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A Relation of gravity of a pulse of light to E/c^2

  1. Mar 4, 2017 #1

    Simon Bridge

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    This came up in a discussion in another forum - the context was the gravity of a photon... for which we need quantum gravity. But imagining a pulse of monochromatic light, it was asserted that it's gravity could be well approximated by replacing one of the mass terms in Newton's Law by E/c^2.

    I am of the impression that this is dodgy ... ie that means the gravity is spherically symmetrical in all reference frames. But it was asserted that the result can be derived as an approximation from the stress energy tensor derived here. @pervect et al.

    I confess I've forgotten how to get from the stress energy tensor to a gravitational field for some observer.

    So my question is: can it?

    For the sake of an example: if I put an observer mass m at the origin, and a pulse of light energy E passes on trajectory: ##\vec r = (b,0,ct)^T## then the replacement suggests: $$F =\frac{GEm}{(b^2+c^2t^2)c^2}$$

    I do recall that the approximation works for a container of light travelling at speed v<<c though.

    qv. https://www.quora.com/Does-photons-...energy-generates-gravity/answer/Rick-McGeer-1
     
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  3. Mar 4, 2017 #2

    mfb

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    It would surprise me if that works at any time.

    Let's assume b is large compared to the Schwarzschild radius of our test mass. We can calculate the deflection of light due to the test mass in Newtonian gravity, using the Newtonian force you posted. We know this will underestimate the deflection of light by a factor 2 - one of the famous early results of general relativity.
    I think total momentum should be conserved in this case. If we underestimate the deflection of light by a factor 2, then we should also underestimate the momentum the test mass gets by a factor 2.

    I have seen the same factor 2 derived independently elsewhere, but I don't find it any more. More generally, the factor between relativity and a naive Newtonian calculation is 1+β edit: 1+##\beta^2## if objects at speed ##\beta## move by. This is not a statement about any point in time - it is just about the total deflection. I don't know the time profile.
     
    Last edited: Mar 5, 2017
  4. Mar 4, 2017 #3

    Orodruin

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    This should stick out as dodgy just by the sound of it. Newton's law of gravitation describes the gravitational force of a stationary mass distribution in classical mechanics. Reading the Quora answer, it is clear to me that the poster has no idea what he is talking about and has fundamentally misunderstood the mass-energy equivalence. Another instance of "do not trust what you read on Quora".
     
  5. Mar 5, 2017 #4

    Jonathan Scott

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    The factor is actually ##1 + v^2/c^2##, or ##1 + \beta^2## where ##\beta = v/c##. This is a weak field approximation; for exact accuracy in isotropic coordinates the ##v^2/c^2## term should be multiplied not by the Newtonian gravitational field (which is based on the gradient of the time component of the metric) but by the gradient of the inverse of the space component of the metric, which is slightly different in a strong field.

    In isotropic coordinates for a weak field this factor applies to the transverse component of acceleration of a light beam passing a source at all points along the path. In appropriate units it applies exactly to the rate of change of coordinate momentum in any direction, which is slightly different from the acceleration because the coordinate speed of light varies with potential (i.e. with the time component of the metric).

    If you have a container in which light is being reflected around, the source strength is the same as that for rest mass equal to the energy of the light. There is no factor of 2 in that case. (However, relative to the same isotropic coordinates, the box is slightly bent, and light moving back and forth in it is still being accelerated by twice the Newtonian acceleration relative to those coordinates).
     
  6. Mar 5, 2017 #5

    pervect

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    In the case of ultra-relativistic particles, there's a paper on this that I rather like, though only the abstract appears to be readily available.

    Genralizing this result to light is reasonably straightforwards, one simply takes the limit of a sequence of particles that keep a constant energy and approach "c", the rest mass of the particles in the sequence dropping to keep the total energy ##\gamma \, m \, c^2## constant.

    There are some older papers by Tollman that address the issue for the case of light directly as well. They get the same answer, though I wouldn't really recommend that a modern student study the Tollman papers as the first (and especially not as the only) thing they read on the topic, due to their antiquated approach.

    Measuring the "mass" of a particle by the velocity induced by an ultra-relativistic flyby is not a universal approach to defining mass, but it's a very simple one, easy to describe, relatively simple to communicate the details of. And it's good enough to show that using Newton's law and replacing m with E/c^2 simply doesn't give the same answers that GR does. It's unclear why people think it should.
     
  7. Mar 9, 2017 #6

    Simon Bridge

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    Thanks folks - I suspected as much :)
     
  8. Mar 9, 2017 #7

    PAllen

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    Note, there are papers (I've linked them in other threads, could find them if requested) that show basically the same result as this for anti-parallel beams of dust or light. The same factor, approaching 2 for relativistic beams, and exactly 2 for light, appears.
     
  9. Mar 11, 2017 #8

    Mister T

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    In that June 1989 Physics Today article by Lev Okun he makes the point that people begin to believe that relativstic mass is a genuine relativistic generalization of the newtonian mass. Why that happens is indeed a good question. In a follow up letter to the editor Rindler argues that they don't. Okun responds that they do. In his famous Lectures on Physics Feynman seems to state that it is!

    I think PF participants would largely agree that many people do. I think it's due to the fact that many authors and teachers actually do believe that it is and use it as a crutch to incorrectly explain the mass-energy equivalence by stating that the energy used to speed up a particle makes the particle heavier. Many learners are left puzzled, thinking there's an increase in the particle's quantity of matter. I know that learners are cautioned that they must abandon the notion that mass is a measure of the quantity of matter, but unless the true meaning of the equivalence of mass and rest energy is taught using composite bodies as examples, there is no justification presented for that abandonment. Instead it's just a memorized lesson, along with the equivalence of inertial and gravitational mass.

    From there it's an easy leap to believe that relativistic mass is equivalent to gravitational mass.
     
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