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Homework Help: Relativistic effects

  1. Mar 7, 2014 #1
    1. The problem statement, all variables and given/known data

    The example below illustrates the relativistic phenomenon that synchronicity of events is not absolute but it depends on the reference frames.
    Spaceships A and B, while moving away from each other with a constant speed of v = 0.553c, are watching a competition between spaceships C and D. Spaceship C is heading towards planet C and spaceship D is approaching planet D. The winner is the spaceship that reaches its target planet first.
    The astronauts on spaceship B find, to their great surprise, that the spaceships C and D reached their planets at the same time. At that moment, planet C was at rC = (-250, 130, -130) ls, and planet D was at rD = (160, -290, -170) ls , where the xyz coordinate system is attached to spaceship B and the first, x, axis is parallel to the velocity vector of spaceship B relative to spaceship A.
    According to spaceship A, however, the race has a definite winner. According to spaceship A, how many seconds were between reaching planet C by spaceship C and reaching planet D by spaceship D?

    2. Relevant equations

    Δτ = γ(Δτ' + βΔx')

    Where ΔT is the difference in time, given in light seconds, β = v/c. The ' notation represents the reference frame.

    γ = 1/(√(1-(v2/c2))

    3. The attempt at a solution

    Let the view from spaceship A be reference frame s.
    Let the view from spacesip B bence frame s'.

    xC' = |rC| = √((2502) + (1302) + (1302))

    xD' = |rD| = √((1602) + (2902) + (1702))

    Δτ' = 0

    Δτ = γ(Δτ' + βΔx')
    Δτ = (1/1-(0.533))(βxD' - βxC')
    Δτ = β(1/1-(0.533))(xD' - xC')
    Δτ = 76679.6992

    Which I'm told is incorrect. Could anyone help me out?
  2. jcsd
  3. Mar 8, 2014 #2

    Simon Bridge

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    This is a simultaniety problem.
    Two events simultaneous in the B frame are not simultaneous in the A frame.
    Work through the reasoning.

    Are C and D racing to the same planet?
    Does the direction that A and B are travelling count?
  4. Mar 8, 2014 #3
    The differences between the y and z coordinates of the two events are the same in the A and B frames of reference. Only the differences in the x coordinates and the time coordinates are not the same. Your Lorentz Transformation equations should involve only the x and t coordinates in the two frames. Actually, all you need is the time transformation equation. Also, don't forget to divide by the speed of light to get the time difference in seconds.

  5. Mar 9, 2014 #4
    If I only need the X and T coordinates, does velocity of the two spaceships have no effect?

    C & D are racing to separate planets.
    I'm not sure whether the direction in which A & B are travelling counts. I would assume so but I can't tell what direction that is.

    I'm not sure what you mean by "work through the reasoning".
  6. Mar 9, 2014 #5
    No. The velocities of the two spaceships has no effect. The two events (i.e., C reaches planet C and D reaches planet D) are reckoned using coordinates from A's frame of reference. That's all that counts.
    What does this mean to you: "and the x axis is parallel to the velocity vector of spaceship B relative to spaceship A."

  7. Mar 9, 2014 #6

    Simon Bridge

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    For simultaniety problems, only the relative velocity of the observers counts.
    If C and D arrive at the same time - it does not matter what speed they are doing: they still arrive at the same time.
    If you know an observer where the two events are simultaneous, then you know the order of events for other observers.

    Thank you to Chestermiller here - my answer is the same.
    It is easy to miss important information. Sometimes it helps to rewrite the problem statement so each sentence starts on a new line.

    I was concerned that you answer relied too heavily on memorized formulae.

    One of the things to reason about, for eg. was which of the lengths was important.
    Your work on length contraction should give you a hint: not all lengths are affected equally.

    Have a look through some examples in your notes where simultaneous events are involved and see how the author decides what to do. There are also lots of examples online.
  8. Mar 10, 2014 #7
    The module seems overly focused on deriving formulae and then using them to solve problems. I have very little idea as to what is actually happening in a particular problem. For example, as similar problem was run through in our lecture and it was suggested to us that we simply select the correct formula from the list of formulae we had derived over the past lectures and use the one that allowed us to solve the problem.

    I'm not sure how length contraction plays into this. The problem was giving to us before we had covered any work on length contraction and, as with all other problems giving to us, it is set with the assumption that we have previously covered the required material in lectures.

    Also is there any material where this is covered thoroughly online? Or do you perhaps know of a website where I can find the information? I can't find similar examples.

    So far I have that the distance between the two events in the S' reference frame (the view from B) is Δr = √((-250-160)2 + (130--290)2 + (-130--170)2)) = 588.3026432. This gives me a time of 671.2260574 seconds. (I'm told that the numerical value of ΔT is the same as the numerical value of Δt. I'm not sure whether that's true or not though).
    Last edited: Mar 10, 2014
  9. Mar 10, 2014 #8
    You don't need to have covered length contraction yet. The only thing you need to solve this problem is the Lorentz Transformation. Have you covered that yet?

    Also, as I pointed out earlier, the Δy's and Δz's are irrelevant to the solution. So Δr = √((-250-160)2 + (130--290)2 + (-130--170)2)) = 588.3026432 is inappropriate to use. Only the Δx counts.

  10. Mar 10, 2014 #9

    Simon Bridge

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    ... sadly common.
    Never mind - now we are clear on where you are at... I'll let Chestermiller take the lead to avoid getting you too confused.

    My favorite crash course:
    ... comes in 4 parts, you only need the first 3.
    ... there are others around, it sounds like you will need something from outside your course.
  11. Mar 11, 2014 #10
    Thank you for your replies. I've only part way through part one and I'm already happier.
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