Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
How to compute $L_{\rm eq}$ from temporal pressure data I have pressure vs time data. how can I compute sound equivalence Level $L_{\rm eq}$ in decibels? can this be done from a frequency spectrum assuming that it is constant?
| Based on this website, sound equivalence level is defined as
$$L_{eq} = 10 \log{\left( \frac{1}{p_{ref}^2} \frac{\int_0 ^T p_A(t)^2 dt}{T} \right)}$$
where $p_{ref}$ is a reference pressure (usually $20 \times 10^{-6} \text{ Pa}$), $p_A(t)$ is sound pressure, and $T$ is the time period of integration.
Given a frequency spectrum, you could use the inverse Fourier Transform to find the pressure as a function of time.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/119210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If we say the universe is expanding, shouldn't it be expanding relative to something? I don't understand, if everything in this world is relative to something else, then cannot we essentially say that nothing exists independently? We say that the universe is considered to be the ultimate 'background'. However, if we say the universe is expanding, shouldn't it be expanding relative to something?
| The universe is expanding, in the sense that things in it are getting farther apart. It is not expanding into anything because it already is everything. There simply is nowhere else to expand into.
Lets knock it down one dimension. Your universe is the surface of a balloon. The balloon is slowly being inflated. Your universe is getting bigger but nothing else is getting smaller (remember, you are unable to leave, look from, or perceive anything that is not on the surface). The only thing you can measure is points are further apart than they used to be.
Classical mechanics don't really work at the two extremes: the quantum level and the whole-universe level. If we ever fully understand the whole process I expect we will find that the complete equation applies across the board, but certain factors are negligible at human-perception levels. Motion is a good example here: we don't need relativity to calculate driving times, even though my watch does slow down when I drive to work.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/119312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Redshift 1+z - CMB Temperature lower? I know that $\frac{\lambda_2}{\lambda_1} = 1 + z$
Suppose a galaxy had redshfit $z=3$. Does this mean that the wavelength becomes $4\lambda$?
Then by wien's law where $\lambda \propto \frac{1}{T}$, does this mean that the temperature now observed is $\frac{1}{4} \times 2.73 K$?
| No, it does not. The redshift $z$ is defined, as you already know, by $$z+1=\frac{\lambda_{\text{now}}}{\lambda_{\text{then}}}$$
If we consider a photon that was emitted at redshift $z=3$ the formula yields $$\frac{\lambda_{\text{now}}}{\lambda_{\text{then}}}=4$$ this tells us that the wavelength we observe now is four times as big as the wavelength when the photon was emitted. This does not say anything about the wavelength of the photon as it was emitted (except through $\lambda_{\text{now}}$, which we may be able to determine experimentally).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/119448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Basic of Capacitor I've 2 capacitors; plate area, difference between plates and dielectric is same. Only thing is that the metal used in plates is different. Since the formula $\displaystyle C=\frac{\varepsilon A}{d}$ states, it won't affect but why?
| The only property of metals used in deriving $C=\varepsilon A/d$ is that they are perfect conductors. Ideally, all metals have this property. So even if you change the metal, it should not matter.
But if you use something other than metal, then it will of course change the capacitance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/119652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why doesn't De Broglie's wave equation work for photons? Well, as I am learning about quantum physics, one of the first topics I came across was De Broglie's wave equation. $$\frac{h}{mc} = \lambda$$
As is obvious, it relates the wavelength to the mass of an object. However, what came to my mind is the photon. Doesn't the photon have zero mass? Therefore, won't the wavelength be infinity and the particle nature of the particle non existent? Pretty sure there is a flaw in my thinking, please point it out to me!
| What you have there isn't actually de Broglie's equation for wavelength. The equation you should be using is
$$\lambda = \frac{h}{p}$$
And although photons have zero mass, they do have nonzero momentum $p = E/c$. So the wavelength relation works for photons too, you just have to use their momentum. As a side effect you can derive that $\lambda = hc/E$ for photons.
The equation you included in your question is something different: it gives the Compton wavelength of a particle, which is the wavelength of a photon that has the same electromagnetic energy as the particle's mass energy. In other words, a particle of mass $m$ has mass energy $mc^2$, and according to the formulas in my first paragraph, a photon of energy $mc^2$ will have a wavelength $\lambda = hc/mc^2 = h/mc$. The Compton wavelength is not the actual wavelength of the particle; it just shows up in the math of scattering calculations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/119729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Why is the constant velocity model used in a projectile motion derivation? I was re-studying university physics last week, I'm now in the chapter about kinematics in 2 dimensions and specifically the one treating projectile motion. In page 86 of his book (Serway - Physics for scientists and engineers) he derives the equation of the range of the projectile motion to be: $$R=\frac{{v_i}^2\sin2\theta_i}{g}$$
But I don't know why he used one of his assumptions
$\color{red}{\bf Question1:}$ Why $v_{xi}=x_{x\rlap\bigcirc B}$? Where $\rlap\bigcirc {\,\sf B}$ is the time when the projectile stops.
$\color{darkorange}{\bf Question2:}$ Why did he use the particle under constant velocity model to derive that formula, whereas here we deal with a projectile under constant acceleration?
Any responses are welcome, I'm disappointed a lot about those matters!
| If we assume there is no wind and air, the only force acting on a projectile in the air is the force of gravity. The force of gravity acts in vertical direction, thereby affecting the vertical component or Y component of velocity only.
Newton's 1st law says that a particle will continue to be in its state of rest or motion if no external force is impressed on it. Can you think of any force acting on the projectile in horizontal direction? There is none if we assume no wind/ air resistance etc. Thus there is no reason for velocity of the projectile to change in X direction. Hope this will make the above derivation simpler for you to understand.
You can also watch this video I made, in case it is still not clear
Analysing Projectile Motion in X and Y direction
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/119778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Help needed to interpret question - Spin States of electron pair in Helium?
For the last part, I'm not sure what they mean by "explain how to form eigenstates of the total spin $\hat S^2$ and $S^z = S_1^z + S_2^z$. Are they simply referring to the spin singlet and tripplet states?
I know for tripplet states total spin is 1, the states are $|S^2,m_{s1} + m_{s2} \rangle$: |1,1>,|1,-1> and |1,0> and for the singlet states total spin is 0, the only state is |0,0>.
| You know that the total spin operator is given by
$$ {\bf S}^2 = ({\bf S_1} + {\bf S_2})^2 = {\bf S_1}^2 + {\bf S_2}^2 + 2 {\bf S_1}\cdot {\bf S_2}$$
and the most general state is given by
$$\left|s,m\right> = a\left|\uparrow\uparrow\right> + b\left|\downarrow\downarrow\right> + c\left|\uparrow\downarrow\right> + d\left|\downarrow\uparrow\right>$$
You need to show that you can choose coefficients $a,b,c,d$ such that
$${\bf S}^2\left|s,m\right> = \lambda \left|s,m\right>$$
This is turns out to be the singlet and triplet states you mention above. To answer the question it might be enough to take these states and show that they are indeed eigenstates of ${\bf S}^2$ and $S_z$ by explicitly calculating ${\bf S}^2\left|s,m\right>$ and $S_z\left|s,m\right>$. Or you can calculate ${\bf S}^2\left|s,m\right>$ on the most general state and show that the only way it can be an eigenstate is if its the singlet or the triplet state.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/119850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Constraint and Applied forces In D'Alembert principle forces are classified into constraint and applied forces? Is this classification different from internal-external forces?
| Yes, they are different.
One must define what the system under study is. Usually, it consists of a number of sub-systems. A marble, for example, has as sub-systems many many atoms. Two point masses connected by a massless spring has two sub-systems. Internal forces are forces between sub-systems: the interatomic forces in a marble, and the spring force for the two point masses. External forces would be the force applied by agents external to the system: the earth in the case of gravity, my thumb in the case of a marble.
For the applied force, the agent and the functional form of the force (or potential) is specified. For the constraint, the agent might be known (a ramp, perhaps) but the functional form is not. Typically we limit the path or parameter space of the system in some way. We simply say that the constraint agent can provide whatever force is necessary to achieve the constrained motion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/119937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is the speed of light arbitrarily the limit? I know Einstein was great and all. Why is it that exactly at the speed of light is where infinite energy is required to accelerate any object with mass? Is it simply because the math of relativity checks out and explains most of everything? Are there any physicists who disagree with Einstein's theory?
| It's not actually. It's not like light has some special status in the Universe. It's just that there is a maximum speed, and light, among other things, tends to get very close to that limit. In practice light is slowed down by its environment, and there was even speculation at one point, that light mad a bit of mass and so neutrinos might move a tiny bit faster. This was retracted, but it goes to show that the thought is possible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/120067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Why would we need to ground an AC source I'm new to this field hence this weird question. Why would we need to ground an AC source? Why wouldn't it be enough to have just one pole to get an AC current going? I understand why it wouldn't work in DC case where current is flowing in one direction. However, in case of AC source where the current is not flowing anywhere but rather just oscillating back and forth it's not that clear to me why connecting load to only one pole wouldn't work? Thanks.
| If you have only one wire, leaving its end open, you still have current flowing in the conductor, but the efficiency will not be optimal due to impedance mismatch. Since the aim is to transport power, we need the other wire to optimize the efficiency.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Concrete example of a two-dimensional harmonic oscillator I am a student of mathematics and some time ago I showed in general that for a two-dimensional harmonic oscillator one can apply the recurrence theorem. So far so good.. now I would like to have a concrete example of a two-dimensional harmonic oscillator. Can you giuve me one?
| The harmonic oscillator is so incredibly important in physics as a whole because of the following consideration. For a more or less smooth potential $V(x)$ with a local minimum at position $x_0$, one can taylor-expand around that minimum:
$$V(x)\approx V(x_0)+\frac{V''(x_0)}{2}(x-x_0)^2+\mathcal{O}((x-x_0)^3).$$
The linear term is zero, otherwise it wouldn't be a minimum. Also, the coefficient in front of the quadratic term has to be positive for the same reason. Often, potentials are symmetric around local minima due to the symmetry of the problem, in which case all the odd-power terms will vanish as a rule, so the approximation is even better. The pendulum mentioned in the comments is an example for this. There, the potential has a cosine dependence on the displacement.
Due to energy considerations it is immediately clear that the region of applicability of the harmonic approximation (that's what it's called by the way) will not be left. Therefore, it is safe to replace the full potential by the quadratic ("harmonic") approximation.
For the desired two-dimensional harmonic oscillator, the argument is exactly the same. If it is supposed to be isotropic, meaning that the restoring force's magnitude only depends on the distance from the equilibrium point, not on the direction, the potential is rotationally symmetric around it's minimum (at least up to second order).
Exact harmonic oscillators are quite rare in physics. They're almost always approximations following the above scheme.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Is there a reaction force on the ball in a vertical circular motion? You have a light string. At one end of the string there is a ball modelled as a particle. The string has negligible mass. The other end of the string is fixed at a point and the ball is undergoing circular motion.
At the bottom, the centripetal force = mg - T.
At the top, the centripetal force = T + mg.
m = mass of ball
g = gravitational field strength
T = tension
If an exam asks you: What is the reaction force at the bottom?
What is the reaction force at the top?
Ball goes in a vertical circular path. Resistive forces = 0.
Is the reaction force always 0? The ball isn't in contact with anything but the string. Isn't the centrifugal force fictitious? Thanks.
Edit: Velocity is not constant due to conservation of energy, so the centripetal force is not constant. I was wondering if there is a reaction force acting on the ball, and if so, what is the value of that reaction force.
| A ball in a circular path is constantly accelerating(just a change in motion) so using $$F=ma$$
We can tell that if the ball has mass, and is moving in a circle, it will have some force exerted on it.
Since the ball is tied to the string, the string pulls on the ball with the same amount of pull, as the ball is pulling on the string. The force at the top of the string should be the same as the force at which the ball is pulling on the string.
I hope that helps.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Symmetry factor of $n$-point one-loop diagram If we have a one-loop diagram in $\phi ^ 3$ scalar field theory with $n$ external lines, then what is its symmetry factor?
I have drawn the diagram I am looking for, but instead of $6$ external lines, I want the diagram to have $n$ external lines. Please ignore the arrows in my diagram and assume that the external points are held fixed.
| Cheng and Li's appendix gives the generic symmetry factor $S^{-1}$ with
$$S=g\prod_{n\geq 2}2^{\beta}(n!)^{\alpha_n},$$
where $\alpha_n$ are the number of pairs of vertices connected by $n$ identical self-conjugate lines, $\beta$ is the number of lines connecting a vertex with itself, and $g$ is the number of permutations of vertices that leave the diagram unchanged with fixed external lines.
For your diagram, as long as the number of vertices $N>2$, all of the $\alpha_n=0$ (I suppose $\alpha_1=N$, but this doesn't affect the symmetry factor). You also have no tadpoles, so $\beta=0$. Finally, $g=1$ since you can't permute the vertices without changing the connectivity of the external lines. So the symmetry factor of the diagram is just one.
That is not to say that there aren't many ($(N-1)!$ in fact) other diagrams with the same kinematic structure that might need to be included in a final calculation of scattering amplitudes, just with permuted vertices.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What interaction is responsible for the 21 cm Hydrogen line transition? The 21 cm Hydrogen line is from the transition between the hyperfine levels of the ground state of the hydrogen atom.
So, what interaction is coupling the hyperfine levels?
I suspect that it is not the usual EM interaction, which causes the 1s-2p transition.
| The 1s energy level is split into two levels, one where the electron spin and nuclear spin are parallel, and the other where the electron spin and nuclear spin are antiparallel.
The 21 cm line is the transition between these two 1s levels.
The transition is a magnetic dipole transition.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How is sweating a pipe an example of capillary action? I learned how to sweat a pipe today from my father. If you're not familiar with the process, this might help.
One thing that jumps out at me is this line (from the above link, as well as my father's explanation)
Solder, which melts at low temperatures, wicks into a joint by capillary action and bonds with copper at the molecular level.
This doesn't seem to be quite right to me. I always thought of capillary action as something like putting a piece of paper vertically into a dye and watching the dye rise up the paper. I also didn't think the solder bonded with the copper at the molecular level, just that it melted and filled in the gaps really well. Wikipedia seems to agree with my definition of capillary action
Capillary action ... is the ability of a liquid to flow in narrow spaces without the assistance of, and in opposition to, external forces like gravity.
When sweating a pipe you place the solder above the pipe, letting it drip down (ie with the assistance of gravity) which makes it not capillary action... right?
Is there something I'm just not understanding here?
| Since you need to use flux on the pipe first the flux "primes" the connection so that when you heat the fitting and the flux burns off the solder is pulled into the joint to replace it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Why don't we consider electrostatic energy of the pair in the case of pair production? I have seen this Wikipedia article and many others, but in none of them I find any mention of the electro-static energy of the generated pair. Why?
I mean, the energy conservation should be written as
$h\nu = E_+ + E_- + Electrostatic\ Energy$
| It depends what you mean by "electrostatic energy".
When we are talking of pair production we are talking of physics at the quantum mechanics framework.
FEYNMAN DIAGRAMS for pair production by a gamma ray (left) or an electron (right). These represent the processes in the preceding sketch.
Lets take the simplest diagram on the left: a photon interacts with the electrostatic field of a nucleus , Z, by scattering off the field an electron and a positron appears to conserve lepton quantum numbers. All three outgoing (Z, e+, e-) conserve energy and momentum. The electrostatics are taken care of in this balanc
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Why doesn't a wall move when you push it if there's space behind it?
In the first screen you can see that if a person were to push a wall within a typical household the wall would not move while keeping themselves tractioned to the floor. If you push hard and do not traction yourself, you move back.
In screen two if you push a person with no space behind them they will move back, always, or at least have the potential. A 10 year old kid could push a 7'4", 680 lb. person back, but the wall doesn't move despite nothing being behind it. Here are the official questions:
*
*Why or how can the wall push you back?
*Why doesn't the wall move if there's nothing behind it to stop it?
*Why will a person always move when you push them, and they don't push back?
| The force of friction is greater in heavier objects. Moreover the object is attached to the ground. The force applied by us is not strong enough to pull it out. That's why the wall didn't move
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Do radio waves travel around the Earth or through it? Whenever you hear someone illustrating/describing the transmission of radio waves they always make it seem like they'd travel perfectly around the Earth to another distant location. For example, a radio wave sent by alternating current to be received from an intercepting device on the other side of the planet.
You'd think of something like:
Top frame shows how most media works, etc., would illustrate it: radio waves illustrated to go perfectly wrapped around Earth like a straight line until it gets to the other side.
Bottom frame shows how it should work (supposedly) if radio waves travel at all directions at the speed of light: will go right through the ground to the other side of the planet.
Basically, it will travel in all directions, but measuring how it gets there it should really appear like this, no?
To put the question in perspective, will radio waves go right through the ground and reach the other point near instantly, all on the other side of the Earth?
| Amateur "ham" radio operators who communicate with HF (3.0 to 30 MHz) frequencies can hear their own signal as it has circumnavigated the globe. This almost only happens with operators using Morse Code (CW) where the distinct signal can be heard and detected with sub-second intervals.
Also, ham operators make a distinction between short path (SP) and long path (LP) contacts. Short path is basically the shorter portion of the line (geodesic) between two points and LP is the longer portion of the line. Due to propagation changes such as due to the movement of the sun where the skip distance can change plus other effects that cause QSB (cyclic fading) some QSOs actually will use SP and LP at various times. Oh, a QSO is a communication conversation.
These effects of SP and LP comms along with hearing your own signal are well documented and even performed via lab quality instrumentation. It is though rare as the skip zones rob power from the signal so to do LP comms requires higher power (usually).
But, traveling through the Earth -- No.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Why is the outside run of high presure line on a ductless mini-split airconditioner insulated? On every ductless mini-split air conditioner I've ever seen, both the high and low pressure lines are insulated between the compressor and the building.
It seems like the liquid refrigerant coming out of the condensing coil can never be cooler than the outdoor ambient air temperature because the outdoor air is what is cooling it in the condenser coil.
It can, however, be warmer than the outdoor ambient temp. Therefore, it seems like leaving that high pressure line coming out of the compressor uninsulated would at worst save costs for some insulation, and at best give the high pressure refrigerant some additional cooling before it gets back to the indoor unit.
What am I missing here?
NOTE: I can think of some reasons why you'd want to insulate that line on the inside of the building where the high pressure liquid refrigerant would be warmer than the ambient indoor air temp.
| I can see the outside of an airconditioner, bought more than ten years ago. It is not an inverter, but it does heat in the winter. There is one well insulated tube entering the wall ( and a water tube coming out).
For heat pumps it makes senseto insulate well both lines, since the use is reversed in winter, and I do not think they any longer sell non heat pump air conditioners.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Fermion as a mixture of particle and antiparticle The solution to the Dirac equation (in the Dirac basis) are 4 coupled fields. The first 2 of them represent a particle (spin up/down), the other 2 fields are the antiparticle (spin up/down). When the particle is observed from its rest reference frame, the antiparticle solutions are zero. However once the particle is moving, all the 4 fields become coupled.
Does it mean that a moving electron is a little bit of a positron at the same time (in the given reference frame)?
| Strictly speaking, a "particle" is only a quantum notion which must be understood in the context of quantum field theory, as a asymptotic state "in" or "out", in some interaction.
So, classicaly, stictly speaking, there is no "particle"
You are speaking of the classical Dirac equation, which is a classical field equation.
The quantum field version of the Dirac equation, is an equation, where the fields are operators, and these operators apply on states. The "particles" are only some particular asymptotic states, in interactions, being on mass-shell. The field operators are mixing creation operators of particles, and destruction operators of anti-particles, because you cannot separate the two. So the classical version of the Dirac equation, which is a approximate view of the quantum version of the Dirac equation, is mixing too the degrees of freedom of pseudo-particles and pseudo-anti-particles. But keep in mind that only a quantum treatment is correct, where particles are states, and fields are operators.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Black hole thermodynamics in a time dependent metric For a time dependent space time metric, to get the thermodynamics, does the standard procedure of Wick rotating the time, and then calculating the free energy, work ?
| In principle it should not. The problem with the Wick rotation is that what you are doing is embedding the Lorentzian manifold in a complex manifold of which it is a slice, and then looking for a different slice with Riemannian signature. In general there is no such Riemmanian slice, and even if it does exist it need not be unique.
There is an old paper from Wald showing that for globally static spacetimes, such as Schwarzschild, everything goes smoothly, there is a unique slice and you can go on.
I'm not familiar with further improvements on this, besides this paper which gives necessary conditions for a spacetime to admit the Riemannian counterpart (very strong conditions indeed, the spacetime must posses totally geodesic three dimensional submanifold).
So you see that in general you have no reason to expect the Wick rotation to still work. But it may be the case that one can construct an example of time-evolving black hole that still has the Riemannian section. In any case it will propably not mantain the KMS condition (the periodicity in time of Green's functions) so that there is no well-defined temperature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/121980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Can a fundamental particle black hole with conserved charge emit Hawking radiation? Let's says there is a fundamental particle:
*
*That is so massive that it is a black hole by itself (Compton wavelength < Schwarzschild radius)
*That carries a conserved quantum number (e.g. charge of an exotic interaction) which no lighter particle carries
Would it be able to emit Hawking radiation? If not, does it contradict with the classical arguments (entropy analogy, pair creation at the horizon etc.) regarding the origin of Hawking radiation?
| This is really a comment, but it got too long for the comment box.
The problem is that the Hawking calculation is semi-classical. That is, it assumes the spacetime curvature is given by the (classical) Einstein equation. Once the radius of the event horizon decreases into the quantum regime the approximations Hawking used are no longer valid. You would need a proper quantum gravity calculation to make any progress.
In fact I'm not sure that your concept of a particle would be a valid description either. The only discussion I've seen of this type of physics was a string theory talk describing the final stages of black hole evaporation $^1$. This was some time ago, and to be honest I understood very little of the talk anyway, but my recollection is that the final stage of the black hole evaporation left behind a string in a highly excited state. The point is that at these sorts of energies it may well be that quantum field theory is not an adequate description of matter, so you can't postulate a particle with well defined properties. That would make your question as it stands somewhat devoid of physical meaning.
$^1$ Whether string theory is the correct description or not no-one knows - as I recall the presenter of the talk admitted it was highly speculative.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Will two perpendicular orbits settle into a disc? Scenario:
*
*one "fixed" object (like the sun...) of mass X
*two "planets" (P1 and P2) of mass Y
*P1's orbit is perpendicular to P2's orbit, and the sun is the center of both orbits
*P1 and P2 will never collide
My question is: Will this setting (P1 and P2 orbits describe a perfect circle in moment 0) ever be permanently "disc"-like (as we observe in galaxies, and not transitory)?
Is there a way to simulate this? Like a program or maybe something like wolfram-alpha?
| Note that circular orbits never exist in nature, and that the scenario that you described would be very rare. The reason why most solar systems have planets with coplanar orbits is because these planets were formed from an accretion disk, which as the name implies, is a very thin (but dense) cloud of dust and debris orbiting a star. That being said, it does stand to reason that the gravitational interactions between the planets would gradually (i.e. millions of years) pull them into coplanar orbits. I would test this hypothesis using the Universe Sandbox, a downloadable planetary physics simulation. If the $10 cost is off-putting, it should not be too hard to code your own simulation.
Note: For further information, see Two orbiting planets in perpendicular planes
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Concept of separation of charges in lightning clouds I have read in an article that when lightning strikes in the clouds the cloud disintegrates into two parts of which the negatively charged part is bigger than the positively charged one. Why does this happen?
Here is the screenshot of that article which is a textbook paragraph.
| I think you just missunderstood the textbook article. It says,
There are ice particles in the clouds, which grow, collide, fracture
and break apart. The smaller particles acquire positive charge and the
larger ones negative charge.
Not the clouds grow, collide, fracture and break apart, but the ice particles. In fact, the article is a bit simplistic on how this leads to a separation of charges. Upon further research, you'll find that it is highly non-trivial and there seem to be a variety of effects at work. This paper explores them in great detail. One of the major contributions seems to be due to supercooled water:
Updrafts in the thunder cloud transport small droplets of water upwards. As the droplets rise they cool down but cannot freeze yet. They become supercooled. Meanwhile, big "ice particle", i.e. hail, forms in the cold upper regions of the cloud and falls downwards due to gravity.
When the falling hail collides with the supercooled droplets, the latter freeze instantly, releasing their latent heat to the hail. Thus, the descending hail is always a bit warmer than its environment pushing it towards melting. The hail becomes also supercooled / "soft" and is then referred to as graupel.
Now, when the falling graupel collides with other rising droplets of water, it rips off electrons from the droplet. This site refers to it as a critical phenomenon, in. A more thorough description may be found here. However, as something related to the phase transition and the accompanying supercritical states of matter, it is bound to be complicated.
Bottom line is that the graupel acquires an additional negative charge due to the excess electrons. The droplet which were on their way up are missing it, so the upper part of the cloud is charged positively.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Melting and Boiling Points of Odd Materials In Chemistry, I was taught that there are three main states of matter: solid, liquid, and gas, and that heat and pressure determine that state. For some substances, the line is blurry between them.
Some materials don't seem intuitively to do this--nor have I been able to find data on them. For example, what is a reasonable estimate of a melting point for brick? What is the boiling point of paper? When will a carpet sublimate?
The common theme seems to be that these are all composite materials. Certainly all the elements have melting points (as applicable) and boiling points. Many compounds do too. However, something like cardboard is a mixture of fiber, glue, pigment, and possibly other things. Each of these might be made up of several compounds, with each compound having its own boiling point.
My suspicion is that for composite materials, individual compounds would exhibit properties roughly individually--so to melt wood, the water would boil off first, and then maybe it would start melting into a glucose-protein slag. Is this truly the right idea for what happens?
| I'm not sure I would focus on the the liquid to gas phase change. Sublimation, seems better behaved. I've heard that some alloys do not have well defined melting points. I don't know if something like that occurs in sublimation. But imagine a crystal lattice consisting of two fairly different substances. In this case it seems the material that sublimes at a lower temperature will induce the other to sublime at a much lower temperature. Since, after all the two substances would hold each other in the lattice.
One starts to wonder what exactly is the cause for a solid to gas, phase transition to occur. Just a small change in temperature causes an abrupt change of state. Apparently, entropy of the system and its surroundings is maximized by this phenomenon. In other words sublimation at an abrupt point is natures best way to maximize entropy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Why can't the work done by a non-conservative force be zero? Why can't the work done by a non-conservative force be zero? The displacement along a closed path is always zero. So, whatever be the type of force, variable or constant, the work has to be zero. Why do we need to calculate the work done for individual paths?
This is a non-conservative force that starts from $A$ moves via Path 1 to $B$ and then back to $A$ via Path 2. Since the displacement is anyways going to be zero, why can't work done be zero?
| For forces that change along the way, displacement is not the thing to calculate work with. Let $\gamma : [0,1] \rightarrow \mathbb{R}^3$ be the (closed or open) path that the particle the force is exerted on follows. Then, the work done along that path is
$$ W[\gamma,F] = \oint_\gamma \vec{F}(\vec{x})\cdot \mathrm{d}\vec{x}$$
which is a line integral. If $\vec{F}$ is conservative, there is a function $V(\vec{x})$ such that $\nabla V(\vec{x}) = \vec{F}(\vec{x})$, then we can apply Stokes' theorem (or, less fancy, the fundamental theorem of calculus) to calculate the work by
$$\oint_\gamma \vec{F}(\vec{x})\cdot \mathrm{d}\vec{x} = \int_{\partial\gamma} V(\vec{x}) = V(\gamma(1)) - V(\gamma(0))$$
For closed paths, $\gamma(1) = \gamma(0)$, so this is zero. If there is no potential with $\nabla V = \vec{F}$, we cannot apply this argument and have to actually calculate the line integral, which may be anything, especially not zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
How quickly should a fluid come to hydrostatic equilibrium? Let's say I'm holding a one-liter water bottle, full of water, which I then drop.
Before dropping the water bottle, the equilibrium is for there to be a pressure gradient in the water canceling the gravitational force on the water. While the bottle is in free fall, the new equilibrium is constant pressure everywhere. Should I expect the water to come to this new equilibrium in the few tenths of a second it takes the water bottle to fall?
I expect the answer is basically yes, because density changes (and therefore pressure changes) should propagate at around the speed of sound, and p-waves might bounce around a few times while exponentially dying away (depending on boundary conditions created by the material of the bottle?), at the end of which we have equilibrium. So for a 30-cm bottle with sound speed 1500 m/s, I might guess the time is a few times .02s, which is longer than the ~.5s it takes for the bottle to fall from my hand to the ground.
Does this sort of reasoning make any sense? How can I justify it in a less handwavy manner?
| You could start with a pack of cards and ask how long does it takes for the whole pack to free fall after
*
*the bottom card supporting the rest of the pack is released
*all cards are individually held from the sides, and then released at
the same time.
From this, I think the answer to your original question is that it depends upon
*
*the shape of the bottle which influences how it initially supports
the liquid
*the speed of the shock wave through the sides of the
bottle when it's released from rest
*the speed of sound in the liquid
I'd expect a 'v' shaped bottle to reach equilibrium in a "significantly" different way to one with straight sides.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
Uniformity in a solenoid I know the magnetic field strength increases as the number of turns in the solenoid increases.
However, I've learnt the field inside the solenoid is usually nearly uniform.
So, does the number of turns in the solenoid effect the uniformity of the field inside the solenoid? Does the field gets closer to uniform as the number of turns increases?
| It is all relative depending on one's measurement scale. A small homogeneous area may be found in a relatively short (length) to diameter (width) solenoid but the homogeneous area will be minute and perhaps not useful or measurable as homogeneous with a 'standard' Tesla/gaussmeter probe due to the probe's relatively large sensors to a relatively small homogeneous volume.
As commented above, from some MRI description and from measurements with two teslameters earlier today, the flux density (field strength) near the 'poles' of a solenoid vary wildly as one moves between the axis and the edge of the coil.
It is similar to variances on the surface of the poles of permanent magnets measured yesterday; The flux density at the centre is weaker than at the edges. The flux density on the surface increases the closer one measures to the edge of a magnet, and decreases towards and at the centre on the pole. The magnets/solenoids I have measured to date have been up to 20mm thick (v.short) and from 75mm sq to c.100mm diameter blocks, discs and coils. Thus as Floris says, a longer relative to diameter coil could give a larger (but might still be relatively small) completely uniform (homogeneous) flux density area inside the coil.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Could Legolas actually see that far? The video “How Far Can Legolas See?” by MinutePhysics recently went viral. The video states that although Legolas would in principle be able to count $105$ horsemen $24\text{ km}$ away, he shouldn't have been able to tell that their leader was very tall.
I understand that the main goal of MinutePhysics is mostly educational, and for that reason it assumes a simplified model for seeing. But if we consider a more detailed model for vision, it appears to me that even with human-size eyeballs and pupils$^\dagger$, one might actually be able to (in principle) distinguish smaller angles than the well known angular resolution:
$$\theta \approx 1.22 \frac \lambda D$$
So here's my question—using the facts that:
*
*Elves have two eyes (which might be useful as in e.g. the Very Large Array).
*Eyes can dynamically move and change the size of their pupils.
And assuming that:
*
*Legolas could do intensive image processing.
*The density of photoreceptor cells in Legolas's retina is not a limiting factor here.
*Elves are pretty much limited to visible light just as humans are.
*They had the cleanest air possible on Earth on that day.
How well could Legolas see those horsemen?
$^\dagger$ I'm not sure if this is an accurate description of elves in Tolkien's fantasy
| In the spirit of your question, having two eyes and assuming you can use them as an array (which requires measuring the phase of the light-something eyes don't do) allows you to use the distance between them for $D$ in the resolution equation. I don't know the spacing of an elf's eyes, so will use $6 cm$ for convenience. With violet light of $\lambda = 430 nm$, we get $\theta \approx 1.22\frac {430\cdot 10^{-9}}{0.06}=8.7\cdot 10^{-6}$. At a distance of $24 km$, this gives a resolution of $21 cm$. You can probably distinguish horsemen, but height estimation is very hard.
The other issue is curvature of the earth. If the earth radius is $6400 km$ you can draw a right triangle with legs $24, 6400$ and discover the other is $6400.045$, so he only needs to be on a $45 m$ high hill. Ground haze will be a problem.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "167",
"answer_count": 9,
"answer_id": 3
} |
Why does a mirror reverse polarization of circularly polarised light? A glass mirror (with metal backing layer) will reverse the polarisation of circularly polarised light upon reflection.
A polished piece of metal will also reverse the polarisation of circularly polarised light upon reflection. (I have tested and confirmed this for myself).
wikipedia states the reason a mirror will reverse the polarisation of circularly polarised light is:
...[A]s a result of the interaction of the electromagnetic field with the conducting surface of the mirror, both orthogonal components are effectively shifted by one half of a wavelength.
However, my understanding of mirrors is that only a polished piece of metal will phase shift a wavelength by half a wavelength, whereas a glass mirror (with metal backing layer) will not produce a phase shift. For example wikipedia which states:
According to Fresnel equations there is only a phase shift if n2 > n1 (n = refractive index). This is the case in the transition of air to reflector, but not from glass to reflector
| While rob is correct about the quantum mechanical picture I think that this case is at least as easy to understand as in the classical description.
Classically circular polarization can be described in terms of a time-varying linear polarization, so let's just look at two points on a wave.
I'm going to chose a beam in the $+z$ direction to examine two points on the wave: one where the polarization currently points along $+\hat{x} - \epsilon \hat{y}$, and a very short time later where the polarization is in the $+\hat{x} + \epsilon \hat{y}$ direction. The wave has right-handed circular polarization.
Now we let the beam bounce off a mirror in the $x\text{--}y$ plane. This reverses the direction of propagation but leaves the time-order in which are two points of interest pass any given point unaffected. A little thought suffices to show that the reflected wave has left-handed circular polarization.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Black and white matters. But why and how? I know black conducts heat while white reflects it.
But they are colors after all.
If a metal is painted black, it conducts more heat or at a rapid speed than it would do before it was coated.
But, as far as I know, colors don't have any special "substance" in them, which might trigger the sudden absorption of heat or reflection of the same.
What is the physics behind this? Are colors by themselves, some catalyst kinda thing?
| I believe that there are some incomplete/incorrect assumptions in this question: the bulk thermal conductivity of a metal will not be affected by a surface coating; it's response to radiative heating will be affected though.
This part of the difference in thermal response to radiative heating is based on the equality of emissivity and absorbtivity: white materials absorb little, and thus emit little radiation, conversely darker materials absorb and emit more electromagnetic radiation. You can see that this must be the case by considering putting the objects inside a "black-body cavity" held at a fixed temperature. Once the object reaches the equilibrium temperature -- for each amount of energy absorbed from the black body radiation filling the oven, and equal amount of energy has to be emitted. Thus absorbtivity=emissivity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
"answer_id": 2
} |
Work done in pulling a rope from a boat
A boy sitting on a boat pulls on a rope with a constant force $F$ over a duration of time $t$. The other end of the rope is either tied to a bridge or to another freely floating boat of equal mass. Does the boy do more work in the case of the bridge, or in the case of the other freely floating boat?
Here's my attempt: Work is defined as $\int F dx$. Since $F$ remains constant for both cases, we only need to analyze the difference in displacement. This is where I am confused. In the case of the bridge, only the boy is moving (the bridge is fixed). In the second case, the boy is moving towards the other boat, and the other boat is also moving towards the other boy. The relative speed of approach would seem to be greater in second case (exactly twice), and since $x = \int v dt$, the boy does more work in the second case.
Is my analysis sound? Is there a better way to deduce the result?
| You are mostly correct. Since you are only asked for a qualitative result, you can simplify what you said:
If the boy pulls on rope with force $F$ for time $t$, it will move a certain distance $d$. If the other end of the rope is fixed, the work done is $F\cdot d$. If the end of the rope is moving towards him, the total length of rope he reels in is greater than $d$ so more work is done.
Or from conservation of energy: the second boat ends up moving as well - work is needed to effect that motion. The only one doing any work in the system is the boy. So if he is moving two boats he must be doing more work.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/122952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What's the conserved quantities correspond to the generator of conformal transformation What's the conserved quantity corresponding to the generator of conformal transformations?
| The conformal charge in CFTs is a special case of any other standard treatment of classical field theory (e.g. chapter 1 of Peskin & Schroeder).
In flat space with $(1, n)$ signature and spacetime $(t,x_1, ..x_n)$ coordinates, we consider a conformal killing vector field with components $\epsilon^\mu$. We assume our field theory yields some energy-momentum tensor with components $T_{\mu\nu}$.
The current
$
j_\mu = T_{\mu\nu} \epsilon^\mu
$
satisfies $\partial^\mu j_\mu = 0$
As usual the conserved charge is
$ Q = \int dx^n j_0 $
is the conserved charge, where the integral is taken over the spatial dimensions.
In CFTs we often choose different coordinates in which these equations will not take these exact forms. Discussion of conformal charges are found in standard CFT books, for example Blumenhagen gives a concise and to the point discussion.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/123026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Black holes and Time Dilation at the horizon What is the difference between proper time and the observer time?
Whilst thinking about Black holes, when we see the Schwarzschild metric
$$c^2\tau ^2 = \left ( 1 - \frac{r_{s}}{r} \right )c^2t^2 - \frac{r^2}{1-\frac{r_{s}}{r}} - r^2d\Omega ^2,$$
and compare it with Einstein's special relativity equation,
$$c^2\tau ^2 = c^2t^2 - x^2,$$
we find that at the horizon of a black hole or at the schwarzschild radius for any infinitesimally small time spend by any object at the horizon the observers time tends to infinite
Why and how is this so? it doesnt make sense whilst trying to imagine it?
| This is essentially the same effect that you get in special relativity as the velocity approaches the speed of light.
If you take a clock and accelerate it towards the speed of light then it will run slowly. If you could get the clock to the speed of light (which you can't of course) then it would stop completely. To use your words for any infinitesimally small time spend by any object approaching the speed of light the observers time tends to infinity.
Indeed there is a sense in which any object falling into a black hole crosses the horizon at the speed of light. Suppose you are hovering just outside the event horizon at some radial coordinate $r$. We call this type of observer a shell observer. If you now watch an object falling freely from infinity then the speed the object passes you is:
$$ v = - \left( \frac{2M}{r} \right)^{1/2} $$
where I'm using geometric units so $c = 1$ and the event horizon is at $r_s = 2M$. As $r \rightarrow r_s$ the speed the falling object passes you approaches $c$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/123189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Fork and Sheet Lightning I've noticed that during British lightning storms, I have only ever seen sheet lightning
However, on holiday in other countries, I frequently see Forked Lightning
Q) Is this just caused by cloud, or is it two distinct forms of lightning?
Q) Is there a reason Britain only seems to get the sheet kind?
| As a long time resident of the UK, I can confirm that forked lightning is not uncommon in the UK, so I think that the OP has just been unlucky. Of course this is just anecdote.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/123284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why ONLY Maxwell's equations are the basic equations of electromagnetism? In electromagnetism we say that all the electromagnetic interactions are governed by the 4 golden rules of Maxwell. But I want to know: is this(to assume that there is no requirement of any other rule)only an assumption, a practical observation, or is there a deeper theoretical point behind it? Could there be a deeper theory behind assuming that there is not requirement of rules other than Maxwell's equations?
| I appreciate @ACuriousMind's comments on my question and thank him/her for pointing out the link that he/she has pointed out. I also apologize for being somewhat reluctant about mathematics in my comments when I posted the question.
The question was whether Maxwell's equations are the only equations that govern the electromagnetic interactions. With the assumption that the Lorentz force law is given (that is to say that it has been determined experimentally beyond any doubt), the question reduces to asking "Whether Maxwell's equations determine the electromagnetic fields completely?". Now, I think the answer is obvious and is answered by the famous and beautiful Helmholtz theorem. With all due respect, I wonder why no other answers chose to mention this simple and conclusive response and instead some chose to patronize the OP about how science works and what physics is about.
The Helmholtz theorem states that if $\nabla \cdot \vec{M}= U$ and $\nabla \times \vec{M}=\vec{V}$, $U$ and $\vec{V}$ both go to zero as $r \to \infty$ faster than $\displaystyle\frac{1}{r^2}$, and $\vec{M}$ itself goes to zero as $r \to \infty$ then $\vec{M}$ is uniquely and consistently determined in terms of $U$ and $\vec{V}$.
Once identifying $\vec{M}$ with $\vec{E}$ and then with $\vec{B}$, it is clear that the Helmholtz theorem dictates that the electromagnetic field is uniquely determined by Maxwell's equations and thus, there cannot be any additional new law of electrodynamics. Because there is nothing left to be described.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/123348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Analytical problems with Green's function I have a question about the right definition of the Green's function in physics. Why do we introduce (or not) an infinitesimal, positive number $\eta$ to the following definition:
$$\left[ i\hbar\frac{\partial}{\partial t} - \hat{H}(\mathbf{r}) \pm i\eta\right]G(\mathbf{r},t;\mathbf{r'},t') = \delta(\mathbf{r} - \mathbf{r'})\delta(t-t')$$
| I know complex analysis and the residuum theorem. This kind of thinks are introduce to evaluate some integrals of the form $\int\limits_{-\infty}^{+\infty}$ with some function which have a pole in $0$ and then You add an infinitesimal to evaluate an integral using residuum theorem. My question is why can't You just go through zero (where the pole is) make a small circle and then the Cauchy principal value limit ? There are some special problems with analytical properties of those Green's functions if I don't intrduce $\eta$?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/123587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
How to calculate required energy to displace a pendulum? How can one calculate the amount of energy needed to displace pendulum with given mass m and string length L to $\alpha$ degrees from resting position when acceleration due to gravity is known?
| By considering the variation in the potential energy $V=mg\Delta y$ due to the vertical displacement $\Delta y$.
A displacement of an angle $\alpha$ would move the mass of the pendulum horizontally by $\Delta x=L \sin\alpha$ and vertically by $\Delta y=L(1- \cos\alpha)$.
Therefore the energy required for a displacement of an angle $\alpha$ amounts to $mgL( 1-\cos\alpha)$.
If the pendulum is a bar or more generally an object of any shape, of course, you replace L by the distance from the pivot to the centre of mass.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/123660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How is heat transferred to a thermometer? Quick question. I can't seem to find a satisfactory answer online. How does a thermometer measure the average kinetic energy of atmospheric air? I assume that the energy is transferred by molecular collisions, and this somehow raises the temperature of the alcohol by doing work on the thermometer. Is this correct? Somehow a thermometer acts as a speedometer right?
| Taking out your last analogy about the speedometer (which I don't find useful but it might work for you), I would add that in a sealed thermometer, thermal equilibrium between the external media and the alcohol is mostly reached by exchange of electromagnetic radiation (photons). But heating or cooling or the glass molecules by atmospheric gas and then from the glass to the alcohol also plays a role, albeit minimal in most circumstances.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/123723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Derivation of (2.45) in Peskin and Schroeder I'm having trouble understanding the step
$$\left[\pi (\vec{x},t),\int d^{3}y ~(\frac{1}{2} \pi (\vec{y},t)^{2}+\frac{1}{2}\phi (\vec{y},t)(-\nabla^{2} +m^{2})\phi (\vec{y},t)) \right]$$ $$ =\int d^{3}y ~(-i\delta^{(3)}(\vec{x}-\vec{y})(-\nabla^{2} +m^{2})\phi (\vec{y},t)) $$
I've tried using the relations $$[\phi (\vec{x},t), \pi (\vec{y},t)] = i\delta^{(3)}(\vec{x}-\vec{y})$$ and $$[A,BC] = [A,B]C + B[A,C], $$ but run into $$[\pi (\vec{x},t), (-\nabla^{2} +m^{2})\phi (\vec{y},t)] ,$$ which I don't know how to evaluate.
Any help would be appreciated.
| This is really straight forward, once you get used to the notation. (Don't you hate it when people say that?)
$$[\pi (\vec{x},t), (-\nabla^{2} +m^{2})\phi (\vec{y},t)] ,$$
Here you need to remember that $\nabla^2$ acts on the $\phi(\vec{y},t)$ only, so $\pi$ can pass right through this wave operator. Now when you evaluate the commutator you'll end up with something like $\phi (\vec{y},t)(-\nabla^{2} +m^{2})\delta^{(3)}(\vec x-\vec y)$, after which you use "self-adjointness" of $\nabla^2$ (really, integration by parts), to make the wave operator act on the first $\phi$. You might need to relabel variables afterwards.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/123848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Can matter be created from energy?
The small, hot, dense early universe the size of an atom was made up entirely of energy, it wasn't until after the expansion began and the universe cooled down some of that energy began converting into the first atomic nuclei.
This quote seems a little dubious but I think this is worth asking anyway:
Can someone explain the atomic process, if it even exists, of how this would work to convert energy in to matter, and what form of energy was initially present, and what is required to cause this change?
| They are actually trying to do so in the lab. They need a very potent laser (I believe they didn't reach the critical power yet).
The idea is simple, inside the lase cavity you generate a very powerful electromagnetic field, powerful enough such that the photos have enough energy to transform a virtual electron-positron pair into a real one. See this link for more detail:
http://www.kurzweilai.net/scientists-discover-how-to-turn-light-into-matter-after-80-year-quest
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/126154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can two single particles interfere with each other? Groups of particles can interfere with one another; In the double slit experiment when measuring single photons at the screen each one arrives at the screen in a random manner and they only show the interference pattern once several particles are detected.
Obviously two waves can obviously interfere with one another, but can two single particles interfere with one another? Cohen-Tannoudji writes that
light simultaneously behaves like a wave and like a flux of particles
But do two particles constitute a flux of particles? I doubt that this could be tested experimentally but if it were so would this constitute a measurement for each particle?
Here was my thought (Disclaimer:I do not have a good idea of what the interference of two single particle states is):
To have interference of two single particles you would have to know something about there position to be able to describe their interference pattern hence the measurement.
| Classical particles such as electron, proton and neutron of course interfere with each over. No doubt. Photons interfere too. See https://en.wikipedia.org/wiki/Photon_bunching.
If you ask about the double slit experiment with single photons the common answer is yes. A single photon interfere at the position of the slits with itself and that is the reason why on an observation screen appears fringes.
But please take in attention that even a single slit and every edge too produces fringes near the shadow. Every edge interfer with particels in the region between the edges material and the free space. A second fact for the influence of edges to the particles is the fact that if you place two crossbred polarizers no light is coming through. Now place a third one under 45° between the overs and you see light behind all the polarizers.
The double slit experiment was made with single electrons too. As result we get the fringes too what was interpreted as interference of the electron with itself. This is a little bit strange because there are the sciences about QED etc. where the fields between the particles (and the edges are made in first line from electrons ergo particles) is part of the calculations.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/126269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Would computers accelerated to high speeds compute "faster" from our point of view? I woke up to this thought yesterday:
Lets say Computers A and B have exactly the same specifications and at time T both are set to process an algorith that would normally take exactly 1 year and exactly at T the A computer is accelerated to 0.5c (or anything c). Both are set to automatically broadcast the result to a central computer.
*
*Would A finish processing first from my point of view?
*Would we be able to receive the broadcast from A?
*Would it matter if A were travelling at a straight line or orbiting a
planet or even a star system?
*Would it be feasible to accelerate a computer to "compress time" on a
Machine on the likes of the LHC.
Excuse me part: I'm sorry if this question is not fit for this StackExchange , I'm sure someone asked that , but I don't know how to look for this- I'm really new to all things physics.
| 1) No, because it's actually going slower from your perspective. In special relativity, "the fastest wristwatch is always your own".
2) Yes, but remember that it's farther away from us now, so it will take some time to get to us (if it was travelling at 0.5c it will take 50% longer to get to us).
3) Mostly in that as an observer the redshift effect would be different.
4) It would be feasible to accelerate to dialate time, but that wouldn't be useful.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/126479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 2
} |
Trying to combine red, green and blue to produce white I tried to mimic the mechanism of typical screens to produce white color out of red, green and blue.
What I did is displayed the attached image on the screen, and moved far away as to let the diffraction effects take place, so that the three colors appear as if they're coming from the same point.
Nonetheless, quite paradoxically, what I have seen was black instead of white.
I don't know if this question fits this place, so excuse me.
| While David Hamman's answer is correct, I wanted to expand a little bit on his answer:
When you use a CRT, you are looking at emitted light. In the case of emission, there is no "absolute" white - something will only look gray in comparison to something else with the same color ratio but brighter. When you turn up the brightness on your monitor, white is still white - just "whiter". In a dark room, even a CRT whose intensity is turned down a lot appears to render white as white.
By contrast when you have a sheet of paper with colors on it, what is really happening is that you have absorption of certain wavelengths / colors. A blue piece of paper absorbs everything except blue; ditto for red, and green. So what you have on your three rectangles is three pieces of paper that absorb 2/3 of the incident light:
red green blue
red 100 0 0
green 0 100 0
blue 0 0 100
When you illuminate the entire card with 100% white light, you get a red reflection from just 1/3 of the card - and the same for green and blue. If you have these three cards against a white background, the background will appear approximately 3x brighter as every part of the background reflects all three colors.
If you made the background gray (which is really just a "darker shade of white"), and shone a bright light onto the setup, you could actually persuade yourself that the background, and the three color sample, were white.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/126536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 1
} |
Ground state Phase Diagram of Bose-Hubbard Model The Hamiltonian of Bose-Hubbard model reads as
$$H=-J\sum\limits_{<i,j>}b_i^{\dagger}b_j+h.c.+\frac{U}{2}\sum\limits_{i}n_i(n_i-1)-\mu\sum\limits_in_i~~~~~~~~~(1)$$
For this we plot phase diagram in ( $J/U$, $\mu/U$ ) space.
Same way if I want to plot phase diagram of Hamiltonian which looks like
$$H=-J\sum\limits_{<i,j>}b_i^{\dagger}b_j+h.c.+\frac{U}{2}\sum\limits_{i}n_i(n_i-1)~~~~~~~~~(2)$$
How to get phase diagram of such hamiltonian? I am solving this model Numerically by Exact Diagonalisation.
| In principle, it is very simple and straightforward.
The problem is to map out the region where the integer filling state is the ground state. Suppose you have $L$ sites. Take $N=L$ particles, find its ground state energy, which is denoted as $E_g(L)$. Note that here the Hamiltonian does not contain the $\mu $ term. Do it again for $N=L+1$, the ground state energy is $E_g(L+1)$. Then, you know below the line
$\mu_+ = E_g(L+1)- E_g(L)$
the $N=L$ state is the lower state with respect to the full hamiltonian containing $\mu$.
Do it once again with $N=L-1$, then you know above the line
$\mu_- = E_g(L)- E_g(L-1)$
the $N=L$ state is the lower state.
Therefore, between the two lines, the $N=L$ state is the lowest state. In this region, the unity filling state is the ground state. This is the first Mott lobe.
The idea is simple, but i really doubt you can get accurate results with ED. You had better do it with DMRG.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/126584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does a wine glass with less water resonate at a higher frequency? In this video https://www.youtube.com/watch?v=hWwM7F-zaHs, Professor Lewin showed that for the tube, the less water there is, the longer the effective length of the tube and therefore, the lower the frequency.
He then demonstrates an opposite effect for a wine glass. Namely, an empty wine glass resonates at a higher frequency than a filled one. Why is that so?
| For the pipe, it is the air that is vibrating. When the column of air is shorter the frequency is higher.
In the case of the wine glass, the glass (not the air) is vibrating. Add water and you increase the inertia of the glass, which lowers the frequency of the resonance. The air may also resonate - but for something the size of a wine glass the frequency is very high - inaudible compared to the vibration of the glass.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/126832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Is there a difference between "average acceleration" and centripetal acceleration? Question adapted from Examkrackers MCAT prep book:
A particle moves along a half circle (diameter=$10\text{ m}$) at a constant speed of $1\text{ m/s}$. What is the average acceleration of the particle as it moves from one side of the half circle to the other side?
A. $0$
B. $0.2/\pi$
C. $0.4/\pi$
D. $1$
The book says C is correct. Acceleration is change in velocity divided by time. Initial velocity is $1\text{ m/s}$ up; final velocity is $1\text{ m/s}$ down. The change in velocity is therefore $2\text{ m/s}$. The time is found from speed equals distance divided by time. Distance is $2\pi r/2$. Thus
$$a= \frac{2}{(2\pi(5)/2)/1} = \frac{2}{5\pi} = 0.4/\pi$$
I thought all of the answers were wrong because I thought they should've used the centripetal acceleration equation: $a= v^2/r$
SO my question like the title: Is there a difference between "average acceleration" and centripetal acceleration?
I searched for a couple hours and couldn't find this issue directly addressed.
| Centripetal acceleration is a type of acceleration but Average acceleration is the calculation of acceleration. They are different.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/126970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Is there a metatime required for space-time to change? Space-time is thought to curve and ripple. Is a kind of metatime required in or during
which such events take place?
|
Space-time is thought to curve and ripple.
Space-time, i.e. the set of all events under consideration (specificly: coincidence events), together with all relations between these events (primarily: by listing who, among all principal identifiable participants, took part any one coincident event), is thought to be not necessarily homogenous and/or isotropic by definition, but (possibly) to consist of distinct regions which (may) differ in terms of suitable measures, such as curvature.
Is a kind of metatime required in or during which such [differences and distinctions] take place?
No: at least some measures of (possible) differences are defined intrinsically, appealing only to participants and the coincidence events in which they took part. Some examples are sketched
*
* here ("Which causal structures are absent from any “nice” patch of Minkowski space?") and
* here ("Can the vanishing of the Riemann tensor be determined from causal relations?").
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Sum of Green's functions in condensed matter I am working on the Ginzburg-Landau model for Charge density waves, and I am carrying out the sum of Green's functions to calculate the terms in the GL model.
Is the sum's order over $ \vec{k} $ (or eventually $ \vec{r} $) and $\omega_n$ important? Mathematically the question is the following,
$$
\sum_{\vec{k}} \sum_{\omega_n} \stackrel{?}{=} \sum_{\omega_n} \sum_{\vec{k}} \, .
$$
If it is not, when does it happen or under which conditions there is a difference?
| If the whole summation converges, then the two summations commute.
For example, the following summation diverges, so the two summations do not commute,
$$\sum_{i\omega}\sum_k\frac{1}{(i\omega-k)^2}.$$
However if we consider the following convergent summations, you can change the summation order freely.
$$\sum_{i\omega}\sum_k\frac{1}{((i\omega)^2-k^2)^2}.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Ampere's law and Biot-Savart law gives different terms for magnetic field in middle of a current running in a loop
I would like if someone could clarify this issue for me:
When dealing with a current $I$ running in a loop with radius $R$ and looking for the magnetic field in the middle of the loop.
By using Ampere's law, I know that the current $I$ runs through a loop with the same radius $R$, we get that:
$$\oint_c\vec{B} \cdot d\vec{l} = \mu_0 I_{enc} $$
$$B= \frac{\mu_0 I}{2\pi R}$$
and when using Biot–Savart we get that
$$d\vec{l} \cdot \vec{r} = |d\vec{l}||\vec{r}|\sin(\frac{\pi}{2})$$ obtaining:
$$B = \frac{\mu_0 I}{2R}$$
Which is not the same result as with Ampere's law.
I obviously miss something, maybe I can't use Ampere's law?
Anyway, if someone could help me out here I would really appreciate it.
Thanks.
| Your calculation of $B$ using Ampere's Law is not correct. The integration surface should have its area perpendicular to the current and should have one of the sides go parallel to $\vec{B}$. The choice of surface, when using that form of Ampere's Law, is usually a square or rectangle. That's why using it for a single loop does not work, because $\vec{B}$ is not uniform at the center of a single loop, meaning it does not point in the same direction along a line. Ampere's Law is very useful if you want to calculate $\vec{B}$ inside a solenoid, because it is approximately uniform in this case. See this example.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Why are most accurate reflective surfaces not white? Since white is our perception of reflection of light, why are mirrors and other metallic objects that are more grey or silvery capable more reflective than a white object?
This is somewhat related: What is the difference between a white object and a mirror? but I'm wondering why more white objects aren't mirrors.
| This is similar to why a glass of water and a glass of milk look different. We can use geometric optics (used in 3D animation ray tracing) to explain this.
When a light ray (of any color) hits a mirror, it reflects as another light ray. But for a white nonreflective surface, it is scattered in all directions.
From the perception side, a eye gathers incoming light rays to form image on the retina. From a mirror, a light ray on a point originated from a single light ray from elsewhere. From a white surface, a light ray originates from contributions from multiple directions, which usually averaged out to white, i.e., combination of wavelengths that stimulates all color pigments in the retina.
Note that I'm assuming our eye is a pin hole camera, so a point in the retina image corresponds to a single light ray direction. The purpose of the lens though is gather multiple light rays (in multiple directions) eminating from a point back to a point again. So if your eyes are unfocused, colors can get blended.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How would an X-ray scanner identify a mirror? A mirror is under normal circumstance used to reflect Electromagnetic radiation also known as photons (light) and in airport security or medical facilities, they use X-rays to detect anomalies inside objects or bodies to detect narcotics or injuries. However, I always wonder what if I add a mirror inside the luggage or put a mirror in front of me during scanning?
That in mind, how would an X-ray scanner see the mirror? Would it be invisible? I am sure I am not the first one to think of this, as a lot of security and criminals thought of this, however I never got an answer, so can someone tell me please?
If there are X-ray reflecting mirrors? Why don't Airport security ban these items and Mirrors all together? Would X-Ray mirror look like a normal mirror? Do they reflect visible light spectrum as well?
| dmckee points out that an ordinary mirror doesn't reflect X-rays, but if you could find an X-ray mirror and put it in your case it would just appear black.
When you look at yourself in the mirror you're seeing light from your skin/clothes that hits the mirror and is reflected back towards your eyes. But airport X-ray machines work by passing X-rays through the luggage. The X-rays are emitted on one side and detected on the other. An X-ray mirror would reflect the transmitted X-rays and stop them passing through your case, so the X-ray detector would just see it as a black sheet.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 3,
"answer_id": 1
} |
My basis set isn't orthonormal? I'm implementing a little QM calculation just for fun and to make sure I understand how it works (calculating the helium ground state energy). My problem is that my basis set doesn't seem to be orthonormal. I'm using the spherical harmonics for the angular part and the Slater type radial function. If I integrate just the angular part
$$\int_{\phi = 0}^{2\pi}\int_{\theta = 0}^\pi Y_l^{m*}(\theta, \phi) Y_{l'}^{m'}(\theta, \phi) \sin(\theta)\ d\theta \ d\phi = \delta_{ll'}\delta_{mm'}$$
then the result satisfies the requirement for orthonormality. But if I include the radial Slater component:
$$\int_{\phi = 0}^{2\pi} \int_{\theta = 0}^\pi \int_{r = 0}^{\infty} C r^{n-1} e^{-\alpha r} Y_l^{m*}(\theta, \phi) \cdot C r^{n'-1} e^{-\alpha r}Y_{l'}^{m'}(\theta, \phi) \cdot r^2 \sin(\theta)\ dr \ d\theta \ d\phi = \delta_{ll'} \delta_{mm'} \frac{(n + n')!}{\sqrt{(2n)!(2n')!}}$$
with
$$C = \left((2\alpha)^n\sqrt{\frac{2\alpha}{(2n)!}}\right)$$
I get some extra factor instead of what should be $\delta_{nn'}$. I've double-checked my references and can't figure out what I'm doing wrong.
Thanks!
| Indeed, the Slater-type orbitals (radial wave functions) are not orthonormal – they are not even orthogonal to each other. The $\delta_{n,n'}$ Kronecker delta symbol doesn't appear in the inner product and it can't because the $r$-dependent integrand is positively definite and there is no room for cancellation.
Their not being orthogonal physically means that the orbitals for different $n$ are not mutually exclusive. For a given molecule, one is supposed to use one value of $n$ only.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
"Lack of inversion symmetry" in crystal? Apparently (first paragraph of this article) the lack of inversion symmetry is some crystals allows all sort of nonlinear optic phenomena.
Now. Does anyone know of an intuitive or just physical explanation as to why this is the case?
What does inversion symmetry mean and what is so special about it?
| In nonlinear optics, the typical approach seems to be: take the relation between the polarization and electric field $P=\epsilon_0 \chi E$ and start adding correctional terms based on the Taylor series.
$$P=P_0+\epsilon_0 \chi^{(1)} E+\epsilon_0 \chi^{(2)} E^2 +\epsilon_0 \chi^{(3)} E^3 +...$$
This particular phenomena, second harmonic generation, can only occur from the even terms in the series. Spacial inversion occurs from making the replacement $\textbf{r} \rightarrow -\textbf{r}$, which implies $P\rightarrow-P$ and $E\rightarrow -E$ (since they're vectors). If spacial inversion is symmetric (aka the relation between $P$ and $E$ is unchanged after inversion), then this implies the even terms must be 0, $\chi^{(even)}=0$.
To my understanding, there isn't a deep meaning about it. Inversion symmetry is just a jargon way of saying the even terms are 0. Edit: Reading the other answers, there is some meaning behind it. The inversion symmetry is based on whether the crystal structure is symmetric under inversion.
Source: Non-linear optics isn't really my area. I found the answer on page 21 of the book Extreme Nonlinear Optics: An Introduction, on Google Books.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 0
} |
Proof of Liouville's theorem: Relation between phase space volume and probability distribution function I understand the proof of Liouville's theorem to the point where we conclude that Hamiltonian flow in phase-space is volume preserving as we flow in the phase space. Meaning the total derivative of any initial volume element is 0.
From here, how do we say that probability distribution function is constant as we flow in the phase-space?
What's the relation between phase space volume and the density function, which instantaneously tells us the probability of finding the system in a neighborhood in phase-space?
| I do not think probability distributions are preserved by the Hamiltonian flow...consider a probability distribution that is a $\delta$-function on the phase space at initial time (you have just one point with probability one), so it is a particle with fixed coordinate and momentum. If you evolve in time by the Hamiltonian flow, you will find yourself at the phase space point corresponding to the evolved position and momentum of the particle. It corresponds again to a $\delta$-function probability distribution, but on a different point, so different from the starting one.
I think you should characterize the evolution of probability distributions with the aid of measures. A probability distribution has a mathematical meaning as a probability measure $\mu$ on the phase space $\mathscr{Z}$. You have that $\mu(\mathscr{Z})=1$ (total probability is one). Given that the initial distribution is the measure $\mu_0$, and calling $\Phi(t)$ the Hamiltonian flow, you should get the measure at time $t$ as the pushforward of the initial measure by the flow: $\mu_t=\Phi(t)_*\mu_0$.
However, I am not completely sure about that, I hope to get some feedback and eventual corrections from someone more expert on classical statistical mechanics ;-)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Why is the pressure inside a soap bubble higher than outside? Apparently, the air inside a soap bubble is under higher pressure than the surrounding air. This is for instance apparent in the sound bubbles make when they burst. Why is the pressure inside the bubble higher in the first place?
| The increased pressure is caused by the surface tension between the soap and the surrounding air. This can be seen by a simple equilibrium energy argument. The total energy of the system reads
$$
E = E_i + E_o + E_s \;,
$$
where $E_i$ is the energy associated with the air inside the bubble, $E_s$ is the interfacial energy, and $E_o$ denotes the energy associated with the air outside of the bubble. Importantly, the contribution of the surface energy is given by $E_s = 2 \gamma A$, where $\gamma$ is the surface tension and $A$ is the surface area of the bubble. The factor of 2 emerges, since there are actually two interfaces (one facing the inside of the soap bubble and one facing the outside).
In equilibrium, the total energy will be minimal. We thus analyze the total differential of the energy. Here, the differentials of the partial energies of the air can be approximated by the ideal gas law, which yields $dE_i = -p_i dV_i$ and $dE_o = -p_o dV_o$.
Next, we have to discuss the degrees of freedom of the system. Generally, the soap bubble wants to keep its spherical shape to minimize the surface area (and thus the surface energy $E_s$) at a given volume. This leaves us with a single parameter, the radius $R$ of the bubble, which can be varied in any process. The volume differentials then become $dV_1 = 4\pi R^2 dR$ and $dV_2 = -4\pi R^2 dR$. Additionally, the surface area changes by $dA = 8\pi R dR$. The differential of the surface energy thus reads $dE_s = 2\gamma 8\pi R dR$, provided that the surface tension stays constant.
Now we got everything and can express the differential of the total energy as
$$
dE = -p_i 4 \pi R^2 dR + p_o 4\pi R^2 dR + 2\gamma 8\pi R dR \;.
$$
Dividing by $4\pi R^2$ and noting that $dE/dR$ vanishes in equilibrium, we thus arrive at
$$
p_i - p_o = \frac{4\gamma}{R} \;.
$$
This expression shows that the pressure inside the bubble is large than outside. The pressure difference is two times the Laplace pressure $2\gamma/R$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 4,
"answer_id": 2
} |
Is speed of light and sound rational or irrational in nature? Just as circumference of circle will remain $\pi$ for unit diameter, no matter what standard unit we take, are the speeds of light and sound irrational or rational in nature ?
I'm talking about theoretical speeds and not empirical, which of course are rational numbers.
| It depends on the unit you want to express it.
If you choose c/100 as the speed unit, c will be expressed with a rational number. If you choose c/π, you'll have an irrational one.
That depends on measure, not on nature.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 0
} |
Which ball touches the ground first? This is a very well known problem, but I can't find an answer in the specific case I'm looking for.
Let's consider two balls :
*
*Ball 1 weighs 10 kg
*Ball 2 weighs 1 kg
*Balls have identical volumes (so Ball 1 is much more dense)
*Balls have identical shapes (perfect spheres)
Let's drop them from a rather important height, on earth, WITH air. (That's the important thing, because all the proofs that I browse take place in a vaccum).
I am arguing with a colleague. He thinks that ball 1 will fall faster in air, and that the two balls will fall at the same speed in a vacuum. I think that the identical shapes and volumes make air friction identical too and that the vaccum has no importance here. Could someone tell who's right and provide a small proof?
| This problem can be easily solved by the formula “F=ma”.
You must be familiar with the reason why it would fall at the same rate in vacuum.
But if we talk about the free fall in atmosphere, as you said there will be friction off course, and as the objects have the same shape, it’ll be same.
As the force of friction is the same on the two bodies, the one with the larger mass will have a smaller (negative) acceleration , and the one with the smaller mass have a bigger (negative) acceleration. So the ball with smaller mass will be slowed down by a great extent ( than the ball with larger mass).
ALWAYS remember, F=ma. Force depends ONLY on the mass and NOT on the density!
PS - I don’t know why are others making the problem so complicated with those formulas!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
"answer_count": 6,
"answer_id": 5
} |
Is it possible to produce images of pair production in home-made cloud chamber? There are some nice pictures on the web showing the counter-spiralling paths of an electron positron pair produced in a bubble chamber with a uniform magnetic field, for example:-
(source: bigganblog.com)
Would it be practical to produce a cloud chamber and [Helmholtz coils] capable of imaging such events in an ordinary domestic garage? What sort of source would I need? How often might such events be detected?
| These electron-positron pairs are created by gamma rays. I don't know anything about how to make a cloud chamber, but detecting cosmic gamma rays at the surface of the Earth is very very rare. The atmosphere is very opaque to gamma rays (Source). Cosmic gamma rays burst are commonly detected on satellites orbiting the Earth, but very few make it to the surface.
These images of pair productions are likely from gamma rays created at particle accelerators.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Calculate water flow rate through orifice I'm not very good with fluid physics, and need some help. Imagine the following setup with water contained in-front of a wall with an opening on the bottom:
How do I calculate the water flow $Q$?. I have made some re-search and found I need to (partially) calculate the pressure across the opening (orifice). But I don't know the pressure on the back side of the orifice. Can this be solved in any way?
Note: I'm not saying "please give me the solution, I'm lazy". I want to figure it out myself. But since, in this case, I only found formulas involving calculating pressure drop, I canno't use them to solve the problem. Therefore I'm turning my face to you, to see if there's another way to solve this problem.
Update: The "tank" holding the water is actually a big lake, and the opening is how much the water gate have opened. I need to very precisely calculate how much water flows through the opening.
| The NCEES: FE Reference Handbook has some good material on fluid flow through a submerged orifice in its fluid mechanics section. You can search for it online. NCEES will provide you with one free of charge.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/127954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
About the definition of expectation value in quantum mechanics In quantum mechanics, the expectation value of a observable $A$ is defined as
$$\int\Psi^*\hat A\Psi$$
But in probability theory the expectation is a property of a random variable, with respect to a probability distribution:$$E(X):=\int X\;d\mu$$
I can't see how probability theory can be adapted to quantum mechanics. Observables are associated with linear operators, not measureable functions, so how can we talk about the expectation of a linear operator? And quantum mechanics textbooks use expectations and variances without mentioning underlying probability spaces. Does quantum mechanics use something other than ordinary probability theory?
| Since you want a bit of mathematical rigor:
A quantum state is a self-adjoint positive trace class operator on a Hilbert space with trace 1. This is called density matrix $\rho$. In its simplest form, given $\psi\in \mathscr{H}$, $\rho$ is the orthogonal projector on the subspace spanned by $\psi$. Let $E_\rho(\cdot):D_\rho\subset\mathcal{A}(\mathscr{H})\to \mathbb{R}$ be the map defined as:
$$E_\rho(A)=\mathrm{Tr}(A\rho)\; ,$$
where $\mathcal{A}(\mathscr{H})$ is the space of self-adjoint operators, $\mathrm{Tr}$ is the trace on $\mathscr{H}$ and
$$D_\rho=\{A\in \mathcal{A}(\mathscr{H})\; ,\; \mathrm{Tr}\lvert A\rho\rvert<+\infty\}\; .$$
The map $E_\rho(\cdot)$ has all the properties of an expectation in probability theory. I don't know if it is possible to characterize the measure $\mu$ associated to it (maybe by means of the projection valued measures associated to $\rho$ by the spectral theorem, but it is not straightforward at least for me).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Recent missed opportunities à la Freeman Dyson There is an excellent paper by Freeman Dyson from 1972 (here) and therein the author cites old talks by Hilbert (here) and Minkowski (chapter 2 here) speaking about similar topics, namely how opportunities for discoveries could be better if mathematicians and physicist worked more closely.
I wanted to ask if there are recent publications (roughly > 2000) of that type?
| There was an FQXi essay competition on this subject in Spring 2015:
"Trick or Truth: the Mysterious Connection Between Physics and Mathematics"
Here is the home page with links to winners and other entries:
http://fqxi.org/community/essay/winners/2015.1
The competition is meant to encourage an informal style, readable by non-experts, but with some scientific and mathematical education. A few authors ignore the advice and jump straight into more advanced language (e.g. Peter Woit), but there aren't many equations and on the whole they are very accessible.
Dyson's paper is cited by many of the entries, alongside the other classic in the genre, Wigner's "Unreasonable Effectiveness of Mathematics in the Natural Sciences" (1960) [pdf].
I also meekly suggest that since Ed Witten descended to earth, progress on the mathematics of physics has been remarkable (see also Atiyah, Maldacena, ...), but perhaps the monstrous edifice of String Theory has temporarily poisoned the well. See the popular books:
*
*"Not Even Wrong" by Peter Woit
*"The Trouble With Physics" by Lee Smolin
No mathematcial physics resource list would be complete without a reference to John Baez's blog "This Week's Finds in Mathematical Physics" [home], which contains many multi-disciplinary tales of mystery and imagination.
Kat
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Why does light travel at the same speed when measured by a moving observer? I know the hypothesis that the light speed is constant is retained by experiments. But is there any theory explaining why the light speed is constant no matter how an observer moves relative to light?
My question is, specifically: Suppose an observer $O$ launches a light and $O$ starts to move at the same time with a uniform velocity $v$ in the same direction that light points. Then why $c$ is still the light speed that $O$ will measure rather than $c-v$?
| A personal point of view is that you may consider that Lorentz transformations apply primarily on momenta, and not primarily on (infinitesimal or not) space-time coordinates.
This is, of course, a "strong" postulate.
If you assume (some additional postulates are needed there) that transformations are linear, and that there is a rotation invariance, you are going to study "boost" transformations : $\begin{pmatrix} p'_z\\E'\end{pmatrix} = A(v) \begin{pmatrix} p_z\\E\end{pmatrix}$. You may show that, because $A(v)A(-v)=1$, $det A(v)=1$. Supposing a group structure, you finally are looking at the one-dimensional subgroups of $SL(2, \mathbb R)$, which are :
$$\begin{pmatrix} \lambda&\\&\lambda^{-1}\end{pmatrix}\quad \begin{pmatrix} 1&v\\&1\end{pmatrix}\quad \begin{pmatrix} 1&\\v&1\end{pmatrix}\quad \begin{pmatrix} \cos \theta&-\sin \theta\\\sin \theta&\cos \theta\end{pmatrix}\quad \begin{pmatrix} \cosh \theta&\sinh \theta\\\sinh \theta&\cosh \theta\end{pmatrix}$$
If you add additional postulates that, in a boost transformation, energy and momentum must change, that there exist a transformation which puts the momentum to zero , and that, if the energy is positive for an observer, energy will be positive for all observers, the first $4$ one-dimensional subgroups of $SL(2, \mathbb R)$ are excluded, and the last dimensional subgoup corresponds to a Lorentz transformation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How would an X-ray mirror work? I was wondering if light can be reflected how can someone reflect X-ray of what material does it need to be made of and is its design completely different to that of our original mirrors? Does this mean during long-space voyages in which radiation is an problem why can scientists not develop large panels of X-ray mirrors and Gamma-Mirrors and simply reflect the radiation off rather than worry about that?
| There are a lot of X-ray mirrors in space and around the world. Applications are X-ray astronomy, synchrotron sources, some medical and litographical application. So it is not something theoretical.
All these mirrors have in common is to reflect light at grazing incidence, they are shaped in way to focus the light in a same way that happens for visible light with visible optics, however this is the reason why mirrors focus, not if and why they reflect X-rays.
Several physical principles are used to reflect light depending on the energy (metallic coating, multilayers or crystals). However the question asks if these mirrors can be used to deflect the radiation for human space travel, and the answer is no, because when radiation hit at a large angle they don't act as mirrors any more, but act more like shields.
At this point is more practical to directly use shields (absorbing material), depending on energy it can mean to carry a good amount of extra-weight in space.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Who proposed the bulk-edge correspondence principle? Who proposed the bulk-edge correspondence principle?
The principle is often quoted in counting the number of zero energy states localized on the interface between two insulators with distinct band topology. However, I could not retrieve who was the first to say that.
| Maybe it is R. Jackiw and C. Rebbi (Phys. Rev. D 13, 3398)
When explaining the quantum Hall effect, Hasan and Kane (Rev. Mod.Phys. 2010, 82: 3045–3067) said "This interplay between topology and gapless modes is ubiquitous in physics and has appeared in many contexts. It was originally found by Jackiw and Rebbi (1976) in their analysis of a 1D field theory" in pages 3048-3049.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Has anyone experimentally shown the quantized thermal hall conductivity in Quantum Hall systems? For background:
In a $D=2+1$ state with edge modes described by a chiral $( c_L \neq c_R )$ CFT there is a predicted thermal Hall conductance associated with the gravitational anomaly at the edge. This is the case for integer and some fractional quantum hall systems. For just free bosons on the edge the conductance is proportional to the difference in number between left and right movers as well as the temperature: $K_H \propto (c_L - c_R) T$
This seems like the sort of thing someone with a lab and a fancy set of probes might be able to measure. What's the state of experimental progress in this direction?
I'm familiar with a paper by Kane and Fisher which describes the difficulty a bit and proposes an experimental set-up but since the paper is almost 20 years old I was curious about the modern state of things. A review of some of the physics that goes into making this measurement would be appreciated!
| The experiment on kxy has not been done yet, to my knowledge, but there are several (recent) experiments which are closely related and probe the heat flow through the edge channels of a Hall bar in the fractional regime: http://www.nature.com/nature/journal/v466/n7306/abs/nature09277.html or http://www.nature.com/nphys/journal/v8/n9/full/nphys2384.html?WT.ec_id=NPHYS-201209 are two examples. Measuring temperatures on a 2D sheet of material embedded in a substrate (as in GaAs/GaAlAs heterostructures) is a challenging task, and the signals for kxy are usually smaller than the ones for "simple" kxx measurements.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 1,
"answer_id": 0
} |
Why does Pluto's orbit cross Neptune's orbit? Why does Pluto's orbit cross Neptune's orbit? Is this a fault in Newton's law of gravitation?
| Pluto's average distance from the Sun is larger than Neptune's but Pluto's orbit has a higher eccentricity – the elliptic orbit is more squeezed, less uniformly circular, and such ellipses simply do intersect each other.
The elliptical orbits with properties first identified by Kepler's laws do follow from Newton's laws of gravity. I stress that the orbits are general ellipses, not necessarily circles.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
In a Big Crunch, would there be more mass than at the Big Bang? I found multiple questions where it is stated that dark energy increases as the universe expands. Assuming a big crunch scenario, will this dark energy "go away" again as the size of the universe decreases again, or will there be more energy (=mass) at the Big Crunch than at the Big Bang?
| Point A) The same mechanism that causes the amount of dark energy to increase as the universe expands will cause the amount of dark energy to decrease if the universe contracts.
Point B) A universe that includes an increasing amount of dark energy due to the same mechanism that exists in our universe (in theory) makes it nearly impossible for the universe to begin to contract (not completely impossible in general, but it is for the universe we think we live in). Thus, there will $\textit{probably}~^1$ never be a Big Crunch (unless we're talking about the chocolate bar. I'm led to believe that already exists).
$^1$(probably allows for the possibility that the universe we think we live in is not the universe we actually live in)
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Double rainbows In my garden, when I'm watering the plants I sometimes see a rainbow or two. How did two rainbows appear? Why can't I see three rainbows then, or how can I see three rainbows?
| The two rainbows that are formed are the primary and secondary rainbows respectively, in order of their intensity or brightness, as you may call it. A primary rainbow is formed as a result of a three- step process: Refraction with dispersion, followed by total internal reflection and then refraction.
The secondary rainbow is formed due to a four- step process: Refraction with dispersion, followed by total internal reflection(twice in this case) and refraction again.
Check out the following:
It is found that in case of the primary rainbow, violet light emerges at an angle of 40 degrees relative to the incoming light and red light at an angle of 42 degrees; thus we see the primary rainbow with red at top and violet at bottom.
In case of the secondary rainbow, emergent angles are 50 degrees and 53 degrees with respect to the incoming light, for red and violet colors respectively. Thus, the violet color is at the top while red is at the bottom.
The intensity of the light is reduced at the second internal reflection, and hence the secondary rainbow is very faint in the sky.You may take a look at the following:
A third rainbow even if it is formed as a consequence of successive total internal reflections, will be too dim to be visible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
How to calculate the drag coefficient using terminal velocity? I was wondering if it were possible to calculate the drag coefficient by allowing an object to reach terminal velocity. Can you rearrange the terminal velocity formula to give the drag coefficient?
| Following on Spaderdabomb answer,
the drag force $F_D$ acts on the body and, as such, it balances with the other forces and the inertia of the body ($\sum\vec{F} = m\tfrac{d\vec{u}}{dt}$).
When you consider a situation where the body is at terminal velocity, this means its velocity is constant (at least in that direction) and its acceleration is null.
Consider the case of a parachute free falling at terminal velocity $\vec{u} = \{0,\,0,\,-w\}$, thus at constant $w$. As such $\tfrac{d\vec{u}}{dt} = 0$ and the force balance becomes $\sum\vec{F} = 0 \Leftrightarrow F_D^z - m g = 0$, noting that it is the drag that is sustaining the parachuter. From the perspective of the parachuter, the flow speed is $\vec{u}_\infty = \{0,\,0,\,w_\infty\}$ where $w_\infty = w$. As the drag is given by $F_D = C_D \tfrac{\rho}{2} w_\infty^2 A $, substituting in the force balance returns Spaderdabomb answer, $C_D = \tfrac{2 m g}{\rho w_\infty^2 A}$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why is baryon or lepton violation in standard model is a non-perturbative effect? The baryon number B or lepton number L violation in the standard model arise from triangle anomaly. Right? Triangle diagrams are perturbative diagrams. Then why the B or L violation in Standard model is said to be a non-perturbative effect? I'm confused.
| It is a non-perturbative effect because it is 1-loop exact.
The triangle diagram is actually the least insightful method to think about this, in my opinion. The core of the matter is the anomaly of the chiral symmetry, which you can also, for example, calculate by the Fujikawa method examining the change of the path integral measure under the chiral transformation. You can obtain quite directly that the anomaly is proportional to
$$\int \mathrm{Tr} (F \wedge F)$$
which is manifestly a global, topological term, (modulo some intricacies) it is the so-called second Chern class and takes only values of $8\pi^2k$ for integer $k$. It is, by the Atiyah-Singer index theorem (this can also be seen by Fujikawa), essentially the difference between positive and negative chiral zero modes of the Dirac operator. This is obviously a discontinuous function of $A$ (or $F$), which is already bad for something which, if it were perturbative, should be a smooth correction to something, and it is also the number describing which instanton vacuum sector we are in, see my answer here. Since perturbation theory takes place around a fixed vacuum, this is not a perturbative effect, since it is effectively describing a tunneling between two different vacuum sectors.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/128971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Energon: is it possible? I'm always quite curious about the "Energy cube" in Transformers, or namely Energon.
Is it really possible to store energy, such as electricity, into such a compact form? safe to distribution, and seems nothing left after being consumed?
ps. Wikipedia has a page for Spark, which is more for transformer's soul. I'm not asking for that yet.
| It is possible to capture positrons (antiparticle of electron) in a magnetically confined plasma - the repulsive forces get very large unless you do something to equalize the charge. The energy density that could be achieved is stunning.
This was the principal plot line behind Dan Brown's "Angels and Demons" - this plasma (made at CERN, that den of mad scientists plotting to destroy the universe) was intended to be released (the power source containing it would become exhausted) underneath the Vatican, destroying the "center of the Evil Church" (the opinion of the perpetrators...). Of course the hero saves the day.
In principle a few grams of antimatter contains as much energy as the fuel tanks of the space shuttle...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Kinetic Energy of a Particle Consider a particle of mass $m$ in $6$ dimension. Its coordinate w.r.t. origin $\left(0,0,0,0,0,0\right)$ is given as $\left(x,y,z,\dot{x},\dot{y},\dot{z}\right)$. If we denote $r = \sqrt{x^2+y^2+z^2}$, then which of the following two is the kinetic energy of this particle:
*
*$T = \frac{1}{2}m\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)~?$
*$T = \frac{1}{2}m\left(\frac{dr}{dt}\right)^2 = \frac{1}{2}m\frac{\left(x\dot{x}+y\dot{y}+z\dot{z}\right)^2}{r^2}~?$
| The kinetic energy is $T = \frac{1}{2} m (\frac{d \vec{r}}{d t})^2$
$$\vec{r} = x \vec{i} + y \vec{j} + z \vec{k}$$
The first exprssion is right.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How much of the sky is visible from a particular location? From a particular point how much of the sky can be observed. For simplicity sake let us assume the particular point is the head of a 6 foot tall man floating in the middle of the ocean with no visible barriers between him and the horizon. If he stays in exactly the same location being able to turn 360 degrees how much of the sky is visible?
Would we calculate the circumference of the circle with the distance to the horizon as the radius? The surface area of a portion of a sphere?
Taking it further what percentage of the total sky can be seen at any given time. The answer above divided by the total surface area of earths atmosphere?
| Visible objects in low-Earth orbit (such as the International Space Station) take about 90 minutes—5,400 seconds—to complete a single orbit. Unequivocally far beyond the mesosphere at 240-250 miles above Earth, such objects might provide a rough proxy for measuring the portion of blue "sky" that is visible.
Now when viewed from their backyard, most people will estimate that an object in low-Earth orbit passes directly over a fixed location in about five to ten seconds. By this measure, from a single point on Earth you can usefully “see” 0.185% (about 2 tenths of one percent) of the local sky.
The nice thing about this method is that you can fine-tune the approximation yourself to tailor it to your specific location, situation, visual acuity, or other circumstance(s).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
} |
Is it possible to 3D print a mirror to create a high quality telescope? Is it possible to 3D print a mirror with todays available materials?
If so, would there be a reduction in image quality?
| The highest resolution 3d printers I know of are around 1600dpi, which is a resolution of about 15$\mu m$. Telescope mirrors have to be smooth to fractions of a wavelength of light, so the resolution of current printers is nowhere near good enough.
Whether 3D printers could one day be good enough is a different question, but given that the improvement in resolution required is at least a factor of 1,000 I think it's not likely because 3D printers are designed to address quick manufacture rather than precision manufacture. In any case, making mirrors is a well established procedure. The difficulty is making them large, and it's not obvious how 3D printers would help with this.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 1
} |
Differences between probability density and expectation value of position The expression $\int | \Psi\left(x\right)|^2dx$ gives the probability of finding a particle at a given position.
If wave function gives the probabilities of positions, why do we calculate "expectation value of position"?
I don't understand the conceptual difference, we already have a wave function of a position. Expectation value is related to probabilities.
So what is the differences between them? And why do we calculate expectation value for position, although we have a function for probability of finding a particle at a given position?
| Let $\Omega\subseteq \mathbb{R}^n$; then $\int_\Omega \lvert\psi(x)\rvert^2dx$, for a normalized function $\psi\in L^2(\mathbb{R}^n)$ gives the probability that the particle is in the region of space $\Omega$, but does not give any further information on its position. If you want to obtain a quantitative information on the latter (within the limits of quantum indeterminacy), you have to calculate the expectation value $\int_{\mathbb{R}^n} x_j\lvert\psi(x)\rvert^2dx$, for each component $x_j$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 1
} |
What is $c + (-c)$? If object A is moving at velocity $v$ (normalized so that $c=1$) relative to a ground observer emits object B at velocity $w$ relative to A, the velocity of B relative to the ground observer is
$$ v \oplus w = \frac{v+w}{1+vw} $$
As expected, $v \oplus 1 = 1$, as "nothing can go faster than light".
Similarly, $v \oplus -1 = -1$. (same thing in the other direction)
But what if object A is moving at the speed of light and emits object B at the speed of light in the exact opposite direction? In other words, what is the value of $$1 \oplus -1?$$
Putting the values in the formula yields the indeterminate form $\frac{0}{0}$. This invites making sense of things by taking a limit, but $$ \lim_{(v,w)\to (1,-1)} \frac{v+w}{1+vw}$$ is not well-defined, because the limit depends on the path taken.
So what would the ground observer see? Is this even a meaningful question?
Edit: I understand $1 \oplus -1$ doesn't make sense mathematically (thought I made it clear above!), I'm asking what would happen physically. I'm getting the sense that my fears were correct, it's physically a nonsensical situation.
| As pointed out by zakk, any proper time of objects moving at light speed would be zero.
That means physically that there does not exist any point of time where you could perform the addition c + (-c). Example: A photon is emitted at A and absorbed at B. There is no point of time A < t < B where the photon is actually moving, the age of the photon at A is the same as at B, and even the hypothetical proper distance between A and B would be zero (even if observers are observing a time lapse between A and B, this observed time lapse does not correspond to any real point of time A < t < B).
This is why it is impossible to subtract (-c).
By the way, this result is confirmed by the second postulate of special relativity: the velocity of light is never (-c), it is always observed at +c.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 4
} |
How to approximate acceleration from a trajectory's coordinates? If I only know $x$- and $y$- coordinates of every point on a trajectory without knowledge of time information, is there any way to approximate Cartesian acceleration angle at each point? Time interval between every two points is very small, ~0.03 second.
| From the path you need to find the radius of curvature $\rho$ at each point. This would be kind of noisy unless you have really precise data. Your best bet into input all the x and y points into cubic spline in order to get what the derivatives $x'$ and $y'$ are (in units of length per frame). In addition, you need to get the kinematic accelerations $x''$ and $y''$.
Then
$$\rho = \frac{ \left( x'^2 + y'^2 \right) ^ {\frac{3}{2} }}{x''\,y' - x'\,y''} $$
You can also estimate the speed by $$ v = \dot{s} \sqrt{ x'^2+y'^2 } $$ where $\dot{s}$ is the sample rate (frames/second).
The tangent acceleration to the path is $$ a_{T} = \dot{v} = \dot{s} \frac{x'\,x'' + y'\,y''}{\sqrt{x'^2+y'^2}} $$
and the transverse acceleration is $$a_{N} = \frac{v^2}{\rho} = \frac{ \dot{s}^2 \left({x''\,y' - x'\,y''}\right) }{ \sqrt{ x'^2+y'^2 } }$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/129701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to determine the sign of the s-wave scattering length? I guess it is relatively easy to determine the magnitude of the scattering length $a$.
We just need to measure the scattering cross section. In this way, we can determine the value of $a^2$.
But how to determine its sign?
| One way to measure the magnitude is by looking for a density-dependent energy shift. If you transfer atoms between, say, two hyperfine states with a microwave transition, the resonant frequency of this transition will change due to the mean-field shift from the interactions. If you start with an interacting mix of states 1 and 2, and you transfer from state 2 to state 3, you will find that
$\Delta f=\frac{2\hbar}{m}n_1(a_{23}-a_{13})$
This equation comes from the PhD thesis of Cindy Regal, where you may find more information in section 5.3. She used this technique to characterize a Feshbach resonance, for which it is very well-suited.
So this depends on the sign of the scattering length, but clearly it won't always be helpful: one would mainly use it if one scattering length is already known, and if they are identical it is completely useless!
I think the more general answer is rather boring: it is determined by a combination of experimental measurements of the magnitude from scattering experiments plus detailed numerical modeling. In some cases, as Mark Mitchison has said, one might tell from states of matter that depend on the sign, but I'm not aware of any cases in which this has actually been used.
I don't really know if this is the whole story or not, so I welcome any other comments.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/130371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Is interstellar flight possible in near future in a way that would keep our civilization alive? Is interstellar flight possible in the near future in a way that would keep our civilization alive?
I mean is it practically possible to obtain technology that would enable us to travel to nearby habitable earth-like planets to keep our civilization alive?
For example, consider this design for NASA’s Star Trek-style Space Ship, the IXS Enterprise. Do you know any good site which goes into detail on both latest practical and theoretical development of this interstellar flight?
| It's easy after we rebuild ourselves first. We need to replace biological brains by digital brains. The problem we now have is that if we travel, we carry a biological machine that is much more advanced than the most powerful supercomputer we can build today, physically to the point of destination. This biological machine must always be kept at the right temperature, it needs to be at the right atmospheric pressure etc. etc.. Obviously, it would be far more practical to have a digital version of these biological machines and then simply upload the data to a machine at the point of destination.
The question is then how to bring machines with the necessary infrastructure to the next inhabitable planet? This may be done using nanotechnological means, you can imagine that only need to send a microscopically small device which only needs a small rocket to a distant planet and then the device will grow all by itself and we'll have the machines ready to receive us via radio transmissions (or perhaps laser transmissions which have a far lager bandwidth).
Another possibility is to send messages to distant civilizations. While two way communications take a long time, to upload ourselves to another civilization only requires one way communications. All we need to do is to repeatedly transmit our messsages that catch the attention of some distant civilization followed by the data containing all the information needed to be able to run our brains. It may be that our civilization will be long destroyed when the messages are still on their way. If millions of years later we are received by a civilization in the Andromeda galaxy and they manage to run our programs, we'll not perceive this time lag. To us it would feel like we've arrived there at an instant.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/130450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Angular Momentum Conservation Definition Did I missed something in angular momentum definition?
Two identical bodies rotate around mass center.
Now I invented anti-gravity and turning gravitational switch off.
Those two bodies will move now in straight line with constant velocity and angular momentum conservation is compromised.
Turning the gravitation off does not provide any external torque to the system.
I also could not find any example when angular momentum is conserved and no internal forces (i.e. gravitation, Coulomb, tension) exist in the system.
The definition of angular momentum does not said that existing of internal forces are necessary.
| One does not need to switch "gravity off" to make a system like that. Replace gravity with a simple string, and cut the string. Physically that's completely equivalent to your problem, as far as I can see.
And while this may seem counterintuitive, the angular momentum in a system of a mass rotating at the end of a string is, indeed, conserved, when you cut the string and the mass flies away along the tangent of its circular trajectory!
The easiest way to see this is to look at the vector definition of angular momentum, which is the cross product of the radial vector with the momentum vector.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/130509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What's the relationship between uncertainty principle and symplectic groups? What's the relationship between uncertainty principle and symplectic groups? Does the symplectic groups mathematically capture anything fundamental about uncertainty principle?
| Yes, of course, symplectic groups describe generalized situations that reveal the uncertainty principle.
The reason for the relationship is that the symplectic groups are defined by preserving an antisymmetric bilinear invariant,
$$ M A M^T = A $$
where $M$ is a matrix included into the symplectic group is the equation holds and $A$ is a non-singular antisymmetric matrix.
Where does the uncertainty principle enter? It enters because $A$ may be understood to be the commutator (or Poisson bracket) of the basic coordinates $x_i,p_i$ on the phase space. If we summarize $N$ coordinates $x_i$ and $N$ coordinates $p_i$ into a $2N$-dimensional space with coordinates $q_m$, their commutators are
$$ [q_m,q_n] = A_{mn} $$
with an antisymmetric matrix $A$. Consequently, the symplectic transformations may be defined as the group of all linear transformations mixing $x_i,p_i$, the coordinates of the phase space, that preserve the commutator i.e. all the uncertainty relations between the coordinates $q_m$.
Curved, nonlinear generalizations of these spaces are known as "symplectic manifolds" and nonlinear generalizations of the symplectic transformations above are known as "canonical transformations".
I think it doesn't make sense to talk about this relationship too much beyond the comments above because the relationship is in no way "equivalence". One may say lots of things about related concepts but they're in no way a canonical answer to your question – they don't follow just from the idea of the "relationship" itself. I just wanted to make sure that a relationship between mathematical structures on both sides, especially the antisymmetric matrix, certainly exists.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/130562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Is our universe an emulation? I was watching one of Neil Degrasse Tyson talks and there was a scientist (can't recall his name sorry) who was talking about a recent discovery:
"Doubly-even self-dual linear binary error-correcting block code" has been discovered embedded within the equations of superstring theory.
Is this for real? Does it imply that our universe just a sophisticated emulation running on some supercomputer?
EDIT:
Would it be possible to set up an experiment that would be able to test this hypothesis?
I am thinking about World of Warcraft, for example, how would an elf in WoW test if his world is an emulation or not? Is it even possible?
| If you were a simulated person inside an emulation, would you ever be able to tell the universe was emulated? the answer is no, not if it is set up correcty.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/130734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If MOND theory doesn't explain gravitational lensing, couldn't well placed black holes explain it? When I first heard about star velocity as a function of its distance from the center of the galaxy and the difference between prediction and observation I immediately thought "there must be a threshold where gravity reacts differently". Then I discovered MOND theory and that it can't explain gravitational lensing.
But what if observed gravitational lens is only well placed black holes ? After all we detect black holes with their gravitational influence.
So, why dark matter instead of MOND + more black holes ?
| You should not take claims about the failure of MOND to reproduce strong gravitational lensing too seriously. In this review article the authors, Benoît Famaey and Stacy S. McGaugh, state:
Due to the fact that all the above models were using the Bekenstein μ-function (α = 0 in Eq. 46), and that this function has a tendency of slightly underpredicting stellar mass-to-light ratios in galaxy rotation curve fits, it was claimed that this was a sign for a MOND missing mass problem in galaxy lenses. While such a missing mass is indeed possible, and even corroborated by some dynamical studies of galaxies residing inside clusters (i.e., the small-scale equivalent of the problem of MOND in clusters), for isolated systems with well-constrained stellar mass-to-light ratio, the use of the simple μ-function (α = 1 in Eq. 46) has, on the contrary, been shown to yield perfectly acceptable fits in accordance with the lensing fundamental plane.
(I have suppressed the references in the above paragraph).
Thus the claim that MOND has problems with gravitational lenses is wrong (in my opinion), or at worst disputed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/130963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given a slanted pipe Question: I have a pipeline that is tilted. I know the length of the pipeline, and the pressure (90psi) that is felt at the very bottom. I'm trying to find out how much liquid has leaked out if the pressure at the bottom decreases to 80psi.
Is the pressure felt at the very bottom directly proportional to the amount of liquid that is in the pipeline?
| The pressure at the bottom of a pipe, or any other column of liquid, for that matter, is directly proportional to the height of the liquid.
The pressure of the liquid is equal to:
$\rho h$, where $\rho$ represents the density of the liquid, and h represents the height.
So, assuming that the cross-sectional area of the pipe is constant, then the pressure felt at the bottom is directly proportional to the amount of liquid in the pipe.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/131031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the density of gold crown using archimedes principle Question: As shown in diagram below the crown has a mass of 14.7kg when measured above water and 13.4kg when measured in water. Is the crown made of gold?
I have this following solution provided:
The apparent weight of the submerged crown, $w'$,equals the actual weight, $w$, minus the buoyant force, $F_B$.
$w' = w - F_B = w - \rho_{fluid}gV $
$ w - w' = \rho_{fluid}gV$
$ w = \rho_{object}gV$
I still agree with the answer up to this point as the only difference before submerging and after is the upthrust that is equivalient to the amount of water being displaced
However, it takes the following ratio between $w$ and $w-w'$ like this:
$w/(w-w') = (\rho_{object}gv)/(\rho_{fluid}gv) = \rho_{object}/\rho_{fluid} = 14.7 kg/ (14.7kg - 13.4kg) = 11.3 $
And it use this value 11.3 as the density of the object. How does this ratio of weight before and after(weight loss because upthrust) is equivalent to the density of the object?Can anyone please enlighten me?
| Consider $w$; it's the volume of the object times its density times $g$.
Consider the buoyant force; it's the volume of the object times the density of the fluid times $g$
So, what's the ratio of the weight to the buoyant fore, when you cancel out the constant factors?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/131106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is entropy really? On this site, change in entropy is defined as the amount of energy dispersed divided by the absolute temperature. But I want to know: What is the definition of entropy? Here, entropy is defined as average heat capacity averaged over the specific temperature. But I couldn't understand that definition of entropy: $\Delta S$ = $S_\textrm{final} - S_\textrm{initial}$. What is entropy initially (is there any dispersal of energy initially)? Please give the definition of entropy and not its change.
To clarify, I'm interested in the definition of entropy in terms of temperature, not in terms of microstates, but would appreciate explanation from both perspectives.
| In terms of the temperature, the entropy can be defined as
$$
\Delta S=\int \frac{\mathrm dQ}{T}\tag{1}
$$
which, as you note, is really a change of entropy and not the entropy itself. Thus, we can write (1) as
$$
S(x,T)-S(x,T_0)=\int\frac{\mathrm dQ(x,T)}{T}\tag{2}
$$
But, we are free to set the zero-point of the entropy to anything we want (so as to make it convenient)1, thus we can use
$$S(x,T_0)=0$$
to obtain
$$
S(x,T)=\int\frac{\mathrm dQ(x,T)}{T}\tag{3}
$$
If we assume that the heat rise $\mathrm dQ$ is determined from the heat capacity, $C$, then (3) becomes
$$
S(x,T)=\int\frac{C(x,T')}{T'}~\mathrm dT'\tag{4}
$$
1 This is due to the perfect ordering expected at $T=0$, that is, $S(T=0)=0$, as per the third law of thermodynamics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/131170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "46",
"answer_count": 14,
"answer_id": 8
} |
Physical applications of matrices and determinants Other than notation devices, I don't see any direct application of matrices/determinants in physics. For example, they are just a different way to write a partial derivative and determinants find if they can be explicitly solved if written down as simultaneous equations. Calculus, for instance, can be directly applied to physical problems, but I don't know of any other application of matrices other than representing equations in a different notation. And in most of the cases like vector products, you just realise that a huge term can just be written down as a determinant, so it is essentially a notational tool. They are used in tensor calculus, but for similar reasons.
Can someone please guide me on more applications with good sources?
| Lie groups are fundamental for talking about anything related to symmetries in physics on a level of some rigor, and every finite-dimensional Lie group is a matrix group. Consequently, the trace as a basic matrix operation shows up anywhere where invariance on the adjoint action of the group is needed, and the matrices are everywhere.
The Slater determinant is what multi-fermion wave-functions are, and this is not a notational trick, since that wave function is actually the n-fold wedge product of basis vectors on some space, which is (up to normalisation) also what the determinant really is.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/131530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Hugh Everett's MWI According to Hugh Everett's many worlds interpretation, all the possibilities of one action can happen at the same time in other parallel universes, so how come we can't see these worlds? now I bumped into something called the quantum decoherence but I can't seem to get how this decoherence work?
| To "see another World" would require doing a measurement that involves (partially) reversing the effect that led to the split. In practice this is impossible to realize because the observer is a macroscopic object itself and it will decohere very fast. Decoherence means that the system becomes correlated with the environment and that poses a big problem here because you would have to reverse the effects of all this. This is similar to why entropy always increases and gives rise to irreversible phenomena on the macroscopic scale.
But you can still contemplate thought experiments and demonstrate how one can perform such measurements in principle. David Deutsch has proposed the following thought experiment to prove the existence of parallel World. Suppose that we create a virtual observer inside a quantum computer. The observer prepares a spin polarized in the x-direction. He then measures the z-component of the spin. The z-component can be found to be 1/2 or -1/2 with equal probability.
According to the MWI, there are two Worlds were both possibilities are realized. Suppose that we then reverse the act of measurement, but such that the observer will keep the memory of having performed the measurement (a complete reversal would necesaarily mean that the observer's memory has been restored to what it was initially). Now it's easy to show that this can be realized by a unitary transform, so it is an operation that can in principle be performed on a quantum computer.
In that final state, the spin is restored in the orgina state where it is polarized in the x-direction and the observer can verify that this is the case by doing additional measurements on it. However the observer also knows that he had measured the z-component of the spin before, but if only one World really exists after the measurement, then the initial state of the spin could never have been restored by that unitary transform.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/131663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is there an alternativee method of transmitting wireless electricity? I have currently been researching a lot lately about wireless transmission of power.
Currently the only methods I have seen that is viable is magnetic induction, and high voltage discharge (Tesla coil and Van de Graff generators).
The 2 methods above are not viable and not as efficient, and they require a lot of input voltage(especially for the Tesla coil and VdG generators).
Are there any other alternatives known? Even if it is ineffecient.
| You can build microwave antennas of any desired size and directivity. The thermodynamic efficiency of a properly designed microwave link should be around 50%, even though the cost would be horrendous. And if you really need lots of remote power, you can simply get yourself a nuclear power plant at that location. Not that I can see any use for that...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/132003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Question about infinite sum in quantum field I read from some books of number theory that $$\sum_{n=1}^{\infty}\frac{1}{n^s} = -\frac{1}{12}\text{,when } s=-1.$$
Now is there such a result $$\sum_{n=1}^{\infty}\frac{1}{n^s} = \pi \text{,when } s=1,$$or $$\sum_{n=1}^{\infty}\frac{1}{n^s} = c \pi \text{,when } s=1 \text{,where } c \text{ is a rational number ?}$$
I get a similar result in mathematics by analogue, I suspect the result may have some interpretation in physics.
| Valter's answer is completely correct, but I'll just briefly expand on it to address the specific values you ask about. The place to go, really, is the Wikipedia page Particular values of Riemann zeta function, which lists mosts of the values of $\zeta(s)$ (which, as Valter explained, equals
$$\zeta(s) := \sum_{n=1}^{+\infty} \frac{1}{n^s}$$
when $\operatorname{Re}(s)>1$) that can be expressed without the use of series, or using simpler ones.
For example, the value $\zeta(2)$ is well known to be $\pi^2/6$, and the other positive, even integers have zeta values which are rational multiples of a power of $\pi$.
On the other hand, the value $s=1$ is rather different, because the zeta function has a pole there. This means that there is no way to make the series
$$
\zeta(1) := \sum_{n=1}^{+\infty} \frac{1}{n}
$$
mean anything other than $\infty$. This series is of course the harmonic series, which is probably the most famous example of a divergent series, and there are multiple simple proofs of why its value must be infinity.
However, if you ask why we insist that
$$1+\frac12+\frac13+\frac14+\cdots$$
is infinite, but we're OK with assigning a finite (and negative) value to
$$1+2+3+4+\cdots,$$
then I would say that the second series is just a handy way of expressing something else, and should not really have been used in the first place.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/132066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Bekenstein bound for electron? Using the Wikipedia version of the Bekenstein bound, and substituting the Wikipedia values for electron mass and radius, one obtains 0.0662 bits. Does this really mean that a system, any system, placed inside a sphere the size of an electron, and weighing no more than an electron does, is almost determinate? How about an electron itself? Wouldn't one need at least a few bits to characterise the behavior of an electron in magnetic space?
(I am a professional mathematician but I know very little about physics, I'm sure I'm missing something obvious here...)
| One can't take results like that too seriously at the scale at which an electron would apply. In particular, the classical general relativistic model, applied naively to a point mass electron would tell you that the electron has too large a charge and angular momentum to have a black hole horizon, and would instead be the exotic type of object called a naked singularity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/132166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The Effect of Tortoise Coordinates Referring particularly to
http://arxiv.org/abs/hep-th/9909056
in regard to the wave equation for Schwarzschild-AdS black holes (p.4), I'm trying to understand tortoise coordinates.
So starting with the 4-dimensionalSchwarzschild-AdS metric in the general form
$$ds^2=-f(r )dt^2+\frac{dr^2}{f(r )}+r^2(d\theta^2+sin^2 \theta d\phi^2),$$
if I want to find the wave equation $\Box \phi=0$ in the Schrodinger-like form. This is done by introducing the separation of variables
$$\phi=\frac{\psi(r ) Y(\theta,\phi)e^{-i\omega t}}{r}$$
and then using the tortoise coordinate $dr_*=\frac{dr}{f(r )}$ to get
$$(\partial_{r_*}^2+\omega^2-V(r_*))\psi=0.$$
But I don't fully understand what this tortoise coordinate really does. In fact when I go through these calculations myself, I use the transformation
$$\psi'(r ) \to \frac{\psi'(r )}{f(r )}$$
and (fortunately) get the Schrodinger like form as in the paper above. However, they never explicitly state the potential and what I find is
$$V(r_*)=\frac{-\ell(\ell+1) f(r )+rf'(r )}{r^2}.$$
where $\ell$ is the angular momentum mode. But note, in my transformation, I never mentioned $r_*$ and hence why my $V(r_*)$ doesn't actually mention an $r_*$. This is where my confusion lies.
Is my potential right if I just replace the $r$ by $r_*$? i.e
$$V(r_*)=\frac{-\ell(\ell+1) f(r_* )+r_*f'(r_* )}{r_*^2}?$$
(I highly doubt it.) And if not, how do I recover $V(r )$ from here?
P.s. It would actually also be extremely helpful if someone knew $V(r )$, i.e. potential in original coordinates, for the Schwarzschild-AdS black hole.
| The form of $V$ does not change. What you mean by $V(r_*)$ is $V(r(r_*))$ so you would've to explicitly find $r$ in terms of $r_*$ (which is impossible) to fully express your potential in terms of $r_*$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/132397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Do charges flow from strong insulator to weak insulator? Do static charge in a strong insulator flow to a weaker insulator when both stay in contact with each other? For example, when an insulator weaker than air placed in a medium of air, would the static charges on the insulator be absorbed to the air slowly and finally the insulator becomes neutral? If so then what is the rate of flow from the insulator to air?
| The Triboelectric effect is the process through which materials can become electrically charged through friction when they come in contact with other different materials.
These materials do not have to be insulators for this effect to take place however if they are good conductors the charge will usually flow away.
There is a series of materials ranging from those that become positively charged to those like to remain neutral to those that become negatively charged. This series is independent of a materials conductivity. With aluminium, steel and silver being good conductors and wool and polystyrene being good insulators in between them on the list.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/132494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Unequally charged hanging pith balls I have done a problem which asks us to find the charge on two equally 'massed and charged' pith balls which are left hanging on a string with a certain length that repels each other and attains an equilibrium point making an angle (the string) with the vertical.
The problem was quite clear, what left me wondering was: Won't the angle be the same if the balls are unequally charged (but equally massed) ? Or does it deviate from the initial case?
| As long as your product $|q_{1}||q_{2}|$ remains the same as in the case where you had the equally charged spheres, then yes, you will get the same value for the angle (provided the masses are equal). This is because the electrostatic force acts equally on both charged spheres.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/132982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Scientists observe the laws of the physics but, Where do they come from? Has anyone ever considered how the laws of physics that we study came into being.
| Based on what we know, both from science and philosophy, these are both ill-phrased questions. You can get as many answers to them as you like, and they will all be equally meaningless, because none will bare any logical relationship to known facts about the universe we live in.
Now, there are much better question along these lines, which deal with the reason that we can find laws, at all. "How do we know that nature is reproducible?" and "Why do we assume that there is an objective physical reality?" would be some of these. Maybe you want to think about those for a while and rephrase your question?
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/133151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A step in zeta function regularization I'm just wondering about the mathematical step
$$\sum_{n=1}^\infty n\exp[-\epsilon n\sqrt x]=\frac1{\epsilon^2 x}-\frac1{12}+\mathcal O(\epsilon).$$
Why is this equality so? I see that
$$\sum_{n=1}^\infty n\exp[-\epsilon n\sqrt x]=-\frac1{\sqrt x}\frac{\partial}{\partial\epsilon}\frac{1}{1-\exp[-\epsilon\sqrt x]}\simeq-\frac1{\sqrt x}\frac{\partial}{\partial\epsilon}\frac{1}{\epsilon\sqrt x}=\frac1{\epsilon^2x}.$$
But how about the $-\frac1{12}$?
| The geometric series formula tells us
$$
\sum_{n=1}^{\infty} e^{-\epsilon n\sqrt{x}} = \frac{e^{-\epsilon \sqrt{x}}}{1-e^{-\epsilon\sqrt{x}}}.
$$
The derivative with respect to $\epsilon$ of the left hand side gives $-\sqrt{x}$ times your sum. Therefore your sum is equal to
$$
-\frac{1}{\sqrt{x}}\frac{\partial}{\partial\epsilon}\frac{e^{-\epsilon \sqrt{x}}}{1-e^{-\epsilon\sqrt{x}}}
$$
which I believe is equal to
$$
\frac{1}{4\sinh^2(\epsilon\sqrt{x}/2)}.
$$
At small $\epsilon$ this diverges like $\frac{1}{x\epsilon^2}$.
Multiply it by $\epsilon^2$ to get something that is finite as $\epsilon\to 0$, do a Taylor series in $\epsilon$, and then divide back by $\epsilon^2$.
Doing that: Let
$$
f(\epsilon) = \frac{\epsilon^2}{4\sinh^2(\epsilon\sqrt{x}/2)}
$$
$f(\epsilon)$ is $\epsilon^2$ times the sum you want to compute. Since the denominator goes to zero proportionally to $\epsilon^2$ for small $\epsilon$ (by the small angle approximation for $\sinh$), $f(\epsilon)$ is finite in the limit $\epsilon\to 0$. In fact, with the definition that $f(0)$ is equal to $\lim_{\epsilon\to 0}f(\epsilon)$, $f$ is analytic at $\epsilon=0$. So let's do a Taylor series for $f(\epsilon)$, expanding it in increasing powers of $\epsilon$:
$$
f(\epsilon) = f(0) + \frac{1}{2}\epsilon^2 f''(\epsilon)|_{\epsilon=0} + O(\epsilon^4)
$$
(There are no odd powers since $f$ is even under $\epsilon\to - \epsilon$.)
The value of $f(0)$ is
$$
f(0) = \lim_{\epsilon\to 0} \frac{\epsilon^2}{4\sinh^2(\epsilon\sqrt{x}/2)} = \frac{1}{x}
$$
by the small angle approximation or L'Hopital's rule, whichever you prefer. The second derivative of $f$ is (according to Mathematica)
$$
f''(\epsilon) = \frac{-2+2\epsilon^2 x + (2+\epsilon^2x)\cosh(\epsilon\sqrt{x})-4\epsilon\sqrt{x}\sinh(\epsilon\sqrt{x})}{8\sinh^{4}(\epsilon\sqrt{x}/2)}
$$
This looks like it diverges as $1/\epsilon^4$, but you can check that the numerator is in fact fourth order in $\epsilon$ as well:
$$
-2+2\epsilon^2 x + (2+\epsilon^2x)\cosh(\epsilon\sqrt{x})-4\epsilon\sqrt{x}\sinh(\epsilon\sqrt{x}) = -\frac{x^2}{12}\epsilon^4 + \frac{x^3}{90}\epsilon^6 + O(\epsilon^8)
$$
so the second derivative is finite at $\epsilon=0$ (If it wasn't, I would have been lying when I said $f$ was analytic) and given by
$$
f''(0) = \lim_{\epsilon\to 0} f''(\epsilon) \\
= \lim_{\epsilon\to 0} \frac{-\frac{x^2}{12}\epsilon^4}{8\sinh^{4}(\epsilon\sqrt{x}/2)}\\
=\lim_{\epsilon\to 0} \frac{-\frac{x^2}{12}\epsilon^4}{\frac{1}2 \epsilon^4x^2}\\
=-\frac{1}{6}
$$
So the Taylor series for $f$ is
$$
f(\epsilon) = \frac{1}x - \frac{1}{12}\epsilon^2 + O(\epsilon^4)
$$
Remember that $f(\epsilon)$ was $\epsilon^2$ times your divergent sum. So the expansion in powers of $\epsilon$ of your divergent sum is
$$
\frac{f(\epsilon)}{\epsilon^2} = \frac{1}{\epsilon^2 x} - \frac{1}{12} + O(\epsilon^2)
$$
"Thus" the dimension of spacetime for bosonic strings is $26$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/133237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Could we tell the difference between population I and II (or even III) neutron stars? This question is related to thoughts I was having about the mass-radius relationship for neutron stars. Is it unique? Is there a single relationship between $P$ and $\rho$ or is there any chance that the NS "remembers" the composition or type of star from which it was born?
So, suppose I were able to measure the masses and radii of a number of (isolated) neutron stars in the solar neighbourhood. Most of these would presumably be population I, but would I be able to spot a population II/III interloper?
| The distinction between populations is one of
*
*what fraction of nucleons are neutrons and
*how the nucleons are grouped (i.e. atomic species)
In the bulk of a neutron star nucleons are no longer grouped in atoms and find their equilibrium by transforming from proton to neutron and back again, so the naive answer is immediately "No, you shouldn't be able to tell."
Now we leave the part of the physics I am sure about and move on to the bits for which we really need an expert.
The important weasel word above is "bulk". As I understand it there are crusts of electron-degenerate (and non-degenerate atomic matter in a metallic crystalline lattice?) on the outside of a typical neutron star. If this crustal material does not exchange nucleons with the interior much then it might still retain some memory of the original composition of the star. In that case one can then ask how you could get access to that information, but at this point I am completely stuck: the process would have to be spectroscopic but we don't know much (anything experimentally) about the spectrum of materials under those conditions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/133324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
What does a supernova look like at its peak luminosity? I know that in some types of supernovae, the cause of the increased luminosity is the radioactive decay of certain elements ejected during the explosion, so a question came to my mind.
If the ejected material carrying the isotopes that decay to give the electromagnetic radiation is expelled at velocity of say 5% the speed of light, and given the fact that some supernovae stay extremely luminous for more than 4 weeks, then by that time the radioactive isotopes will have traveled more than 30 billion kilometers from the exploded star.
So, does that mean a supernova at 4 weeks can be expected to look like a star with a radius of 30 billion kilometers and luminosity of $10^8$-$10^9$ times the solar luminosity ?
Or I am getting the idea of the radioactive decay as the source of the supernova luminosity wrong ?
| Your math does check out:
\begin{align}
r&=vt \\
&=0.05\cdot2.9979\times10^{10}\frac{cm}s\cdot4\cdot604800\,s\\
&=3.63\times10^{15}\,cm\\
&=36.3\times10^9\,km\\
&=0.012\,pc
\end{align}
When a supernova explodes, it enters the free expansion phase, it's position is linear in time ($r=vt$, as used above). It stays in this phase for a few hundred years (depends heavily on the ambient density); assuming 200 years, then
$$
r_{fe}=9.45\times10^{18}\,cm=3\,pc
$$
After this point, the Supernova Remnant (technically speaking, SNe is the explosion while SNR is the result of the material after said explosion) continues expanding, though at a reduced rate (because it has swept-up ambient material this entire time, building up a thick shell of thickness $w\sim0.1\,pc$) for many thousands of years.
SNe theory says a normal Type Ia produces about $0.5\,M_\odot$ of nickel-56 which then decays to an excited state of cobalt-56, which then emits an X-ray photon:
\begin{align}
\,^{56}{\rm Ni}+e^-&\to\,^{56}{\rm Co}^*+\nu_e \\
\,^{56}{\rm Co}^*&\to\,^{56}{\rm Co}+\gamma
\end{align}
The cobalt-56 (lifetime around 100 days) then decays to iron-56 which also decays with some X-ray photons. Until SN 2014J, we had only observed the iron-56 decay line due to the fact that the lifetime of the above reaction is about 9 days and the ejecta are opaque to these lines due to Compton scattering in this same time-frame. SN 2014J provided $\gamma$-ray and X-ray emissions due to the cobalt-56, proving the theory correct.
Note that the shell remains very thick during this whole time. Wikipedia provides an image of SN 1006 (exploded in the year 1006, so it's now 1008 years old) that shows the expansion of the shell:
This shell is measured to be between 0.04 and 0.2 pc, which are roughly $1.2\cdot10^{12}$ km and $6.2\cdot10^{12}$ km thick, which is just shy of 1 lightyear. And after all this time, it is strong in radio, X-ray & $\gamma$-ray emissions (from this site):
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/133387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Why should an area vector point normal to the surface? Why is it that the direction of an area vector should be always along the normal drawn to the surface? Can't it also be some other angles with the plane?
| This convention is extremely convenient when doing things that physicist often like or need to do, such as computing a flux through a surface. When the area vector is chosen normal to the surface, one can simply take use an dot product to get what you're looking for.
In the context of differential forms, this also turns out to be the natural definition, since - at least in 3-space - the surface vector is essentially a cross product of two vectors spanning the surface.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/133466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
No saved queries yet
Save your SQL queries to embed, download, and access them later. Queries will appear here once saved.