<p>Photons always travel at the speed of light, wherever they are. This is a consequence of them being massless particles.</p>
<p>Although &quot;light&quot; travels slower than <span class="math-container">$c$</span> in a refractive medium, the individual photons making up a light beam do not.</p>
<p>e.g. <a href="https://physics.stackexchange.com/questions/19203/what-is-the-mass-of-a-photon-in-non-empty-spaces">https://physics.stackexchange.com/questions/19203/what-is-the-mass-of-a-photon-in-non-empty-spaces</a></p>

CC BY-SA 4.0
2020-10-15T07:15:30.403Z
2020-10-15T07:25:42.637Z
2020-10-15T07:25:42.637Z
ProfRob
<blockquote>
<p><strong>Brief context</strong>: when studying <a href="https://en.wikipedia.org/wiki/Stellar_structure" rel="nofollow noreferrer">Stellar Structure</a> one of the main goals, aside of determining equations that describe quantities of interest of the system, is to also determine equations of state that allows us to approximate solutions for the coupled differential equations system we find.</p>
</blockquote>
<p>An important step is to find equations of state, e.g., an expression for the Pressure <span class="math-container">$P$</span> of the system. Following common derivations on this subject (e.g., see <em>Rose, Advanced Stellar Astrophysics</em> or <em>Clayton, Principles of Stellar Evolution and Nucleosynthesis</em>) an integral for <span class="math-container">$P$</span> over a cone of particles with momentum <span class="math-container">$p$</span> would be</p>
<p><span class="math-container">$$P =\int\limits_0^{\pi/2}\int\limits_0^\infty 2 \, v_p \,\cos^2\theta \, n(\theta,p)\,d\theta\,dp$$</span></p>
<p>where <span class="math-container">$n(p$</span>) is the gas' distribution w.r.t <span class="math-container">$p$</span>. An example could be <strong>an ideal monoatomic gas, in the non degenerate, non relativistic case</strong>. In particular:</p>
<p><span class="math-container">$$n(p) \, dp = 4\pi p^2 n \frac{\exp\left(-\frac{p^2}{2mkT}\right)}{(2\pi m kT)^{3/2}}\,dp$$</span>
and <span class="math-container">$p = m \, v_p$</span> so</p>
<p><span class="math-container">$$P = n k T$$</span></p>
<blockquote>
<p><strong>The problem</strong>: let us find then this same <span class="math-container">$P$</span> for a Photon Gas (radiation pressure) considering that now <span class="math-container">$n(p) \,dp$</span> is given by a <a href="https://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_statistics#Bose%E2%80%93Einstein_distribution" rel="nofollow noreferrer">Bose-Einstein Distribution</a>. In particular <span class="math-container">$$n(p)\,dp = \frac{8\pi \, p^2\,dp}{h^3}\frac{1}{e^{pc/kT}-1}$$</span> where <span class="math-container">$p = h\nu/c$</span> and <span class="math-container">$v_p \overset{!}{=} c$</span>.</p>
</blockquote>
<p>The procedure is very straightforward:</p>
<ol>
<li><p><span class="math-container">$n(p)\,dp$</span> can be written as <span class="math-container">$$n(\nu)\,d\nu = \frac{8\pi}{c^3}\frac{\nu^2 d\nu}{e^{h\nu/kT}-1}$$</span> so the integral becomes <span class="math-container">$$P = \frac{8\pi h}{3c^3}\left(\frac{kT}{h}\right)^4 \int\limits_0^\infty \frac{x^3\,dx}{e^x-1}$$</span> where <span class="math-container">$$x \equiv \frac{h\nu}{kT}$$</span>. Here the <span class="math-container">$\int$</span> (e.g., related to <a href="https://math.stackexchange.com/questions/3228573/i-need-some-help-with-the-integral-int-0-infty-fracx3ex1-dx/3228589#3228589">this question</a>) can be either solved by using <a href="https://es.wikipedia.org/wiki/Funci%C3%B3n_polilogar%C3%ADtmica" rel="nofollow noreferrer">Polylogarithms</a>,  <a href="https://en.wikipedia.org/wiki/Riemann_zeta_function" rel="nofollow noreferrer">Riemann zeta and Gamma functions</a>, complex analysis, etc.</p>
</li>
<li><p>It's <em>easy</em> to check that <span class="math-container">$$P = \frac{8\pi k^4}{3c^3h^3} T^4 \zeta(4)\Gamma(4)$$</span> then <span class="math-container">$$P = \frac{8\pi^5 k^4}{45c^3h^3}T^4 = \frac{4\sigma}{3c} T^4$$</span> which matches evenly.</p>
</li>
</ol>
<blockquote>
<p><strong>The question</strong>: if we're dealing with Stellar structure, potentially <em>dense</em> environments (the interior of a star!) then EM radiation is not necessarily propagating in vacuum (in fact, I guess it's not the case!). Then would be <span class="math-container">$v_p \overset{!}{=} n \, c  \leq c$</span> but this is not usually considered on calculations. Why is this? <strong>Do you know any resource where anybody have examined this and see if star interiors are affected? (it might be negligible)</strong></p>
</blockquote>
<p>Thanks in advance,</p>

CC BY-SA 4.0
2020-10-15T07:12:07.367Z
What does your strange symbol ! over an equals sign mean? Is this $\neq$ ?
ProfRob
CC BY-SA 4.0
2020-10-15T20:13:25.523Z
@RobJeffries with $\overset{!}{=}$ I meant to say that it is equal but watch out (!) because I don't buy it blindly. Thanks a lot for asking
nuwe
<astrophysics><stellar-astrophysics>
photon gas pressure and Independence on refraction index
nuwe
# photon gas pressure and Independence on refraction index

(Asked by: username_0 on 2020-10-14)

> **Brief context**: when studying [Stellar Structure](https://en.wikipedia.org/wiki/Stellar_structure) one of the main goals, aside of determining equations that describe quantities of interest of the system, is to also determine equations of state that allows us to approximate solutions for the coupled differential equations system we find.

An important step is to find equations of state, e.g., an expression for the Pressure $P$ of the system. Following common derivations on this subject (e.g., see *Rose, Advanced Stellar Astrophysics* or *Clayton, Principles of Stellar Evolution and Nucleosynthesis*) an integral for $P$ over a cone of particles with momentum $p$ would be

$$P =\int\limits\_0^{\pi/2}\int\limits\_0^\infty 2 \, v\_p \,\cos^2\theta \, n(\theta,p)\,d\theta\,dp$$

where $n(p$) is the gas' distribution w.r.t $p$. An example could be **an ideal monoatomic gas, in the non degenerate, non relativistic case**. In particular:

$$n(p) \, dp = 4\pi p^2 n \frac{\exp\left(-\frac{p^2}{2mkT}\right)}{(2\pi m kT)^{3/2}}\,dp$$
and $p = m \, v\_p$ so

$$P = n k T$$

> **The problem**: let us find then this same $P$ for a Photon Gas (radiation pressure) considering that now $n(p) \,dp$ is given by a [Bose-Einstein Distribution](https://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_statistics#Bose%E2%80%93Einstein_distribution). In particular $$n(p)\,dp = \frac{8\pi \, p^2\,dp}{h^3}\frac{1}{e^{pc/kT}-1}$$ where $p = h\nu/c$ and $v\_p \overset{!}{=} c$.

The procedure is very straightforward:

1. $n(p)\,dp$ can be written as $$n(\nu)\,d\nu = \frac{8\pi}{c^3}\frac{\nu^2 d\nu}{e^{h\nu/kT}-1}$$ so the integral becomes $$P = \frac{8\pi h}{3c^3}\left(\frac{kT}{h}\right)^4 \int\limits\_0^\infty \frac{x^3\,dx}{e^x-1}$$ where $$x \equiv \frac{h\nu}{kT}$$. Here the $\int$ (e.g., related to [this question](https://math.stackexchange.com/questions/3228573/i-need-some-help-with-the-integral-int-0-infty-fracx3ex1-dx/3228589#3228589)) can be either solved by using [Polylogarithms](https://es.wikipedia.org/wiki/Funci%C3%B3n_polilogar%C3%ADtmica), [Riemann zeta and Gamma functions](https://en.wikipedia.org/wiki/Riemann_zeta_function), complex analysis, etc.
2. It's *easy* to check that $$P = \frac{8\pi k^4}{3c^3h^3} T^4 \zeta(4)\Gamma(4)$$ then $$P = \frac{8\pi^5 k^4}{45c^3h^3}T^4 = \frac{4\sigma}{3c} T^4$$ which matches evenly.

> **The question**: if we're dealing with Stellar structure, potentially *dense* environments (the interior of a star!) then EM radiation is not necessarily propagating in vacuum (in fact, I guess it's not the case!). Then would be $v\_p \overset{!}{=} n \, c \leq c$ but this is not usually considered on calculations. Why is this? **Do you know any resource where anybody have examined this and see if star interiors are affected? (it might be negligible)**

Thanks in advance,

**Tags:** astrophysics, stellar-astrophysics

> **Comments:**
> **username_1**: What does your strange symbol ! over an equals sign mean? Is this $\neq$ ?  
> **username_0**: @RobJeffries with $\overset{!}{=}$ I meant to say that it is equal but watch out (!) because I don't buy it blindly. Thanks a lot for asking  

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## 1 answers

### Answer by username_1 on 2020-10-15. Score: 1 | Accepted answer

Photons always travel at the speed of light, wherever they are. This is a consequence of them being massless particles.

Although "light" travels slower than $c$ in a refractive medium, the individual photons making up a light beam do not.

e.g. <https://physics.stackexchange.com/questions/19203/what-is-the-mass-of-a-photon-in-non-empty-spaces>


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