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Light and Relativistic Mass

  1. Mar 17, 2012 #1
    Hi, a while back i had ran into the equation for dynamic mass [M = m/√(1-(v^2/c^2))] and had immedietly taken an interest in it in that it seems to imply that for an object with mass moving at the speed of light its mass would become, virtually, infinite. Is this an erroneous assumption? If not, what would the implications for light itself be? Could the massless photon be explained by the idea that, because it moves at the speed of light, its mass has increased to such a point that the photon can no longer be distinguished as a discreet entity in so far as the measure of its mass is concerned?
     
  2. jcsd
  3. Mar 17, 2012 #2
    Well, most physicists would call that quantity the energy of the particle. As you know, photons have a finite energy and move at the speed of light, so you would be completely right to conclude that they must have zero (rest) mass.
     
  4. Mar 17, 2012 #3
    If photons have a finite amount of energy, wouldn't that require a non-zero quantity for its mass? and if it has a measurable mass, wouldnt it be restricted from travelling at the speed of light?
     
  5. Mar 17, 2012 #4
    The photon's rest mass has not increased. It s rest mass is zero and is not subject to the relativistic mass increase that you mention above. The concept of relativistic mass is not encouraged these days as it is confusing. For example from the point of view of a muon passing the Sun, the apparent mass of the Sun is enough for it to be a black hole but that is not the case. Relativistic mass is just a measure of how much energy is required to accelerate a given mass from a rest state to its current velocity in a given reference frame . Active gravitation appears to be a property of rest mass and this does not increase with relative velocity. Rest mass can only be defined for an object that has a definable rest reference frame so this does not apply to a photon. Curiously, although a single photon has no rest mass, a pair of photons going in opposite directions can have a definable centre of momentum frame and thus have a combined rest frame. I imagine this is why a cloud of photons can in theory be a source of active gravitational mass. Photons are also curious in that they have a definable momentum and can impart momentum to particles with rest mass when they collide with them. Photons also have a very clearly defined total energy that is a function of their frequency and this total energy is nowhere near infinite.
     
  6. Mar 17, 2012 #5

    jtbell

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    Staff: Mentor

    The general relationship between energy, momentum, and mass is

    $$E^2 = (pc)^2 + (mc^2)^2$$

    where m is the invariant mass (often called the "rest mass"), which is zero for a photon. So a photon has E = pc.
     
  7. Mar 17, 2012 #6
    but if momentum is

    $$p^μ=mv^μ$$

    wouldnt a massless particle render the entire expression as 0? or does p represent something other than momentum?
     
  8. Mar 17, 2012 #7

    HallsofIvy

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    That formula is not correct for relativistic speeds.
     
  9. Mar 17, 2012 #8

    jtbell

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    Assuming you mean the four-momentum and four-velocity here, what's the four-velocity of a photon? :wink:
     
  10. Mar 17, 2012 #9
    The expanded expression for total energy E is:

    [tex]E^2 = \left( \frac{m v c}{\sqrt{1-v^2/c^2}} \right)^2 + (mc^2)^2 [/tex]

    For a photon v = c, so this becomes:

    [tex]E^2 = \left( \frac{0}{0} \right)^2 + (0)^2 = \frac{0}{0} [/tex]

    which is undefined (so not equal to zero).

    This is resolved by using the alternative expression for momentum:

    [tex]p = hf/c[/tex]

    where h is the Planck constant and f is the frequency. Apparently this expression applies to both massless and massive particles via the DeBroglie wavelength relationship.

    Anyway, the total energy of a photon using the above momentum expression is E = hf.
     
  11. Mar 26, 2012 #10
    Thank you yuiop for the thorough replies. this is definitely something i will have to think on before completely grasping it, but you have answered my question.
     
  12. Mar 26, 2012 #11
    [itex]p^\mu = mv^\mu[/itex] is most definitely correct at relativistic speeds, assuming this is what you were responding to.
     
  13. Mar 27, 2012 #12

    rbj

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    the reason that photons are massless (that their rest mass or "invariant mass" is zero), is that their speed relative to any reference frame is [itex]c[/itex]. turn your equation around a little:

    [tex] m = M \sqrt{1 - v^2/c^2} [/tex]

    if [itex]v=c[/itex], then [itex]m=0[/itex] no matter what the photon's momentum or inertial mass [itex] M = p/c [/itex] is.
     
  14. Mar 28, 2012 #13
    Im curious though, if an object other than a photon would reach the speed of light, (and this is purely hypothetical), would its mass become infinite?
     
  15. Mar 28, 2012 #14
    hypothetical yes.
     
  16. Mar 28, 2012 #15

    jtbell

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    Staff: Mentor

    And impossible.
     
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