Can Electric Fields Affect the Unruh Temperature of Vacuum Fluctuations?

[jcap]

The Unruh temperature is given by
$$T=\frac{\hbar\ a}{2\pi c k_B}.$$
As I understand it one can think of pairs of electrons and positrons popping out of the vacuum and then annihilating. Imagine that we apply an electric field $\vec{E}$ to a region of space. Each electron and positron would be accelerated in opposite directions parallel to the field so that they will take slightly longer to annihilate. Thus I could imagine the properties of the vacuum changing in such a circumstance. We could naively use Newton's 2nd law to find the equation of motion of an electron/positron:
$$\vec{E}\ e=m_e\ \vec{a}.$$
Substituting the magnitude of the acceleration $a$ into the Unruh formula we get
$$T=\frac{\hbar}{2\pi c k_B}\frac{E\ e}{m_e}.$$
If we take the applied electric field strength $E=1\ \hbox{MV/m}$ then the Unruh temperature is
$$T\approx 10^{-2}\ \hbox{K}.$$
Perhaps this temperature could be measured?

 

[Demystifier]

In the absence of electric field, it's not really that electron-positron pairs pop out of the vacuum. In the presence of the field the pairs can be created, but then it is the Schwinger effect, not Unruh effect. See e.g. https://www.physicsforums.com/threads/schwinger-effect.53064/

 

[jcap]

In the absence of electric field, it's not really that electron-positron pairs pop out of the vacuum. In the presence of the field the pairs can be created, but then it is the Schwinger effect, not Unruh effect. See e.g. https://www.physicsforums.com/threads/schwinger-effect.53064/

I was thinking that if the field strength is less than the Schwinger field of $10^{18}$ V/m then maybe just photons are produced rather than electron/positron pairs. I'm not sure where the energy would come from to produce these particles - maybe from the vacuum itself?

 

[PeterDonis]

As I understand it one can think of pairs of electrons and positrons popping out of the vacuum and then annihilating.

That's not a good heuristic description of the Unruh effect. Remember that only an accelerated observer detects Unruh radiation. The "source" of the radiation is ultimately whatever energy source is producing the acceleration--for example, the observer's rocket engine. From the standpoint of the quantum field, the radiation is produced by the interaction of the accelerated detector with the field.