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Charged particles and gravity

  1. Oct 27, 2012 #1
    Hello ladies and gentleman! I have a question that I have been wondering about for some time now. Does a particle with mass m and charge q have a larger gravitational field than a particle that is the same with mass m but no charge? I am assuming that the particles are static. I thought that charged particle would have a larger gravitational field because of the stress produced from it's electric field. Thank you for your time guys:smile:
     
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  3. Oct 27, 2012 #2
    It depends on what particles you are talking about. If you are talking about something like an atom, then a negatively charged particle would have more gravity (because of excess electrons) and a positively charged particle would have less (missing some electrons) compared to the neutral atom.

    If you are comparing sub-atomic particles like electrons and positrons, they would both have the same gravity, as the mass is the same irrespective of charge (at least I believe so).
     
  4. Oct 28, 2012 #3

    Mentz114

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  5. Oct 28, 2012 #4
    I am talking about subatomic particles. Sorry about that:blushing: And I have checked Wiki for charged blackholes. What I am asking is whether a subatomic particle with an electric field has a greater gravitational field than one without. Which one has a greater gravitational field?
     
  6. Oct 28, 2012 #5
    Theoretically, a charged subatomic particle has the same magnitude gravitational field as an uncharged subatomic particle with identical inertial mass. Experimentally, it is unknown. If there were a difference then the equivalence principle would be violated.
     
    Last edited: Oct 28, 2012
  7. Oct 28, 2012 #6

    Simon Bridge

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  8. Oct 28, 2012 #7
    I found this article in Wiki talking about a "black hole electron". It was an idea created by Einstein. In the article, the Schwarzschild radius of the electron is computed without the charge of the electron and then (this is what I think they did) with the charge of the electron using the Reissner-Nordstrom metric. The radius computed by the Reissner-Nordstrom metric is larger than the radius computed without the charge of the electron. How can this be? Doesn't that violate the equivalence principle?
     
  9. Oct 28, 2012 #8
    In addition to an equivalence principle violation, a differing gravitational charge would also result in a third law violation and a violation of the conservation of momentum.
     
  10. Oct 28, 2012 #9
    The "Black hole electron" is what the article is called, in case anyone wanted to see it themselves. It's on Wiki:smile:
     
  11. Oct 28, 2012 #10

    Simon Bridge

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  12. Oct 28, 2012 #11
    It adds the two metrics right?
     
  13. Oct 28, 2012 #12
    I mean event horizons.
     
  14. Oct 28, 2012 #13
    :blushing: lol once again
     
  15. Oct 28, 2012 #14
    Wait, doesn't that imply that the gravitational field of a charged subatomic particle will have a greater gravitational field than the same subatomic particle without a charge anyway?
     
  16. Oct 28, 2012 #15
    why are you relating charge with it.If you are thinking in terms of mass then you should know that neutron mass is greater than proton.
     
  17. Oct 28, 2012 #16

    Dale

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    You should read the links that Simon Bridge and Mentz114 posted about the RN metric. That is really all you can say about the topic according to GR. The field is definitely different, but since there are 10 independent components it is hard to characterize as simply stronger or weaker. The charged one does have more curvature since it is not a vacuum solution.
     
  18. Oct 28, 2012 #17
    So the equivalence principle and the conservation of momentum do not apply with GR at the subatomic level? The OP seems to be stating that both particles have the same inertial mass.
     
  19. Oct 28, 2012 #18

    Dale

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    GR is classical, so it does not make any distinction between subatomic and other levels, and objects are treated as continuums, not quanta. However, the equivalence principle (properly stated) does hold in GR, and in GR momentum is locally conserved. Neither of those facts contradict what I wrote.
     
  20. Oct 28, 2012 #19
    Ok, I was taking your statement "The charged one does have more curvature since it is not a vacuum solution." to mean the charged one will have greater active gravitational mass (gravitational charge). But if that is not what you mean, then I do not understand. Do you mean the charged particle will be physically smaller than the uncharged particle, giving it a greater curvature? But that doesn't make much sense because it would only apply for distances smaller than the radius of the uncharged particle. I will admit that I am looking at this from strictly a classical mechanics point of view. So if it takes greater understanding of GR then I will probably just have to take your word for it. :)
     
  21. Oct 28, 2012 #20

    Dale

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    I mean that the Schwarzschild metric (uncharged) is a vacuum solution so it has a Ricci curvature tensor of zero everywhere outside the object. In contrast, the RN metric is not a vacuum solution so the Ricci curvature tensor is non zero.
     
  22. Oct 28, 2012 #21
    Well, I set myself up for that one. :) All I can say is, I hope it answers the OP's question, because I have no idea what you're talking about or how it applies. But that's only because of my ignorance.
     
  23. Oct 28, 2012 #22

    Dale

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    I'm sorry, I know this stuff, but not very well so I am afraid that I don't know how to simplify it very well. What it means physically is that if you have a dust cloud of a certain shape falling outside an uncharged mass then tidal forces will cause it to stretch radially and compress longitudinally, but its volume will be the same. On the other hand, the same dust cloud falling outside a charged mass will change its volume also.

    So the gravitational field outside a charged mass is different, but there isn't really a way to simply characterize it as stronger or weaker. It is more complicated than that.
     
  24. Oct 28, 2012 #23
    So then charge will make a difference under certain circumstances? I felt like it would since there is stress in the electric field of our subatomic particle. And from what I know, energy fllux and density, stress, and momentum fllux and density are sources for the curvature of spacetime.Correct me if I am wrong please because I figured that for a charged particle q with mass m would have a sort of "combined" stress-energy momentum tensor. What I mean by "combined" I mean the combination of the stress energy tensor of the subatomic particle regarding no charge and the stress enrgy tensor of the particle regarding only it's charge. That is why I am asking this question:smile: I am no good at tensors so I really need help.
     
    Last edited: Oct 28, 2012
  25. Oct 28, 2012 #24
    Do you mean it will produce more curvature or less curvature?
     
  26. Oct 28, 2012 #25

    Dale

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    Yes. The dust cloud is one example.

    Yes, this is the distinction between a vacuum and non-vacuum spacetime that I mentioned above. For the charged particle the field is a gravitational source, which is absent for the uncharged particle.

    Unfortunately, again, it is not so simple. The EFE are non-linear, so you cannot just add two fields to get a combined field. All I can do is recommend that you read the links provided above about the RN metric.
     
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