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Does a sphere remain a sphere

  1. Jan 11, 2008 #1

    cornfall

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    You watch a sphere pass by at a fast steady speed, does it remain a sphere?
     
  2. jcsd
  3. Jan 11, 2008 #2

    Mentz114

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    According to special relativity it will appear to be deformed along one axis. You can get more details by looking up the 'Lorentz contraction' or 'Special Relativity'.
     
  4. Jan 11, 2008 #3

    DaveC426913

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    If that "steady speed" is a large fraction of c, it will be flattened in the direction parallel to its path.
     
  5. Jan 11, 2008 #4

    A.T.

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    If you watch a sphere, you will see a sphere. But there is a difference between what you see, and what you measure. The sphere will be flattened when measured, but due to different light travel times it will look round. A very good explanation is here

    Since the OP asked about watching I would put that way: It will appear to be round but in fact it is flattened in the observers frame of reference.
     
    Last edited: Jan 11, 2008
  6. Jan 11, 2008 #5

    Mentz114

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    A.T.:
    Now you are being kind. I was wrong, strictly speaking. Do you happen to know the OP ?
     
  7. Jan 12, 2008 #6
    A.T., that is very interesting, I'd never been told that if I were LOOKING at an object traveling at relativistic velocities that it would look the same as if it were at rest (well, except for blue/red-shift) and only that it would MEASURE differently. It's very interesting that this should be different. Thankyou.
     
  8. Jan 12, 2008 #7

    A.T.

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    It doesn't really look the same. The texture on the moving sphere looks distorted and rotated. More detail:
    http://www.spacetimetravel.org/bewegung/bewegung5.html
     
  9. Jan 12, 2008 #8

    Shooting Star

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    Last edited by a moderator: Jan 12, 2008
  10. Jan 12, 2008 #9
    Hello all.

    Regarding the link in the last post by A.T. showing the rotation of a cube at relatavistic speeds we see that the cube is rotated so that the rear of the cube becomes visible from the front. This mechanism is explained by a photon emitted in some direction towards the observer having "components" being considered in the direction of motion of the cube and perpendicular to it and so these components have speeds of less than c. Is this a valid construction (or deconstruction). I have never seen it before.

    Matheinste
     
  11. Jan 12, 2008 #10

    Shooting Star

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    Hi Matheinste,

    Are you having problems with the math involved, or the concept itself?
     
  12. Jan 12, 2008 #11
    Hello shooting Star.

    I am unable to access the link in your post.

    Referring to the link in the post by A.T. i have trouble with the concept of a photon having components in more than one direction.

    Thanks for your reply.

    Matheinste.
     
  13. Jan 12, 2008 #12

    Doc Al

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    I fixed it.
     
  14. Jan 12, 2008 #13

    Shooting Star

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    Open a blank browser window, then copy the whole address and paste it on the address bar, write "http://" before it and press enter.

    math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html
     
  15. Jan 12, 2008 #14

    Shooting Star

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    What is the way to fix it?
     
  16. Jan 12, 2008 #15

    JesseM

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    Have you looked at this thread? Any photon moving diagonally relative to your x and y axis (and z axis) will have a component along more than one axis, but the total speed [tex]\sqrt{v_x^2 + v_y^2 + v_z^2 }[/tex] is always equal to c.
     
  17. Jan 12, 2008 #16
    Thanks Doc Al i have given it it a quick read but it obviously deserves more attention but i am not sure if it helps with the problem i have with the photon component concept.

    Thanks Matheinste.
     
  18. Jan 12, 2008 #17
    Thanks JesseM.

    I saw yuour post after i read that of Doc Al. I must take some time to digest these links and see where my error lies.I will try to get back to you soon.

    Thankyou all Matheinste.
     
  19. Jan 12, 2008 #18

    Shooting Star

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    But what has that got anything to do with the shape of the moving sphere? That is just basic decomposition into mutually perpendicular components of any vector.

    EDIT: I wrote this before I saw the last post by Matheinste. So, just ignore it for the time being.
     
    Last edited: Jan 12, 2008
  20. Jan 12, 2008 #19
    Hello Shooting star.

    The link referred to shows a cube not a sphere but this may be irrelevant.

    JesseM i see that the velocity of the photon can be resolved into components satisfying the usul vector law. But the argument in the referred to link seems to rely on the observer of the cube seeing only the component of the photon which is travelling in his direction at less than c, but surely the photon itself, viewed as a particle, does not have resolvable components although its velocity does. Also viewed as a wave surely the wavefront has a speed of c in all directions.

    If you look at the diagram in the referred to link it shows the point of emission of the photon being, as it were, left behind. But i always thought, perhaps mistakenly, that the point of emission of the photon or, if you like, the light flash remains central to the propagating sphere of light (or the expanding sphere of photons) no matter what the motion of the emitter, and so the cube in the diagram cannot "outrun" the source of emission, which in the diagram is at the rear of the cube.

    Matheinste.
     
  21. Jan 12, 2008 #20

    A.T.

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    This is correct. Light travels with the same speed in all directions, regardless the speed of the emitter.

    The source of emission is a static point in space, where the rear of the cube was, at the time of emission. I see no problem in outrunning a non moving point.
     
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