Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

SR question

  1. Aug 17, 2009 #1

    In this image, we have a bar, on the lower mark we have placed a lamp and on the upper a mirror as shown. At some point the lamp goes on and the light reaches the mirror, it reflects and goes back from where it started. The events A = light emition and B = return at starting point are two events which take place in the same reference frame (Σ). For Σ the light travelled a distance equal to 2(AB) = 2d with speed C. So Δt = 2d/c

    Observer Σ along with the bar, move with velocity v in relation to Σ' to the right. Therefore Σ' will notice that the light follows the track Α'Β'Γ' with velocity c............

    Can somebody please explain to me, why the light follows the track Α'Β'Γ', I can't seem to "feel" the above image.

    P.S. If you have any links that will help me with the understanding of special relativity or have some exercises (With solutions or answers) i would be very thankful.
  2. jcsd
  3. Aug 17, 2009 #2


    User Avatar
    Science Advisor

    If the emitter emits light straight up in our frame when it's at rest in our frame, then it must continue to emit light straight up in its new rest frame if it's given some constant velocity relative to our frame--if this wasn't true it would violate the first postulate of relativity, which says all the laws of physics should work the same in each inertial frame, so two observers in closed rooms moving at constant velocity relative to one another should get the same result if they each do the same experiment in their own room.

    And of course, if the emitter continues to shoot light straight up in its rest frame when it's moving in our frame, that implies that in our frame the light will have to move diagonally (so it always remains directly above the emitter even as the emitter moves).
  4. Aug 26, 2009 #3
    Thanks for the answer it helped me understand some concepts of special relativity.
    I have another question if you can help me :

    It's from serway's book.
    I am having a hard time understand the bolded text, I have a feeling it says that because the space traveler is inside the spacecraft but I cannot justify it exactly. If someone can clarify this one for me :)
  5. Aug 26, 2009 #4

    Doc Al

    User Avatar

    Staff: Mentor

    Realize that from the spacecraft's frame, it's the stars that are moving while the spacecraft remains at rest. So the passages of the stars occur at the same location in the spacecraft's frame. (If spacecraft is located at x = 0 in the spacecraft frame, then both passages occur at location x = 0.)
  6. Aug 27, 2009 #5
    Hey thanks a lot for that, it helped me.

    This is a problem from serway again.
    We calculate the times goslo and speedo needs to land.

    The problem now is their age difference. I have some thoughts about that but my head is currently close to blowing!
    Is the age difference the time according to the Earth observer (or planet-x for that matter) that goslo needs to reach the planet minus the one that speedo needs minus the time that speedo is at rest on planet-X and goslo is travelling?

    Thanks in advance

    EDIT : Ah I finally got it thanks
    Last edited: Aug 27, 2009
  7. Aug 27, 2009 #6

    Doc Al

    User Avatar

    Staff: Mentor

    I'll bet you figured it out already, but they want the difference in the twins ages--which is the time that they record in their bodies and their wristwatches. Not the time elapsed according to an Earth observer.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook