# Stefan-Boltzmann Law

1. May 29, 2004

### Michael2003

I've been reading about the Stefan-Boltzmann law, but there is one thing that I don't understand. Why is it T^4? I can't think of anywhere that the 4 is coming from and can't find anything about this with Google searches.

2. May 29, 2004

### Hydr0matic

3. May 29, 2004

### Janitor

I know it can be derived from Planck's blackbody law, but historically that is doing things in reverse. I remember some author expressed surprise that quantum mechanics got its start from the study of blackbody radiation rather than some other phenomenon that would have allowed easier analysis, such as the photoelectric effect or specific heat or atomic spectra.

Last edited: May 29, 2004
4. May 29, 2004

### quartodeciman

Joseph Stefan induced this relation from some experimental data (1884). His student, Ludwig Boltzmann, derived the 4th power of absolute temperature T relationship from radiation thermodynamics and electromagnetic waves. That is why his name is attached to Stefan's discovery.

I can't follow all of this, but here is a bare outline of how it goes.

-----------

TdS = dU + pdV

{combined 1st and 2nd laws of thermodynamics in differentials, T = absolute thermodynamic temperature of radiation, S = entropy for a reversible process, U = internal energy of radiation, p = radiation pressure against container walls, V = volume of radiation container, thermal equilibrium between radiation and container walls, treat radiation as if it is a gas}

T(S,V) = (U,V) + p

{(S,V) means partial derivative of S wrt V, (U,V) means partial derivative of U wrt V, and so on, of course (V,V)=1}

--

H = U + pV

{H is enthalpy function}

dH = dU + (dp)V + pdV = (dU + pdV) + (dp)V = TdS + V(dp)

{substitute from combined thermodynamics laws}

But,

dH = (H,S)dS + (H,p)dp

{partial derivatives of H}

(H,S) = T and (H,p) = V

{substitutions from prior result for dH, dH is an exact differential}

((H,S),p) = ((H,p),S)

{partial second derivative}

(T,p) = (V,S)

{substitutions from preceding equations}

(S,V) = (p,T)

(inverse partial derivatives}

T(p,T) = (U,V) + p

{substitution into the 2nd equation from the top}

--

p = ë/3

{ë = energy density of radiation, *some detailed electromagnetic derivation of waves reflecting off container walls producing pressure, the pressure is isotropic}

U = ëV

(internal energy of radiation is energy density times interior volume of container}

(p,T) = (ë,T)/3
(U,V) = ë

{partial derivatives for preceding two equations}

T(ë,T)/3 = ë + ë/3 = 4ë/3

{substitutions into preceding differential equation result}

(ë,T)/ë = 4/T

{separate the variables for purpose of integration}

(ln ë,T) = 4/T = 4(ln T,T)

ln ë = 4(ln T) + constant = ln(T4) + constant

ë = (econstant)T4

{e is base of natural logarithm function}

{DONE}

So, radiation density is proportional to fourth power of absolute thermodynamic temperature of radiation under these conditions.

-----------

I got this from looking at Longair,Theoretical Concepts in Physics,Cambridge(1984),section 8.2.3 and earlier sections as cited

5. May 30, 2004

### Janitor

Thank you, Quartodeciman.

One of your equations has coefficients of:

1/3 on the left side

1 + 1/3 = 4/3 on the right side.

Multiplying through by 3 gave the coefficient of 4 which wound up being the power of 4 in the Stefan-Boltzman law.

One step involved "interior volume of container." Can you tell if the three-dimensionality of space carries through the derivation (without cancelling out at some point) so as to effect the power to which absolute temperature is raised?

6. May 30, 2004

### quartodeciman

Three dimensionality figures into the equation

p = ë/3

.

Radiation energy density is assumed to be uniformly homogeneous throughout the interior of the enclosure (I forgot to state that!) and the resultant radiation pressure at the wall is directed along just one of the three dimensions. But I find this part of the demonstration a bit too tough to follow in detail. The pressure equation is at least plausible. The energy density is manifest always in all three dimensions, while the pressure at any wall is registered only along the normal direction, one dimension. (I'm kinda repeating myself!) Since pressure and energy density are treated as dimensionally the same (force/area and force*length/volume=force/area), the equation has a rough plausibility about it, IMO.

I don't know what happens to the energy density and pressure in case space dimensionality were N. Is it p = ë/N ? What would happen to volume V as a thermodynamic variable in this case?

7. May 30, 2004

### Janitor

I believe you are right. I wonder if you would get, for an N-dimensional space, 1/N on the left side, and 1 + 1/N on the right side. Multiplying through by N, you would derive that energy density goes as T to the power of N+1.