Cavity black-body radiation aboard a relativistic rocket

In summary, the homework statement says that if we have a cavity filled with black-body radiation, an this is placed onboard a relativistic rocket (uniform linear motion). I want to yield the transformation laws (from the proper frame S' joint to the rocket and moving with constant velocity V in respect to a coordinate frame S) for energy flux phi, energy density epsilon, energy E, pressure p, heat Q, and entropy S.
  • #1
Methavix
38
1

Homework Statement


If we have a cavity filled with black-body radiation, an this is placed onboard a relativistic rocket (uniform linear motion).
I want to yield the transformation laws (from the proper frame S' joint to the rocket and moving with constant velocity V in respect to a coordinate frame S) for energy flux phi, energy density epsilon, energy E, pressure p, heat Q, and entropy S.

Homework Equations


I know these equations expressed in the proper frame:

phi' = sigma'*T'^4

epsilon' = (4/c)*phi' = (4/c)*sigma'*T'^4

E' = epsilon'*psi' = (4/c)*sigma'*T'^4*psi'

p' = epsilon'/3 = (4/c)*(sigma'*T'^4)/3

dQ' = (16*sigma'*T'^4*dpsi')/(3*c)

S' = (16*sigma'*T'^3*dpsi')/(3*c)

where

sigma' = Stefan-Boltzmann constant in the proper frame
T' = temperature in the proper frame
psi' = volume of the cavity in the proper frame

The Attempt at a Solution


Being (it is a known formula):

E = (E’+p’*psi’*beta^2)/sqrt(1-beta^2)

by subsituting, i yield:

E = [(1+(beta^2)/3)/sqrt(1-beta^2)]*E’

And, if I want to transform pressure? I know in general that:

p = p’ ----> (4/c)*(sigma*T^4)/3 = (4/c)*(sigma'*T'^4)/3 ----> sigma*T^4 = sigma'*T'^4

then:

sigma = sigma’*gamma^4

but this implies also:

epsilon = epsilon’

and for the Boltzmann constant k:

k = k’*gamma (by the definition of sigma, if we suppose invariant the Planck constant, is it right?)

The heat transforms as follows:

dQ = (16*sigma'*T'^4*dpsi')/(3*c*gamma)

Now, there is a problem. In fact, if a calculate E directly from epsilon (considering right its transformation):

E = epsilon*psi = epsilon’*(psi’/gamma) = E’/gamma

But this is in disagree with the transformation known of the energy.

Where is my error?
Thanks
 
Last edited:
Physics news on Phys.org
  • #2
First, get a book on relativistic thermodynamics. Second, i hope you don't insinuate that sigma' is any different from sigma.

Daniel.
 
  • #3
i don't say that sigma' is different from sigma, my calculations said this. in fact at the end of my post i wrote: "where is my error" ? :)
 
  • #4
Hi,
Where do you have this relationship?

dQ' = (16*sigma'*T'^4*dpsi')/(3*c)

It is saying that the change of heat in the BB is related to the change of volume?
Thanks
David
 
  • #5
I think this is not the right way of using the formula:
E = (E’+p’*psi’*beta^2)/sqrt(1-beta^2)
Where ever you got the second term from, for sure in the rest frame E' there is no division by sqrt(1-beta^2).
the Energy-Momentum states: E^2(moving frame) = E'^2(rest frame) + c^2.p^2.
 

1. What is cavity black-body radiation?

Cavity black-body radiation refers to the thermal radiation emitted from a small, enclosed space, such as a cavity or box, that is in thermal equilibrium with its surroundings. This type of radiation is a result of the thermal energy of the particles inside the cavity and is independent of the material of the cavity walls.

2. How does cavity black-body radiation behave aboard a relativistic rocket?

In a relativistic rocket, the velocity of the rocket causes a Doppler effect on the emitted radiation, leading to a shift in the wavelength of the radiation. This means that the radiation emitted from the cavity will appear to have a higher energy and shorter wavelength from the perspective of an outside observer compared to an observer on board the rocket.

3. What is the significance of studying cavity black-body radiation aboard a relativistic rocket?

Studying cavity black-body radiation aboard a relativistic rocket allows scientists to better understand the effects of high velocities on thermal radiation. This has implications for fields such as astrophysics, where objects can reach extreme velocities, and can also provide insights into the behavior of matter at high temperatures and energies.

4. Does the size or shape of the cavity affect the behavior of black-body radiation aboard a relativistic rocket?

Yes, the size and shape of the cavity can affect the behavior of black-body radiation aboard a relativistic rocket. This is because the size and shape of the cavity can impact the distribution of thermal energy within it, leading to variations in the emitted radiation. However, these effects are typically small and can be accounted for in calculations.

5. Can cavity black-body radiation aboard a relativistic rocket be observed or measured?

Yes, cavity black-body radiation aboard a relativistic rocket can be observed and measured using specialized equipment and techniques. These measurements can provide valuable data for understanding the behavior of thermal radiation in extreme conditions and can also be used to validate theoretical predictions.

Similar threads

Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
11
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
804
  • Advanced Physics Homework Help
Replies
3
Views
2K
  • Advanced Physics Homework Help
Replies
4
Views
414
  • Advanced Physics Homework Help
Replies
2
Views
3K
  • Advanced Physics Homework Help
Replies
6
Views
1K
  • Advanced Physics Homework Help
Replies
6
Views
2K
  • Advanced Physics Homework Help
Replies
8
Views
2K
  • Atomic and Condensed Matter
Replies
0
Views
787
Back
Top