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Photon mass?

  1. Feb 12, 2010 #1
    I've been reading some of these posts and it sounds like they are saying photons have zero mass. I was under the impression that light is bent or attracted to stars as it passes by- because of the mass of the star-or in other words isn't gravity at work here? and if so isn't that saying a photon has mass? And in the example of black holes if light can't escape then isn't that because of gravity? Thanks for any input!
     
  2. jcsd
  3. Feb 12, 2010 #2

    ZapperZ

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    Please read our FAQ in the General Physics forum.

    Zz.
     
  4. Feb 12, 2010 #3
    To elaborate slightly on the FAQ, the famous equation [itex]E_0 = m_0 \, c^2[/itex] (rest energy equals rest mass times the square of the speed of light) only applies to particles, or systems of particles, with non-zero mass, because its derivation from the more general equation

    [tex]E^2 = (pc)^2 + (m_0c^2)^2[/tex]

    applies specifically to the case where momentum, p, is zero. We can't talk about the rest energy of a photon because a photon has no rest frame (it isn't possible to choose a spacetime coordinate system in which the photon is not moving because in any frame we do choose, the photon will be moving with the speed c). Since the photon has no (rest) mass, we can simplify the above equation for the photon to

    [tex]E = pc.[/tex]

    (On the other hand, a system of photons, not all moving in the same direction, does have a rest mass because we can choose a reference frame (spacetime coordinate system) in which there's no total velocity, e.g. if we have two photons travelling in opposite directions, we can pick a point between them in which the vector sum of their velocities, and momenta, in opposite directions is zero. Velocity, momentum and acceleration are all vectors; they have a magnitude and direction, whereas speed is a scalar, just a number.)

    But acceleration due to gravity (the rate at which the velocity of a particle changes, i.e. how much its path is bent) doesn't depend on the mass of the particle being accelerated. If, like whichever of the Apollo astronauts it was, you drop a feather and a rock on the moon (where there's no air to impede the fall of the feather) they hit the moondust at the same time. Gravity is an unusual force in this respect, and, because of this, in general relativity isn't even considered a force, but rather the "curvature" of spacetime. The mass of the sun curves spacetime, and whatever moves through space in time (regardless of its own mass) moves along a curve called a geodesic, which is the "next best thing" or "most direct route" through curved spacetime.

    To get an intuitive idea of what a geodesic means in general, it's best to start with a 2-d surface such as the surface of the earth. If you go in what looks at the small scale like a straight line on the earth, you're following a geodesic; these are the paths which on the surface of a sphere will--if you follow them far enough--bring you back to where you started (but that's not a general property of geodesics, it just happens to be true for geodesics on a sphere).
     
  5. Feb 12, 2010 #4
    Oh the mind numbing world that is general relativity... Tregg if I were you I would read more into general relativity, though start off by reading special relativity.
     
  6. Feb 13, 2010 #5

    Dale

    Staff: Mentor

    I don't know why this concept shows up so often. Yes, you do need GR to get the numerically correct answer, but even in Newtonian physics the acceleration due to gravity is independent of mass.
     
  7. Feb 13, 2010 #6
    Yet it requires an object to have mass.
     
  8. Feb 13, 2010 #7
    To include the case where m = 0, I'm guessing you'd define acceleration due to gravity as the limit

    [tex]\lim_{m\rightarrow0}\frac{\mathbf{F}}{m}=\lim_{m\rightarrow0}\frac{GMm}{mr^2}\hat{\mathbf{r}}=\lim_{m\rightarrow0}\frac{GM}{r^2}\hat{\mathbf{r}}=\frac{GM}{r^2}\hat{\mathbf{r}},[/tex]

    where G is the gravitational constant, M the mass whose influence is being calculated, and m the mass of whatever it is that you're calculating M's influence on, r being the distance between the centre of mass M and what is being influenced, and [itex]\hat{\mathbf{r}}[/itex] a position vector of unit length pointing from the centre of mass M towards the thing being influenced.
     
    Last edited: Feb 13, 2010
  9. Feb 13, 2010 #8

    Dale

    Staff: Mentor

    How so? The passive acceleration due to gravity is well-known to be independent of mass (e.g. the hammer and feather demo from the moon landing). If x is independent of y then why would it be "required"? I don't know why this error is so common.
     
  10. Feb 13, 2010 #9

    Dale

    Staff: Mentor

    Exactly. Also, note that by Newton's second law: F = ma, if m=0 then f=0 regardless of a. So the fact that there is no force is consistent with a finite acceleration of a massless object.
     
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