A light speed gedanken

Discussion in 'Physics & Math' started by CANGAS, Apr 28, 2006.

  1. CANGAS Registered Senior Member

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    Foreword: there is no intention here to present a trick question or anything of the sort. This is intended to be an analysis of photon motion on a path perpendicular to the motion of its source on theoretical physics grounds.

    This gedanken thought science experiment consists of a straight and practically rigid pipe .99 meter in diameter and 9,999 meters long, coated inside with material that absorbs any photon and capped on one end with a photon detector and counter. On the other end is a Xaser ( a hard X-ray laser ) adjusted to fire single photons. During manufacture the Xaser is aimed dead center on the end cap. The pipe is tightly packed full of pure vacuum. The entire assembly is painted with special pigment that completely absorbs any external photon or any substance that could create a photon inside.

    The assembly is mounted on an Enterprise class starship with its major axis perfectly perpendicular to the fore and aft axis of the starship. It looks like the starship has a long skinny pipe for a wing.

    The science experiment modus operandi is for the starship to execute a high speed fly by, on a straight path, past the Earth at a closest approach of 99,999 kilometers while stationary obversation station area observers on the ground observe it. The starship will fly on impulse only but will activate the forward asteroid deflector shield to ward off any and all disturbance from oncoming photons, cosmic rays, or anything else. It will build up speed and make a long fly by at .9999c, according to the observers. The experiment has been rehersed repeatedly to obtain timing parameters and the team is confident that that programming of automatic controllers on the starship will give perfect coordination.

    During the fly by, when the starship is nearest the Earth, the controllers will trigger the Xaser to fire one photon. To be doubly and unnecessarily repetetively redundant, the pipe is exactly perpendicular to the starship flight path.

    When the starship lands at the stationary observer station area, the stationary observers will probably rush over to to it at an average, or drift, velocity of about 9.9 kilometers per hour and anxiously examine the photon detector and counter on the end cap.

    How many photon strikes will the counter show for the result of this experiment and why?
     
  2. Pete It's not rocket surgery Moderator

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    One, assuming everything works as planned.
    Because the photon strikes the end cap.
     
  3. Tom2 Registered Senior Member

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    Yes, if you fire one photon then one photon strikes the detector. This is true in every frame.

    Your setup doesn't reflect that. The motion of the photon is at right angles to the axis of the ship in the ship's frame. But in the ship's frame the ship isn't moving. In the Earth frame the motion of the photon is not at right angles to the motion of the ship. To examine what you say you want to examine you would have to fire the photon from Earth.
     
  4. Pete It's not rocket surgery Moderator

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    Or place the pipe at the right angle in the ship.
     
  5. Tom2 Registered Senior Member

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    Yes, that would work too. Another way to do it would be to abandon the Earth frame and try to find a frame in which the motion of the ship and the photon are at right angles. I was suggesting the fix that would be the least work.

    But no matter how you slice it, there's only one photon hit.
     
  6. Pete It's not rocket surgery Moderator

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    That could be interesting... I think I'll try that as an exercise when I have time to play.
     
  7. Tom2 Registered Senior Member

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    No need to waste your time, I've just figured out that you can't find such a frame. Start from the 3D version of the velocity addition formula.

    u=[&gamma;v+u'+((&gamma;-1)/v<sup>2</sup>)(v<sup>.</sup>u')v]/[&gamma;+(&gamma;/c<sup>2</sup>)v<sup>.</sup>u')

    Let u= velocity of the photon in the new frame (call it S).
    Let u'= velocity of the photon in the ship frame (call it S').
    Let v= velocity of S' relative to S.

    Let the ship axis lie along the y' axis and the photon tube lie along the x' axis. Then u'=ci. The problem is to find v such that v is perpendicular to u, that is, such that v<sup>.</sup>u=0.

    Take the dot product of both sides of the velocity addition formula with v, set the left side equal to zero, and plug in u'. You will get the following.

    0=v<sup>.</sup>v --> v<sup>2</sup>=0 --> v=0

    In other words the condition u<sup>.</sup>v=0 cannot be satisfied unless you are in the frame of the ship.
     
    Last edited: Apr 28, 2006
  8. Tom2 Registered Senior Member

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    And just because I'm SUCH a sweetheart :D ...

    To determine the angle use the velocity addition formula from my last post. The problem is to find an angle &theta; through which to rotate the photon tube such that the photon moves at right angles with the ship in the Earth frame.

    Let S= Earth frame and S'=ship frame.
    Let u= velocity of photon with respect to Earth = ci
    Let u'=velocity of photon with respect to the ship = c cos(&theta; )i+c sin(&theta; )j
    Let v=velocity of the ship with respect to the Earth=vj

    Plugging the above vectors into the velocity addition formula and equate coefficients. You will find the following.

    x-component
    (1): [c cos(&theta; )]/[&gamma; (1+(v/c)sin(&theta; )]=c

    y-component
    (2): [v+c sin(&theta; )]/[1+(v/c)sin(&theta; )]=0

    From (2) we have that sin(&theta; )=-v/c. The pythagorean theorem tells us then that the cos(&theta; )=(+ or -)1/&gamma;. The restriction that u=+ci tells us that we need the (+) sign.
     
    Last edited: Apr 28, 2006
  9. Pete It's not rocket surgery Moderator

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    Something feels wrong. If we can set the pipe at an angle so that the photon moves at perpendicular to the ship in Earth's frame, then it seems that we should be able to find a frame in which the motion of the ship and the photon are perpendicular...

    Because rather than moving the pipe, we could adjust the angle of the whole ship, right? But then, rather than adjusting the angle of the ship, why not adjust our frame?

    I'll work through the numbers and see what I can figure.
     
  10. kevinalm Registered Senior Member

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    The problem is that earthbound observer and the shipboard observer will disagree on the angle of emision of the photon, even though they do agree on the fact that the pipe is perpendicular to the line of travel of the starship. As viewed by the earhbound observer, the emiter is moving, and will have a forward angle emission, even though it is not angled forward.
     
  11. DaleSpam TANSTAAFL Registered Senior Member

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    For any "mounting angle" you can always find a frame where the motion of the photon is perpendicular to the axis of the ship. But the axis of the ship is not necessarily related to the motion of the ship, so it can be that the "perpendicular" frame is the rest frame of the ship where the ship has no motion.

    -Dale
     
  12. Tom2 Registered Senior Member

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    That's right. If we turn the photon tube then we can use the velocity addition formula to find a frame such that u<sup>.</sup>v=0, and that frame will not correspond to v=0.

    Exactly. There are an infinite number of frames in which the motion of the photon is perpendicular to the axis of the ship. If the velocity of the ship as measured from Earth is v=vi, then an observer in any frame whose velocity (again, as measured from Earth) is w=vi+wk (doesn't matter what w is) is going to say that the photon moves at right angles to the ship's axis. But he won't say that the photon moves at right angles to the ship's motion. That's because both the ship and the photon have a velocity component in the same direction.
     
  13. CANGAS Registered Senior Member

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    Gentleman:

    How plainly does it need to be stated that we have stationary observers on Eath who are keeping tabs on these goings-on? Did the thread starter say anything about observers on the starship? It could be a ballistic lump of metal.

    The pipe was not postulated to be on a swivel. Any reasonable person would have to assume that it was welded in place during manufacture. Is there a little bit of desperation being shown by someone who fears that there may be a connection here with something inimical to SR?

    Can a photon source which is traveling at very near c, according to a stationary observer, emit a photon at a right angle to its motion, according to a stationary observer, which then moves at its usual velocity of c in its emitted direction and also at a velocity of .99c in the direction of its source's velocity, all according to the stationary observers?

    If so, what is the photon's resultant velocity, according to a stationary observer?

    If we have a right triangle, with one side .99c and the other side 1.oc, what does the hypoteneuse calculate to be? Less than c? Or FTL?
     
    Last edited: Apr 30, 2006
  14. DaleSpam TANSTAAFL Registered Senior Member

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    What desperation are you talking about? The first response, by Pete, answered your question succinctly and completely.

    The rest of the discussion is about your mistaken setup. If "This is intended to be an analysis of photon motion on a path perpendicular to the motion of its source" then you did not accomplish your goal. We are discussing how you could have accomplished it. How is that desperation?

    -Dale
     
  15. Tom2 Registered Senior Member

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    We got that, thank you.

    This is irrelevant. There are two events in this problem: the emission of the photon, and the absorption of the photon. Those events have spacetime coordinates in the frame of the ship, whether there is someone there or not. Those coordinates have to be known so that they may be transformed to the frame that the observers do occupy.

    Furthermore, you are the one who specified the problem from the point of view of the ship. The 9999 meter length, the 0.99 meter diameter, and the perpendicularity of the pipe and ship axis are all determined from the ship's frame.

    We know that. See my first post in this thread, and see Dale's last post in this thread. Adjusting the angle of the pipe was one way to fix your broken thought experiment, which does not have the photon motion at right angles to the ship's motion, as determined from the Earth.

    Puh-leeze.

    For the second time: In the Earth frame the motion of the photon will only be perpendicular to the motion of the ship if you do not mount the pipe such that it is perpendicular to the ship in the ship's frame. You would have to adjust the mounting angle in the way I described earlier.

    In my earlier analyses I let the y-axis lie along the ship axis, and I let the photon tube lie along the x-axis. If you still insist that the photon tube be perpendicular to the ship axis in the ship's frame, then the resultant velocity of the photon in the Earth's frame will be the following, according to the velocity addition formula I quoted.

    u= (c/&gamma; )i+vj

    I'll leave it to you to verify that the magnitude of this velocity vector is in fact c.

    The speed is exactly c, in any frame. That should be perfectly obvious before any calculations are done because:

    1.) SR assumes that the speed of light is invariant, and
    2.) SR is deductively valid, and therefore internally inconsistent.

    It should come as no surprise then that in any gedanken involving velocity addition that you get out what was put in: the speed of light postulate.
     
  16. Pete It's not rocket surgery Moderator

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    That's not really relevant.

    The point is that if you can orient the ship so that the motion of the photon is perpendicular to the ship's motion, then you can also orient your reference frame so that the motion of the photon is perpendicular to the ship's motion.

    Why place that restriction on the other observer's velocity? What if w = ui + wk (u not equal to v)?

    One of us is making a mistake. Assuming your equation is true*, I get:
    v.u' + v&sup2; = 0
    v<sub>x</sub>c = -|v|&sup2;

    Giving us two frames for any given velocity, and an infinite number of frames altogether (even without considering the z direction).

    * Not that I'm doubting you (much :)), I just haven't seen it before (so I thank you), and havent bothered verifying or deriving it.
     
  17. Neddy Bate Valued Senior Member

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    This experiment seems to lack any mirrors. The usual SR examples rely upon a reflection in order to complete one full cycle of the time-clock (symmetrically). Consider two half-mirrors at each end of the tube, and a required round-trip for the beam of light, and then it would have to be postulated that the mirrors are parallel to each other in every frame.

    This discussion about maintaining a 90 degree angle between light-ray and line-of-motion is leading people to consider a tube which is tilted toward the rear of the moving spacecraft. In this case, the light-ray would be at a 90 degree angle to the line-of-motion, but only in one direction of the light beam's travel. Had there been parallel mirror, the reflected beam would be at an angle in the 'other' direction which would be very different from perpendicluar.
     
    Last edited: May 1, 2006
  18. Tom2 Registered Senior Member

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    I pointed it out because I thought maybe you were confusing the idea that the photon is moving at right angles to the ship's motion with the idea that the photon is moving at right angles to the ship's axis. If that's not the case then never mind my remark.

    Well, I did look for that frame and I showed you what I found. More at the end of this post.

    I placed that restriction on the other velocity because I didn't feel like doing any more calculations, and I could do that one in my head. I haven't worked out the case you suggested, so I don't know what will happen.

    The dot product v<sup>.</sup>u' is equal to zero. The photon in the ship's frame is perpendicular to the relative velocity vector v. That's why I only find one frame in which v<sup>.</sup>u=0.

    By the way, the equation is true. I'll refer you to the following online textbook: Modern Relativity - SR. The 3D boost found in Equations 1.1.3 (this is the inverse of Equations 11.19 in Jackson's Classical Electrodynamics, 2ed). Put the boost in differential form, then divide dr by dt. Then on the left side divide the top and bottom by dt'. Set u=dr/dt and set u'=dr'/dt'. I simplified the result by substituting in the definition &beta;=v/c.
     
    Last edited: May 1, 2006
  19. Pete It's not rocket surgery Moderator

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    Thanks, Tom!

    OK, so you only looked at frames with velocity parallel to the ship's axis.

    So, there is no frame with velocity parallel to the ship's axis in which the photon is moving at right angles to the ship's motion.

    But, I think that there are two frames with velocity in the ship-Earth (x-y) plane in which the photon is moving at right angles to the ship's motion, and infinite frames with arbitrary z-axis velocity in which the photon is moving at right angles to the ship's motion.
     
  20. Pete It's not rocket surgery Moderator

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    I don't think it needs to get any more complicated than it is, Neddy... at least not until we sort out with CANGAS what's happening. Maybe later.
     

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