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If I shine a light on a parabolic mirror parallel to optic axis, all of the rays will be reflected through its focal point. What happens if I heat it? Is heat going to be radiated through the focal point or will it be spread in all directions?

Edit: The question is about direction of heat emissions. Is heat emitted in only one direction, perpendicular to the surface or in all directions from each of the points on the mirror? Mirror is heated internally, it is not reflecting anything.

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    $\begingroup$ Note that there are different kinds of concave mirrors, not all of which actually have a focal point. Even for those that do, it's not generally true that if you shine light on the mirror it'll be reflected through the focal point. That's just a special case, e.g. for a parabolic mirror with parallel on-axis incoming light, or for an elliptic mirror light coming from the second focal point. Light coming from elsewhere may still be focused at least somewhat, but not at the focal point. $\endgroup$ Jul 5 at 9:10
  • $\begingroup$ Good point, I've added parallel to optic axis. $\endgroup$ Jul 5 at 12:33
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    $\begingroup$ @LukaSantrić: The "parallel to the optic axis" => "through the focal point" relation holds for parabolic mirrors. But there are infinitely many concave shapes other than parabolic. $\endgroup$
    – MSalters
    Jul 5 at 12:49

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Is heat emitted in only one direction, perpendicular to the surface or in all directions from each of the points on the mirror?

Thermal radiation is emitted in all directions at each point on the surface. It is not coherent in either frequency or direction. That said, a good reflector like a mirror will not be a particularly good emitter of thermal radiation. It will emit substantially less than a black body at the same temperature. Some of that emitted radiation will reflect off the concave surface, at odd but non-random angles.

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    $\begingroup$ Equally, it's not a very good absorber either. Since the surface is concave, some emitted photons will hit another spot on the mirror. When these photons hit, they will likely be reflected instead of absorbed. As a result, at a macroscopic distance from the mirror, the angular distribution will depend on the shape of the mirror. $\endgroup$
    – MSalters
    Jul 5 at 11:39
  • $\begingroup$ @MSalters very good point, thanks. I will update my answer $\endgroup$
    – Dale
    Jul 5 at 11:44
  • $\begingroup$ Makes sense. Is energy of emissions uniformly distributed over all directions? Is this general property of all the materials. Can a material which focuses more energy in specific direction without changing the shape of the mirror exist? $\endgroup$ Jul 5 at 12:27
  • $\begingroup$ @LukaSantrić: That's a rather complicated question. One of the reasons is your word "material". The best mirrors are typically made from stacks of alternating materials, i.e. not from a single material. These materials are semi-transparent, so you will have thermal photons originating not from the surface, but from the interior. And on their path to the surface, they'll interact with the stacks of alternating materials, which will cause a wavelength-dependent distribution. $\endgroup$
    – MSalters
    Jul 5 at 12:46
  • $\begingroup$ @LukaSantrić I think that you need to ask a separate question for that as it is too big to respond to properly in comments $\endgroup$
    – Dale
    Jul 5 at 13:02
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You seem to be thinking of a macroscopic mirror, one whose dimensions are much larger than the wavelength of the emitted black-body radiation. Also, the mirror appears to be smooth enough to have a well-defined focal point. In other words, it is locally flat (almost) everywhere.

In this simple model, the thermal radiation that's locally emitted is going to be uniformly distributed - there's simply no preferred direction. But on a non-local scale, some of the emitted photons may hit the mirror. In turn, some of these may be absorbed, while others are reflected.

Now note that the photons that are absorbed again do not cause local heating. The reason is fairly simple once you realize that light paths are reversible. If a thermal photon emitted at point A hits point B, then thermal photons emitted from B can also travel back to A.

Still, the photons that are emitted at A and reflected at point B will be subtracted from the original uniform distribution, and have a new direction dependent on the angle between the vector AB and the surface normal at point B. For the effects of a single reflection, you'd have to double integrate this over all points A and all points B visible from points A. And of course a photon could be reflected more than once - consider a concave mirror that's shaped like a tube - that's quite similar tp a glass fiber.

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Heat as part of the electromagnetic spectrum, in terms of energy, as light or infrared rays, acts as a waves (classically). Say, the heat is in the infrared spectrum instead of light. If it is directed towards the concave mirror, it will still reflect to the focal point, the same as light, but you would not be able to see it without an infrared camera.

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    $\begingroup$ Heat is not directed toward concave mirror. It is emitted by it. I'm heating the mirror itself let's say internally by flowing current through it. The question is about direction of this emission. Is it emitted in only one direction, perpendicular to the surface or in all directions? $\endgroup$ Jul 4 at 21:22
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    $\begingroup$ @Luka Santrić: Difficult to say. I think if you heat the concave mirror it would emit radiation in all directions, as it's not directed towards the mirror's reflective surface. It's not being reflected. $\endgroup$ Jul 4 at 21:32
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    $\begingroup$ It's not difficult to say at all! $\endgroup$ Jul 5 at 10:57

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