How to relate the radiation pressure to the internal energy?

In summary, the relationship between radiation pressure and internal energy of an object can be derived using the Maxwell equations for equilibrium radiation and the kinetic theory approach. The pressure is equal to one third of the average Poynting energy density, and for photons, the momentum can be replaced by the photon energy per unit volume. This results in a factor of 1/3, similar to that for molecular gases. Different derivations exist, with some using integration to consider the grouping of molecules by velocity.
  • #1
Outrageous
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  • #2
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.
 
  • #3
See how you like this kinetic theory approach...

For N molecules of mass m in a container of volume V the pressure p is [tex]p = \frac{1}{3}\frac{N}{V} m \left\langle c^2 \right\rangle.[/tex]
For a gas whose particles all travel at the same speed, we have simply
[tex]p = \frac{1}{3}\frac{N}{V} m c^2 .[/tex]
[itex]\frac{1}{6}\frac{N}{V} m c[/itex] 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. [itex]\frac{N}{V}[/itex] is the number of molecules per unit volume.

Now, for photons, we can replace [itex]\frac{1}{6}\frac{N}{V} m c[/itex] by [itex]\frac{1}{6}\frac{u}{c}[/itex] in which u is the photon energy per unit volume. This because the momentum of an individual photon is [itex]\frac{h}{\lambda}=\frac{hf}{c}[/itex] = photon's energy /c.
Hence we finish up with [itex]p = \frac{1}{3}u.[/itex] The factor is [itex]\frac{1}{3}[/itex] rather than [itex]\frac{1}{6}[/itex], 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.
 
  • #4
Philip Wood said:
The factor is [itex]\frac{1}{3}[/itex] rather than [itex]\frac{1}{6}[/itex], 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.
 
  • #5
Indeed I do. Or, better still: 1/6 - (-1/6) = 1/3.
 
  • #6
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
 

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  • #7
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.
 

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  • #8
Really thank you. One thing to ask why do the total momentum need to times prism volume / V ?
 
  • #9
Can you please upload the integration one? Thank you.
 
  • #10
(prism volume)/(container volume) is the fraction of the Nu,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 [itex]\Delta t .[/itex]

Yes, I'll post the integration method soon.
 
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  • #11
As promised...
 

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  • #12
Thank you for uploading .
We integral the theta from 0 to pi/2 because we consider that strip volume can be a semi sphere?
 
  • #13
Exactly so, except that it's area, not volume.
 
  • #14
Yup, should be area. Really thank you.
 

1. How does radiation pressure affect the internal energy of a system?

Radiation pressure is a measure of the force exerted by electromagnetic radiation on a surface. This force can transfer energy to the particles in the system, increasing their internal energy. The higher the radiation pressure, the more energy is transferred and the greater the impact on the system's internal energy.

2. Can radiation pressure be directly related to the internal energy of a material?

While radiation pressure does contribute to the internal energy of a system, it is not the only factor. Other forms of energy, such as thermal energy, also play a role in determining the total internal energy of a material. However, the amount of radiation pressure can be used to calculate the change in internal energy over time.

3. How is radiation pressure calculated and measured?

Radiation pressure is typically calculated using the formula P = E/c, where P is the pressure, E is the energy of the radiation, and c is the speed of light. It can be measured using specialized instruments such as a radiation pressure balance or a laser power meter.

4. What factors can affect the radiation pressure and its impact on internal energy?

The radiation pressure on a material can be influenced by several factors, including the intensity and wavelength of the radiation, the material's reflectivity and absorptivity, and its physical properties such as density and temperature. These factors can also affect how much of the radiation energy is converted into internal energy.

5. How does radiation pressure relate to other forms of energy, such as thermal energy?

Radiation pressure can be considered a form of energy transfer, as it can cause an increase in the internal energy of a material. However, it is distinct from thermal energy, which is a measure of the average kinetic energy of particles in a system. Radiation pressure and thermal energy can both affect the internal energy of a material, but in different ways and through different mechanisms.

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