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A Shkadov thruster is a megastructure which reflects all light from a star in a given direction.

What reflects more radiation, a white or a mirrored surface?

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closed as off-topic by sphennings, Vylix, L.Dutch, MichaelK, Secespitus Oct 19 '17 at 7:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about worldbuilding, within the scope defined in the help center." – sphennings, Vylix, L.Dutch, MichaelK, Secespitus
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ You're effectively asking "What is more reflective; a white surface, or a mirrored one?" Besides the fact that for visible light the answer is quite obvious, as written this question has little to do with worldbuilding besides the fantastic nature of a Shkadov thruster. $\endgroup$ – sphennings Oct 19 '17 at 4:22
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    $\begingroup$ I guess space or physics stack exchange might be a better place to ask, as this is quite well established theory, and what you need is sound physics rather than creativity. It might be on topic here, but still. $\endgroup$ – Mołot Oct 19 '17 at 5:16
  • $\begingroup$ "White" and "mirrored" are qualities that only applies to visible light. So what qualities the surface has towards all other forms of radiation is unknown, since you have not specified that. Hence the question is unanswerable. If is kind of like saying "At my zoo, how well does it work for the birds, dolphins, otters and tigers if my bear pens are fences or plexi-glass; which is best?" $\endgroup$ – MichaelK Oct 19 '17 at 7:12
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Short answer: It depends.

There is no such thing as a perfect mirror. Depending on the material, it will absorb and retransmit some frequencies, while reflecting others (with some absorption loss). If we are talking about all incoming frequencies, you'd need the know the breakdown of the spectrum in order to design a suitable mirror.

A white surface absorbs all (in a manner of speaking) light and retransmits it in the visible spectrum. If your incoming spectrum is mostly UV and above, that's a major loss in terms of energy. If incoming light is mostly visible, the loss may be insignificant. EDIT: This assumes that the surface is perfect white, i.e., retransmits all visible frequencies in equal proportions. Again, no such material is known.

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  • $\begingroup$ Before all, white surface radiates in all directions while mirror reflects. So white surface is less effective as thrust provider $\endgroup$ – Vashu Oct 27 '17 at 4:32
  • $\begingroup$ @Vashu: White surfaces absorb and retransmit at lower frequencies. Simply saying it radiates in all directions is an oversimplification. $\endgroup$ – nzaman Oct 27 '17 at 5:08
  • $\begingroup$ We are talking about thruster, so what we care about is impulse of photons. Their frequency does not matter to us. But direction does - if incoming light is perpendicular to the surface then mirror would get double impulse, and white surface less than that - 15 percent less I think. $\endgroup$ – Vashu Oct 28 '17 at 7:37
  • $\begingroup$ @Vashu: When a photon collides with a stationary body, and is completely reflected from it, then the body receives an impulse p = 2hf/c, where f is the frequency of the photon. the body should then be moving with a speed of v = p/m, which in this case is v = 2hf/mc sciforums.com/threads/impulse-of-a-photon.43433 $\endgroup$ – nzaman Oct 28 '17 at 16:35
  • $\begingroup$ Frequency of photon is proportional to its energy. If a photon was absorbed and then re-radiated with lower frequency then where does that extra energy goes? It is radiated later in random direction. If a photon is simply absorbed completely then it's energy is emitted as infrared in random direction - exactly the same way. So we can ignore the change of frequency and only account for imperfection ratio. If we have 97% perfect white surface and 97% perfect mirror, then those 3% of imperfection behave the same. And 97% refection of white would give 1/6 less thrust for perpendicular light. $\endgroup$ – Vashu Oct 28 '17 at 23:48

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