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Is it possible to "see" atoms? As per my knowledge, atoms are small beyond our imaginations. But there is an image on Wikipedia that shows silicon atoms observed at the surface of silicon carbide crystals.
The image:
How can we see these distinct atoms if they are so small?
|
This is an image of a Sc2O3 nanocrystal obtained from an abberation corrected scanning transmission electron microscope.
The left image is recorded by measuring only electrons that have been bent/deflected by passing through the material (in this case we dont see the oxygen atoms very well)
The image on the right measures all the electrons that pass through the material. (In this case we see quite clearly oxygen and scandium columns - which, in this case, are columns of 5 atoms or so)
In this case we see columns of atoms but tomographic STEMs exist and can reproduce the 3D locations of individual atoms in a material
STEMs operate by sending electrons into a sample and recording how those electrons are scattered, absorbed, or transmitted entirely analogous to how light microscopes work only electrons have a MUCH smaller wavelength than light.
We cant see atoms using light because atoms are much much smaller than the wavelength of light.
But electrons have a much smaller wavelength allowing us to probe much smaller features than light could hope to allow
This image has a resolution of about 70 picometers (0.07nm) and atoms have "diameters" roughly of 0.1nm...10^(-10) meter. More than enough resolution to see atoms
Contrary to the previous answer we can, in fact, image atoms very well using STEMs and TEMs
Furthermore modern STEMs can chemically identify atoms based on how the electron beam deflects through the sample.
More electrons in the atoms => greater deflection.
So not only can we see atoms we can also study their chemistry and physical properties while we look at them!
Below is an image of a Nd3+:Sc2O3 nanocrystal. The brighter dots correspond to the Nd atoms (due to their much greater number of electrons)
David B. Williams and 1 more
Transmission Electron Microscopy: A Textbook for Materials Science (4 Vol set)
Is a very thorough and complete source on all thing electron micriscooy
Images recorded with a JOEL ARM200F and fourier space filtered and analyzed with gatan
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What are the quantum numbers of Majorana neutrinos? I have a question about majorana neutrinos.
Majorana particles are particles that are their own antiparticle.
From this I would argue that they need to have all quantum numbers equal to zero. My question is: what about the weak isospin? This cannot be zero, as neutrinos partecipate into weak interactions.
Thank you for any help on this.
| You say about Majorana fermions:
"From this I would argue that they need to have all quantum numbers equal to zero."
which is not true. Charge conjugation is defined on Dirac spinors as $\psi^c := \mathrm{i}\gamma^0\gamma^2\bar\psi^T$. Being Majorana means $\psi^c = \psi$. While this would imply the spinor has zero electric charge (and zero all other $\mathrm{U}(1)$ charges, if there were any, probably), this does not imply anything about the representations of the other gauge groups it transforms in.
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Resonant vs. Non-resonant Raman What does it mean to say that the conventional Raman effect is non-resonant? And, how/why does resonant Raman give a stronger signal than the non-resonant type?
| The normal nonresonant Raman scattering happens when a photon interacts with a molecule; the molecule absorbs the photon momentarily and re-emits it with slightly less energy.
In an energy diagram, that looks like this.
The frequency of the incoming photon is $\omega_i$, and the frequency of the scattered photon is $\omega_s$. The thick lined level is the ground state of the molecule, and the thin solid lined levels are vibrational states that are only slightly above the energy of the ground state.
The dotted line is a virtual energy state, which doesn't actually exist; but since the photon is re-emitted such a short time later, that's OK.
Most of the time, this doesn't happen; it's very uncommon compared to Rayleigh scattering, where $\omega_s=\omega_i$. That's why the signal from nonresonant Raman scattering is so low. Only a tiny, tiny fraction of scattered photons have their energy changed due to Raman scattering.
Now, resonant Raman scattering is exactly the same thing as nonresonant Raman scattering, except that $\omega_i$ is such that the virtual state's energy level is very close to one of the molecule's excited states (which are not virtual, but very real indeed.)
Being so close to an excited state makes the Raman process much more likely to occur. That's why the signal is so much stronger and can be used to detect much lower concentrations of a substance than conventional Raman can.
For more information, read the Wikipedia page.
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Is there an infinite amount of wavelengths of light? Is the EM spectrum continuous? The electromagnetic spectrum is a continuum of wavelengths of light, and we have labels for some ranges of these and numerical measurements for many.
Question: Is the EM spectrum continuous such that between two given wavelengths (e.g. 200nm and 201nm) there is an infinite number of distincts wavelengths of light? Or is there some cut-off of precision with which light might exist (e.g. can light only have wavelengths of whole number when measured in nanometers, etc.)?
| Formally there are an infinite number of different wavelenghts. However, any given physical system can only be found in a finite number of distinct physical states. To create a light source with a wavelength $\lambda$ that is well defined up to some resolution $\delta\lambda$, requires observing it within a system of size of the order of $\lambda^2/\delta\lambda$. So, the smaller we make $\delta\lambda$, the larger the system must be before we get physically distinct states within each such smaller interval.
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Approximations of the kind $x \ll y$ I have an expression for a force due to charged particle given as
$$F=\frac{kQq}{2L}\left(\frac{1}{\sqrt{R^2+(H+L)^2}}-\frac{1}{\sqrt{R^2+(H-L)^2}}\right) \tag{1}$$ where $R$, $L$ and $H$ are distance quantities.
Now I want to check what happens when:
*
*$H\gg R,L$
*$R,H\ll L$
How can I work out the approximation of this force?
Do I have to write it slightly different into form (2) to get it right?
$$ ~F=\frac{kqQ}{2LR}\left(\frac{-1}{\sqrt{1+\left(\dfrac{H+L}{R}\right)^2}}+\frac{1}{\sqrt{1+\left(\dfrac{H-L}{R}\right)^2}}\right) \tag{2}$$ (which is the same expression just written out differently). Any explain about this subject would be very helpful.
| For $H\gg R,L$ And for $L\gg R,H$ you get pretty much the same thing.
First off, $(H+L)^2\sim H^2$ and the same goes for $(H-L)^2$. That means that $(H+L)^2+R^2\approx (H+L)^2$. However, $(H+L)\not\approx H$, which means that $\sqrt{R^2+(H+L)^2}\approx H+L$. This makes the first approximation have $\frac{1}{H+L}-\frac{1}{H-L}$ in it.
The second approximation proceeds much the same way. $R$ is approximated away and you are left with the root of $(H\pm L)^2$. The resulting approximations should be identical
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100°C = 100 K =? I'm in first year. Our class is in lesson " Heat and Thermodynamics". While solving a numerical problem of a reversible engine he told us that 100 degree Celsius is equal to 100 kelvin. I inquired but could not get satisfactory answer. Pleas help me understand it.
Here is the numerical, please consider it: A reversibe engine works between two temperatures whose difference is 100c. If it absorbs 746J of heat from the source and rejects 546J to the sink, calculate the temperature of the source and the sink. Ans (100°C, 0°C)
| The important part is that it works "between two temperatures whose difference is 100°C". Celsius and Kelvin are not the same, but their degrees measure the same.
You can see that by direct substitution $T(ºC)=T(K)+273.15$
therefore
$T_2(ºC)-T_1(ºC)=T_2(K)+273.15 -(T_1(K)+273.15)=T_2(K)-T_1(K)$
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What do spacelike, timelike and lightlike spacetime interval really mean? Suppose we have two events $(x_1,y_1,z_1,t_1)$ and $(x_2,y_2,z_2,t_2)$. Then we can define
$$\Delta s^2 = -(c\Delta t)^2 + \Delta x^2 + \Delta y^2 + \Delta z^2,$$
which is called the spacetime interval. The first event occurs at the point with coordinates $(x_1,y_1,z_1)$ and the second at the point with coordinates $(x_2,y_2,z_2)$ which implies that the quantity
$$r^2 = \Delta x^2+\Delta y^2+\Delta z^2$$
is the square of the separation between the points where the events occur. In that case the spacetime interval becomes $\Delta s^2 = r^2 - c^2\Delta t^2$. The first event occurs at time $t_1$ and the second at time $t_2$ so that $c\Delta t$ is the distance light travels on that interval of time.
In that case, $\Delta s^2$ seems to be comparing the distance light travels between the occurrence of the events with their spatial separation. We now have the following definitions:
*
*If $\Delta s^2 <0$, then $r^2 < c^2\Delta t^2$ and the spatial separation is less than the distance light travels and the interval is called timelike.
*If $\Delta s^2 = 0$, then $r^2 = c^2\Delta t^2$ and the spatial separation is equal to the distance light travels and the interval is called lightlike.
*If $\Delta s^2 >0$, then $r^2 > c^2\Delta t^2$ and the spatial separation is greater than the distance light travels and the interval is called spacelike.
These are just mathematical definitions. What, however, is the physical intuition behind them? What does an interval being timelike, lightlike or spacelike mean?
| Spacelike separation means that there exists a reference frame where the two events occur simultaneously, but in different places. Timelike separation means that there exists a reference frame where the two events occur at the same place, but at different times. Lightlike means that, well, light could travel between those points.
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How can you have odd nuclear spin angular momentum but positive parity or vice versa? How does it happen that you can get states like $J^\pi=3^+$ or $J^\pi=2^-$? I think this could be because $\pi=(-1)^l$ so you could have an even state in $l$ but the $J=l+s$ sum could be an odd number?
| Your guess is essentially correct.
If you want to think about the nucleus in a shell-model sort of way, you can say that
the nuclear spin $J$ is the vector sum of the spins $S_i$ and orbital angular momenta $L_i$ of all the nucleons in the nucleus.
In the deuteron, for instance, the nucleus must be antisymmetric under exchange of the "identical" proton and neutron. In that special case we know from the non-existence of the stable diproton or dineutron that the deuteron has isospin zero, and therefore unit spin. In order to have positive parity, the deuteron wavefunction must have even $L$. (And experimentally, the deuteron is mostly $s$-wave, though there's a $d$-wave component due to the tensor part of the nuclear force.)
There are a few places in the chart of nuclides where you can look at the spin-parities of adjacent isotopes and ascribe them to the motion of the "extra" particle. For example, lithium-6 is $1^+$, and lithium-7 is $3/2^-$: you can say that the "extra" neutron has $L=1$. Or compare oxygen-16, $0^+$, to oxygen-17, $5/2^+$: that's consistent with the extra neutron having $L=2$. In large or non-magic nuclei, though, the many-body interaction is usually too complicated to make simple statements like that.
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What is the mass of a sphere? A solid sphere of mass M is rotating along an axis.
We can consider it as a collection of large number of point masses, every point mass is moving with respect to center of mass with velocity which depends on its radius from rotating axis.
Then, according to relativity, the mass of every point increases and consequently the mass of the sphere increases.
But if we consider the overall sphere, it is not moving at all and its mass remains the same. which produces a contradiction.
Please tell me where I am wrong.
| The mass of a moving body increases only for the observer in a rest frame, not for the moving body, as it feels itself as being in rest.
The rotating sphere is a different story, there are other forces and the point on moving sphere feels them, you cannot play a simple paradox game here. The situation is more difficult. I suppose you dont have any gravitation forces in the example.
However, as @Danu commented, you put a kinetic energy $K_{rotation}$ to the sphere to rotate it, so your total energy must increase by this. I expect that $E=mc^2$ contains all types of internal energies, so the rotation should come there also and your mass is increased.
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Why are stars white? That is may be an easy question, but I am not a professional. The Sun is a star, and when I look at the Sun it is usually yellow. Why are stars in the sky at night white?
I suppose it could be due to their distance. What is the explanation?
| Our sun is actually white. Sun seen from space. It's just that when the sun rays enter the earth, our atmosphere scatters the white light resulting in different colors.
The reason why the Sun appears to be white sometimes is usually because it's directly overhead. Then the rays coming from the sun have to travel the least distance and hence encounter less and less atmosphere which is not enough to scatter the white light of the Sun.
Coming to other stars. Most stars in the night sky appear to be white when they really are not. Many of them are red, blue, green, orange etc. But because they are light years away from us, our eye is hardly ever able to distinguish the color and we see it as either white or slightly blue.
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What does it mean for a state to have a negative partial decay width? I don't understand what it means when a particular decay mode has a negative partial decay width. I'm guessing the total decay width for a particular system must always be positive (now that wouldn't make much sense to me), but I'm not quite sure how the negative partial decay width plays in.
What does it mean when a particular decay mode has a negative partial width? Have I misunderstood anything?
| The negative sign for the partial decay widths denote the sign of the corresponding reduced width amplitude in the R-Matrix formalism, which is what the author of this paper (and everybody else analyzing cross sections of low-energy nuclear reactions) is using to do his analyses. He's using the R-Matrix code SAMMY, which sadly has very little documentation.
The partial decay width of a state is parameterized by the square of the reduced-amplitude for that particular state. This leaves the sign of the reduced-width amplitude undetermined. More details on the R-Matrix formalism are given here.
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If I bend a rod, will its moment of inertia change? In the first picture, there is a homogeneous metal rod of length $2L$ and mass $M$. If it rotates around a normal axis passing by $O$ (which is the center of gravity), then its moment of inertia is:
$$I_{1}=\frac{M(2L)^2}{12} = \frac{ML^2}{3}.$$
After bending the bar on $O$, will the moment of inertia for rotations around a normal axis passing by $O$ be the same, or will it dependent on the angle $\theta$ that the two segments of length $L$ make?
| The answer depends on the direction of the axis of rotation.
If the axis is normal to the plane, then you have the same amount of material the same distance from the axis of rotation as before - and thus the moment of inertia about that axis would be unchanged.
However, if the axis of rotation you consider is in the plane of the paper, the answer will change. For the vertical axis in the plane, the projected mass per unit length will increase while the apparent length of the rod is shortened: in other words, looking at the setup from the top of the V, it looks like you have a shorter rod with more mass per unit length and the moment of inertia about that axis will decrease; similarly, the moment of inertia about the horizontal axis in the plane (originally along the axis of the unbent rod) will increase (since by the perpendicular axis theorem, their sum must equal the moment of inertia about the third, normal-to-the-paper axis).
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What does it mean to "uplift" a supergravity solution to higher dimensions? What does it mean to "uplift" a supergravity solution to higher dimensions? This is a common term used in the literature but I cannot understand it. A very common example is "uplifting d-dimensional solution to 11-dimensional supergravity or M-theory". Is there some simple example?
| An uplift is the opposite of a dimensional reduction. Take for example the relation between (the low-energy limit of) M-theory and type IIA supergravity: the former is eleven-dimensional, while the latter lives in ten dimensions. If you find a solution of M-theory, you can get its equivalent in type IIA by Kaluza-Klein reduction. For example, the eleven-dimensional metric $G_{MN}$ gives rise to the ten-dimensional metric $G_{mn}$, a gauge field $G_{m,10}$ and a scalar $\Phi=G_{10,10}$ (the dilaton). Since both descriptions are completely equivalent, one can also start in ten dimensions, solve type IIA supergravity and then construct the eleven-dimensional fields. This is referred to as "uplifting". While the details of course vary from case to case, the principle is always the same.
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Is there a difference in the energy output of a nuclear fission reaction as opposed to fusion? For example, if I split a Helium atom will I get the same amount of energy as when I fuse Hydrogen into Helium? If there is a difference, what will be the difference (in general not according to Helium/Hydrogen), and why?
| Splitting a helium atom requires energy, whereas fusing two deuterium atoms into helium liberates energy.
As it can be seen from this graph:
the energies you were talking about will be the same (since they involve the same number of nucleons), but the sign will be different. Note that for small nuclei, energy is released by fusing them, while for large nuclei it is released by splitting them.
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Do photons with a frequency of less than 1 Hz exist? A photon with a frequency of less than 1 Hz would have an energy below
$$
E = h\nu < 6.626×10^{−34} \;\rm J
$$
which would be less than the value of Planck's constant. Do photons with such a low energy exist and how could they be detected? Or does Planck's constant give a limit on the amount of energy that is necessary to create a single photon?
| The frequency $\nu$ is in seconds$^{-1}$, which is purely human-based unit having a relation to rotation of the Earth. Thus no reason why 1 Hz was a limit. Planck $h$ value is also not massless unit and it's value has relation to SI system.
Existence: while I don't see a principal reason for non-existence of such a photon, neither I see a physical process, that would generate such a radiation. You would need some anthena of length $\sim 10 ^8$ meters and some process that would correlate a charge across such a distance.
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How to heat water to 80°C? I'm trying to figure out an easy water to get 80°c to consistently make the perfect cup of coffee. Let's assume room temperature water is 22°c.
What is the ratio of boiling water to room temp water to achieve 80°? What is the formula with room temp being variable?
| Your post seems an awful lot like a homework question, so you should probably tag it as a homework-like question.
If by "cup" you mean 250 mL on the dot, then we can say that the ratio of boiling water (100 Celsius) to room temperature water (22 Celsius) to achieve 80 degrees is
$$ \frac {(T_1)(V_1) + (T_2)(V_2)}{V_1 + V_2}$$
In your case, we have:
$$ 80^oC = \frac {(100^oC)(x mL) + (22^oC)(250 - x mL)}{250 mL}$$
And you can solve for x to find the amount of boiling water you need to add to a certain initial amount of room temperature water.
That should be enough to get you started, we can't give away the full answer for the sake of academic integrity.
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Dissipated Energy from Falling Object using Lagrangian
A plate of mass $M$ moves horizontally with initial speed $v$ on a frictionless
table. An object of mass $m$ is dropped vertically onto it from the height
$h$ and smashes. How much energy is dissipated in this process?
I have been taught E-L formulas, but don't see how they could be applied to the question at hand.
In the y direction, the falling object gains energy due to gravitational potential and eventually smashes on the plate. Since the table is frictionless, I imagine that energy in the x-direction would be conserved. Thus, I first imagined that the dissipated energy came only from mgh. Of course, this answer doesn't make use of the E-L formula and I was told that I am overlooking something.
I have absolutely no idea how to approach this and hate to be so vague, but I'm at a complete loss. Any push in the right direction would be appreciated!
| You are overthinking, it looks like a simple inelastic collision.
Before:
Vertical (potential energy):
$E_{plate} = 0$, $E_{object} = mgh$
Horizontal (kinetic energy):
$E_{plate} = \frac{1}{2}Mv^2$, $E_{ball} = 0$
After:
Vertical:
The object smashes, not bounces, so its entire potential energy dissipates
Horizontal:
The plate now has a combined mass of $M+m$, the momentum is conserved
$Mv = (M+m)v'$
Kinetic energy of "Plate with smashed object" $E' = \frac{1}{2}(M+m)v'^2$
The total dissipated energy is the difference between "before" and "after", $ \frac{1}{2}v^2\left(\frac{Mm}{M+m}\right) + mgh$
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Chern-Simons action in 4 dimensions I can not understand why we do not have a Chern-Simons action for four or even forms?
And why is it not a good theory for (3+1) dim?
| *
*By definition, the Lagrangian form $\mathbb{L}$ of Chern-Simons (CS) theory (wrt. a Lie algebra valued one-form gauge field $A$) is a CS form, i.e. the CS action reads $$S[A]~=~\int_M\mathbb{L}.$$ The exterior derivative $\mathrm{d}\mathbb{L}$ of a CS form is (also by definition) the Lie algebra trace of a polynomial of the 2-form field strength $F$. In the other words, $\mathrm{d}\mathbb{L}$ must have even form-degree, or equivalently, $\mathbb{L}$ must have odd form-degree, and hence the dimension of spacetime $M$ must be odd.
*Of course, one could introduce a new definition of generalized CS theory. More generally, there is e.g. the notion of TQFT. TQFTs can exist in any dimensions.
*In particular, we should mention that there exists a generalized 4D CS theory by Costello, Witten & Yamazaki defined on $\mathbb{R}^2\times\mathbb{C}$, cf. e.g. this Phys.SE post.
References:
*
*M. Nakahara, Geometry, Topology and Physics, 2003; Section 11.5.
*Higher CS theories on nLab.
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Test whether a glass prevents you from getting your Vitamin D portion? I work in an office with glass, which I believe filters the UV radiation of the sunlight.
Is it possible to test if exposure to the light coming through the glass will supply Vitamin D for me as a mammal?
| The wavelengths that stimulate vitamin D production are between 280nm and 320nm, which is called UVB. You would need to use a detector capable of measuring light in this wavelength.
However there is no need, because normal windows are made from soda-lime glass and this transmits no wavelengths shorter than about 350nm. Some Googling will find you the relevant absorption spectrum. I found this example here:
The plot shows that soda-lime glass doesn't transmit any UVB. Unless your office uses borosilicate or pure quartz windows (which is highly unlikely) you'll need to get outside to get your vitamin D fix.
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Hamiltonian from a Lagrangian with constraints? Let's say I have the Lagrangian:
$$L=T-V.$$
Along with the constraint that $$f\equiv f(\vec q,t)=0.$$ We can then write:
$$L'=T-V+\lambda f. $$
What is my Hamiltonian now? Is it
$$H'=\dot q_i p_i -L'~?$$
Or something different? I have found at least one example where using the above formula gives a different answer then the Hamiltonian found by decreasing the degrees of freedom by one rather then using Lagrange multipliers.
| The Hamiltonian is defined by
$$ H = \sum_{i=1}^n \left( \frac{\partial L}{\partial \dot q_i} \dot q_i \right) - L $$
So in your case: $ H' = H - \lambda f $
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Can someone clarify what should and should not be an operator in my verification of the 1D solution to the SE for a free particle? I just worked out the 1D free particle solution to the Schrödinger equation.
My wave function was
\begin{equation}
\psi(x,t) = Ae^{i(px-Et)/\hbar}
\end{equation}
So I plugged this into both sides of the time-dependent Schrödinger equation and started to verify. I did LHS and RHS separately.
I then ended up with
\begin{equation}
i\hbar\frac{\partial}{\partial t}\psi = \frac{1}{2}\frac{p^2}{m}\psi
\end{equation}
which looks like the correct form for the free particle solution.
My Confusion
I don't understand where the operators went. Usually when I see the Hamiltonian defined in the time-dependent SE it reads
\begin{equation}
i\hbar\frac{\partial}{\partial t}\psi = \hat{H}\psi = \frac{1}{2}\frac{\hat{p}^2}{m}\psi
\end{equation}
But my answer seemingly is hatless. Above I defined $p = \hbar k$ which is the de Broglie relationship. But the article I got the original wave function from didn't say I needed to make the $p$ in $ \psi(x,t) = Ae^{i(px-Et)/\hbar}$ an operator. So I'm confused what should and should not be an operator.
My Question:
Can someone clarify what should and should not be an operator in my verification of the 1D solution to the SE for a free particle?
| Your solution is right. What you get verifying it is that $\psi$ is also an eigenfunction of the momentum operator, which means
$$\hat p\psi=p\psi,$$
where $\hat p=-i\hbar\nabla$ is momentum operator, and $p$ is its eigenvalue.
Now, applying $\hat p$ twice and dividing by $2m$, you can get
$$\frac1{2m}\hat p^2\psi=\frac1{2m}p^2\psi,$$
which is just another form of time-independent Schrödinger's equation for free particle:
$$\frac1{2m}\hat p^2\psi=E\psi.$$
Here $\hat T=\frac1{2m}\hat p^2$ is the kinetic energy operator, and $E$ is its eigenvalue, i.e. energy of the particle in this eigenstate $\psi$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why most of physics is somehow related to light? It seems that for the past 200 years, every physicist is concerned about light. For example : Newton's particle model, Young experiment, Photo-Electrict effect and Einstein's formula, Special Relativity (constant speed of light), Bohr's atom model (using Photons to emit electrons), Double-slit experiment from the quantum aspect etc.
My question is - What's so special about light and why do we care so much about it? I'm currently at 12th grade and all I've learned so far in physics (exept classical mechanics of course) is always related to light.
| Instead of the word light it would be better to use the word electromagnetism. Newton and Young were fascinated about the decomposition of white light into it colors and about fringes behind edges. Since Maxwell it was obvious that light was only a small part of the electromagnetic spectrum. Later were discovered the weak and the strong nuclear power. But in our life were only the two forces electromagnetism and gravitation we feel every day. About gravitation is only to say that we are subjected to it and that we speculate about the curvature of the space in dependence of masses.
About electromagnetism we know a lot and we use it every day. We know that EM radiation is the result of accelerated particles. Since Lorentz we know how to bind together the three phenomena moving charge, magnetic field and mechanical force. the resulting devices are the electric generator, the electrodrive and the magnetic coil. Laser and LED are EM phenomena too. For the understanding of chemical compounds it is helpful too to know electromagnetism.
There is a lot of physics where you don't need knowhow of EM. but every time you go deeper into some phenomena you end with it. Try it out.
| {
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Microscopic definition of dynamic pressure of fluids So we have a moving fluid and we know from Bernouli's equation that there is a term called dynamic pressure(not to be confused with the hydrostatic pressure of the fluid).
So,what exactly is it and how can it be explained microscopically?
Note:do not involve relativity please!
| Dynamic pressure of fluids is the kinetic energy per unit volume of the fluid. Its unit are the same as pressure, and Bernoulli's equation,
$0.5\rho v^2 + \rho gy + p = constant$
can be written as
$P_{dynamic} + \psi_{gravity} + P = constant$, where $P_{dynamic} = 0.5\rho v^2$ and $\psi_{gravity}$ is the force potential of gravity.
As for microscopic explanation, it is the total kinetic energy per unit volume of a fluid (average sum of KE of particles per unit volume).
Update: Microscopic explanation in more detail
Consider all the particles in a (tiny) region with constant flow speed $v$. Consider one particle. This particle has an arbitrary velocity, $v_i$, and mass $m_i$. Hence, the kinetic energy of this particle is $0.5mv_i^2$. The total kinetic energy in the region is
$\Sigma 0.5m_iv_i^2 = 0.5\Sigma m_iv_i^2 = 0.5 \Sigma m_iv_i \cdot v_i = 0.5\Sigma m_iv_i \cdot \Sigma v_i = 0.5m_{total}v_{cm} \cdot v_{flow} = 0.5m_{total}v_{cm}^2$, using the fact that the flow velocity is equal to the center of mass velocity in a region of constant flow velocity (so, any tiny region in a fluid).
Hence, the total kinetic energy per unit volume of a fluid is equal to the dynamic pressure term, explaining the motivation for the dynamic pressure term. Note: This term comes from an energy analysis, not a force analysis.
| {
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relationship of number of standing waves with Temperature? If we have let us say fixed air column of length 'L', in a open-closed column problem, lamba is equal to 4*L/(2n-1).
n = number of nodes / anti nodes in air column
How does 'n' changes with respect to the temperature? Is the "number of nodes / anti nodes" at room temperature equal to "number of nodes/anti nodes" in same air column at let us say T =1200C? or do they increase or decrease with temperature?
As speed = frequency * wavelenght
and speed and temperature have reciprocal relationship, that means
increase Temperature will increase speed, which will result in increase of frequency but is there any relationship that after certain frequency(or temperature indirectly) the quarter standing waves turn into 3 quarters ?
| The wavelength is fixed by the dimensional length of the tube, but since wavelength is equal to the speed of sound divided by the wave frequency, temperature will affect the frequency because temperature determines the speed of sound.
| {
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Can I blow soap bubbles (of air) inside a vacuum chamber? When I blow soap bubbles from a liquid dish soap mixed with distilled water at atmospheric pressure at ground level both internal and external air pressure nullify and the tension of the bubbles holds, can soap bubbles be formed inside a vacuum chamber with all air pumped out?
| Unfortunately, you can not make a bubble with water and soap in a high vacuum.
as you can see in this phase diagram of pure water, if the pressure is lower than 611.657 Pa, liquid water will not exist no matter what the temperature is.
Maybe you should try some other material, but not H2O.
| {
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What's my $\mathrm dM$? Gravitational Potential inside a circle of mass
I'm trying to find the gravitational potential for an arbitrary point within a ring of uniform mass density. The point is constrained to be in the same plane as the ring.
So we start with:
$$\Phi=\int G\frac{\mathrm dM}{r}$$
Let's assume that the point of interest is along the $x$ axis $r$ away from the origin (which is at the center of the ring). An arbitrary point on the ring lies at:
$$a\cos\phi\hat{x}+a\sin\phi\hat{y}$$
And of course the point of interest is:
$$r\hat{x}$$
The distance between the point of interest and an arbitrary point on the ring is then:
$$\sqrt{r^2-2ar\cos\phi+a^2}$$
Back to the integral above, we get:
$$\Phi=\int G\frac{\mathrm dM}{\sqrt{r^2-2ar\cos\phi+a^2}}$$
Cool. I'm pretty happy up to this point, but what do I do about the $\mathrm dM$? Were I at the center of the circle, I would use $\mathrm dM=r\mathrm d\phi$. But I feel like it shouldn't be that simple if the center of my integration isn't the center of the circle. Should I use $$\sqrt{r^2-2ar\cos\phi+a^2}\mathrm d\phi~?$$ Am I completely off base here?
| I think the following diagram should help:
It is perfectly legal (and makes the math simpler) to use the center of the circle as the "center of integration", as long as you use the right value of $d$ for the distance to the mass element $dM$.
So your equation for the potential should use $d=\sqrt{a^2+r^2-2ar\cos\phi}$, and then you can express everything in terms of the angle $\phi$ and the total mass of the ring, $M$:
$$\begin{align}\Phi &= \int G\frac{dM}{d}\\
&= \int G\frac{M d\phi}{2\pi a ~d}\\
&= \int_0^{2\pi} G\frac{M d\phi}{2\pi a ~\sqrt{a^2+r^2-2ar\cos\phi}}\end{align}$$
| {
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Magnetic field in a wire with constant current
I assume I have a wire parellel to the $z$ axis and with radius $R$. A constant current $I$ flows through it in the $z$ direction. I want to know the magnetic field inside the wire at distance $r<R$.
In the figure, the pink dots represent the flow of electrons in the z direction for $r<R$. The red dots represent the the flow of electrons in the z direction for $r>R$. I added the green magnetic field $B_o$, so that one might be wrong.
The formula I came across is:
$$ B=\frac{\mu_{0} I r}{2 \pi R^2} = \frac{\mu_{0} I_{enc} }{2 \pi r} \text{ for } r<R$$
Why does the magnetic field (blue in figure) at $r$ only depend on the magnetic field caused by the enclosed current $I_{enc}$ (the pink dots) ?
Why isn't there an influence by the magnetic field caused by the remainder of the current [$ I-I_{enc}$] (red dots)? For example, from $r \angle 40$ radially towards $R \angle 40$ there is current flowing in the $z$ direction (3 consecutive red dots in figure) which at $r \angle 40$ causes a magnetic field (green $B_o$ in figure) that is opposite to the one caused by the enclosed current (blue in figure), or am I wrong?
| Ampere's Law. That magnetic field that you calculated depends on $r$ and the enclosed current is $j*ds$ where $j$ is the current density and $ds$ is surface differential. So if you want to know the magnetic field outside of a wire you should take a $r>R$.
| {
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Could I break the sound barrier using circular motion? (And potentially create a sonic boom?) Ok, Lets say I get out my household vaccum cleaner, the typical RPM for a dyson vaccum cleaner reachers 104K RPM, Or 1.733K RPS. In theory, this disc would be travelling with a time period of 0.00057692307 seconds, If we take the speed of sound to be 344.2 metres per second, a breach in the sound barrier is easily possible for an item on the edge of the disc.
One question remains: For an extremely strong disc , could an item stuck onto it break the sound barrier, and create a sonic boom?
| It's very possible to spin something fast enough to create a sonic boom, but engineers usually try very hard to avoid it.
Several aircraft have been built, on purpose, with supersonic propellers. One production aircraft is the Tupolev TU-95 Bear long-range bomber. Eight counter-rotating 4-blade props turn fast enough so the prop tips are supersonic. Rumour has it that the plane is loud enough to be heard by submarines - be happy you don't live near one of their bases.
Another is the XF-84H Thunderscreech, an experiment back in the 1950s. The prop tips moved at Mach 1.1 even on the ground, it was apparently audible 25 miles away and the shock wave from the prop tips was not only dangerous to the ground crew but actually visible far from the aircraft.
Note that you don't get a boom in either case, you get a constant very very loud noise.
So, if you want to break out your Dyson's motor and extend the rotor a bit it could very well go supersonic. Highly unlikely the motor has enough power, and you will need to balance the shaft very, very carefully. Wear earplugs! (and maybe body armour)
| {
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Binomial expansion of non-commutative operators I would like to determine the general expansion of
$$(\hat{A}+\hat{B})^n,$$
where $[\hat{A},\hat{B}]\neq 0$, i.e. $\hat{A}$ and $\hat{B}$ are two generally non-commutative operators. How could I express this in terms of summations of the products of $\hat{A}$ and $\hat{B}$ operators?
| There is a nice formula that provides the result in terms of the binomial expansion plus terms related to the noncommutative algebra
https://arxiv.org/abs/1707.03861
| {
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Feynman Diagram in QED non-relativistic limit I am following Peskin & Schroeder to read a Feynman diagram. But in this given image, they used the non-relativistic limit to write the incoming fermion gamma mu product. How did he derive the bottom formula? Any explanations in detail?
|
How did he get it?
How did he get what? Which part of the calculation it is that's confusing you. You have listed multiple equalities, which one is at issue?
Eq 3.55 of the pdf online version of Peskin and Schroder shows that:
$$
u^\dagger u=2E_p\xi^\dagger \xi
$$
and since
$$
\gamma_0^2=1
$$
we know that
$$
\bar u\gamma_0u=u^\dagger u
$$
and, in the non-relativistic limit (i.e. $p<<mc$)
$$
E_p\to m\;.
$$
(N.b., $c=1$)
| {
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movement of particles in electric field I am confused about a homework problem. Let's assume we have two electrically charged particles of which we know the charge and mass respectively. Let's say that at first they are fixed at some distance $r_1$ and then released simultaneously. I want to find their velocities at distance $r_2$.
Due to conservation of energy, we should have the equation $$ \frac{m_1 v_{1}^{2}}{2} + \frac{m_1 v_{1}^{2}}{2} = \int_{r_1}^{r_2} F \; dr $$
Where $$F = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_1 Q_2}{r^2}$$.
Now I obviously need another equation. I was thinking that by the law of conservation of momentum, I'd get (as the momentum equals 0 when both particles are still fixed in position)
$$m_1 v_1 + m_2 v_2 = 0$$
But this is where I get confused: Consider the case where one particle remains fixed and we let go of the other one. Wouldn't we get $m_1 v_1 = 0$ by conservation of momentum and something not equal to 0 by conservation of energy in the same way I obtained the first equation above?
| If one particle is fixed, some force is keeping it fixed, and in the presence of an external force, conservation of momentum doesn't apply. Your second equation is then $v_1 = 0$ (assuming particle #1 is the one that is fixed), not $m_1 v_1 + m_2 v_2 = 0$.
| {
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Does the electric force on a charged particle in a uniform electric field increase? If I have a proton in a uniform field between two parallel oppositely charged plates and the proton accelerates, the electric force acting on it stays constant seeing it is a uniform field and as a result the acceleration of the particle is also constant.
But does the electric force on the proton increase as it nears the negative plate?
I would think yes seeing the distance is getting smaller (cf. Coulomb's law) so the electric force (which originates from Coulomb interactions between the charged particle and the plates) must be getting bigger. But this conflicts with the fact that the force is constant. Any help?
| Your brain/mind might be processing the information in the wrong way. As field is uniform, force remains constant and acceleration remains constant. You see, the acceleration remains constant, but velocity doesn't remain constant, as the time to which your proton is accelerated increases, your proton's velocity also increases. So, your proton will be drifting at higher and higher velocity as it goes nearer and nearer to your plate, even though force remains constant. Blame your brain for misdirection!
| {
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If the Earth is a good conductor of electricity, why don't people get electrocuted every time they touch the Earth? Since the Earth is a good conductor of electricity, is it safe to assume that any charge that flows down to the Earth must be redistributed into the Earth in and along all directions?
Does this also mean that if I release a million amperes of current into the Earth, every living entity walking barefooted should immediately die?
| Firstly we are not the best conductors, so current might be having a relatively hard time getting through us.
But I believe the real reason is that you also need a high potential difference in order to get current flowing through you.
Like lightning which needs a huge potential difference between the clouds and earth (so big that most of times a neutral earth does not give this high difference but needs some accumulation of opposite charge from that on the cloud).So without that potential difference nothing happens.
(Note: the Earth actually causes voltage difference at about 200V per meter, so we do have that big potential difference, but Earth tends to make every object that it is touching neutral--to have the same potential as its surface-- so we distort the equipotential lines that the earth produces. For more on this, check Feynman Lectures Vol2)
| {
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Experimental evidence for Z boson coupling to right handed fermions I do have a question about electro-weak interactions.
I know the Z boson is an admixture of two fields, one that couples only to the left-handed part of the fermions (the neutral field introduced to make SU(2) a local symmetry) and one that couples to both left and right handed components of the fermions (the field introduced to make U(1) a local symmetry). My question is, what is the experimental evidence telling us that the Z boson actually couples to both left and right handed components?
Thanks for any help!
| In addition to TwoBs' comments, in the 90s, there had been many collisions at SLAC (USA) with polarized electron/positron beams of the SLC collider running at the $Z$ pole. Therefore, the right handed component of the electroweak interaction has been extensively tested.
See the wikipage:
http://en.wikipedia.org/wiki/SLAC_National_Accelerator_Laboratory
| {
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Error calculation in parallel resistances This is the question:
There are two resistors with resistance values $R_1=100\pm3$ ohm and $R_2=200\pm4$ ohm. Find the equivalent resistance of parallel combination.
According to what I've learnt, in any expression of multiplication or division, the percentage errors of each term are added up to find the equivalent percentage error. That is, if $$y=\frac{\text {AB}}{\text C}$$ then $$\%\;\text{error in y}=\%\;\text{error in A}+\%\;\text{error in B}+\%\;\text{error in C}$$
For the above problem, let $R_s$ denote series combination. Then $R_s=300\pm7$ ohm.
Let $R_p$ denote parallel combination.
$$\therefore R_p=\frac{R_1R_2}{R_1+R_2}=\frac{R_1R_2}{R_s}$$
Ignoring errors, we get $R_p=\frac{200}{3}$ ohm $=66.67$ ohm
$\%\;\text{error in R}_1=3$, $\%\;\text{error in R}_2=2$, $\%\;\text{error in R}_s=\frac73$
Hence, $\%\;\text{error in R}_p= 3+2+\frac73=\frac{22}{3}$
So, error in $R_p$ will be $\frac{22}{3}\%$ of $\frac{200}{3}$, which is approximately $4.89$.
Hence, I got $R_p=66.67\pm4.89$ ohm.
However, the book used the formula described and proved here and arrived at the answer $R_p=66.67\pm1.8$ ohm.
So, is the percentage error method wrong?
| My calculation dR/R = dR1/R1 + dR2/R2 -dR1/(R1+R2) -dR2/(R1+R2)
(Come from the derivative of R = (R1R2)/(R1+R2))
Do manipulation dR/R = [dR1/R1 -dR1/(R1+R2)] + [dR2/R2 -dR2/(R1+R2)]
dR/R = (dR1/R1)[R2/(R1+R2)] + (dR2/R2)[R1/(R1+R2)]
dR/R = (3/100)[200/300]+(4/200)[100/300] = 0.0267
dR=0.0267R= 0.0267(66.7) = 1.78
| {
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$\nabla^{\mu}\nabla_{\mu}$ in general relativity I am trying to work out $\square=\nabla^{\mu}\nabla_{\mu}$ in the metric
$
ds^{2}=-A(r)dt^{2}+B(r)^{-1}dr^{2}+r^{2}d\Omega^{2}
$$
My work:
when applying $\square$ to a scalar $\phi$, then
$
\square\phi=\nabla^{\mu}\nabla_{\nu}\phi=\nabla^{\mu}\partial_{\mu}\phi=g^{\mu\nu}\nabla_{\nu}\partial_{\mu}\phi=g^{\mu\nu}(\partial_{\nu}\partial_{\mu}-\Gamma^{\lambda}_{\mu\nu}\partial_{\lambda})\phi
$
Christoffel symbol
\left(
\begin{array}{ccc}
\left\{0,\frac{A'(r)}{2 A(r)},0\right\} & \left\{\frac{A'(r)}{2 A(r)},0,0\right\} & \{0,0,0\} \\
\left\{\frac{1}{2} B(r) A'(r),0,0\right\} & \left\{0,-\frac{B'(r)}{2 B(r)},0\right\} & \{0,0,-r B(r)\} \\
\{0,0,0\} & \left\{0,0,\frac{1}{r}\right\} & \left\{0,\frac{1}{r},0\right\}
\end{array}
\right)
substituting the metric and affine values in the equation above, my answer came to be
$$
\square=-A(r)^{-1}\frac{d^{2}}{dt^{2}}+B\left(\frac{d^{2}}{dr^{2}}+\frac{1}{r}\frac{d}{dr}\right)+\frac{1}{2}\left(B^{\prime}+\frac{B A^{\prime}}{A}\right)\frac{d}{dr}
$$
However, the answer happens to be
$$
\square=-A(r)^{-1}\frac{d^{2}}{dt^{2}}+B\left(\frac{d^{2}}{dr^{2}}+\frac{2}{r}\frac{d}{dr}\right)+\frac{1}{2}\left(B^{\prime}+\frac{B A^{\prime}}{A}\right)\frac{d}{dr}
$$
Could someone please show me where the third comes from in the second term?
| I'll prove a formula that is probably easier to use for this.
\begin{equation}
\begin{split}
\frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} g^{\mu\nu} \partial_\nu \phi \right) &= \frac{1}{\sqrt{-g}} \partial_\mu \left( \sqrt{-g} \right) g^{\mu\nu} \partial_\nu \phi + \partial_\mu \left( g^{\mu\nu} \partial_\nu \phi \right) \\
&= \frac{1}{2g} \partial_\mu g g^{\mu\nu} \partial_\nu \phi + \partial_\mu g^{\mu\nu} \partial_\nu \phi + g^{\mu\nu} \partial_\mu \partial_\nu \phi \\
&= - \frac{1}{2} g^{\alpha\mu} g^{\nu\beta} \left[ \partial_\mu g_{\alpha\beta} + \partial_\alpha g_{\mu\beta} - \partial_\beta g_{\alpha\mu} \right] \partial_\nu \phi + g^{\mu\nu} \partial_\mu \partial_\nu \phi \\
&= g^{\mu\nu} \left( \partial_\mu \partial_\nu - \Gamma^\lambda_{\mu\nu} \partial_\lambda \phi \right) \\
&= \nabla^\mu \nabla_\mu \phi
\end{split}
\end{equation}
| {
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Waves on a guitar string Can there be traveling waves on a string that is attached at both ends (like a guitar string), or does such a configuration allow only standing waves ?
| Of course there can be (and are) traveling waves on a string - when you pluck anywhere except at the center, you are generating an asymmetrical impulse that will travel up and down the string. The fact is that your "pluck" consists of many different frequencies, and as these travel back and forth two things happen:
*
*the highest frequencies will be damped most quickly due to dissipation (to the air, the supports, and internal friction in the string)
*waves that do not "fit" tend to cancel after a few round trips (interference) leaving just the standing wave pattern.
It is only the time evolution of the traveling wave that turns it into two waves of equal amplitude and discrete frequencies that travel in opposite directions - and that you recognize as standing waves.
A detailed mathematical analysis can be found in this paper
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Magnetic force between two point charges I tried to derive the magnetic force between two point-charges for iterative computation. Starting out with Lorentz force and Biot–Savart law for a point charge. $$ \vec F = q_2( - \Delta \vec{v} \times \vec{B})$$ $$\vec B = (\vec \Delta v \times \vec \Delta x) ( \frac {q_1}{ ||\Delta \vec x||^3}\frac{\mu_0}{4\pi}) $$ And got this for the answer by direct subtitution: $$ \vec F = q_2(-\Delta \vec{v} \times (\vec \Delta v \times \vec \Delta x) ( \frac {q_1}{ ||\Delta \vec x||^3}\frac{\mu_0}{4\pi}))$$
It does not seem to be right for several reasons.
*
*It seems as if electron would be attracted to the nucleus by both magnetic and electrostatic force. Considering hydrogen atom.
*It is possible to show that between non-moving particle and a non-moving wire with current in it should exist magnetic force. (Magnetic force acts between charge carriers in the wire and the point-charge. Non-moving particles in the wire do not have influence on the magnetic force)
Where have I gone wrong and how to find the correct expression for the magnetic force between two point-charges? Does the equation hold if $ \Delta v << c$?
If this equation proves to be wrong, what would be the correct approach?
| It appears that you have used the Biot-Savart law which assumes no static electric field. Instead, you should use the Lorentz force law, $$ F =q(E +Δv × B),$$
which accounts for both magnetic and electrostatic interactions.
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Special relativity mirror clock experiment inconsistency Say I set up a relativistic mirror clock experiment in which a spaceship contains a set of mirrors with a photon bouncing between them. Say the photon's motion is parallel to the direction of motion of the space ship, and the spaceship is moving at near to the speed of light, say 0.99c.
From the perspective of someone inside the ship, the photon bounces back and forth as normal, at a rate of one bounce per second or whatever it is. No problems there.
From the perspective of someone at some "stationary" point, the speed of the photon is c and the speed of the plates is 0.99c, so the photon must "race" to approach the front mirror and then when it bounces of the front mirror it should almost immediately hit the back mirror and start racing forward again.
So the observer on the ship sees regular periods but the observer outside the ship sees the first "half" of the second take much longer than the second "half". This seems to me like a paradox or a contradiction or something.
Where did my logic go wrong?
| That's a very long ship.
Your logic didn't go wrong. The concept of simultaneity is broken with special relativity. Although the clock doesn't appear symmetric to the stationary observer, the ticks (a complete trip) remain consistent with the speed of light when you account for length contraction. And yes, to the stationary observer, it would seem like the trip going in the direction of the ship's velocity would take longer. That's because the light is essentially covering more distance. The trip back would shorter for the same reason.
I would suggest the ladder problem to clear this up: http://en.wikipedia.org/wiki/Ladder_paradox
| {
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What exactly is the mass of a body? What determines it? The term "mass" is very common. But what does it depend on? How is it known?
| In classical physics mass has two definitions:
*
*It measures the amount of inertia that you have. In order to accelerate something you have to apply a force to it. The heavier your thing is, the less it will accelerate,
$$ a = \frac{F}{m} \, .$$
If you know the force and can measure the acceleration, you have access to the mass.
*In the physics of gravity, the mass is the gravitational charge. Just like charged particles exert a force on each other that is proportional to the product of their charges, two massive particles attract each other with a force that is proportional to the product of their masses,
$$ F = m_1 m_2 G /r^2 \, .$$
What is fascinating about this (and lead to the discovery of general relativity) is the fact that these two definitions coincide. Inertial mass and gravitational masses are the same!
| {
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Compact Disc Optics - Why use a linear polariser and a quarter wave plate? I just came across this website about the application of a quarter wave plate. Link: Compact Disc Optics.
My question is why does the beam need to be linearly and then circularly polarised before sending to the compact disc? And the returned beam undergoes the same before reaching the detector? Is it related to power loss?
| What you may not realize from that website is that the same optical element is shown on the left and on the right. Laser emission and detection happen simultaneously, but for clarity they are drawn separately.
See how circular polarization changed after the reflection from the CD. Now, when it comes through the quater-wave plate it has polarization perpendicular to the incoming laser beam and gets reflected towards the detector. Neat!
| {
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Mass dimensions and weak interaction The Fermi constant has a mass dimension of $-2$ and a value of $10^{-5}GeV^{-2}$.
How can I infer from this information that the mass scale of the weak interaction is about $10^2 GeV$?
|
The Fermi constant has a mass dimension of -2
The Fermi constant, $G_F$ itself has (by defintion) the dimension "energy times volume".
You are apparently asking about the "reduced" quantity $\frac{G_F}{(\hbar~c)^3}$,
(which is considered to be more practical in some applications) and
which has the dimension of "inverse energy squared";
symbolically: $$\frac{G_F}{(\hbar~c)^3} = (S_E)^{(-2)} = \frac{1}{S_E^2} = \frac{1}{(S_m~c^2)^2},$$
for the corresponding value $S_E$ of "energy scale", or the corresponding value $S_m$ of "mass scale".
and a value of $10^{-5}~GeV^{-2}$.
(Approximately, of course.)
How can I infer from this information that the mass scale of the weak interaction is about $10^{2}~GeV$ ?
Well, just rearrange the above formula; i.e. solve for $S_E$:
$$S_E = \frac{1}{\sqrt{\frac{G_F}{(\hbar~c)^3}}},$$
and insert the (approximate) value of Fermi constant:
$$S_E = \frac{1}{\sqrt{10^{-5}~GeV^{-2}}} = \sqrt{10^{5}~GeV^{2}} = \sqrt{10^{5}}~\text{GeV} \approx 300~\text{GeV}.$$
That's of course only a rough "order of magnitude" approximation of $10^{2}~\text{GeV}$ as "mass scale of the weak interaction".
In the Wikipedia page linked above you also find the relation
$$\frac{G_F}{(\hbar~c)^3} = \frac{\sqrt{2}}{8}~\frac{g^2}{(m_W~c^2)^2},$$
in terms of $m_W$, the mass of an actual particle (the $W$ boson) which "actually mediates charged weak interactions". The experimentally determined value of $$m_W~c^2 \approx 80\text{GeV} $$
is (incidentally) considerably closer to $10^{2}~\text{GeV}$.
| {
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Do magnetic fields cause ionisation of gases? I am doing my final year engineering project on Magnetic Field Assisted Combustion and was curious to see what people thought about it.
Companies sell rare earth magnetic arrangements to be attached to fuel lines of gas burners and they are said to improve combustion efficiency but why exactly?
I have performed a number of experiments using a standard butane/propane gas burner with some magnets manufactured by one of said companies and have had some contradicting results. With lower strength magnets, heat transfer unexpectedly slowed down but with a much stronger arrangement, heat transfer rate was increased.
Also, the burn out time of the same amount of gas took 8 minutes less with the magnets in place around the fuel line.
I have read a number of journals on similar subjects but even within these, the actual reason for the increase in heat output is still not known.
Any thoughts on the subject would be massively appreciated and possibly give me some other areas to investigate that I have not already thought of.
| I don't know about ionizing gas but I do know about magnetophoresis. Oxygen is paramagnetic so oxygen concentrations should be higher around magnets than in ordinary air. This could make it easier for fuels to combust.
| {
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Torque on puck moving on plane without friction
We have two pucks moving on a plane without friction. On one of them a force is applied on it's center of mass. On the second a force of equal magnitude is acting tangential to the puck and at a distance equal to it's radius. Which one will be faster?
Now if I try to solve the problem and take only translation in account then I would say that both forces are acting in the CM (center of mass) and both will have the same acceleration. But if I think in terms of energies, then the force acting on the CM will convert all work into translational kinetic energy while the other force would convert a part of it's work into rotational energy. So the pucks will not have the same acceleration of their center of mass.
|
We have two pucks moving on a plane without friction. On one of them a force is applied on it's center of mass. On the second a force of equal magnitude is acting tangential to the puck and at a distance equal to it's radius. Which one will be faster?...
So the pucks will not have the same acceleration of their center of mass.
Yes, your intuition is correct: the puck on which the force acts on its CM will be faster, because all energy will produce linear momentum, whereas in the other puck energy will distributed between linear and angular velocity/momentum.
| {
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Consideration of centrifugal force during descent If we imagine an object falling from a height h above the surface of the earth. We can go into a rotating frame and therefore introduce Coriolis and centrifugal forces.
Using the Coriolis force the deflection in the East-West Plane can be calculated. However, I am wondering does the centrifugal force influence the trajectory of the particle? Therefore, introducing some displacement in the North-South Plane.
To summarise my question is:
Does the centrifugal force cause objects to deflect when they fall , when considered in the rotating frame?
I have never seen it introduced into this type of question and therefore I am tempted to say it doesn't. However, I can not think of why this would be true.
| You're presumably talking about objects dropped away from the equator. At the equator it causes a lower effective gravity. Away from the equator it causes a deflection towards the equator.
| {
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Why does it seem as if big vehicles "attract" mine when I drive close to them? When I drive a car at high speed and when I am near to another big car (like a van, or transport vehicle) I feel an attraction to or something push me toward the other big car. What's the physics in this case?
| No, not at all. It doesn't have anything to do with gravity. The masses of the cars are so very small that you can't even think of the gravity between them. There is gravity, but it is negligible.
And one person answered with pressure. It's not that, either. Yeah, if you put two pieces of paper against each other and blow in the middle of them, the papers look like they attracting to each other, but the thing is, the air pressure in the middle is removed and only the pressure of outside is applying force to the papers. However, in our case, the air pressure is not that much to push two cars! If you blew between two pieces of metal, will they bend toward each other? Of course not!
But your answer – it's probably mental. Maybe the fear of hitting to that car or anything else force you to think that you are getting closer to the car but it's not like that. Actually, I've read an article some years ago about when you are sitting in a height and the feeling that you want to push the other person or yourself down, even if you don't want to. But anything it is, it doesn't have to do with a physical phenomenon.
| {
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How does "contamination" through (radioactive) radiation work? Physically, what does it mean when people or objects are contaminated with radiation? Is it because they actually carrying heavy metal particles?
| There are two distinct ways for previously non-radioactive material to become radioactive.
*
*Contamination refers to any behavior where existing radioactive material sticks to or is incorporated in an previously non-radioactive object or body (this can include a tract of land). This requires you to go to where there is mobile radioactive material or for that material to come to you.
This is the more common way.
*Activation refers to materials in a body getting transmuted by exposure to radiation. Most kinds of ionizing radiation can cause activation under the right circumstance, but it is usually a very slow process. Rare outside of labs and nuclear power facilities.
In the case of a nuclear power facility accident the scary issue is almost entirely contamination where the containment is lost material from the core (and nearby activated structural material, but mostly the core) which get carried into the wider environment by the action of water and steam. Chernobyl was special because the antique design of the reactor allow a significant conventional explosion to launch portions of the core out of the building directly.
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Limits of Integration Trig, Mag Field Infinite Length Wire I don't understand how the limits of integration should be defined when doing basic integrals of trig functions. It seems like it's an arbitrary decision, I don't understand it.
Here's the set up: For the field near a long straight wire carrying a current $I$, show the Biot-Savart law gives the same result as Ampere's law.
Now intuitively, for me at least, with the way that $\theta$ is defined, I would view the angle as becoming smaller as $y$ moves toward negative infinity. So the limits of integration make sense in that regards. But then the cosine doesn't make sense anymore. As $y$ becomes more negative, which corresponds to an angle between $0$ and $\pi/2$, then cosine should always be positive. But because cos=adj/hyp, then $\cos\theta=y/r$, and $y$ would be negative, even though the corresponding angle is between $0$ and $\pi/2$?
I know I'm misunderstanding something fundamental, hopefully somebody can help me so I can move on. I've been struggling with this for so long because it's easy enough to arbitrarily assign limits to get the answer you're looking for, but I want to know the right way, and more importantly, why it's the right way.
| Over the length of the wire from -infinity to +infinity the angle theta varies from
-pi/2 to +pi/2 and cos(theta) ranges from 0 through 1 and back to 0. It's never negative and the distance from any point on the wire to P is always positive. I think you create an unnecessary problem by treating the line from wire segment dl to P as a directional vector which it cannot be, for otherwise a point P at the center of a round current loop would experience no magnetic field.
| {
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How many atoms are in a piece of paper? How many atoms are there in a common sheet of paper?
The paper is A4, i.e. $210 \, \mathrm{mm} {\times} 297 \, \mathrm{mm}$ $\left(8.27 \, \mathrm{in} {\times} 11.7 \, \mathrm{in}\right).$
| You can estimate number of atoms by finding out average molar mass of paper and mass of one sheet of paper.
If we assume that paper is mainly composed of cellulose, we can neglect other components as insignificant. Then we find out that cellulose's molar mass is approximately $162.14 \, \mathrm{g}/\mathrm{mol}$. Mass of one sheet of paper is about $4.5 \, \mathrm{g}$.
Doing the math you get that one sheet of paper is approximately $0.02775 \, \mathrm{mol}$. With Avogadro's constant, we can calculate that there are$$
0.02775 \, \mathrm{mol} \times 6.02214 \cdot {10}^{23} \, \frac{\mathrm{molecule}}{\mathrm{mol}} ~=~ 1.07 \cdot {10}^{22} \, \mathrm{molecule}.$$
As in one cellulose molecule there are 21 atoms, the number of atoms is $2.247 \cdot {10}^{23}$ atoms.
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Drying clothes with the sun's heat, without any air Will my wet clothes dry if I hang them under the sun, and if there is no air around the clothes? In other words, do I need both air and heat to dry wet clothes, or is heat alone (in the imagined absence of any air) enough to dry wet clothes? Related question : will wet clothes dry with only the suns heat, but when placed in a vacuum? Please note - I am trying to dry my clothes differently on earth, and not in outer space.
| No air means no vapor too. So without air your clothes will dry more easily, because the wetness will vaporize more easily.
| {
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Why are neutron stars mostly composed of neutrons? I understand that it is due to electron capture $(p + e \rightarrow n + v_e)$. My precise question is:
What are the conditions needed for a star core to start undergoing this process at a large scale?
| The minimum mass for one to form is 1.44 solar masses.
Energy barrier:
Neutron mass - 1
Proton mass - 0.99862349
Electron mass - 0.00054386734
$p+e→n+v_e$
Assuming that the neutrino mass is negligible, we get the difference between the neutron mass and the electron-proton mass to be 780 keV, meaning this is the energy barrier.
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Electric flux of a point charge in 2D I am working in two dimensions and have a point charge in the $x$-$y$ plane at the origin. The charge is surrounded by a square walls, $-0.5<x<0.5$; $-0.5<y<0.5$, which are grounded. If I have to calculate the electric flux through the boundaries of the square, should I just calculate $\int \vec{E}\cdot \mathrm{d}\vec{\ell} $ where $\mathrm{d}\vec{\ell}$ runs through the closed boundary? I ask this because until now, flux made sense to me, but only when talked about passing through some 2D region.
| Flux in two dimensions into a curve $\mathcal{C}$ should be
$$
\Phi=\int_{\mathcal{C}} \vec{E}\cdot\hat{n}\, d\ell
$$
Where $\hat{n}$ is the unit normal vector to the curve.
This for instance is consistent with gauss's law $\Phi=\oint_{\mathcal{C}} \vec{E}\cdot\hat{n}\, d\ell=q_{in}/\epsilon$, when you notice that poisson equation in two dimensions for a point particle gives a potential $\varphi(r)\propto$ log $r$, which gives $E \propto 1/r$
| {
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Are there any scales other than temperature that have different zero points? For most physical measurements, zero is the same regardless of the units used for the measure:
$0 \mathrm{mi} = 0 \mathrm{km}$
$0 \mathrm{s} = 0 \mathrm{hr}$
but for absolute temperatures, different systems have different zeros:
$0 ^\circ\mathrm{C} \neq 0\,\mathrm{K}$
Are there any other physical, measurable quantities (other than temperature) that have different zero points?
I'm looking for measurable quantities that are applicable anywhere -- things like voltage or temperature, not local quantities like "distance from the Empire State Building".
| Motion.
Velocity is obviously relative, and no "absolute rest" frame is known to exist.
Even acceleration, which is in a sense absolute, is sometimes specified relative to a local inertial frame (ie. freefall), sometimes relative to the distant stars (so you can talk about the acceleration of astronomical bodies due to gravity), and most commonly in everyday life, relative to the Earth. You think you are sitting still at your desk, but relative to a local inertial frame you are actually being accelerated upwards at about $9.8 m/s^2$ due to the pressure of the chair on your ass!
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How can static friction do work?
By definition, the work done by a force is $W = F\cdot d$, so how can static friction do work?
Can this force move the body a distance of $75~\text{m}$?
|
By definition, the work done by a force is $W = F\cdot d$, so how can static
friction do work ?
Can this force move the body a distance of $75~\text{m}$ ?
Friction does negative work on the truck, slowing it down and does not move it forward.
What does positive work on the truck, accelerates it and makes it translate $75~\text{m}$ is the engine of the truck.
The 80-Kg crate does negative work on it, because it is opposing the truck which is trying to push it forward, and acts over a distance of 75m.
Knowing that in $75~\text{m}$ of space, it reaches the velocity of $72~\text{Km/h}$, you can find the acceleration of the crate/truck. Using the weight of the crate and the coefficients of friction, you can find out the negative work done by the crate on the truck, subtracting energy and slowing it down.
Static friction locks the crate to the truck and prevents it from slipping back and off the truck's bed.
yes, this is the work done by the friction force on the truck not on
the crate. I need to understand the work done by the friction force on
the crate . – mech.eng
As I said, friction just locks the crate, it is an interface, like the clutch on a motor-car. The amount of positive work on the crate equals the amount of negative work $-W = +W$ on the truck. But the work on the crate is actually done by the engine of the truck.
The black arrow to the right shows the truck pulling the crate (thanks to friction) speeding it up to $v=20m/s$ and therefore giving it $W = E =16,000J$. I hope it is clear now.
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How to calculate the focal length of spherical shell (zero meniscus lens) I can get my head around the optics of a spherical shell (a meniscus lens that neither diverge nor converge, being neither positive nor negative etc).
I don’t see how I can describe in a formula how rays would diverge or converge towards some focus through such a lens. The lens maker formula 1/f = (n-1)(1/R1 - 1/R2) certainly suggests a focal length when plugging in R1 and R2 being inner and outer radius of the spherical shell.
I guess my question is if the lens maker formula is valid? And also is it so that rays effectively translate as opposed to change direction, i.e., all parallel rays entering also exists in parallel?
| I know that this post is quite old. However, the answer is maybe interesting for you. Directly applying lensmaker's equation doesn't work due to the air inside the lens.
I would rather suggest the following:
Since the middle of such a shell consists of air we can conceptually cut it in two halves.
These two halves are two identical lenses (just turned by 180°).
The focal length of one half is:
$$ \frac{1}{f} = (n-1) \left(\frac1{R_1} - \frac1{R_2} + \frac{(n-1) d_\text{lens}}{n R_1 R_2}\right)$$
Note that both $R_1$ and $R_2$ are positive. In our special case $R_2 < R_1$ and so this lens acts as a diverging lens. $d_\text{lens} = R_1 -R_2$ is the thickness of our lens.
Now we've got a lens system of two such lenses and we have to combine those. This done via:
$$\frac{1}{f_{res}} = \frac1{f_1} + \frac1{f_2} - \frac{d}{f_1 f_2}$$
where $d$ is the distance between this lenses (two spherical shell halves). In our case it is $d=2\cdot \frac{R_1+R_2}{2}$.
To finally answer your question. In the case of a zero meniscus lens we get $R_1 = R_2$ and also $d_\text{lens} = 0$. This means that the focal length is 0. So you won't see any effect of the spherical shell.
edit: this whole calculation neglects spherical aberrations. So this is only valid for rays close to the center of the shell.
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In special relativity, is it correct to say the speed of a body B relative to A is the same as measured by every observer? Here, two objects A and B are moving relative to each other. I use the Einstein velocity addition formula $v = {v_1 + v_2\over 1 + {v_1 v_2\over c^2}}$ to calculate the relative speed between A and B, where $v_1$ (for A) and $v_2$ (for B) are as measured by each observer in their own inertial frames and are as relative to the observer (say there is an observer C, then $v_1$ and $v_2$ are measured by C relative to C). The question is will the observers all calculate the same value $v$ in their own frames? In particular, will A and B agree on the same value $v$? Let's say one calculates $v = 0.1c$, will others also get $0.1c$?
Another thought is if A and B are stationary to each other as measured by C, will another observer D also see A and B stationary to each other?
Update
Is this question related to the principle of relativity, i.e.(copied from wikipedia), if a system of coordinates K is chosen so that, in relation to it, physical laws hold good in their simplest form, the same laws hold good in relation to any other system of coordinates K' moving in uniform translation relatively to K.
So in the above exmaple, if C sees A and B are stationary to each other, but D doesn't agree, wouldn't this be a violation of the principle of relativity?
| Let's say you have 3 systems. $B$ moving relative to $C$ with velocity $u$ and $A$ moving relative to $C$ with velocity $v$, all along one axis.
$A$ will "measure" for the velocity of $B$:
$$ u' = \frac{u-v}{1-\frac{uv}{c^2}} $$
While $B$ will "measure" for the velocity of $A$:
$$ v' = \frac{v-u}{1-\frac{uv}{c^2}} = -u' $$
It holds true that the velocity of $B$ relative to $A$ is the same in terms of magnitude but opposite in terms of direction than the velocity of $A$ relative to $B$. If this were not true, you couldn't define a proper transformation for both systems, since the transformation $A\rightarrow_v B \rightarrow_{\tilde{v}} A'$ would not yield $A'=A$. The systems $A$ and $B$ should obey the same unique transformation rules. There is nothing special about them.
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Is there any such thing as a change in acceleration (ex: 3 m/s/s/s)? If there exists something like that, then in $distance/time/time/time$, how is it expressed?
| An actual example in which there is a non-zero change in acceleration, that is, jerk, occurs is a spring. A spring's motion is described by a sinusoidal function. The derivative of a sinusoidal function is just another sinusoidal function. As a result, you can differentiate such a function infinitely many times, and will never have a derivative that's 0/a constant. So not only is there there a non-zero jerk in the motion of a spring, every single derivative of position is non-zero.
| {
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Why do the $u$ and $d$ quark not have an associated quantum number? All the other quarks ($c$,$s$,$b$ and $t$) have quantum numbers of charmness, strangeness, bottomness and topness that are conserved in strong interactions.
This allows, among other things, flavour changing neutral currents in $K^0$, $B^0$ and $D^0$ mesons as I discussed in this question but prevents them in pions $\pi^0$.
Is there any physical meaning as to why there are no quantum numbers of 'upness' and 'downness'?
| You certainly can say that up quarks have +1 upness and down quarks have -1 downness. See Griffiths' particle book, 2nd ed., p. 49. It's just not very useful. The only quark with $S = C = B = T = 0$ and electric charge of $+\frac{2}{3}$ is the up quark. The down quark has $-\frac{1}{3}$ charge.
You could also say the up quark has baryon number of $\frac{1}{3}$ and the third component of isospin $+\frac{1}{2}$. You see, there are already plenty of ways to describe the up and down quarks, and these other quantum numbers predated the quark model.
| {
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Do the copper connectors we used in measuring the resistance of a wire contribute to the said measurement? We measured the resistance of a wire by setting the multimeter to ohmmeter mode and connected the ends of the ohmmeter to the ends of the wire; and we also calculated the theoretical value using a table of resistivity. The given wire is made of nichrome material. But the measured value deviates about 40% of the theoretical one. Do the connectors made of copper contribute to the measured resistance?
| As noted by Floris, the best way to measure small-value resistance is to use a four-point Kelvin connection; unless the current drawn by the voltage meter is significant (in which case there are other problems) then provided that current-source probes are either the inner two or the outer two (as opposed to being interleaved with the voltage-reading ones) the device will measure the resistance of the portion of the device under test which is between the two inner probes, regardless of how much resistance the connections have or well-matched they are.
If the connection resistance is unknown but the resistances of at least two of the connections can be presumed to be well-matched, it's possible to use three wires rather than four by using a circuit which uses a zero-current wire to sense the voltage drop across one of the high-current wires and subtracts twice that value from the voltage drop observed between the two current-carrying wires. This approach may be good if a significant portion of the resistance between the measurement apparatus and the device under test will be in the form of wire resistance and will always be well-matched.
| {
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Why do we use capacitors when batteries can very well store charges? Can batteries be used instead of capacitors? I am trying to figure out a basic, superficial and any obvious difference between the two.
|
Why do we use capacitors when batteries can very well store charges?
There's an important point that, so far, I don't see in other answers.
Neither of these devices store charge!
A "discharged" battery or capacitor contain the same net quantity of electrical charge as a "fully charged" battery or capacitor.
What they are "charged" with is energy, not electrical charges. The verb "charge" here is used in the same sense as when you are invited to charge your glass with champagne at a celebration. The verb "charge" and the noun "charge" are not, in this context, as related as their spelling suggests.
Actual use
Batteries are used for storing energy over long periods of time (typically hours, days, months or years) and for then supplying that energy to a device for a period of operation that may be minutes but is more likely hours.
Capacitors are more typically used for purposes for which batteries are unsuitable
*
*filtering.
*smoothing.
*oscillating.
*etc.
These generally involve much shorter timescales and much higher frequencies.
There are some, relatively small, areas of overlap. For example Supercapacitors can be used in some automotive applications in place of batteries.
| {
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Are circles stronger than triangles? I've often heard in engineering that, "there is no shape stronger than a triangle." I also recall that arches are also very strong shapes, which can be crudely described as a perpendicularly-symmetrical half-an-ellipse; Which can be simplified to half a circle.
If there were no conventional complications with designing structures to utilize circles; Which shape is stronger?
Given a simple two-dimension-like application such as simple bridges or trusses as an example, of obvious visualizations.
| I'm glad that my head is spheroid instead of pyramidal: then the circle is favored.
The eggs have an extremely strong shape: then the circle is favored.
The insects are extremely strong and they favor the round shape.
The engineering/construction by humans are easier with rectilinear elements and the triangular shape are favored.
If I were an explorer of new continents (I'm recalling the conquest of the west of america and the confrontation with the natives(Indians)) I'd put the wagons in a circle to strengthen the defense. The circle is again favored.
/_ A corner of a triangle _\ is easier attacked from the two exterior sides, i.e. there are three weaker regions in the triangle (compared to the central region) .
In the circle there are no weaker regions and the limiting perimeter is minimal irt the protected area/volume.
The towers of the castles are more often than not with a circular pattern. This surface shape offers more options of reflection of any non perpendicular stroke.
| {
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Does it take more energy to open a door when applying force close to the hinge?
Assuming an ordinary hinged door (without any springs), would it take more energy to open it when applying force in the middle of the door (point b), rather than at the end of the door (point a), where the door knob is?
"Opening the door" should be interpreted as accelerating the door to a certain rotational speed.
My own answer is no, since the change in force would be proportional to the distance required to open the door and therefore the total energy would remain the same.
| Torque=(R) x (F)
Energy req. to rotate=(T).(Theta),
Ta=Tb,
ONLY Fa is less than Fb. & T at Hinge=0,
it'll not rotate there. The force required will increase from A to hinge point. Energy needed hence is const from A to hinge(except hinge point).
[Ta means torque applied at A]
| {
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Photon: speed and mass Is it correct to think that the speed of light does not depend on the speed of light source because photons have no mass, so they have no the kind of inertia that is associated with mass, so they can not "feel" (acquire) the speed of the light source?
| Is it correct to think that the speed of light does not depend on the speed of light source because photons have no mass, so they have no the kind of inertia that is associated with mass, so they can not "feel" (acquire) the speed of the light source?
No. Photons have an energy E=hf or E=hc/λ where f is frequency and λ is wavelength. The frequency and wavelength are there because photons have a wave nature. The speed of a wave doesn't depend on the speed of the emitter, it depends on the properties of the medium. Space is such a medium, with properties such as permittivity and permeability, wherein the speed of light is given as $c_0={1\over\sqrt{\mu_0\varepsilon_0}}$, see Wikipedia. This expression is somewhat similar to shear wave velocity $v_s = \sqrt{\frac {G} {\rho} }$, again see Wikipedia
As an aside, note that a photon has no rest mass because it isn't at rest. However it does have a non-zero "inertial mass". This is a measure of energy rather than a measure of mass per se, because when unqualified, mass is assumed to mean rest mass. See the last line of Einstein's E=mc² paper: "If the theory corresponds to the facts, radiation conveys inertia between the emitting and absorbing bodies". Radiation conveys inertia, which is why the photon has a non-zero inertial mass.
| {
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How does force transmit through a solid block? If you take a solid block, say, a cube of side length $l$ with density $\rho$, you place it upon a solid floor, and you apply an external pressure of $p(x,y)$ on the upper face ($x$ and $y$ define a rectilinear coordinate system for a horizontal plane), what pressure, $q(x,y)$, does the block exert on the ground?
First of all, I realise that one component of $q(x,y)$ will be the weight of the block, which attributes a uniform pressure distribution of $q_{grav}(x,y)=\rho l g$. So what I'm really interested in is the difference, $q_0 = q-q_{grav}$.
If you assume that $q_0 = p$, then this seems plausible as net moment is zero (block won't rotate) and it seems intuitive if the pressure above applied is uniform, the pressure below should be uniform. However, this seems unrealistic for applying point loads onto the top surface: if you poke the top face with a pin, will the bottom face really exert a sharp point force onto the ground?
So, next, I hypothesis that the block might act like a sort of force diffuser, where the distribution of $p$ spreads and blurs to the extent you go down the block, until you get to the bottom face. So $q_0$ might look like $p$ having undergone a blur transformation, so that a point load applied on the top face might look like a bell distribution on the bottom face.
I also hypothesise that the height of the block determines the extent of the blurring. This seems intuitive in the cases for applied point forces on a flat plate, and on a tall column.
However, this is hypothetical, and have no means, so far, or determining this. Can this be determine mathematically? Or more practically, can this be tested by experiment? (e.g. A floor that changes colour depending on yielding to pressure exerted)
| Your force will transmit through the cube to the bottom and it will spread like a projection from your pressure point. having sand underneath it will give away a pression mark the size of your cube showing how your force transmitted. the height will no matter, being a solid cube and your force will not deformed it and lose itself in the process.
| {
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Constants of motion in quantum mechanics What is the meaning of a constant of motion in quantum mechanics (an observable-operator that commutes with the Hamiltonian) in contrary with classical mechanics?
| According to the Wikipedia article "Constant of Motion"
A quantity $A$ is conserved if it is not explicitly time-dependent and
if its Poisson bracket with the Hamiltonian is zero
That is to say, if both
$$\frac{\partial A}{\partial t} = 0 $$
$$\{A, H\} = 0 $$
then
$$\frac{dA}{dt} = \frac{\partial A}{\partial t} + \{A, H\} = 0, \, \Rightarrow A \mathrm{\;is\; a\; constant}$$
For an operator $\hat A$ corresponding to the classical observable $A$, if the operator is not explicitly time-dependent, and the commutator with the Hamiltonian is zero, the expectation of the operator $\hat A$ is a constant.
That is to say, if both
$$\frac{\partial \hat A}{\partial t} = 0 $$
$$[\hat H, \hat A] = 0 $$
then
$$\frac{d\hat A}{dt} = \frac{\partial \hat A}{\partial t} + \frac{i}{\hbar}[\hat H, \hat A] = 0, \, \Rightarrow \langle \hat A\rangle \mathrm{\;is\; a\; constant}$$
| {
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Why is charge of the electron negative? How did scientists figure out that the charge of the electron was indeed negative? I know how the cathode ray tube experiment works, but how did Thompson know that the plate that the cathode ray beam was attracted to was positive, meaning the cathode ray was negative? What is the history behind positive and negative charges leading up to the cathode ray experiment. Ben Franklin postulated that a body with excess electricity was positive as in a surplus of electricity, and that surplus flows from positive to negative. But how and when was it discovered that electricity flows from negative to positive?
| As Zeldredge said the name is arbitrary and does not matter electron could have been positive and protron negative just the name.
| {
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At what energy consumption would we get a 1 degree rise in the Earth's temperature? If energy consumption continues to rise at (say) 4% per year, how long before the heat dissipation seriously impacts climate?
| *
*Global energy consumption is $5\times10^{20}\ J/yr$
*Assume it is all used to power incandescent lightbulbs, so 95% goes to heating the atmosphere
*The mass of the atmosphere is $5\times10^{18}\ kg$
*The heat capacity of air is $1\times10^{3}\frac{J}{kg\cdot °C}$
Assuming all the heat goes to the atmosphere and stays there, using the definition of heat capcity $C$:
$$Q = mC \Delta T$$
$$\Delta T = \frac{Q}{mC}=\frac{\left(95\%\right)\left(5\times10^{20}\ J/yr\right)}{\left(5\times10^{18}\ kg\right)\left(1\times10^{3}\frac{J}{kg\cdot °C}\right)} = 0.095 °C/yr$$
To answer your question about a rise of 4% let's calculate how many years it would take before the atmosphere was heating at 1 °C/yr... oh compound interest formulas, how I've missed you:
$$n=\frac{ln(FV)-ln(PV)}{ln(1+i)}=\frac{ln(1)-ln(0.095)}{ln(1+0.04)}=60$$
In other words, if the rate of energy consumption goes up at 4% per year and nearly all that energy ends up as heat (using lightbulbs as a model), in 60 years the atmosphere would heat up at 1 °C/yr. Please note these calculations are extremely naive and have little bearing on reality in that they ignore 1) factors that actually control the climate, such as gas composition, greenhouse effect, transfer of heat between air-oceans-land, feedback mechanisms, etc. 2) the economics of steady 4% energy consumption increase 3) the fraction of energy consumption that heats the atmosphere.
| {
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In aircraft design why are light materials preferred to heavy ones? Especially given the relative cost between (say) steel and carbon composites. After all, I assume most fuel is consumed overcoming drag not accelerating mass. Once an aircraft reaches cruising speed it should not matter how heavy/dense the aircraft is as long as the lift/drag remains constant.
| Building on Carl's comment:
The way any heavier than air aircraft works is that the wings exert a downward force on the air - increasing the downward momentum of the air results in a net lift force in the wing.
Now if you are heavier, you need to either move more air down per unit time, or move it down faster, in order for you to generate sufficient force.
You can move more air by flying faster - increasing drag; or you move the air down further by changing the attack angle of the wing - increasing the projected area and increasing the drag.
A simplistic calculation suggests that if you have a wing span $\ell$ and velocity $v$, you sweep an area $\ell v$ per unit time. If air density is $\rho$ and the attack angle is $\theta$ you move the air down with a velocity of $v \tan\theta\approx v\theta$ for small \theta$.
Now the projected area of the wing scales with the angle $\theta$ - it follows that if you want to generate more lift, whether by flying faster or increasing the angle of attack, you need more power.
The penalty is surprisingly large. Over the operational life of a plane, one kilo of additional mass can result in hundreds (thousands) of dollars of additional fuel costs. I did the math once - I will see if I can find it.
| {
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What really is the self-adjoint extension? Going through the Quantum mechanics book by Capri, am time and again held with some stupid doubts on this topic of self-adjointness. We have for the momentum operator in finite domain,
$$
p = -i\hbar \frac{\partial }{\partial x} \\
D_p = \big\{f(x),f'(x)\in \mathrm{L_2}(0,L) , f(0) = f(L) = 0 \big\}
$$
Now we go on to define, the adjoint of the $p$ operator, with the larger domain
$$
p^{\dagger} = -i\hbar \frac{\partial }{\partial x} \\
D_{p^{\dagger}} = \big\{f(x),f'(x)\in \mathrm{L_2}(0,L) , f(0) = \mathrm e^{i\theta}f(L)\big\}
$$
Now, he says, we can see that $p^{\dagger}$ is self-adjoint, which I believe means that $p^\dagger = p ^{{\dagger}^{\dagger}}$ ($D_{p^\dagger} = D_{p ^{{\dagger}^{\dagger}}}$). So we have a self-adjoint extension of $p$. So my question now is that, is it $p ^{{\dagger}^{\dagger}}$ that is the extension of p or what is it that is its extension ?
But he also goes on to say that a symmetric operator $A$ is essentially self-adjoint if $ A^{{\dagger}^{\dagger}} $ is self-adjoint. But is not this true for the above case, given that $p^\dagger = p ^{{\dagger}^{\dagger}}$ and a $p^\dagger $ is self-adjoint ? But however the above case is not essentially self-adjoint since there are infinite ways of choosing $\theta$.
| A closed extension $A_c$ of an operator $A$ is an operator whose action is the same as $A$, the domains satisfy $D(A_c)\supset D(A)$ and $A_c$ is closed.
Given that, the smallest closed extension of a symmetric (densely defined) operator is its double adjoint $A^{**}$. We call it the closure of $A$, and denote it by $\overline{A}$.
An operator $A$ is called essentially self-adjoint if its closure $\overline{A}$ is self-adjoint. This also implies that $\overline{A}$ is the unique self-adjoint extension of $A$ (a self-adjoint extension is a closed extension of $A$ that is self-adjoint). This is a nice feature, since in general a symmetric, densely defined, operator may have zero, one or infinite self-adjoint extensions.
In the case above, it depends on the definition of $D_{p^*}$. If it is dependent on a fixed $\theta$, then something seems not right, but maybe the domain is of functions that satisfy the condition for any $\theta$, and then it could be effectively self-adjoint (and therefore $p$ ess. s-a).
| {
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The general equation for a wave packet derivation? On Wikipedia it gives the general equation for a wave packet (and therefore for a wave?) to be:
$$u(x,t)=\frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty}A(k)e^{i(kx-\omega t)} dk$$
I have been trying to derive this but have had no luck. The closest I get is:
$$f(x,t)=\sum_{j=1}^{\infty}A_j\int_{-\infty}^{\infty}\hat{u}_j(\omega)e^{2\pi i\omega x}d\omega\int_{-\infty}^{\infty}\hat{v}_j(\omega)e^{2\pi i\omega t}d\omega$$
But I don't know if this is right or where to go from here? So how can I derive the first equation?
| A solution to the Schrödinger equation for a free particle is a plane wave, and because any combination of solutions is also a solution we can construct solutions by summing up plane waves.
The equation you quote is constructing a solution by Fourier synthesis. Since the plane wave function $e^{i(kx-\omega t)}$ is a solution we can use Fourier synthesis to sum up an infinite number of these solutions to produce any wavepacket we want. The function $A$ is just the Fourier transform of our desired wavepacket function $u$.
The equation is the most general form possible because we have complete freedom to choose $A$ so we can build any function $u$ that we want i.e. we can make the wavepacket envelope any shape we want. We can be confident that $u$ must be a solution to the Schrödinger equation because it is constructed from plane wave solutions to the Schrödinger equation.
| {
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Why do $S_x$ and $S_y$ flip up/down spin states but $S_z$ does not? By using the notation $S\lvert s,m_s\rangle$, such that $\bigl\lvert\frac{1}{2},\frac{1}{2}\bigr\rangle=\lvert+\rangle$ and $\bigl\lvert\frac{1}{2},-\frac{1}{2}\bigr\rangle=\lvert-\rangle$ we can obtain following the eigenvalue equation for the components of the spin vector:
$$\begin{align}
S_x\lvert\pm\rangle&=\frac{\hbar}{2}\lvert\mp\rangle \\
S_y\lvert\pm\rangle&=\pm i\frac{\hbar}{2}\lvert\mp\rangle \\
S_z\lvert\pm\rangle&=\pm\frac{\hbar}{2}\lvert\pm\rangle
\end{align}$$
However, I am struggling to interpret the result. If we consider the $x$ and $y$ component of spin, then there seems to be flipping of the state function, but it's not the same for the $z$-component. Why is that?
| You chose the $\lvert \pm \rangle$ to be an eigenvector of $S_z$ with eigenvalue $\pm\frac{1}{2}$ - that's what the $m_s$ is: The eigenvalue of the state w.r.t. the $z$-spin.
Since $S_x$ and $S_y$ do not commute with $S_z$, $\lvert \pm \rangle$ is not an eigenvector of them, hence the state cannot stay the same after they are applied to it.
That the spin is precisely flipped comes from the form of the commutation relations
$$ [S_i,S_j] = \mathrm{i}\epsilon_{ijk}S_k$$
and you can see that the spin is flipped by evaluating $S_z$ on $S_x\lvert \pm \rangle$ and $S_y\lvert \pm \rangle$.
| {
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Is it possible to derive the angular frequency of a simple harmonic oscillator using total energy?
I want to show that
$$\omega=\sqrt{\frac{k}{m}}$$
using the fact that
$$E=K+U=\frac{1}{2}mv_x^2+\frac{1}{2}kx^2=\frac{1}{2}kA^2.$$
The issue is that I have derived a formula that isn't correct:
I first took the time derivative of the energy equation,
$\dfrac{dE}{dt}=mv_x\dfrac{dv_x}{dt}+kx\dfrac{dx}{dt}=kA\dfrac{dA}{dt}$
$\dfrac{dE}{dt}=mv_xa_x+kxv_x=kAv_x$.
So we now have
$mv_xa_x+kxv_x=kAv_x$
$ma_x+kx=kA$
Substituting $a_x=-\omega^2x$ gives
$-m\omega^2x+kx=kA$
$m\omega^2x=k(x-A)$
$\omega^2=\dfrac{k}{m}\dfrac{(x-A)}{x}$
$\omega=\sqrt{\dfrac{k}{m}\dfrac{(x-A)}{x}}$.
This is very nearly correct but not quite...
Is it possible to prove this formula in this way? Also, if I haven't done something that's completely incorrect, then is this a standard formulae that I just haven't seen before?
| Whenever you want to prove something, you need to get your hypotheses straight. I understand that you want to prove that a mass $m$ on a spring with constant $k$ moves harmonically with frequency $\sqrt{k/m}$, but you seem to be assuming that result in your "proof":
First off, writing $E = \frac12 k A^2$ already implies that the motion is harmonic with frequency $\sqrt{k/m}$, partly because you need to know $x(t)$ and $\dot{x}(t)$ to plug into $E$, but also because $A$ is the amplitude, a notion which doesn't make sense if the motion isn't harmonic!
Also, $dA/dt$ isn't $v$ (because $dx/dt$ is $v$), it's zero: the amplitude is a constant. Using that you can get the equation of motion, but there's that same problem again: stating that $\ddot{x} = -\omega^2 x$ already assumes that $x$ is a simple harmonic motion.
The way to prove what you want is to guess a solution of the form $x=\cos(\omega t)$ (you could also use $\sin(\omega t)$) and plug it either into the energy (and then demand that the energy be a constant, that is, $dE/dt = 0$), or into the equation of motion $\ddot{x} + \frac{k}{m}x =0$. You will find that for the equations to be satisfied, it must be that $\omega^2 = \frac{k}{m}$.
| {
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What compounds or elements only have one phase or two phases? Wood appears to be one. I think gases like helium and hydrogen cannot exist in the solid state under normal pressures, correct? And why do those "phase cheaters"-- those elements/compounds which sublimate directly, skipping a phase, or "procrastinators"-- elements/compounds which just never reach the phase-- why do they do that?
| Iodine - solid to vapor when heated under normal condition. In this context the word you need to search for is "sublimation"
| {
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Escape velocity for Schwarzschild metric I can't fill in the gaps in my solution to this and assistance or a reference would be appreciated.
The question begins with the straightforward derivation of the EoM for a massive particle orbiting in the equatorial plane, as
$$ \left( \frac{du}{d\phi}\right)^2 = \frac{c^2 k^2}{h^2} - \alpha \left( \frac{c^2}{h^2} + u^2 \right) $$ where $u = \frac{1}{r}$, $ h, k$ are constants arising as $ \alpha \dot{t} = k$ and $\dot{\phi} = h u^2$, and $\alpha = 1-\frac{r_s}{r} $ where $r_s$ is the Schwarzschild radius.
It then says a stationary experimenter at radius $a > r_s$ projects a massive particle with speed $v$ normal to the radial direction, and asks me to show that in the case $h^2 > 3 r_s^2 c^2$ the particle will be ejected if $v$ exceeds an escape velocity similar in form to the Newtonian one.
Clearly the above condition restricts to the case of three real roots, and I think that the condition I want is that the smallest root of the above cubic (there's an extra $u$ in the $\alpha$) is $\leqslant 0$, though I'm not entirely sure why that's necessary/sufficient. Given that, I obtain the result $ v \geqslant \sqrt{\frac{2GM}{a}} $.
Is this result correct? And could someone explain why that condition is the right one?
| Let $f(u)$ be the third degree polynomial, so that
$$\left(\frac{du}{d\phi}\right)^2 = f(u)\tag{*}$$
The experimenter starts at $u=1/a$ and must reach infinity, $u=0$. The crucial point is that if $f(u)$ is negative somewhere in the region $0<u<1/a$, then the equation of motion $(*)$ prevents crossing the negative region, so you can't reach infinity. In other words if $u(\theta)$ solves $(*)$ then $f(u)>0$; since $u$ is continuous, you can't connect $1/a$ to $0$ if $f$ is negative somewhere in between.
Now it's a matter of ordinary calculus to determine the shape of $f$: we learn that it has exactly one root $u_\ast$ in the range $u<r_s$ and that $f<0$ for $u<u_\ast$ and $f>0$ for $u_\ast<u<1/r_s$. Since $f$ must be positive for $0<u<1/a<1/r_s$, we must have $u_\ast<0$.
Whether that's the right equation depends on what you mean, i.e. see this question.
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How does mechanical energy conservation work? I am a little confused about how mechanical energy conservation operates when it comes to things like predicting velocity. I know that if conservative forces are the only forces acting on a body, then we can say that mechanical energy is conserved. This is simple to see when we have lateral up and down motion, but when it comes to predicting the velocity of a pendulum or a roller coaster (neglecting all friction) I'm not sure how the law operates. For example, given the initial peak height of the roller coaster, I can predict the velocity at any point, despite the fact that there are various loops and curves. And for a pendulum, the motion is in an arc. Despite these complexities, the same equations used for these situations are used for simple free-falling situations. Could someone give me a deeper understanding of how these equations are able to make predictions about velocity and such in complex situations like riding a roller coaster?
| When someone moves in a field then the work generated by field forces is independent of motion and related to the starting and ending positions. The quantity $U$
$$\int_{r_1}^{r_2} \mathbf F( \mathbf r) \cdot d \mathbf r = [U ( \mathbf r)]^{r_2}_{r_1}$$ is called potential energy and is associated with a potential $V$.By work energy thm we get $$dK = W = W_0 - dU$$ and when $W_0$ is 0 then the quantity $E=K+U$ is conserved.
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Can we find the exponential radioactive decay formula from first principles? Can we find the exponential radioactive decay formula from first principles? It's always presented as an empirical result, rather than one you can get from first principles. I've looked around on the internet, but can't really find any information about how to calculate it from first principles. I've seen decay rate calculations in Tong's qft notes for toy models, but never an actual physical calculation, so I was wondering if it's possible, and if so if someone could link me to the result.
| Any population whether human, animal or atomic nuclei, will with no other complications
change proportional to the amount already there. Yielding a very simple differential equation.
$$
\frac{dP(t)}{dt} = k\,P(t)
$$
where $k$ is a constant with a negative sign for exponential decay and plus sign for exponential increase.
i.e. the solution is
$$
P(t)=P(t=0)\exp(kt)
$$
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Does it mean the molecules of all matter above absolute zero temperature are moving? According to my knowledge, heat is the energy that is stored in form of kinetic energy of molecules in Brownian motion. However, in a macroscopic view, a rigid body seem to be "stable" but still store heat. I am wandering does it mean in a microscopic view, the molecules of solid matter are still moving?
| Let us understand Brownian motion in liquids before we look at the motion in solids. If you observe a glass of water at rest on a table, it "appears" to be motionless. However, all we need is a magnifying glass to observe the random, incessant motion of water on the surface. This random motion is a manifestation of heat.
The same thing happens in a solid. A block of steel "appears" to be at rest. But its atoms are constantly vibrating. They cannot wander off too much, unlike water molecules, because they are bound by the lattice. These vibrational modes are always present at room temperature. As you cool the solid, they will gradually vanish. However, even at absolute zero, you will still observe "zero-point vibrations".
Please note that atoms/molecules in a solid at rest are always exchanging energy with their surrounding. They absorb photons and emit them. They are always "active" and in motion.
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Can a brane have infinite volume? I'm asking this within the framework of brane theory. My intuition is that branes are objects with finite volume, but then how can they accomodate an Universe that is infinite in extent?
| Branes typically have infinite volume. As an intuitive example of a brane, imagine an infinite plane in 3D, which is a 2-brane. The brane is specified by (for example), $z=0$ if $(x,y,z)$ are the usual Cartesian coordinates. An infinite line is a 1-brane and is specified by (again, for example) $y=z=0$. If the branes are Euclidean and wrapped up, then it is possible that they have finite volume.
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Why does the Magnetic Flux Density B, saturates in a ferromagnetic material with increasing H? I understand that the magnetization must saturate as more and more domains are aligned. But $B$ is still directly proportional to $H$, and hence it must increase linearly with $H$.
But every book that teaches $B-H$ curve, says that it saturates after some time.
How can this be so?
| No, your understanding is wrong: $B$ isn't proportional to $H$, the relationship is $\vec{B}=\mu_0 (\vec{H} + \vec{M}(\vec{H}))$ where $\vec{M}$ is the magnetization (see Wiki page of this name). And $\vec{M}$ saturates for precisely the reason you state: its maximum value is reached when all the magnetic dipoles in a medium are perfectly aligned with the ambient field $\vec{H}$. At very high fields, when the material saturates, $\vec{B}$ keeps increasing with increasing $H$ owing to the "freespace" $\mu_0\vec{H}$ term in $\vec{B}=\mu_0 (\vec{H} + \vec{M}(\vec{H}))$, but for many materials, the gradient $\mu_0$ is extremely small compared to the rate of change of $\vec{M}$ in the unsaturated region, so the $B-H$ curve looks as though it flattens out even though its gradient has fallen to $\mu_0\ll \left.\mathrm{d}_H M\right|_{H=0}$.
| {
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What is precisely the energy scale of a process? Coupling constants run with the energy scale $\mu$. But what is exactly this energy scale. My question is, if I have a physical process, how do I compute $\mu$?
| You go to the center of mass frame to find that $\sum_i \vec{p}=\vec{0}$, and the total momentum four vector is thus $$P_{\text{tot}}^{\mu}=\left(\frac{1}{c}\sum_{i}E_i^{\text{COM}}, \vec{0}\right)$$
then we define the energy scale covariantly as $\mu=\sqrt{-s}$ where $s$ is the mandelstam variable $s\equiv P_{\text{tot}}^{\mu}P_{\text{tot}\mu}$ in units of $c=1$, so that in COM frame
$$
\mu = \sum_i E_i^{\text{COM}} = \sum_i \frac{m_{oi} c^2}{\sqrt{1 - v_i^{\text{COM }2}/c^2}}
$$
| {
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Can electricity flow through vacuum? People say yes, and give a wonderful example of vacuum tubes, CRTs. But can we really say that vacuum (..as in space) is a good conductor of electricity in a very basic sense?
| The conductivity of the vacuum is not a very trivial issue. In fact, depending on how you look at it, it behaves in two different ways.
Firstly, there is no retarding force on any charged particle with constant velocity in vacuum. To this extent, no extra work is required in maintaining a constant current through any surface in vacuum.
In stark contrast however, is the presence of free charges in conductors. Normally, when an electric field $\mathbf{E}$ is applied across a conductor, we get a current density due to the 'internal' charge flow, given by:
$$\mathbf{J} = \sigma\mathbf{E}$$
where $\sigma$ is the conductivity. Clearly, $\sigma = 0$ in a vacuum - electric fields do not spontaneously cause currents to flow. Thus, in this sense, the vacuum is not a conductor at all. Even everyday insulators have low but non-zero values of $\sigma$.
Thus, the resistance of the vacuum is in fact, infinite, as long as we define resistance in terms of the response of the charge carriers of a material. In this sense, we might say that it is an insulator - there are no charge carriers.
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Why does light "expand"? The "opening" for light to travel out of a flashlight is circular. Why does the light shine on the objects the flashlight is pointed at over a much larger surface area than the surface area of the aperture? Why doesn't it just illuminate the whole room or wherever the flashlight is?
| Light is a wave , an electromagnetic wave classically. When leaving a point source a wave expands isotropically in angle, spherically, and its intensity falls like 1/r^2 where r is the distance from the source
If a point source lamp is set in a room, it will illuminate spherically all of it, as happens with the lamps hanging from the roof.
A flash light has two extra hardware. The light from the point source is focused with a lens that focuses the point source ahead, i.e. does not allow light to disperse spherically . The other is the aperture that also constrains the start of the dispersion of energy to a beam.
Even without the lens the light would start dispersing spherically from the aperture. The lens concentrates the intensity to a beam.
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What does magnetic field vector mean? I am trying to understand what a magnetic field vector tells us about the magnetic field.
I understood that a vector is just a representation of a point and how much it is moved in x,y and z direction from its origin.
But how can this explanation apply to the magnetic field?
source: http://www.cprogramming.com/tutorial/3d/rotationMatrices.html
second paragraph
| Generally, a magnetic field is produced either by a moving charge or a magnet. This magnetic field's strength and direction at a point are described by magnetic field vector.
The closer we move the point to magnet, the denser is the field at the body. Thus, the length of magnetic field vector increases.
magnetic field vector is also used in determining the force on a moving charge.
$$\vec{F} = q(\vec{V} \times \vec{B})$$
where q = charge in coulumb,
v = velocity of the charge
B = magnetic field vector.
In this way, there are many applications of magnetic field vector.
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Why can't I harness normal force? Lets say I have my palm flat with a book resting on top of it, and I have my feet on the ground. I extend my arm so that now it's kind of difficult to keep the book up. Why doesn't my hand just produce normal force on the book, cancelling out the force of gravity, and costing me no effort whatsoever?
| Name has a really great answer, but I'm gonna try to clarify it a little. If you hold the book, its very unlikely that our arm is perfectly at 90 degrees, perpendicular to the ground. That means you will have to counteract the book's torque. But even if you held your arm perfectly at 90 degrees you would still get tired. It's not like the book is perfectly above your center of gravity, and that is also causing some torque. Also, think about all the joints in your body from the top of your arm to your foot, they are not all perfectly aligned so the muscles need to do work so that you don't collapse.
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The amount of potential energy at the height of h When we lift an object upwards with a constant velocity for a distance of $ h $ the work that we've done is $mgh$ and the work done by the force of gravity is $-mgh$. So the net work on the object is zero and it doesn't gain any energy. how its potential energy will be converted to kinetic energy when we drop that object while it hasn't gained any energy?
| Gravitational weight is a conservative force so work done against it is entirely equal to the change in potential between the start and end points (in the absence of dissipative forces like friction). The internal energy of the mass does not change as it is moved upwards and your assessment of the energy considerations is correct.
Consider alternatively the forces before conversion to work:
The weight of the object is $F_g = mg$
Since you're raising the mass at a constant velocity, the force you apply to lift it is equal and opposite to the above, i.e. $F_l = -F_g = -mg$
This produces zero net force on the mass and so zero net work. It continues to move at constant velocity and its internal energy does not increase.
When you stop applying your lifting force $F_l$ the mass experiences only its weight $F_g$, producing a non-zero resultant force, which then performs work on the mass as it moves from the area of high potential to the area of low potential (i.e. falls downwards).
The work done by this force up to a given moment in its fall is $W_{fall} = F_g \Delta h$ (where $\Delta h$ is the distance the object as fallen since being released), and is equal to the kinetic energy of the object at that exact moment.
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What is the uncertainty principle? I looked on Wikpedia for information on the uncertainty principle, but after reading it I still had no idea. I know it has something to do with how many things you can hold at some spot for some amount of time (maybe?). This is inspired by this question.
| Let's forget physics for a moment and just talk about the mathematics of waves.
The uncertainty principle is a property of waves. Think of a single, narrow pulse traveling along one direction. The pulse is narrow, and so the position of the pulse at any given time is easy to quantify. But this is a single, non-periodic pulse. You can build up such a pulse from a lot of sinusoids, but in doing so, the wavenumber (frequency) of such a pulse becomes ill-defined. It's a combination of a lot of wavenumbers, so no single wavenumber describes it.
Conversely, think of a pure sinusoidal wave. It has a single wavenumber, but what is the position of such a wave? It a periodic thing; it doesn't have a single position. You can describe it in terms of magnitudes at a lot of positions, but that's all.
That's the uncertainty principle, a fundamental property of waves. You can localize a wave in terms of its position or wavenumber, but not both at the same time (at least, not below a certain limit). The quantities $\sigma_x$ and $\sigma_k$ describe the distribution of positions and wavenumbers for a given wave (smaller $\sigma_x$ means that the wave is more like a localized spike at one position; larger $\sigma_x$ means that it's smeared out more like a sinusoid), and their product is constrained by the uncertainty principle.
The uncertainty principle is usually phrased in terms of position and momentum, instead of wavenumber as I've written. This is just a little bit of physics: we say that momentum is proportional to wavenumber by the constant $\hbar$.
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Where do forces point in an equilibrium system
I have the system above, with three identical balls of weight $W$ and radius $r$. The angle joining the centres is $\theta$, and the coefficient of friction between the balls and the plane is $\mu$ and between the balls $A/C$ and $B/C$ is $\mu '$. The system is in equilibrium. Denote $S$ the point of contact between balls and $P$ the point of contact between ball and plane.
My problem is that I don't know where to put the forces from which to do the calculations. Consider $A/C$: At $S$, there is a reaction force $R_S$ towards the center of both balls. But my question is, are there two forces or one more force at $S$? There are three options that I can think of:
Both a frictional force $F_{FS}$ and a force created by the weight of the ball $C$, $F_S$?
A more general force $F_S$ that includes both the mentioned above. But where would it point?
Or simply the only force that exists at $C$ is the frictional force $F_{FS}$, and the reaction force $R_S$ encompasses the force that pushes the ball $A$?
| Okay, the system is in equilibrium so all the forces must be balanced.
First consider the weights
We have $W$ downwards from the centre of each of the 3 balls (assuming their centres of mass are at their geometric centres).
We also have $N_A = N_B = \frac{3}{2}W$ upwards from the points of contact between the plane and balls $A$ and $B$ respectively. These are the 2 reaction forces and sum to match the weights of all 3 balls.
Now consider the points of contact between balls
At each of the 2 points of contact (between balls $A/C$ and $B/C$) you have 2 equal and opposite normal reactions $N$ which (as you correctly noted) will be directed away from the points of contact towards the centres of the balls.
Using simple trigonometry to match the weight of ball C, which only acts vertically, we see that $N$ satisfies $\frac{W}{2} = N\sin(\theta) $, which rearranges to $N = \frac{W}{2\sin(\theta)} $
Then the horizontal component of the reaction force at each point is simply $N\cos(\theta) = \frac{W\cos(\theta)}{2\sin(\theta)} = \frac{W}{2\tan(\theta)}$
Now consider the friction
Since the system is in equilibrium (i.e. unmoving) we only have to consider static friction, which ranges from $0$ up to a maximum value of $F_{max} = \mu N$.
The horizontal components of the normal reaction at the contact points between balls $A/C$ and $B/C$ are pushing balls $A$ and $B$ outwards, with a force equal to $\frac{W}{2\tan(\theta)}$ as above. To be resisted by static friction at the bases of balls $A$ and $B$, this force must be less than or equal to $F_{max} = \mu N_A = \mu N_B = \frac{3}{2}\mu W$ (assuming the coefficients of friction you supplied were the coefficients of static friction, since coefficients of kinetic friction are irrelevant to a system at rest).
This friction produces a torque on each of the balls $A$ and $B$ equal to $T = r \times F_{max} = \frac{3}{2}\mu Wr$. This requires static friction between $A/C$ and $A/B$ equal to $ F_{max}' = \frac{T}{r} = \frac{3\mu Wr}{2r} = \frac{3}{2}\mu W $ (i.e. the same friction acting at the bases of the balls, which makes sense because $r$ is the same). The static friction formula is again simply $F_{max}' = \mu' N = \frac{\mu' W}{2\sin(\theta)}$ using $N$ from above.
Hence, for the system to remain in equilibrium, we require that
$F_{max} = \frac{3}{2}\mu W \geq \frac{W}{2\tan(\theta)}$
i.e. we need $\mu \geq \frac{1}{3\tan(\theta)}$ (otherwise the plane is too slippery and the balls will slip apart)
and also we need $F_{max}' \geq F_{max} $ or $\frac{\mu' W}{2\sin(\theta)} \geq \frac{3}{2}\mu W$
i.e. we need $\mu' \geq 3\sin(\theta)\mu $ (otherwise the balls are too slippery and $C$ slips down while $A$ and $B$ roll apart).
Illustration via a special case
$\theta$ reaches its minimum possible value when balls $A$ and $B$ actually touch and their radii form an equilateral triangle of side length $2r$ and $\theta$ is 30 degrees. In this case the coefficients of static friction must satisfy
$\mu \geq \frac{1}{3\tan(30)} = \frac{\sqrt{3}}{3} \approx 0.57735 $
and
$\mu' \geq \frac{3\sin(30)}{3\tan(30)} = \cos(30) = \frac{\sqrt{3}}{2} \approx 0.86603$
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What is the state of water at exactly 0°C? Theoretically speaking, what is the state of water at bang on 0°C - not any lower or higher?
Any lower would make it a solid whereas any higher would make it a liquid. But what about bang on 0°C?
Thanks in advance
Edit: I understand that other factors are involved, such as pressure and temperature which would shift the equilibrium, but in a 'theoretical perspective', what would occur - assuming that all of the particles are at exactly the same temperature with the same kinetic energy?
| Temperature defines the amount of energy and vice versa. However, supplying or taking small amounts of energy away from water strictly at zero degrees Celsius, is not going to change the temperature! First, this energy will be used to make transition between liquid and solid form, until you get just one kind of form, either all liquid or all solid. Only after that the temperature will change. In other words, to say that the temperature is zero, is not sufficient to define the phase of water. It could be any! Solid, liquid or a mix.
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Relative motion at almost the speed of light My title might not be precise but I didn't know how to decribe it better.
I've been reading about the special and general theories of relativity for a short while and I started wondering. First, every movement is only relative to some other object/observer, there is no absolute movement/speed, correct? Assuming so, if an object moves at almost the speed of light relative to me, do I also move at almost the speed of light relative to it? If so, am I affected by all the "weird" effects that follow moving at such speeds (increased mass, different time flow, whatnot...)?
From everyday observation I'd have to say not but in that case what defines "moving at almost the speed of light"? Does it have to be relative to something (and what)? If not, is there any absolute measure of speed?
I know I might be talking nonsense but I'm new to these concepts and I'm trying to understand.
| The key point here is the "weird" change in the values you measure depends on what frame you observe them in relative to the object. There is no experiment you can conduct to determine an absolute value for your speed. Speed only makes sense relative to an object. So if a near light speed object relative to us was heading towards us, from its perspective they would measure very different values for the length of objects and time for events to occur etc, and likewise we would measure these values for it as different than in its rest frame.
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Magnetic monopoles and special relativity I was thinking about magnetism as a product of special relativity and the result of this approach to the magnetic monopoles.
So if magnetism is a product of electricity(like electricity from another point of view),then why do we need monopoles to exist?I know that many theories predict the existence magnetic monopoles but i am referring specifically to the concept of classical relativistic magnetism and magnetic monopoles,so do not give me answers that are mainly based on what other theories predict.
EDIT: why do we need electric AND magnetic monopoles to describe electromagnetism if the two are the same thing from another moving frame of reference?And if we do not NEED magnetic monopoles,why is there even a place for them to exist in relativistic electromagnetism?
EDIT: I know that the mathematics of the theory allow for magnetic monopoles, but the essence of the question is the following:
If I work from one frame of reference and change to any other frame of reference, there are no sources of magnetism that can be related to magnetic monopoles?
| Magnetic field appears in Relativity not in one step, but in several.
At first, one looks at two electrons in rest. They are subjects for Coulomb repulsion. The repulsion force is, say $F$ and it would cause acceleration of $a = F/m_e$
Now, one looks at the same two electrons from the moving frame.
According to relativity, they should repulse weaker, because time should flow slower inside the pair.
But according to electrodynamics, electrons are just moving. They have the same $q$ and should repulse at the same rate.
So, who it right? We know, that relativity IS right, then we should FIX electrodynamics and add magnetic field to it.
Moving electrons not only repulse electrically, but also attract magnetically.
Magnetic field -- is a patch to electrodynamics, which conserves relativity.
Since it is natural thing, it is not obliged to work only at the patching site -- it can work elsewhere. For example, it can belong to some particles -- which are monopoles.
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Electric field in a conductor Is it always true that the electric field in a conductor is zero?
What would happen if I put a very big charge inside an ungrounded hollow conducting sphere like this image?
The charges inside the conductor are supposed to rearrange so as to cancel the field created by the big charge.
So in that case, the electrons (I believe only electrons can move freely ..?) will move towards the inner surface (because they're attracted by the big positive charge) and it creates a field in the opposite direction that somewhat cancels the existing field. But what happens if there is still a field when all the electrons are on the inner surface ?
What if it's not enough to cancel the field created by the big charge?
| Then there would be electric field inside that shell. Well if it is still intact. Because you need an immense positive charge to do that to a typical shell, more than 100000 Coulombs (assuming you have 1 mole of conductor with about 50 electrons per atom). To bring that much charge together requires immense force and energy. In fact potential energy of 10 cm sphere with that much charge is more than $10^{20}$ Joules. It take years for a 1GW nuclear power plant to produce that much energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Do quarks violate quantization of charge? Does existence of various kinds of quarks like up, down, strange, charm, top, bottom violate the quantisation of charge or just redefine it as up quark have charge +2/3 and have -1/3. Or do things get even complex for unified theories like the proposed string theory?
| Quarks do not violate quantization of charge, it's simply that $\frac{1}{3}e$ instead of the electron charge $e$ is the smallest unit of electric charge.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Particle Horizon and CMB If particle horizon is the maximum distance we can see presently in the universe, how come we are able to see CMB? CMB is radiation from surface of last scattering happened at t~380k years.
Suppose universe is expanding at a constant rate ( i.e. no acceleration), will we be able to see CMB again??
| The particle horizon is the distance from which light emitted at the moment of the Big Bang will just now be reaching us.
The CMB was emitted 380,000 years after the Big Bang. So the CMB radiation we see has been travelling for less time than the light emitted at the Big Bang, and therefore the CMB radiation has travelled a shorter distance than the particle horizon.
In other words, the CMB horizon is always inside the particle horizon, so we will always be able to see it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Finding $p(z), \rho(z)$ of an ideal classical gas in a box We have a classical ideal gas of particles of mass $m$ at fixed chemical potential $\mu$ and fixed temperature $T$. We have a potential energy $U(z)=mgz$ and the gas is in a rectangular box of height $h$ and base area $A$. How do we calculate quantities like the pressure $p(z)$ and density $\rho(z)$?
I have calculated the grand partition function to be
$$\mathcal{Z}(T,\mu,\mathbf{x})=\exp \left( e^{\beta\mu}\left( \frac{2m\pi}{\beta \hbar^2} \right)^{3/2}A(1-e^{-h}) \right)$$
So the grand potential will be
$$\Phi=-k_BT\left[e^{\beta\mu}\left( \frac{2m\pi}{\beta \hbar^2} \right)^{3/2}A(1-e^{-h}) \right]$$
I thought this may be possible to do from $d\Phi=-SdT-Nd\mu+\mu dN-pdV$ as we have that
$$\frac{\partial \Phi}{\partial V}=-p$$
but this doesnt seem very computable from the given and I have no reason to believe that $p$ will be a function of $z$ alone.
| The density is directly given by
\begin{equation*}
\begin{split}
\rho(z)&=\frac{\int dpAe^{-\beta(p^2/2m+mgz)}}{\int dp\int_0^hdzAe^{-\beta(p^2/2m+mgz)}}\\
&=\frac{e^{-\beta mgz}}{\int_0^hdze^{-\beta mgz}}=\frac{\beta mg}{1-e^{-\beta mgh}}e^{-\beta mgz}
\end{split}
\end{equation*}
since the momentum is homogeneous throughout the system.
Suppose there are N particles in total, then starting from a height $z$ till the top $h$, the total gravitational force would be given by
$$F(z)=\int_z^h dz' A\rho(z')Nmg$$
after which you can obtain the pressure to be $F(z)/A$.
EDIT: normally when we talk about chemical potential, we need to have a "particle bath", but in this case it's more favorable to think of it as a closed equilibrium system and therefore the grand partition approach is not so convenient.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How can I tell if the spectrum of an operator in QM is degenerate? I know that the collection of all the eigenvalues of an operator $\hat{Q}$ is called its point spectrum, and sometimes two or more linearly independent eigenfunctions share the same eigenvalue, and in this case the spectrum is said to be degenerate.
My question is how I can find out if this spectrum is degenerate, for example for $\hat{Q}=i\frac{d^2}{d\phi^2}$?
| As Phoenix87 said, you need to solve the asociated eigenvalue problem. $\lambda$ is an eigenvalue if and only if $\hat{Q}|\Psi>=\lambda|\Psi>$. If you have a matrix expression for the operator $\hat{Q}$, then the usual way to solve the problem is writing the above equation as Phoenix87 did, writing everything in the left side:
$$(\hat{Q}-\lambda\hat{I})|\Psi>=0\iff\det(\hat{Q}-\lambda\hat{I})=0$$
Using the above equation you can calculate the $\lambda$ values, simplifying the determinant and making the result equal 0. The spectrum will be degenerate if you find something like $(\lambda-1)^2(\lambda-2)=0$ because solution $\lambda=1$ appears twice. (And the associated subspace will be 2-dimensional).
If the operator has a differential expresion, then I recommend to write $|\Psi>$ as a function and solve the eigenvalue problem as a differential equation. In the example you've used,
$$\hat{Q}|\Psi>=\lambda|\Psi>\Rightarrow i \frac{d^2\Psi(\phi)}{d\phi^2}=\lambda\Psi(\phi)$$
Remember that the function must be bounded:
$$\int_{-\infty}^{+\infty}{|\Psi(\phi)|^2d\phi<+\infty}$$
Also, you may want to apply boundary conditions. This usually leads to a quantization of $\lambda$, which are the eigenvalues. Once you have the eigenvalues, you have to take a look to the eigenfunctions and check the dimension of the subespaces associated with every $\lambda$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/182978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Rayleigh's Criterion I was just wondering, when two different objects have have light coming from them to our eyes, do the intensities of each diffraction interfere? For example, in the picture below:
Do the intensities interfere with each other?
| Rayleigh's criterion gives a quantitative measure of the angle ($\theta$), being a function of wavelength ($\lambda$) and diameter of the lens ($d$), necessary to distinguish between two light sources. In particular we have,
$$
\theta = 1,220\cdot \frac{\lambda}{d}
$$
The two light sources are non coherent which means they can not interfere with each other. If the light bulbs are to be replaced by two coherent laser sources, they would interfere with each other. Note that the light from light bulbs comes from photons emitted by a wire which can be seen as a statistical process.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there a definite boundary between a powder and a fluid? Given a powder of solid substance what will happen if we make the granules smaller and smaller mechanically? Will this eventually make a liquid or gas from the powder?
Can there be gaseous substanse made not from single molecules but from microscopic granules of a solid?
Also, I wonder whether making a power finer will decrease its angle of repose to zero.
| Why is solid not big gas particles. Where is the boundary between metals and non metals. Their properties? Their form? When does one change into the other. When does something stop and start being an antenna?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 4
} |
Why isn't ice a good electrical conductor? Water can conduct electricity, and some solids can conduct. Why can't ice? Are ice molecules too packed together to let valence shell electrons bounce across each other to create electrical charge? Does ice stop conducting completely at absolute zero?
| Conduction in water is mostly ionic - for pure water you have a very small fraction of ionized molecules (about 2 parts in 10$^{-7}$), so conductivity for pure water is poor. Add a little electrolyte (for example NaCl) and conduction improves. But in an ice crystal, the molecules / ions cannot move, so the main conduction mechanism is disabled. In that case you rely on occasional conduction band electrons - but there aren't many of those around. The band gap is about 7.8 eV source which means that the number of electrons excited into the conduction band at 0C will be extremely low - the fraction given by the Boltzmann factor $e^{-E/kT}=e^{-322}$
Contamination and impurities can bring the band gap down significantly - but pure ice is a good insulator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Expansion of a ket-physical interpretation of coefficients Consider I have a state represented by the Ket:
$$|\psi\rangle=\sum_i a_i |\phi_i\rangle$$
What are the physical interpretations of the coefficients $a_i$? My guess is that $|a_k|^2$ represents the probability that the 'paritcle' is in the state $ |\phi_k\rangle$ is this right? If so does it only hold for orthonormal basis or any basis?
| It's convention to normalize the wave function, but this is purely something done to make calculations easier. States multiplied by a c number(complex number) are completely equivalent.
This is why we choose to normalize states (magnitude 1). It makes the calculations for things like the transitions probability between states easier. If you include your constant your just have to divide by the norm anyways in this kind of calculation, which is equivalent to having a normalized state to begin with.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Does lithium-6 "decay" when hit by a neutron? I am talking about the nuclear reaction
$$
^6\text{Li} + n \rightarrow\ ^4\text{He} +\ ^3\text{H} + 4.78\text{MeV}
$$
A neutron hits a lithium-6 nucleus and together they form an alpha and triton particle. Is it valid to say that the lithium nucleus "decays" when hit by a neutron? Is there any other verb which better describes the change of the lithium nucleus?
I am interested in the correct terminology.
| Decays happen to individual nuclei ( particles). When more than one nucleus(particle) are involved it is called an "interaction". In this case neutron Li scattering
Neutron capture by a nucleus is a possibility, in this case there is an intermediate nucleus formed , which can then decay.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/183391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
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