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Homework Statement:
 There are two mirrors which act as grey bodies which are a short distance away from one another. There is no transmission at each of the mirrors. What is the net heat transfer between the mirrors?
Relevant Equations:

[itex] q = \epsilon \sigma T^4 [/itex]
[itex] \alpha + \tau + \rho = 1 [/itex]
Hi,
So there is already a written solution which I have, but this is more a question about why we omit reflection that come back to the same mirror?
Method:
Let us consider one of the mirrors, we know it will emit a heat flux given by: [itex] q_{1} = \epsilon_{1} \sigma T_{1}^4 [/itex]. Given that we are dealing with grey bodies, then we know that [itex] \alpha = \epsilon [/itex] and we are told that [itex] \tau = 0 [/itex]. If we want to consider the heat that reaches and is absorbed by the other mirror, then we get the following sum (as a result of successive reflections):
[tex] q_{12} = \epsilon_{2} q_{1} + \rho_{2} \rho_{1} \epsilon_{2} q_{1} + \left( \rho_{2} \rho_{1} \right)^2 \epsilon_{2} q_{1} + \left( \rho_{2} \rho_{1} \right)^3 \epsilon_{2} q_{1} + ..... = \frac{\epsilon_{2} q_{1}}{1  \rho_{2} \rho_{1}} [/tex]
By symmetry, we can find that [itex] q_{21} = \frac{\epsilon_{1} q_{2}}{1  \rho_{2} \rho_{1}} [/itex].
Then we can find [itex] Q_{net} = Q_{12}  Q_{21} = \frac{A \epsilon_{1} \epsilon_{2} \sigma (T_{1}^4  T_{2}^4) } {1  \rho_{2} \rho_{1}} [/itex] (assuming areas are equal and that T_1 is greater than T_2). This is the answer that I am 'supposed' to get.
However, I was wondering why we don't also consider the rays that are emitted from a mirror and are reflected back towards itself (i.e. the [itex] \epsilon_{1} \rho_{2} q_{1}, etc... [/itex])? I think including these terms would change the q12 term to be: [itex] q_{12} = \frac{\epsilon_{2} ( q_{1} + \rho_{1} q_{2})}{1  \rho_{2} \rho_{1}} [/itex]
Thanks in advance.
So there is already a written solution which I have, but this is more a question about why we omit reflection that come back to the same mirror?
Method:
Let us consider one of the mirrors, we know it will emit a heat flux given by: [itex] q_{1} = \epsilon_{1} \sigma T_{1}^4 [/itex]. Given that we are dealing with grey bodies, then we know that [itex] \alpha = \epsilon [/itex] and we are told that [itex] \tau = 0 [/itex]. If we want to consider the heat that reaches and is absorbed by the other mirror, then we get the following sum (as a result of successive reflections):
[tex] q_{12} = \epsilon_{2} q_{1} + \rho_{2} \rho_{1} \epsilon_{2} q_{1} + \left( \rho_{2} \rho_{1} \right)^2 \epsilon_{2} q_{1} + \left( \rho_{2} \rho_{1} \right)^3 \epsilon_{2} q_{1} + ..... = \frac{\epsilon_{2} q_{1}}{1  \rho_{2} \rho_{1}} [/tex]
By symmetry, we can find that [itex] q_{21} = \frac{\epsilon_{1} q_{2}}{1  \rho_{2} \rho_{1}} [/itex].
Then we can find [itex] Q_{net} = Q_{12}  Q_{21} = \frac{A \epsilon_{1} \epsilon_{2} \sigma (T_{1}^4  T_{2}^4) } {1  \rho_{2} \rho_{1}} [/itex] (assuming areas are equal and that T_1 is greater than T_2). This is the answer that I am 'supposed' to get.
However, I was wondering why we don't also consider the rays that are emitted from a mirror and are reflected back towards itself (i.e. the [itex] \epsilon_{1} \rho_{2} q_{1}, etc... [/itex])? I think including these terms would change the q12 term to be: [itex] q_{12} = \frac{\epsilon_{2} ( q_{1} + \rho_{1} q_{2})}{1  \rho_{2} \rho_{1}} [/itex]
Thanks in advance.