- #1
Elias1960
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[Moderator's note: New thread spun off from previous discussion due to more advanced subject matter being discussed.]
There is, in fact, a quite good argument that Hawking radiation cannot be derived by semiclassical theory.
It is the comparison with the scenario where the collapse stops some ##\epsilon## above the Schwarzschild radius. In this case, Hawking radiation stops once the collapsing star becomes stable. And the time for this is very short.
Now the question is how all this depends on the ##\epsilon##. Let's assume it is quite small, say, ##\epsilon=10^{-100} l_{Planck}##. The number does not really matter, the time is quite short because the dependence is logarithmic. Let's make it even smaller, ##\epsilon=10^{-1000} l_{Planck}##. This will add some seconds, but not more.
Now, where does the last Hawking radiation particle come from? Once it would not have been created for ##\epsilon=10^{-100} l_{Planck}##, it has been created during the collapse from ##\epsilon=10^{-100} l_{Planck}## to ##\epsilon=10^{-1000} l_{Planck}##. And what has caused the creation was the difference between the two solutions. But this difference is localized completely below ##\epsilon=10^{-100} l_{Planck}## from the Schwarzschild radius. Which is, therefore, the region where it has been created.
What is the time dilation, and, therefore, the corresponding redshift the particle obtains moving from that place? There will be a ##10^{100}## factor for moving up to Planck length and then even more from Planck length to infinity. So, what resulted in Hawking radiation of some ##10^{-8}K## has, at the origin at the surface, a quite solid energy. That for such energies the semiclassical approximation is applicable is not plausible at all.
There is, in fact, a quite good argument that Hawking radiation cannot be derived by semiclassical theory.
It is the comparison with the scenario where the collapse stops some ##\epsilon## above the Schwarzschild radius. In this case, Hawking radiation stops once the collapsing star becomes stable. And the time for this is very short.
Now the question is how all this depends on the ##\epsilon##. Let's assume it is quite small, say, ##\epsilon=10^{-100} l_{Planck}##. The number does not really matter, the time is quite short because the dependence is logarithmic. Let's make it even smaller, ##\epsilon=10^{-1000} l_{Planck}##. This will add some seconds, but not more.
Now, where does the last Hawking radiation particle come from? Once it would not have been created for ##\epsilon=10^{-100} l_{Planck}##, it has been created during the collapse from ##\epsilon=10^{-100} l_{Planck}## to ##\epsilon=10^{-1000} l_{Planck}##. And what has caused the creation was the difference between the two solutions. But this difference is localized completely below ##\epsilon=10^{-100} l_{Planck}## from the Schwarzschild radius. Which is, therefore, the region where it has been created.
What is the time dilation, and, therefore, the corresponding redshift the particle obtains moving from that place? There will be a ##10^{100}## factor for moving up to Planck length and then even more from Planck length to infinity. So, what resulted in Hawking radiation of some ##10^{-8}K## has, at the origin at the surface, a quite solid energy. That for such energies the semiclassical approximation is applicable is not plausible at all.
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