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Photon beaming

  1. Aug 20, 2010 #1
    From what I have learned so far, it appears that a light emission orthogonal to motion acts exactly like a Newtonian massive particle with conserved longitudinal momentum. Is this correct???
    If so it would seem to be a cosmic coincidence of monumental proportions.
    Having given it some thought all I have arrived at is: Possibly Newton was right and photons do have mass, Then the massless staus would appear to be a conventional calibration of the metric and a lack of rest photons.
    In this case the independence of light from the motion of the emitter would seem to only strictly apply wrt emissions parallel to motion where the physical cap of c would prevent momentum conservation while all other angles of emission would retain coordinate speed of c but would have different velocity vectors
    Is there any conceptual mechanism to explain this??
    Am I totally misinformed regarding beaming???
  2. jcsd
  3. Aug 20, 2010 #2


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    What do you mean by "conserved longitudinal momentum" here?

    For what it's worth, the energy-momentum four-vector (E/c, p_x, p_y, p_z) transforms from one frame to another (via a Lorentz transformation, of course) in exactly the same way as the time-position four-vector (ct, x, y, z). This is true for both massive and massless particles.

    In particular, for a relative velocity in the x-direction (only), the Lorentz transformation affects only p_x, leaving p_y and p_z unchanged, just as it affects only x, leaving y and z unchanged.
    Last edited: Aug 20, 2010
  4. Aug 20, 2010 #3
    I mean that a photon emitted transversely in a moving system and reflected within that system bounces straight up and down just like a Newtonian ball. It retains the forward motion of the system. i.e. conserved forward momentum. It is not independant of the motion of the source except with regard to emissions along the path of motion.
    As I understand it, this is true as observed from other inertial frames as well. Photon beaming.
    So if the frame is moving in x and the photon is emitted with a pure y vector relative to that frame , as observed in other frames the photon would have an x velocity component as well. Is this incorrect???
    Unless I am misunderstanding what I have been reading.
  5. Aug 20, 2010 #4
    Consider a hydrogen atom with the electron emitting a 2p-1s transition photon (~10.2 eV) in the y direction transverse to its motion with velocity βc in the x direction. What is the energy, direction, and mass of the photon in the moving frame?

    Use Lorentz transform (in thumbnail)

    In hydrogen atom rest frame, the photon is:

    E = 10.2 eV =pyc, and pxc = pzc = 0

    The photon mass is M2 = E2 - (pyc)2 = 0

    In moving frame:

    E* = γE

    pxc*=- βγE

    pyc* = pyc = E
    pzc* = 0

    Photon mass in moving frame is

    (M*)2 =(E*)2 - [(pxc*)2 +(pyc*)2]

    = (γE)2 -[(βγE)2 + E2 ] = γ2(E22E2) - E2 = 0
    Photon has no mass in moving frame.

    angle in moving frame is θ = tan-1(pyc*/pxc*) = tan-1(1/βγ)

    Bob S

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    Last edited: Aug 20, 2010
  6. Aug 20, 2010 #5


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    Thanks, Bob S. Good answer.
  7. Aug 20, 2010 #6


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    1. The scalar speed of light is invariant (the same in all frames) but the vector velocity of light is not. It can change direction when you change frames.
    2. Photon momentum behaves like any other kind of momentum. The total momentum of a system is conserved (remains unchanged over time when measured in the same frame throughout).
    3. Momentum is conserved but not invariant.
    4. Momentum of a massive particle relative to a frame is given by [itex]p = \gamma mv[/itex].
    5. Momentum of a photon relative to a frame is given by [itex]p = E / c = h \nu / c = h / \lambda [/itex]
  8. Aug 21, 2010 #7
    From these posts I seem to understand that my initial grasp was basically correct as far as the behavior if not on any implications of mass.
    As a self test:
    1) A massive particle accelerated transversely changes vector magnitude as well as direction relative to an observing frame for both velocity and momentum??
    2) A photon emitted transversely changes direction of the velocity vector but not the magnitude??
    Changes both the direction and magnitude of the momentum vector with the magnitude decreased relative to the magnitude in the emitting frame??

    3) With regard to a massive particle say a bullet that is rotating around the axis of motion: in a relative frame the forward momentum would result in a sideways drift relative to the spin??
    Wrt a photon this is not such a comfortable picture. If we consider a photon as a traveling waveform this would mean it would be out of phase along an orthogonal front relative to its linear motion, yes???
    It would seem to be more realistic to view the angle derived from the vector sum of conserved momentum as an actual propagation angle with the wave front in phase, relative to travel, does this make any sense??

    4) Given the picture I have so far it would seem impossible for any photon to have a straight linear path orthogonal to the emitter motion.
    That if there is a light detector at the end of a long light absorbtive tube aligned directly perpendicular to the source motion that it would seem impossible for a photon to reach the detector??
    Whats wrong with this picture???

    Thanks to all for your help in elucidating this question
  9. Aug 21, 2010 #8


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    My first thought on this is that from the point of view of the moving emitter, the tube will appear to be tilted towards the emitter, giving a point on its trajectory when a photon can be fired down the tube.

    I'll try the calculation tomorrow unless someone (please :smile:) beats me to it.

    The scenario is shown in the picture. The tube has end coordinates (0,0) and the (0,L). The emitter travels along (V t, yE ). We can define the tube in 4 dimensions by imagining a light beam emitted at the bottom end and sent to the top end. This gives a 4D position vector of the event when the light reaches the top ( L/c, 0, L ). From the emitter frame this transforms

    [tex] ( L/c,0,L) \rightarrow ( \gamma L/c, \gamma \beta L/c, L)[/tex]

    The x-coord of the top moves left, tilting the tube. I've cheated by choosing axes so the bottom doesn't move, but it shows the effect.
    The important thing is that the bottom and top of the tube have different relative velocities with the emitter, and so the change in the x-coords will be greater at the top than the bottom, giving the tilt. As the emitter passes over the tube, it appears to the emitter that the tube becomes vertical then starts tilting the other way.

    Because every point of the tube has a different relative velocity with the emitter frame, the tube will appear bent and tilted.

    I could have got this wrong but it seems right.

    Attached Files:

    Last edited: Aug 21, 2010
  10. Aug 21, 2010 #9
    Look at it this way. Consider an emitter in the CM system emitting a photon in the direction perpendicular to its direction of motion at velocity βc with respect to an observer in the Lab system. The photon is moving down a long straight absorptive tube, at rest in the CM system. Regardless of the velocity of the emitter and the absorptive tube relative to the observer, the photon will always exit the end of the tube, independent of the emitter's velocity in the Lab system. The observer sees the photon exiting the absorptive tube, at an angle θlab = tan-1(1/βγ) relative to its direction of motion. (see Post #4).

    So why does the absorptive tube appear to be rotated in the Lab system? It is due to the Terrell effect:


    Also see visual effects in

    http://www.anu.edu.au/physics/Searle/ [Broken]

    also see Post #9 in thread https://www.physicsforums.com/showthread.php?t=19860&highlight=terrell+relativistic

    Bob S
    Last edited by a moderator: May 4, 2017
  11. Aug 22, 2010 #10


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    This is embarassing, I came back to delete my post #8 but I'm too late.
  12. Aug 22, 2010 #11
    If in the rest frame of the emitter, the photon reaches a detector at the end of the tube, then it follows that in any other frame the same must be true, in relativity and in Newtonian physics. Considering a vertical tube in a train. A ball is dropped in the top of the tube by an observer on the train and the ball appears to follow a path straight down to the bottom of the tube. To an observer outside the train, the ball follows a diagonal path that is combination of the balls vertical path due to gravity and trains horizontal motion. In the Newtonian context, the ball follows a diagonal path while staying within the confines of the vertical tube at all times. In the relativistic case, it is pretty much the same. If the tube is exactly orthogonal to the motion, then it does not rotate or bend. If the tube is tilted at some angle other than orthogonal or parallel to the motion, the tube angle relative to the motion appears different in the different frames and appears to be tilted differently (but not bent). Either way the ball (or a photon) stays inside the tube in both frames and hits the target or detector in both frames. I gave an equation for calculating the velocity of a particle or photon that is emitted at an angle to the relative motion of two frames, in a similar thread: https://www.physicsforums.com/showthread.php?t=423338
  13. Aug 22, 2010 #12
    Hi Bob S ....I am afraid I was not clear enough in my post.
    The tube and detector are both at rest in the lab frame. There is clearly no problem if the tube is in the emitter frame otherwise a light clock could not work.
    As for Terrell rotation I am very interested , even if it is an optical effect that would not really apply to this situation. I read about the effect years ago and cant remember the source , From what I do remember the effect only involved length contraction and the optical effects resulting from light propagation time from vatious parts of the object.
    I have searched for more detailed links other than the one you cited but without success.
    I will try the thread you referenced , thanks.
    BYW thanks for your post wrt momentum and mass. Some of the notation was unfamiliar to me but I think I got the gist.
    Last edited by a moderator: May 4, 2017
  14. Aug 22, 2010 #13
    Ooops I was too ambiguous . In this case it is dropping a ball from the train down a vertical tube in the track frame. Thanks
  15. Aug 22, 2010 #14
    Looking at your post I don't exactly get the source of any embarrassment.
    Perhaps you could explain in detail just how you screwed up:tongue:
  16. Aug 22, 2010 #15
    Is it perhaps the different relative velocities wrt the emitter that you are unhappy with???
  17. Aug 22, 2010 #16
    If the tube and detector are both in the Lab (observer's) frame, then the tube tilt angle is

    θlab = tan-1(1/βγ)

    as shown in my earlier post #4. The tube is not bent in the lab frame, nor is the observed trajectory of the emitted photon. If the photon reaches the end of the tube and the detector in the lab frame, then it will also, when observed in the CM frame.

    Bob S
  18. Aug 23, 2010 #17
    Hi Bob I understood that the angle in post #4 was the vector sum i.e. direction of the photon relative to the lab frame.
    If the tube (in the lab) is orthogonal to the path of the emitter where does the tilt come from??? The terrell effect is purely optical as far as I know.
    SO the question is; can a photon follow a path directly perpendicular to the motion of the emitter???
    If not then how could it travel through the tube???
    Thanks for your responce
  19. Aug 23, 2010 #18
    The tube in the lab cannot be orthogonal to the direction of motion; it has to be at an angle

    θlab = tan-1(1/βγ).

    Bob S
  20. Aug 23, 2010 #19
    What do yu mean "can't" As in " if its not at an angle the photon won't make it through "

    or that somehow it is not possible to align a tube orthogonal to the motion of the emitter???
  21. Aug 24, 2010 #20
    If the tube is stationary in the lab frame, then it has to be aligned at an angle θlab = tan-1(1/βγ) relative to the direction of motion.. If the tube is stationary in the rest frame of the emitter, it has to be aligned orthogonal to the direction of motion.

    Bob S
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