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Moving light bulb sphere of photons

  1. Apr 1, 2009 #1
    If you have a moving light bulb and an observer at rest wrt the light bulb. The observer is tiny and he is inside the light bulb. The light bulb is travelling along the x axis. The light is then turned on what will the observer see..a sphere of photons travelling outwards from the observer?

    Keep in mind if you are in a space ship travelling in the direction of the x axis and there is a a laser trained on a spot along the Y axis . Regardless of the velocity of the frame of reference the laser will always remain trained on the spot.

    so if the frame goes faster the photons from the laser will "bend" more so they remain on the spot

    Therefore photons travelling perpendicular to the direction of travell MUST bend more than non perpendicular photons. The amount a photons bends is propriotional to the angle the photons makes wrt the x axis. Photons travelling along the x axis will not bend at all.

    Therefore the photons cannot form a sphere around the observer as they are not all bending by the same amount. The shape formed wiil be a skewed sphere or a shapes that is NOT an sphere... is there something I am missing here?
  2. jcsd
  3. Apr 1, 2009 #2


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    Yes, you are missing that photons propagate at the same velocity to all observers (assuming that the inside of the light bulb is a vacuum, i.e. there is no argon, neon or nitrogen inside).
  4. Apr 1, 2009 #3
    If you are on the light bulb, you see an expanding sphere of photons.
    On the other hand, if you are any other observer, you see an expanding sphere of photons.

    You are missing that the speed of light is c in all reference frames.

    In a frame in which the light bulb is moving, you will also observe doppler shifting and relativistic beaming, but the wavefront is always a sphere.
  5. Apr 1, 2009 #4
    It's not only frequence to be different because of the moving frame, but amplitude too, so I can't understand how the wavefront can stay spherical.
  6. Apr 1, 2009 #5
    The wavefront remains spherical because the speed of light is c. Frequency and amplitude aren't speed.
  7. Apr 1, 2009 #6
    Yes, it's correct, sorry.
  8. Apr 1, 2009 #7


    Staff: Mentor

    The equation of the sphere of light in the stationary frame is:
    c²t² = x² + y² + z²

    Transforming to the moving frame is:
    (ct'γ-vx'γ/c)² = γ²(x'-vt')² + y'² + z'²

    Which simplifies to:
    c²t'² = x'² + y'² + z'²
  9. Apr 1, 2009 #8


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    My animations at
    http://physics.syr.edu/courses/modules/LIGHTCONE/LightClock/#circularlightclocks [Broken]
    might help you visualize the situation.

    The intersection of
    the light rays emitted by an event on an inertial observer's worldline
    and the worldtube traced out by a circular array of mirrors around that observer
    is, according to that observer,
    a set of simultaneous events that are equidistant in space from the observer
    (i.e. a circular wavefront).

    The analogous situation is constructed for
    an inertial observer moving with respect to this first observer.
    Last edited by a moderator: May 4, 2017
  10. Apr 1, 2009 #9
    Could he see a sphere of photons travelling outwards?
  11. Apr 2, 2009 #10


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    ... which is rather obvious, because the transformation
    ct = ct'γ-vx'γ/c
    x = γ(x'-vt')
    is derived precisely from the requirement that the speed of light be constant to any observer (i.e. from c²t² = x² + y² + z² holding in any Lorentz-equivalent frame).
  12. Apr 2, 2009 #11
    Thanks, DaleSpam, robphy, CompuChip.
  13. Apr 2, 2009 #12
    Assume we have the stroy above with the globe moving along the x axis yadda yadda yadda

    so you have the observer at the centre of photons moving out in all directions. Imagine the photons as lines of equal length raditiang from a central point. Imagine for now just in 2D. A perfect analogy is a wagon wheel with lots of spokes. But some of the spokes are straight( the photons moving along the x axis) and some of the spokes are bent with the amount of bending being propotional to the angle the spoke makes with the x axis. Spokes at 90 degrees to the x axis bend the most.

    With some of the spokes straight and others bent, and they are all the same lenght they will NOT form a circle. Rotate this malformed wagon wheel thru 360 degrees around the x axs and it is not a sphere
  14. Apr 2, 2009 #13
    the lines will be equal length as all photons are moving at C. The wagon wheel analogy is what the photons will look like at a snap shot in time say T0
  15. Apr 2, 2009 #14
    When I say that the spokes are bent I mean they will not be radiating out from the central point, perpendicular to a tangent of that central point.

    if you draw a circle around the central point with the central point as the centre of the circle, the radiating lines will not be perpendicular to the tangent of that circle.

    Only the photons moving along the x axis will be perpendiular to a tangent to the aformetioned circle.
  16. Apr 2, 2009 #15
    the only way the spokes of a wagon wheel, bike wheel etc, with equal length spokes, can form a circle is if all the spokes are perpendicular to the tangent to a second circle, where the tangent is taken at the point where the spoke intersects the second circle. With the centre of the second circle being the centre of the spokes.

    Move any spoke the tiniest amount away from perpendicular and the spokes will no longer form a perfect circle
  17. Apr 2, 2009 #16


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    If all the spokes radiate out from a single center point, they will automatically be tangent to a circle with that point at its center, no? There's no way you could have a bunch of equal-length spokes that all radiate from a single point that don't all end in points on a circle. After all, every radius of a circle is just a straight line segment going from the center to a point on the circle, it's impossible to draw a straight line going from the center that doesn't lie along one of the radial lines.
  18. Apr 2, 2009 #17


    Staff: Mentor

    Yes it is a sphere; I derived it above.

    Since your conclusion is demonstrably wrong, why don't you go back and see if you can spot the mistake.
    Last edited: Apr 3, 2009
  19. Apr 4, 2009 #18
    sorry I apologise you are all right

    sorry to labour the point but another thought occured to me

    what will change in a moving frame of reference is the photon density

    take my wagon wheel analogy from before
    place the axis of the wagon wheel on the origin of a cartesian coordinate system in the xy plane

    The wagon wheel is now broken into 4 quadrants by the coor system +x+y,-x+y,-y-x and -y+x

    if the spokes of the wagon wheel are evenly spaced in a stationary frame ie 1 spoke every 2 degrees so 180 spokes for the entire circle and each spoke is evenly spaced. So the densitty of the spokes in any given quadrant will be 45 spokes per quadrant. If the spokes were photons same deal

    the same wagon wheel in a moving frame the spokes will not be evenly spaced, thier density in each quadrant will not be the same.

    In a moving frame the photons (spokes) rotate. The photons must rotate to keep the laser on the spot from my riginal post. The photons in the +y half of the circle rotate clockwise, the photons in the -y half of the circle rotate counter clockwise

    so for any quadrant spoke(photon) density in stationary frame will be 45
    so for any quadrant spoke(photon) density in moving frame will not be 45

    I draw this conslusion from the fact that the photons moving perpendicuar to the direction of movement of the frame rotate more than photons at a lesser angle. The amount of rotation will be proprtional to the angle the photon makes with th x axis.

    I suspect the rotation is a function of sin(x) as maximum rotation will occur at 90 degrees. Photons moving along the x axis will not rotate at all

    or am I wrong?
  20. Apr 5, 2009 #19
    Why would some of the spokes be bent? The path of a photon is always a straight line as seen from an observer at rest with the source or in uniform relative motion.
  21. Apr 5, 2009 #20
    for the observer within the space ship to observe the laser on the spot at any velocity the photons from the laser must rotate in the direction of travel. If they didnt then the observer within the space ship would see the laser move off the spot and he would know he was moving
  22. Apr 5, 2009 #21


    Staff: Mentor

    Can you derive the equation that describes this rotation?
  23. Apr 7, 2009 #22
    no, but i cant derive the equation that describes the sun ether but its there

    the photons must rotate otherwise given the laser spot scenario in my inital post it would be trivial for an observer in a MFR at rest wrt that MFR to prove that they were in a MFR. I know it has somthing to do with conservation of momentum and maxwells symmetry
  24. Apr 7, 2009 #23
    as I said before it will be a function of the velocity of the MFR and sin(N) where N is the angle the photon makes with the x axis
  25. Apr 7, 2009 #24
    Hello Dreads.

    Think of it this way. The centre of the expanding sphere of photons is at rest in EVERY frame. So for ANY observers present at the point of emission, no matter what their relative velocities, the expanding sphere of photons behaves as if any and all of the observers are, and remain, central to it. It is non intuitive and makes no "common sense" but it is due to the speed of light being the same in all inertial frames and is at the very heart of relativity.

  26. Apr 7, 2009 #25


    Staff: Mentor

    Hi Dreads,

    The reason I asked is because I don't know what you even mean by the rotation of a photon. Do you picture a photon as some extended rigid body object that spins about some axis, or is it rotating about some external axis, or what?

    I was hoping you could describe that mathematically (you don't need to derive it from first principles) so that I could understand what you meant by a photon rotating.
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