This is an improved version of the argument in Electromagnetic Unruh effect?

In the quantum vacuum particle pairs, with total energy $E_x$, can come into existence provided they annihilate within a time $t$ according to the uncertainty principle $$E_x\ t \sim \hbar.$$ If we let $t=x/c$ then we have $$E_x \sim \frac{\hbar c}{x}$$ where $x$ is the Compton wavelength of the particle pair.

Let us assume that there is a force field present that immediately gives the particles an acceleration $a$ as soon as they appear out of the vacuum.

Approximately, the extra distance, $\Delta x$, that a particle travels before it is annihilated is $$\Delta x \sim a t^2 \sim \frac{ax^2}{c^2}.$$ Therefore the particle pairs have a new Compton wavelength, $X$, given by $$X \sim x + \Delta x \sim x + \frac{ax^2}{c^2}.$$ Accordingly the energy $E_X$ of the particle pairs, after time $t$, is related to their new Compton wavelength $X$ by \begin{eqnarray} E_X &\sim& \frac{\hbar c}{X}\\ &\sim& \frac{\hbar c}{x(1+ax/c^2)}\\ &\sim& \frac{\hbar c}{x}(1-ax/c^2)\\ &\sim& E_x - \frac{\hbar a}{c}. \end{eqnarray} Thus the particle pair energy $E_X$ needed to satisfy the uncertainty principle after time $t$ is less than the energy $E_x$ that was borrowed from the vacuum in the first place. When the particle pair annihilates the excess energy $\Delta E=\hbar a/c$ produces a photon of electromagnetic radiation with temperature $T$ given by $$T \sim \frac{\hbar a}{c k_B}.$$ Thus we have derived an Unruh radiation-like formula for a vacuum that is being accelerated by a field. If the field is the gravitational field then we have derived the Hawking temperature. By the equivalence principle this is the same as the vacuum temperature observed by an accelerating observer. But this formula should be valid for any force field.

Let us assume that the force field is a static electric field $\vec{E}$ and that the particle pair is an electron-positron pair, each with charge $e$ and mass $m_e$. The classical equation of motion for each particle is then $$e\ \vec{E}=m_e\ \vec{a}.$$ Substituting the magnitudes of the electric field and acceleration into the Unruh formula gives $$T \sim \frac{\hbar}{c k_B}\frac{e|\vec{E}|}{m_e}.$$ If we take the electric field strength $|\vec{E}|=1$ MV/m then the electromagnetic Unruh/Hawking temperature is $$T\approx 10^{-2}\ \hbox{K}.$$ If this temperature could be measured then one could experimentally confirm the general Unruh/Hawking effect.

Is there any merit to this admittedly non-rigorous argument or can the Unruh/Hawking effect only be analyzed using quantum field theory?


closed as unclear what you're asking by ACuriousMind Nov 22 '17 at 11:32

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    $\begingroup$ Once again, I remind you that peer review, non-mainstream physics and/or check-my-work questions are off-topic here. Please ask a specific question about established physics instead of asking us to comment on ideas. Please stop posting questions where you are not actually asking a specific question about established physics. $\endgroup$ – ACuriousMind Nov 22 '17 at 11:33

You aren't really asking a question but here is my assessment of your argument.

The Unruh effect states that if one were to couple a detector to a quantum field, the detector would detect a thermal excitation as it is being accelerated. More generally, however, this excitation has to do with the thermal character of the vacuum and not necessarily the coupling of a detector. So the acceleration argument is not exactly necessary. In fact (due to an argument by Sciama), the necessary and sufficient condition is that the vacuum be stable and stationary from the perspective of a uniformly accelerated frame.

Your argument is very hand-wavy. There is a confusion of frames, there is no reference to a thermal density matrix, you have not constructed a boost Hamiltonian, you have not addressed the subtleties of the "quantum" equivalence principle, I don't know what metric you're talking about and so on.

  • $\begingroup$ Thanks for the feedback. I guess one would have to do a proper QFT analysis of the effect of a static electric field on the ground state of the electron field. Well beyond my abilities! Maybe Schwinger did that sort of analysis when he derived the QED breakdown field limit. $\endgroup$ – John Eastmond Nov 22 '17 at 11:32

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