1
$\begingroup$

Should a black hole, in the absence of all external electromagnetic forces, generate its own magnetic field? Would all particles in the vicinity of the black hole by effected by this magnetic field--i.e., are there any particles that do not respond to some magnetic force?

$\endgroup$
  • 1
    $\begingroup$ What does the second part of your question have to do with black holes? A magnetic field exerts no force on a particle with no charge (e.g. a neutrino). $\endgroup$ – Rob Jeffries Jun 11 '17 at 7:07
  • $\begingroup$ Related: physics.stackexchange.com/q/39161/2451 and links therein. $\endgroup$ – Qmechanic Jun 11 '17 at 10:34
4
$\begingroup$

A non-rotating black hole can have an electric field, but the difference between electric and magnetic fields depends on the restframe. An electric field in motion generates a magnetic field and vice-versa. Hence, black holes can be said to have magnetic fields, in particular if they rotate. Electric and magnetic fields always come together.

You may instead ask why a black hole has electric charge but no magnetic charge. That's because in Maxwell's theory of electrodynamics, there are no magnetic monopoles. If there were magnetic monolopes, black holes could also have magnetic charge. (These things are known as 'dyonic black holes'.)

As to why the electromagnetic field of a black hole has the form it has, that's simply what you get if you solve the Maxwell-equations coupled to General Relativity.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ "Black holes can be said to have electric field"? Who says that? They can have mass, spin and charge. $\endgroup$ – Rob Jeffries Jun 11 '17 at 7:04
  • 1
    $\begingroup$ If they have charge they have an electric field. $\endgroup$ – WIMP Jun 11 '17 at 9:52
  • $\begingroup$ Well you addressed this yourself. Electric field is frame-dependent; charge isn't. $\endgroup$ – Rob Jeffries Jun 11 '17 at 14:13
  • $\begingroup$ Invariants built with the electromagnetic field will be non zero and frame independent. $\endgroup$ – Rexcirus Jun 12 '17 at 13:12
  • $\begingroup$ As Rexcirus says. $\endgroup$ – WIMP Jun 19 '17 at 11:17
-1
$\begingroup$

The definition of a black hole describes a body that's information is hidden behind an event horizon if the black hole has a magnetic field the field must be limited to within the event horizon as otherwise the field could have a causal effect on events outside the horizon and violate the definition

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ I think you should read a bit about the Reissner-Nordström black holes. In essence, yes the BH can have electric field out of the EH. $\endgroup$ – peterh - Reinstate Monica Jun 11 '17 at 1:14
  • $\begingroup$ Surely without reading the advised info the event horizon separates an external observer from any event that occurs within it . knowledge of those events would predispose the nature and logic of an horizon it's simply we do not know the nature of events it may seem like an intellectual Dodge like a lot of answers but an event horizon is simply another way of saying we cannot describe with any certainty and our knowledge extends beyond the boundary and events external to it. $\endgroup$ – 8Mad0Manc8 Jun 11 '17 at 1:28
  • $\begingroup$ The event horizon prevents information from inside the event horizon from reaching the universe outside the event horizon. So the gravitational field and electromagnetic field that the collapsing star has as it collapses into a black hole gets essentially frozen into a "fossil" field. See this old Usenet FAQ article How does the gravity get out of a black hole? There are also relevant pages here on SE.Physics. $\endgroup$ – PM 2Ring Jun 11 '17 at 1:34
  • $\begingroup$ @PM2Ring without reading the link does the description of fossil field include the limitations of its spacial influence to be within the event horizon and can it only effect the contents within the horizon as in my reasoning or is a black holes event horizon more loosely defined. $\endgroup$ – 8Mad0Manc8 Jun 11 '17 at 14:57
  • 1
    $\begingroup$ Yes, that's exactly what I'm saying. In pure general relativity the EH is a causal boundary so the fields outside the EH are no longer affected by the matter / energy inside the EH. Quantum mechanics modifies this picture slightly. Roughly speaking, the Heisenberg uncertainty principle means that the position of the EH is not exact, instead it's slightly "fuzzy". $\endgroup$ – PM 2Ring Jun 11 '17 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.