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Thermal radiation of various gravitational backgrounds*

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1 Thermal radiation of various gravitational backgrounds*
Douglas Singleton, CSU Fresno DPF2006 November 1st, 2006 *Work in collaboration with E.T. Akhmedov (ITEP), V. Akhmedova (CSU Fresno), Terry Pilling (North Dakota State University)

2 Hamiltonian-Jacobi Equation
The Hamiltonian-Jacobi equation can be used as the starting point for a quasi-classical study of whether or not a gravitational background emits radiation or not. Consider a scalar field With One arrives at the HJ equations We are looking for solutions of the form

3 Tunneling/Decay Rate and Temperature
An imaginary part to S0(x) corresponds to tunneling/decay of the background in question This in turn allows on to obtain the temperature of the Planckian radiation

4 Black Hole Radiation: Schwarzschild Frame
In Schwarzschild coordinates the HJ equation for S0(x) is The solution is ± are ingoing (outgoing) particles

5 Black Hole Radiation: Schwarzschild Frame
S0 (x) has an imaginary part which can be found by contour integration (a semi-circle going around the pole from left to right  r-2M=εeiθ) The result is [E.T. Akhmedov, et. al, PLB, 642, 124 (2006)] This is twice the temperature originally obtained via QFT inspired methods

6 Black Hole Radiation: Isotropic Frame
This discrepancy might be attributed to the “badness” of Schwarzschild coordinates. So try other coordinates – isotropic coordinates giving a better behaved metric The solution for S0 is now For the semi-circular contour from the previous contour this appears to give Im(S0)=±4πME and the original temperature  T=ħ/8πM [M. Angheben et. al JHEP, 0505:014 (2005); M. Nadalini J. Phys. A 39, 6601 (2006)]

7 Black Hole Radiation: Isotropic Frame
However the contour also needs to transformed using This transforms the contour The semi-circle goes to quarter circle and iπ x Residue  iπ x Residue/2 and Im(S0)=±2πME giving the same temperature as in the Schwarzschild frame

8 Black Hole Radiation: Painlevé Frame
The Painlevé-Gulstrand coordinates appear to give the correct answer [G.E.Volovik, JETP Lett. 69, 662 (1999) M.K. Parikh and F. Wilczek, PRL, 85,5042 (2000)] This solution is regular (no horizon for incoming particles). The solution for S0 is The two contributions have the same magnitude. For incoming (+) particle Im(S0)=0 and for outgoing (-) particles Im(S0)=-4πME, which give the temperature of the original calculation.

9 Canonical Invariance There is a problem with this: 2Im(S0)=2Im∫ p dr is not canonically invariant so that Γ~exp[2Im(S0)] is not a proper observable [B.D. Chowdhury, hep-th/ ] is canonically invariant. A proper, observable decay rate is then For Schwarzschild, Isotropic frames For Painlevé-Gulstrand All frames give twice the originally calculated temperature

10 Unruh radiation For an observer undergoing linear, constant acceleration on can transform Minkowski spacetime into Rindler. This has a horizon at x=-1/a and solving the HJ equation in this case yields Which gives a Planckian spectrum with temperature T=aħ/2π. Note here we can write

11 Circular Unruh Effect: Rotating Case
The metric in this case is Now the action has the form S=Et+μφ+S0(r) and the HJ equation is The action can have imaginary parts but these correspond to tunneling through the centripetal barrier, not Unruh/Hawking radiation [Coordinate change r=(1/ω)+x]

12 Circular Unruh Effect: Orbiting Case
The analysis is essentially the same for the case of an orbiting observer. Now the metric is Now taking μ=0 there are two solutions to the HJ equations Neither have imaginary parts and so there is no thermal circular Unruh radiation in the tunneling picture. Unlike linear Unruh there is no barrier/horizon to prevent photons from reaching the rotating/orbiting observer.

13 Summary and Conclusions
Tunneling picture using the HJ equation allows one to quickly/easily determine if a particular gravitational background emits radiation. There are some discrepancies between the simple tunneling calculation and QFT calculations (Hawking temperature, circular Unruh radiation). In the circular Unruh case the resolution maybe that there is radiation, just not thermal. In Hawking radiation case either tunneling picture is wrong (in detail) or temperature maybe twice as large [‘t Hooft, J. Geom. Phys. 1, 45 (1984); R. Brout and Ph. Spindel, Nucl. Phys. B, 348, 405 (1991)]

14 Acknowledgments The work supported by the following grants : Russian Foundation for Basic Research grant (RFBR) and a CSU Fresno CSM International Activities Grant


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