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Can somebody please explain why radiation in a cavity (assuming thermal equilibrium has been reached) is isotropic ? I came across an argument involving a heterogeneous body placed in a cavity. The argument is as follows. A small heterogeneous body ( i.e. a body made of different materials) is placed inside a cavity until thermal equilibrium is reached. Then the body is rotated arbitrarily and it is observed that the system is still in equilibrium. This shows that the radiation is isotropic. I do not understand this argument. There is an inherent anisotropy in the heterogeneous body. So how can an anisotropic object serve as a benchmark ? I would argue using a homogeneous body i.e. a body that is isotropic. If a isotropic body in a cavity is rotated arbitrarily and it is noted that the rotation keeps the thermal equilibrium intact, then the radiation must be isotropic. Could somebody please comment on this ?

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Black-body spectra depends solely on the temperature of the body, not the material or the shape. It is in equilibrium with the radiation, so it absorbs as much as it emits

Consider your case: A small heterogeneous body ( i.e. a body made of different materials) is placed inside a cavity until thermal equilibrium is reached. Then the body is rotated arbitrarily and it is observed that the system is still in equilibrium.

First we know that the body is in equilibrium. Because it is in equilibrium it must absorb as much energy as it radiates.

Now the objetc is rotated in an arbitrary direction. The incomming radiation is independent of the position of the object, so a given point will receive as much as energy as the point placed there previously is emitting. But since the body is still in equilibrium, the emision of every point in this position must be the same af the emision of the point placed there before. Extending the argument to an arbitrary rotation, every point must emmit the same amount of radiation, hence black-body radiation must be isotropic.

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Radiation in a cavity is not necessarily isotropic, but then it wouldn't be blackbody radiation because some part of the cavity would be hotter than another.

Blackbody radiation results from something in equilibrium at a definite temperature. An anisotropic radiation field would transfer heat from one place to another leading to temperature changes/differences.

The example you give of a heterogeneous object is a good one. You imagine a sphere with one side black, the other a mirror, but all at the same temperature. Only one side absorbs and emits radiation. To come into equilibrium, the object must emit as much as it absorbs, but if the radiation field were anisotropic that would depend on its orientation. It would need to be at a higher temperature if it absorbed more radiation. Only in an isotropic radiation field will its temperature be independent of its orientation.

Now if we make the sphere homogeneously black, whatever orientation it is in it will absorb the same amount of radiation, and its temperature will not change if it is rotated. This is regardless of the isotropy of the radiation field.

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  • $\begingroup$ Thanks a lot for answering my question. I think I was confused because I was thinking of absorption coefficient and emissivity as independent quantities. However, I now realize that these two quantities are not independent. When thermal equilibrium is maintained, then a body with a higher absorption coefficient must also have a higher emissivity. If it absorbs more, then it must emit more. In effect, I think heterogeneity of a body does not matter due to this. $\endgroup$ – Tirthankar Mar 22 '17 at 12:50

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