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Time Dilation and Length Contraction

  1. Feb 6, 2014 #1
    I had a quick question about Time Dilation and Length Contraction.

    Are the two just different ways of measuring/describing the same effect? Or rather they both follow as a consequence from one another?

    i.e. I can find how much a length is contracted by finding the dilated time interval and multiplying by velocity.

    Or I can how much a time interval is dilated by finding how much the length is contracted and then dividing by the velocity?
     
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  3. Feb 6, 2014 #2

    Mentz114

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    They both depend on ##\gamma##, so the most practical thing to do is measure the relative velocity.
     
  4. Feb 6, 2014 #3
    I know how to find them, but that's not really what I'm asking. I'm asking, COULD each be found directly from the other, given relative velocity?
    Or more generally:
     
  5. Feb 6, 2014 #4

    WannabeNewton

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    They are both a consequence of relative simultaneity but they describe two different phenomena.
     
  6. Feb 6, 2014 #5

    Mentz114

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    They are different things and both come from the postulates of special relativity.

    How would you find the length contraction ?
     
  7. Feb 6, 2014 #6
    Find the proper length and multiply 1/##\gamma##.

    But I mean that's just me applying the formula from the book. I'm trying to understand it further than that.

    The book says the two(time dilation and length contraction) are closely related, but that's it, as far as I can tell, that's all it does is just state that they are related and then they don't go into anymore detail about. So this is my effort at trying to figure how they are related.
     
  8. Feb 6, 2014 #7
    It's possible if you know one of the effects beforehand. For instance (v is the relative velocity, γ the Lorentz factor):

    Deriving length contraction from time dilation:
    Transport one clock along a rod of proper length L_0 during time T (measured with clocks at rest relative to the rod). The clock's proper time is thus shorter by T_0=T/γ due to time dilation. In the rest frame of the clock, the length of the rod is thus simply given by T_0·v=T/γ·v. Thus the moving rod is shorter by γ.

    Deriving time dilation from length contraction:
    Let there be a rod of proper length L_0. A clock moves along the rod in time L_0/v=T. However, in the clock's rest frame, the rod is contracted by L=L_0/γ due to length contraction, therefore the time with which the rod passes by is T_0=L/v=L_0/γ/v. This time interval is indeed shorter by γ than in the rod's rest frame.
     
    Last edited: Feb 6, 2014
  9. Feb 6, 2014 #8

    stevendaryl

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    Length contraction, time dilation and relativity of simultaneity are all three necessary to insure that the laws of physics look the same in any inertial coordinate system.

    A way to see that length contraction and time dilation must go hand-in-hand is by considering the round-trip time for a light signal from one end of a traveling spaceship to the other and back.

    If the rocket were stationary, and its length were [itex]L[/itex], then the round-trip time would be simply [itex]T = \frac{2L}{c}[/itex] where [itex]c[/itex] is the speed of light.

    Now, consider the same rocket ship moving forward at speed [itex]v[/itex]. If the rocket did not undergo length contraction or time dilation, then the round-trip time would be:

    [itex]T = \frac{1}{1-\frac{v^2}{c^2}} \frac{2L}{c}[/itex]

    Why is that: because if you send a light signal from the rear of the rocket toward the front, then the front of the rocket is moving away from the light signal, making it take longer to reach the front. Specifically, the time to get to the front would be: [itex]T_1 = \frac{1}{1-\frac{v}{c}} \frac{L}{c}[/itex]. For the return trip, from the front of the rocket to the rear, the time would be shorter, since the rear would be moving toward the light signal. The time would be: [itex]T_2 = \frac{1}{1+\frac{v}{c}} \frac{L}{c}[/itex]

    The total time for the round-trip would be [itex]T = T_1 + T_2 = \frac{1}{1-\frac{v^2}{c^2}} \frac{2L}{c}[/itex]

    In terms of [itex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/itex], the round-trip time would be:

    [itex]T = \gamma^2 \frac{2L}{c}[/itex]

    This is without length contraction or time dilation. If we assume that (as measured in the original stationary frame) moving clocks slow by a factor of [itex]\gamma[/itex], then in terms of the moving clocks, the time would be less by a factor of [itex]\gamma[/itex], so clocks aboard the rocket would measure [itex]T' = \frac{T}{\gamma}[/itex] for the round-trip time for light. So they would measure:

    [itex]T' = \gamma \frac{2L}{c}[/itex]

    If we further assume that moving rockets are length-contracted by the same factor of [itex]\gamma[/itex], then that would cut down the measured round-trip time further. Instead of

    [itex]T' = \gamma \frac{2L}{c}[/itex]

    the measured time would be

    [itex]T' = \gamma \frac{2L'}{c}[/itex]

    where [itex]L' = \frac{L}{\gamma}[/itex]. Therefore, the measured time for a round trip would be

    [itex]T' = \frac{2L}{c}[/itex]

    the same as for a rocket that is stationary.
     
  10. Feb 6, 2014 #9
  11. Feb 6, 2014 #10
    Certainly in an empirical setting they are not closely related they are exactly related, and different ways of expressing the same thing. For example a light year is a distance and a time period.

    Look at it this way, distance has no meaning without time, absent time any distance you think of as relative to each other, i.e. the length of a pencil or the distance to another galaxy, would instead be infinite. In other words without time there is no ability to change position, or to observe any distance they all become infinite. The flip side, without distance time has no meaning, is also true in an empirical setting, but less intuitive and usually more difficult for people to see. Simple algebra may help demonstrate this as t = x/s, or s = x/t (where t is time, x is distance, s is speed), the numerator for both is distance, time or speed are different ways of expressing the same thing (a relationship to distance), and of course you could write x=t * s.

    I know it’s a rather simplistic explanation, but it does show that time and distance are not closely related, rather they are exactly related and necessary to define each other. If you examine the function of clocks you'd see they all use an incremental change of position to define a duration, what are the properties of that, distance, and a magnitude of change in position of that distance (commonly described as a speed), and a similar increment like a light year is a speed, duration, and distance all in one.
     
    Last edited: Feb 6, 2014
  12. Feb 6, 2014 #11
    Consider the postulate that c is invariant regardless of comparative motion. Consider that c is a measure of velocity which requires both a length & time measurement. That's closely related.
     
  13. Feb 6, 2014 #12
    That makes sense, and that's where my head is at right now, but I'm still unclear on how exactly that relationship works in relativity.

    And I'm still a bit confused. I was reluctant to do this, but I'm going to post the problem that led to this confusion in the homework section.

    https://www.physicsforums.com/showthread.php?t=736914
     
    Last edited: Feb 6, 2014
  14. Feb 6, 2014 #13

    ghwellsjr

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    No.

    No. They both follow from the Lorentz Transformation process which is the easiest way to understand them.

    I don't understand how you can make these two things work. Here is a spacetime diagram showing the two ends of a rod of 15 feet. I offset them like that just so that the next diagram wouldn't be so large that it wouldn't fit on the page. The dots mark off 1 nanosecond increments of Proper Time for both ends of the rod. I'm defining the speed of light to be 1 foot per nanosecond:

    attachment.php?attachmentid=66360&stc=1&d=1391721726.png

    Now I transform to a speed of 0.6c (so that the rod is moving to the left):

    attachment.php?attachmentid=66361&stc=1&d=1391721726.png

    This shows the length contraction of the rod from 15 feet to 12 feet and if I divide that by 0.6 feet per nanosecond, I get 20 nanoseconds which is simply the time it takes for the end of the rod to traverse 12 feet at 0.6c but I don't see anything relating to time dilation as indicated by the dots on either end of the rod.

    And I have no idea what you have in mind for the other calculation where you multiply the dilated time interval (where is that on the diagram?) by the velocity to get the contracted length.

    Please explain exactly how your two formulas work.
     

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  15. Feb 6, 2014 #14
    It was a question of IF that would work, not a declaration that it would work.

    And after playing with the equations I did see that it would not work.
     
  16. Feb 9, 2014 #15
    When matter moves, light interactions involve longer distances. Since light speed is constant and independent of its source,

    a. em field strength changes and matter contracts in the direction of motion when accelerated by a factor of 1/γ,[1]

    b. the light component within the light clock completes a cycle at a rate of c/γ.

    There are two different phenomena resulting from the same two factors, motion and constant light speed.
    Their complementary nature, was as explained by stevendaryl, results in a scaled copy of a universal 'rest' frame, thereby inheriting the same (x, t) relations, i.e. the rules in a moving frame are the same as in an absolute frame (if you could find one).

    ref:
    1. research: Oliver Heaviside, H. A. Lorentz, George Searle
     
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