Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
Converting heat into energy I'm currently building a custom desk. In this desk I will also build-in a small part with a custom electronic panel to provide power to USB devices.
As my gaming pc generates a lot of heat and the panel will generate some too, I was thinking about doing something with this heat. I read another post on this physics part of stackExchange where they say it isn't efficient and you would need a alot of cold air supply.
As I live in Belgium, most time of the year there is enough of cold air, just outside of my window. Another thing is, it doesn't have to be super efficient. If I could just generate enough current to give power to one or more USB connectors or store it in a battery, I would be happy.
Is it possible to create enough energy for my USB, if I got a hot air flow of 50°C and a cold air reservoir of 0-10°C? Follow up question, is it practically possible?
USB works at 5 Volt at a current from 0.5 to 5 Amps.
Other (possible?) options I found was using thermocells and thermocouples.
| Ok, never mind. After checking some videos of homemade thermo electric generators, it's obvious my situation will never provide enough current and voltage for my USB. One created a voltage of arround 0.5V with a flame but with hardly enough current. Maybe I will create one for fun to check if it could light up some LEDS. Thanks!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/90548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why is a vacuum cleaner not as good heater as an electric radiator? I've read this question and answer: How efficient is an electric heater?
, but still don't understand.
If I have an electric radiator it heats the room with 1000 Watts of power. And I feel the room's getting warmer.
In contrast, if I turn on a vacuum cleaner which consumes 1000 Watts as well as the radiator, it doesn't seem to heat the room as well.
Why? Won't all kind of energy transform into heat ultimately?
| A more creative answer involves knowledge of statistical physics:
Using the vacuum cleaner will transfer the system (your room) in a state of less entropy (if the rising temperature is neglected) as the dust is compressed in a smaller volume (just like the mixed state of two gases has more entropy then the state where the gases are separated). This transfer costs energy (that follows from the second law of thermodynamics) and thus the vacuum cleaner needs to use some of the energy for this transition reducing the heating power.
The heater doesn't reduce the entropy in any way.
Therefire the heater is a tiny bit more efficient.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/90612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
At what frequency does the medium of air change from acoustic to light? I'm calculating the max doppler frequency of a fading channel in an in-room environment and looking at different carrier frequencies. Calculated as follows:
F = vf/c
Where
F -> max doppler shift
v -> Velocity of object
f -> carrier freq
c -> speed of waves in medium
At lower frequencies (ultrasound), I use the speed of sound. At higher frequencies (GSM range), I use the speed of light. How do I know when the acoustic medium ends and the light medium begins?
| Acoustic and electromagnetic waves are totally different. There is no overlap whatsoever. You can obviously have very long EM waves, but best example I can think of would be old long wave radio. In my country they still transmit at 225 kHz which is quite easy to achieve with sound wave too, but their nature is different. EM wave propagates in vacuum while the sound wave does not. EM is transverse wave while sound is longitudal wave (mostly).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/90743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Observations in the cathode ray tube experiement
1.One of the observations I learned was that the glass tube begins to glow with a brilliant green light. Many websites I read through refer to a fluorescent material. However, as shown in the above diagram there was no fluorescent material in the experiment carried out first on the cathode ray tube. So where does the green glow come from. Is this the color of the radiation itself?
2."Cathode rays travel in straight lines. That is why, cathode rays cast shadow of any solid object placed in their path. The path cathode rays travel is not affected by the position of the anode." I just can't seem to understand this explanation of the one of the observations.Also, different websites analyses this observation differently. For example, " The cathode rays consist of material particles because they produced shadow of objects placed in the way"
3.Two of the conditions of the experiment were air at very low pressure and secondly a very high potential difference. Could someone please tell me why these conditions were necessary?
I know the questions are very silly but because different websites refer to different things, I am becoming confused with something that should be simple to understand.
| In this experiment discharge glass tube was taken and at both end of glass two metal plate were placed. they were connected with high voltage battery(10000 volt.) the H2 WAS filled at high pressure and at anode end a layer of ZnS WAS placed.
1) At high pressure i.e. 1atm when high voltage current passed then no change seen in the tube. it is because gas is poor conductor of electricity
2) At low pressure i.e. 1torr when high voltage current passed through tube then a greenish glow is seed on ZnS screen. it is because the molecules get ionized and some negatively charge particle travels from cathode plate to anode plate in form of invisible ray called cathode rays>
3) At very low pressure when high voltage current passed thorgh tube then tube became dark. it is because at very low pressure the scattering of light does not take places.
charge/mass ratio of cathode ray is 1.74*10 power 8 c/gram
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/90837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Photons traveling backwards in time? Imagine that two widely separated charged particles $A$ and $B$ exchange a photon.
Because they are far apart one can imagine that there is a major contribution to the photon propagator that travels at the speed of light from $A$ at a time $T_0$ to $B$ at a time $T_1$ where $T_1 > T_0$.
But in that case is there also a major contribution to the photon propagator that travels backwards in time at the speed of light from $B$ at time $T_1$ to $A$ at time $T_0$?
The forwards-in-time photon imparts momentum to particle $B$ whereas the backwards-in-time photon imparts a reaction momentum back to particle $A$.
| Suppose $A$ is at the space-time origin $0$, and $B$ is at space-time event $x$. You suppose that a real photon could go from $A$ to $B$, so this means that $A$ and $B$ are separated by a light-like interval, that is $x^2 = (x^0)^2- \vec x^2=0$. This means that $x^0>0$, too.
Now, the propagator $D_{\mu\nu}(x)$ represents the amplitude for a photonic field perturbation to go from $A$ to $B$ (implicitely you have electronic sources $J(A)$ and $J(B)$)
The (Feynman) propagator may be written (skipping polarizations indices for simplicity):
$D(x) = -i\int \frac{d^3k}{(2\pi)^3 2\omega_k}[\theta(x^0)e^{-i(\omega_k x^0- \vec k.\vec x)}+\theta(-x^0)e^{+i(\omega_k x^0- \vec k.\vec x)}] \tag{1}$
where $\omega_k = |\vec k|$, is a positive value.
Now, with your hypothesis ($x^0>0, x^2=0$), equivalent to $x^0=|\vec x|$, the propagator may be written :
$D(\vec x, |\vec x|) = -i\int \frac{d^3k}{(2\pi)^3 2\omega_k} e^{-i(\omega_k |\vec x|- \vec k.\vec x)} \tag{2}$
However, even with this expression, the propagator is still a field perturbation which "propagates" from $0$ to $x$, and you cannot consider it as a particle. A possibility, in this very special case, would be to consider the propagator as a "kind-of" sum of contributions (with a weight) of pseudo-classical-real-particles, with momentum $|\vec k|$ a and positive energy $\omega_k =|\vec k|$, and , "supposed" going from $0$ to $x$ (it would be "possible" because $x^2=0$). But I don't think this is a good idea, because this pseudo-pattern is no more applicable for $x^2>0$ and $x^2<0$, so it is better to think at the propagator as representing a field perturbation which may take different representations following the sign of $x^0$ and/or the values of $x^2$, and clearly this field perturbation cannot be considered as a particle.
In fact, the term "propagator" is not the best one, one should better think of $D(x)$ as a correlation amplitude between the sources $J(0)$ and $J(x)$.
For instance, a analogy is to think about entanglement, you may have spatially separated sub-systems which could be however correlated.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/90953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Angular Momentum of a rigid, extended object: When we see a rotating object, is the state of rotation is totally relative? Angular momentum of an object is a physical quantity that depends on the chosen point about which to calculate the angular momentum.
It is often said that an object that has been thrown up in the air and is rotating, is physically rotating about the center of mass. I don't think that is what is physically happening. We chose the center of mass as the point of rotation because it is convenient mathematically (it makes the separation of translation and rotational energy easier).
We could choose any other point (inside, outside the object, in motion or at rest relative to the object) and calculate the rotation about that arbitrary point.
It is not a physical fact that the free object rotates about the center of mass.
Even an object that is constrained to rotate about a fixed axis, we could describe the rotation about any point, not necessarily points on the fixed, constrained axis.
|
So when we see an object rotating, its state of rotation is totally relative, as it happens for many other physical quantities...
Is that correct?
The state of rotation is not $totally$ relative; for example, the angular velocity of rotation is the same for all points of reference. It is true that you can use different point of reference for the rotation from the center of mass, and sometimes this is useful, especially if that point moves with constant velocity. But often it is the center of mass that moves with constant velocity, hence it is said
It is often said that an object that has been thrown up in the air and is rotating, is physically rotating about the center of mass.
because in inertial reference frame, only the center of mass can be viewed as non-rotating.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/91181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
What does 99.9% speed of light mean when there is no absolute velocity? So when people say: 'I am approaching the speed of light, and to get to 100% light I would need infinite energy' they are essentially saying that this situation is impossible?
I read this in Hawking's book and confused me because I assume when he says 99.9% speed of light, he means 99.9% speed of light in relation to someone outside observing?
I just cannot understand this notion of needing more and more energy to get closer to light as absolute velocity does not exist? (in that it is a purely relative concept). Surely the ability to accelerate further cannot possible be impeded because speed is all relative, there should be no limit to acceleration? If I 'accelerate' a further 50MPH, will I get to the destination exactly 50 miles early?
From what I can gather you 'can' accelerate FTL (sort off) but instead space bends towards you so you will get to your destination 'ftl' but only due to the curvature in space? So in effect, you can go light years in seconds (lets forget the practicals for a second), but from anyone observing, this will ALWAYS take light years.
Also, if for me I am going 'FTL', does outside observers see me as going light speed, or is it 99.999%.. is there a specific number?
| It means 99.9% of the speed of light in a vacuum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/91233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Quantum eraser double slit experiment In the quantum eraser double slit experiment, does the photon (or wavefunction) pass through one slit or both slits when different polarizers are placed over the slits?
| Quantum mechanics is a theory that can only predict probability distributions. It cannot predict trajectories. It is ruled by differential equations which have as solutions the wavefunctions, and the complex conjugate square of the wavefunction gives the probability of a specific, photon, electron, to be at (x,y,z,t) given the boundary conditions of the problem.
The importance of the boundary conditions is demonstrated clearly in this experiment.
In the experiment you refer to, both the boundary conditions and the initial states are changed during the experiment and it is not surprising that the interference disappears when the boundary condition constrains the slit through which a photon passed. The fact that by further manipulation an interference pattern appears, is again due to the wavefunction's probabilistic waving.
Change of boundary conditions changes the wavefunctions, the boundary conditions are not only "photon scatter from two slits" , but "photon scatter from two slits and all paraphernalia like polarizers upstream and downstream". If one really wanted to solve for the specific quantum mechanical problem all these should be taken into account when picking the wavefunction that describes the probability of the photon scatter.
Single photon double slit experiments for classrooms can be seen here.
It is the photon wavefunction that has a probability for the photon to pass through a slit, and has a sinusoidal dependence that gives interference patterns in the accumulation. The photon, when detected is detected as a point particle interaction with a screen or a ccd or a photomultiplier. One has never seen experimentally the signal of an elementary particle spread in space. Thus it is not the photon that is waving, but its probability of manifesting at (x,y,z,t).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/91317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is the spin 1/2 rotation matrix taken to be counterclockwise? The spin 1/2 rotation matrix around the $z$-axis I worked out to be
$$
e^{i\theta S_z}=\begin{pmatrix}
\exp\frac{i\theta}{2}&0\\
0&\exp\frac{-i\theta}{2}\\
\end{pmatrix}
$$
Is this taken to be anti-clockwise around the $z$-axis?
| For your example, we have $e^{i\theta S_z}\mathbf{S}e^{-i\theta S_z}=\begin{pmatrix}\cos\theta & -\sin\theta&0\\\sin\theta & \cos\theta&0\\0&0&1\end{pmatrix}\mathbf{S}$, with $e^{i\theta S_z}=\begin{pmatrix}e^{i\frac{\theta }{2}} & 0\\ 0 & e^{-i\frac{\theta }{2}}\end{pmatrix}$ and $\mathbf{S}=\begin{pmatrix}S_x\\ S_y\\ S_z\end{pmatrix}$ representing the spin-1/2 operators.
Comments:
In fact, for the most general spin rotation, we have $$U\mathbf{S}U^\dagger=A\mathbf{S}\rightarrow (1)$$, where $U$ represents the general spin rotation operator
$U=e^{i\alpha S_z}e^{i\beta S_y}e^{i\gamma S_z}=\begin{pmatrix}\cos{\frac{\beta }{2}}e^{i\frac{\alpha + \gamma}{2}} & \sin{\frac{\beta }{2}}e^{i\frac{\alpha - \gamma}{2}}\\ -\sin{\frac{\beta }{2}}e^{i\frac{\gamma-\alpha}{2}} & \cos{\frac{\beta }{2}}e^{-i\frac{\alpha + \gamma}{2}}\end{pmatrix}\in SU(2)$, and
$A=\begin{pmatrix}\cos\alpha \cos\beta\cos\gamma-\sin\alpha\sin\gamma& -\sin\alpha \cos\beta\cos\gamma-\cos\alpha\sin\gamma &\sin\beta\cos\gamma\\ \cos\alpha \cos\beta\sin\gamma+\sin\alpha\cos\gamma & -\sin\alpha \cos\beta\sin\gamma+\cos\alpha\cos\gamma&\sin\beta\sin\gamma\\-\cos\alpha\sin\beta&\sin\alpha\sin\beta&\cos\beta\end{pmatrix}$ $\in SO(3)$ with the three Euler angles $\alpha,\beta,\gamma$.
Eq.(1) gives the map from $SU(2)$ to $SO(3)$ and the relation $SO(3)\cong SU(2)/Z_2.$
Remarks:
$e^{i\theta S_x}=\begin{pmatrix}\cos{\frac{\theta }{2}} & i\sin{\frac{\theta }{2}}\\ i\sin{\frac{\theta }{2}} & \cos{\frac{\theta }{2}}
\end{pmatrix},e^{i\theta S_y}=\begin{pmatrix}\cos{\frac{\theta }{2}} & \sin{\frac{\theta }{2}}\\ -\sin{\frac{\theta }{2}} & \cos{\frac{\theta }{2}}\end{pmatrix},e^{i\theta S_z}=\begin{pmatrix}e^{i\frac{\theta }{2}} & 0\\ 0 & e^{-i\frac{\theta }{2}}\end{pmatrix}.$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/91483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Placatory motion of more than two bodies Recently in a documentary I heard that Newton's law of gravitation very well explains motion of two bodies such as sun and earth. And when applied to three bodies the answer is chaotic and not stable.
Then how do physicists explain the motion of higher number of bodies such as the solar system or just the three bodies of sun, earth and moon?
| Here is an example of a (relatively) recent paper to deal with the many-body gravitational problem in regards to the orbit of Mercury. The upshot is that big masses are only slightly perturbed by smaller masses, but small masses can be strongly influenced by the behavior of larger masses. So think of it this way: the details of Jupiter's orbit are pretty messy because of the other planets in the solar system, but it can be approximated very well without worrying about higher-order perturbations (like the pull of Venus - as one example). On the other hand, Mercury's orbit is much messier because the other planets (Venus and Jupiter especially) have a big affect on its orbit.
Poincare famously showed that even in the limit of two bodies undergoing mutual gravitation without any perturbation, a test particle introduced into the system would exhibit (what would later become known as) chaotic motion.
So the answer is this: Newton's Law of Gravitation works fine for three or more bodies, but the system will generally have exponential sensitivity to its initial conditions (i.e. be chaotic). This is strongly evident in the orbit of Mercury, but less obvious in the orbit of, say, Jupiter.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/91533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is it possible to "cook" pasta at room temperature with low enough pressure? It is known fact, that boiling point of water decreases by decreasing of pressure. So there is a pressure at which water boils at room temperature.
Would it be possible to cook e.g. pasta at room temperature in vacuum chamber with low enough pressure?
Or "magic" of cooking pasta is not in boiling and we would be able to cook pasta at 100°C without boiling water (at high pressure)?
| Starches in the durum wheat flour will only activate at boiling temperature. Hydration is irrespective and can be achieved under vacuum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/91685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 7,
"answer_id": 6
} |
Why does a "speed of sound" exist? I've recently read that wind cannot be faster than the speed of sound (german source).
But why is there a speed of sound? I understand (well, mostly accept to be honest) that the speed of light in vacuum is a maximal speed for all matter. And I understand that you need more energy the more you accelerate particles. But why can't you make wind faster?
For example, the sound produced by my voice, by a semi truck and by an explosion all seem very different: it seems amazing that they must move at precisely the same speed. Why is there a single speed of sound in air?
|
I've recently read that wind cannot be faster than the speed of sound
This is false. Wind can go faster than Mach 1. However, the wind present in nature usually doesn't, so the Fujita scale doesn't cross Mach 1.
But why is there a speed of sound?
Sound is, after all, a propagating vibration of air molecules. Molecules do not move around instantaneously; their speed is affected by the density and elasticity of the material — which in turn affect the speed of vibrations and thus sound.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/91747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Realistic calculation of heat loss for pipe Good day everyone,
I am new on this site and I hope to find here help, since I am not going anywhere with the literature I have found.
I try to calculate realistically the heat loss of a hot, uninsulated pipe. Let's say,
it is $170\,\,C$, 1 meter long, $8" (=0,203\,\,m)$ of diameter. I want to calculate the whole losses for the situation of no wind (only normal convection) and $20C$ of outside temperature.
The $170C$ is the temperature at the outside surface of the pipe, the surface is known, I also know the Stefan–Boltzmann constant ($\sigma$) and I take $\alpha=5$ for the convective coefficient of air at the surface of the pipe as well as $\epsilon=0.85$ for the steel pipe emissivity.
My problem is, that in literature, some parameter is always being rejected as negligible due to the fact, that most of the times the example is some kind of exercise of a heat transfer class. I want to calculate the real thing and then decide what is negligible and what is not.
So my main question is: do I just add the radiation losses to the convective losses of the pipe ?
$$Q_{loss} = Q_{conv}+ Q_{rad}$$
$$= (\alpha*Α*\Delta T) + \epsilon*Α*\sigma*((T_{pipe})^4-(T_{air})^4) \tag{with T in K}$$
According the above, I get $Q_{conv} = 479\,\,W$ and $Q_{rad} = 958\,\,W$.
Is there any mistake in my way of thinking or is it truly that simple ?
Thanks in advance.
Marcus
| The natural convective heat transfer coefficient can be estimated through correlations and depends on geometry and its orientation.
There is a calculator which does this for a horizontal pipe here:
https://www.poppyi.com/app_design_form/public_render/free%20convection-%20horizontal%20cylinder
Hope this helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/91792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
General relativity in terms of differential forms Is there a formulation of general relativity in terms of differential forms instead of tensors with indices and sub-indices? If yes, where can I find it and what are the advantages of each method? If not, why is it not possible?
| It was Cartan who developed General Relativity in his book "ON MANIFOLDS WITH AN AFFINE CONNECTION AND THE THEORY OF GENERAL RELATIVITY " relying only on "Affine Connections", it is not clear to me what to be called a "formulation of General relativity in terms of differential forms", but I take it granted from the question that one is trying to develop a theory using index free notation and keeping in mind "Christoffel Symbols" are fundamental building blocks in deriving Field equations. Actually "bundle of linear forms" - what Cartan mentioned, gives rise to a variant of Christoffel symbols (hence in some loose sense FORMS), and torsion of the Geometrical space considered (actually Cartan gives more, he actually predicts "spin" like systems that are absent in Einstein's formulation of GTR as Einstein considered Manifolds with ZERO torsion and torsion is a FORM also, made precise in the above mentioned text), this is also the birth place of modern day "Fiber Bundle" theory ( see his book "Riemannian Geometry in an Orthogonal Frame") this fiber bundle theory has the power to accommodate Yang-Mills theory into solid Geometrical Ground and brings GTR and YM having a common mathematical background, both of this book contains enough material to satisfy the approach of the seeker. Now this theory is called Cartan-Einstein theory, and Einstein's theory is contained in this Cartan-Einstein theory as sub-theory.
One can consult the following letters (Elie Cartan - Albert Einstein Letters on Absolute Parallelism 1929-1932 ) between Einstein and Cartan, to have a taste of how far this theory can be treated as "physical"
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/91867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Is it possible to estimate the speed of wind by the sound emitted by a cable of an overhead power line? I was near ($\approx40m$) an overhead power line and I heard a sound coming from the cables of the power line; I think the sound was made by the vibrations of the power cables due to the wind but I am not sure. The wind was very light.
The sound was not the "buzz" asked about here.
My question is: assuming the sound was generated by the wind, is it possible to estimate the speed of wind from the sound properties (i.e. its spectrogram) and the mechanical properties of the cable?
If yes, how accurate will be the estimate?
If yes, can you provide some back-of-the-envelope calculation?
| In the design of aeolian vibration dampers the frequency of oscillation is given empirically by $$ f = 3.26 V/d $$ where $f$ is in $\rm Hz$, $V$ wind speed in $\rm mph$ and $d$ the cable diameter in $\rm in$. The problem is that beyond $15 \,{\rm mph}$ the wind is too choppy to excite one frequency and the vibration amplitude (and hence sound) drops. Only across flat terrain (sand, river crossing, snow cover) the vibration can be sustained up to about $25 {\rm mph}$.
Thus the vibration frequency can only be used for low speed, and steady winds.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/91978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How can space and time arise from nothing? Lawrence Krauss said this on an Australian Q&A programme.
"...when you apply quantum mechanics to gravity, space itself can arise from nothing as can time..."
Can you elaborate on this please?
It's hard to search for!
| Space and time, as ordinarily understood, came into being after the Big Bang and after the temperature of the universe decreased to the point that the particles (matter) created could not revert back to energy.
When two "permanent" particles were created, space (distance between the particles) was created, and the relative motion between them, created time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/92037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 7,
"answer_id": 5
} |
Does a sound at 50dB at 1m have the same intensity of a sound of 51dB at 10m? Does a sound at 50dB at 1m have the same intensity of a sound of 51dB at 10m, and also the same intensity of a 52dB sound at 100m?
| The scale of sound pressure (decibel) is logarithmic
$$
L_p=20\log_{10}\left(\frac{p}{p_{ref}}\right)
$$
With $p_{ref}$ a reference pressure, with a commonly used value of $20\mu Pa$ according to Wikipedia, because it is roughly the threshold of human hearing.
Due to this definition the intensity/pressure roughly doubles for every 6 dB ($20\log_{10}(2)=6.020599913...$).
A wavefront of sound, when not obstructed, propagates like a sphere. This means that whenever you double your distance from sound producing source, you will hear the sound half a loud.
$$
L_{p2}=L_{p1}+20\log_{10}\left(\frac{r_1}{r_2}\right)
$$
From this you should be able so derive yourself that what you are asking will not be true (assuming that all sounds are measured from the same reference distance).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/92088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What exactly is a bound state and why does it have negative energy? Could you give me an idea of what bound states mean and what is their importance in quantum-mechanics problems with a potential (e.g. a potential described by a delta function)?
Why, when a stable bound state exists, the energies of the related stationary wavefunctions are negative?
I figured it out, mathematically (for instance in the case of a potential described by a Delta function), but what is the physical meaning?
| It means the same thing it means in classical mechanics: if it is energetically forbidden to separate to arbitrarily large distance they are "bound".
The Earth is gravitationally bound to the Sun and the Moon to the Earth. Electrons in a neutral atom are electomagnetically bound to the nucleus. A pea rolling around in the bottom of a bowl is bound.
By contrast the Voyager probes are (barely) unbound and will fly (slowly) off into the galaxy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/92244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 6,
"answer_id": 3
} |
Lax-Pair for principal chiral model This question concerns Eq. (2.10) of the paper https://arxiv.org/abs/hep-th/0305116 by Bena, Polchinski and Roiban.
In section 2.1 they are showing that the infinite number of conserved quantities for the principal chiral model
\begin{equation}
L = \frac{1}{2\alpha_0} \mathrm{Tr}(\partial_\mu g^{-1}\partial_\mu g)
\end{equation}
are given by the fixed-time Wilson lines $U(\infty,t;-\infty,t)$
where
\begin{equation}
U(x;x_0) = \mathrm{P}\, e^{-\int_{\mathcal{C}}a}
\end{equation}
and $a$ is a 1-parameter family of flat connections given by Eq. (2.3).
My question is what becomes of the last two terms (i.e. $-a_0a_1 +a_1a_0$) in the second line of Eq. (2.10). Do they cancel? I don't see why the should because the $a$'s are non-commuting (Lie algebra-valued).
| To simplify, take the notation : $U_y(x)= U(y,t;x,t)$, $U^{-1}_z(x)= U(x,t;z,t)$, $a_i(x) = a_i(x,t)$
Note that you have (on the spatial choosen path $ C = \int dx^1 = \int dx $) :
$\partial_x U^{-1}_z(x)=-a_1(x) U^{-1}_z(x)$, and $\partial_x U_y(x)= U_y(x)a_1(x)$
The minus sign difference can be understood because $\partial_x (U_y U^{-1}_y)=0$
Now, the last line of $2.10$ is :
$a_0(y,t)U(y,t;z,t) − U(y,t;z,t)a_0(z,t)$
With our notations, we have :
$a_0(y) U^{-1}_z(y) - U_y(z)a_0(z) \\=
U_y(y)a_0(y) U^{-1}_z(y)- U_y(z)a_0(z)U^{-1}_z(z)
\\ =[U_y(x)a_0(x) U^{-1}_z(x)]_z^y
\\ = \int_z^y ~dx~\partial_x(U_y(x)a_0(x) U^{-1}_z(x))
\\ = \int_z^y ~dx~(\partial_x U_y(x))a_0(x) U^{-1}_z(x) +
\int_z^y ~dx~U_y(x)(\partial_x a_0(x)) U^{-1}_z(x) + \int_z^y ~dx~ U_y(x)a_0(x) (\partial_x U^{-1}_z(x))
\\ = \int_z^y ~dx~U_y(x) (a_1(x) a_0(x) + a'_0(x) -a_0(x)a_1(x)) U^{-1}_z(x)$
So, this is the second line of $2.10$. There is a global minus sign difference, I think it is because here the integration is going from $z$ to $y$, while the first line of $2.10$ uses the integration from $y$ to $z$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/92316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Collision between a photon and a massive particle Just a small question regarding collisions.
Imagine a head-on collision between a photon and a particle with mass that moves with a non-relativistic speed, the particle was on its ground state, completely absorbs the photon, and moves to its next energy level. Is it always the case that the particle ends up with a non-relativistic speed after the collision?
Something more specific:
To study the properties of isolated atoms with a high degree of precision they must be kept almost at rest for a length of time. A method has recently been developed to do this. It is called “laser cooling” and is illustrated by the problem below. In a vacuum chamber a well collimated beam of Na23 atoms (coming from the evaporation of a sample at 103 K) is illuminated head-on with a high intensity laser beam (fig. 3.1). The frequency of laser is chosen so there will be resonant absorption of a photon by those atoms whose velocity is v0. When the light is absorbed, these atoms are exited to the first energy level, which has a mean value E above the ground state and uncertainty of (gamma). Find the laser frequency needed ensure the resonant absorption of the light by those atoms whose kinetic energy of the atoms inside the region behind the collimator. Also find the reduction in the velocity of these atoms, ∆v1, after the absorption process.
Data
E = 3,36⋅10-19 J
Γ = 7,0⋅10-27 J
c = 3⋅108 ms-1
mp = 1,67⋅10-27 kg
h = 6,62⋅10-34 Js
k = 1,38⋅10-23 JK-1
|
Is it always the case that the particle ends up with a non-relativistic speed after the collision?
No - it depends on the total energy of the particle after the interaction, which (given that the particle is initially non-relativistic) depends on the mass of the particle and the energy of the photon. Compare the energy of the photon to the mass of the particle (in equivalent energy units) and this should help you decide how to proceed with your calculation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/92392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Elementary particles That's what Wikipedia says about Elementary Particle:
In particle physics, an elementary particle or fundamental particle is a particle whose substructure is unknown, thus it is not known to be composed of other particles.
Assumed the above sentence:
Can we ever know the structure of an elementary particle "Z" if it is an agglomeration of other n sub-particles "T" that, toghether, cancel their actractive forces creating our singular Z particle?
An answer could be to collide 2 "Z" particles until they break and show us the "T" particles, but again, the same question can be formulated for the "T" particle. We can demonstrate that a particle is not elementary, but we can't say that it is NOT composed of other particles.
This leads to another concept:
suppose we make and demonstrate an M-Theory called "X" which affirm that "n" different particles can stick making all other known particles with the same known properties , we're not able to demonstrate that there's no other M-Theory "Y" that generalize the "X" one.
Furthermore can even make sense to formulate tens of different "Y" M-Theories which logically generalize the "X" one and that, simulated on a utopian machine, generate the "X" environment.
The core question is, will we ever know when we are done?
| Well it turns out that all particles in the standard model are maybe made of particles called "preons". However this is not a well supported theory. The theory that is trying to answer what particles are really made of is string theory. They say that all particles are made of one dimensional "strings".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/92623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the Schwarzchild radius of a star of solar mass 30 I am currently trying to determine the Schwarzchild radius of a star with solar mass 30. I am calculating it both with respect to solar mass, and w.r.t kilograms, however I am getting conflicting answers. (of a factor of 10)
$$
1 \text{ solar mass} \sim 1.9891 \cdot 10^{31}\,\text{kg}
$$
so I calulated
$$
30\,\text{SM}\sim 5.97 \cdot 10^{32}\,\text{kg}
$$
Using the formula for the Sch Radius:
$$
R_s =\frac{2GM}{c^2}
$$
I determined that you can calculate this using both the solar mass, and the kg mass to confirm.
Using given proportionality constants for $2G/c^2$:
$$
= 2.95\,\text{km/solar mass}\\
= 1.48 \cdot 10^{-27}\,\text{m/kg}
$$
Using the formula above, I have obtained:
$$
\text{using solar mass: }R_s=88.5\,\text{km}\\
\text{using kg: } R_s=883\,\text{km}
$$
If someone could work this out and help me clarify I would be very grateful!
| Your method is correct, but you've got lost in the numbers. This is a good opportunity to use some neat web tools.
Google: 30 solar masses
Answer:
30 solar masses =
5.9673 × $10^{31}$ kg
So you have miscalculated your solar masses in kg.
Secondly, there is a neat WolframAlpha tool:
Given Mass = 30 solar masses
Answer: 88.59 km
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/92990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to get a $\mathcal{N}=2$ SuperYang-Mills Lagrangian from a quiver How can one write down the $\mathcal{N}=2$ SuperYang-Mills Lagrangian given a quiver graph?
For concreteness consider the quiver
$$(2)-(4)-[6]$$
where the node $(2)$ corresponds to a $U(2)$ factor of the gauge group, the $(4)$ node is a $U(4)$ factor and the $[6]$ node is a $SU(6)$ global symmetry (flavour).
The lines correspond to two $\mathcal{N}=2$ hypermultiplets charged in the fundamental representations of one node and the antifundamental representation of the other node into which the line ends.
In particular, how can one read the superpotential from this graph?
How can one read the Kahler potential? Is this latter one always assumed to be canonical?
In case you would like to choose an easier quiver to make an easier example it is perfectly fine.
| A quiver is a neat way of representing the field content of a large class of supersymmetric field theories. One also needs to specify the amount of supersymmetry to understand what a node or an edge stands for.
$\mathcal{N}=2$ supersymmetry implies that that the complete action can be written in terms of a holomorphic function which simultaneously determines both the Kahler and superpotential. The quiver doesn't give you this holomorphic function. With some assumptions on the form of the action in a derivative expansion, you can write out the function.
The simplest examples are those studied by Seiberg and Witten (arXiv:hep-th/9407087 and arXiv:hep-th/9408099) and in my opinion, the best starting point from a pedagogical viewpoint. The first paper discuss the case of a single node i.e., the quiver (2) and the second paper discusses the quiver $(2)-[N_f]$ for $N_f=1,2,3,4$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/93170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Do electromagnetic waves always move in straight lines? When we send an electromagnetic short wave to the sky, it reflects due to the ionosphere effects. But if we send it horizontally, is it correct that it moves around the surface of the earth, and if it has enough energy, it can return to its first position?
If yes, then how could that happen?
| Due to the refraction index dependence on the air density, optical (and radio) rays bend in the atmosphere. It turns out that an atmospheric layer with temperature inversion (temperature increasing with height) can create conditions where the curvature of the ray would match the curvature of the Earth surface. This is called "circulating rays" and these web-pages contain explanations and references on this phenomenon:
http://mintaka.sdsu.edu/GF/explain/simulations/ducting/duct_intro.html,
http://mintaka.sdsu.edu/GF/explain/atmos_refr/bending.html#circulating
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/93243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Basic geometric optics question - how come we don't have to have exact focus to capture objects clearly? The top frame of the image below shows an image formed on the screen (at right) of an object (pencil on the left) located at some distance $D$ from the lens. The lens focuses all the light rays hitting it from the top of the pencil onto the corresponding point on the screen.
The bottom frame shows the same setup, except with the top of the pencil missing. The point on the screen which previously had only light rays from the top of the pencil impinging on it, now has rays from other points to the left of the pencil (for instance, the triangle,square,circle, etc.) hitting it. Those points are all completely different light sources (different frequency, phase, etc.), their combination at the screen will not be coherent or of any single frequency.
This means that a lens focused on an object a distance $D$ from it (to photograph the bottom half of the pencil), should show make a screen image with the bottom half of the pencil sharp, but anything behind its (missing) top half completely blurry.
But then how come in reality, when we take regular photographs, even though we set the focus to capture an object some specific distance away, the objects (not too far) behind it still come out mostly properly (correct colors) and sharply? In other words, how come it is almost never the case that when we focus a camera or our eyes on some object, everything else in front or behind it is blurry?
| This is because usually you are taking pictures of objects that are in much greater distance from the lens than the focal length. In this case the photographed objects are all almost in "infinity" and the rays do not diverge that substantially.
In your drawings here you have the object in the same distance as the focal length, so it is more like a macro photography, where the depth of field is really very small and objects out of focus are very blurry.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/93306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Do centripetal and reactive centrifugal forces cancel each other out? In order for a body to move with uniform velocity in a circular path, there must exist some force towards the centre of curvature of the circular path. This is centripetal force. By Newton's Third Law, there must exist a reactive force that is equal in magnitude and opposite in direction. This is the reactive centrifugal force.
My question is simple, and it is probably the result of lack of common sense but here it goes:
In uniform circular motion, why don't these forces simply cancel each other out? If they did, how would we know they exist in that situation?
When I swing a rock tied to a rope, I feel the centrifugal force, but not the centripetal force. In this situation how can the reactive force be greater than the force itself?
| NO, They do not cancel out each other, while centripetal (center seeking force) is generally provided by some other agency/force, like for revolution of planets it is provided by gravitational force, centrifugal force(outward force) is a pseudo force which is felt in the reference frame of the revolving/rotating body. Clearly since the two forces belong in different frames, they do not cancel out each other in your frame i.e. from the viewers frame they cancel out only in the frame of reference of body as the body does not move in that frame.
When you are rotating a stone/ball tied to a thread you seem to think that you are feeling an outward/centrifugal forcre, but it is actually the tension of the thread, see at the end of the ball tension is directed towards the centre of rotation and is hence centripetal force, but the same tension at the point/centre of rotation is directed towards the ball, therefore you feel an outward force but it is not centrifugal force.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/93599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 2
} |
Will density of a metal increases during forging? This question is metallurgical engineering, but I had a similar doubt regarding density of liquids and what causing it.
Forged parts refines defects, dislocations will be moved strengthening the metal. But will the density of forged metal change?
My earlier question was, what causes liquids to have different densities?
| Forging inevitably compresses the mass and therefore increases the density. In a perfect metal, forging would make little or no difference but no metals are perfect and most contain unwanted substances and voids. During the forging process voids are compressed and it is also possible that the impurities also are diminished in size depending on what they are. The difference in density may well be very small.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/93745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Could quarks be free in higher-dimensional space than 3D? Reading this answer, I now wonder: if quarks are confined by $r^2$ potential, could their potential allow infinite motion in higher-dimensional space?
To understand why I thought this might be possible, see what we have with electrostatic potential: in 3D it is proportional to $r^{-1}$. This is just what Poisson equation tells us for point charge. If we solve Poisson equation in 2D space, we'll see potential is proportional to $\ln\frac r {r_0}$, and in 1D it's proportional to $r$. We can see that it only allows infinite motion starting form 3D.
Could the same hold for quarks, but with some higher than 3D dimension? Or is their potential of completely different nature with respect to space dimensionality?
| If you take the classical analogy of a charge generating field lines then the force at some point can be taken as the density of field lines at that point. In 3D at some distance $r$ the field lines are spread out over a spherical surface of area proportional to $r^2$ so their density and hence force goes as $r^{-2}$ - so far so good.
The trouble with the strong force is that the interactions between gluons cause the field lines to attract each other, so instead of spreading out they group together to form a flux tube or QCD string. In effect all the field lines are compressed into a cylindrical region between the two particles so the field line density, and hence the force, is independant of the separation between the quarks.
This means it doesn't matter what the dimensionality of space is, because the field lines will always organise themselves along the 1D line between the quarks. The quarks woould be confined in any dimension space.
Annoyingly I can't find an authoritative but popular level article on QCD flux tubes, but a Google will find you lots of articles to look through.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/94054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
If a material is built to handle tension, would removing the tension damage it? If an object is designed to cope with large forces such as tension, would removing these forces risk damaging the object?
For example: The neck of a guitar is built to handle the tension of steel strings (~800 Newtons),
if you removed/reduced the tension (removed the strings) for a long period of time would this risk damaging the guitar neck?
| This answer is specifically about guitars because I have guitar building and repair experience.
The strings and the truss rod are under tension so the neck itself is mainly under compression. There is some tension on the back side of the neck due to neck relief (forward bow of neck) but not much. Necks are made from wood from the trunk, a material 'built' to handle the compressive stress from the weight of the tree.
The safety factor (failure load/service load)of guitar necks are very large.
A very basic calc. to demonstrate this:
Compressive yield strength parallel to grain for the most common guitar neck wood (maple) = 21.5 M Pa
Approximate Cross sectional area = 0.0007 m ^2
F = PA = (21.5 M Pa) * (0.0007 m^2) = 15 Kilo Newtons.
SF = (failure load/service load) = (15 kN)/(800 N) = 18.75
Basically, guitar necks are super strong and only fail due to impact (see Gibsons broken headstock syndrome) or warp due to poorly seasoned wood or extreme humidity or temp. changes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/94222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Potential energy curve for intermolecular distance
(source: a-levelphysicstutor.com)
I'm trying to understand this curve better, but I can't quite figure out what "negative potential energy" means.
The graph should describe a molecule oscillating between $A$ and $B$, however where I'm stuck in reasoning this is that the PE is equal in $A$ and $B$, but then why does this mean $r$ will increase in $A$ (repel) and decrease in $B$?
| It seems you are equating value of potential energy and repulsion/attraction. Identical values of potential energy do not mean identical behavior of the particle/system in question. Recall that force $F_r=-dU/dr$. If you want to find out whether there is attraction ($F_r<0$) or repulsion ($F_r>0$), you should look at the negative derivative of potential energy, not the value of potential energy.
You also claimed that you were unsure about "negative potential energy." In classical mechanics, there isn't really in physical meaning to it, since only changes in potential energy matter. For example, it's the derivative that determines force, not the value. As another example, recall $U_\text{grav}=mgy$, where one can set $y=0$ anywhere; here, again, the particular value doesn't matter.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/94281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
During reflection does the emitted photon have same properties? When light (photon) is reflected the the original photon is absorbed by an electron and then emitted again. Does this "new" photon have the same wavelength, frequency etc. as the original?
|
During reflection does the emitted photon have same properties?
During reflection the color does not change and the phases do not changes otherwise the reflected images would be fuzzy.
I will complete the other answers by examining the mirror material.
Not all materials reflect. Reflecting materials are materials where the microscipic domains in the solid are coherent forming crystal structures. A highly polished metal mirror, for example reflects classically because of the very high reflection coefficient. That is why all mirrors are glass backed be some type of metal in highly reflective mode. (Glass itself, as water too, have a transmission coefficient which reduces the reflection coefficient to a very small number except at certain angles).
Photons are a different framework, a quantum mechanical framework . In quantum mechanics we have solutions of equations with boundary conditions for each given problem. The problem of "a photon hits a highly reflective metal surface and is reflected" has a very specific solution with the given boundary conditions. As you know quantum mechanics has specific energy states that ensure that there is no dissipation of energy at that energy state. Thus, if the photon is reflected and its energy(color) is still the same it means that a total state ( photon/ metal crystal) gave the solution of an elastic scatter where no energy was lost ( within the heisenberg uncertainty principle). Nature is very good at analogue solving of problems :). Good reflectors have many states that can accommodate a solution for reflection,( metals mainly because of the bands of electrons not attached to individual molecules and atoms).
So both classically and quantum mechanically there is consistency: the photon does not lose its energy in interacting with the whole crystal/domain structure of the reflector.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/94359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Angular speed of the planets Do all the planets in our solar system have the same angular speed? Physics teacher says yes, my research is not crystal clear. I want to make sure I have the right information for future reference.
| I would guess there is something missing in this question. Are you sure you have the teacher's statement correct? If the planets all had the the same angular speed (about the Sun) they would all complete an orbit in the same amount of time.
If it is angular speed about their axis, again clearly not true. Mercury rotates once in a Mercury year and a Mars day (what they call a Sol in the Mars mission groups - I wonder what they will call it on other planets?) is a little longer than an Earth day.
Plus the planets are in elliptical orbits and their angular velocity changes with their distance from the sun - the equal times, equal area rule of Kepler. In fact, Kepler's three laws clearly state the relation between the time to orbit for bodies at two different distances from the Sun.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/94597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Gravitational collapse and free fall time (spherical, pressure-free)
A very large number of small particles forms a spherical cloud. Initially they are at rest, have uniform mass density per unit volume $\rho_0$, and occupy a region of radius $r_0$. The cloud collapses due to gravitation; the particles do not
interact with each other in any other way.
How much time passes until the cloud collapses fully?
(This was originally from a multiple-choice exam - I solved the problem via dimensional analysis on the options then. I'm wondering how it might be solved directly now).
The answer is $$t = \sqrt{\frac{3\pi}{32G\rho_0}}. $$
| A partial answer only.
Given a spherical collapse, and ignoring relativistic effects, the time is the same as the time taken for a particle at the edge of the cloud to fall to the centre.
As all the mass is inside the edge, we can determine the mass pulling that edge in as the volume of the sphere of radius $$r_0$$ times the density.
I.e., the key idea is that the cloud is not important, just the total mass.
So the falling particle behaves as it if is falling to a point mass.
But this is just a special kind of orbit, very ellipsoidal. The period of the orbit:
http://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body
depends on the cube of the radius.
$$t = 2\pi\sqrt{\frac{a^3}{GM}}$$
As Wikipedia shows, this actually means that the period is independent of the radius:
$$t = \sqrt{\frac{3\pi}{G\rho_0}}$$
http://en.wikipedia.org/wiki/Orbital_period#Orbital_period_as_a_function_of_central_body.27s_density
The in-fall time is only the same as a quarter of an orbit. So the total time should be:
$$t = \sqrt{\frac{3\pi}{16G\rho_0}}$$
So I'm missing a factor of $$\sqrt{\frac{1}{2}}$$ here - hence the 'partial answer'.
What have I missed?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/94746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
What is the center of Earth's landmass? I have read, several times, that the Great Pyramid is located at "the center of Earth's landmass".
How do we define "Earth's landmass"? And once defined, how do we find the center of it?
| This Wikipedia article has a quick discussion and finds the center to be near Ankara, Turkey. I would think a better point would be well inside the earth, the CM of the thin shell discussed in the article. Finding an average point on the surface of a sphere(oid) is problematic. It depends on your definition.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/94863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
White light diffraction I have a hard time understanding why light waves of different wavelengths diffract in a different manner. According to Huygens' principle, every point on the wavefront is a source of a secondary wave. So if we have a white light going through, say, a single slit (light rays parallel to each other and perpendicular to slit's plane), all what's supposed to happen is a plain diffraction, just like of any other wave. That is, the wave will progress spherically, but it will still be a white light. Why instead we get a splitting of different wavelengths? In other words, how does light color affect diffraction geometrically?
| Diffraction effects depend on the wavelength of the light. Considering a single narrow slit with monochromatic light, light with wavelengths much larger than the slit will not be transmitted and light with wavelengths much shorter than the slit will be transmitted without significant diffraction effects, but light with wavelengths comparable to the slit will show significant diffraction effects.
The reason that diffraction effects are able to split white light into its different colors is because white light is composed of an incoherent combination of many different wavelengths of light. The different wavelengths get diffracted by different amounts, and the effect you see is that the white light gets split into its spectrum of colors. Additionally, since the light is incoherent, you don't see dark and bright spots like you would with monochromatic light.
How do we understand from Huygen's principle that light with wavelengths much shorter than the slit do not diffract very much? This is because points near the middle of the slit and points near the edges of the slit, which are both emitting spherical waves will interfere destructively except for in the direction straight ahead.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/94967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Measuring background radiation We tried to measure background radiation using a geiger counter for a experiment at school. The meter showed $0.12$-$0.21$ microSv/h during the day averaging at about $0.14$ mcSv/h.
As we tried to see ways how to shield incoming radiotion nothing seemed to work. Taking a cue from nuclear power plant we put the meter into a watertigt glasscontainer and sank it in the gym pool so that the container had a good three feet of water all around. The measurement did not budge at all. We tried all settings fast/slow/auto. The geiger meter should be okay as it had recently been calibrated.
Can someone explain.
| If you are not living close to a uranium mine or other natural sources of radioactivity the main radioactivity that your geiger counter measures outside is cosmic radiation. At sea level this is composed mainly by muons , which are weakly interacting and will not be stopped by a meter of water. This is also true for cosmic origin neutrons . ( neutrons from reactors are of lower energy).
Electrons and gammas will be stopped but possibly your counter is not good for measuring electrons which can be stopped by a metal cover.
You could try measuring in a second basement, low energy muons will be absorbed by the ground above. On the other hand there is natural radioactivity in the ground and the cement of the basement. Still you might see a controllable difference. Have a look at this list of natural radiation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/95129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Can we calculate the frame dragging force of the Earth? Although clearly this force would be significantly greater with a rotating black hole, is it still possible to calculate this drag for say a satellite orbiting the Earth?
| This is really an add-on to the excellent answer by Pulsar. Sarah Jayne never got her units worked out fully because she had the wrong "g" and used "r" in km instead of meters. I solve the equation that she has posted using the universal constant G = 6.67e-11 m3/kgs2 as:
radian/second = 6.67e-11*8.02e37*7.29e-5/(2*3e8*3e8*7013000^3) = 6.27e-15 rad/s
Then Pulsar showed the conversion to mas/yr
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/95361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Centre of instantaneous rotation problem Is there a point of Centre of Instantaneous Rotation (CIR) for every type of motion or only for cases of rolling?
| I assume you are talking about a rigid body in motion in a plane.
Consider any two different points on the body, A and B.
At any point in time, each one has a velocity vector $\vec{v_A}$ and $\vec{v_B}$ (assuming neither one is, itself, the center).
Consider the line normal to $\vec{v_A}$, call it $n_A$, and likewise $n_B$.
Where these two lines intersect is the instantaneous center.
If the two lines are parallel, the motion is pure translation.
If you want to extend it to 3 dimensions, $n_A$ and $n_B$ are planes normal to $\vec{v_A}$ and $\vec{v_B}$.
Where they intersect is a line, an "axle" if you like.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/95486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Is the frequency/wavelength of light modified when multiple light sources are combined? Let's say I light a wall with two spotlights: One red and one green one. Where they overlap, I'll see a yellow area at the wall.
My question is, whether this is caused by an modification of the frequency/wavelength or simply by my eye combining the two incoming lights.
Light is "added", wavelength is modified:
The eye combines two separate lights:
| The bottom picture is correct. Your eye is only made up of red, blue, and green cones. This is what your eye uses to see all colors. When a yellow light enters your eye, the red and green cones activate, and you perceive yellow. This is the same as if red and green light enters your eye. Your mind perceives this as yellow. The wavelength of both red and green are not combined into yellow before they enter your eye. They simply cause your eye to react the same way as if it was seeing yellow light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/95559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Can a liquid insulator be electrically charged by touching a charged conductor?
Can a liquid insulator be electrically charged by touching a charged
conductor?
I understand that solid insulator will only be charge on the surface where it is touch, but the case is different from liquid which it circulates, so I imagine that slowly liquid insulator will be fully charge due to circulation.
| It is certainly possible to charge insulators. I have personally charged ping pong balls and rubber balloons, and a certain Robert A. Millikan earned himself a Nobel prize by charging oil droplets.
But to get to the point of your question, you are correct that a liquid object (like Millikan's oil drops) will disperse the charge over its surface because it can flow. Even if the liquid surface is a perfect insulator the charge will eventually disperse over the surface.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/95634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is $U(\Lambda)^{-1} = U(\Lambda^{-1})$ for a unitary representation? This is from the beginning of Srednicki's QFT textbook, where he writes (approximately):
In QM we associate a unitary operator $U(\Lambda)$ to each proper orthochronous Lorentz transformation $\Lambda$. These operators must obey the composition rule
$$U(\Lambda'\Lambda) = U(\Lambda')U(\Lambda).$$
So far OK.
But where does he get the following from?
$$U(\Lambda)^{-1} = U(\Lambda^{-1})$$
| I'd like to add a something to V. Moretti's (correct) answer. You may be wondering
Where does the property $U(I) = I$ come from?
This property and the original one you asked about, generally hold for any group representation and, in fact, for any group homomorphism.
Let $\mathrm{SO}(1,3)^+$ denote the proper, orthochronous Lorentz group, and let $U(\mathcal H)$ denote the group of unitary operators on the Hilbert space $\mathcal H$ of the theory. The representation $U:\mathrm{SO}(1,3)^+\to U(\mathcal H)$ about which you are asking, is a group representation which, in particular, is a group homomorphism. A group homomorphism from a group $G$ to a group $H$ is a mapping $\phi:G\to H$ such that
\begin{align}
\phi(g_1g_2) = \phi(g_1)\phi(g_2)
\end{align}
for all $g_1, g_2\in G$. This property implies some interesting things. Let $I_G$ be the identity of $G$ and $I_H$ be the identity of $H$. Then notice, for example, that
\begin{align}
\phi(I_G) =\phi(I_GI_G) = \phi(I_G)\phi(I_G) ,
\end{align}
and multiplying by $\phi(I_G)^{-1}$ on both sides gives
\begin{align}
\phi(I_G) = I_H.
\end{align}
In other words
Group homomorphisms map the identity to the identity.
V. Moretti's answer can then be adopted in general to show that they also have the property $\phi(g^{-1}) = \phi(g)^{-1}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/95779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How do photons 'connect' during wireless connection? So wireless router broadcasts a signal and then your device searches. So what actually happens when the photons 'meet' it's kind of like saying, 'ah your one of us, so we will follow you, show us the way'
It's so bizarre, how do photons connect during wireless connection?
| I'm not 100% sure, but I suppose photons are identified by its frequency.
WiFi routers have usually 12 or 14 channels, whose frequency depends on the local laws for telecommunications. They are supposed to be the only photons in each channel.
Any photon of a different frequency just will not be absorbed by the antenna.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/95949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If one is travelling at a significant fraction of $c$, will the length of the trip be shortened? Assuming two stars are 1 light year apart and a traveler is travelling at 0.75 of $c$, from the point of view of the traveler what would be the observed time en route? Also, if a vehicle is constantly accelerating, will it reach 0.75 of $c$ within a reasonable amount of time? What would the Lorenz transforms look like for these situations?
Be gentle, I'm not a physicist. I'm writing book, and I'm trying to devise a scenario whereby a vehicle with a practically unlimited amount of energy can travel between stars. I don't care about the actual travel time only that experienced by the travelers.
| Suppose to have an observer at rest with respect to the star and assume that their distance remains constant (1 light year). A spaceship traveling at $0.75 \, c$ with respect to this observer will cover 1 light year in $1.33$ years (16 months). However, for an observer inside the spaceship the length is contracted by a factor
$$
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = 1.51
$$
and so the distance of the two stars for this observer is
$$
d = \frac{1 \;light \; year}{\gamma} = 0.66 \; light \; years.
$$
The spaceship travels at $0.75 \, c$, so the travel, from the point of view of the observer inside the spaceship, will take
$$
t = \frac{0.66 \; light \; years}{0.75 \, c} = 0.88 \; years \approx 10 \; months.
$$
Of course, going faster than $0.75 \, c$ will reduce as much as you want the time felt by the travelling observer to go from one star to another.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/96006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Wavefunction of a Baryon How to write the total wavefunction of a Baryon including space part, spin part, isospin part and color part such that the net wavefunction is antisymmetric? What is the difference in wavefunctions of two different baryons but of same quark content say proton $p$ and $\Delta^+$ baryon?
| You can think of $\Delta^+$ as just the energized state of $p$, due to the spin configuration of the three quarks, and therefore different total spin ($3/2$ vs. $1/2$). This is much the same as the hydrogen atom, where different angular momentum states lead to different energy states.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/96046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Energy of an inductor I know that for an inductor having self inductance $L$ energy stored in its steady state when a current $I$ has been established is given by $U = \frac{LI^2}{2}$.
But after this current has been established, if we suddenly cut the wires attaching the inductor to the potential source or short the circuit, what happens to the energy ?
It must not be stored anymore as $\frac{LI^2}{2}$ as there can be no $I$, could not have decayed as heat because we cut off the wires and did not have any circuit which may have allowed for reverse flow of current.
I have one thought that it might have gone as EM radiation but I am not sure.
| The place where you cut the wire acts as a temporary Capacitor where a huge potential difference is formed. This potential difference causes an intense electric field to develop, which is where the energy is initially stored. If the potential difference developed exceeds the dielectric breakdown voltage of the intervening medium, the charges are lost as a spark discharge which dissipates energy as EM waves and heat.
But usually the current is never cut down this abruptly, providing enough time for the energy to dissipate as the normal safe resistive heating. If not, then the energy will be lost by the aforementioned discharge which is very intense and might damage the equipment under consideration. Hence the use of a parallel capacitor with a large inductor, which allows slow dissipation of energy as LC oscillations (EM waves) and normal resistive heating.
EDIT
The said capacitance ceases to exist only if a spark discharge dissipates the gathered charge or, the instantaneous back emf is slowly reduced by resistive heating (the circuit is not cut-off). (i.e. if we assume the cessation of current occurred instantaneously, the developed field would exceed the breakdown field leading to a spark, or if we assume that the change is slow enough so that no spark is developed, then the finite time it takes for the current to die down, the resistances of the circuit would dissipate the energy in that time). The sudden stopping of the current is only an ideal occurrence and does not occur in practice.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/96240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Phonon-phonon interaction I have been told that phonon-phonon Interaction is an anharmonic effect so only arises if terms of third and higher order in the displacement of the ions the Hamiltonian for the nuclii is taken into account.
So how do I see the there is no phonon-phonon interaction in the harmonic approximation?
| When you solve for such Hamiltonian for the harmonic oscillator you get a set of eigenstates which by definition are orthogonal and thus you have phonons that don't interact.
When you include a non harmonic term to the Hamiltonian and treating it as a perturbation you get new eigenstates that are a mixture of those the simple harmonic oscillator. This mixture is what it's called phonon-phonon interaction.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/96388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why does a picture of a person seem to be looking in the same direction irrespective of the angle of observation? If you observe a picture of a person hanging on a wall who seems to be looking directly towards you always seems to be looking at you even though you change your angle of observation to the extremes.
The same can be observed in a television. If a television is watched by many people from different angles all observe that a person on the screen is looking at them.
Why does it happen like that?
Update:
I First thought that it may be because of some data being lost due to conversion of 3D to 2D. But same is observed in a theater while watching a 3D movie.
| A picture has only two dimensions. Ultimately all depth created through perspective and light and shadow is a trick, similarly eyes in a painting following you is also an illusion. The light, shadow and perspective depicted in a painting are fixed, meaning they don't shift. in a painting also it don't change, they look pretty much the same no matter from what angle you look at it. So if a person is painted to look at you, he or she will continue to look as you move. If a person is painted looking away from you, the light, shadow and perspective shouldn't allow him or her to ever look at you, even if you move yourself to the point where the person has been painted looking toward.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/97694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Which electron gets which energy level? Electrons sit in different energy levels of an atom, the farther the higher energy is. Every electrons have the same structure, they can gain energy from environment, electrons which gained energy could jump to a higher energy level and will finally fall back again.
I'm wondering why some electrons have the "right" to "store" that high energy since every electron is the same. Why do those electrons can have more energy and sit in higher energy level than other electrons?
| Pauli's exclusion Principle requires no two electrons to occupy the same quantum state. Based on spin, it is decided which electron 'sits' where it does. As far as the 'jumping' to the higher energy is concerned, it depends on the way the electron gains energy. If say, light of energy which matched the energy difference between two energy level is incident, then the electrons 'jump' to that energy level.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/97908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Branching lemon drop "smoke rings" This might be hard to ask, but here goes nothing.
I recently poured a cup of water into a black coffee cup. There was a light source--not very bright--above the cup. Anyways, I was squeezing a lemon into my water mindlessly, and to make sure I got every last drop of lemon juice into the water, I watched the lemon juice hit the water. Then I looked closer as something pretty neat was happening.
When a single lemon drop hit the water, it dissipated into a shape that many would describe as a "smoke ring." Knowing some physics/fluids, I understood what was happening here was nothing out of the ordinary. But then I kept watching. As the ring dissipated, it eventually broke off into about 5 other smaller "smoke rings." I couldn't see further down, so who knows if it continued; but I would assume that the 5 smaller rings would turn into 5 + X amount more, etc.
At what point would they stop breaking up into smaller rings? Whats causing this to happen? Does this movement/shape have a scientific term? Does anything else do this that can be easily seen?
I'm more interested in what the movement/pattern that is happening here, not the chemistry aspect.
| The smokering is the epitomy of perfect turbulence as is the 'smokering' described in this question. The fluid through which the smokering is passing flows laminarly against the circumference of the rotating ring as the smoke ring travels. The energy imparted to the molecules of the ring is stored in the circular rotation of the ring. A drop falling into the water is molecularly identical to a huff of air forming a smoke ring.
Part dieux. Surface tension will eventually 'gather' the molecules of the ring into 'pieces' and each piece will be 'required' to preserve the rotational energy of the original ring. This results in dAughters of the mother ring with the same rotatioal characteristics as the mother.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/97967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
} |
Propagator of Maxwell-Chern-Simons theory I need to compute the "topologically massive photon" propagator.
I've started with:
$$
\mathcal{L}=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu} + \frac{\mu}{4}\epsilon^{\mu\nu\lambda}A_\mu\partial_\nu A_\lambda
$$
$$
=A_\mu\underbrace{[\frac{1}{2}g^{\mu\lambda}\partial^2+\frac{\mu}{2}\epsilon^{\mu\nu\lambda}\partial_\nu]}_{(\Delta^{-1})^{\mu\lambda}}A_{\lambda}
$$
So how can I invert The under braced part which will yield the topologically massive photon propagator?
| The following is a rough calculation. If the operator,
$$\triangle^{\mu\lambda} = \eta^{\mu \lambda} \partial^2 + \mu \epsilon^{\mu\nu\lambda}\partial_{\nu},$$
is the one you wish to invert, then we must solve the differential equation
$$ \triangle^{\mu\lambda} G = \left[\eta^{\mu \lambda} \partial^2 + \mu \epsilon^{\mu\nu\lambda}\partial_{\nu} \right] G = -i\delta^{(4)}(x-y)$$
where $(-i)$ is by convention. I think the momentum space equivalent is
$$\left[ \eta^{\mu \lambda}p^2 + i\mu \epsilon^{\mu\nu\lambda}p_{\nu}\right]\hat{G} = -i\mathrm{e}^{-ip\cdot y}.$$
To obtain $\hat{G}$ in position/physical space, you must perform an inverse fourier transform. Let me know what you get, I'm curious!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/98040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Admixtures of longitudinal and timelike photons! In the quantization of electromagnetic field the physical states $|\psi\rangle$ are found to obey the following relation:
$[a^{(0)}(k)-a^{(3)}(k)]|\psi\rangle=0$
It is explained as the physical states are admixtures of longitudinal and timelike photons. What do longitudinal and timelike photons physically mean? Why the polarizations, $\epsilon^{(0)}$ and $\epsilon^{(3)}$, timelike and longitudinal photons, are called unphysical?
| It seems to me that longitudinal photons are not unphysical. They are responsible for the Coulomb interaction between charged particles.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/98173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Frequency of rotating coil Given a coil initially in the x-y plane, rotating at angular frequency $ \omega $ about the x-axis in a magnetic field in the z-direction. This uniform time varying magnetic field is given by $B_z (t)=B(0)cos(\omega t) $ I am required to show that there is a voltage of frequency $2\omega $ across the loop. Clearly when t=0 the flux is at a maximum, but I dont understand how to relate to the frequency?
If the frequency is just the inverse of the period then $f=\omega / 2\pi $ ?
Clearly I am not understanding something. How does the voltage affect the frequency?
| The angle $\theta$ between the normal($\hat n$) to the surface of the coil is given by $\theta=\omega t$ at any instant $t$. Also The magnetic field $B_n$ in the direction of $\hat n$ is given by $$B_n=B_z(t) \cos\omega t$$. calculation of flux through the coil of area $A$ is easy.
$$\Phi = B_n(t)A$$
$$\Phi = B_z(t)cos(wt)A$$
It is given that $B_z(t)=B(0)cos(wt)$
so $$\Phi = B(0)A cos^2(wt)$$
now voltage induced is given by $-\dfrac{d\Phi}{dt}$
$\dfrac{d\Phi}{dt}= \omega B(0)A(-\sin2\omega t)$
So induced emf is $e=\omega B(0)A\sin2\omega t$
Let $\omega B(0)A=E_0$
then $e=E_0\sin(2\omega t)$
We can say that output voltage varies with twice the frequency as that of the input magnetic field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/98316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What makes a rainbow happen? A rainbow is formed when a raindrop refracts light, but why then does the whole sky not become a huge rainbow when it rains? Would the light not be dispersed into ordinary white light? What causes it to look as if each end is nearly touching the ground?
| The best is probably to give you an insightful link. You will find there an applet which illustrates the following, which is your intuition:
*
*Coming from the sun, light rays hit the droplet and enter it with refraction air to water
*They reflect internally in the droplet
*They come to your eye, following a second refraction water to air
Since rays from the sun have all possible values of andle of incidence between 0 and $\pi$ when hitting the water sphere which is a droplet, you could expect the eye to see refracted rays coming from all directions as well, thus forming a coloured disk for all colours, hence a "sky color" disk, i.e. nothing special at all.
The trick is that when a combination of refractions and reflection like the one just described happens, then the reflection and refraction coefficients come into play. You may find the details of Snell's law and Fresnel's formula here, but the crux of the matter is that such coefficients are dependent on both angle of incidence and wavelength. As a consequence the final intensity which gets to your eye is a highly dependent function of the angle of incidence of the ray of light (and a function of wavelength).
It so happens that this function displays a sharp maximum for a certain value of angle of incidence (which itself depends on the wavelength of the light). Therefore, for each colour you do see a disk, but this disk is much brighter at the edges of the disk, where such maximum occurs. Hence, you see an annulus instead of a disk. Moreover, the diameter of each disk is different because it is dependent on the wavelength, therefore your eyes perveives a set of annuli of different colours, next to another, i.e. a rainbow.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/98375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Is using a swing an example of normal or of parametric resonance? Parametric resonance is a situation where the driving frequency is a multiple of the eigenfrequency. Various people say that using a swing and propelling it oneself is such a case, with the driving frequency being the double of the eigenfrequency. But when I use a swing, as I did today, my own motion has the same frequency as that of the swing, not twice the frequency.
An example description is http://www.hk-phy.org/articles/swing/swing_e.html which claims that self-propelled swinging is normal resonance.
The opposite is said here: Wikipedia entry
What is the correct view: is self-propelled swinging normal resonance or is it parametric resonance?
| In the case of a father pushing a child, it's normal resonance.
In the case of a child driving the motion itself it's parametric resonance. The Wikipedia article you mentioned states: For example, a well known parametric oscillator is a child pumping a swing by periodically standing and squatting to increase the size of the swing's oscillations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/98448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Why can you make two repelling positively charged rods touch? Shouldn't the Coulomb force become infinite? For a physics lab on the Triboelectric effect, we rubbed two rods with fur which gave both of them a positive charge.
We then brought them close together, and they obviously repelled. We then held one rod down firmly and touched the rods together. We were able to do so.
Why?
Coulomb's Law says that as the distance between two like charges become zero, the force becomes infinite. Using superposition on all the charges in the rod, shouldn't the infinite force have prevented us from touching the rods together?
| The reason is that Coulomb's law is only directly valid for point charges, i.e. for charges, sizes of which are much smaller than distance between them. For particular symmetry reasons it appears to also be applicable to spherically symmetric balls of charge - but note that the distance you should take is not between the surfaces - it's between the centers of the balls. Thus, even if you make the balls touch, distance between their centers will still be twice the radius, so the Coulomb force will be finite.
Your objects are not even balls. The actual Coulomb force is a bit trickier to compute, and at small distances it will grow not that fast as for spherical charges as you place them closer. Although this does make it a bit easier for you to make the rods touch, their finite size in all directions is the core reason for the force to remain finite, similarly to the case of balls.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/98541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
When to use the Boltzmann distribution and the chemical potential? How do you know when to use the Boltzmann distribution for a particular problem?
I have many polymers connected together in many different possibilities by connector agents. All are in a solvent. I wrote the partition function of the system and used the Boltzmann distribution, with the chemical potential. However, some problems do not require it. How do you identify the situations that do require the chemical potential?
| The chemical potential is needed if you have an open system, that is able to exchange matter with another system. For instance, when water condensates, the liquid phase can exchange matter with the gaseous phase.
In polymer solutions, one usually needs to take the chemical potential into account because the monomers can leave the polymer and create a monomer phase. Moreover, in your case you have the connecter agents that can attach and detach. If you have very few of them, they will prefer to remain free (to gain entropy), if you have a lot of them, they will be connected in great majority (to minimize energy). This shows that their action depends on their concentration, therefore you will need the chemical potential of the connecter agents in solution to solve your problem.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/98599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Is crystal momentum an operator? My teacher has for Bloch waves the notation $\langle \vec{r}|\vec{k} \rangle = e^{i\vec{k}\cdot \vec{r}}u_{\vec{k}}(r)$ and uses it consistently. However, does this not assume that there is an operator that has eigenstates $|\vec{k} \rangle$? If so, how would such an operator be defined?
| I was confused when I wrote that question... The answer, trivially, is that $|\vec{k} \rangle$ is just a state, not necessarily an eigenstate of any operator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/98927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Can zinc nitride be used as light emitting material? I cannot manage to find any journal papers about the applicability of zinc nitride as active layer of an light emitting diode (LED). But certain papers got mention that zinc nitride with a direct bandgap can be fabricated with potential applications in optoelectronics.
Can anyone give me some opinions on this subject?
| In principle yes. But to make a good $\text{Zn}_3\text{N}_2$ homojunction LED you need the capability to incorporating both p-type and n-type dopants (normally oxide materials are naturally n-type) which might not be possible.
From what I have read, this material has been proposed as a way of making p-type ZnO (which is naturally n-type) by a post growth annealing step. This means that via $\text{Zn}_3\text{N}_2$ you maybe able to make a p-ZnO/n-ZnO homojunction LED.
More info here, http://pubs.rsc.org/en/content/articlehtml/2013/ra/c3ra46558f.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/99122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Definition of quantum anharmonicity I have been reading research papers in mathematical physics for some months now, and I've seen the the term "anharmonic oscillator" quite frequently. At first I assumed that given a Schrodinger equation
$$\frac{d^2u}{dx^2}+(E-V(x))u=0$$
where $E$ is the energy, and $V(x)$ is the potential function. If $V(x) = x^2 +$ higher order polynomial terms, then this gives rise to the anharmonic oscillator since the higher order terms ensure that the potential will deviate from the "harmonic path". However, I've recently seen potentials of the form $$V(x) = \frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}$$ described anharmonic oscillator as well. I just wish to know what is a good definition of anharmonic oscillators and anharmonicity?
| I've always treated anharmonic oscillators to mean the potential has the form
$$
V(x)=\gamma x^2 + \beta_ix^i
$$
with $i$ being any value except 2, including negative values as well. Anharmonicity then follows as the deviation of the eigenvalue of $V(x)$ above from the harmonic solution.
For example, the paper you link above, Case 1 has an energy eigenvalue of
$$
E_n'=\hbar\omega\left(n+\frac32+\frac{c}{\sqrt{2d}}\right)
$$
So then the anharmonicity would be
$$
\Delta E=\left|E_n - E_n'\right|=\hbar\omega\left(1+\frac{c}{\sqrt{2d}}\right)
$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/99276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Do photons and cosmic rays radiate energy through gravitational waves? If not, why not? Due to the mass-energy equivalence, both matter and EM radiation bend spacetime, and both are capable of forming singularities (black hole, white hole/kugelblitz). In light of this, why do photons traveling from the most distant reaches of the observable universe not lose energy due to the gravitational radiation they must emit? Furthermore, mustn't cosmic rays (e.g. protons) slow down or stop as they lose energy through the same mechanism?
| Photons or cosmic rays don't (normally) emit gravitational waves.
Consider the comparison with radio waves. A moving electron doesn't emit radio waves. It has to be accelerating to emit EM radiation. Specifically radio waves are only emitted when there is a changing dipole moment.
So you wouldn't expect a particle moving at constant velocity (photon or otherwise) to emit gravitational waves, and in fact unlike EM even an oscillating gravitational dipole won't emit gravitational waves. For gravitational wave emission you need an oscillating quadrapole moment. In principle a photon whose trajectory is bent by a gravitational potential could emit gravitational waves, but in practice the intensity of the radiation emitted would be so small that you could never measure the energy loss.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/99486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
While holding an object, no work done but costs energy (in response to a similar question) I read the answer to Why does holding something up cost energy while no work is being done?
and wanting to know more, I asked my teacher about it without telling him what I read here. Instead of referring to muscle cells and biophysics, he answered my question in terms of entropy. He told me that while my arm muscles are stretched when I hold the object, they are more ordered. When my arm is at rest and muscles are not contracted, the muscles are less ordered (more entropy). So his conclusion was that the energy is required to keep the system (my arm muscles) from going to a state of higher entropy.
However, the answer in terms of muscle cells doing work on each other (i.e the answer to the hyper-linked question) made more sense to me. Could someone please give me some intuitive sense to my teacher's answer or explain the link between the two answers if there are any...
| Put the object on a table. Nothing happens, table or object. Your meat is metabolizing about 2000 Calories/day basal metsbolism, 8.4 million joules/day. "There is only about 0.1 mole of ATP in the body, but daily energy needs require 100 to 150 moles, or about 50 to 75 kg. The average human adult will use their body weight in ATP each day – so ATP must be recycled as it is used."
http://gasexchange.com/notes/metabolism/
The physics is wholly independent of the biology.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/99616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Number of Parameters of Lorentz Group We embed the rotation group, $SO(3)$ into the Lorentz group, $O(1,3)$ : $SO(3) \hookrightarrow O(1,3)$ and then determine the six generators of Lorentz group: $J_x, J_y, J_z, K_x, K_y, K_z$ from the rotation and boost matrices.
From the number of the generators we realize that $O(1,3)$ is a six parameter matrix Lie group.
But are there any other way to know the number of parameters of the Lorentz group in the first place?
| It's the same way you know there are three parameters in $SO(3)$. The equation $\Lambda^T \eta \, \Lambda = \eta$ has $(n^2+n)/2$ independent scalar equations. To see this, write the equation in component form: $\Lambda^{\mu\nu} \Lambda_\mu{}^\rho = \eta^{\nu\rho}$. Now we see there are $n^2$ scalar equations equations, but because $\eta$ is symmetric and the left hand side is symmetric in $\nu$ and $\rho$ as well, the equations related by switching $\nu$ and $\rho$ are the same. Thus we have established that there are $(n^2+n)/2$ independent scalar equations.
Since $\Lambda$ has $n^2$ components, we get $n^2-(n^2+n)/2 = n(n-1)/2$ degrees of freedom. In $3D$ this comes out to $3\cdot 2 /2=3$, and in $4D$ this comes out to $4 \cdot 3/2=6$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/99906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How do wind turbines cause lightning? I saw this paper about Lightning discharges produced by wind turbines
More lightning strikes wind turbines than comparable non-moving structures.
What would be the physical cause of this? Does it depend upon the motion of the wind turbine blades or is it because these are connected to the electric power grid?
| I don't think that the fact that they are connected to the grid has anything to do with that. In fact, the generator part of the turbine is one of the most expensive parts, so it is protected against impacts from the outside.
The reason is simple: the turbine blades are being charged as a result of rubbing against the humid air. As a result we can observe electrostatic interactions between cloud and blades.
I also found that: "(...)the researchers think the motion of the blades allows them to outrun their corona—the ionized air that surrounds a charged object. For a stationary object, that corona acts as a sort of buffer that dampens the electric field. By escaping the sheath of ionized air, moving objects become more likely to experience an electrical discharge."
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Hydrogen atom: potential well and orbit radii I happened to open up an old solid-state electronics book by Sah, and in it he says:
"it is evident that the electron orbit radius is half the well radius at the energy level En"
The orbit radius is $r_n=\frac{4\pi\epsilon_0 ℏ^2 n^2}{mq^2}$ and the potential well $V(r_n)=\frac{−q^4m}{(4\pi\epsilon_0)^2ℏ^2n^2}$
Of course the orbit has to be confined in the well, but it's not obvious to me why it should be exactly half the well radius? This isn't something I recall seeing before either in any other text.
Thanks
| Express V and E as explict functions of r.
$$V = \frac{-q^2}{(4\pi\epsilon_0)r}$$
$$E = \frac{-q^2}{2(4\pi\epsilon_0)r}$$
Also, the previous page of Sah emphasizes that 2E = V.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Are the axial electric field lines of a dipole the only ones that extend to infinity? Consider an electric dipole and its electric field lines.
There will be many field lines that do not extend to or originate from infinity, but rather begin at the positive charge and loop back around to terminate at the negative charge.
However, the field lines exactly on the dipole axis will extend out to infinity (or originate from infinity).
But are these the only field lines that extend to infinity, or are there other off-axial lines that also go to infinity? How can one prove this?
I have tried writing out the differential equations for the field lines, but they seem impossible to solve. No heuristic argument has satisfied me either. One can assert that all off-axial electric field lines must curve away from the axis as one goes to infinity, but does this curvature ultimately cause them to curve all the way back or do they reach some sort of linear asymptote?
| Dipole
$\def\vp{{\vec p}}\def\ve{{\vec e}}\def\l{\left}\def\r{\right}\def\vr{{\vec r}}\def\ph{\varphi}\def\eps{\varepsilon}\def\grad{\operatorname{grad}}\def\vE{{\vec E}}$
$\vp:=\ve Ql$ constant $l\rightarrow 0$, $Q\rightarrow\infty$.
\begin{align}
\ph(\vr,\vr') &= \lim_{l\rightarrow0}\frac{Ql\ve\cdot\ve}{4\pi\eps_0 l}\l(\frac{1}{|\vr-\vr'-\ve\frac l2|}-\frac{1}{|\vr-\vr'+\ve\frac l2|}\r)\\
&=\frac{\vp}{4\pi\eps_0}\cdot\grad_{\vr'} \frac1{|\vr-\vr'|}\\
&=\frac{\vp\cdot(\vr-\vr')}{4\pi\eps_0|\vr-\vr'|^3}\\
\vE(\vr)&=-\grad\ph(\vr)\\
\vE(\vr) &= \frac{1}{4\pi\eps_0}\l(\frac{-\vp}{|\vr-\vr'|^3}+3\frac{\vp\cdot(\vr-\vr')}{|\vr-\vr'|^5}(\vr-\vr')\r)
\end{align}
Field Lines
We calculate the field lines with $\vp=\ve_1$ and $\vr'=\vec0$.
For calculating the field lines we may multiply the vector field with any scalar field called integrating factor.
We choose $4\pi\eps_0|\vr|^3$ as integrating factor and get the differential equation
\begin{align}
\begin{pmatrix}
\dot x\\
\dot y
\end{pmatrix}
&=
\begin{pmatrix}
-1+3\frac{x^2}{x^2+y^2}\\
3\frac{xy}{x^2+y^2}
\end{pmatrix}
\end{align}
for the field lines. In the complex representation $x(t)+i y(t)=r(t)e^{i\phi(t)}$ this reads as
\begin{align}
\dot x + i\dot y &= -1 + 3\frac{x(x+iy)}{x^2+y^2}\\
&=-1 + 3\cos(\phi)e^{i\phi}\\
\dot r e^{i\phi} + ir e^{i\phi} \dot\phi &= -1 + 3\cos(\phi)e^{i\phi}\\
\dot r + i r\dot\phi &= -e^{-i\phi} + 3\cos(\phi)
\end{align}
Written in components:
\begin{align}
\dot r &= -\cos(\phi) + 3\cos(\phi) = 2\cos(\phi)\\
r\dot \phi &= \sin(\phi)
\end{align}
If we divide these two equations we get
\begin{align}
\frac{\dot r}{r\dot\phi}=2\cot(\phi)
\end{align}
We re-parameterize such that $\phi$ becomes the parameter
\begin{align}
\frac{r'}{r} = 2\cot(\phi)
\end{align}
Integrate:
\begin{align}
\ln\l(\frac{r}{r_0}\r) &= \int_{\phi_0}^{\phi} 2\cot(\phi) d\phi
= 2\ln\l(\frac{\sin(\phi)}{\sin(\phi_0)}\r)\\
\frac r{r_0} &= \l(\frac{\sin(\phi)}{\sin(\phi_0)}\r)^2
\end{align}
As soon as you have some $\phi_0\neq 0$ for $r_0>0$ you get a bounded curve.
Ling-Hsiao Lyu gets the same formula (only that she uses $\phi_0=\frac\pi2$ and $r_0=r_{\rm eq}$) in Section "Dipole Magnetic Field Line" of her Lecture Notes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why is the moment of inertia (wrt. the center) for a hollow sphere higher than a solid sphere (with same radius and mass)? Why is the moment of inertia (wrt. the center) for a hollow sphere higher than a solid sphere (with same radius and mass)? I have completely no idea and I am inquiring about this as it is an interesting question that popped in my head while doing physics homework.
| A hollow sphere will have a much larger moment of inertia than a uniform sphere of the same size and the same mass.
If this seems counterintuitive, you probably carry a mental image of creating the hollow sphere by removing internal mass from the uniform sphere. This is an incorrect image, as such a process would create a hollow sphere of much lighter mass than the uniform sphere. The correct mental model corresponds to moving internal mass to the surface of the sphere.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 8,
"answer_id": 0
} |
Center of rotation and trajectory of a rigid body in a plane with applied *fixed* forces This is my first question so please excuse me if my format is a bit off.
Given a 2D rigid body with forces applied to it in such a way that the angle the force vector makes with the surface of the object remains constant (think of a spaceship with fixed rockets attached to it), I have problems mapping out it's trajectory and angle it has rotated respect the starting position over a certain period of time.
The net force from the rockets applied on the centre of mass of this object is easy to calculate on a local reference solitary to said object, but given the existence of external torque, it is a reference possessed by angular acceleration, which unsettles me as I do not know if this requires any extra considerations when translating that to a general reference.
Determining the centre of rotation is also difficult, as from what I have researched, it's bound to be the centrer of mass, wherever the forces I've applied are at, although this seems unintuitive to me.
| The problem as is stated is somehow ambiguous, but using some simplifications we can manage to get something: if we assume that forces don't change depending of the angle (i.e. there is no "correcting trayectory rocket" that acts depending of its orientation), and that the center of mass is fixed, then you can express net force F' and net torque T' with respect to the ship frame as functions only of time. Let's call $\theta$ the angle between the ship frame and the "ground" frame (some external frame at rest), then:
$ \frac{d^2 \theta}{dt^2} = \frac{T'(t)}{I_0(t)}$,
where $\ I_0$ is the moment of inertia with respect to the center of mass, posibly a function of time. you can find the angle $\theta$ by double integration and then express the force in the ground frame in terms of the force in the ship frame using the matrix rotation
$ R_\theta = \begin{bmatrix} cos\theta & -sin\theta \\ sin\theta & cos\theta \end{bmatrix}$
Then, wrt to the ground, $ \bar{F} = R_\theta \bar{F'} $ and $\ \bar{T} = \bar{T'} $, hence
$ \frac{d^2\bar{r}}{dt^2} = \frac{R_\theta}{m} \bar{F'} $
In the simplest case, with $ \bar{F'}, \bar{T'}, I_0, m $ constant, and the ship starting at rest, you get
$ \theta(t) = \frac{T}{2I_0} t^2 $, and
$x(t) = \frac{F_{x'}}{m} \iint cos(wt^2) dt^2 + \frac{F_{y'}}{m} \iint sin(wt^2) dt^2 $
$y(t) = -\frac{F_{x'}}{m} \iint sin(wt^2) dt^2 + \frac{F_{y'}}{m} \iint cos(wt^2) dt^2 $
for $ w = \frac{T}{2I_0} $.
Unfortunately those integrals are not doable, see this wikipedia entry. Interestingly, this kind of fuctions are used in freeway design, where forces between vehicles and the road at constant speed are also perpendicular to the trayectory(see Euler spiral).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Relativistic fomulae for energy and momentum? I know that the relativistic formulae for energy and momentum are:
$E = \gamma mc^2$ and $\textbf{p} = \gamma m\textbf{v}$;
Can we derive these formulae?
If yes, where from?
| Once we have the position 4-vector
$$x^\mu=
\left(
\begin{array}{c}
ct\\
\vec{x}\\
\end{array}
\right)
$$
It is natural to define the momentum and energy in a fashion which is analogous to the Newtonian case (and reduces to it in the frame of the particle itself, when $\vec{v}=0$:
$$
p^\mu \equiv
m\frac{d}{d\tau}
\left(
\begin{array}{c}
ct\\
\vec{x}\\
\end{array}
\right)
=
\left(\begin{array}{c}
E/c\\
\vec{p}\\
\end{array}
\right)
$$
where $\tau=\frac{t}{\gamma}$ is the eigentime. This immediately leads to your formula's
$$\vec{p}=\gamma m\vec{v}\hspace{2cm}E=\gamma mc^2$$
I hope this semi-first principles version is satisfactory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What is decay associated spectra? What is decay associated spectra?
Suppose we measure the fluorescence intensity over different wavelengths and over time, we get:
$$I(\lambda,t) = \sum_i^n \alpha_i(\lambda) \exp(\frac{-t}{\tau_i}).$$
The assumption is that there are n component,species, in the $I(\lambda,t)$. If we fit the right hand to the experimentally obtained $I(\lambda,t)$, and get $\alpha_i$ and $\tau_i$, then people call $\alpha_i$ the decay associated spectra.
Now, if we integrate over time we get the steady state emission spectra. The thing that I cannot understand is the decay associated spectra. What does it mean? If it is the steady state spectra of species i, then why does it become negative sometimes? People, say that when it becomes negative, it indicates energy transfer between species. Could someone please elaborate this concept more?
| Sure it can be negative. It simply means that there is a dynamics (reaction, energy transfer, ...) between species, so that there is an exponential rise at some wavelengths
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Explain the microscopic nature of Electric current? Explain the microscopic nature of Electric current?i.e
What is is average current and Instantaneous current? A microscopic view what really happens?
| From a microscopic point of view you can image metal (conductors) in a lot of different ways. The easiest model is the Drude model in which atoms are fixed in the space and everyone have one or two (in a metal) free electrons. When you apply an external electric field this particles move as a consequence of Coulomb force. It's important to say that electric field permeated space with light velocity and so in everyday situation electrons along all the wire (for instance) start "immediately" to move. So the current is the charge variation in time in a fixed point along the wire. The concept of current is intrinsically connected to time. You can measure current average simply consider in a point the charge variation for the wanted time interval. In order to understand instantaneous current think of the difference between instantaneous and average velocity in classical mechanics. Put your time interval very small you reach in practise the instantaneous current value.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Linearized mass conservation equation I'm working on global seismology and I'm currently facing troubles understanding how an equation is obtained. The equation concerned is the following one :
$$
\rho^{E1} = -\nabla \cdot (\rho^0\mathbf{s})
$$
(From the book Theoretical Global Seismology, Princeton Press : Google Book Link equation (3.46) p.65)
Where $\rho^{E1}$ is the first order perturbation of the density and $\rho^0$ the density in the reference state. $\mathbf{s}$ is the displacement field.
It comes from the mass conservation equation which has been integrated in time :
$$
\partial_t\rho^E + \nabla \cdot (\rho^E \mathbf{u}^E) = 0
$$
and the following decomposition :
$$
\rho^E(\mathbf{x}, t) = \rho^0(\mathbf{x}) + \rho^{E1}(\mathbf{x}, t)
$$
Where $\rho^0$ is independent of time and $\mathbf{u}$ is the velocity field. How do you obtain the first equation from the second one ?
If someone could help me out on this one it's be really nice !
Cheers
| If you substitute the decomposition in, you get:
$$ \partial_t \rho^0 + \partial_t \rho^{E1} + \nabla \cdot (\rho^0 \mathbf{u}^E) + \nabla \cdot (\rho^{E1} \mathbf{u}^E) = 0 $$
Typically the decomposition used assumes that $\rho^0$ is constant in time and that $\rho^{E1}$ is random in time, such that it's mean value is 0. Therefore, $\partial_t \rho^0 = 0$.
So when you integrate over all time, the value of $\overline{\rho}^{E1} = 0$. Which means $\int_t \nabla \cdot (\rho^{E1} \mathbf{u}^E) = 0$ and $\int_t \nabla \cdot (\rho^0 \mathbf{u}^{E}) = \nabla \cdot (\rho^0 \int_t \mathbf{u}^E)$ and of course the integral of the velocity is just the displacement field. So the final expression is:
$$ \rho^{E1} = -\nabla \cdot (\rho^0 \mathbf{s})$$
Edit:
Based on the comments, there are two things to clarify. First is the decomposition itself. Typically when a term is split as done here, it is split into a mean and a fluctuating component. The fluctuating component has a zero mean. This has to be true because $\rho^0$ is defined as the mean component (brackets indicate taking the mean):
$$ \left< \rho^E \right> = \left< \rho^0 + \rho^{E1} \right> = \rho^0 $$
which means $\left< \rho^{E1} \right> = 0$ by definition. Otherwise $\rho^0$ wouldn't be the mean component.
For why the term $\int_t \nabla \cdot (\rho^{E1}\mathbf{u}^E)$... Perform the same decomposition on $\mathbf{u}^E$, $\mathbf{u}^E = \mathbf{u}^0 + \mathbf{u}^{E1}$. When you substitute that in, you get:
$$\int_t \nabla \cdot (\rho^0 \mathbf{u}^0 + \rho^0 \mathbf{u}^{E1} + \rho^{E1} \mathbf{u}^0 + \rho^{E1} \mathbf{u}^{E1})$$
So now, we transpose the operators: $\int_t \nabla \cdot () = \nabla \cdot \int_t ()$ because we consider them to be linear operators. If we consider those terms and group the first two back together into $\rho^0 \mathbf{u}^E$, then we see $\rho^0$ is independent of time and moves outside the integral while $\int_t \mathbf{u}^E = \mathbf{s}$ and you get that term.
Now you're left with two terms: $\nabla \cdot \int_t \rho^{E1} \mathbf{u}^0$ and $\nabla \cdot \int_t \rho^{E1} \mathbf{u}^{E1})$. Since we are interested in linear effects only, that second term is considered a higher order term and is neglected as insignificant. If it is significant, than our linear assumption breaks down but we're assuming things are linear.
So then the term $\nabla \cdot \int_t \rho^{E1} \mathbf{u}^0$ is easy to see that it's zero. The $\mathbf{u}^0$ moves outside the integral, and the time integral of $\rho^{E1}$ is zero based on the decomposition given above.
All of that said, this is very standard procedure for linearizing equations. Any textbook in engineering that talks about linear assumptions or stability of solutions will cover this procedure in detail, whether it's structural mechanics, turbulence, aerodynamics... Learn the procedure, practice it, because it's an essential part of the analysis of differential equations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Bulk flow of air in a long tube between Antarctica and Australia? I have a 5km diameter clear plastic tube which is open at each end and runs from the center of Antarctica to Lake Eyre in Australia. The tube is on the ground where it can be and at sea level on the ocean.
Will there be bulk flow of the air in the tube? If so, which way will the air flow?
| any flow is driven by pressure gradient, which must overcome the friction of flow through the tube. It's a very long way, hence huge pressure drop relative to the initial pressure gradient. No flow
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
States of Matter and Equilibrium Can I say that, matter generally when cooled decreases in volume because, when it is cooled,i.e., we lower the temperature of the surroundings, then the avg. energy of our sample will be higher than the surroundings and hence, the molecules coming out will be more, therefore, the volume decreases...And when equilibrium is reached the molecules coming out and going in will be same and that's when they are at the same temperature.
| No, you should not refer to molecules coming out or going in, unless there is more than one phase (such as in evaporation, sublimation, condensation, etc.). The volume decreases (or increases) because the space between molecules or atoms decreases (or increases), even though the number of molecules or atoms is constant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/100961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Do exact beta functions exist in (super)gravity theories and string theory? An exact beta function exists for Super-Yang-Mills theories in 4D without matter - the so-called NSVZ beta function.
Does a similar exact beta-function exist in gravity or supergravity theories? In string theory? Please provide references too.
| @Ten The NSVZ beta function exists for theories with matter as well. Just read the scholarpedia article carefully. What happens is that the NSVZ beta function for the gauge coupling constants depends on the anomalous dimensions of the matter fields.
A very nice example is to consider $\mathcal{N}=4$ SYM theory and write it out as a $\mathcal{N}=1$ theory -- the spectrum then consists of one $\mathcal{N}=1$ vector multiplet and three chiral multiplets. Deform the superpotential into the most general cubic superpotential. Leigh and Strassler use the vanishing of the NSVZ beta function for the gauge coupling imposes an additional constraint that anomalous dimensions of the chiral scalars should vanish leading to a theory which is conformal with $\mathcal{N}=1$ supersymmetry. This theory generalises the beta-deformation of $\mathcal{N}=4$ SYM. There are many more examples in their paper. Two more papers by Arkani-Hamed and Murayama would provide further examples. Paper 1 and Paper 2.
(@Ten I realise that you want examples from string theory/supergravity but it is important to note the generality of the NSVZ beta function. So I will post my answer. I hope you don't mind. If so, let me know and I will delete my answer. Of course, one knows that any two-dimensional CFT has a vanishing beta function.)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Neutrinos and anti-neutrinos in the Standard Model In standard model neutrinos and the left handed electron forms SU(2) doublet.
*
*What about the anti-neutrinos in the standard model? Do they also form some doublet?
*If neutrinos have tiny masses will it not imply indirectly and conclusively that right-handed neutrinos must exist in nature?
EDIT : Neutrinos will have Majorana mass term if they are Majorana fermion. Is that right? Now, if neutrinos are Majorana fermions, will they have definite handedness? For example, does $\nu_M=\begin{pmatrix}\nu_L\\ i\sigma^2\nu_L^*\end{pmatrix}$ have definite handedness? Therefore, doesn't it imply that if neutrinos are massive then a right-handed component of it $\begin{pmatrix} 0\\ i\sigma^2\nu_L^*\end{pmatrix}$ must exist? Although we are not using $\nu_R$ to construct this column, does it imply $\nu_M$ do not have a right handed component? It is the column $\nu_M$ which we should call a neutrino. Then it has both the components. However, one can say that a purely right-handed neutrino need not exist if the neutrino is a Majorana fermion. Therefore, it seems that if neutrinos are massive a right handed component of it must exist (be it a Dirac particle as well as a Majorana particle). Correct me if I am wrong.
| The antineutrinos do indeed form a doublet. The particle-antiparticle conjugation operator is usually denoted by $\hat{C}$ and is defined through:
\begin{equation}
\hat{ C}: \psi \rightarrow \psi ^c = C \bar{\psi} ^T
\end{equation}
where $ C \equiv i \gamma _2 \gamma _0 $. So given a neutrino you can always get its complex conjugate with this operator:
\begin{equation}
\nu _L ^{\,\,c } = i \gamma _2 \gamma _0 ( \overline{\nu _L} ) ^T
\end{equation}
Its easy to check this that this antineutrino is actually right handed, by applying a left projector onto it.
The antineutrino forms a doublet with the antileptons:
\begin{equation}
\left( \begin{array}{c}
\nu _L ^{\,c } \\
e _L ^{ \, c}
\end{array} \right)
\end{equation}
With regards to your second question, no having neutrino masses does not imply that there exist right handed neutrinos. This is because neutrinos could have Majorana masses ($\frac{m}{2} \nu _L \nu _L +h.c. $) as well as Dirac masses $m( \overline{\nu_L} \nu_R + h.c.)$. Majorana masses could arise if for example there exists a heavy Higgs which is a triplet under $SU(2)_L$ (which can be rise to what's known as a type 2 See-saw mechanism).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Confused about Impulse Encountered a problem that involves impulse while studying for my exam and I'm not sure how to even approach it. I know that momentum is conserved, but I'm not sure how to relate that to avg force. Maybe someone can help point me in the right direction? I know that it's in quadrant III, through intuition, but I can't come up with a provable explanation
Relevant equation: $J=F_{avg}\Delta T$
| Alternatively, and qualitatively, think about the components of velocity (in the x y directions) have changed. Along the x axis, velocity has reduced, so the re has been a force in the -x direction. In the y axis, velocity has changed sign, so there must have been a force in the -y direction. Hence the total force is down and to the left, ie quadrant III.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Why does foam dull knives? I have recently taken up prop making and just started my first foam-built costume from a video game. These kinds of costume armour builds are often built out of the various foam floor mats you can buy in say Home Depot or BJ's or Five Below for around $1 for 4 sq. foot tile.
The one thing that puzzles me is why foam dulls a sharp steel blade so easily and so quickly.
I have tried x-acto knives, those snap-off knives, even surgical scalpels. Regardless, after cutting several linear feet (maybe in the neighborhood of 5-10?), the knife blade begins to dull rapidly and no longer produces nice cuts, but begins to tear the foam instead. You can always sharpen your blade on a stone, but in short order it will again dull and tear.
I generally do all my cutting on one of the green, self-healing cutting mats. This could explain the tip dulling, but it is actually the whole length of the blade exposed to the foam that gets dull.
It happens across types, thicknesses and densities of foam. The floor mats are about 0.5
" and medium density. The craft foam from Michaels is 6mm, high density and rigidity. The roll I have is 0.25" low density, low rigidity
It absolutely baffles me how soft squishy foam just destroys the edge of a metal knife.
I am guessing it has something to do with the molecular organization of the foam or something, but would love to know if anyone has a good explanation for why and how this happens.
| My guess is that the foam has some type of sillica in it which may stick to the knife having a detremental effect on its cutting ability possibly cleaning the blades regularly with alcohol could help this but i'm no chemist, hopefully this is helpful
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Which angle should it be? in the formula
$$dB = \frac{\mu_0l ~|dl \times r|}{4 \pi r^3} $$
and the image
where dl is in y-z plane and dB is in x-y plane. the ring conductor is in y-z plane carrying current I in direction as mentioned
EDIT: also point p can move in the circular ring
EDIT 2:To clear the confusion...The dl vector is having (L alphabet) and current is I (i alphabet).
I want to know that is the angle between dl and r is 'Theta' ? how?
| The angle between $\vec {dl}$ and $\vec r$ is $2n\pi \pm\dfrac{\pi}{2}$ because the angle between them is the angle between the x-r plane and y-z plane.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Does the equation of continuity hold for turbulent flows? My textbook mainly deals with laminar flows. The book derives the equation of continuity, which states that the cross-sectional area times the velocity of a flow is always constant. But nowhere in the derivation does the textbook explicitly assumes that the flow is laminar. So, does the equation hold for turbulent flows too?
| In order to have such a relation, your flow needs to be be stationary, which is never the case for turbulent flows.
The conservation of the mass gives you the local continuity equation.
$$\partial_t \rho+ \nabla . (\rho \vec{v})=0 $$
For a stationary problem without sources, Ostrogradsky's theorem allows you to reach:
$$ \oint_S \vec{v}.d\vec{S}=0 $$
But this last step is not possible for turbulent flows.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Does a mirror help a near-sighted persion see at a distance clearer? A near-sighted person without eye-glasses can not clearly see things at distance.
If he takes a photo of the things at distance, he can see the things from the photo much clearer, because he can place the photo much closer to his eyes.
If he turns his back at the things at distance, and holds a mirror close to his eyes in a position so that the mirror reflects the things at distance behind him, will he see the things much clearer than if he looked at the things at distance directly?
| Edit: Reading other people's answers, I forgot to mention I assumed a flat mirror.
Excellent question, but the answer is no. The reason is because the object (in the strict optics meaning) in the case of the photograph is actually on the paper whereas in the case of the mirror it is still at the same place, far behind: the rays of light coming from it are reflected on the mirror but still require tuning from the eye muscles to get the focal point right (the eyes are in fact sort of a tunable lens with a detector similarly to cameras).
I would recommend you draw a diagram of the light rays coming from the edges of the objects into the eye for both cases to get a better idea.
Example
As an example, consider the following: if the mirror has dust on it, try looking first at the dust and then at an object that the mirror displays behind you. You will have to strain your eyes to focus on the dust, which proves that what the mirror displays is not actually at its surface (the image, again in the optics strict sense, is behind the surface of the mirror).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 8,
"answer_id": 4
} |
Wrong positioned Ampere-meter and Voltmeter I'm dealing with a problem here and even that I'm trying to solve it i can't
It says:
In what figures the voltmeter and ampere-meter are wrong positioned?
I think that all the the others are correct except the second one.
Can anyone help me?
| The ammeter actually creates a short across the battery in the second figure (so that one is certainly wrong). Very likely, this would cause damage to the ammeter, or at least result in a blown input protection fuse.
I think the other diagrams are OK, depending on what it is you are trying to measure (which is open to interpretation).
This answer assumes that the components marked as an X within a circle is an unknown component, and not a "mixer", which is what this symbol actually stands for, and which is almost never probed using either an ammeter or a voltmeter, because such a measurement would tell you almost nothing of interest.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Would there be fringe pattern in this arrangement? Figure shows a standard two slit
arrangement with slits S1, S2. P1, P2 are the two
minima points on either side of P.
At P2 on the screen, there is a hole and behind P2
is a second 2- slit arrangement with slits S3, S4
and a second screen behind them.
Would there be a fringe pattern on the second screen?
I think no because the only way light can get to the slits S3 and S4 is through the hole at P2. But, its a minima there so no light passes through S3 and S4 and thus no fringe pattern on the second screen.
Is my thinking correct?
| In interference and diffraction, light energy is redistributed. If it reduces in one region, producing a dark fringe, it increases in another region, producing a bright fringe. There is no gain or loss of energy, which is consistent with the principle of conservation of energy.
If you consider a point where there is destructive interference, there is a dark fringe. So, "no more light energy" is passing through it, provided amplitude of light waves undergoing destructive interference is same or width of both the slits is same. Thus, we can't expect fringe pattern in the second screen.
If amplitudes of light waves undergoing destructive interference at the point is not same, or if width of two slits is not same, intensity of light at the point of destructive interference will not be zero, then we can expect fringe pattern in the second screen.
Good comment by Carl Witthoft made me to stop answering this post for a while. In talbot effect there is only one diffraction grating. Later at regular distances from the grating, the light diffracted through it forms a nearly perfect image of the grating itself. But there are diffraction grating in between. It is totally a different wonderful concept. If interested I would suggest one to read this article: Rolling out the (optical) carpet: the Talbot effect.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
The best way to cool the teapot My professor today in the class made us a question:
"Lets say we have a teapot with water in it.The water is hot.Now we want to cool the water.
Will it cool faster if we put an ice cube above the teapot or under the teapot."
My answer was the it will cool faster if we put the ice cube above it because the warm air stays up and the ice cube will melt faster.
He didn't tell me if I was right or not.
Can anyone help me ?
| Put the ice cube atop. Poorer air and water conductive and convective heat transfer kinetics are more than compensated by meltwater and cooled air dribbling down the pot, further cooling it by heat capacity as they warms from 0 C. If the ice cube is touching the bottom, meltwater and cooled air flow away from the pot.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Valley meaning explanation for foreigner English is not my native language and I have some hard time translating this word.
I was searching in couple dictionaries(both paper and online) and could not find it.
Could anyone provide me definition of word "Valley".
example usage:
Suggest using only the valley degree of freedom
valley and spin degeneracy
Valley relaxation
Thanks in advance!
| Sounds like graphene physics or something similar. You won't find it in a dictionary.
In the band structures of many materials, it is common to find multiple similar points in reciprocal (momentum) space. For example, in silicon's band structure there are six distinct conduction bands that all have similar behaviour. These six points came to be known as valleys. Among other details, the "valley degeneracy" of 6 is an important factor to take into account when calculating electronic properties of silicon.
Graphene's band structure has two distinct bands. These are often known as the K and K' valleys, and they are centered around the K and K' points in reciprocal space. They are very symmetric and also are closely related to a spin-like property of the electrons in the graphene, known as pseudospin. In essence, you can imagine the K valley as being "pseudospin up" and the K' valley as being "pseudospin down". An electron can also be in a superposition of pseudospin up and down, so you can imagine "pseudospin left", "pseudospin right", and so on. The resemblance with spin comes from the fact that scattering processes between the two valleys are fairly rare, and so pseudospin is conserved over some distance. Also, it means that besides the normal factor of 2 degeneracy from real spin, the electrons in graphene have a further factor of 2 degeneracy from the valleys / spin. On the other hand, pseudospin is not associated with a magnetic moment, unlike real spin.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Deriving commutation relations in second quantisation I am trying to start from:
\begin{align*}
[\phi(x),\pi(x')] = i\hbar\delta(x-x') \\
[\phi(x),\phi(x')] = [\pi(x),\pi(x')]=0
\end{align*}
to derive:
\begin{align*}
[a(k),a(k')^\dagger]=\delta_{kk'}\\
[a(k),a(k')]=[a(k)^\dagger,a(k')^\dagger]=0
\end{align*}
So starting with:
\begin{align*}
\phi(x) = \sum_k \left(\frac{\hbar c^2}{2\omega_k}\right)^\frac{1}{2}[a(k)u_k(x)+a(k)^\dagger u_k(x)^*]
\end{align*}
where $u_k(x) = \frac{1}{\sqrt{V}}e^{i(k \cdot x - \omega_k t)}$ and $\pi(x) = \frac{1}{c^2}\dot{\phi}(x)$
\begin{align*}
&[\phi(x),\pi(x')] \\
&=-i\sum_{k,k'} \frac{\hbar}{2}\sqrt{\frac{\omega_k}{\omega_{k'}}}\left([a(k)^\dagger,a(k')]u_k(x)^*u_k(x')-[a(k),a(k')^\dagger]u_k(x)u_k(x')^*\right)
\end{align*}
I'm not sure how to continue...
| I believe in the last line, the plane-wave functions $u_k(x)$ should carry different coordinates and momenta, e.g
$$ [a(k)^\dagger,a(k')]u_k(x)u_{k'}(x') $$
You may note that the commutator $[\phi(x),\pi(x')]=i\hbar\delta(x-x')$ holds if one choses $[a_k,a_{k'}^\dagger]=\delta_{kk'}$. However, this indirect reasoning is no proof that this choice is unique. It also won't tell you that $[a_k,a_{k'}]=0$
I recommend inverting the relation, that is expressing $a_k$ and $a_k^\dagger$ in terms of $\phi(x)$ and $\pi(x)$. Then you may check the CCR by direct calculation, because the commutation relations
$$[\phi(x),\pi(x')]=i\hbar\delta(x-x')\\
[\pi(x),\pi(x')]=[\phi(x),\phi(x')]=0$$
are know.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Aerosol size distribution When the size distribution of particles in an aerosol is obtained, there is always a bimodal distribution. The small peak accounts for the particles that are in sub-nanometer size range and the larger peak accounts for the particles that are in nanometer to micrometer size range. Why does it show a bimodal distribution? What is the physical reason for this occurrence?
| There may be several reasons for this.
In general the bimodal character of atmospheric aerosols is the owed to different lifetime of particles. These are dependent on the particle size and loss mechanisms predominantely associated with the respective sizes.
Usually atmospheric aerosols consist of four size modes. From smallest to biggest median, these are:
*
*Nucleation Mode: Particles originate by means of homohenous/heterogenous nucleation triggered by gas phase reactions. Particles within this mode rapidly grow out of this mode due to condensation, coagulation and especially diffusion (Diffusion is highest for small particles).
*Aitken Mode: Particle grow out of this mode due to coagulation. (soot particles)
*Accumulation Mode: Usually has the longest liftime of all modes ranging from a few days to a few years (dependent on the height (atmospheric layer)). Predominant loss mechanism is wet deposition. (e.g. smoke particles)
*Coarse Mode: Due to their size, these particles are usually heavy and drop out ambient air by means of sedimentation quiet quickly (e.g. Mineral dust, vulcan ashes, sea salt).
Dependent on the circumstances a mode may be more dominant than others, while another may hardly be present resulting in a bimodal character.
Concerning Sub Nanometer particles. I am not quite sure wheter these particles are considered to be aerosol particles still, as they approach molecular dimensions and physics may be different there.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
How does a half-life work? Carbon-14 has a half-life of 5,730 years. That means that after 5,730 years, half of that sample decays. After another 5,730 years, a quarter of the original sample decays (and the cycle goes on and on, and one could use virtually any radioactive isotope). Why is this so? Logically, shouldn't it take 2,865 years for the quarter to decay, rather than 5,730?
| The mass of radioactive materials follows the ordinary differential equation:
$$
m'(t)=-am(t),
$$
where $m$ is the mass and $a$ a positive constant - i.e., constant relative rate of decay.
This implies
$$
m(t)=m(0)\mathrm{e}^{-at}. \tag{1}
$$
If $T_h$ is half life, then $$m(T_h)=m(0)\mathrm{e}^{-aT_h}=\frac{1}{2}m(0),$$
which implies that
$$
T_h=\frac{\log 2}{a},
$$
and hence $(1)$ can be written also as
$$
m(t)=2^{-t/T_h}m(0).
$$
So the quarter-life is $T_Q$, for which $m(T_Q)=\frac{1}{4}m(0)$ or
$$
m(T_Q)=2^{-T_Q/T_h}m(0)=\frac{1}{4}m(0),
$$
which holds only if $T_Q=2T_h$!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 13,
"answer_id": 9
} |
Prove identity of partial derivatives I can not do the following problem:
Prove the identity:
$$\left( \frac{\partial x}{\partial y} \right)_{z}\left( \frac{\partial y}{\partial z} \right)_{x}\left( \frac{\partial z}{\partial x} \right)_{y}=-1$$
State the properties that must be $x=x(y,z)$, $y=y(x,z)$, $z=(x,y)$.
The truth is I do not know how to start, and they do not know how to interpret the functions $x$, $y$, $z$. Any help or explanation will be the most grateful.
| The way I always thought about this was to pick one of the variables to be thought of as the dependent variable. Here I will pick $z$. Then we think of $z(x,y)$ to be a function which has partial derivatives $\partial_x z = \frac{\partial z}{\partial x}$ and $\partial_y z = \frac{\partial z}{\partial y}$.
Now we must compute $$\left( \frac{\partial x}{\partial y} \right)_{z}\left( \frac{\partial y}{\partial z} \right)_{x}\left( \frac{\partial z}{\partial x} \right)_{y}.$$
Let's look at each term individually. The strategy will be to write each term in terms of the "regular" partial derivatives $\partial_x z$ and $\partial_y z$. These are "regular" in the sense that they are partial derivatives of the dependent variable with respect to the independent variable.
The third term is the easiest. It is just already a regular partial derivative $\partial_x z$.
The second term is more foreign. If we are thinking about $z$ as a dependent variable, then it looks like we are taking the derivative of a independent variable with respect to a dependent variable. However, I just think of this as a shorthand for $1/\partial_y z$.
The first term is the most complicated. Here it looks like we are taking the derivative of an independent variable with respect to another independent variable. You might think this would be zero, but $z$ is supposed to be held fixed. So the question is "If I change $y$ how much do I have to change $x$ to keep $z$ fixed?" Let's imagine we change $y$ by an amount $dy$. Then $z$ will change by an amount $\partial_y z dy$. To compensate we must change $x$ by an amount that will cause the opposite change ($-\partial_y z dy$). The correct $dx$ is given by the equation $\partial_x z dx = -\partial_y z dy$, so the amount we must change $x$ is $dx=\frac{-\partial_y z dy}{\partial_x z}$. Then $dx / dy = -\partial_y z / \partial_x z$.
Now putting our three terms together we have $$\left( \frac{\partial x}{\partial y} \right)_{z}\left( \frac{\partial y}{\partial z} \right)_{x}\left( \frac{\partial z}{\partial x} \right)_{y}=-\frac{\partial_y z}{ \partial_x z}\frac{1}{\partial_y z} \partial_x z =-1. $$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why are free electrons free? This is what I understand so far: in a conductor, the ions have a weak pull on the valence electrons. So when an electric field is applied, the free electrons are able to easily move about. Makes sense.
In a neutral conductor with no electric field, the free electrons aren't bound to any ions. Why? I understand that the ions have a weak pull on the electrons, but what makes electrons leave the ion and stay free?
| The intuition is that the valence electrons are so far away from their nucleus that when they combine to form metals, they feel the attraction of all the other nuclei as strongly as from theirs.
In a more rigorous description, the orbitals for the valence electrons fully overlap with their neighbouring atoms, so their "play field" extends all over the material.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Naive visualization of space-time curvature With only a limited knowledge of general relativity, I usually explain space-time curvature (to myself and others) thus:
"If you throw a ball, it will move along a parabola. Initially its vertical speed will be high, then it will slow down, and then speed up again as it approaches the ground.
"In reality, the ball in moving in a straight line at constant velocity, but the space-time curvature created by the Earth's gravitation makes it appear as if the ball is moving in a curved line at varying velocity. Thus the curvature of space-time is very much visible."
Is this an accurate description, or is it complete nonsense?
| You have the right basic idea. But it gets simpler to visualize if you just drop the ball, or throw it vertically. Then there is just one spatial dimension to consider, and you can directly compare the paths in space and in space-time, like shown here:
http://www.youtube.com/watch?v=DdC0QN6f3G4
But note that this doesn't involve any intrinsic space-time curvature. Such curvature is related to tidal effects (geodesic deviation), which are negligible over small distances (like the height of an apple tree or a ball throw). However, over larger distances intrinsic curvature is inevitable, as shown in the last link of the video description:
http://www.adamtoons.de/physics/gravitation.swf
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Auto-refrigeration I'm reading about the auto-refrigeration effect and can't find a really good explanation. Is the idea that when you have your condensed liquid, and then release it into a low pressure environment, some of it evaporates instantly, leaving behind a cold liquid?
Let me explain what I'm thinking again: pressurized coolant is in a tube. It was pressurized by the compressor and cooled down moderately by the condenser until it barely became a liquid. Now, it reaches the throttling valve, which lets a little bit of this liquid out at a time into a lower pressure environment. Some of the liquid evaporates instantly (due to the lower pressure and thus lower boiling point), but some of it is left behind as a liquid. The part that evaporates instantly is hotter than the pressurized liquid, while the liquid left behind is colder than the pressurized liquid. The part that evaporated instantly moves through the evaporator quickly while the super cold liquid is left to slowly evaporate and move through the evaporator. This means the super cold liquid lingers in the evaporator and we can then run a fan over it and get nice AC or refrigeration.
The key point I'm getting at is that when the pressurized liquid moves through the throttling valve, the auto-refrigeration effect is really a way of splitting the hot vapor "part" away from the cold liquid "part". The vapor and liquid combined equal the same temperature as the source pressurized liquid, but the hotter part is instantly separated from the colder part (via vapor), and we take advantage of the colder part (that remains a cold liquid for enough time to be useful to us).
Am I right? Thanks!
| If you want to separate hot and cold gas streams from a common stream, use a Ranque-Hilsch vortex tube,
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A paradox to Lenz's law I have read that in simple words, Lenz's law states that:
The direction of current induced in a conductor is in such a fashion, that it opposes its cause.
This validates law of conservation of mass-energy.
I arranged the following thought experiment:
Let there be a pendulum with its bob being a small bar magnet. The pendulum is oscillating in a direction parallel to the horizontal axis of the bar magnet on which the North and South poles lie. Also, the pendulum is in complete vacuum. (But gravity is there to make the pendulum oscillate.)
At one of the extreme positions of the pendulum, we keep a solenoid, ends of which are connected to a load resistance.
As the North pole of the bar magnet approaches the solenoid, current is induced in the solenoid in such a fashion that a North pole is formed at the end near to the bar magnet's North pole, and the bar magnet gets repelled towards the other side.
The bar magnet then goes to the other end and then comes back (as a pendulum does) and again the same process is repeated. This should go on forever, and current should keep appearing across the load resistance.
How does the law of conservation of energy hold here?
| As the magnet approaches the solenoid, a current is induced. The current generates a magnetic field. The field repels the magnet, slowing it's approach. The amplitude of the oscillations diminish.
If there was no resistance, this would work in reverse as the magnet receded from the solenoid. The magnetic field would accelerate the magnet. The magnet would induce a current in the other direction, reducing the current to 0. This would reduce the field of the solenoid to 0. The amplitude would not diminish.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
Why do we have a TeV scale? When model building we don't want to introduce any new scales into our theory. We usually try to have new particles at the Higgs (TeV) scale (to solve the hierarchy problem), at the GUT scale, or at the Planck scale.
However, if the Higgs VEV already gives us a new scale, why would there not be new particles at some intermediate scale, for example say at $10^{5}$ TeV? In other words, what is unnatural about adding in new scales beyond the 3 that we are used to?
| If you want the new physics to solve the hierarchy problem, it's best if it is close to the weak scale, or else you will be left with a residual little hierarchy.
You are describing the "big desert" between the weak and GUT scales. I think it was motivated by the idea that SUSY lived at the weak scale, solving the hierarchy problem and insuring gauge coupling unification. Any physics between those scales would be unnecessary, could spoil unification or induce FCNC and proton decay.
I don't think the big desert hypothesis is particularly compelling, especially in light of the LHC results, but it is plausible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
electron levels in a high voltage conductor what is the electron energy level in a 300,000 volt power line? This voltage is way above the ionisation potential but electrons are not emitted from the wire.
| The electron energy in the line is $300kV$ * the electron charge $q$. Whether or not electrons are emitted depends on the gradient of the voltage, i.e the electric field, at the surface of the wire, which determines the force on the electron. See the wikipedia article on field electron emission.
A well-designed high-voltage line keeps the peak electric field low enough that emitted power (power that is wasted) is small enough to be economical.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Schrödinger's Equation and its complex conjugate I would like to know why there is a minus sign on the right-hand side of the Schrödinger's complex conjugate equation, whereas in the Schrödinger's equation there isn't. I know it is a simple question, but I don't know where this comes from.
$$
-\frac{\hbar^2 }{2m}\frac{\partial^2\psi}{\partial x^2} + V(x)\psi = i \hbar \frac{\partial \psi}{\partial t}
$$
$$
-\frac{\hbar^2 }{2m}\frac{\partial^2\psi^*}{\partial x^2} + V(x)\psi^* = -i \hbar \frac{\partial \psi^*}{\partial t}
$$
| I personally (maybe wrongly) see this feature as an early sign of the $CT$ symmetry where $C$ is the charge conjugate symmetry operation and $T$ is the time-reversal symmetry operation. Having no explicit charge in your equation, the charge conjugate symmetry operation would be simply taking the complex conjugate of the wave function while the $T$ operation would transform $t$ into $-t$.
You can, as a matter of fact, notice that the minus sign you are bothered with disappears if you perform this $T$ transformation.
Hence the Schrodinger equation is invariant under $CT$ operation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Bogoliubov transformation with a slight twist Given a Hamiltonian of the form
$$H=\sum_k \begin{pmatrix}a_k^\dagger & b_k^\dagger \end{pmatrix}
\begin{pmatrix}\omega_0 & \Omega f_k \\ \Omega f_k^* & \omega_0\end{pmatrix} \begin{pmatrix}a_k \\ b_k\end{pmatrix}, $$
where $a_k$ and $b_k$ are bosonic annihilation operators, $\omega_0$ and $\Omega$ are real constants and $f_k$ is a complex constant.
How does one diagonalise this with a Bogoliubov transformation? I've seen an excellent answer to a similar Phys.SE question here, but I'm not quite sure how it translates to this example. Any hints or pointers much appreciated.
| The Hamiltonian can be written as
$\sum_k \psi^\dagger M \psi$
where $\psi=\begin{pmatrix}a_k \\ b_k\end{pmatrix}$ and $M=\begin{pmatrix}\omega_0 & \Omega f_k^* \\ \Omega f_k & \omega_0\end{pmatrix}$.
We introduce a new set of operators $\phi=\begin{pmatrix}c_k \\ d_k\end{pmatrix}$, via $\psi=U \phi$ where $U$ is neccesarily a 2x2 matrix. This gives us
$\psi^\dagger M \psi = \phi^\dagger N \phi$
where $N = U^\dagger M U$. We wish for this new form of the Hamiltonian to be diagonal. aka we wish for the matrix $N$ to be diagonal. As per the standard process of diagonalising a matrix, a matrix $M$ is diagonalised by $M \rightarrow U^\dagger M U$ where $U$ is the matrix with the eigenvectors of $M$ as its columns.
Therefore, first we find the eigenvectors of $M$, substitute those as columns into a 2x2 matrix $U$, diagonalise $M$ so that $N=U^\dagger M U$, then our diagonalised Hamiltonian is
$H=\sum_k\phi^\dagger N \phi$
where $\phi=U^{-1} \psi$.
Thanks to @luming and @Vladimir for the pointers.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Electric field near surface of a conductor?
If the electric field on a positive charge is non-zero, then the charge accelerates in the direction of the field.
The field at the surface of a conductor is perpendicular to the surface. Why don't the surface charges accelerate away from the surface of the conductor then?
| In a conductor the electric field within itself is always equal to 0 and therefore all of the charge is found on the surfaces of the conductor, forming an equipotential. We are left with a surface charge density $\sigma$ and an electric field (close to the surface):
$$\vec{E}\approx \frac{\sigma}{2\varepsilon_{0}}\hat{n}$$
We are left with the natural question: Why do the surface charges not try to escape the boundary of the material they are contained in?
The answer is that they do, but there is a potential barrier (dependent upon the work function $\Phi$). A good way to think about this is imagining that an electron does manage to momentarily leave the surface of the object. The object will then be temporarily positively charged and the electron will experience an acceleration back towards the object it left (you can think of this using the method of image charges). Therefore, unless the electron leaves the surface with sufficient kinetic energy (as in the photoelectric effect) it will remain trapped within the confines of the conductor.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/103049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Superconducting wire in a Magnetic Field? A superconducting wire($SC$) is moved rapidly in a magnetic field( $1$ $Tesla$), what would happen to the wire? Are there any forces induced of attraction or repulsion?
In a typical conductor, we know that if it is moved around a magnetic field $-V$ is induced within the wire based on Faraday's law, however, with the condition of the $SC$ what could happen if $R = 0$ $ohms$?
Will Faraday's law still be applied to that wire with no resistance?
moving a $SC$ in a magnetic field will not induced $EMF$?
| The induced current will flow in such a way that the flux produced will tend to cancel the change in flux. According to traditional classical electrodynamics, the magnetic field does not do any work and it is the electric field and the charge carriers which do the work and ultimately limit Faraday's law in extreme cases.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/103104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Step in a proof that $\textrm{div} \ \mathbf{B} = 0$ from Biot-Savart's law Notation: The magnetic field $\mathbf{B}$ generated by a point charge $e$ moving with velocity $\mathbf{v}$ is given by Biot-Savart's law
$$\mathbf{B} = \frac{\mu_0 e\ \mathbf{v} \wedge \mathbf{r}}{4\pi r^3}$$ where $\mathbf{r}$ is the vector from the charge to the point at which the field is measured, $r = \left| \mathbf{r} \right|$, and $\wedge$ denotes vector product.
Question: According to my book: Since $\mathbf{r}\, / \, r^3 = - \,\textrm{grad} \left(1\, /\, r \right)$, we have $$\textrm{div} \left( \mathbf{v} \wedge \frac{\mathbf{r}}{r}\right) = \mathbf{v} \wedge \textrm{curl} \left( \textrm{grad} \frac{1}{r}\right) = 0.$$
What I have is $$\textrm{div} \left( \mathbf{v} \wedge \frac{\mathbf{r}}{r}\right) = \mathbf{v} \, . \, \textrm{curl} \left( \textrm{grad} \frac{1}{r}\right) - \left( \textrm{grad} \frac{1}{r} \right) \, . \, \textrm{curl} \ \mathbf{v},$$ so I think there is a typo in the book ($ \, \wedge$ should be $\, . \,)$. However, I still don't know how to go from my equation to the correct one. Is it because $\textrm{curl} \ \mathbf{v} = 0$? If so, why? I understand that the curl of a gradient vanishes identically.
| $\textrm{curl} \ \mathbf{v} = 0$, since $\mathbf{v}$ is not a function of the position at which the field is measured, i.e., it is not a function of the variable with respect to which we are differentiating.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/103226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Effect of linear terms on a QFT I was told when first learning QFT that linear terms in the Lagrangian are harmless and we can essentially just ignore them. However, I've recently seen in the linear sigma model,
\begin{equation}
{\cal L} = \frac{1}{2} \partial _\mu \phi _i \partial ^\mu \phi _i - \frac{m ^2 }{2} \phi _i \phi _i - \frac{ \lambda }{ 4} ( \phi _i \phi _i ) ^2
\end{equation}
with $m ^2 =-\mu^2 > 0$, adding a linear term in one of the fields $\phi_N$, does change the final results as you no longer have Goldstone bosons (since the $O(N)$ symmetry is broken to begin with).
Are there any other effects of linear terms that we should keep in mind or is this the only exception of the "forget about linear terms" rule?
| Linear terms are important. But in a Poincare covariant QFT, one can always remove them by shifting the field by a constant computed as a stationary point of the Lagrangian.
If there is only one stationary point, it must be a minimizer (to have the energy bounded below), then this gives a unique normal form without linear terms.
If there are multiple stationary points, not all of them are minimizers, but to get physical results, one must shift by a minimizer (usually the global minimizer). This is not always unique.
Thus linear terms can be ignored ''without loss of generality'' in the theory.
But in any particular model they cannot be ignored but must be properly removed by shifting the field, and this field shifting affects all other constants in the Lagrangian. Simply ''forgetting'' the linear terms would give wrong results.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/103328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.