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Space time curvature caused by fast electron

  1. Nov 7, 2011 #1
    Hi everybody!

    what happens if an electron passes by with a speed of, say, 99.999999999...% of the speed of light (relative to me). Its mass will then be enormous. Will this electron cause a relevant curvature of spacetime? Can it be so fast that it acts like a black hole?

    I guess not. But why?
  2. jcsd
  3. Nov 7, 2011 #2
    Hi, although I think that you are right that a very high energy electron will cause a lot of spacetime curvature, it cannot turn into a black hole. For that you need a high enough rest energy density.
    That's easy to see if you transform to a frame in which the electron is in rest: in that frame its kinetic energy is zero, and so it can never become a black hole.
  4. Nov 7, 2011 #3
    thanks harrylin,

    what I don't understand is when to use rest energy and when to use the energy with respect to the actual reference frame (e.g., me).

    Let's assume the electron moves horizontally. I understand that its horizontal inertia is really high given that it cannot further be accelerated. But what about vertical inertia/gravity?

    If it is only rest energy that decides if an object turns into a black hole, why isn't rest energy all that counts for spacetime curvature?
  5. Nov 7, 2011 #4


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    No, it won't. The spacetime curvature caused by any object is a frame-independent, invariant quantity. Since the object's velocity is not a frame-independent, invariant quantity, the curvature the object causes can't depend on the velocity. (Same argument for the kinetic energy.)

    You don't use either. You use the stress-energy tensor, which takes into account both the object's rest energy and the object's motion (and pressure and internal stresses in the object) in such a way that the curvature caused by the object is frame-invariant. The equation that expresses the relationship is the Einstein Field Equation (EFE).

    By "frame-invariant" I mean that you can write down the EFE in any frame you like to get answers to questions about actual physical observables, like whether or not an object can become a black hole. So if you write down the EFE in a frame in which the object is at rest, the stress-energy tensor in that frame (for an object like an electron that has no internal pressure or stresses) does contain *only* the object's rest energy density, and as harrylin says, you can then predict whether the object will become a black hole just by looking at its rest energy density.

    (Note: For a more realistic object that could collapse into a black hole, such as a collapsing star, the stress-energy tensor in the object's rest frame will also contain components representing the pressure inside the object, which can significantly affect whether the object will become a black hole. That's why I put in the qualifier above about the object having no internal pressure or stresses.)
  6. Nov 7, 2011 #5
    Matthias: Several years ago Dr Greg was kind enough to explain two types of spacetime curvature in response to a question I posted.....

    The key is that gravitational curvature IS observer independent (as already noted) and is reflected as curvature of the spacetime manifold ("graph paper" as described below). Frame dependent curvature (observer dependency) is a variable overlay on top of this fixed background curvature,

  7. Nov 7, 2011 #6
    When we speak of a high energy electron, what we mean is the kinetic energy; thus I don't understand what you mean with "same argument for the kinetic energy". And now I have a similar question as the OP, for the rest energy and mass of a plasma with high speed electrons is in principle higher than that of a plasma with slow speed electrons. How can that not affect spacetime curvature? Here below you discuss how to calculate the amount of curvature in a frame independent way and you seem to agree that the electron's kinetic energy ("motion") contributes to curvature.


    PS this seems to be related to the citation by Naty, but some more elaboration could be helpful. :smile:

  8. Nov 7, 2011 #7
    I understand why you say this but is the matter really so black and white?

    For instance take the rest mass of a gold atom, a lot of that is contributed by electrons moving at relativistic speeds.
  9. Nov 7, 2011 #8
    an "accelerated observer ... in the absence of gravitation"???

    isn't that a key assumption of general relativity that what an accelerated observer experiences cannot be distinguished from gravitation?

    Anyway, thanks for the citation, it looks very instructive, I'll just have to read it a few more times ;-)
  10. Nov 7, 2011 #9


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    The OP was talking about a single electron; for that case the rest energy density of the electron is the only thing that causes spacetime curvature. The kinetic energy is frame-dependent, just as the velocity is; in the electron's rest frame it is zero, and we can predict all physical observables, like whether the electron forms a black hole, by solving the EFE in the electron's rest frame.

    Now you are talking about a different case, where we have a bunch of electrons all moving in different ways, but there is some average "rest frame" for the system as a whole. To model this in GR, you need to assign a stress-energy tensor to the system as a whole, and that stress-energy tensor will contain components describing fluid pressure in the system's rest frame. That pressure is due to the internal motions of the parts of the system, in this case electrons. The pressure makes an additional contribution to the spacetime curvature caused by the system, and it increases with the energy (more precisely, with the temperature) of the plasma, so you are correct that a plasma with higher temperature (and hence higher-speed electrons in it) will cause more spacetime curvature than a low-temperature one. But that's a different case than a single electron.

    Not sure what you're referring to, but if you mean my description of the stress-energy tensor for a *single* electron, in a frame in which the electron is not at rest, that tensor will contain components describing the electron's momentum, which are not there in the electron's rest frame, but the energy component will also be different than it is in the electron's rest frame, so that the final result, the predicted spacetime curvature, is the same.

    If you are referring to a stress-energy tensor for a system with many particles, in the "average" rest frame of the system, see above for how it incorporates the motion of the individual particles.
  11. Nov 7, 2011 #10


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    Wouldn't the spacetime curvature caused by a fast moving spherical mass be described by Lorentz contracted version of the Schwarzschild metric? Or do you mean that the curvature measures would not be affected by the uniform contraction, because they depend on the second derivates of the metric?
  12. Nov 7, 2011 #11


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    See my post in response to harrylin just now. I'm not sure if anyone has ever written down a stress-energy tensor for an atom that tries to actually model the internal motions of the electrons as pressure, but the SET for a white dwarf does the same thing with the "electron fluid" in the white dwarf (and the SET for a neutron star does the same thing for the "neutron fluid" in the neutron star). So a gold atom at rest does cause more spacetime curvature due to its relativistic electrons than it would if its electrons were at rest inside it. But a gold atom moving as a whole does not cause any more spacetime curvature than a gold atom at rest.
  13. Nov 7, 2011 #12


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  14. Nov 7, 2011 #13


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    Locally, yes. If you look at a small piece of spacetime near you, it's very hard to tell whether it's curved or flat. The curvature, or lack of curvature, won't become noticeable until you look at a larger region. (In the same way as it's difficult to tell whether we live on a flat Earth or a spherical Earth if you only look around your immediate neighbourhood.)
  15. Nov 7, 2011 #14


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    The numerical values of the components of the curvature (for instance the tidal forces) that you measure would definitely be affected by the velocity of the moving mass for a transverse boost, i.e. if the mass is whizzing by you and not approaching head on. Of course, in an abstract sense it is the "same tensor", just seen from a different viewpoint.

    In the limit of an ultra-relativistic transverse velocity, you get the Aichelburg-Sexl solution. http://arxiv.org/abs/gr-qc/0110032 actually writes down the Riemann tensor for this solution.

    Note that, for a transverse boost, said tensor contains delta functions, i.e. values that approach infinity (the integrals are finite, though). So this is considerably different than the finite non-delta function values for the Riemann curvature tensor of a stationary mass.

    This is the behavior for a transverse boost. For a parallel boost, the outcome is completely different and pretty much as Peter originally stated - the components of the curvature tensor don't change with velocity. This is discussed in the paper I quoted above, and in MTW as well. I believe there was some good reasons for this, but I don't recall what they were offhand. But one of the consequences of this interesting fact is that if you are falling directly into a black hole (with no transverse component of your velocity), the tidal forces don't depend on the velocity of your approach.

    If one is familiar with how the electric field of an electron transforms according to SR, one can gain a lot of insight as to how its gravitational field transforms. For instance, the electric field of a relativistically moving electron is not spherically symmetric, and one can't apply Coulomb's law to derive it.

    YOu can compare the statements about the boost of the Riemann tensor to behavior of the electric field - they're pretty similar. The transverse component of the E-field gets boosted by gamma, the parallel component of the E-field is not affected by the boost.
    Last edited: Nov 7, 2011
  16. Nov 7, 2011 #15


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    I should clarify that I didn't mean to imply that the *components* of the curvature can't change with relative motion. I only meant that the physical invariants calculated from those components don't change. For example, the tidal force experienced by an observer following a given worldline can be expressed as a contraction of the observer's 4-velocity with the curvature tensor, i.e., as an invariant.
  17. Nov 8, 2011 #16
    Thanks for the clarifications. :smile:
    Indeed, it would not make much sense to me if total space time curvature would not be the equal to the sum of the contributions of all the relevant elements - such the OP's electron that passes by at high speed.

  18. Nov 8, 2011 #17


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    Well, of course it isn't the sum. GR is non-linear. Further, any situation where you ask about a rapidly moving massive body's effect on a stationary test body can be transformed to a question about the interaction between a rapidly moving test body and a stationary massive body. The results must be identical, as to any invariant or measured quantity. Thus all observables relating to a rapidly moving massive body can be answered as if the body is stationary. Period.

    Thus, I think Peter is perfectly correct here:

    - The expression of curvature tensor (like any tensor) is coordinate dependent.
    - The 'amount' of curvature as measured by an invariant (e.g. the Kretschman invariant) or by an observable, is not dependent on the speed of the body. Observations may well depend on the relative motion of an instrument and a source of gravity.

    The amount of curvature produced by system of interacting particles is another thing altogether - motion of the parts definite affects the invariant curvature. However, the invariant effect can be analyzed in coordinates where there is no overall motion of the system.

    Another way to look at this: if you have a coherently moving packet of dust (all particles moving in the same direction at the same speed), all observables may be treated assuming you have a stationary packet of dust (including the small contribution of self-gravity causing the dust to coalesce over time). On the other hand, if you have a dust packet with random internal motions, then the KE of the particles contributes intrinsically to the curvature produced. More curvature will be produced than the coherent packet.
  19. Nov 9, 2011 #18
    Thanks for the precision, and sorry if my formulation was sloppy: what I meant is that it would not make sense if in GR the total contribution of zero effects is unequal to zero.

    Thus I appreciated your clarification that indeed the numerical values of the components of the curvature that you measure would definitely be affected by the velocity of the moving mass, if the mass is whizzing by you: that is exactly how I understood the OP.
    Indeed, that's also what I stressed in my first reply.

  20. Nov 9, 2011 #19


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    Idea that result depends only from relative velocities follows from principle of relativity. But how can we check that GR respects principle of relativity? Or maybe we know already that it doesn't?
  21. Nov 9, 2011 #20
    Interesting question, it has been the source of some debates from science historians and Einstein texts specialists. The conclusion seems to be that although Einstein was convinced at first that GR not only respected but generalized to arbitrary motion the principle of relativity, later he changed his mind about this and had to admit GR doesn't generalize the principle of relativity, and it only respects it in a local way (by the equivalence principle), deviating from it as soon as curvature makes an entrance.
    "The Twins and the Bucket: How Einstein Made Gravity rather than Motion Relative in General Relativity." Michel Janssen
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