# How are the clock hypothesis and Unruh effect reconciled.

1. Feb 5, 2012

### Naty1

How do we reconcile the clock hypothesis in special relativity with the Unruh effect?

The first says accelerating clocks tick at the same rate as their instantaneous velocity and the later says an accelerating observer records a temperature rise not observed by an inertial observer.

To keep it simple, consider an accelerating and inertial observer passing right by each other.
Seems like temperature changes caused by acceleration [the Unruh effect] would
cause energy changes and hence [gravitational] clock changes.

Do we just ignore this in SPECIAL relativity? Do we say it's just an 'instantaneous' differnce so it can't be observed? Is the observer instrument not 'warmed' by the background she observes? [That's impossible,,,nothing would register!] or something else?
thanks!

2. Feb 5, 2012

### timmdeeg

The accelerating clock and an accelerating observer will both record Unruh radiation, which is a vacuum effect. An inertial observer will not. He will see the accelerating clock slowing down according to the relative velocity between him and the clock. There is to my opinion no bridge between special relativity and the Unruh effect and nothing to be reconciled.

3. Feb 5, 2012

### Bill_K

The Unruh effect IS special relativity. It is a consequence of quantum field theory in Minkowski space.

4. Feb 5, 2012

### MiljenkoM

Unruh effect is ONLY seen from perspective of accelerating observer. Clock change you are describing (don't really know what you mean by this), described from accelerating observers perspective are due to Unruh effect, and inertial observer is describing it with "effective gravitation" from acceleration.

Unruh effect is necessary for internal consistency of QFT. If you want do describe some effect from non-inertial perspective like accelerating observer, you need to take in account Unruh effect.

Last edited: Feb 5, 2012
5. Feb 5, 2012

### PAllen

The core observation about the OP is simply there is no equivalence at all between accelerated observers and inertial obervers in SR. The clock hypothesis says the time dilation of some object observed by an inertial observer depends only on the speed of object (accelerating or not) measured by that inertial observer. It implies nothing at all about equivalence between momentarily comoving inertial and accelerated frames. They are simply not equivalent for most physics.

6. Feb 7, 2012

### Naty1

The poster is assuming the accelerating observer is moving at a greater velocity than the inertial observer.

I did not explain very learly what I thought I saw as a contradiction .....let me restate it this way, and with the insights from the above posts, I can answer my own question:

Problem:
I was supposing an inertial observer is making a temperature observation. An accelerating observer approaches the inertial observer and overtakes him....during the period of acceleration the accelerating observer will record a higher temperature than the inertial observer....ok.
But as the two approach a common point, the accelerating observer stops accelerating and now the two observers have the same inertial velocity. At that instant, seems like the two temperature observations might be different....

Resolution: If the temperature recording device is instantaneous, then each of the comoving inertial observers will record the same temperature...from the Minkowksi vacuum, inertial space. [I had been thinking any temperature measuring device would have residual heat signature...but even if so, that does not represent any sort of 'different' observation, just different observation times...
thanks....

7. Feb 7, 2012

### timmdeeg

It is probably simpler than you believe.

Inertial observers do not record Unruh radiation, whether comoving or not.

Accelerating observers record Unruh radiation. In the moment they stop acceleration, they are inertial observers.

Consequently, in the moment an accelerating observer passes by an inertial observer, the former records Unruh radiation, the latter doesn't.

Last edited: Feb 7, 2012
8. Feb 7, 2012

### MiljenkoM

I don't agree with bald part. Accelerating observer is constantly accelerating. Whan he is approaching "stacionary" observer, he is slowing down (acceleration), when they are both comoving, accelerating observers still has acceleration since next moment he is traveling faster relative to "stacionary" observer. Accelerating observer is never going to be inertial observer. Inertial observer is defined by having constant velocity.

Actually, inertial observer and accelerating observer will detect different temperatures. And there is no paradox to be reconciled. Main point is that inertial observer has his definition of vacuum and he is calibrating his measuring device with respect to this vacuum. Accelerating observer on the other hand has his own definition of vacuum, and his measuring device is calibrated to measure his definition of particle (or temperature). So there is no conflict between there views.

9. Feb 9, 2012

### Demystifier

The clock-hypothesis is local in time, while the Unruh effect is non-local effect. You must accelerate for a long time to detect "particles" with a thermal distribution.

10. Feb 9, 2012

### timmdeeg

This is misleading. The Unruh temperature is proportional to the acceleration. If e.g. the acceleration is constant, you can accelerate as long as you like, the temperature is still the same.

11. Feb 9, 2012

### Naty1

While Unruh temperature IS proportional to acceleration, Demystifier is correct.....
He is saying in effect, I believe, that in order for Unruh temperature/radiation to be detected the Rindler horizon must be visible.....but I've also read the horizon appears asymptotically so I never quite grasped how we ever are supposed to in theory record such an effect.

12. Feb 9, 2012

### Demystifier

True. But if acceleration is NOT constant, then temperature cannot even be defined.

13. Feb 9, 2012

### Demystifier

Actually, the horizon is not essential. If you accelerate for a long but not infinite time, then you will see particles with a distribution APPROXIMATIVELY equal to a thermal distribution. See also
http://xxx.lanl.gov/abs/gr-qc/0103108 [Mod.Phys.Lett. A16 (2001) 579-581]
for further demystification on the Unruh effect.

14. Feb 9, 2012

### timmdeeg

From that:
No hint and no need to "accelerate for a long but not infinite time". Otherwise kindly clarify.

15. Feb 9, 2012

### Naty1

I understand what the reference is post #13 is claiming but am not sure how widely accepted is such an explanation.

Seems like he is saying either Hawking or Unruh is wrong.....I sure can't parse what is correct or not.

The Wikipedia article here expresses the temperature results which I have seen in other sources:
http://en.wikipedia.org/wiki/Unruh_effect#Explanation

I'm sticking with that result for the time being despite some underlying uncertainties.

16. Feb 10, 2012

### Demystifier

Not really, because Hawking and Unruh study only the case of uniform acceleration.

17. Feb 10, 2012

### Demystifier

An acceleration cannot be uniform if it does not last for a long time.