$5.11.15.$ The photon energy $E$ is related to its momentum $p$by the relation $p = \frac{E}{c},$
where
$c$ is the photon velocity equal to the speed of light. Prove that the photon gas pressure $P$ is related to the energy density $ω$ by the relation $P = \frac{w}{3}.$
### Solution
We consider a mirror-walled box with dimensions $a\times b\times c$.
$N_0$ photons bounce off the walls, making pressure, like in usual gas. Collisions are, of course, elastic.
Gas has isotrophy of directions of photons' speeds, also energy distribution is equal alongside $x, y, z$ axes. This can be shown with Boltzman's distribution in impulse phase space, since photon gas is closest to an ideal gas physical model. (probability of photon collision is $0$).
Since all photons trave with the same speed $c$, we can talk about some average energy $\bar{E}$ that every photon have. Since gas has isotrophy of directions of photons' speeds, we can divide the gas into 3 different independent gases with same amount of photons $N=\frac{N_{0}}{3}$.
@@ -15,7 +15,7 @@Solution
Let's look at those, who moves parallel to $a$.
Once in a time $t=\frac{2a}{c}$ every photon hits the right wall once, giving impulse $\mu=2\frac{\bar{E}}{c}$.
−
Total impulse: $\mu_0=\frac{n_{0}}{3}\cdot2\frac{\bar{E}}{c};$
+
Total impulse: $\mu_0=\frac{N_{0}}{3}\cdot2\frac{\bar{E}}{c};$
Total force: $F=\frac{\mu_0}{t}=\frac{\bar{E}N_0}{3a};$
$5.11.15.$ The photon energy $E$ is related to its momentum $p$by the relation $p = \frac{E}{c},$
$5.11.15.$ The photon energy $E$ is related to its momentum $p$by the relation $p = \frac{E}{c},$
where
where
$c$ is the photon velocity equal to the speed of light. Prove that the photon gas pressure $P$ is related to the energy density $ω$ by the relation $P = \frac{w}{3}.$
$c$ is the photon velocity equal to the speed of light. Prove that the photon gas pressure $P$ is related to the energy density $ω$ by the relation $P = \frac{w}{3}.$
### Solution
### Solution
We consider a mirror-walled box with dimensions $a\times b\times c$.
We consider a mirror-walled box with dimensions $a\times b\times c$.
$N_0$ photons bounce off the walls, making pressure, like in usual gas. Collisions are, of course, elastic.
$N_0$ photons bounce off the walls, making pressure, like in usual gas. Collisions are, of course, elastic.
Gas has isotrophy of directions of photons' speeds, also energy distribution is equal alongside $x, y, z$ axes. This can be shown with Boltzman's distribution in impulse phase space, since photon gas is closest to an ideal gas physical model. (probability of photon collision is $0$).
Gas has isotrophy of directions of photons' speeds, also energy distribution is equal alongside $x, y, z$ axes. This can be shown with Boltzman's distribution in impulse phase space, since photon gas is closest to an ideal gas physical model. (probability of photon collision is $0$).
Since all photons trave with the same speed $c$, we can talk about some average energy $\bar{E}$ that every photon have. Since gas has isotrophy of directions of photons' speeds, we can divide the gas into 3 different independent gases with same amount of photons $N=\frac{N_{0}}{3}$.
Since all photons trave with the same speed $c$, we can talk about some average energy $\bar{E}$ that every photon have. Since gas has isotrophy of directions of photons' speeds, we can divide the gas into 3 different independent gases with same amount of photons $N=\frac{N_{0}}{3}$.
@@ -15,7 +15,7 @@Solution
Let's look at those, who moves parallel to $a$.
Let's look at those, who moves parallel to $a$.
Once in a time $t=\frac{2a}{c}$ every photon hits the right wall once, giving impulse $\mu=2\frac{\bar{E}}{c}$.
Once in a time $t=\frac{2a}{c}$ every photon hits the right wall once, giving impulse $\mu=2\frac{\bar{E}}{c}$.
Total impulse: $\mu_0=\frac{n_{0}}{3}\cdot2\frac{\bar{E}}{c};$
Total impulse: $\mu_0=\frac{N_{0}}{3}\cdot2\frac{\bar{E}}{c};$
Total force: $F=\frac{\mu_0}{t}=\frac{\bar{E}N_0}{3a};$
Total force: $F=\frac{\mu_0}{t}=\frac{\bar{E}N_0}{3a};$
