# How are the clock hypothesis and Unruh effect reconciled.

by Naty1
Tags: clock, effect, hypothesis, reconciled, unruh
 PF Gold P: 5,552 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!
 P: 170 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.
 Sci Advisor P: 3,483 The Unruh effect IS special relativity. It is a consequence of quantum field theory in Minkowski space.
P: 20

## How are the clock hypothesis and Unruh effect reconciled.

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.
 Sci Advisor PF Gold P: 4,504 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.
PF Gold
P: 5,552
 Clock change you are describing (don't really know what you mean by this)
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....
 P: 170 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.
P: 20
 Quote by Naty1 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....
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.

 Quote by Naty1 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....
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.
P: 4,384
 Quote by 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.
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.
P: 170
 Quote by Demystifier You must accelerate for a long time to detect "particles" with a thermal distribution.
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.
PF Gold
P: 5,552
 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.
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.
P: 4,384
 Quote by timmdeeg 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.
True. But if acceleration is NOT constant, then temperature cannot even be defined.
P: 4,384
 Quote by 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.
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.
P: 170
 Quote by Demystifier 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
From that:
 For a uniform acceleration the two approaches (Minkowski, Rindler) agree in the prediction of a thermal distribution for ΔE.
No hint and no need to "accelerate for a long but not infinite time". Otherwise kindly clarify.
PF Gold
P: 5,552
I understand what the reference is post #13 is claiming but am not sure how widely accepted is such an explanation.

 For a uniform acceleration, the two approaches agree in the prediction of a thermal distribution for E. However, even this partial agreement of the two approaches does not generalize when the uniform acceleration is replaced by a more complicated motion [8]. We now see that at least one of the two approaches must be wrong.
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.