In a recent tweet, Dr. Michio Kaku stated that perfect blackness would violate the Heisenberg Uncertainty Principle, i.e. every object must emit some radiation. I have two questions regarding this statement.

  1. Could someone please elaborate on how exactly perfect blackness violates the uncertainty principle? I am somewhat familiar with this principle, so a detailed answer would be appreciated.

  2. Apparently, dark matter doesn't emit any radiation. How isn't this a direct contradiction to Kaku's statement and a violation of the uncertainty Principle?

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    $\begingroup$ Dark matter is already breaking (apparently ) a few physical laws, and a violation of one more, if true, may not be a huge surprise. $\endgroup$ – user214814 Jan 6 '19 at 15:16
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    $\begingroup$ I don't see how this should be true for Dark Matter. Apparently DM could not even be coupled to normal matter in any other way than by gravitation so how should it emit photons in that case? $\endgroup$ – Katermickie Jan 6 '19 at 15:20
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    $\begingroup$ it most likely emits thermal gravitational radiation, barely detectable of course. Plus dark matter is not black, it is transparent, which is quite different, meaning it does not absorb EM radiation, but just "ignores" it $\endgroup$ – Wolphram jonny Jan 6 '19 at 15:55
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    $\begingroup$ @StudyStudy What physical laws does the dark matter break? $\endgroup$ – safesphere Jan 6 '19 at 16:50
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    $\begingroup$ @StudyStudy The absence of dark matter would indeed violate the reverse square law of gravity: en.wikipedia.org/wiki/Modified_Newtonian_dynamics $\endgroup$ – safesphere Jan 6 '19 at 19:14

There's no contradiction, but Kaku is being pretty cavalier.

In both classical and quantum physics, an uncharged point particle can never emit or absorb any light. Nothing about this violates the uncertainty principle. The electromagnetic field is just one of many quantum fields in the universe, and there's no law saying you have to interact with it.

Kaku is talking about something different: black holes, which classically can absorb light but never emit it. This is very different from an uncharged particle, which doesn't interact with light at all. It's true that in the quantum theory, black holes can emit light by Hawking radiation.

However, this is only tenuously related to the uncertainty principle. I assume the relation Kaku had in mind was something like "the uncertainty principle means that particles can spontaneously pop out of thin air. A quantum black hole can radiate because a particle can spontaneously appear right outside its event horizon and leave." This is the standard popsci explanation, but it's rather misleading, in the sense that nothing about Hawking's actual calculation looks anything like this.

Again, Hawking radiation is a statement about event horizons. It has no strict relation with dark matter. Furthermore, dark matter doesn't even have to be perfectly dark. For example, there are theories of millicharged DM where dark matter does have a tiny charge and hence can radiate. Dark matter could also be due to primordial black holes, which do radiate by Hawking radiation. In both cases the radiation is too little to see.

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  • $\begingroup$ "This is very different from an uncharged object, which doesn't interact with light at all" First you say particle, then you say object. Certainly, objects with no net charge still interact with light. And uncharged object can cause gravitational lensing. $\endgroup$ – Acccumulation Jan 7 '19 at 19:20
  • $\begingroup$ @Acccumulation Indeed, I edited, that was inconsistent usage. As for the lensing, I am not counting indirect interactions with light like that, because otherwise everything technically interacts with everything else. Which is true, but beside the point. $\endgroup$ – knzhou Jan 7 '19 at 20:35
  • $\begingroup$ The popular explanation I've always heard for Hawking radiation is slightly more sophisticated, even if it's still an extreme simplification or only slightly related to the technical theory: Even in vacuum, virtual particle pairs can spontaneously form. These usually cancel out again extremely quickly - but if one half the pair goes into a black hole but the other escapes the horizon, the black hole is radiating energy (and slightly "evaporating"). $\endgroup$ – aschepler Jan 7 '19 at 22:46
  • $\begingroup$ @aschepler I know about that one, but it’s equally unrelated to the truth. Any explanation involving anything randomly popping out of thin air is, from the standpoint of the actual physics, totally unsalvagable. But if one accepts it then there are millions of stories one can tell. $\endgroup$ – knzhou Jan 7 '19 at 22:53
  • $\begingroup$ @knzhou Hmm, I've also heard about closed Feynman diagrams being related to a "vacuum energy problem". Is there any merit to that explanation? Or maybe that's just not much related to the Hawking radiation topic? $\endgroup$ – aschepler Jan 7 '19 at 23:22

Every charged object, when accelerated, will emit electromagnetic radiation. This means that as long as atoms are jostling about on the surface of a solid due to random thermal motion, they will emit radiation that we then can detect. But if we cool an object down so its atoms vibrate less, this means that it will emit less radiation- by which means we would conclude that once an object is cooled to absolute zero, its atoms stop moving and therefore they stop radiating.

Now we note that the uncertainty principle tells us that the more certain we are of a particle's position, the less certain its momentum becomes. If the atoms of an object are so cold that they aren't moving about, then we can accurately locate them- but this means that their momenta are very uncertain, and this in turn guarantees that they must remain in motion- and emit some radiation in the process.

In this way, the uncertainty principle requires that all objects radiate.

Dark matter is postulated to not experience the electromagnetic force. This means that it does not emit radiation when accelerated, and this in turn means that it can't be seen with optical or radio telescopes.

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  • $\begingroup$ The atoms not at the surface also emit radiation--it just doesn't escape. I'm pretty sure the inside of the Earth is, contrary to "common sense", a very, very, bright place. $\endgroup$ – JEB Jan 6 '19 at 19:24
  • $\begingroup$ answer edited. -NN $\endgroup$ – niels nielsen Jan 6 '19 at 19:32
  • $\begingroup$ The uncertainty principle tells us that when we measure a particles position; we must impart energy into it; making us less sure of its momentum... If you don't measure it; it could get to 0K and stop radiating.... This would mean that you would know the momentum; but could no longer know if the particle was in the room or not. $\endgroup$ – UKMonkey Jan 7 '19 at 12:50
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    $\begingroup$ @UKMonkey : The uncertainty principle is not "disturbance through measurement". See physics.stackexchange.com/a/87903/50480 for one of many, many explanations. The uncertainty principle is automatically imposed on any two quantities related by a Fourier transform, entirely independently of any measurement. $\endgroup$ – Eric Towers Jan 7 '19 at 17:17
  • $\begingroup$ @erictowers, this is cool, where can I learn more about that fourier connection? $\endgroup$ – niels nielsen Jan 7 '19 at 19:01

The answer to question (2) is just: dark matter is transparent, not black. Perfect blackness, if I understand Kaku correctly, would be something that absorbs radiation but does not emit radiation. Dark matter neither absorbs nor emits (at least not enough to be measurable), it is not 'perfect blackness' because it does not absorb.

An aside: since dark matter interacts with ordinary matter through gravity (which is how it was discovered in the first place), it is expected to emit and absorb gravitational radiation. Which is too weak to be measurable though.

As for question (1), I think Kaku is just wrong. In addition to knzhou's remarks I argue: A system in a stable state (for example a hydrogen atom in an otherwise empty region of space) has quantized energy levels, and there is a lowest of these levels, the ground state. If the system is in that ground state, it will not radiate, even if there is motion (in this example the motion of the electron in the atom). This may be an idealization (a real object is never completely stable, and in a macroscopic object there are always some constituent parts that are not in their ground state) but it is compatible with basic physics.

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  • $\begingroup$ You have a good point. A better name would be "transparent matter". We don't call window glass "dark matter" even though it ideally is pretty hard to observe, at right angles. $\endgroup$ – my2cts Jan 7 '19 at 16:12
  • $\begingroup$ @my2cents I understand your point, but I think "dark" is good terminology, as long as "dark" is not mistaken for "black". $\endgroup$ – Menno Jan 7 '19 at 21:22

Alternative simple answer:

Dark matter is dark because it simply has no interaction whatsoever (as far as we know) with photons.

That doesn't mean it can't gain or lose energy or interact in other ways (for example by definition it interacts with gravity). We don't yet understand dark matter well enough to know what else it can do, state-wise and energy-wise.

But it does mean that when we call it "dark", we don't mean that it is "black" in the sense that ordinary black (or black body) objects might be - low or near-zero energy, photons emitted at low frequencies below the visible spectrum, etc.

Dark matter could be capable of many kinds of high energy or low energy states and fundamental interactions, for all we know. (For example, if other unidentified fields/interactions also existed, which interacted with the known fields and their particles). But it wouldn't affect photons, and without a photon interaction, it won't have any visual or electromagnetic presence at all. It won't be black. It will be transparent, as photons will go straight "through" it in blissful ignorance (so to speak).

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I think the premise of question 2 is wrong. Where have you seen it stated that dark matter must be perfectly black? It's called "dark matter" because it doesn't appear to emit detectable amounts of electromagnetic radiation. Currently, there is no universally accepted explanation of what dark matter is and some of the candidates might weakly emit radiation.

There is a big difference between "perfectly black" and "doesn't emit detectable amounts of radiation".

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