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Does photons have a mass?

  1. Mar 7, 2009 #1
    Been thinking about this for sometimes now, most people say photons doesn't have a mass, and in theory since photons travel the speed of light then it must either have no mass or always been travelling at the light. But if a photon doesn't have a mass then how can things such as blackhole or even just a star with high enough gravitational force to bend light. My personal thinking is that photons might have a mass but so small that we can't measure, but seriously not sure. :P
     
    Last edited: Mar 7, 2009
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  3. Mar 7, 2009 #2

    Dale

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    In Newtonian gravity a particle's acceleration due to gravity is independent of its mass. In GR particles follow geodesics, so the motion is also independent of mass.
     
  4. Mar 7, 2009 #3

    ZapperZ

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    You may want to start by reading the FAQ in the General Physics section, and definitely before you venture into making your own personal theory.

    Zz.
     
  5. Mar 7, 2009 #4
    Photons in my perspective, is the light or radiation emitted by something like a light bulb to our Sun. The smallest mass that we have discovered is the electron that orbits the nucleus of an atom. Anything can be measured in a scientific way, big or small. They may be exact or an approximation to the true value. There are exceptions, but photons are like radiation like the visible light that we see.
     
  6. Mar 7, 2009 #5
    I kinda understand what you are saying but, I don't quite get then how can gravity be calculated using gm/r^2
    and also during the relativistic effect a particle will increases mass dramatically which causes the acceleration to slow down, so i thought there must be a clear link between mass and gravity.. :( hmm..
     
  7. Mar 7, 2009 #6
    what do you mean by mass?
    i think
    E=m.c2
    gives the permission to think about the mass of a photon,but rest mass,is not !

    there is a difficulty m=m0/[tex]\gamma[/tex]
    where [tex]\gamma[/tex]=[tex]\sqrt{}[/tex]1-v2/c2
    if m0=/=0
    then m=infinity for v=c
    i think mass and energy both are same for gravity ....
     
  8. Mar 8, 2009 #7
    I really don't understand from m=m0/[tex]\gamma[/tex]
    m0=/=0, surely since m0 is on the top of fraction and its 0; the answer must be 0. How did you get infinity?
    But yes its true that mass must be either infinite or 0 to actually move at the speed of light, thats why it requires infinite amount of energy to do that.
    But then again I really don't understand how can gravity affect light.
     
  9. Mar 8, 2009 #8
    E=mc^2 is just that, energy and mass is the same thing.

    You don't notice any difference between effective mass and rest mass, its impossible to determine if you trapped 1kg worth of photons within a totally reflective box or just 1 kg of normal matter.

    And on top of that photons also creates gravity fields around themselves which is obvious since they are effected by the fields, otherwise we could create physical absurdities.
     
  10. Mar 8, 2009 #9
    Don't worry now, I realise that I forgotten theory of relativity when I thought of this question.
    But I don't think light does create a gravitational field though, since if the mass of a photon is 0 then gm/r^2 must also equal 0
     
  11. Mar 8, 2009 #10

    Mentallic

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    While energy and mass can be equivalently converted between each other, can someone please confirm this statement true? If we had a internally reflective box filled with [tex]c^2 J = 9.10^{16} J[/tex] of photon energy and another box with 1kg of mass, would these 2 boxes now be equivalent in mass? I am skeptical about this so can someone please confirm/disprove this.
     
  12. Mar 8, 2009 #11

    Dale

    Staff: Mentor

    That is theoretically correct, but since there is no such thing as a perfectly reflective box it is not experimentally testable.
     
  13. Mar 8, 2009 #12

    Dale

    Staff: Mentor

    That is a common misconception that I am always fighting. Let's split our thoughts about gravity into the dynamics (i.e. the forces) and the kinematics (i.e. the motion). And for simplicity let's start with Newtonian gravity and only consider cases where M>>m (i.e. a small particle in the gravitational field of a big planet).

    Dynamics: The particle is acted on by a force F = GMm/r². The mass of the particle does determine the force on the particle, so a particle with m=0 should have no force. But how do we measure this force? It turns out that this force is impossible to measure directly, we can use a scale to weigh the object, but what this actually measures is the normal force not the force of gravity. Since we cannot measure this force then how do we know it is given by F = GMm/r²? We know because of the kinematics, i.e. on a scale we measure the normal force, the object is kinematically stationary, so there must be a gravitational force cancelling out the normal force.

    Kinematics: The particle undergoes an acceleration of a = GM/r² if it is free to move. So a free particle of m=0 should (according to Newtonian theory) undergo the same acceleration as any other free particle. We can measure this acceleration directly, so in some sense, this acceleration is the primary effect of gravity and the force is a secondary or presumed effect. Now, since we can directly measure the acceleration of a massless particle let's work backwards and see what the force should be. We can use F = ma = 0 GM/r² = 0. So from the kinematics, despite the fact that the particle is accelerating, the force is 0, i.e. it does not take any force to accelerate a massless particle, they have no inertia. So this kinematic result is consistent with the dynamics above.

    GR: Now, there is one big problem with all of the above, and that is that all massless objects travel at c where relativistic considerations are obviously important. So it turns out that the Newtonian approximation is off by a factor of 2. I don't want to get too far into the details, but in GR gravity is not considered a real force at all. Instead, the observed kinematics are attributed to curved spacetime rather than to any force. In other words, free particles, like a satellite, travel in (locally) straight lines called geodesics through a (globally) curved spacetime. There is no force of gravity, so there really are no dynamics, and the path of the particle depends on the curvature of the underlying spacetime and not on the mass of the particle. Therefore, massless free particles also travel on these geodesics so their path is (globally) curved just like everything else. Again, the kinematics do not depend on the mass of the particle.
     
  14. Mar 8, 2009 #13
    Its because that formula is wrong though unless you consider effective mass, and photons got effective mass. (Effective mass is just another word for total energy contained)

    As I said energy and mass is indistinguishable except that energy is moving and mass is standing still.
     
  15. Mar 9, 2009 #14
    if rest mass m0 not equals to zero
    then m=infinity for v=c
    so if v=c
    then m0 must be zero
    that's why i write m0=0 for photon
    look at the equation E2=p2 c2+m02c4
    if m0=0 & E not equals to zero
    then p=momentum not equals to zero
    so relativistic momentum can be non-zero even for a zero rest mass
     
    Last edited: Mar 9, 2009
  16. Mar 9, 2009 #15
    Ok now leads me to a few more questions:
    1. If gravity is considered as a force that pulls bodies together, but can it be considered as a force that bend space-time, which then allows bodies to fall into each other?

    2. Many people talked about effective mass and rest mass, but since photons or light are always traveling, so it most of the time are exerting effective mass, can we not just ignore rest mass?

    3. I know that photons does have momentum, hence the maths and experiment. If a massless object have acceleration and momentum, but doesn't have force, then does this formula still applies? F = [tex]\frac{dp}{dt}[/tex], so if there is a momentum over time, then there's a force. This is not a really good question, sorry Im not very good in physics. :(
     
    Last edited: Mar 9, 2009
  17. Mar 9, 2009 #16
    to
    nelsyeung,
    "But I don't understand the last piece of maths"
    which one?
     
  18. Mar 9, 2009 #17
    Oh sorry it suppose to be a statement that I don't understand does F = dp/dt still applies
     
  19. Mar 9, 2009 #18

    ZapperZ

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    Have you even read the FAQ that has been suggested to you? There are at least TWO separate entries in there that are relevant here that might have easily answered your original question. If you start from there and elaborate on which part you do not understand, rather than keeping on making up a rather vague starting point, it might be easier for the rest to address the problem you are having in understanding all this.

    Zz.
     
  20. Mar 10, 2009 #19
    That one applies perfectly, however usually you wont see it as a force acting on the light but instead the light acting with a force on other objects except when dealing with gravity.
     
  21. Mar 10, 2009 #20
    Maxwell in about 1871 deduced that electromagnetic radiation exerted radiation pressure on any surface, and in 1901 Nichols (at Dartmouth) verified it experimentally to 1 or 2% accuracy by reflecting light striking a mirror on a radiometer in a vacuum. The radiation pressure of light is equivalent to momentum transfer. Since light in vacuum always travels at the speed of light in any inertial coordinate system, it is nonsensical to talk about photon rest mass. But if you count photons (quantize light), you can deduce the amount of momentum transfer per photon. So if a photon has energy E = h*w/(2pi), (Planck's constant times frequency) its momentum is E/c or h*w/(2c*pi). But E/c is the same as the momentum of an extremely relativistic particle wth total energy E (consider relativistic electron or neutrino beams, where the rest mass is neglegible). Also think about using a light beam as (very inefficient) thrust for interplanetary vehicle.
    Photons falling in a gravitational field gain energy (and momentum), as proved by the famous Mossbauer Effect experiment at Harvard (Purcell).
     
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