Новое решение

huz0 правка от 10 июн. 2026 г., 14:06

правка #18575 позже →

en/5.11.15.md +22 −0

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		+	### Statement
		+
		+	$5.11.15.$ [Insert the problem statement]
		+
		+	### 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 $N=\frac{N_{0}}{3}$.
		+
		+	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 force: $F=\frac{\mu_0}{t}=\frac{\bar{E}N_0}{3a};$
		+	Pressure: $P=\frac{F}{S}=\frac{F}{bc}=\frac{1}{3} \cdot \frac{\bar{E} \cdot N_0}{abc}=\frac{E_{total-of-the-photon-gas}}{3V}=\frac{w}{3}.$
		+
		+	#### Answer
		+
		+	[Insert a concise answer or boxed result]

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			### Statement

			$5.11.15.$ [Insert the problem statement]

			### 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 $N=\frac{N_{0}}{3}$.

			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 force: $F=\frac{\mu_0}{t}=\frac{\bar{E}N_0}{3a};$
			Pressure: $P=\frac{F}{S}=\frac{F}{bc}=\frac{1}{3} \cdot \frac{\bar{E} \cdot N_0}{abc}=\frac{E_{total-of-the-photon-gas}}{3V}=\frac{w}{3}.$

			#### Answer

			[Insert a concise answer or boxed result]