How to relate the radiation pressure to the internal energy?

[Outrageous]

How to relate the radiation pressure to the internal energy of an object??
How to get that formula P=u/3??


http://en.wikipedia.org/wiki/Stefan–Boltzmann_law#Thermodynamic_derivation

Thank you.

 

[Jano L.]

This relation holds for the EM radiation in equilibrium. It turns out that the Maxwell equations are such that for equilibrium radiation the Maxwell pressure is one third of the average Poynting energy density. For the derivation, look for example into the sections 34,35 of Landau Lifgarbagez "The Classical Theory of Fields", 4nd edition.

 

[Philip Wood]

See how you like this kinetic theory approach...

For N molecules of mass m in a container of volume V the pressure p is $$p = \frac{1}{3}\frac{N}{V} m \left\langle c^2 \right\rangle.$$
For a gas whose particles all travel at the same speed, we have simply
$$p = \frac{1}{3}\frac{N}{V} m c^2 .$$
$\frac{1}{6}\frac{N}{V} m c$ represents the total molecular momentum in a direction normal to and towards a given area of wall, per unit volume of container. This is standard kinetic theory. $\frac{N}{V}$ is the number of molecules per unit volume.

Now, for photons, we can replace $\frac{1}{6}\frac{N}{V} m c$ by $\frac{1}{6}\frac{u}{c}$ in which u is the photon energy per unit volume. This because the momentum of an individual photon is $\frac{h}{\lambda}=\frac{hf}{c}$ = photon's energy /c.
Hence we finish up with $p = \frac{1}{3}u.$ The factor is $\frac{1}{3}$ rather than $\frac{1}{6}$, just as for molecular gases, because the force on the wall is due to the change in momentum of particles when they collide, and the total normal momentum of reflected particles is equal and opposite to that of incident particles.

 

[Outrageous]

The factor is $\frac{1}{3}$ rather than $\frac{1}{6}$, just as for molecular gases, because the force on the wall is due to the change in momentum of particles when they collide, and the total normal momentum of reflected particles is equal and opposite to that of incident particles.

Thank you

Do you mean that 1/6 + 1/6 =1/3.

 

[Philip Wood]

Indeed I do. Or, better still: 1/6 - (-1/6) = 1/3.

 

[Outrageous]

This is the derivation I found. But I don't understand how can the time for the change of momentum is so large? Why is not the moment when it changes momentum but the time between collision.
But I think you may have better derivation , can you please guide?
Thank you

 

[Philip Wood]

I'll deal just with molecular gases, not photons. The molecular gas treatment can be adapted to photons as explained earlier.

1. Dividing by the time for a collision to take place would give you the mean force exerted by a molecule on the wall when it's colliding. But that's not what you want. You need the mean force on the wall all the time, whether or not a molecule is in the process of colliding.

2. I'm not fond of the derivation you give, though it was the standard one given to A-Level students in the UK, in the days when exam boards still required a derivation to be known. Students, quite reasonably, disliked the restriction of the cuboidal container, and the notion of a molecule bouncing back and forth between opposite walls, unimpeded by other molecules.

The derivation in the thumbnails below is, in my opinion, far superior. It is my version of a derivation I first met in a very old book, The Kinetic Theory of Gases by Sir James Jeans. Note that, in the thumbnails, I'm using u to mean x-wise velocity component, not energy per unit volume!

If you're happy with a bit of integration, there's another version of the Jeans argument which you might prefer. It deals differently with the 'grouping' of molecules by velocity.

 

[Outrageous]

Really thank you. One thing to ask why do the total momentum need to times prism volume / V ?

 

[Outrageous]

Can you please upload the integration one? Thank you.

 

[Philip Wood]

(prism volume)/(container volume) is the fraction of the N_u,v,w molecules (with the relevant velocities) in the whole container which are in the prism, and therefore which will hit area A of the wall in time $\Delta t .$

Yes, I'll post the integration method soon.

 

[Philip Wood]

As promised...

 

[Outrageous]

Thank you for uploading .
We integral the theta from 0 to pi/2 because we consider that strip volume can be a semi sphere?

 

[Philip Wood]

Exactly so, except that it's area, not volume.

 

[Outrageous]

Yup, should be area. Really thank you.