Any object can emit and absorb radiation and the power of emission can be represented by the Stefan-Boltzmann law:

$$P=A\epsilon\sigma T^4$$

In many texts the net power radiated is the difference between the power emitted and the power absorbed:

$$P_{net}=A\epsilon\sigma (T^4-T_s^4)$$

where $$T_{s}$$ is the temperature of the surroundings.

Why can the surrounding and the object share the same $\epsilon$ ?

If we try to find out the radiation emitted from the surrounding it should be $P_s=A\epsilon_s\sigma T_s^4$, and if $\epsilon_s<\epsilon$, we will get a strange result that energy radiated from the surrounding is less than the radiation absorbed by the body from the surrounding. What am I missing?


2 Answers 2


The Stefan-Boltzmann law for net power radiated pertains to the object. That is, we're simply asking, how much radiation leaves this object (this depends on the object's emissivity), and how much radiation is absorbed by this object (this depends on the objects absorptivity). The emissivity and absorptivity in the equation you present thus pertain to the object, not the environment. That equation makes some assumptions. I couldn't find a good explanation for why the coefficients are what they are in the net power formula you posted, so I thought I'd take a step back and derive it.

The power emitted per unit area from the surroundings is

$$P_s=\epsilon_s \sigma T_s^4$$

The object will absorb a fraction of that based on its area and absorptivity:

$$P_a=\alpha \epsilon_s \sigma T_s^4$$

The object will emit:

$$P_e=\epsilon \sigma T^4$$

The net power delivered to the object is

$$P_{net} = P_a - P_e = \epsilon\sigma T^4 - \alpha \epsilon_s \sigma T_s^4$$

If the absorptivity and emissivity are equal, and $\epsilon_s = 1$ (blackbody), we get:

$$P_{net} = P_a - P_e = \epsilon \sigma (T^4-T_s^4)$$

So you'd have to assume that the surroundings perfectly emitting, and that the absorptivity and emissivity are equal. The latter is true under thermodynamic equilibrium or local thermodynamic equilibrium. See the Wikipedia page for Planck's law and in particular the section on Kirchhoff's Law.

  • $\begingroup$ The problem is, there still exist a situation that the emissivity of surrounding and the object are different. In the case that emissivity of the surrounding is smaller. Energy radiated from surrounding will be less than energy absorbed by the object if we use Pa=αϵsσTs^4 to calculate the absorption. Even you add a coefficient α the problem is not resolved. $\endgroup$
    – Kelvin S
    Jul 22, 2014 at 5:53
  • $\begingroup$ The above equation assumes that the radiation absorption only depends on the surround temperature and the property of the absorbent. It doesn't include the ability of the surrounding to emit radiation. What am I missing? $\endgroup$
    – Kelvin S
    Jul 22, 2014 at 6:00
  • $\begingroup$ @KelvinS Which equation? Yours? Yours makes some assumptions, I believe. Look at the first two equations in my answer. They include information about the environment and the object. $\epsilon_s$ prescribes ability of surrounding to radiate. Since $\alpha \leq 1$, $P_a \leq P_s$. $\endgroup$ Jul 22, 2014 at 15:17
  • $\begingroup$ Does your equation $$P_{net} = P_a - P_e = \epsilon\sigma T^4 - \alpha \epsilon_s \sigma T_s^4$$ allows heat flow from lower temperature to higher temperature? It seems it is possible that if $$\alpha \epsilon_s $$ is very small there exists a situation that even the surrounding temperature is higher than the object's temperature, energy still flows from the object to the surrounding. $\endgroup$
    – Kelvin S
    Jul 23, 2014 at 1:38

I think the key to the paradox is that you can't ignore the reflection coefficent. Lets say you have a lump of coal inside a shphere of polished steel. Yes, the coal emits much more heat than the steel; but that doesn't mean there is a net transfer of heat from the coal to the steel. Because the steel reflects back the excess heat.

There is a fixed relationship between the absorption, emission and reflection coefficients.


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