Undergrad Charged Particles Radiating in "Free Fall" into Black Holes

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The discussion centers on whether charged particles radiate while in free fall into black holes, with no consensus reached among participants. Research by DeWitt and Brehme indicates that while a charged particle aims to follow a geodesic path, its Coulomb field causes it to deviate, leading to radiation. This raises questions about the differing accelerations of particles like electrons and neutrons in a gravitational field and whether this could result in ionization of atoms falling into black holes. Some participants argue that in a co-moving frame, no radiation occurs, while others highlight the complexities of electromagnetic fields in curved spacetime. Overall, the topic remains unresolved, with ongoing debate about the implications of charged particle behavior near black holes.
sha1000
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Hello everyone,

"Does a charged particle radiate in free-fall?".

I read many threads on this subject and I was surprised to find out that there is no unanimous "Yes or No" answer to this question. Here is an interesting answer from researchgate.net:

The question is widely discussed in the web and seems most definitively have been answered by a work:
Bryce S DeWitt, Robert W Brehme : Radiation damping in a gravitational field. Annals of Physics, Volume 9.2 1960, pp. 220-259.
The answer is yes.
I copy a sentence from the abstract of this work:
"The particle tries its best to satisfy the equivalence principle in spite of its charge. It is only prevented from doing so (i.e., from following a geodetic path) because of the fact that, contrary to the case of flat space-time, the electromagnetic Green's function in a curved space-time does not generally vanish inside the light cone, but gives rise to a “tail” on any initially sharp pulse of radiation."......


......the full statement was that the charge would not radiate if it would follow a geodetic line exactly. As DeWitt and Brehme point out in detail the charge caries with it a Coulomb field which produces reaction forces and prevents the charge from following a geodetic line precisely and by this mechanism let's the charge radiate and changes the answer to yes. Of course, the quantitative aspect of the matter is not clear from this and it may well be that 'yes' may mean 'negligible in all conceivable practical circumstances'.

My question:

Lets suppose that a charged particle radiates in the gravitational "free fall" (does not follow its geodesic line precisely). From this point one may conclude that an electron and the neutron do not accelerate in the same manner in the gravitational field., right? Would this mean that an atom which is falling into the black hole will ionize at some moment (because of the acceleration differences of the elementary particles)? Do we see this experimentally? Or there is no radiation whatsoever and the stars are quietly falling into the black holes?

Thank you.
 
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I think no radiation take place in free fall of a single charge into neutral charged BH. In co-moving frame with a falling charge, there is no reason for photon to be generated. Not a single charge but heated plasma gas falling into BF emit X-ray, but it is another story for you I suppose.
 
sweet springs said:
I think no radiation take place in free fall. In co-moving frame with falling charge, there is no reason of photon generated.
The problem is that the fields associated with the charge aren't pointlike, so neglecting curvature by considering only a small region of spacetime neglects the particle's fields. And the fields must be changing because the particle is moving through regions of spacetime with different curvature. Hence the complexity.

I don't know enough about it to answer the OP's question.
 
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In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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