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Gravitational field of photons

  1. Jan 1, 2010 #1
    According to L. Susskind in his GR lecture series, a photon has a gravitational field around it. This is intuitive to me as well. The photon has energy, therefore it curves space-time. I have searched, but I cannot find any sources which discuss the nature of the field.

    If we call the x-axis in rectangular coordinates the trajectory of the photon, what is its gravitational field?
     
  2. jcsd
  3. Jan 1, 2010 #2

    Dale

    Staff: Mentor

    Since we lack a quantum theory of gravity it is kind of hard to answer this question. However, the general class of solutions modeling the spacetime curvature of massless radiation is called pp-waves.
     
  4. Jan 1, 2010 #3
    I would love to see a diagram of equal gravitational potential around the photon. The effect of gravity can never precede the photon in the direction of travel. And the wake it leaves behind dies-off quickly due to the photon moving on. And the photon is not a point particle (quantum mechanics) it is more a distribution that falls off with distance from the peak. No point no singularity problems.
     
  5. Jan 1, 2010 #4

    Dale

    Staff: Mentor

    I am not expert in this, but my understanding is that gravitational potential energy is generally only defined for static spacetimes, and pp-waves are not static.
     
  6. Jan 2, 2010 #5
    as a photon has no location between the time it is emitted and the time it is absorbed, it seems counterintuitive that there would be any gravitational field associated with it. i believe susskind is incorrect.
     
  7. Jan 2, 2010 #6
    Imagine ideally reflective sphere containing matter and antimatter. Matter is source of gravity. Then matter and antimatter annihilates and it is converted into photon gas inside the sphere. GR insists that gravitational field cannot magically ‘disappear’, hence, photon gas *MUST* create a gravitational field.

    In fact, situation is more complicated:
    Before the annihilation gravity is created by:
    A. Matter and anti-matter
    B. Mass of the sphere

    After the annihilation:
    C. Pressure of the photon gas inside the sphere.
    D. Mass of the sphere
    E. Tension of the sphere, created by the pressure of the photon gas.


    I believe A+B=C+D+E, so the far observer should not notice anything (is it correct?)
     
  8. Jan 2, 2010 #7
    Also, I am not sure what happens if there is no sphere.
    And what is % of gravity created by C, D and E?
     
  9. Jan 2, 2010 #8
    huh? If I have a bench top setup to emit photons on the left and have them travel over the table and be detected on the right I know at all times approximately where the photon is. Let say there is a vacuum over the table to make this easier. The photon moves about 1 foot per nanosecond. We can plot the photons position as a function of time. We know where it is.
     
  10. Jan 2, 2010 #9
    Doesn't the energy of the photons count for something?
     
  11. Jan 2, 2010 #10
    No, energy (=relativistic mass) is not a source of gravity in GR.
    So objects moving close to c do not generate MORE gravity because they are 'heavier'. This is a common misconception (that relativistic mass=energy creates gravity). I was also a victim of that misconception before I came here.
     
  12. Jan 2, 2010 #11
    Energy density is a component of the stress-energy tensor and therefore it contributes to spacetime curvature. This curvature is the source of the gravitational field. This is how I understand it, and Susskind also makes this argument in his lectures.
     
  13. Jan 2, 2010 #12
    Dmitry67 and espen180, am I correct in says that the two of you are holding exact opposite positions on this?
     
  14. Jan 2, 2010 #13
    First off, energy is not relativistiv mass. Relativistic mass is m=γm0, while energy relates to mass through E=mc2. Dmitry seems to be saying that energy is not a source of gravity, but the way I percieve it is that GR does not differentiate between mass and energy density, since they are equivalent through E=mc2.

    Second, Dmitry seems to be saying that the momentum density and momoentum flux is independent of mass, though this is not the case, as p=γmv

    If I have interpreted his posts correctly, then yes, we are at a disagreement.
     
  15. Jan 2, 2010 #14
  16. Jan 2, 2010 #15
  17. Jan 2, 2010 #16

    Hepth

    User Avatar
    Gold Member

    see "frame-dragging", "gravitomagnetism".
     
  18. Jan 2, 2010 #17
    Yes, I am aware of both concepts and their applications. I was speculating about the photon's properties, not the general existance of such a field.
     
  19. Jan 3, 2010 #18
    For photon gas, there is momentum flux, but for photons momentum is NOT p=ymv !
    Forget about y and check the particular case of photon gas with m0=0 and v=c
     
  20. Jan 3, 2010 #19
    You have to know the energy of the photon, that is, its frequency. The momentum is given by p=hf/c.
     
  21. Jan 5, 2010 #20
    edpell - inre location of a photon - no, you do not know where a single photon is. here is a clip from david snoke's paper on photons:

    "Note that the uncertainty principle also is essentially a wave property, and has nothing to do with “wave-particle duality.” This is a fact known to all students who learn the mathematics of Fourier analysis, usually in the sophomore or junior year. It is also stressed in introductory graduate quantum mechanics. The momentum of a wave corresponds to its wavelength. A wave with a single wavelength is by definition an extended wave which fills all space, and therefore can be assigned no single position in space. "
    ref - http://philsci-archive.pitt.edu/archive/00001504/01/photon.doc
     
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