If perfect blackness violates the Uncertainty Principle, how isn't dark matter a violation?

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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).

Answer 5

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".