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What Does Shankar Mean? Intro to Relativity

  1. Mar 23, 2017 #1
    Hope this is a valid question.

    I can't quite understand this very simple thing.

    Prof Shankar's introduction to relativity on Youtube starting at about 50 minutes has a quick calculation that comes up with the Lorentz transform.

    I follow the maths, no problem.

    But he confuses me a bit when he's setting the stage.

    He says two trains, mine and yours, cross at the origin. Yours is moving. As they cross a beam of light is sent forward to the detector, distance x from the origin for me. distance x prime for you.

    And he says there is a distance u for you. So that x prime = x - u.

    That's what confuses me. Where is the distance u from? Is that the distance you will travel during the time it takes for the light beam to reach the detector?

    I am thinking it must be but I'm so much a novice at all this I'd love confirmation or correction if I'm wrong.
     
  2. jcsd
  3. Mar 23, 2017 #2

    Doc Al

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    Staff: Mentor

    "u" is the speed of the primed train with respect to the unprimed (stationary) train. So that equation should be x' = x - ut. (You left out the time.)
     
  4. Mar 23, 2017 #3
    yes, thank you. and that time is the time taken for the light pulse to reach the detector?
     
  5. Mar 23, 2017 #4

    Doc Al

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    Staff: Mentor

    That's right. According to the stationary train the light pulse hits the detector at (x, t). But the other train views that same event as happening at (x', t').

    (Both trains agree that the pulse was emitted at point (0, 0).)
     
  6. Mar 23, 2017 #5
    Thank you. I thought so but was a little uncertain.

    I find Shankar a bit confusing sometimes. In that talk there he seems to be all over the place - talking of firecrackers and bullets and this and that - where the meat of the matter is a mere 6 minutes right at the end. And that's a Yale education!

    Lots of talk of F=MA. And inertial frames of reference. I was just thinking, that'd mean there's no Force in an inertial frame of reference, wouldn't it? Seeing as there can be no acceleration.

    different question. sorry.

    thanks for the help.

    :)
     
  7. Mar 23, 2017 #6

    Doc Al

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    Staff: Mentor

    No. Just because the frame isn't accelerating doesn't mean that objects cannot accelerate with respect to that frame. In fact, F = ma is generally first taught using only inertial reference frames.
     
  8. Mar 23, 2017 #7
    Ah, yes. Thank you.
     
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