Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Photon/particle's energy in General Relativity

  1. Jun 3, 2009 #1
    Hi everybody!

    I am wondering, among other things, whether the Special Relativity relationship E = p for a photon (I am using c = 1 units ) is still valid in General Relativity.

    Let me explain my question in detail. By applying the null geodesic condition with a diagonal metric, we obtain

    [tex] ds^2 = 0 \Rightarrow p^2 = -g_{00}(P^0)^2 [/tex] (1).

    where [tex] P^\mu [/tex] is the four-momentum and [tex] p^2 \equiv g_{ij} P^i P^j = P^i P_i [/tex].

    My question is: in SR we did not have any doubt in identifying P^0 as the energy of the particle. How does it transfer to GR? What is energy and what is momentum? My shot was that, if we want to preserve the E = p relation, then we need that

    [tex] p \rightarrow momentum [/tex]


    [tex] \sqrt{-g_{00}} P^0 \rightarrow energy [/tex].

    Is this correct?

    And what about velocity?
    It is easy to find that

    [tex] v^i = \frac{P^i}{P^0} [/tex],

    where [tex] v^i \equiv \frac{dx^i}{dt} [/tex].

    If we contract both sides of the above expression with [tex] g_{ij} v^j [/tex] and if we define [tex] v^2 \equiv g_{ij} v^i v^j = v^i v_i [/tex], we find that

    [tex] p = (P^0) v [/tex]. (2)

    By comparing (1) and (2) it emerges that

    [tex] v^2 = -g_{00} [/tex].

    In a flat space-time (or in an inertial frame), then [tex] g_{00} = -1 [/tex] and v = c. Otherwise we have that the speed of light is different from c. Does it mean that in GR the concept of "constant speed of light" makes sense only in inertial frames? This is equivalent to say that the speed of light is equal to "c" only locally, since for the equivalence principle we can always find an inertial frame, but it has to be locally defined (in a space-time sense).

    One last question. With respect to Cosmology, one usually defines the comoving distance as the distance covered by a photon in the comoving frame. The particle horizon is the comoving distance from the Big Bang to today; by using a LCDM model it amounts to around 14.000 Mpc. I always thought that this was some sort of non-physical distance since it is the distance the photon traveled in a reference frame decoupled from the expansion. I was thinking that the "physical" particle horizon was just the age of the Universe times the speed of light, i.e. around 4.500 Mpc. Now I see that this is not the case. In fact, there is no such a "physical" frame in which the speed of the photon is constantly equal to "c" along all his path, hence it makes no sense to say that it "physically" travelled speed_of_light X time_of_travel kilometers. Am I right?

    Thank you very much for any answer, and sorry for the long post :)


    Last edited: Jun 3, 2009
  2. jcsd
  3. Jun 3, 2009 #2


    User Avatar
    Science Advisor

  4. Jun 4, 2009 #3
    Hi atyy,

    thank you for your answer!

    The link seems very interesting. However, I was hoping you could point some intuitive physical argument, not a mathematical one.


  5. Jun 4, 2009 #4


    User Avatar
    Science Advisor

    If you want values for physical quantities, make sure to define them independent of coordinates. g00 doesn't have a meaning at all, unless your coordinate system is defined in a specific way.
    For a light wave with wave vector k, the frequency (energy/h) measured by an observer with four-velocity u is [tex]f=-u_a k^a=-g_{ab}u^bk^a[/tex].
  6. Jun 4, 2009 #5
    Hi Ich,

    thank you for your answer.

    This is the point: I would like to know what is the energy of the photon in a generic system of coordinates in which the metric is described by [tex] g_{\mu\nu} [/tex]. Is it [tex]g_{ij}P^iP^j[/tex]? (sum is over spacial indexes)

    Is the implicit summation over 0,1,2,3 or just over the spacial indexes? In the latter case, wouldn't we have f=0 if the observer is still?
  7. Jun 4, 2009 #6

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    With respect to which observer?
  8. Jun 4, 2009 #7
    Hi George,

    you are perfectly right, my question does not makes sense without specifying the observer's motion with respect to the photon. So I guess the correct answer is Ich's:

    [tex] f=-u_a k^a=-g_{ab}u^bk^a [/tex],

    because it takes into account that. However, I am curious about the quantity

    [tex] g_{ij}P^iP^j [/tex].

    How does it relate with the energy of the photon?

    Thank you,

  9. Jun 4, 2009 #8

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is always the square of the photon's energy when indices represent components with respect to the observer's orthonormal frame. Unfortunately, many courses and texts put little or no emphasis on frames. I hope to write more about this (maybe tomorrow).

    For an example of the use of frames for physical speeds, see

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook