# Radiation emission and absorption

1. Jul 21, 2014

### kelvin490

Any object can emit and absorb radiation and the power of emission can be represented by the Stefan-Boltzmann law: P=AεσT4

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

Pnet=Aεσ(T4-Ts4) where Ts is the temperature of surrounding.

Why the surrounding and the object can share the same ε ?

If we try to find out the radiation emitted from the surrounding it should be Ps=AεsσTs4, and if εs<ε we will get a strange result that energy radiated from the surrounding is less than the radiation absorbed by the body from the surrounding. How to solve it?

2. Jul 21, 2014

### 256bits

You would choose to do so if:
1. both are blackbodies where ε = 1,
2. both are a greybody where ε <1, and ε1 = ε2

If ε1 <> ε2 then obviously the emissivity of each is different.

3. Jul 22, 2014

### kelvin490

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 calculate the absorption power by P=AεσTs4.

The problem is, 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?

4. Jul 22, 2014

### Jano L.

The derivation of the formula
$$P = A\epsilon\sigma (T^4 -T_s^4)$$
for energy loss per unit time most probably assumes that the radiation of the surroundings is "black" (produced by body with emissivity =1). Only the radiation produced inside the body is "grey" (with emissivity $\epsilon < 1$).

The Kirchhoff law of radiation

says the absorber absorbs given radiation with efficiency proportional to emissivity $\epsilon$ of the absorber. If the radiation of the surroundings was grey (producing body had emissivity $\epsilon_s < 1$), it would be the intensity of this radiation which would depend on emissivity $\epsilon_s$, so its power would be given by $A\epsilon_s\sigmaT_s^4$ and the power of losses would be

$$P = A\epsilon\sigma (T^4 - \epsilon_sT_s^4)$$

5. Jul 22, 2014

### kelvin490

Does the equation $$P = A\epsilon\sigma (T^4 - \epsilon_sT_s^4)$$ allows heat flow from lower temperature to higher temperature? It seems it is possible that if $$\epsilon\epsilon_s$$ 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.

6. Jul 23, 2014

### Jano L.

Let us think of two bodies that face each other with area $A$. The above formula for $P$ gives energy loss of the colder body. It can be positive (body loses energy) or negative (body gains energy), depending on the value of $\epsilon_s$.

What happens to the warmer body is a different story. This body radiates energy $A\epsilon_s\sigma T_s^4$ and absorbs energy $A\epsilon\epsilon_s\sigma T^4$ per unit time, so total energy loss of the warmer body is

$$P_s = A\epsilon_s\sigma(T_s^4 - \epsilon T^4).$$

Since $T_s > T$ and $\epsilon \leq 1$, this is positive, no matter the values. In this situation, the warmer body always loses energy.

If $\epsilon_s$ is low enough, also the colder body loses energy. How is that possible?

Total loss of energy

$$P+P_s$$

goes into energy of heat radiation in space between the two bodies.

7. Jul 23, 2014

### Jano L.

This is valid description only as long as only the characteristic (from inside the body) radiation of the bodies is prevalent over the reflected radiation. In time, reflected (non-absorbed) radiation of both bodies returns to the body of origin and as soon as this happens, this changes the value of absorbed radiation per unit time so the above formulae are no longer accurate. I've never thought about this in more detail, but I suppose what should happen is that as time goes on, the two bodies approach their temperatures to a common value.