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

Does c change in this case?

  1. Jan 21, 2008 #1
    Assume I am travelling in a spaceship at 0.9c towards a certain star, away from Earth. Then, some guy on Earth shines a really bright LASER light at the same star, causing a beam right beside my ship, so that the light and my ship travel in parallel paths. Because I am travelling at 0.9c, I should see that LASER to be going at 0.1c right? But, when I am moving, the time interval should be smaller (to get from one place to another). Say that The LASER took 1 second to travel 3x10^8 meters. To me, would it be going faster, since less than 1 second has passed for me while it travelled the same distance?

    v = d/t

    Since t is smaller because I'm moving at 0.9c, v becomes bigger. Would c be bigger in my moving frame of reference?
    Also, does time slow down only when you accelerate or when you travel with uniform motion also?
  2. jcsd
  3. Jan 21, 2008 #2


    User Avatar
    Science Advisor

    No, velocities don't add that way in relativity. You have to use the formula for relativistic velocity addition, which says that if A is traveling at speed v relative to B, and B is traveling at speed u relative to C, then A's speed relative to C is (v + u)/(1 + u*v/c^2). In this case, if A is the light beam and B is the Earth, so v=1c, then if C is your ship, so u=-0.9c (negative because in your frame the Earth is moving in the opposite direction as the light beam is moving in the Earth's frame), then you will measure the velocity of the light beam as (1c - 0.9c)/(1 - 1*0.9) = 0.1c/0.1 = 1c. It is in fact one of the fundamental postulates of relativity that each observer should measure light to move at the same speed in his own rest frame.
    To understand why each observer measures a light beam to move at c, you must take into account the fact that each observer sees a moving observer's rulers to be shrunk, their clocks to be slowed down, and their clocks to be out-of-sync with one another. I gave an example showing how all these factors together ensure they both measure the light's speed to be c in post #6 from this thread.
    Keep in mind that there's no "objective" truth about whose clocks slow down--each observer measures the other one's clocks to run slower than their own. However, from the perspective of a given observer, a moving clock will always be slowed down regardless of whether it's accelerating or moving inertially; at any given instant, if its instantaneous velocity is v, its rate of ticking at that instant is [tex]\sqrt{1 - v^2/c^2}[/tex].
    Last edited: Jan 21, 2008
  4. Jan 21, 2008 #3
    ...and always remember that when dealing with relativity and "modern physics", you sort of want to treat it like politics..... take 90% of rational thinking and throw it out the window....

    another suggestion...read the Mr. Tompkins book....
  5. Jan 21, 2008 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Yes, read Mr. Tompkins in Wonderland, its a fun read.

    HOWEVER, it is not rational thinking you need to throw out it is preconceived notions as to how you think it OUGHT to be. Rather then closing your eyes and dreaming, study the math and let the MATH guide you to the correct and rational conclusions.
  6. Jan 21, 2008 #5
    yes you are right...

    I must say, when I first began studying relativity, it was very odd.... like you said, preconceptions on how things "are" from environmental influence vs. how they are after "proving them with math's" can be a totally different thing...

    spacetime still gets me sometimes....
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Does c change in this case?
  1. How does time stop at C? (Replies: 87)