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Wrong Time Dilation

  1. Jul 13, 2011 #1
    Hello !

    Can you tell me what's wrong? I suppose two observers. They measure speed of light. One observer is in rocket of length L. Observer in rocket measured speed
    [tex]C'=\frac{L}{t'}[/tex]
    and t' is time from observer's view. Second observer is outside. He measured
    [tex]C=\frac{L+vt}{t}[/tex]
    where v is speed of rocket. As C' = C, we have
    [tex]\frac{L+vt}{t}=\frac{L}{t'}\quad\Rightarrow\quad t'=\frac{t}{1+\frac{vt}{L}}=\quad\Rightarrow\quad t'=\frac{t}{1+\frac{v}{c}}[/tex]
    But this is not
    [tex]t' = \frac{t}{\sqrt{1-v^2/c^2}}[/tex]
    known from theory. Thanks.
     
  2. jcsd
  3. Jul 13, 2011 #2

    PAllen

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    The first problem I see is that you forgot about length contaction.
     
  4. Jul 13, 2011 #3
    So all bad?
     
  5. Jul 13, 2011 #4

    ghwellsjr

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    This is very confusing. Are the two observers each measuring the speed of light independently of each other or is the outside observer watching the rocket observer measure the speed of light and not actually making his own measurement?

    In order to measure the speed of light, you have to have a timing device located at the source of the light and a reflector a measured distance away. You then time how long it takes for the light to leave the source, bounce off the reflector and come back to the timing device. You calculate the average speed as twice the distance divided by the time interval.

    Each observer needs his own timing device and his own reflector, although if they happen to be colocated when the light flash occurs, they can use a common light source.

    I don't see anything like that in your equations.
     
  6. Jul 13, 2011 #5

    Dale

    Staff: Mentor

    Yes, L is not the same as L'. The easier way to derive this is from the metric (proper time):

    [tex]d\tau^2 = dt^2 - \frac{1}{c^2} \left( dx^2 + dy^2 + dz^2 \right) [/tex]
    [tex]\frac{d\tau^2}{dt^2} = 1 - \frac{1}{c^2} \left( \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2} + \frac{dz^2}{dt^2} \right) [/tex]
    [tex]\frac{d\tau^2}{dt^2} = 1 - \frac{v^2}{c^2} [/tex]
    [tex]\frac{d\tau}{dt} = \sqrt{ 1 - \frac{v^2}{c^2} } [/tex]
     
  7. Jul 13, 2011 #6
    They are using there own timing devices. The time measured by the observer on the rocket using his own clock is t' and the time measured by the outside observer using his own clock is t The reflector is not required. Each observer uses two clocks that are synchronised in their own rest frame. In the rest frame of the rocket the light simply moves the length of the rocket (L) in time (t'). In the rest frame of the outside observer, the light moves the length contracted length of the rocket L*sqrt(1-v^2/c^2) plus the distance the front of the rocket moves in the time it takes for the light to get from the back of the rocket to the front (v*t).
     
    Last edited: Jul 13, 2011
  8. Jul 13, 2011 #7
    O.K. I have figured out how to do it.

    Observer in rocket measured speed
    [tex]C'=\frac{L}{t'}[/tex]
    Second observer is outside. He measured
    [tex]C=\frac{L\sqrt{1-v^2/c^2}+vt}{t}[/tex]
    Now the total distance (x) travelled by the light particle according to the outside observer is

    [tex]x = L\sqrt{1-v^2/c^2} + vt \quad\Rightarrow\quad L = \frac{(x-vt)}{\sqrt{1-v^2/c^2}}[/tex]
    so we can write:

    [tex]C=\frac{x}{t} [/tex]
    and since C' = C

    [tex]\frac{x}{t} = \frac{L}{t'} \quad\Rightarrow\quad t' = \frac{t}{x} L = \frac{t}{x} \frac{(x-vt)}{\sqrt{1-v^2/c^2}} = \frac{t(1-vt/x)}{\sqrt{1-v^2/c^2}} [/tex]

    Now x/t = c and t = x/c (in this case as we are measuring the speed of a light particle) so:

    [tex]t' = \frac{t(1-vt/x)}{\sqrt{1-v^2/c^2}} = \frac{t(1-v/c)}{\sqrt{1-v^2/c^2}} = \frac{t-vt/c}{\sqrt{1-v^2/c^2}} = \frac{t-vx/c^2}{\sqrt{1-v^2/c^2}} [/tex]
    Which is the result we are looking for as this is the Lorentz transformation for time. See http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/ltrans.html#c2

    Simples :tongue:

    P.S. In this case the initial event is at the origin and the final event is at x in the outside frame so the more complicated transform is required. If both events are at the same location (eg a clock that remains at the origin of the outside frame) then x=0 and the equation simplifies to the more familiar time dilation equation:

    [tex] t' = \frac{t}{\sqrt{1-v^2/c^2}} [/tex]
     
    Last edited: Jul 13, 2011
  9. Jul 14, 2011 #8

    ghwellsjr

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    If they don't each use a single timing device and their own reflector, then they are not measuring the speed of light. Instead, they are relying on previous definitions of what a Reference Frame is in Special Relativity and what the speed of light is going between two points.

    I'm sure you are providing the answer that stanley.st is looking for but I'm just pointing out that it doesn't have anything to do with measuring the speed of light as he stated in his original post.
     
  10. Jul 14, 2011 #9
    Thanks for help so much to all. But any of these replies didn't satisfied me, but now I

    solved my problem.

    To yuiop: The problem is, what the hell is L\sqrt{1-v^2/c^2} :-) Of course, I

    know what is it, but we didn't derive it here.

    To PAllen: This was that obstacle :-) I really forgot a length contraction. Let me rewrite
    [tex]C=\frac{\varphi(L)}{\psi(t)}[/tex]
    be a general measurement of rocket observer (using some transformation). Then
    [tex]\frac{\varphi(L)}{\psi(t)}=v+\frac{L}{t}[/tex]
    from my previous notes. So that
    [tex]\psi(t)=\frac{\varphi(L)}{v+\frac{L}{t}}=\frac{t}{\frac{vt+L}{\varphi(L)}}[/tex]
    The same equation must hold for reverse situation. From Rocket observer's view the other

    observer is moving opposite velocity. So that
    [tex]t=\frac{\psi(t)}{\frac{L-vt}{\varphi(L)}}[/tex]
    And combining last two equations we have
    [tex]t\frac{L-vt}{\varphi(L)}=\frac{t}{\frac{vt+L}{\varphi(L)}} \Rightarrow L^2-v^2t^2=\varphi^2(L)[/tex]
    In this way, I have found a function phi
    [tex]\varphi (L)=L\sqrt{1-\frac{v^2t^2}{L^2}}=L\sqrt{1-\frac{v^2}{C^2}}[/tex]
    Next steps will be the same as yuiop's.
     
  11. Jul 14, 2011 #10
    Hi Stanley. I am not totally sure all your equations add up. In your first post you stated:

    In the last step of the above quoted text you have substituted c for L/t but if you look at the steps above you have defined c as L/t' or (L+vt)/t and not as L/t so that is an incorrect substitution. (I am ignoring the fact that length contraction has been left out as you are already aware of that.)

    In your second derivation you attempt to derive the length contraction factor but I am not sure all your steps are sound:

    Who is the "rocket observer"? Before you had an observer in the rocket and an observer outside the rocket. Is the rocket observer inside the rocket or observing the rocket from the outside? I will assume you mean the outside observer because you used C and not C' and I am assuming you are being consistent with your first post. Now you have defined the speed of light as the distance light travels in the outside frame as a function of the proper length L ([itex]\varphi(L)[/itex]) in a given time as measured by the outside observer as a function of t ([itex]\psi(t)[/itex]). Why would you have a function of t or did you mean as a function of the proper time t' as measured by the observer in the rocket?

    Here you have effectively defined:

    [tex]C = v+\frac{L}{t}[/tex]

    because you previously defined:

    [tex]C = \frac{\varphi(L)}{\psi(t)} [/tex]

    This is a mistake because:

    [tex]C /ne v+\frac{L}{t}[/tex]

     
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