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I Question about Time Dilation

  1. May 25, 2016 #1
    Hello. There is a question that I've been trying to understand for about a year. Why is it that the time dilation effect applies equally to all clocks as it does with "light clocks". For example, if John is moving on a space ship with respect to Alex who is on earth, and both are sitting next to a light clock where a beam of light is bouncing up and down, Alex will see John's beam of light moving upwards and horizontally in a diagonal line. Because the speed of light is constant for both, Alex will see John's beam of light taking longer to reach the top than the beam on his own clock, because he sees his own beam is going straight up and down, and he sees John's going up and sideways.

    Time dilation with light clocks is affected by the constancy of the speed of light. Say we chose another type of clock, like a ball bouncing up and down instead of a beam of light. The speed of the ball is not constant for both of them. Therefore, if they both use clocks with a ball bouncing up and down, Alex will see John's ball going up a longer path just like in the light clock example. However, Alex will also observe the ball going up faster to compensate for the extra length. Therefore, although time dilation might occur here to, the amount of dilation should be different. Why then, does time dilate for all clocks on John's ship, in Alex's point of view, by the same factor of gamma? Shouldn't the factor be different depending on what type of clock is being used, and what the Lorenztian sum of the velocity is between the speed of John's ship and the speed of the clock?

    If what I am saying is true, then the effects would be very strange: i.e. someone could go on a voyage at a very high speed with various clocks (including their biological clock) and come back with some of their clocks far behind those on earth, and other clocks only slightly out of sync.

    Thank you.
     
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  3. May 25, 2016 #2

    Orodruin

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    Indeed, but this is precisely the point. The relation between the clocks would be observer dependent. If one observer sees clocks with the same motion tick at the same rate, all observers must.
     
  4. May 25, 2016 #3

    Doc Al

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    If time dilation affected different kinds of clocks differently, then we'd be able to tell our "absolute" speed by comparing two different clocks. Which would violate one of the basic premises of special relativity.

    Realize that light clocks are chosen for analysis because they are easy to analyze! If you performed the correct analysis of a bouncing ball clock (or any kind of clock), you'd find the same time dilation factor at work. But unlike the light clock, that bouncing ball clock would be difficult to analyze.
     
  5. May 25, 2016 #4
    Thank you Orodruin. You said "if one observer sees clocks with the same motion tick at the same rate, all observers must". However, it doesn't seem like Alex would observe John's clocks ticking at the same rate? If both of them started out on earth stationary with respect to one another, and each had their own "light clock" and "ball clock" right next to them, both clocks would be ticking at the same rate. But once John starts moving, doesn't his light clock, according to Alex, start ticking slower than Alex's light clock? And, doesn't John's ball clock, according to Alex, start ticking slower, but only to a lessor degree, than Alex's. If John's light clock started ticking 50% slower than Alex's, for example, couldn't it be true that John's ball clock only started ticking 25% slower than Alex's?
    You also said "the relation between the clocks would be observer dependent". Did you mean the relationship between John's clocks, or the relationship between John and Alex's clocks? Also, did you mean the relationship of how fast the clocks tick, or the relationship of the distances that the path of the clock takes (e.g. the path of the beam of light going up diagonally vs. straight up).
    Just to clarify, is the answer to the question about whether the time on different clocks on a space ship, relative to those on earth, dilate differently depending on what type of clock is used, yes?
     
  6. May 25, 2016 #5

    Orodruin

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    This is only true if you use classical addition of velocities instead of relativistic. The entire point is that you get into this problem only if you presuppose that classical addition of velocities is true. The resolution is of course to introduce relativistic velocity addition. There really is nothing more to it.
    No.
     
  7. May 25, 2016 #6
    Thank you Doc Al. "If you performed the correct analysis of a bouncing ball clock (or any kind of clock), you'd find the same time dilation factor at work."
    It seems logically impossible, though. I must be making some mistake with the math or something? Take the hypotenuse of a right triangle which is the path that Alex sees John's beam of light moving up. That light moves at the same speed for Alex and Bob, but travels a further distance than Alex because he is stationary. In the ball example, the speed of the ball is not constant. Alex can add the velocities of John's ship and of John's ball (using the Lorentz addition). Therefore, even though the hypotenuse will be the same length, Alex will see the ball traveling up faster than John will see it. Am I making some fundamental error in the way I'm thinking about it?

    Also, Orodruin, even if you use Lorenz addition, you are still adding velocities in the ball example, which you can't do at all in the light example which would seem that time should dilate less. I agree that with regular addition velocity, time wouldn't dilate at all.
     
  8. May 25, 2016 #7

    Doc Al

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    Why don't you actually do the calculation? It's not that difficult. Use the Lorentz transformation for velocity and figure out the time it takes for the ball to make a vertical bounce as seen by another observer.

    (The reason that elementary treatments use light clocks --and not bouncing balls-- is that they're simple enough that you don't need the LT to figure them out, just the basic assumption that the speed of light is constant.)
     
  9. May 25, 2016 #8

    Orodruin

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    This is just wrong. If you put the velocity of the object to c in one frame, relativistic velocity addition will just give you c in any frame. It is just a special case of relativistic velocity addition.
     
  10. May 25, 2016 #9
    That makes sense- I wanted to do that. Physics has never been my major, I've just been really interested in the nature of time for a while.

    I will use concrete numbers: Say John has traveled a distance of 4 light years from Earth while Alex has remained. Say both have a light clock that is 3 light years high. Using the Pythagorean theorem, Alex will see John's beam of light traveling a distance of 5 light years. He would also measure the beam of light to take five light years to reach the top when John measured it to take only 3. John's light clock is working 60% as fast as Alex's.

    Now same example except John is traveling at half the speed of light. Instead of a light clock, there is a ball traveling upwards a distance of three light years, and the speed of the ball is half of the speed of light from John's perspective. Say John's ship is also traveling at half the speed of light. Using the Lorentz addition velocity, Alex would calculate the ball to be traveling at .8 the speed of light. He would also see the ball traveling a distance of 5 light years. At .8 the speed of light, this would take 4 light years from Alex's point of view. In this example, John's ball clock is working 75% as fast as Alex's.

    Did I miss a step or miscalculate along the way?
     
  11. May 25, 2016 #10

    Doc Al

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    You are using the wrong velocity addition formula. The velocities are not parallel. (Treat the ball as moving in the y-direction with respect to the ship, and the ship as moving in the x-direction with respect to the other observer.)
     
  12. May 25, 2016 #11

    Doc Al

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  13. May 25, 2016 #12
    Cool thanks! I will check it out.
     
  14. May 25, 2016 #13
    The formula for addition of non parallel velocities, as its stated on the website, is a little tricky for me.

    The formula for addition of parallel velocities is: S = (V + U)/ UV/Csquared

    What is the formula for addition of non parallel velocities using the symbols S, V, U, and C?

    Thank you,

    Joe
     
  15. May 25, 2016 #14
    No, not 0.8 c because the two speeds of 0.5 c are not parallel. Take the component of the ball's speed that's parallel to John's motion and use the relativistic formula for velocity addition. The math is probably more than you want to do, but I assure you that when you do it, it works out that the clock is ticking at the same rate as the light clock.

    The basic issue is this. Light clocks tick at the same rate as all other clocks, so whatever light clocks do, that's what other clocks do, too. The only room left for objection is that clocks don't really measure time. That's a philosophical objection, not a physical objection. As long as you accept that what clocks measure is time, then you must conclude that time dilation is real. Moreover, tests of time dilation have shown that it's real. Moving objects really do live longer than they would if they weren't moving. Or to say that more precisely, when we measure the lifetime of objects at rest relative to us, and then measure that lifetime when those same objects are moving relative to us, we find that the lifetime is greater when the objects move.
     
  16. May 26, 2016 #15

    Doc Al

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    Well, you could do it that way, but it's easier just to stick to the components of velocity. So the only transformation you really need is for the y-component of the velocity:
    $$u_y = \frac{u'_y}{(1 + vu'_x/c^2)\gamma}$$
    Where ##u'_x## and ##u'_y## are the components of the ball's velocity in the ship frame and ##u_y## is the y component of the ball's velocity in the other frame.

    Since, in the ship frame, the ball moves vertically we can say that ##u'_x = 0##, which simplifies the above transformation to:
    $$u_y = \frac{u'_y}{\gamma}$$

    So, given that, imagine that the ball bounces through a vertical distance L. Figure out the time it takes according to both observers and compare.
     
  17. May 26, 2016 #16
    Thank guys! After playing with the math, I am starting to understand it more! I will need to go through more examples, but I feel like I'm making progress.

    Mister T you wrote: "The only room left for objection is that clocks don't really measure time. That's a philosophical objection, not a physical objection." That's a very interesting point! One of the interesting things I've learned from SR is that time from the point of view of physics is, I think, just repetitive motion- e.g. the rotation of the earth around the sun, the ticking of a clock, etc, and that time can be measure by anything that has repetitive motion.
     
  18. Jun 19, 2016 #17
    I realized I'm still confused. Just when I think I'm starting to understand special relativity, I realize I'm far off.

    Doc Al, where does this formula which you've posted come from?: The one where you divided by a number with C squared in the denominator? Are we looking for the Y component of John's velocity from Alex's point of view? Once we get that do we need to plug it in to the velocity addition formula?

    Mister T, you said "Take the component of the ball's speed that's parallel to John's motion and use the relativistic formula for velocity addition." Isn't the ball traveling at speed zero in the X direction for John?

    To make matters worse, I'm not even sure if my example makes sense. I used a 3,4,5 triangle for convenience, and I based the length of the sides based on the height of the light clock and on the distance that John traveled with respect to Alex. It seems like if the triangle is going to illustrate time dilation, the sides should be based on the speed that John is traveling with respect to Alex and on the ball's speed. The 3,4,5 triangle was arbitrary, and the lengths would come out differently based on the speeds of John and the ball.
     
  19. Jun 20, 2016 #18

    Ibix

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    There's a fairly straightforward semi-mindless way of working out what happens in any special relativity problem.

    1. Write down the problem specification in one frame.
    2. Work out the coordinates of interesting events.
    3. Lorentz transform the coordinates into the other frame
    4. Think about what that is telling you.

    So, step 1. I am at rest in a frame called S, which uses coordinates (x,y,z,t). I have a light clock which is two light seconds long pointed along the y axis, and a ball clock which is one light second long, whose ball does 0.5c in my frame, and which is also aligned along the y axis. One end of both clocks is at the origin. Both the light pulse and the ball bounce off this end at t=0. You fly past at speed v=0.8c in the +x direction. What do you see?

    Step 2. Interesting events are when the ball bounces off either end of its clock and when the light pulse bounces off either end of its clock. For the light pulse this happens at ##(x,y,z,t)## coordinates ##e_0=(0,0,0,0)##, ##e_1=(0,2,0,2)## and ##e_2=(0,0,0,4)##. For the ball clock it happens at ##e_3=(0,0,0,0)##, ##e_4=(0,1,0,2)## and ##e_5=(0,0,0,4)##.

    Step 3. Use the Lorentz transforms to work out the ##(x',y',z',t')## coordinates that you would assign to the events.

    Step 4. Now you know the positions and times of the bounces in your frame it's easy to work out the speeds in your frame.

    I'll leave those last two steps to you. There's no substitute for doing it yourself.

    You will find, if you break the velocity into x' and y' components, that the x velocity transforms according to the equation Doc Al gave (##u_x'=(u_x-v)/(1-u_xv/c^2)##). The y velocity transforms according to ##u_y'=u_y/\gamma##. In this particular case ##u_x=0##, which makes the first one easy. You can derive these formulae directly from the Lorentz transforms if you want.

    Does that help?
     
  20. Jun 20, 2016 #19

    Doc Al

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    It's a relativistic velocity transformation: One of the Lorentz transformations for velocity. You can think of it as a version of the 'velocity addition formula' when the velocities are not parallel.

    Yes.

    No. The formula quoted is the velocity addition formula. (What you are calling the 'velocity addition formula' is only valid for the special case where the velocities are parallel.)
     
  21. Jun 20, 2016 #20

    Nugatory

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    We can, but you have to know the angles to do that, and they're frame-dependent so it's not obvious what they will be. Ibix's approach is to calculate the lengths of the sides of the right triangle; from this you could calculate the angles if you needed them, but it's easier to go directly to the x and y components of the velocity. The x component of the velocity is the distance travelled in the x direction divided by the time, and similarly for the y component.
     
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