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

Telling if a reference frame is moving or at rest

  1. Dec 26, 2009 #1
    As stated in SR and the principle of equivalence, there is no privileged reference frame or reference frame at absolute rest. However, I went across the following idea that seems to be able to tell if a reference frame is moving or at rest. Please help me point out what is wrong in my idea:

    Below is a typical diagram in SR for time dilation:


    The top left picture shows a light clock at rest. Note that flash of light bulb emits light towards all directions. Let us say seven rays to be convenient - besides the vertical ray, there are three on the left and three on the right.

    The top right picture shows a light clock moving towards the right at about 1/4 c, so the 5th ray from the left will hit the mirror on the top and bounce back to hit the same position as the bulb.

    Now let us switch to observe it in the reference frame of the moving light clock. Since the 5th ray from the left will bounce up and down. Therefore, there are four rays on the left and only two on the right.

    So, my idea is: By measuring the difference of light ray density (or number of photons in unit area) on the left and right side, I can tell if the reference frame is at rest or moving, and the velocity of moving.

    Please tell me where I am wrong. Thanks!
  2. jcsd
  3. Dec 26, 2009 #2
    Your "idea" sounds suspiciously similar to using the well known doppler effect to determine the relative motion between a light source and observer. Of course the difference in the frequency of the light received will show that the source is in motion relative to the observer, and the observer is in motion relative to the source. In one reference frame, the source is in motion while the observer is at rest. In another reference frame the observer is in motion while the source is at rest.
  4. Dec 26, 2009 #3


    User Avatar
    Science Advisor

    It will if the light clock is moving towards the right at about 1/4 c relative to the mirror. If both light clock and mirror are moving at the same velocity, the situation is exactly the same as in the top left picture. You have still measured relative speed.

    Last edited by a moderator: Dec 26, 2009
  5. Dec 26, 2009 #4
    This sounds similar to the experiments that were carried out to deduce the velocity of the supposed aether. If you choose to carry out your experiment, you will, provided you do it in vacuum, find that you cannot distinguish between the top left and bottom right situation. What you are doing in the bottom right situation is measuring the light's velocity relative to a supposed medium through which light travels. If you were to do a similar experiment using sound waves in a medium, you would have gotten a difference between top left and bottom right, but this is not the case with light in vacuum.
  6. Dec 26, 2009 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I think the error is that in your top right diagram, you assume that the light has equal intensity in all directions. I don't think this is correct. The light will be more intense in the direction of motion. This is similar to the optical aberration effect illustrated in this video: The only difference is that the bulb is emitting the light, whereas the simulated camera in the animated video is absorbing it. In general, observers in different frames of reference do not agree on angles. The angle shown as a 90-degree angle in the top left diagram will not appear as a 90-degree angle in the top right diagram.
    Last edited by a moderator: Sep 25, 2014
  7. Dec 26, 2009 #6

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    First off, why don't you just measure the frequency and be done with it? Scientists do this all the time. It is called redshift (or occasionally, blueshift).

    You are not testing whether the reference frame (what reference frame?) is moving. You are testing whether the light is moving with respect to you.
  8. Dec 27, 2009 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Because he's come across this particular SR paradox, and he wants to understand what's wrong with it.

    He knows that. It's very clear from his original post that he knows there must be a hole in the reasoning, and he just wants to understand where the hole is.

    Let's not bite the newbies. Lots of people seem to be telling this guy that his question is stupid, but not taking the trouble to read his question carefully. If I start a thread saying, "Here's my proof that 2+2=5, can you find the mistake?," it's not really helpful to get responses saying that it's 4, or that I should do it with a calculator, or that I could get the right answer by counting on my fingers.
  9. Dec 27, 2009 #8
    Thanks for the replies. I think bcrowell's statement below especially answers my question:

    So my top right diagram is wrong. Since the light clock (light bulb + mirrors) are moving towards right together as a whole, the observer in the light clock reference frame will see uniformed light intensity distribution, while the observer at rest will see the light is more intense on the right side.

    Based on this theory, if a light bulb is moving away from an observer who is at rest, at high speed (say, v = 9/10 c), so in order to have equal intensity in all directions in the light bulb reference frame, for the observer at rest, the light bulb must emit more light to the direction away from him, leaving very little light to the direction towards him. This results in the observer seeing a very dim light bulb. Is this true?
  10. Dec 27, 2009 #9


    User Avatar
    Gold Member

    yinfyudan, your lower right diagram is incorrect. With the light source and observer both in the same frame of reference, the light will look exactly like your upper left diagram.

    The observer sees ray 4 propogating directly perpendicular to his motion, and sees just as many rays going forward as going backward.
  11. Dec 27, 2009 #10
    Yes, my top right diagram and bottom right diagram are both wrong. The bottom right one should look exactly as the top left one. The top right one should have the 4th ray emitting diagonally upwards and bouncing back to the new position (dashed) of the light bulb.

    Can you take a look at my second post to see if the claim in it is true? Thanks!
  12. Dec 27, 2009 #11


    User Avatar
    Gold Member

    Too much run-on. Must break up into individual thoughts.
  13. Dec 28, 2009 #12
    Yes. A light source going away form an observer will look dimmer and a light source coming towards him will look brighter. This is known as "relativistic beaming".
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook