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The Schwinger effect can be calculated in the world-line formalism by coupling the particle to the target space potential $A$.

My question relates to how this calculation might extend to computing particle creation in an accelerating frame of reference, i.e. the Unruh effect. Consider the one-loop world-line path integral:

$$Z_{S_1} ~=~ \int^\infty_0 \frac{dt}{t} \int d[X(\tau)] e^{-\int_0^t d\tau g^{\mu\nu}\partial_\tau X_\mu \partial_\tau X_\nu},$$

where $g_{\mu\nu}$ is the target space metric in a (temporarily?) accelerating reference frame in flat space and the path integral is over periodic fields on $[0,t]$, $t$ being the modulus of the circular world-line. If the vacuum is unstable to particle creation, then the imaginary part of this should correspond to particle creation.

Since diffeomorphism invariance is a symmetry of the classical 1-dimensional action here, but not of $Z_{S^1}$, since it depends on the reference frame, can I think of the Unruh effect as an anomaly in the one-dimensional theory, i.e. a symmetry that gets broken in the path integral measure when I quantize?

This question would also apply to string theory: is target space diffeomorphism invariance anomalous on the worldsheet?

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I'm a bit disappointed that this hasn't got any responses from people in the know. My first instinct was to question whether you could apply the worldline machinery to the Unruh/Hawking effect at all. The worldline calculations that I've looked at are deeply concerned with the vacuum of an interacting QFT - they make extensive use of the effective action. In contrast, the Unruh effect arises even in a free QFT, just from a Bogoliubov transformation of the vacuum. So is the worldline formalism applicable at all to Unruh style particle creation? –  twistor59 May 11 '13 at 9:49
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I guess the initial thought was that the graviton interaction, even though it is only implementing a change of reference frame, is sourcing the particles. According to some notes by Bastianelli the measure should also change in a covariant way $DX(t) -> \sqrt{g(X(t))} DX(t)$ so perhaps this solves the conundrum: everything changes covariantly so $Z_{S_1}$ is also covariant. However Emparan uses worldline methods for a similar calculation, so I am still a bit confused. –  shouldknowbetter May 11 '13 at 15:53
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Oh I see where you're coming from - the Bastianelli example with the metric minimally (or whatever) coupled to the matter field via the Ricci curvature clarifies the sort of action you had in mind. And you were wondering about whether breaking of diffeomorphism invariance has a role to play somewhere...At least I understand the question now. Thanks! –  twistor59 May 11 '13 at 16:27
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The general diffeomorphism symmetry in the target space is not a symmetry of the world line theory or, analogously, the world sheet theory! A general spacetime diffeomorphism changes the metric tensor $g_{\mu\nu}(X^\alpha)$ which plays the role of the "coupling constants" (coefficients defining the action, e.g. your exponent) in the world line or world sheet theory! If a transformation changes the values of coupling constants, then it's clearly not a symmetry, not even classically.

The isometry, a diffeomorphism that actually preserves the metric at each point, is a symmetry of the world line theory or the world sheet theory both at the classical and quantum level.

I guess that the confusion that led to the question were the omnipresent misleading comments about "background independence". One may be tempted to say that the diff symmetry is there because we may also change the background metric. But if we do, we are changing the rules of the game. The full spacetime dynamics (at least in string theory) ultimately allows us to change the spacetime metric by creating condensates of gravitons in a state etc. But in the world line theory or the world sheet theory, this "emergent" process has a different interpretation: the spacetime background metric has to be considered as a fixed collection of coupling constants and it just happens that we may prove that the "full theory" with one metric field configuration is equivalent to another but that is something else than saying that any particular world line or world sheet theory has a diff symmetry! It doesn't.

Sorry I haven't mentioned the word "Unruh" because I believe that the core of the aforementioned paradox has nothing to do with the Unruh effect.

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Thank you for clearing up part of my confusion. Just to relate it to my original question: if I compute the one-loop world-line integral for a scalar particle in 4d Minkowski space in a non-accelerated reference frame with Cartesian coordinates, I will get the standard divergent integral. But if I now consider the same integral in Minkowski space with the metric for an accelerating reference frame (i.e. Rindler space), I cannot expect to get the same answer, because I have changed the coupling constants of my theory. Is that correct? –  shouldknowbetter Feb 2 at 12:58
    
Yes, I think so. The Rindler space is equivalent, by a coordinate transformation, to a part of the Minkowski space but the fact that we are only considering a "part" has implications for the path integral etc. It really means that some boundary conditions in the Rindler space won't be quite well-defined, and one should entangle the Rindler wedge with the other one and discuss the possible entangled states etc., and some of these effects of the "other" Rindler wedge are also linked to the Minkowski space divergence. –  Luboš Motl Feb 2 at 18:14
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