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Applying force to light

  1. Jan 23, 2014 #1
    Good day all, I am simply curious, so excuse my ignorance, I am not a scientist... I wish to posit a question regarding the speed of light through a vacuum, is it possible to apply a force to a photon in the direction of it's travel? And what happens in that instance? Based on special relativity, is it true to say that the practical mass of a proton along the vector of it's travel is infinite?
     
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  3. Jan 23, 2014 #2

    Nugatory

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    (I'm assuming you meant to say "photon" not "proton" the second time)

    It is not possible to apply a force to a photon in the direction of its travel (or any other direction, for that matter). Even though we call them "particles", photons aren't like very small bullets that can be pushed around by forces and the concepts of force and acceleration aren't meaningful for them. It's like trying to weigh a color or measure the length of an idea.
     
  4. Jan 23, 2014 #3
    Ok thanks, so then light has no mass relativistic or otherwise, any force derived from light is then secondary. So then light changes direction without force? And the only thing that changes the direction of light is space-time distortion as in gravity?
     
  5. Jan 23, 2014 #4

    Nugatory

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    Pretty much, yes. If you can get hold of Richard Feynman's book "QED: The Strange Theory of Light and Matter" it's well worth reading - it's written for laypeople.
     
  6. Jan 23, 2014 #5
    Thanks I will be sure to get a copy. So it's almost as if light does not actually travel and it's not really a particle (otherwise it would have infinite mass), it's almost like its information that propagates as fast as it can, almost as if it is in fact instantaneous, but there is a relationship between space and time, a slight gradient if you like (linear or otherwise).
     
  7. Jan 23, 2014 #6

    PeterDonis

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    Light does have relativistic mass; relativistic mass is just another word for energy, and light has energy. Light has zero rest mass.

    In vacuum, yes.
     
    Last edited: Jan 23, 2014
  8. Jan 23, 2014 #7

    PeterDonis

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    This isn't really correct. There are plenty of contexts in which light does behave like a particle; for example, check out the Compton effect, in which X-ray photons exchange energy and momentum with electrons just like two billiard balls colliding.

    A better way to describe the difference between light and a particle like an electron is to say that light has zero rest mass while the electron has nonzero rest mass, and objects with zero rest mass are fundamentally different, physically, from objects with nonzero rest mass; they obey different rules. The rule that energy (relativistic mass) increases without bound as the speed of light is approached only applies to objects with nonzero rest mass. The rule for objects with zero rest mass is that they always move at the speed of light (when they're in vacuum), but they can have any finite energy.
     
  9. Jan 23, 2014 #8

    vanhees71

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    Well, I'd like to be a bit more picky on that issue. Photons are neither particles nor waves but quanta. They are properly described only with relativistic quantum field theory. They have very little in common with "billiard balls". It's not even possible to define their position in a strict sense, which is true for all massive particles with a spin [itex]\geq 1[/itex].

    An electron is massive and thus admits at least a position operator, but as well as an elementary particle, it is a quantum, described by relativistic quantum field theory, or for small energies and momenta in the classical limit by non-relativistic quantum mechanics.

    Also I don't like to call energies mass. The idea of relativistic masses is an old-fashioned idea from the early days of relativity, but that's just energy divided by [itex]c^2[/itex] and thus not needed in addition to energy, which together with momentum builds a four-vector in Minkowski space. Nowadays mass is always understood as "invariant" mass, which is a scalar quantity. Photons are, to very high precision, massless quanta.
     
  10. Jan 23, 2014 #9

    PeterDonis

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    Yes, this is true; strictly speaking photons are not "particles". But if we're being that picky, neither are electrons. I only meant that in some contexts, like that of the Compton effect, the observed behavior is similar to what you would predict if you thought of the photon and electron as little billiard balls colliding and exchanging energy and momentum. There are plenty of other contexts, such as X-ray diffraction in crystals, where such a model does not work well.

    Photons aren't "massive particles"; they have zero rest mass. Did you mean "massless"?

    Also, I'm not sure what you're referring to here: are you referring to Newton-Wigner localization, and the fact that the operators obtained by this method do not work for massless particles with spin ##\geq 1##? Your next comment appears to suggest that.

    I agree with all this; I don't like the term "relativistic mass" either. But some people still use it; there are even some physicists who still use it.
     
  11. Jan 23, 2014 #10

    ghwellsjr

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    Of course light travels--it's speed is defined to be 299792458 m/s. That definition provides the relationship between space (if by that you mean distance) and time. The definition of our unit of time is based on a real clock using cesium.

    You are trying to apply concepts that only apply to material objects (and by that I mean objects with mass) to light. In the theory of Special Relativity, time goes slower for moving material objects (clocks tick slower) but since you can only build a clock out of material objects, they cannot propagate at the speed of light so time does not apply to light, that is, we can't say that "it is in fact instantaneous". It's not that time for a photon is zero, it's that time doesn't apply for a photon.

    I have no idea what you mean by "a slight gradient".

    By the way, welcome to PF.
     
  12. Jan 23, 2014 #11
    That never made any sense to me. A photon cannot be at rest so it doesn't make any sense to specify any value for its rest mass. Saying its rest mass is zero is saying that if a photon were at rest its mass would be zero.

    I.E. If we violate the laws of physics (allow a photon to be at rest) what do the laws of physics say would happen (when conducting an experiment designed to measure the mass of the photon)?
     
  13. Jan 23, 2014 #12

    WannabeNewton

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    It has zero invariant mass; for time-like particles the invariant mass is the same thing as the rest mass since time-like world lines admit instantaneous rest frames i.e. at any event ##p## on the world line of the particle there exists a frame ##\{e_a\}## such that the future-directed unit tangent ##u^{\mu}## to the world line is given by ##(e_0)^{\mu} = u^{\mu}##. The invariant mass is defined by the on-shell relation ##p_{\mu}p^{\mu} = -m^2##. For null particles, like photons, this obviously vanishes (I'm using the term "particle" in a semi-classical manner of course since photons, or better yet photon states, are normal mode excitations of the photon field).
     
    Last edited: Jan 23, 2014
  14. Jan 23, 2014 #13
    I don't know how to apply force a photon, but if you apply force on a massless particle, it will change its momentum and energy just like massive particles too. So if the force is applied to accelerate the particle, it will maintain its speed of light, but gains more momentum and energy. It get's more "blue". If the force is applied to decelerate the particle, it will again maintain its speed of light, but lose momentum and energy. It get's more "red".
     
  15. Jan 23, 2014 #14

    pervect

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    Well, I see at least two issues here:

    The first issue is how does special relativity treat forces. The simple answer is that f = dp/dt, where p is the relativistic momentum. A slightly more involved answer notes that because t is observer dependent, forces are observer dependent, and that usually people who actually work problems use f = dp / dtau, where tau is proper time, rather than dp/dt, where t is coordinate time.

    Note that dp/dt is not the same as f = m a for several reasons. One of the important points to be made here is that photons have momentum, without having any rest mass, which is clearly different than the f=ma approach. Some students seem to have problems with "unlearning" f=ma, I'm not sure what (if anything) can be done to help them.

    The second issue is how quantum mechanics deals with forces. One way in which this is done is via an exchange virtual particles. But the actual details of making this work isn't something I'm personally familiar with. Another wrinkle here is that when photon interacts with a crystal, one has to deal with quantized lattice vibration (called phonons).
     
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