# B Quantised Inertia

1. Sep 30, 2017

### timmdeeg

https://www.nature.com/articles/srep07195
It is well known that a uniformly accelerated detector which interacts with external fields becomes excited in the Minkowski vacuum. This effect is named as Unruh effect1,2, which indicates the fact that quantum properties of fields is observer dependent3,4,5,6,7,8,9

Hm, but I understand that proper acceleration is invariant. I've to think about it.

2. Oct 2, 2017

### kimbyd

I think the way to resolve the apparent discrepancy is that all observers agree upon what the proper acceleration of a given body is, and they thus agree upon what the Unruh radiation observed by that body would be.

The apparent contradiction of Unruh radiation being observer-dependent would be a true contradiction if different observers would interpret the Earth as interacting with different amounts of Unruh radiation. But as long as proper acceleration is an invariant, this isn't the case.

Though incidentally, proper acceleration is identically zero for an object in free-fall, like the Earth, so that might be the wrench that makes the use of proper acceleration in this context meaningless. While it is true that stars in galaxies experience some friction and thus aren't exactly in free-fall, they're pretty darned close to it, so that proper acceleration would be minuscule.

3. Oct 3, 2017

### timmdeeg

Yes, I think this is the key point which was also mentioned by PeterDonis in #25.

And thus to argue with Unruh radiation.

4. Oct 3, 2017

### Staff: Mentor

Yes, but not for the reason you suggested earlier. Proper acceleration of an object being zero (which means it should observe zero Unruh radiation) is not observer-dependent, any more than proper acceleration being nonzero is.

5. Oct 3, 2017

### timmdeeg

In #28 I refered to the case that proper acceleration is "meaningless" (described by kimbyd in #27). Then its meaningless to argue with Unruh radiation too.

I agree that both, zero and non-zero proper acceleration is invariant (my post #26). What I suggested in #24 was wrong, perhaps you refer to that.

Its confusing I just stumbled apon

https://en.wikipedia.org/wiki/Unruh_effect
The Unruh effect (or sometimes Fulling–Davies–Unruh effect) is the prediction that an accelerating observer will observe blackbody radiation where an inertial observer would observe none.

Could you kindly clarify?

6. Oct 3, 2017

### Staff: Mentor

"Blackbody radiation" is just another way of saying "thermal radiation at some temperature", where in the case of the Unruh effect the temperature depends on the proper acceleration of the observer.

7. Oct 3, 2017

### timmdeeg

Hopefully I got it. If you say "an inertial observer watching the process would describe it as radiation ..." this doesn't imply that he observes it. Correct? Otherwise it seems that Wikipedia is wrong.

8. Oct 3, 2017

### Staff: Mentor

Wrong. The inertial observer does observe radiation--but he observes it being emitted by the accelerated detector, whereas the accelerated observer, moving along with the detector, observes it absorbing radiation (the Unruh radiation). In other words, the detector itself can't avoid changing the state of the quantum field when it detects radiation, and that state change is a "real" change, seen by all observers, though different observers will describe the change differently.

Yes. And this surprises you?

Basically, the Wikipedia article is ignoring the effect of the detector on the quantum field. Many informal descriptions of the Unruh effect do so as well. But that's not really correct. See above.

9. Oct 3, 2017

### timmdeeg

Well, indeed, I had the impression that the inertial observer doesn't see any radiation for quite a wile, due to some sources. The more I'm thankful that you have clarified this matter!

10. Oct 3, 2017

### kimbyd

No, I don't think so. PeterDonis was pointing out that proper acceleration is not, in fact, observer dependent.

I added to that that the proper acceleration for an object in free-fall is identically zero. Which means that the proper acceleration of all stars and galaxies is so close to zero that it can be considered negligible, thus there should be no noticeable Unruh radiation with which to cause the rotation curve effect suggested by McCulloch. He seems to have made the mistake of assuming the Newtonian acceleration would result in the Unruh effect, but this isn't the case.

It is perhaps interesting that using this as an effect leads to results that match observation better than MOND. But the model is probably a non-starter as the premise appears to be impossible. I figured there was such an explanation why this idea wouldn't work, as otherwise the model would probably have gained more traction among theorists. I just wasn't sure what that explanation was early in the discussion, and I think this is it.

11. Oct 3, 2017

### LeandroMdO

It's not wrong but it's a bit ambiguous. What is meant by "where an inertial observer would observe none" is merely "vacuum state". They don't attempt to reconcile the detection events on the accelerated detector with the vacuum state measured by the first. This is done, as was said, by realizing that according to the inertial observer the detector emits radiation and Unruh radiation makes sense. You can think of it as a quantum mechanical analogue of centrifugal force, which is something you have to include for consistency when writing Newton's laws in a noninertial frame.

That said, any phenomenon that you explain in a noninertial frame has to have a corresponding explanation in the inertial one, for consistency. There is no Rindler horizon for the inertial observer, so it can't impose any boundary conditions on the radiation (even if it made sense that horizons did this, and it doesn't). This is a serious problem.

12. Oct 4, 2017

### timmdeeg

Yes. As a layman I was confused by this approach. The observer dependence of Unruh radiation is a bit subtle as I realized here, but that it is due to proper acceleration isn't.

13. Oct 4, 2017

### timmdeeg

They should have mentioned the latter at least shortly. They didn't and I'm the best example what kind of misunderstanding it creates.