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Conceptual problem with time dilation and simultaneity

  1. May 18, 2012 #1

    I am having a problem resolving the common examples that are used to explain time dilation as a consequence of the constancy of the speed of light. I can't help but assume that there is something simple and fundamental that I'm missing here. I apologize in advance for forcing you to use your imaginations, but I am as yet unable to post images.

    In the time dilation example, a pulse of light is issued from an inertial reference frame (perhaps the floor of a boxcar) and travels straight upward to a mirror which reflects it back to the source. The path traveled by the light, according to the observer in that frame, is simply twice the distance between the source and the mirror. This is compared to a longer distance seen by an outside observer due to the relative motion of the frame.

    In the simultaneity example (usually also using train cars) light pulses are issued from equal distances on either side of an observer. The light from the front of the car reaches the observer before the light from the rear of the car, resulting in the lack of observed simultaneity in contrast to an outside observer.

    Here is my trouble...

    In the first example, the only way the light could travel straight up and down with respect to the boxcar is if its horizontal velocity is equal to that of the car. Otherwise, the light would travel toward the rear of the car as it moved upward. So I see this as having the following components :

    Vz = c
    Vx = c + V (isn't this impossible?)

    where V = boxcar, Vx = light horiz., Vz = light vert.

    But in the second example, the only way the light could reach the observer at two different times is if the two pulses were NOT carrying the car's velocity. This means:

    Vfx = -c
    Vrx = c

    where Vfx = forward pulse horiz., Vrx = rear pulse horiz.

    I hope someone could take a few moments to set me straight on this.

    Thank you
  2. jcsd
  3. May 18, 2012 #2

    Here's where you're going wrong. That's where Special Relativity comes from. So the light has to travel at c from the reference frame in which the boxcar is moving, so it takes longer to get back (more distance, same speed), so the clock appears to run slower, so observers relative to you appear to be moving more slowly.

    In the second example, there is a very important distinction. Light must move at c, but it must have the same component of velocity perpendicular to the source's motion as the source. Hope this helps.
  4. May 18, 2012 #3


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    The speed of light is c. That means that with "horizontal velocity equal to that of car" its horizontal component of velocity is V (NOT c+ V), the speed of the car. And then we must have [itex]V_z^2+ V^2= c^2[/itex] so that [itex]V_z= \sqrt{c^2- V^2}[/itex].

    You seem to be thinking that "the speed of light is c" means that each component of the velocity is c. That is not true. If light velocity has components [itex]\left<V_x, V_y, V_z\right>[/itex], then we must have [itex]\sqrt{V_x^2+ V_y^2+ V_z^2}= c[/itex]
  5. May 18, 2012 #4
    A "pulse" of light is often represented as a spherical surface increasing radius at c.

    Both the observers will see the radius increase at c wrt the point of origen in his own frame... which point of origen is for him a non-moving point in his frame, but a point that would appear to be moving wrt the other observer's frame.

    Each sees the point of origen as at rest, and infers that the other observer will see that point of origen in motion in their own frame.

    Since the light being seen is coming from the origen, there is at least a kind of agreement on that. With radial plused light, there may be some disagreement on whether both observers are actually measuring the "same" light - is the light directly emmitted from rest at a particular angle the same light whose horizontal and vertical motion components comprise a resultant at that same angle?

    When the moving observer measures the vertically emmitted light boucing off the ceiling mirror is that the same light that the stationaly observer sees having emmitted at an angle?

    Replace radial light with lasers and place a light proof hollow tube between the mirrors for the light laser to pass through... Now what happens to the observation of the stationary observer?
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