and not the Earth (the body doing the work). Why is it not the same with a spring doing work on a body?
It is my understanding that we can define a potential energy function for all systems being acted on by a conservative force. Since the spring force is conservative, why can't we define a spring (elastic) potential energy for the body? Why is the potential energy defined only for the spring?
| Potential energy like Force occur in pair. If one has some potential energy due to 2nd, 2nd will have the same potential energy as the first. In the Gravitational potential energy equation :
$U = \frac{GMm}{r}$
The potential energy is dependent on both the masses. This value is same regardless whether it is for 1st or 2nd. Both the body can do Same amount of work. But since work is $F.d$ = $m.a.s$, due to larger mass of $M$ it will undergo negligible acceleration and displacement.
Similarly, we can equally say that potential energy due to stretched or compressed spring is equal for body or the spring. Both spring and the body has equal capacity to do work when released. if you release spring of its hinges, or release the force on the object, the object and the spring will do equal amount of work on them respectively.
Hope I didn't confuse you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why isn't the acceleration at the top point of a ball’s journey zero? When I shoot a ball vertically upward, its velocity is decreasing since there is a downward acceleration of about $9.8\,\mathrm{ms}^{-2}$.
I have read that at the top most point, when $v = 0$, the acceleration is still $9.8\,\mathrm{ms}^{-2}$ in the downward direction where $v=0$. That is, the acceleration is still the same.
But at the highest point, the ball is stationary, so it is not even moving. How can it accelerate?
| When you shoot the ball upwardly, gravity acts on it with a force $mg$ where $m$ is the mass of the ball and $g=9.81 ms^{-2}$ the Earth's gravitational acceleration.
If the initial upward velocity was $v_0$ then the instantaneous velocity $v$ is given by:
$v=v_0-gt$, so after some time $t=\frac{v_0}{g}$ the balls's velocity becomes $v=0$.
However, we know the ball will now start falling back to Earth immediately and if we defined $v_0$ as positive then $v=v_0-gt$ then now becomes negative. The acceleration $g$ hasn't changed though because the force $mg$ acts all the time during the trajectory.
The fact that at the apex of its path velocity becomes momentarily $0$ does not mean $g$ changes: it doesn't because the Earth's gravity acts on the ball, regardless of its velocity or elapsed time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 1
} |
Can tidal forces significantly alter the orbits of satellites? I would assume that there are other larger, more significant, forces acting on artificial satellites, but can tidal forces drastically alter the orbit of a satellite over time?
I was thinking this could especially be an issue for a satellite in geostationary orbit, because they have to be extremely precisely positioned. However, I could see this being an issue for satellites in other orbits as well, just not to the same degree.
| No, the movement of water bodies on earth does not significantly influence the orbit of man made satellites. Due to the movement of the water, and the shape of the earth, the center of gravity of the earth shifts slightly. Sometimes this pulls the satellite a bit more to the front, and sometimes a bit more to the back (sideways is also possible). On average, the effect is zero though.
If the orbit of the satellite is exactly synced with the orbit of the moon, there would be an effect over time. In that case, extra fuel would need to be brought to keep it in orbit. See here: https://en.wikipedia.org/wiki/Tidal_acceleration
There are minor perturbations of the orbit due to tidal forces though: https://link.springer.com/chapter/10.1007/BFb0011470 They are in the order of centimeters, whereas geostationary height is 35,786 km.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Contradiction in a simple torque/rotation problem - two ways of calculating external force don't agree Suppose we have a rigid system of two point objects (both of unit mass) connected with a massless rod, with the objects horizontal on the x axis, at a distance $r_1$ and $r_2$ from the origin, respectively. A vertical force $F1$ is applied to the first object, and causes an angular acceleration $\alpha$ of the system around the origin.
The total force on the first object is $ma = 1 * (r_1 \alpha) = r_1 \alpha$, and similarly the total force on the second object is $r_2 \alpha$. So the total force on the system is :
$r_1 \alpha + r_2 \alpha = F1_{total} + F2_{total} = (F1_{int}+F1_{ext}+F2_{int}+F2_{int}) = F1_{ext}$, since $F2_{ext}=0$ and the internal forces cancel out. So
$F1_{ext} = r_1 \alpha + r_2 \alpha. \tag{1}$
On the other hand, we know that $T_{ext} = I \alpha$. The moment of inertia $I$ here is $m r_1^2 + m r_2^2 = r_1^2 + r_2^2$, and so $F1_{ext} \ r_1 = T_{ext} = (r_1^2 + r_2^2) \ \alpha$, from which follows that
$F1_{ext} = \alpha(r_1^2 + r_2^2)/ r_1 \tag{2}$.
But this is incompatible with $(1)$ above, since $r_1 \alpha + r_2 \alpha = \alpha(r_1^2 + r_2^2)/ r_1$ only when $r_1 = r_2$.
What is wrong with the above?
| The 2nd solution you wrote down appears to be the correct solution. Offhand, I see two problems in the first solution. First, I think that a problem with the first attempted solution is that you made a subtle mistake in assuming that F=ma means that $F=mr_1α$. That seems like a plausible step at first but if you examine this step more closely you'll realize that it can't be correct to say that the acceleration of an object in this case is simply equal to the distance from the point of rotation times the angular acceleration. Suppose that the angular acceleration is zero (e.g., the masses are spinning around the origin at a constant angular velocity). According to your reasoning that would mean that the acceleration of the object is also zero. But you do know that an object spinning around a origin at a constant angular velocity doesn't have a zero acceleration, right? Rather, it has an acceleration vector which is of constant magnitude and always pointing toward the origin.
Another problem I see is that these are not freely moving masses being acted upon by only a force $F1_{ext}$. There are also the "hidden" forces of constraint which are forcing the masses to revolve around the point of origin.
Approaching this problem in terms of torques, moments of inertia, and angular acceleration looks like the most straightforward way of approaching this problem, and I believe that your 2nd solution is correct.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Physical intuition about the inertia tensor I'm studying Mechanics on Goldstein's book (Classical Mechanics) and Spivak's book (Physics for Mathematicians) and I'm in doubt about the physical intuition about the inertia tensor. On both books, the inertia tensor appears naturally when computing the angular momentum $L$ of a rigid body which, for simplicity, is only rotating.
The inertia tensor is then defined as the linear operator $I : \mathbb{R}^3 \to \mathbb{R}^3$ given by
$$I(\phi) = \sum_{i} m_i b_i \times (\phi \times b_i),$$
where $b_i\in \mathbb{R}^3$ are the initial positions of the particles of the body, and $m_i$ their masses. With this definition, it is shown that
$$L = I(\omega),$$
being $\omega$ the angular velocity of the rigid body. All of that, from the mathematical point of view, is fine.
Now, what is the physical intuition behind this? The linear operator $I$ allows one to relate, in a linear way the angular velocity and the angular momentum. This looks much like mass relates in a linear way velocity and momentum. But on the latter case, mass is a scalar while $I$ is a linear transformation.
What is, then, the best way to physically understand the inertia tensor?
| I have also encountered the same problem until recently I understood the meaning behind the indices.
Although there are a lot of definitions of a tensor, we are left to decide which one is palatable for the kind of context we are in.
$I_{xy}$ means how much the 3D object would be accelerated in the $y$ axis when I apply the torque in the $x$ axis.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Why are there only two linearly independet quartic Higgs terms for the adjoint $24$ in $SU(5)$ GUTs? I've read the statement in countless papers, for example, here Eq. 4.2 or here Eq. 2.1 without any further explanation or reference, that the "most general renormalizable Higgs potential" for an adjoint (=24 dimensional) Higgs $\phi$ is
$$ V(\phi) = -\frac{1}{2} \mu^2 Tr(\phi^2) + \frac{1}{4} a (Tr(\phi^2))^2 + \frac{1}{2} b Tr(\phi^4) $$
where I neglected the cubic term for brevity.
In group theoretical terms this potential can be written as
$$ V(\phi) = -\frac{1}{2} \mu^2 (24\otimes 24)_{1_s} + \frac{1}{4} a (24\otimes 24)_{1_s}(24\otimes 24)_{1_s} + \frac{1}{2} b ((24\otimes 24)_{24}(24\otimes 24)_{24})_{1_s} $$
Nevertheless, pure group theory tells us there are several other quartic invariants possible. We have
$$ 24\otimes 24 = 1_s \oplus 24_s \oplus 24_a \oplus 75_s \oplus 126_a \oplus \overline{126_a} \oplus 200_s ,$$
where each representation is denoted by its dimension and the subscripts $s$ and $a$ denote symmetric and antisymmetric respectively. Naively, I would say we have 7 quartic invariants:
$$ ((24\otimes 24)_{1_s} (24\otimes 24)_{1_s} )_1 + ((24\otimes 24)_{24_s} (24\otimes 24)_{24_s} )_1 +( (24\otimes 24)_{24_a} (24\otimes 24)_{24_a})_1 + ((24\otimes 24)_{75_s} (24\otimes 24)_{75_s} )_1 +( (24\otimes 24)_{126_a} (24\otimes 24)_{126_a} )_1 +( (24\otimes 24)_{\overline{126_a}} (24\otimes 24)_{\overline{126_a}} )_1 +((24\otimes 24)_{200_s} (24\otimes 24)_{200_s})_1 ,$$
because
$$ 1_s \otimes 1_s = 1 \quad 24_s \otimes 24_s =1 \quad 75_s \otimes 75_s =1 \quad etc. $$
An thus my question: Why are all these other products missing in the "most general renormalizable potential"? Maybe only two of these seven terms are linearly independent, but, at least for me, this is far from obvious. And in addition, then why are exactly these two a suitable linearly independent choice?
| The easiest answer to your question can be gleaned from a nice book by F. Iachello, Lie Algebras and Applications, Lect. Notes Phys. 708 (Springer, Berlin Heidelberg 2006), DOI 10.1007/b11785361 , ISBN-10 3-540-36236-3
SU(5) (~ A4) has rank 4, and thus 4 independent Casimir invariants (your φ transforms like the adjoint generators of the Lie algebra.) The invariants are quadratic, cubic, quartic, and quintic. The last one would provide a non-renormalizable Higgs interaction. The cubic one was rejected by fiat (the BEGN discrete iso-parity symmetry imposed)basically to simplify the model and the analysis. You thus have to take it as a mater of trust in the book that the quartic invariant is, in fact, independent of the quadratic. Unfortunately, however, you have completely garbled it.
What is multiplied by b in your second formula should have been, in your idiosyncratic language,
$$
\frac{1}{2} b Tr(\phi^4) \mapsto
\frac{1}{2} b(24\otimes 24 \otimes 24\otimes 24)_{1_s} .$$
That is, your four 24s need not compose pairwise to 24s, which then compose to a singlet!
Your first and second terms are the quadratic invariant and its square, but the last one is the quartic invariant, which you could convince yourself is independent of it, but, for slick detail you must hit the book(s). You might contrast this to the adjoint of SU(2), rank 1, where there is only one invariant, so, necessarily, $Tr(\phi^4)\propto (Tr\phi^2)^2$, easy to check by diagonalizing φ, as in your Ruegg reference.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is there a special name for thermodynamic process during which no work is performed? Let $W$ denote the work done on a system during a thermodynamic process. Is there a commonly-accepted, dedicated term for a process during which $W=0$?
| Not in general, no. In the special case of no volumetric work being done, we call the process isochoric. Volumetric work occurs when there is a change in the volume of the system, whether due to an external agent or due to the system itself. There are plenty of ways, however, that work could be done on a system without impacting its volume -- examples include rotating a rigid paddle in the system, passing a small electric current through it, etc.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Primitive unit cell of fcc When I consider the primitive unit cell of a fcc lattice (red in the image below) the lattice points are only partially part of the primitive unit cell. All in all the primitive unit cell contains only one single lattice point.
My question is how much each point at the corners of the red primitive unit cell contributes? At every corner a point is only partially inside the red primitive unit cell such that all parts together form a single point. How big are these individual parts?
In principle it should be possible to calculate that, but I hope there a known results in the literature. Unfortunately I can't find no such thing...
| The counting of 1/8 (corner) or 1/2 (face) is only relevant to the shared cubic cells. For the primitive cell, all atoms are shared by 8 primitive cells, thus, 8 x 1/8 = 1, which can be seen by the drawing of physicopath.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/210963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Central force law An object has an orbit in polar coordinates as $r(\theta) = a\theta^2$ (where $a$ is constant).
Assuming the central force is directed towards the origin $r=0$, how can I know which central force law lead to such an orbit? And how to find $r$ and $\theta$ as function of time?
| I don't want to give away the answer directly. So I will provide some hints.
A central force in polar coordinates has to be of the form: $$\vec{F} = m\vec{a} = m(\ddot r - r \dot \theta^2)\hat r$$
Now try to mess around with your $r(\theta) = a\theta^2 $
I believe you need to specify $\dot \theta$ in order to solve the full equation of motion. So pick for your self. A linear equation $\theta = kt $ may be a good starting point.
Good luck!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/211105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How does color (or reflection in general) work? I'm confused, does the absorption and emission determine the color of something? Or does that only happen when something is emitting energy?
When light hits an object, the photons get absorbed, then emitted with a different wavelength right?
| Not really as in your assumption, if that happens, you can have (for example) either fluorescence or raman. The color you see is determined by the light which is reflected from the object, i.e. by the light which is missing compared to the white light. Each material only reflects certain wavelengths, and absorbs others, influencing the color you can see.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/211388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why does light bend? I read about the dispersion of light by a prism and a block (slab), but I don't understand why light bends at all.
I know that red light has the longest wavelength and that energy is inversely proportional to wavelength, hence red light contains the least energy. I also know that it bends the least. But why? Why does red light not bend as much as violet light?
Please don't use Snell's law in your answer.
| To answer this question first you need to understand what prisms are made of, usually glass, that is silica (SiO2).
Now the atomic size of for example an atom inside the prism is 60 pm, that is 0.06 nm.
Now this size is very small compared to visible light photons' wavelength which is about 400-700 nm.
When the photon's wavelength is much bigger then the atom's size they interact with, the interaction can be described (and in the case of glass is best described) by elastic scattering (Rayleigh), by the way this is the reason why the sky is blue.
is the predominantly elastic scattering of light or other electromagnetic radiation by particles much smaller than the wavelength of the radiation.
https://en.wikipedia.org/wiki/Rayleigh_scattering
Now this causes the photons with shorter wavelength to interact with the atoms more (higher probability), causing the angle to change more in the case of shorter wavelength.
Just like the sky is blue, that is, the shorter wavelength photons get scattered more (higher probability), and change angle more into our eyes to make the sky look blue, analogously the shorter wavelength photons will interact with the atoms in the prism more and scatter more and change angle more.
As you say, the red light photons contains the least energy, have the longest wavelength (in the visible range), and interact with the atoms the least, thus, they follow an almost straight path through the prism.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/211473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Stability of a system of Brownian particles with non-physical collision A few months ago I wrote this simulation of a system of circles bouncing off each other. It's a two-dimensional box with elastic balls in it that bounce off each other. I came back to it and noticed that I didn't sanitize the input for the "elasticity" constant (which multiplies the resultant velocities after performing momentum transfer on collision) and I started putting in values above 1. Of course, this makes no sense physically, but it was still interesting nonetheless.
I noticed that there seems to be a certain threshold of elasticity beyond which the total energy of the system becomes unbounded and grows exponentially instead of slowing to a halt due to friction (dynamic friction from moving). Can the system be simplified to a point where this threshold can be easily calculated as a function of the friction, the initial energy of the system, and the size of the spheres?
I found this question (Collision time of Brownian particles) which is very much related, but doesn't quite answer the general case with many particles.
| Since you programmed the simulation, you should know or otherwise have access to the underlying equations simulating the dynamics of your particle collisions. If this dynamic equation is a linear difference equation, you can transform the equations into the z domain and express the equations as a transfer function. As a transfer function you can solve for the roots of its denominator, and if any of the roots exist outside the unit circle, the system will behave in a divergent manner and be unstable. This would be the analytical version of the 'smoking gun' to explain the growing energy you observed.
It really depends on the structure of your equations, but as a very simple explanation for feedback in a linear discrete systems, if the feedback (loop) gain is greater than 1, your system is unstable. In terms of physical parameters, you might be modeling say the parameter of viscosity that adds energy to the system rather than dissipates it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/211974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What happens to mass during beta decay? Sorry for being ignorant, but I'm in high school and our chemistry teacher barely went over beta decay. I decided to do some research and learned that in β+ decay, positrons are emitted from protons in order to turn it into a neutron. But positrons have mass, so where does that mass come from? Do protons "give it mass"? If so, why wouldn't protons lose mass, and how could they become neutrons if neutrons are more massive?
| In beta decay, the mass difference between the parent and daughter particles is converted to the kinetic energy of the daughter particles. For instance, in the decay of the free neutron,
$$
\rm n \to p + e^- + \bar\nu_e, \tag{$\beta^-$ decay}
$$
the difference between the mass on the left and the mass on the right is about $0.78\,\mathrm{MeV}/c^2$, and this is the energy liberated in the decay. (If you're a chemistry person, an eV is a useful energy unit; the $E=mc^2$ conversion is roughly $1000\,\mathrm{MeV}\approx 1\,\mathrm{amu}\times c^2$.) Equivalent processes like
$$
\rm p + \bar\nu_e \to n + e^+ \tag{neutrino capture}
$$
don't occur unless the kinetic energy on the left side is already large enough to account for the extra mass on the right side. Since the electron/positron mass is about $0.51\,\mathrm{MeV}/c^2$, neutrino capture on protons at rest is impossible for neutrinos with less than $1.80\rm\,MeV$ kinetic energy. This means, among other things, that neutrinos emitted from neutron decay at rest will never have enough energy to cause positron emission on protons at rest elsewhere.
You get $\beta^-$ decay from free neutrons because free neutrons are heavier than free protons. However it's not the case for all nuclei that the more positive isobars are less massive. For instance, the mass difference between postassium-40 and argon-40 is about $1.50\,\mathrm{ MeV}/c^2$, with potassium (19 protons) heavier than argon (18 protons), so the decay
$$
\rm ^{40}_{19}K \to {}^{40}_{18}Ar^- + \beta^+ + \nu_e + 0.48\, MeV
$$
is allowed (though rarer than some other branches) and merrily proceeding inside the bananas on your kitchen counter.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/212168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How proof the Lorentz algebra using the Poincaré algebra?
Show that
$$[J_{i},J_{j}]=i\varepsilon_{ijk}J_{k},\quad [K_{i},K_{j}]=-i\varepsilon_{ijk}J_{k}, \quad [J_{i},K_{j}]=i\varepsilon_{ijk}K_{j},$$
using $$[M_{\mu\nu},M_{\rho\sigma}]=ig_{\nu\rho}M_{\mu\sigma}+ig_{\mu\sigma}M_{\nu\rho}-ig_{\mu\rho}M_{\nu\sigma}-ig_{\nu\sigma}M_{\mu\rho}. $$
Hint: $M_{\mu\nu}=-M_{\nu\mu}$, $M_{0i}=K_{i}$ and $M_{ij}=\varepsilon_{ijk}J_{k}$.
This doubt is in the page 4 of following reference: http://arxiv.org/abs/hep-th/0101055
| Just to take one term: letting $\mu = \rho = 0$ and $\nu = i \in \{1,2,3\},\;\sigma = j \in \{1,2,3\}$ we can immediately write$$[M_{0i},M_{0j}] = [K_i, K_j] = i g_{i0} M_{0j} + i g_{0j} M_{i0} - i g_{00} M_{ij} - i g_{ij} M_{00}.$$
Antisymmetry means $M_{00} = 0$ and the fact that $g$ is diagonal [more specifically, $g = \operatorname{diag}(1, -1, -1, -1)$ as far as I can tell] means that $g_{0j} = g_{i0} = 0$ as these are off-diagonal terms. That leaves just $g_{00} = 1$ and $M_{ij} = \epsilon_{ijk} J_k,$ proving:$$[K_i, K_j] = -i \epsilon_{ijk} J_k.$$Other terms are left as an exercise for you.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/212714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For a diatomic molecule, what is the specific heat per mole at constant pressure/volume? At high temperatures, the specific heat at constant volume $\text{C}_{v}$ has three degrees of freedom from rotation, two from translation, and two from vibration.
That means $\text{C}_{v}=\frac{7}{2}\text{R}$ by the Equipartition Theorem.
However, I recall the Mayer formula, which states $\text{C}_{p}=\text{C}_{v}+\text{R}$.
The ratio of specific heats for a diatomic molecule is usually $\gamma=\text{C}_{p}/\text{C}_{v}=7/5$.
What is then the specific heat at constant pressure? Normally this value is $7/5$ for diatomic molecules?
| "At high temperatures, the specific heat at constant volume $C_v$ has three degrees of freedom from rotation, two from translation, and two from vibration." I can't understand this line. $C_v$ is a physical quantity not a dynamical system. So how can it have a degrees of freedom?? You can say the degrees of freedom of an atom or molecule is something but it is wrong if you say the degrees of freedom of some physical quantity(like temperature, specific heat etc.) is something. Degrees of freedom is the number of independent coordinates necessary for specifying the position and configuration in space of a dynamical system.
Now to answer your question, we know that the energy per mole of the system is $\frac{1}{2} fRT$. where $f$= degrees of freedom the gas.
$\therefore$ molar heat capacity, $C_v=(\frac{dE}{dT})_v=\frac{d}{dT}(\frac{1}{2}fRT)_v=\frac{1}{2}fR$
Now, $C_p=C_v+R=\frac{1}{2}fR+R=R(1+ \frac{f}{2})$
$\therefore$ $\gamma=1+ \frac{2}{f}$
Now for a diaatomic gas:-
A diaatomic gas has three translation(along x,y,z asis) and two rotational(about y and z axis) degrees of freedom. i.e. total degrees of freedom is $5$.
Hence $C_v=\frac{1}{2}fR=\frac{5}{2}R$ and $C_p=R(1+ \frac{f}{2})=R(1+ \frac{5}{2})=\frac{7}{2}R$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/212986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Where does the energy go in a rocket when no work is done? While playing Kerbal Space Program, I wondered where my chemical energy would go when fired at 90° to the motion. It would do no work on the rocket, but all that energy has to go somewhere, right? Anyway, my question is, where does the energy go?
| If a train is moving along the tracks at 60 mph and there's a crosswind at 30 mph the crosswind, in fact, does not do any work on the train. That's because the train's motion is constrained by the rails so that it cannot move in the direction that the wind is pushing it. If you remove the constraint its motion will change because the wind is doing work on it. That's why, for example, if you row a boat straight across a river you end up downstream from where you started: the current moved you downstream. In order to end up directly across from where you started you have to aim the boat upstream, to compensate for the current. Either way, the current is doing work on the boat. Same thing for your rocket.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/213279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Time derivative of Noether charge I understand that the Noether charge can be written as
$$ Q= \int~\mathrm d^3 x ~J^0$$ and the time derivative of the Noether charge is zero $$ \dot Q=0 $$
but how would you explicitly calculate it?
| While Sebastian nailed it in his comment, I'll post the same as a proper answer.
The four-vector current $J^{\mu} = (J^0, J^i) \equiv (J^0, {\vec J})$ obeys the continuity equation
$$\partial_{\mu} J^{\mu} = 0,$$
or in other words:
$$\frac{\partial J^0}{\partial t} + {\vec \nabla}\cdot{\vec J} = 0$$
This represents the statement of local conservation of the Noether charge (whatever that may be: electric charge, baryon number, or simply any Noether charge).
The next bit is global conservation. The same follows from the above by integrating both sides over entire space, which yields:
$$\frac{\partial}{\partial t} \left(\displaystyle\int~\mathrm d^3 x ~J^0 \right)+ \left(\int ~\mathrm d^3 x \ {\vec \nabla}\cdot{\vec J} \right)= 0$$
The second term on the right reduces to a surface integral with the use of the divergence theorem, and the first term in simply $\partial Q/\partial t$. Now, since the surface integral is on the surface enclosing the volume, and we have taken the integration over entire space, we are talking about a surface integral at $r \to \infty$. Thus, the behavior of the surface integral $\displaystyle\int \left({\vec J}\cdot {\hat n}~ \mathrm dS \right)$ depends on the behavior of ${\vec J}$ as a function of $r$. In fact, by substituting for the area element, $\mathrm dS = r ~\mathrm dr ~\mathrm d\theta$, it may be verified that this surface term would $\to 0$, as $r \to \infty$, provided $J$ falls faster than $1/r^2$. In this case, and only in this case, the global conservation of the Noether charge holds:
$$\frac{\partial Q}{\partial t} = 0\,.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/213400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Gradient, divergence and curl with covariant derivatives I am trying to do exercise 3.2 of Sean Carroll's Spacetime and geometry. I have to calculate the formulas for the gradient, the divergence and the curl of a vector field using covariant derivatives.
The covariant derivative is the ordinary derivative for a scalar,so
$$D_\mu f = \partial_\mu f$$
Which is different from
$${\partial f \over \partial r}\hat{\mathbf r}
+ {1 \over r}{\partial f \over \partial \theta}\hat{\boldsymbol \theta}
+ {1 \over r\sin\theta}{\partial f \over \partial \varphi}\hat{\boldsymbol \varphi}$$
Also, for the divergence, I used
$$\nabla_\mu V^\mu=\partial_\mu V^\nu + \Gamma^{\mu}_{\mu \lambda}V^\lambda = \partial_r V^r +\partial_\theta V^\theta+ \partial_\phi V^\phi + \frac2r v^r+ \frac{V^\theta}{\tan(\theta)} $$
Which didn't work either.
(Wikipedia: ${1 \over r^2}{\partial \left( r^2 A_r \right) \over \partial r}
+ {1 \over r\sin\theta}{\partial \over \partial \theta} \left( A_\theta\sin\theta \right)
+ {1 \over r\sin\theta}{\partial A_\varphi \over \partial \varphi}$).
I was going to try
$$(\nabla \times \vec{V})^\mu= \varepsilon^{\mu \nu \lambda}\nabla_\nu V_\lambda$$
But I think that that will not work. What am I missing?
EDIT: The problem is that the ortonormal basis used in vector calculus is different from the coordinate basis.
| I made two YouTube videos explaining how to due precisely these problems.
The first one explains how to use standard covariant derivatives (what you are using) to compute the divergence and gradient in spherical coordinates:
https://www.youtube.com/watch?v=jEvPY6-ISUI
And the other explains how to compute the curl in spherical coordinates using covariant derivatives:
https://www.youtube.com/watch?v=ZatyvboG58Q
They show the explicit calculation for all three operators, and explain the principles behind the process so that it can easily be applied for other cases in the future. They literally answer precisely your question.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/213466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
Maintaining symmetry? Minkowski metric is found to be
$$ds^2=-dt^2+dr^2+r^2d\Omega^2$$
where $d\Omega^2$ is the metric on a unit two-sphere.
Why should we keep track of the $d\Omega^2$ so that spherical symmetry holds well?
| See your own question here: Why do people put exponentials there
You can multiply by functions of $r$ on each of the terms. What you cannot generally do is "muck" with the $r^2 d\Omega$ part itself. That's what gives you the part of the distance that corresponds to the angular directions on the sphere. If you change that, then depending on where you are, you'll get different proper distance traveled for different amounts of angle traversed. I'm being a little sloppy in the language here, but this is the basic idea.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/213641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Modes inside a cavity and black body radiation Consider a perfect conductor that encloses a spatial volume such as a parallelepiped or cylinder. If we solve Maxwell's equations inside that volume, seeking solutions that depends on time with a dependency of the form $e^{-i\omega t}$, we find that only TE and TM modes can exist inside of that volume (and no TEM modes). However both TE and TM modes have a cutoff frequency. This seems to imply that any EM wave inside the cavity cannot have any frequency and that it should be greater than a threshold.
However if we look at the problem from another perspective, the one of a black/grey body at a temperature $T > 0K$, we'd think that the walls are emitting EM waves without any cutoff frequency (and with a continuous spectrum).
I understand that the sum of two solutions to Maxwell's equations in the cavity is also a solution and I think that I could write any allowed EM wave as a sum of TE and TM modes, but if both TM and TE modes have a cutoff frequency, I don't see how I could obtain an EM wave with a lower frequency that the cutoff one.
Hence I don't see how to reconciliate the blackbody radiation with TE and TM modes inside a cavity. Where do I go wrong?
| It's true that a hollow conductor has a minimum cutoff frequency.
However, a hollow conductor is not a black body.
A black body has perfect absorption of radiation at all frequencies, while a perfect conductor perfectly reflects all radiation.
A black body emits radiation according to the Planck law.
Since the black body absorbs all incoming radiation, but then re-emits it according to the Planck law, it must be capable of converting from e.g. a single high energy photon to several lower energy ones.
The hollow perfect conductor can't do this because it reflects all radiation without ever absorbing and "processing" it.
For more details I strongly recommend reading the other Physics Stack Exchange post Why is black the best emitter? and the associated answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/213860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Thermal state vs equibilibrium state Could someone explain what's the difference between a thermal state and an equilibrium state? Or is it even the same?
| Basically there are two types of equilibrium in thermodynamic context :
*
*Thermal equilibrium. It's when in system there are no heat flow :
$$ {\frac {d Q}{d t}}= 0 $$
*Thermodynamic equilibrium. No net flow of matter or energy in a system.
So for example take a look at this diffusion process of dye in a water. In this case system is in thermal equilibrium, but not in thermodynamic equilibrium, because there exist net macroscopic flow of dissolved dye particles in a vessel. (However I bet that there may be even a small heat flow too, due to dye particles absorbing water molecules kinetic energy, but this needs to be tested). So in this process, there is a matter flow, thus breaking thermodynamic equilibrium. At other cases, this type of equilibrium can be broken by energy flow. It's when in system there exist a net radiation or other means of energy transfer. For example, when you bump with a metal stick on a railroad track,- you generate sound wave which transfers vibrational energy across track, thus breaking track's thermodynamic equilibrium.
EDIT
If thermodynamic equilibrium (TE) to be formalized, then :
$$ \frac {d E}{d t} + c^2 \frac {d m}{d t}= 0 $$
There are two solutions to this :
*
*$\frac {d E}{d t} = 0\,,\,\frac {d m}{d t} = 0$ $\to$ inflow of energy and mass zero
*$\frac {d E}{d t} = - c^2 \frac {d m}{d t}\,\to$ inflow of energy equals to outflow of mass (or in reverse).
As an example of this case imagine laser pulses heating gas effectively converting it to a plasma, at the same time pushing heated plasma out of gas container. In this case container will be in TE.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/213971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Will this rope break due to the tidal forces or not? While I was thinking about how tidal forces can make objects float at the surface of a planet orbiting a massive object like a black hole, the fact that any material on the Earth isn't held together by gravity only, but also by chemical bonds which give it its tensile strength came into my mind.
The scenario I thought of is as following:
1- A 10 kg metal sphere is tied to a 1 meter long thin rope and this rope is nailed into the surface of the Earth.
2- The tensile strength of the rope is 10 N. (So the metal sphere has to accelerate at 1 $m/s^2$ to break the rope)
3- The Earth is orbiting a black hole at its Roche radius, and so our metal sphere on the surface of the Earth is effectively weightless and floating, but it is still held by the rope.
Here is a simple picture to summarize:
Now, if we move the Earth to orbit the black hole even closer until the tidal forces on Earth due to the black hole become $\Delta a$ = 10.8 $m/s^2$ and so the metal sphere is pulled by the difference between $\Delta a$ and the Earth's gravitational acceleration (9.8 $m/s^2$) which is 1 $m/s^2$ towards the BH, will the rope break or not ?
In other words, if a is the gravitational acceleration towards the BH, in order to break the rope, which one do we need ? :
1- $a_{Point A}$ - $a_{Point B}$ = 1 $m/s^2$
OR
2- $a_{Point A}$ - $a_{Point C}$ = 1 $m/s^2$
| Between points A and C. The effect of having the Earth there is that it is providing a link between the center of mass of the Earth (C) and the test sphere (A). Effectively, there is an infinitely strong, rigid rod going from B to C, which is then connected to the rope. The tensions in that rod (the Earth) must match that of the rope.
Another way to think about it, is that if you removed the Earth entirely from the problem, and wanted to see if the rope was strong enough to resist the tidal forces --- then you would consider the difference in acceleration between points A and B.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
The Speed of Sound Just a question about Physics I'm doing at school.
If the speed of sound is inversely proportional to the density of a material, why does sound travel faster in solids (it is the most dense).
I have read that it takes more energy for sound to travel in dense materials so it takes longer but then neighbouring molecules are closer so sound does not have to travel that far, making it faster. This doesn't make any sense because it says the more dense a medium is, sound is both faster and slower.
Also, how does bulk modulus affect the speed of sound.
| short answer: wave speed (indeed, its square) is the ratio of rigidity and mass. For (most) solids and liquids, as compare to gaz, the point is not that they are denser, but that they are near-incompressible: numerator wins.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Killing Vectors in Schwarzschild Metric Given the Schwarzschild metric with $(-,+,+,+)$ signature,
$$\text ds^2=-\left(1-\frac{2M}{r}\right)dt^2+\left(1-\frac{2M}{r}\right)^{-1}dr^2+r^2(d\theta^2+\sin^2\theta\,d\phi^2)$$
the lack of dependence of the metric on $t$ and $\phi$ allow us to read off the Killing vectors $K_1=\partial_t$ and $K_2=\partial_{\phi}$. These vectors, in their coordinate representations, are given by
$$K_1=\left(-\left(1-\frac{2M}{r}\right),0,0,0\right)$$
$$K_2=\left(0,0,0,r^2\sin^2\theta\right)$$
How does one immediately read off those vector components for $K_1$ and $K_2$? What is the logic behind reading them off? How would I "read off the Killing vectors" if I, while maintaining no explicit dependence on $t$ or $\phi$, added some off-diagonal terms to the metric? Please help me intuitively understand what's going on here.
| If all components of the metric are independent of some particular $x^\nu$, then you have the killing vector $\vec{K}$ with components $K^\mu = \delta^\mu_\nu$. That is, the contravariant form just has a constant in the appropriate slot and zeros elsewhere. In Schwarzschild, you have $K^\mu = (1, 0, 0, 0)$ and $R^\mu = (0, 0, 0, 1)$ ($\vec{K}$ and $\vec{R}$ being your $K_1$ and $K_2$, respectively).
To find the covariant forms, simply lower with the metric. In Schwarzschild we have
\begin{align}
K_\mu & = g_{\mu\nu} K^\nu = g_{\mu t} = \big({-}(1-2M/r), 0, 0, 0\big) \\
R_\mu & = g_{\mu\nu} R^\nu = g_{\mu\phi} = \big(0, 0, 0, r^2 \sin^2\!\theta\big).
\end{align}
This is where off-diagonal terms would come in. For example, in Boyer-Lindquist we also have no $t$-dependence, so we have $K^\mu = (1, 0, 0, 0)$ and
$$ K_\mu = g_{\mu t} = \big({-}(1-2Mr/\Sigma), 0, 0, -(2Mar/\Sigma)\sin^2\!\theta\big), $$
where the fourth component is precisely $g_{t\phi}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Why are HCP materials brittle while FCC materials are ductile? Why are hexagonal close packed materials brittle, While face centered cubic is ductile. Is it related to crystal planes?
| please learn a few definitions
Slip plane – is the plane of greatest atomic density.
Slip direction – is the close-packed direction WITHIN the slip plane
Slip system = slip plane and slip direction TOGETHER
THEN;
5 independent slip systems are necessary to make a polycrystalline
material ductile.
HCP - Has three slip systems (one plane and three directions, giving 3x1= 3 slip systems, we know that minimum 5 independent slip systems are necessary to make a polycrystalline material ductile.therefore HCP is brittle.
FCC - has 12 slip systems (three {111} family of planes and four <110> family of directions, giving 3x4 =12 slip systems, which is more than 5 independent slip systems therefore FCC is ductile.
BCC -has 48 slip systems and expecting better ductile but it is brittle (six {110} family of planes and two <111> family of directions =6x2 = 12 slip systems + six {211} family of planes and two <111> family of directions =6x2 = 12 slip systems + six {321} family of planes and four <111> family of directions =6x4 = 24 slip systems; grand total 12+12+24 = 48 slip systems)
BCC lattice structure has too much of slip systems(48), here slip systems are INTERFERE OR MUTUALLY OBSTRUCT each other therefore slip movement in BCC is made very difficult thus BCC is brittle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to find the direction of an eddy current? Suppose there is a magnetic field going from left to right. Suppose a thin sheet of metal conductor (e.g. a 1m*1m square) is dropped through the magnetic field such that the plane of the conductor is PERPENDICULAR to the magnetic field.
Now I know that by Faraday's Law, there will be an induced emf that will induce eddy currents in the conductor which oppose the motion of it by Lenz's Law. However, I have no idea which way the eddy currents are flowing, i.e. clockwise or anticlockwise.
Does there exist a simple hand rule which can predict the direction of eddy current?
| An eddy current is generated due to Lenz's law, so the current will produce a magnetic field in order to oppose the change that created it right.
So for example, You move a metal sheet into a magnetic field, a current will be created so as to OPPOSE this force moving it into the field. So the force that the eddy current creates will be to the left. Using your right hand palm rule. Fingers point into the page since magnetic field is into the page, and palm points to the left....so your thumb points up. Now eddy currents whirl around in a circle, so imagine the current like a circle. Your thumb pointing up means that the current is going anticlockwise.
You may get confused as to whether your thumb is pointing up at the 3 o clock position or the 9 o clock position of the circle. But since the metal sheet is moving from left to right, it is the 3 o clock position that the current is pointing up in.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Single Slit Diffraction I am trying to derive the intensity variation function for a single slit diffraction.
Sorry for the poor diagram...
So I decided to take the amplitudes of the waves originating from the slit on the left (wherein the variable that denotes distance within the slit is $l$) and integrate the amplitudes over the entire slit width, taking some point at a distance $x$ on the screen to achieve the resultant amplitude of the waves that strike the screen. With this function, I decided I would use the standard expression for intensity (i.e. $I=\kappa A^2)$
The amplitude for a wave originating from a point on the slit should be:
$$ y=a\sin{kr}$$
where $r$ is the distance between the point of origin on the slit and point of contact on the screen (and $k$ is the angular wave-number).
So:
$$ r^2=D^2+(x+l)^2$$
and on approximating:
$$ r\approx D+\frac{1}{2D}(x+l)^2$$
So I took the amplitude function (for the screen) as $A(x)$ and:
$$ A(x)=a\int_{-l/2}^{l/2}\sin{kD+\frac{k}{2D}(x+l)^2} dl$$
substituting $k(x+l)/2D=u$ (ignoring limits for now):
$$ A(x)=a\sqrt{\frac{2D}{k}}(\sin{kD}\int_{l_1}^{l_2}\cos{u^2}du+\cos{kD}\int_{l_1}^{l_2}\sin{u^2}du)$$
I looked these integrals up so I know that they are Fresnel Integrals, but more importantly that they are transcendental functions.
So my questions are:
*
*Are my assumptions flawed?
*Is there a flaw somewhere in the procedure?
*If what I've done is correct, how shall I proceed?
| It is not possible to write a closed form equation for the Fresnel diffraction pattern. Usually one will use the Cornu spiral to evaluate problems like this.
The Cornu spiral is a graphical tool that maps the phase / amplitude contribution of a infinitesimal element of the aperture.
(image by R. Nave, from http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cornu.html#c2 )
To get the phase and amplitude at a particular point on the screen, you need to determine the position of the left and right hand edge of the screen in terms of normalized parameters $v$, which represents the phase difference from the point on the aperture to the point on the screen. In your case, you have a plane wave incident, and the parameter $v$ is
$$v_± = \frac{\sqrt{D^2 + (x±\frac{\ell}{2})^2}-D}{\lambda}$$
You then draw the line from $v_-$ to $v_+$ to get a line that represents both amplitude and phase of the wave at a given point on the screen.
A derivation of the shape of the curve (which is a representation of the Fresnel integrals, as you correctly found) can be found here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Are chemical bonds matter? So it recently blew my mind that chemical bonds have mass. And that a spring that's wound up similarly weights a little more.
But there is a distinction between mass and matter.
I believe that a chemical bond, even though it has mass, is not considered matter and is instead a form of energy.
If I'm getting any of that wrong, I'd love to hear the rational.
| I have to say you have this backwards. Energy is released when atoms form bonds and therefore a decrease in mass takes place.
$E_{unbondedsystem}$ < $E_{bondedsystem}$ $therefore$ $M_{unbondedsystem}C^2$ < $M_{bondedsystem}C^2$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/214964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Streamlines and line of flow of fluid particles
*
*Can the streamlines of a fluid particle show the position of the particle at a time(using the streamlines)?
*I know that streamlines cannot intersect because at a specific instant the particle reaching the intersection will have two different directions of motion (in the steady flow of particles), but can I say the same thing in turbulent flow for the line of flow of the particles, why or why not?
*Is it necessary that a streamline will be a perfect straight line(to a good approximation)? If no why is it called a line then?
| *
*On a steady flow, streamlines correspond to the trajectory of "fluids particles" or parcels. (Not to be confused with the one of the real "particles" that are the molecules.)
*But if it's not a steady flow, it's wrong. You might even see appearant source and sink in the lines that do not exist as a flow.
*Steady flow streamlines cannot cross because of basic continuity and conservation laws: otherwise a particle should teleport itself on the other side of the obstacle that is the crossing flow ;-)
*A streamline is a curve, not especially a straight line.
*More on wikipedia.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What it sounds like when I'm travelling at the speed of sound totally hypothetical here:
lets say a man is playing a song on a guitar and I begin travelling quickly away from the guitar, if I were to reach the speed of sound, what will I hear? (my assumption is that I will hear a single note humming in a constant state...like pressing a key on a synth).
assume im not in a vehicle and the sound of air wizzing past me isn't involved...not a practical situation, just hypothetical.
total noob here, my apologies.
and to take it a step further...if i can speed up or slow down (move forward or backward) ever so slightly from the current note "im in", then back to the speed of sound at another note, would this be possible?...to move from one note of the song to another?
| You would hear Whoosh... and Boom! The sound of air rushing past your ears traveling at the speed of sound and the shock waves would entirely drown out the sound of any guitar even with the most powerful amps! ;)
And if you say well then remove the air and do the experiment in space - you will not hear any sound since it requires the air to propagate. ;) ;)
But assuming you could somehow overcome the more practical issues, the answer is you would not hear the sound at all as Bob Bills says. Even if the guitar started playing before you took off - when you reached sonic speed the guitar sound waves would be Doppler shifted to zero frequency.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Significance of Time constants in LR and RC circuits What is the significance of a time constant in a circuit? Is there any use of knowing the time at which the charge in a capacitor is 1-1/e times its total charge?
What is special about the capacitor having 63% of total charge?
(I know how to find time constant)
| As you may know, it takes infinite time to charge a capacitor. So, the time when the capacitor is 100% charged never comes. Thus, we require a Time Constant to help us understand the time when the capacitor has got a decent amount of charge and after which the rate of charging becomes really slow and thus charging further is not of much use.
You may also think of it in another way :
All Electrical or Electronic circuits or systems suffer from some form of “time-delay” between its input and output, when a signal or voltage, either continuous, ( DC ) or alternating ( AC ) is firstly applied to it. This delay is generally known as the time delay or Time Constant of the circuit and it is the time response of the circuit when a step voltage or signal is firstly applied. -ElectronicsTutorials (source).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Raw data acquirable from amateur astrophotography What raw data can I possibly acquire from an 8" Classical Dobsonian Telescope, and a DSLR? Could anything eye-opening to amateur astronomers be computed or calculated first-hand with such equipment? I'm sure scientists must've considered this equipment "advanced technology" at some point in history not too far back...Could I rediscover or calculate some Laws (like Kepler's laws) or some other things amateur astronomers would be amazed to calculate themselves (like the distance to a planet) using this equipment?
| I just stumbled upon your question. If you go to astrobin.com and put Dobsonian in the search field. You'll see a huge number of excellent images taken with Dobsonian telescopes. Most are of the moon and planets but a few are of the brighter deep sky objects such as the Great Orion Nebula. There are better instruments to use for astrophotgaphy but if you have access to a Dobsonian and an iphone, you may be surprised what you can get just holding it to the eyepiece! It will help if you check out your local astronomy club. They will be happy to teach you and some even lend out equipment.
Cheers,
BB
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
integrability and area-preservation property of dynamical systems Suppose we have a map defined on a plane, $x_{1}=f(x_{0})$, where $x \in \mathbb{R}^{2}$. Assume it is integable: there exists a function $I$ of the phase space variable $x$ such that $I(x)=I(f(x))$. I do not assume global integrability in the sense of Liouville-Arnold - just that the phase space is foliated by invariant curves of the type $I(x)=c$, where $c$ is constant.
Must this map be necessarily area-preserving?
I do not know either how to prove this, or come up with a counter example. Similarly, for maps defined for $\mathbb{R}^{n}$...
| No.
How about, e.g., the Hénon map with $a=1.25$ and $b=0.3$? It has an attractor, so it's not area-preserving, and this attractor is periodic, so the system is integrable.
Unless our definitions don't match, any integral map in $\mathbb{R}^n$ that has an attractor is a counter-example to integrability and area-preservation being linked.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is Earth really a magnet? I am a student of class 9. When I was going through magnetism and read that an earth is a magnet I got some doubts. My question is: is earth really a magnet? Doesn anyone have any proof that earth is a magnet? Is there a magnetic core at the center of the earth? Has anyone reached the core of the earth?
| The difference between a magnetic substance and non-magnetic substance is their difference in spin alignment or we may call it spin polarity. That is why we have ferromagnets, diamagnets and antiferromagnets. You can google for more information on source of magnetism.
The Earth behaves as a magnet because it affects those magnetic substances (e.g. compass or any bar of magnet) and its magnetic field is ordered regularly. There is no doubt that the Earth is like a bar of magnet! The core is a dynamo of electrical current. And electricity and magnetism are two faces of the same coin. You can read Maxwell's unification theory. It is simple and understandable with your current knowledge because algebra is enough to understand it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/215747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 4,
"answer_id": 3
} |
How to start an artificial gravity? I understand how artificial gravity in space stations works. It is by normal force the wall exerts on the foot.
But I wonder how to start it in the first place. I just learned about centrifugation in a centrifuge. To start, the side-wall of the tube produce a tangential acceleration. Because of the inertia (tendency to go tangentially) of the material contained, normal force is thus needed to keep the material from going through the tube and keep it rotating in a circle.
But in the space station, there is no friction, so there is no way to create that tendency that produces the need for normal force in the first place.
| You are quite correct that if you have items floating freely inside your space station they won't experience any artifical gravity as the station starts spinning. The artificial gravitational acceleration of an object is a consequence of its tangential velocity $v$ and is given by:
$$ g = \frac{v^2}{r} $$
where $r$ is the distance to the axis. The freely floating objects will initially have $v = 0$ and therefore $g = 0$.
As several comments have said, when you start spinning the station the air inside will start spinning as well, and that will produce aerodynamic forces on the floating objects that will eventually propel them towards the outer wall of the space station.
But assuming we're talking about the real world, you don't leave things floating about in your space station especially if you're applying any forces to it. You'd secure everything to the station walls, so everything in the station would start spinning along with the station.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Conformal compactification and the use of spinors (Twistor theory) I was reading the book from Huggett and Tod "An introduction to twistor theory" and as the book evolves they reach to the necessity to "found" a Lie derivative of a spinor respect to a conformal vector field. Then, they explain the problem that one has when one wants to work with conformal vector fields, because they don't have a complete representation in the Minkowski space (a type of conformal transformations, defined by this vector fields, sends the light cone to the "infinity"). In order for everything works well, they Compactify the Minkowski space and then all conformal transformation can be represented in this other space without worries.
Done this, they continue working with the spinors of the Minkowski space. They found the Lie derivative of a spinor, and proceed to found the way the zero-rest field equation and twistor equation changes under a conformal re-scaling, and also how their solutions change under conformal transformations (for this, they occupy the Lie derivative).
$\textbf{So here is my question:}$ Why they compactified the Minkowski space if they would still working with the geometry of Minkowski space (which is filed in the spinor representation)?. I mean, if they still working with the spinors of the Minkowski space then the problem that is in doing a conformal transformation still remains, right?
| Twistor equations are invariant under conformal rescaling of metric and all geometric notions used in twistors are essentially conformally invariant. Twistors are reduced spinors of the pseudo-orthogonal group O(2,4) which happens to be locally isomorphic with conformal group of Minkowski space. So the description of space-time that you get using twistor coordinates enjoys the fifteen parameter conformal symmetry group, and not just the ten parameter Poincare group. This explains why we need a compactified Minkowski space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What is Pressure Energy? While deriving Bernoulli's Theorem, our teacher said that the sum of KE, PE and Pressure Energy per unit volume remains constant at any two points.
$$P + \rho g h + \frac{\rho v^2}{2} = \text{Constant}$$
In this, he stated that the first term is Pressure Energy per unit Volume. What exactly is meant by Pressure Energy?
I know we can write:
$$P = \dfrac{F}{A} = \dfrac{F\cdot d}{A\cdot d} = \dfrac{W}{V} = \dfrac{\text{Energy}}{V} $$
What is physical significance and expression of pressure energy?
| Consider a gas that apparently moves in the $x$-direction with speed U. The molecules are also in random motion with velocities (u',v',w'), and they also have internal energy $i$ per unit mass, coming from rotation or vibration.
The energy of an individual molecule is $$(m/2 )((U+u')^2+v'^2+w'^2+i)=(m/2)(U^2+u'^2+2Uu'+v'^2+w
'^2+i)$$ On average the term $2Uu'$ will cancel (because on average $u'=0$), so the average of the molecular energy per unit volume is $$(\rho/2)(U^2+(u'^2+v'^2+w'^2)+i)$$ displaying three kinds of energy. The term $U^2$ represents gross kinetic energy. The term $
(u'^2+v'^2+w'^2)$ is proportional to pressure. The term $(u'^2+v'^2+w'^2)+i$ is proportional to temperature. When a fluid is in motion in a way that does not lose energy these forms of energy get traded. Bernouillis equation specifies how the "pressure energy" gets traded with kinetic energy. The exact form of the trade can be derived from $F=ma$.
Also, it is the random collisions of molecules with the walls of the container that exert the force per unit area that we call pressure. This is given by the "pressure energy" per unit volume. So it makes perfect sense to regard pressure as a force per unit area, or as form of energy per unit volume, with the same dimensions in each case. A similar explanation can be given for a liquid, although the molecules have less freedom of movement.
This is, I admit, an oversimplified account, meant to do no more than remove some confusion in terminology, and give some intuitive understanding to the OP.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 8,
"answer_id": 6
} |
Splitting molecule The photon reacts with the binding electrons orbiting the two atoms.
The photons have the 'correct' wavelength for Bond Dissociation Energy (BDE).
'Splitting' the molecule involves applying the photon wavelength to separate the electron from the molecule.
With the photons being applied between the binding electrons in between the two oxygen atoms, does 'splitting' occur when there is one photon reacting with one binding electron, even when there are two binding electrons?
| When a molecule absorbs a photon it reaches to an excited state and there are various mechanisms in which the molecule can relax. Dissociation of the molecule is just one of the possibilities.
It is not necessary to ionize (to separate the electron from) the molecule for dissociation to occur. What is necessary is to excite a bonding electron, that is, an electron in the molecule which is involved in bonding. This electron can still be bound to the molecule after absorption of the photon and of course it can leave the molecule. This is to be determined by the binding energy of the electron and the energy of the exciting photon.
One side information, except certain cases (dense laser light) in the process of absorption of light by mater only one photon and one electron is involved. One electron absorbing two photons or one photon exciting two electrons is not common.
After this prelude the answer to your question is that: one photon can excite one bonding electron and as a result a molecule can dissociate. Yes, even there are two bonding electrons this happens. Because as soon as one electron is excited the molecule has a hole in a bonding level and therefore in an unstable state.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the definition of linear momentum? Every where and book I search I get that the definition of linear momentum is the amount of speed (quantity of motion) contained in it or simply it is mass $\times$velocity? So, what is an appropriate definition of linear momentum? What did Newton think when he discovered it? He certainly did not think it as the amount of speed in a body.
| I suppose Newton may have devised the momentum equation to numerically express how objects of exact speeds (but different densities) would create different effects upon impact and perhaps how much energy would be needed to move such objects to a given speed. Consider the following:
*
*a wood ball (25g, 33.5 cc) hurled at a sheet metal target at an average speed of 9 m/s. p = 225
*a lead sphere (380g, 33.5 cc) hurled at a sheet metal target at an average speed of 9 m/s. p = 3,420
Most people would probably intuitively know which of those objects would cause the greatest impact. We can calculate that the lead sphere needs over 15 times the momentum (force) to get to the same average velocity as the wood ball--which also means more force or energy is needed to move the lead to that average speed. It also means it's impact force is greater.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Why light moves sideways? Greatings!
I'm trying to understand special relativity and have one question bugging me.
In almost every book or video about the subject there is a thought experiment with moving light clock. I hope I need not elaborate on the sutup and the outcome of the experiment.
So the question is this: When a light source on the MOVING light clock emits a photon of light the path of the photon is triangular.
How this can be?
I thought since light speed is constant in every reference frame the movment of the "emmiter" can't affect the movment of the light so light should shoot right up (from the point of stationary observer) and thus move backward and up from the point of moving observer.
So you see my logic is: light speed is invariant => you can't make it move UP & FORWARD only UP. i.e. you can't change it's direction once it's emited in UP direction. Why the hell light moves sideways then?
And I'm puzzeled. There is flaw in the logic but I don't see it.
Thanks for help.
| There are many ways to convince yourself that light will travel sideways w r t the ground observer. One simple way to see this is to check how transverse velocities add in relativity. The vertical component of emitted photon ( you can check this!) In the ground frame is $c/\gamma=\sqrt{c^2-v^2}$ where $v$ is the speed of the moving frame. But the speed of light wrt to ground should still be $c$ ( First postulate), this implies there must be a horizontal component equal to v so that total magnitude of the velocity is still $c$. Therefore, the fact that light travels diagonally is a consequence of relativity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
How does inhalation work? In school, we learn that during inhalation, the diaphragm expands, causing air to get sucked into our lungs. You can feel this suction by putting your hand over your mouth while inhaling.
Why is that? Does the expanded capacity of the lungs cause the air outside my body to rush into my body to, shall we say, keep the lungs full?
| From what I've gathered, I think my initial guess is correct. Air tries to maintain a constant pressure. According to the ideal gas law, there are two ways to maintain the same amount of pressure with an increasing volume: 1) increase the amount of gas, and 2) increase the temperature of the gas.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/216918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Why does a ruler continue to slide after toppling? I was playing with a small ruler by repeatedly toppling it, see diagram below:
The ruler, standing $h$ tall and approximately a regular bar, is prevented from free-rotating by a fixed ridge (a book, usually) in point $O$ and then allowed to topple until it hits the $xz$-plane. My desk and the ruler both have quite smooth, hard surfaces and I can't help noticing that after impact the ruler slides along the surface in the $x$-direction for a bit until it is halted by friction.
This suggests the ruler had momentum in the $x$-direction, after impact with the desk's surface. But I can't work out where it comes from. Is some of the rotational kinetic energy the ruler acquired during toppling converted to translational kinetic energy on impact? If so, how?
The forces acting on the centor of gravity of the ruler are shown in the right hand top corner of the diagram. Obviously it's the moment $\frac{Th}{2}$ that's causing rotation around $O$ and the equation of motion is:
$$\frac{Th}{2}=I\ddot{\theta}.$$
With:
$$T=mg\sin\theta.$$
And:
$$\ddot{\theta}=\omega\frac{\mathrm{d} \omega}{\mathrm{d} \theta}.$$
So:
$$\frac{mgh}{2}\sin\theta \mathrm{d}\theta=I\omega \mathrm{d}\omega.$$
Integrated between $0,0$ and $\frac{\pi}{2},\omega$, we get:
$$K=\frac{I\omega^2}{2}=\frac{mgh}{2}.$$
$\frac{mgh}{2}$ is of course simply the amount of potential energy that has been converted to kinetic energy by lowering the CoG from $h/2$ to $0$.
Now let's look at point $O$:
In $O$, $mg\cos\theta$ has to be countered to prevent the ruler from moving radially (from $O$ to the CoG or vice versa).
Decomposing we get:
$$F_x=mg\cos\theta \sin\theta.$$
But at $\theta=\pi /2$, $F_x$ vanishes, so it can't be that force that's responsible for any horizontal motion.
Clearly I'm missing something here, but what is it?
| As noted by John Rennie in the comments, there will be a point as the ruler falls where it loses contact with the ridge and begins to slide to the right. The idea here is that if the ruler were to keep its pivot point fixed, then at some point, the force applied by the pivot point would have to switch from having a component to the right to having a component to the left (i.e., pulling the CM back in rather than pushing it out.) Since the "ledge" specified in the OP can only exert a force to the right, this will be the point that the base of the ruler begins to slide away from the ledge. (This is similar in spirit to the classic "disc slides down a frictionless hemisphere" problem.)
To prove this, we use conservation of energy to find the ruler's angular velocity as a function of $\theta$. This becomes
$$
\frac{1}{2} I \omega^2 = mg \frac{h}{2} ( 1- \cos \theta) \quad \Rightarrow \quad \frac{1}{3} h^2 \omega^2 = gh (1 - \cos \theta) \quad \Rightarrow \quad \omega^2 = \frac{3g}{h}(1 - \cos \theta).
$$
Taking the derivative of both sides with respect to time, we get
$$
2 \omega \alpha = \frac{3 g}{h} \sin \theta \omega \quad \Rightarrow \quad \alpha = \frac{3gh}{2} \sin \theta
$$
The linear acceleration of the center of mass is therefore
$$
\vec{a} = \frac{h}{2} (- \omega^2 \hat{r} + \alpha \hat{\theta}) = - \frac{3g}{2}(1 - \cos \theta) \hat{r} + \frac{3g}{4} \sin \theta \hat{\theta}
$$
using polar coordinates (with $\theta = 0$ at the vertical and increasing clockwise.) In terms of the Cartesian components, we have $\hat{r} = \cos \theta \hat{y} + \sin \theta \hat{x}$ and $\hat{\theta} = \cos \theta \hat{x} - \sin \theta \hat{y}$, so all told this becomes
\begin{align*}
\vec{a} &= - \frac{3g}{2}(1 - \cos \theta) (\cos \theta \hat{y} + \sin \theta \hat{x}) + \frac{3g}{4} \sin \theta (\cos \theta \hat{x} - \sin \theta \hat{y}) \\
&= \frac{3g}{2}\left((\cos \theta - 1) + \frac{1}{2} \cos \theta \right) \sin \theta\hat{x} + \frac{3g}{2}\left((\cos \theta - 1) \cos \theta - \frac{1}{2} \sin^2 \theta \right) \hat{y}.
\end{align*}
We see that $a_x = 0$ when $\frac{3}{2} \cos \theta - 1 = 0$, or $\cos \theta = \frac{2}{3}$, or $\theta \approx 48.2^\circ$. Thus, once the ruler falls past this angle, the net force on center of mass must be to the left to keep it moving in a circular arc. On a perfectly frictionless table, the ruler would leave the "ledge" at this point, since the ledge is unable to provide a force in this direction. In reality, friction might be able to hold the bottom of the ruler in place for a bit longer than this, making the angle at which the ruler leaves the ledge much closer to the horizontal.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Sterile neutrino constraints (summary of results) Constraints on sterile neutrinos come from an amazingly wide range of different sources. From astrophysical observations (review1, review2) to particle physics experiments (MiniBooNE, Daya Bay, LSND, MINOS)
I have been reading about the astrophysical observations (e.g. XMM-Newton), but they rarely go much into detail about the constraints and searches for sterile neutrinos via experiments.
Having little experience in particle physics, I would truly appreciate it if someone could document here, in a sentence or two, what each experiment attempts to constrain, and how.
| First of all, one should distinguish the mass scale of the sterile neutrino. This review talks a lot about the keV sterile neutrino and even heavier cases, that however are not very interesting from the point of view of neutrino oscillations, because their mixing is typically very small. The keV sterile neutrino has possible implications as a dark matter candidate and to explain some astrophysical measurements, however.
A different case is the one of the "light sterile neutrino", with a mass around 1 eV, that could explain the so-called Short BaseLine (SBL) anomalies, which include the results of LSND, MiniBooNE and others.
To finally verify the existence of this light sterile neutrino, there are currently ongoing dedicated experiments such as NEOS and DANSS, while many others are planned or under development. It would be very hard to resume here all the experiments and the details, but I think you should be able to find most of the significant information on the light sterile neutrino in the following papers:
*
*a review on theory and constraints from neutrino oscillations, beta and neutrinoless double beta decay and cosmology;
*the most recent global fit that considers the constraints on the light sterile neutrino;
*another interesting comparison of NEOS, DANSS and Daya Bay results.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What would be the implications for current theories if gravitational waves are not detected? Let's assume that scientists trying to detect gravitational waves get a huge raise in funding, design even better experiments, and run them for decades, but just can't find any gravitational waves whatsoever. How would theories be changed to cope with this?
We're pretty confident that gravity propagates at the speed of light. So what other explanation would there be?
| In many ways it is almost inconceivable that they don't exist in some form or other as it would be difficult to reconcile the absence of gravitational waves with special relativity. It's only when you have an infinite speed of propagation, such as in Newtonian gravity, that you would not expect to see gravitational waves, however infinite propagation speeds are pretty much impossible to square with special relativity (the empirical evidence for special relativity is so vast, that any rival theory would have to almost exactly re-create all the predictions of SR).
It is possible that general relativity is not the correct relativistic theory of gravity, but even if that wasn't the case you'd expect very similar predictions to GR in weak fields, including similar predictions for gravitational waves.
Unfortunately gravitational waves are very difficult to detect directly by their own nature and the lack of detection, where we might expect it, points more to our lack of understanding about the physical situations in which they are produced than our lack of understanding about the waves themselves.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How many photons does my remote control garage opener emit? Every time I drive up to my house I imagine all the photons spitting out of the remote control garage opener when I press the button. And I imagine the door opener in the garage receiving them. There must be a ton of these particles going in all directions if the door always opens, right?
I'm just curious:
How many photons does the remote control send out, roughly?
And how many photons must my garage door opener receive to know it is time to open the door? Just one?
| Most likely, your garage door opener operates at a frequency of 315 MHz. Multiplying by Planck's constant, that means each photon has energy of about $2\times 10^{-25}$ joules.
Most likely, your garage door opener operates at about $1/10$ of a watt (or less, per comments below). So each second, it emits 1/10 of a joules of energy.
That's $(2/10) \times 10^{25}$ photons per second or (very roughly) $5\times 10^{24}$. In other words, $5,000,000,000,000,000,000,000,000$ photons per second.
To see how many of those hit the receiver, let $r$ be the distance from the transmitter to the receiver. The surface area of a sphere of radius $r$ is $4\pi r^2$. So if the reciever has area $A$, a fraction $A/(4 \pi r^2)$ of the photons hit the receiver. That's going to be a pretty small fraction, but still a whole lot of photons.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Does turning a spoon in water raise the temperature? I read about Joule's experiment proving the transformation of mechanical work into heat. But say I have a bowl with some water, and I start turning a spoon in it very fast, thus doing work — the water won't get hotter! What am I missing?
I think maybe the work I put is simply kinetic, and won't turn into heat. But then how do you explain Joule's experiment?
| Well first you have the energy in the form of kinetic energy of the spinning water. Once you let that water settle, it DOES get hotter.
The only problem is that water has a high specific heat (it takes a LOT of energy to heat up water), so you don't notice the water getting hotter since the amount it's heating up is not very noticeable. Coincidentally, it is this property of water that makes the earth a habitable planet--we have moderate temperatures compared to other planets because our oceans, bays, and lakes can absorb or release large amounts of heat to moderate the atmospheric temperatures.
If you want a more observable experiment, try taking a piece of metal (maybe a paper clip?) and bending it back and forth a lot of times. Although it'll eventually break, you should be able to notice it getting hotter
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
What is the distance between two objects in space as a function of time, considering only the force of gravity? What is the distance between two objects in space as a function of time, considering only the force of gravity? To be specific, there are no other objects to be considered and the objects in question are not rotating.
For instance, say you have two objects that are 6 million miles apart. One is 50,000 kg and the other is 200 kg. Say I want to know how much time has passed when they are 3 million miles apart. How would I go about doing that?
EDIT: Looking at the other question I am having trouble following David Z's steps in his answer. Intermediate steps would be helpful. In particular I don't see how the integration step works. I also don't understand why the initial r value, ri remains as a variable after it's derivative has been set to 0, wouldn't the integral of that derivative (i.e. the function ri) be 0 + C? I also don't see how you wind up with a term that includes 2 under a square root sign.
I can not ask for the intermediate steps on the question itself because I do not have the reputation points.
I think it probably answers my question or will once I understand it, but I am not sure.
EDIT: I can sort of understand the integration step. But it seems like he is integrating with respect to two different variables on both sides, the variables being r on the left and the derivative of r on the right. There must be something I'm missing here.
| This is the elliptic case of the radial Kepler problem, the equation for time as a function of position is
$$ t(r) = \sqrt{ \frac{d^3}{2 g} } \left( \arccos\left( \sqrt{ \frac{r}{d} } \right) + \sqrt{ \frac{r}{d} \left(1 - \frac{r}{d} \right) } \right) $$
where t is the time, r is the position, d is the initial (maximum) separation, and g=G(m1+m2).
In this case the the two masses will take 14.93 billion years to move from 6 million miles apart to 3 million miles apart, and then another 3.32 billion years to move from 3 million miles apart to zero miles apart (collision). Gravity is a very weak force.
The the solution to the inverse problem (finding the distance as a function of the time) is:
$ r(t) = d \left( y - \frac15 y^2 - \frac{3}{175} y^3 - \frac{23}{7875} y^4 -
\frac{1894}{3931875} y^5 - \frac{3293}{21896875} y^6 \cdots \right) \Bigg|_{y = \frac{1}{d} \left(\frac92 g \right)^{1/3} (t_\rm{freefall}-t)^{2/3} } $
where $ t_\rm{freefall} = \frac{ \pi}{2} \sqrt{ \frac{d^3}{2g} }$, and t is the time.
For more info see my website here, and here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Difference between convexo-concave and concavo-convex lenses? What is the difference between concavo-convex and convexo-concave lenses? We dont say convexo-plane for plano-convex. Does that mean concavo-convex and convexo-concave are essentially the same?
| The most important thing to remember is that a CONVEX lens (meniscus if you like) is thicker in the middle than at the edges, no quibbles, no exceptions. It accordingly causes a beam of light passing through to converge, and is said to have positive optical power.
Conversely for CONCAVE.
A plano-CONVEX lens still is a CONVEX lens, only a special kind with one side flat. But it remains conVEX, because its middle is thicker than its edge.
A concavo-CONVEX lens similarly is a special kind of CONVEX lens because its middle is thicker than its edge. The word after the hyphen is the type, whatever you put before the item is just part of its description.
A convexo-CONCAVE lens is still a concave lens, just one that has a convex surface somewhere. Being concave, it is thinner in the middle than at the edge.
And whether a lens is thicker in the middle or the edge, is the thing that make concavo-convex different from convexo-concave.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Direction of $\textbf{H}$ and $\textbf{B}$ inside and outside a bar magnet I seem to have encountered a contradiction when thinking about the directions of $\textbf{H}$ and $\textbf{B}$ inside and outside a bar magnet.
Suppose that a bar magnet has a roughly constant magnetisation M pointing along the positive z direction. Suppose also that the magnet is made from an isotropic material, i.e. $\textbf{M}=\chi_m
\textbf{H}$ (1). Therefore $\textbf{H}$ should be pointing in the same direction as $\textbf{M}$, provided that $\chi_m$ is positive, which should be the case for most magnetic materials? From the definition $\textbf{H}=\frac{1}{\mu_0}\textbf{B}-\textbf{M}$ and substituting using (1), we get $\textbf{B}=\mu_0(1+\chi_m)\textbf{H}=\frac{\mu_0(1+\chi_m)}{\chi_m}\textbf{M}=\frac{\mu_0\mu_r}{\mu_r-1}\textbf{M}$. Unless $\mu_r$ is less than 1, $\textbf{B}$ should also be parallel to $\textbf{M}$. Therefore so far we get that both $\textbf{H}$ and $\textbf{B}$ should point in the same direction as direction as $\textbf{M}$ inside the magnet, that is along the positive z direction.
We know that $\nabla\cdot\textbf{B}=0$ and thus applying boundary conditions on the top and bottom surfaces of the magnet (normal to the z axis) using an infinitesimally thin gaussian pillbox, we get that $\textbf{B}^\perp_\textrm{outside}=\textbf{B}^\perp_\textrm{inside}$. Therefore $\textbf{B}$ immediately outside the top surface should also point in the positive z direction. Assuming that there are no free currents near the magnet, $\nabla\times\textbf{H}=\textbf{J}_\textrm{free}=\textbf{0}$. Applying boundary conditions to the side surfaces of the magnet (parallel to the z axis), $\textbf{H}^\parallel_\textrm{outside}=\textbf{H}^\parallel_\textrm{inside}$. Therefore $\textbf{H}$ immediately outside the side surfaces must also point along the positive z direction. However, the field lines must also curl around and meet the top surface of the magnet, where $\textbf{H}$ will therefore need to point in the negative z direction. Since $\textbf{M}=\textbf{0}$ outside, $\textbf{H}=\frac{1}{\mu_0}\textbf{B}-\textbf{M}=\frac{1}{\mu_0}\textbf{B}$ and $\textbf{H}$ needs to be parallel to $\textbf{B}$ but we just argued that they point in opposite directions due to the boundary conditions. How do we resolve this contradiction?
| $B$ actually behaves as you explain, however there's a problem
with $H$.
You say:
"However, the field lines must also curl around and meet the top surface of the magnet, where $H$ will therefore need to point in the negative z direction."
Why the field lines need to meet the top surface of the magnet?
Outside the magnet $B$ and $H$ are proportional so we can use the picture.
If you take a line in the upper half of the magnet, the field lines
go away from the magnet, then curl and point in the
minus z direction, then go near to the magnet and curl.
If you take a line in the lower half of the magnet
(the figure doesn't apply), the field
goes into the magnet and because we have
$div \vec{H} \ne 0$ there is no continuity of the field lines.
Therefore $H$ doesn't point in the minus z direction at the top surface.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Tension in string question
Problem: A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire that passes without slippage over a frictionless pulley (the figure (Figure 1). The pulley has the shape of a uniform solid disk of mass 2.10 kg and diameter 0.520 m.
What are the horizontal and vertical tensions?
The answers are:
$T_h = 32.6\:\mathrm N$, $T_v = 35.4\:\mathrm N$.
Complete guide to problem is here.
*
*I don't understand why the tensions are different. For a rope, isn't the tension uniform?
*Can you also explain why using $F_\mathrm{net} = (m_1+m_2)a$ wouldn't work here? $a$ would turn out different if I did this.
*Also, why wouldn't you add the moment of inertia provided by the mass to the pulley moment of inertia? I remember doing a problem where you had to account for the added the $I$ of objects.
| *
*some of the tension goes into accelerating the pulley rotationally. Specifically $\Delta T \;R = I \frac{ \dot{v}}{R}$
*Again you have to account for the mass moment of inertia of the disk. You could write $F_{net} = (m_1+m_2+\frac{I}{R^2}) a$
*The masses do not rotate so their mass moment of inertia is of no use.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/217984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Veterans Day Memorial in the U.S For the ones who know about it:
How many more times will the Veterans Day Memorial stones' holes "align" perfectly?
Some info from this site:
At precisely 11:11 a.m. each Veterans Day (Nov. 11), the sun's rays pass through the ellipses of the five Armed Services pillars to form a perfect solar spotlight over a mosaic of The Great Seal of the United States.
I know that there are probably several factors which could affect it (like tectonic motion or earth's orbit maybe), but if the designers hadn't been somewhat sure about its permanence, then I guess they wouldn't have used it for something of such importance to the U.S. So my question is essentially, for how long have they made sure that it works (from a physicist's standpoint)?
| The answer to your question can be found in the description of the engineering of the monument
You can read the whole story at that link; I will just quote the most pertinent statement:
Using the statistical mean of the 100-year data, the altitude and azimuth angles for the structure were adjusted to provide time/error fluctuation of plus or minus 12 time seconds from the International Atomic Time mark of 11:11:11 a.m. That small time difference allows for additional compensation of the variations that you mention. We also checked the variance 500 years out, and if the structure is still standing, it will work.
I added the bold in the quote above. So the answer to the exact question you posed is: "They made sure it works for 500 years" - but probably longer.
Incidentally, you can check this yourself by putting numbers through the NOAA online calculator to see how much the azimuth and elevation change from year to year. The answer is: very little.
Here is a snapshot of the calculation for yesterday:
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/218066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.