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Question about time in Lorentz Tranformation

  1. Jul 27, 2011 #1
    I recently started self-learning special relativity. I am having lots of trouble convincing myself about the Lorentz transformations. I realize that I must be doing something really stupid, but I just can't pinpoint it.

    I get that the equation for x' follows from the fact that "rulers" used in the reference frame of x' is shorter than the rulers used in x by a factor of sqrt(1-(v/c)^2), so that the x' coordinate must be equal to the x-coordinate shifted by how far the two frames have moved relatively and then divided by sqrt(1-(v/c)^2).

    But for the time equation, I tried using a similar reasoning. The person in the x' frame measures time with a clock which, according to the person in the x frame, is slower. That means that if in x, an event happened t seconds after some time coordinate, the event happened after a shorter interval of time the x' frame. Doesn't that mean that t' should be divided by the time dilation factor?

    So basically, my understanding is that clocks are sort of like time-rulers. But length is contracted, while time is dilated. How can it be that the equations for t' and x' have the save factor in the denominator?

    Thanks in advance for any help!
     
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  3. Jul 27, 2011 #2

    BruceW

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    Let's say the time between two ticks of a clock is t as measured by jack who is at rest relative to the clock. And x is the difference in position of where those two ticks happened. Obviously, x is equal to zero, since jack is at rest relative to the clock.
    Now, let's say some other observer (james) moving relative to jack also measures the time between the two ticks (of that same clock), which we denote t'. We can use the Lorentz transformation to see what james will measure:
    [tex] ct' = \gamma (ct - \beta x) [/tex]
    and since x is zero, we have:
    [tex] t' = \gamma t [/tex]
    Clearly, anyone not in the same reference frame as jack will measure a greater time difference, since [itex] \gamma [/itex] is always greater than 1. (Which is why its called time dilation). So a clock always ticks most frequently when measured by someone which has no relative movement to it.

    EDIT: to make it clear, [itex] \gamma [/itex] is equal to:
    [tex] \frac{1}{1 - \frac{v^2}{c^2} } [/tex]
    Another edit:
    [tex] \beta = \frac{v}{c} [/tex]
     
    Last edited: Jul 27, 2011
  4. Jul 27, 2011 #3

    BruceW

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    The derivation of length contraction is a little bit more difficult, but here I go.
    Lets say jack is at rest relative to a ruler. And james is moving relative to jack. We must define the length of the ruler according to james to be the difference of the positions of the endpoints of the ruler when measured at the same time (simultaneously) by himself. For jack, the length can be the difference of positions of the endpoints measured at any two times, since the ruler isn't moving relatively to him, these positions don't change with time.
    So we say the difference of positions of the endpoints of the ruler (i.e. the length) as measured by jack is x and the length measured by james is x' and the time between the two position measurements according to james is t' (which is equal to zero, as I've said above). The backward Lorentz transform is:
    [tex] x = \gamma (x'+ \beta ct') [/tex]
    and t' is equal to zero, so we have:
    [tex] x' = \frac{x}{ \gamma } [/tex]
    Which means that the ruler is longest when measured by someone who is not moving relatively to it. (which is why its called length contraction).
     
  5. Jul 27, 2011 #4

    Mentz114

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    This confusion is because clock rate ( frequency) is the reciprocal of time with dimension T-1. So you have to invert the time equation to get the inverse [itex]\gamma[/itex].
     
  6. Jul 29, 2011 #5
    Thank you both for responding! Unfortunately, I still don't really get it... From BruceW's post, it seems that I don't even understand the definition of t'. Is the time transformation equation according to the person in the t reference frame, or the person in the t' reference frame? It seems that the answer would be different, since each person sees the other's clock as slower than his own.

    Also, Mentz114, I don't really understand what you mean by inverting the time equation for the inverse lorentz factor (where does this inverse lorentz factor appear?). Could you please clarify?

    Thanks again! I really appreciate you experts taking the time to help.
     
  7. Jul 29, 2011 #6

    BruceW

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    The transform equation isn't according to only one or the other. It is simply relating measurements made in the t frame to measurements of the same events made in the t' frame.

    Yes, each person will see the other person's clock ticking slower than their own. It seems nonsense, but in fact there are no paradoxes.

    Also: probably the best way to learn relativity is by going to a course of lectures on it, while doing examples along the way. Or if not that, then by getting a book on relativity (which gives examples to practice) or watching a lecture course online. How have you been learning so far?
     
  8. Jul 29, 2011 #7
    Ok, I will try to learn from the basics. So far I have been reading Feynman's lectures, but he does not go very in depth (since he only has a few chapters devoted to relativity).
     
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