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Easy special relativity

  1. Jan 11, 2006 #1
    Easy special relativity :(

    I have this problem with special relativity, its really confusing for me ;) I have this problem that seems very easy but I am of course not getting the right answer. A planet X is 12 light years away as measured from earth.
    If a rocket leaves from earth and flys to planet X in 7 years by its onboard clock how fast was it going. Seems easy but I am not doing something right. I did the following:
    t'=[tex]gama*t_{earth}[/tex] , where t' = 7 years and [tex]t_{earth}[/tex]=12 light years/v. It seems to be that this would be right but I guess not :confused:
     
    Last edited: Jan 11, 2006
  2. jcsd
  3. Jan 11, 2006 #2

    Doc Al

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    It looks like you are trying to apply the time dilation formula, but incorrectly. Instead ask: According to the rocket: What distance does he travel? (Think length contraction.) Then apply the usual Speed = Distance / Time.

    [StatusX beat me too it; we are saying essentially the same thing.]
     
    Last edited: Jan 11, 2006
  4. Jan 11, 2006 #3

    StatusX

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    Well I deleted what I had before because I thought I was giving away the answer. But to give another hint, keep in mind that the earth (and so also the planet) will have the same velocity as measured by the ship as the ship has as measured by earth. So the ship measures the planet to be 7v light years away at the start.
     
    Last edited: Jan 11, 2006
  5. Jan 11, 2006 #4
    Thanks for the help, I see the above part is right with length contraction. the part I dont get I guess is why it doesn't work with time dilation also. If we have:

    t' = 7 years(the time measured on the craft)
    t = 12 light years/v = the time it takes the ship to get to planet X from earths time frame
    v= velocity of the craft

    then use t'=gama* t and solve that equation for v? I know im missing something stupid/basic but as I am just starting out can't find it. I would guess I just don't understand the equations themselves. Thanks for the help.

    Edit: I had t=4 instead of 12
     
    Last edited: Jan 11, 2006
  6. Jan 11, 2006 #5

    George Jones

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    An alternative method for solving this problem, and one which I much prefer over Lorentz contraction and time dilation, is invariance of the interval.

    The key concept in special relativity is invariance of the interval.

    The first thing to do is to indentify the key events. In this case, key event 1 is the coincidence of the rocket and the Earth, and key event 2 is the coincidence of the rocket and planet X.

    There are 2 frames of reference - the frame of reference of the Earth and planet X, and the frame of reference of the rocket. Arbitrarily label one of the frames as unprimed and one as primed - it doesn't matter which is which. Take, as you do, the Earth-X frame as the unprimed frame.

    In the Earth-X frame, the spatial distance between the 2 key events is [itex]\Delta x = 12[/itex], and the elapsed time between the events is [itex]\Delta t = \Delta x/v[/itex] since [itex]\Delta x = v \Delta t[/itex].

    In the rocket's frame, the spatial distance between the 2 key events is [itex]\Delta x' = 0[/itex] since the rocket is coincident with both events, and the rocket doesn't move in its own frame. The elapsed time between the events is [itex]\Delta t' = 7[/itex].

    Invariance of the spacetime interval, a fundamental property of spacetime gives

    [tex]\left( \Delta x \right)^2 - \left( \Delta t \right)^2 = \left( \Delta x' \right)^2 - \left( \Delta t' \right)^2.[/tex]

    From the above,

    [tex]\left( \Delta x \right)^2 - \frac{\left( \Delta x \right)^2}{v^2} = - \left( \Delta t' \right)^2.[/tex]

    Solve for [itex]v[/itex], and plug in the numbers.

    In the above I have used c = 1 lightyear / 1 year = 1.

    I feel that time dilation should be introduced via the light clock (after clock sychronization) before the invariance ot the interval, but at that point, time dilation should not emphasized as a method of problem solving. Then invariance of the interval should be introduced. After students are comfortable using invariance of the interval to solve problems, then time dilation and Lorentz contraction, and the circumstances under which they apply, should be explained carefully. Only then (and maybe not even then) should they be used as problem solving techniques.

    This is the way topics are covered in Thomas Moore's wonderful little book A taveler's Guide to Spacetime: an Introduction to the Special Theory of Relativity.

    The fact that, in one of the frames, the spatial distance between the key events is zero shows that time dilation is very appropriate. In fact, [itex]\Delta t = \gamma \Delta t'[/itex], which is the other way around from what you wrote. Use this together with the spatial displacements and elapsed times that I listed above, and the problem can be solved without invaiance of the interval.

    Maybe this is more detail than you wanted.

    Regards,
    George
     
    Last edited: Jan 11, 2006
  7. Jan 11, 2006 #6

    Doc Al

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    Actually it works perfectly well with time dilation, if you apply it correctly. (I should have been clearer before; your problem was that you had the time dilation formula backwards.)

    The time dilation formula applies to a single moving clock as observed by a "stationary" frame. In this case, the moving clock is the rocket's clock. If the rocket clock measures [itex]\Delta t'[/itex] and the earth observers measure [itex]\Delta t[/itex], then:

    [tex]\Delta t = \gamma \Delta t'[/tex]

    Note that moving clocks run slowly.
     
  8. Jan 11, 2006 #7

    StatusX

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    Actually, this problem is a good example of how time dilation and length contraction are really two ways of looking at the same phenomenon. First, use length contraction to solve the problem. In the earth frame, the distance the ship travels is x=12. In the ship frame, X covers a distance x'=x/[itex]\gamma[/itex]=12/[itex]\gamma[/itex]. It does this at a speed v, and so in a time t'=12/[itex]\gamma[/itex]v. From this you can solve for v by setting t'=7.

    Now time dilation. In the earth frame, the ship covers a distance 12 at a speed v, and so in a time t=12/v. The time dilation formula is t'=t/[itex]\gamma[/itex] (note the similarity). But this can also be found using the t' we found before. Length contraction implies time dilation and vice versa.

    Remember that since t' is the time of the moving observer, it ticks slower so that less time elapses, explaining why t' is smaller than t ([itex]\gamma[/itex] is always greater than or equal to 1).
     
    Last edited: Jan 11, 2006
  9. Jan 11, 2006 #8

    robphy

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    I think that this is true for THIS problem, and maybe a few simple problems. However, "time dilation and length contraction are really two ways of looking at the same phenomenon" is certainly not true in general. Geometrically speaking, length-contraction deals with the spatial separation between a pair of parallel worldlines, and time-dilation deals with a pair of intersecting worldlines, projecting one unit tangent vector onto the other line.
     
  10. Jan 11, 2006 #9

    StatusX

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    By "the same phenomenon," all I was referring to was the invariance of the spacetime interval. But, yes, in this problem the analogy is more direct than in general because of the constant velocity, relating space and time intervals by a constant factor.
     
    Last edited: Jan 11, 2006
  11. Jan 11, 2006 #10
    Thanks for the help, it's starting to come together now.
     
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