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Closure Speeds

  1. Jul 11, 2010 #1
    Using drudkh's example cited above and using Einstein's velocity addition formula, closure speed between a particle moving to the right at v and light coming from the right and moving left at c would still be c.

    Why? Let's use v as the velocity of the particle expressed as a fraction of c and therefore c = 1

    closure velocity = (c + v)/(1 + vc/c2) = (1 + v)/1 + v) = 1

    If, instead of a particle moving to right at v, we substitute a photon or a light wave moving at c or 1 and still light comes to the left at c or 1, the formula becomes:

    closure velocity = (1 + 1)/(1 + 1) = 1

    Ain't life grand? The max speed attainable is 1 or c.

    Can we ever have a closure velocity > c? Not according to the above.

    Someone show me and drudkh wrong!
     
    Last edited: Jul 11, 2010
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  3. Jul 11, 2010 #2

    Fredrik

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    The velocity addition formula tells you the velocity of that light in the particle's rest frame, but you asked about the closing speed. It's defined as the rate of decrease of the coordinate distance between the two objects, i.e. as minus the time derivative of the coordinate distance. Your two objects are located at a+vt and b-ct for some a and b with b>a, so the coordinate distance between them at time t is (b-ct)-(a+vt)=b-a-ct-vt, and the closing speed is therefore

    -d/dt(b-a-ct-vt)=c+v.
     
    Last edited: Jul 11, 2010
  4. Jul 11, 2010 #3
    Closing velocity or mutual velocity describes the rate of change in a single frame of a vector connecting two objects. It is just the normal Newtonian addition of velocities and can be as high as 2c. Of course it is only the distance between the objects that is changing at the closing velocity, nothing material is moving at this velocity.

    The relative velocity is the velocity which one of the objects ascribes to the other and is calculated using the relativitydition formula for velocities and cannot exceed c.

    Matheinste.
     
  5. Jul 11, 2010 #4
    .....0..................................(a + vt).....>.......................<....(b - c-t)

    Help me out here... If the origin of one FOR was at A (a + vt and moving right at a steady speed) and the origin for a second frame of reference was at B (b - c-t) where "c- means "just below the speed of light but greater than v," then doesn't the closure speed mean that closing speed is c- + v which is > c.

    How is that possible? How do you "leap frog" > c? Relative to the FOR at A any particle moving left at c- would be booking > c and I thought that was impossible in any FOR. Even looking at it from B being steady and A coming from the left, a particle would be moving to the right at c- + v which would be > c (both cases if v were large enough but still , c.)

    This is a major new concept for me to understand. I understand it from Galilean or Newtonian point of view - that's easy.
     
  6. Jul 11, 2010 #5
    Matheinste or Fredrik - if you just dropped the velocities of both objects to c- or speeds just below light speed, say 29,900,000 m/sec, which allows for the movement of "real" particles what is the difference here? How, in the real world which SR does apply, would one ever achieve a closure velocity or relative velocity of > c. I am conflating closure velocity with relative velocity and therein may lie my misconception, but then when do we use closure velocity versus relative velocity?

    stevmg
     
  7. Jul 11, 2010 #6

    Fredrik

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    Yes. (But you don't need to mention several frames of reference. You can just talk about the positions of two "objects", and you don't have to let that speed be slightly less than c. If one of the "objects" is light, then the speed is exactly c).

    Why wouldn't it be?

    No, because if you want to calculate the speed in frame A of that particle, you have to use the velocity addition formula to get (v+c)/(1+vc/c2). If you want to calculate the closing speed, you just add their speeds (assuming that they're moving towards each other) to get v+c.

    Then you understand it in Minkowski spacetime too. There is no difference.
     
  8. Jul 11, 2010 #7

    Fredrik

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    Just have Alice shoot a beam of particles with speed 0.6c towards Bob, and at the same time have Bob shoot a beam of partices with speed 0.6c towards Alice. The closing speed is 1.2c. The velocity of one of the beams in the rest frame of the other is (0.6c+0.6c)/(1+0.6*0.6)=1.2c/1.36<c.

    I agree.

    I don't know any situation where closing speed is an interesting concept.
     
  9. Jul 11, 2010 #8
    No, because if you want to calculate the speed in frame A of that particle, you have to use the velocity addition formula to get (v+c)/(1+vc/c2). If you want to calculate the closing speed, you just add their speeds (assuming that they're moving towards each other) to get v+c.

    That is where you lose me. I understand your words but not the concept.
     
  10. Jul 11, 2010 #9
    It is used routinely in explaining MMX from a frame of reference outside the lab.
     
  11. Jul 11, 2010 #10

    Fredrik

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    I think you're just assuming that there's something tricky here that needs to be understood. There isn't. If you're driving 50 mph relative to the road and the cars going in the opposite direction are also going at 50 mph, the closing speed is 100 mph because it's defined to be the number of miles that the distance to the other car is decreasing each hour. It doesn't involve a second coordinate system in any way. The only coordinate system we're talking about is the one in which the road is stationary. So you have no use for the Lorentz transformation or the velocity addition formula, which both involve two inertial frames.
     
  12. Jul 11, 2010 #11
    Fredrik, Here is where I got "messed up." This was a beautiful explanation of the twins paradox solution (just using SR) by JesseM which he wrote to my questions months ago. I have highlighted in magenta where he used the term "closing speed" but on review now, I see he put it in quotes and he did initially use the term "relatavistic velocity addition formula."

    As illustrated in his solution, JesseM really used the "relatavistic velocity addition calculation" and NOT closing speed. Now that you have stated there is sort of no use for closure speed, things now fit in place. With reference to starthaus's "MMX"
    I have no understanding. As an old Air Force officer he may be referring to a missile closing in on a target which is clearly "outside the lab." I also see MMX as some technology with new chips.

    A little help in that area by starthaus or you would be helpful.
     
    Last edited: Jul 11, 2010
  13. Jul 11, 2010 #12
    MMX - some new Intel microprocessor technology which, I presume, increases speed of "calculations." I guess there are two stacks reaching out towards each other, or something like it and there is a "closure speed," if you will.

    If he meant the missile, then that would be an MX missile, not MMX.

    If I know starthaus, it is neither, and it will be months before you guys let me in on your secret.

    I am clueless and pulseless.
     
  14. Jul 11, 2010 #13

    Fredrik

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    If it makes you feel better, I had to look it up too. Wikipedia has a disambiguation page that says:

    • MMX (instruction set), a single-instruction, multiple-data instruction set designed by Intel
    • MMX Miner, a Brazilian mining company
    • Michelson-Morley Experiment, the 1887 experiment attempting to find evidence of the luminiferous aether
    • Mega Man X, a series of video games produced by Capcom and the eponymous main character
    • Malmö Airport, an airport in Sweden (IATA airport code)
    • MMX, an upcoming album by rapper Xzibit
    • 2010, in Roman numerals
    I'm assuming it's the third one. I don't think I've heard it called "MMX" before.
     
    Last edited: Jul 11, 2010
  15. Jul 11, 2010 #14

    jtbell

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    MMX as an acronym for "Michelson Morley experiment" is sometimes seen on Internet forums like this one that discuss relativity intensively, and maybe on a few Web sites, but nowhere else as far as I know.
     
  16. Jul 12, 2010 #15
    Thanks, gentlemen,

    That's all I need now - to become another Sherlock Holmes!

    stevmg
     
  17. Jul 12, 2010 #16
    yes, Michelson-Morley Experiment

    BTW, closing speed figures prominently in the Einstein gedank experiment with the train and the track in explaining relativity of simultaneity.
     
    Last edited: Jul 13, 2010
  18. Jul 13, 2010 #17
    Thus, "closing speed" is a Galilean concept

    "relativistic velocities" is the relativity analogue: a light beam shining directly at another light beam still go at each other at c but the closure speed, which has no bearing on reality, is 2c

    Nice shot, starthaus, with "MMX" - you had me. I even saw the list in Wikipedia which included Michelson-Morley and blew it before the other folks did get it.
     
  19. Jul 13, 2010 #18
    Not really, it is a general concept, applies equally in SR.


    Closing speed has a lot of bearing on reality. I just showed you that, without it, you can't explain Einstein gedank, nor can you explain MMX as viewed from a frame different from the lab.
     
  20. Jul 13, 2010 #19
    Getting late... wife's in bed and so will I be.

    Will go over your post in detail later.

    stevmg
     
  21. Jul 14, 2010 #20
    are closure velocities invariant?
     
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