Box of Radiation

 Box of Radiation

Consider a box mirrored on the inside that contains disordered radiation. The radiation will then have the stress-energy-momentum (SEM) tensor as a perfect fluid. See diagram

The SEM tensor of a fluid of radiation in a zero momentum frame is   

(1)

 where p = r₀c²/3. Let k₀  = rest mass density of walls. The SEM tensor for the sides of the box (i.e. the sides parallel to the x-axis) is

 (2)

 Transforming The SEM tensor for the sides of the box from O to O' moving in the x-direction with respect to O

 (3) T^m'n'= L^m'_aL^n'_bT^ab

 gives

(4) k = T^0'0'/c² = g²(k₀  + b²P_xx /c²)

The mass of the walls normal the x-axis transform normally, i.e. in the same way as a point particle. Since the walls are completely reflective the force on each wall is 2pA and therefore P_xx = 2pA/da 

(5) k = g²(k₀  + 2b²pA /c²da) 

Letting B = total mass of walls in O’ and B₀ = total mass of walls in O

(6) B = (1/3)gB₀ + (Lda/g)[g²(k₀  + 2b²pA /c²da)] = (1/3)gB₀ + Ldag(k₀  + 2b²pA /c²da)] = (1/3)gB₀ + gk₀ Lda + 2gb²pLA /c²

Substituting p = r₀c²/3, A = L²

(7) B = (1/3)gB₀ + gk₀ Lda + (2/3)b²r₀L³ /g = gB₀ + (2/3)b²r₀L³ /g

Similarly the mass density box's radiation in O’ is  

(8) r =  g²(r₀  + b²p /c²) =  g²(r₀  + b²r₀ /3)

 and therefore the mass of the radiation in O’ is

 (9) U = r V = (L³/g)[g²(r₀  + b²r₀ /3)] = gL³(r₀  + b²r₀ /3)

(10) U = gr₀L³  + gb²r₀ L³/3

Let U₀ = r₀L³ = mass of radiation in O. Let M  = total mass of Radiation + Box in O’. Then

(11) M = B + U = gB₀ + (2/3)b²r₀L³ /g + gr₀L³  + gb²r₀ L³/3

(12) M = B + U = gB₀ + gU₀ + gb²r₀ L³

 (13) M = g(B₀ + U₀) + gb²r₀ L³

(14) M (v) = gM₀ + gb²r₀ L³

 Where M₀ = B₀ + U₀ = M(0)