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Doesn't the Unruh effect violate EP?

Since the temperature of an absolutely accelerating reference frame would be different from an inertial one according to this effect, then one finds a criterion to distinguish inertial frames from non-inertial ones.

Bastam Tajik
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2 Answers2

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The equivalence principle is not an equivalence between non-inertial and inertial frames; it is an equivalence between one inertial frame and another. To be precise, it asserts that physics relative to a frame in freefall in any region where one might expect gravitation, such as near a massive body, is the same as physics relative to a frame in freefall far from any massive body, and this in turn is the same as the special relativistic limit of G.R.

To make this mathematically precise, one may bring in some statements about locality: one is really asserting that the metric is Minkowski up to the first order terms in departures from some event (those first order terms being zero in the Minkowski metric).

The Unruh effect concerns a case where some other, non-gravitational, force is causing acceleration. The accelerating body experiences internal excitations which are like those which would happen if it were bathed in thermal radiation at the Unruh temperature. If one analyses this in the accelerating frame, then one can do so, but one is not then adopting a frame in freefall and the equivalence principle has nothing to say about the comparison of such a frame with an inertial frame. But what it does say is that the physical effects associated with the acceleration relative to a LIF are what they are, locally, and it does not matter where the events may be relative to other things such as gravitating bodies, because the local spacetime will still be Minkowski to first approximation. So one can predict from this that the Unruh effect will be seen by one standing still relative to the ground on the surface of a planet, just as it would by one accelerating in a rocket not near to any planet.

One can use a sort of close cousin of the equivalence principle to connect the Unruh and Hawking effects. In a non-inertial frame held at fixed distance from a massive body one can approximate the local metric using the Rindler metric, and then the Unruh calculation applies, and one can deduce that one expects Hawking radiation from a black hole horizon. Such a calculation does not capture all aspects of Hawking radiation but it can predict the temperature correctly.

Andrew Steane
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  • To my knowledge, EP implies that there should be no way to discriminate acceleration with respect to the gravitational field from, movement with respect to it(locally). Imagine you are in the interstellar space accelerating with respect to the local inertial frame determined by the gravitational field, you measure the temperature, and once you are on the surface of a planet with a gravitational acceleration of the same magnitude of the aforementioned spacecraft, you measure the temperature. EP says the two temperatures should be the same, but the Unruh effect predicts the opposite! wrong? – Bastam Tajik Feb 27 '21 at 11:17
  • I guess even in my case, one can say there's still no way to understand if one is in a spacecraft or on a planet, we merely read different numbers, but we don't know which number relates to which observer. they are only different. True? – Bastam Tajik Feb 27 '21 at 11:23
  • @BastamTajik I agree the observation about accelerating rocket compared to standing on a planet is pertinent and correct. I would say it follows from the EP but is not itself the EP. But I think Unruh effect will be observed by one standing on a planet with local accel $g$, and they will get same temperature as one in a rocket with that same acceleration $g$ relative to local inertial frame. – Andrew Steane Feb 27 '21 at 12:45
  • @BastamTajik I just extended by answer a bit to account for this observation. – Andrew Steane Feb 27 '21 at 13:02
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Unruh effect tells us that the number of particles in an accelerated referential (with respect to us) viewed from our referential is non zero, even if either us and the accelerated referential are in the vacuum : \begin{equation*} N_{acc/us}=\langle 0_{us} | a^\dagger_{acc} a_{acc} |0_{us}\rangle \neq 0 \end{equation*} But one has the following equalities : \begin{equation*} N_{acc/acc}=\langle 0_{acc} | a^\dagger_{acc} a_{acc} |0_{acc}\rangle = 0 \end{equation*} \begin{equation*} N_{us/us}=\langle 0_{us} | a^\dagger_{us} a_{us} |0_{us}\rangle = 0 \end{equation*} This means that the temperature of the accelerated referential is the temperature of the vacuum, which is also equal to our temperature.

Jeanbaptiste Roux
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