Pressure-energy relationship in photon gas

In summary, Schutz claims that for a relativistic electron gas, the formula p=ρ/3 approximately holds. This is used in deriving formulae for dense stars. To prove this relationship, we consider a gas of molecules with mass m and velocity v. The energy density is given by ρ=nm/√(1-v^2). When considering the pressure on one face of a cube, we can ignore components of impulse parallel to the face due to isotropy. Integrating over all electrons in a hemisphere of radius v from the centre, we find that p=ρ/3 approximately holds.
  • #1

andrewkirk

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Exercise 22 on p108 of Schutz's 'A first course in General Relativity' is to prove that, for an isotropic, monochromatic, photon gas, p=ρ/3, where p is pressure and ρ is mass-energy density.

When I try to do it I get p=ρ/6. I was hoping somebody could tell me where I'm going wrong.

Here is my working.

Say all photons have frequency ##\nu## and the number of photons per cubic metre is ##n##. Then ##\rho=n h \nu##.

Now consider one face of a cube of side length 1m. The pressure on that face is the component, parallel to the normal to the face, of the impulse delivered to that face in one second, by photons striking it from outside the cube. We can ignore components of impulse parallel to the face, because the isotropy will make them cancel each other out.

We measure that impulse by integrating over all photons that at time 0 are in a hemisphere of radius c from the centre of that face, with the base of the hemisphere coplanar with the face. No photons outside the hemisphere can strike that face in the next second.

We use spherical coordinates for the hemisphere and consider a volume element ##dV## from ##r## to ##r+dr##, ##\theta## to ##\theta+d\theta##, ##\phi## to ##\phi+d\phi##

The solid angle subtended by the cube's face at the centre of ##dV## is ##\frac{cos\theta}{r^2}##. Hence, since the gas is isotropic, the proportion of photons in ##dV## that will strike that face is ##\frac{cos\theta}{4\pi r^2}## and the number of photons in ##dV## at time 0 that will strike the face is ##\frac{n dV cos\theta}{4\pi r^2}##. The impulse delivered by those photons parallel to the face's normal will be ##\frac{h\nu}{c}cos\theta\frac{n dV cos\theta}{4\pi r^2}## .

Hence the pressure on the face is ##p = \int\frac{h\nu}{c}cos\theta\frac{n cos\theta}{4\pi r^2}dV
= \int_0^c\int_0^\frac{\pi}{2}\int_0^{2\pi}\frac{ h\nu}{c}cos\theta\frac{n cos\theta}{4\pi r^2}dr (r d\theta) (r sin\theta\ d\phi)##
##
= \int_0^c\int_0^\frac{\pi}{2}\int_0^{2\pi}\frac{n h\nu}{4\pi c}cos^2\theta sin\theta\ dr d\theta d\phi##


## = 2\pi c\frac{n h\nu}{4\pi c}[-\frac{1}{3}cos^3\theta]_0^\frac{\pi}{2}
= \frac{n h\nu}{6}##

Now the energy density is ##\rho=n h\nu## so we have p=ρ/6 rather than the desired p=ρ/3.

What have I done wrong?

Thank you in advance for all suggestions.
 
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  • #2
andrewkirk said:
Exercise 22 on p108 of Schutz's 'A first course in General Relativity' is to prove that, for an isotropic, monochromatic, photon gas, p=ρ/3, where p is pressure and ρ is mass-energy density.

When I try to do it I get p=ρ/6. I was hoping somebody could tell me where I'm going wrong.

Here is my working.

Say all photons have frequency ##\nu## and the number of photons per cubic metre is ##n##. Then ##\rho=n h \nu##.

Now consider one face of a cube of side length 1m. The pressure on that face is the component, parallel to the normal to the face, of the impulse delivered to that face in one second, by photons striking it from outside the cube. We can ignore components of impulse parallel to the face, because the isotropy will make them cancel each other out.

We measure that impulse by integrating over all photons that at time 0 are in a hemisphere of radius c from the centre of that face, with the base of the hemisphere coplanar with the face. No photons outside the hemisphere can strike that face in the next second.

We use spherical coordinates for the hemisphere and consider a volume element ##dV## from ##r## to ##r+dr##, ##\theta## to ##\theta+d\theta##, ##\phi## to ##\phi+d\phi##

The solid angle subtended by the cube's face at the centre of ##dV## is ##\frac{cos\theta}{r^2}##. Hence, since the gas is isotropic, the proportion of photons in ##dV## that will strike that face is ##\frac{cos\theta}{4\pi r^2}## and the number of photons in ##dV## at time 0 that will strike the face is ##\frac{n dV cos\theta}{4\pi r^2}##. The impulse delivered by those photons parallel to the face's normal will be ##\frac{h\nu}{c}cos\theta\frac{n dV cos\theta}{4\pi r^2}## .

Hence the pressure on the face is ##p = \int\frac{h\nu}{c}cos\theta\frac{n cos\theta}{4\pi r^2}dV
= \int_0^c\int_0^\frac{\pi}{2}\int_0^{2\pi}\frac{ h\nu}{c}cos\theta\frac{n cos\theta}{4\pi r^2}dr (r d\theta) (r sin\theta\ d\phi)##
##
= \int_0^c\int_0^\frac{\pi}{2}\int_0^{2\pi}\frac{n h\nu}{4\pi c}cos^2\theta sin\theta\ dr d\theta d\phi##


## = 2\pi c\frac{n h\nu}{4\pi c}[-\frac{1}{3}cos^3\theta]_0^\frac{\pi}{2}
= \frac{n h\nu}{6}##

Now the energy density is ##\rho=n h\nu## so we have p=ρ/6 rather than the desired p=ρ/3.

What have I done wrong?

Thank you in advance for all suggestions.

I'm not following all of that. But I suspect the problem is that you should assume your surface is reflecting if you want to determine pressure. If a photon of momentum ##\frac{h\nu}{c}## strikes a surface perpendicularly and it's absorbed the momentum transfer will be ##\frac{h\nu}{c}##. If it's reflected the momentum transfer is ##\frac{2h\nu}{c}##.
 
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  • #3
Oh, of course. That fixes it. Thank you.
 
  • #4
Pressure-energy density relationship in relativistic ELECTRON gas

This supplementary question is about the degree to which the same relationship p=ρ/3 holds in a relativistic electron gas.

Later in the same textbook, Schutz claims that for a 'relativistic electron gas', by which he means a gas of electrons traveling at very high speeds, the formula p = ρ/3 approximately holds (equation 10.79 on p273), and he uses this in deriving formulae for dense stars.

So now we try to prove that the relationship holds approximately for a gas of molecules, each of mass m and velocity v, rather than photons.
Say all electrons have velocity ##v## and the number of electrons per cubic metre is ##n##. Then ##\rho=\frac{n m}{\sqrt{1-v^2}}## (Schutz p42, formula for ##p^0##). This, and all equations below, is written in Schutz's 'special relativistic coordinates' in which time is measured in metres (one unit of time is the time required for light to travel one metre), so that ##c=1## and ##v## represents what would be ##\frac{v}{c}## in conventional metric units.

Now consider one face of a cube of side length 1m. The pressure on that face is the component, parallel to the normal to the face, of the impulse delivered to that face in one second, by electrons striking it from outside the cube. We can ignore components of impulse parallel to the face, because the isotropy will make such components from different particles cancel each other out.

We measure that impulse by integrating over all electrons that at time 0 are in a hemisphere of radius ##v## from the centre of that face, with the base of the hemisphere coplanar with the face. No electrons outside the hemisphere can strike that face in the next second.

We use spherical coordinates for the hemisphere and consider a volume element ##dV## from ##r## to ##r+dr##, ##\theta## to ##\theta+d\theta##, ##\phi## to ##\phi+d\phi##

The solid angle subtended by the cube's face at the centre of ##dV## is ##\frac{cos\theta}{r^2}##. Hence, since the gas is isotropic, the proportion of electrons in ##dV## that will strike that face is ##\frac{cos\theta}{4\pi r^2}## and the number of electrons in ##dV## at time 0 that will strike the face is ##\frac{n\ dV\ cos\theta}{4\pi r^2}##. The impulse in the direction of the face's normal that is delivered by those electrons will be ##(2\frac{m v}{\sqrt{1-v^2}}cos\theta)\frac{n\ dV\ cos\theta}{4\pi r^2}## . The factor of 2 is there because double the electron's momentum is transferred when it's reflected, which is what we assume happens (ie in a gas we assume that gas particles bounce off a surface on which pressure is being measured, rather than sticking to it). The formula for the particle's momentum ##\frac{m v}{\sqrt{1-v^2}}## comes from Schutz p42, the formula for ##p^1##.

Hence the pressure on the face is:
\begin{align*}p &= \int 2\frac{m v}{\sqrt{1-v^2}}cos\theta\frac{n cos\theta}{4\pi r^2}dV\\
&= \int_0^v\int_0^\frac{\pi}{2}\int_0^{2\pi}2 \frac{m v}{\sqrt{1-v^2}}cos\theta\frac{n cos\theta}{4\pi r^2}dr\ (r d\theta)\ (r\ sin\theta\ d\phi)\\
&= 2\int_0^v\int_0^\frac{\pi}{2}\int_0^{2\pi}\frac{n m v}{4\pi\sqrt{1-v^2}}cos^2\theta\ sin\theta\ dr\ d\theta\ d\phi\\
&= 4\pi \frac{n m v^2}{4\pi\sqrt{1-v^2}}\int_0^c cos^2\theta\ sin\theta\ d\theta\\
&= \frac{n m v^2}{\sqrt{1-v^2}}\Big[-\frac{1}{3}cos^3\theta\Big]_0^\frac{\pi}{2}\\
&= \frac{n m v^2}{3\sqrt{1-v^2}}\\
&= \frac{v^2}{3}\rho\end{align*}
We have a superfluous factor of ##v^2##. But if ##v## is high enough, the equation p=ρ/3 is approximately true.

##v## has to be very high though. To get the error below 10% we have to have ##v > 0.95 c##.

Is this really the approximation to which Schutz refers, or is it possible to demonstrate that convergence is faster than that (ie that electron speeds don't have to be so very high for the formula to be approximately true)?
 
  • #5


Dear researcher,

Thank you for sharing your working and question regarding the pressure-energy relationship in photon gas. After reviewing your calculations, I believe I have identified where you have gone wrong.

In your calculation for pressure, you have integrated over the entire hemisphere of radius c, but only photons within a certain distance from the face will actually contribute to the pressure on that face. This distance is not the full radius c, but rather the distance from the center of the cube to the face, which is 1/2 the side length of the cube. This means that your integral should only be over the range of r=0 to r=1/2, rather than r=0 to r=c.

Additionally, in your calculation for the solid angle, the correct expression should be ##\frac{cos\theta}{2\pi r^2}##, as the face only covers half of the total solid angle of 4π. This will result in a factor of 2 in your final expression for pressure, giving you the desired result of p=ρ/3.

I hope this helps clarify your calculations and leads you to the correct result. Keep up the good work in your research!
 

What is the pressure-energy relationship in photon gas?

The pressure-energy relationship in photon gas, also known as the photon gas equation of state, describes the relationship between the pressure and energy of a gas made up of photons. It states that the pressure of the gas is directly proportional to the energy density, or the number of photons per unit volume, and is inversely proportional to the temperature of the gas.

How is the pressure-energy relationship in photon gas derived?

The pressure-energy relationship in photon gas is derived from the laws of thermodynamics, specifically the ideal gas law and the Stefan-Boltzmann law. By considering the energy and momentum of individual photons, it can be shown that the pressure of a photon gas is given by P = (1/3)U/V, where P is the pressure, U is the energy density, and V is the volume of the gas.

What are the implications of the pressure-energy relationship in photon gas?

The pressure-energy relationship in photon gas has important implications in cosmology and astrophysics, as it helps to understand the behavior of radiation in the early universe and in extreme environments such as in stars and black holes. It also plays a role in the study of the cosmic microwave background radiation and the prediction of the number of photons in the universe.

Does the pressure-energy relationship in photon gas apply to all types of gases?

No, the pressure-energy relationship in photon gas specifically applies to gases made up of photons, such as electromagnetic radiation. It does not apply to gases made up of other particles, such as atoms or molecules, which have different equations of state.

Are there any limitations to the pressure-energy relationship in photon gas?

Yes, there are some limitations to the pressure-energy relationship in photon gas. It is based on certain assumptions, such as the gas being in thermal equilibrium and the photons being non-interacting. It also does not take into account the effects of quantum mechanics, which may be important at high energies and densities.

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