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Eisteins theory of relativity

  1. Feb 14, 2013 #1
    Is this statement true? The speed of light (about 300,000,000 meters per second) is the same for all observers, whether or not they're moving.

    If so could someone please explain it to me, because it was my understanding that if an object was moving at a constant velocity it would appear to be travelling slower to an observer which is stationary than an observer travelling at any constant velocity.

    Found on http://science.howstuffworks.com/warp-speed2.htm

  2. jcsd
  3. Feb 14, 2013 #2


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    I didn't see anything in that link that says that light or anything else "would appear to be travelling slower to an observer which is stationary than an observer travelling at any constant velocity". Can you please copy and post the text in that link that says anything like that?
  4. Feb 14, 2013 #3
    The statement is false. The speed of light is constant (299,792,458 m/s) regardless of reference frame and velocity of an observer.
  5. Feb 14, 2013 #4
    That wasn't a quote, that was what I thought, as i said "because it was my understanding that...".
    read the question more thoroughly please.
  6. Feb 14, 2013 #5


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    But then you put a link right after that statement. It is naturally implied that you learned that from that link!

    So you need to thoroughly read your post and make sure you convey your message clearly.

    IF that link wasn't your reference, what is the purpose for it being presented there?

  7. Feb 14, 2013 #6
    This doesn't answer my question you just re quoted the statement with a more accurate speed of light and different context.
  8. Feb 14, 2013 #7


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    That statement is true (aside from AmanDhaliwal's point about it being 299,792,458 m/s instead of "about 300,000,000" m/s).

    Your understanding is wrong - speeds do not add and subtract that way. Let's say that you're at rest, somebody ("Slowguy") passes by you at speed u; and then a moment later "Fastguy" moving at some faster speed v passes by you in the same direction. You're expecting that Fastguy's speed relative to Slowguy will be (v-u). It's not; it's [tex]\frac{v-u}{1-\frac{uv}{c^2}}[/tex]
    Two interesting things about this formula:
    1) For speeds that are small compared with the speed of light, it comes out REALLY close to the v-u that you expected; that's why you've never noticed.
    2) If v=c (that is, fastguy is actually a pulse of light) the formula comes out to give fastguy moving at speed c relative to both you and slowguy, which makes the statement that you found on the web consistent.
  9. Feb 14, 2013 #8


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    In a way, it did answer your question. It is one of the fundamental postulate of SR that the speed of light is a constant in ALL reference frame.

    The reason why you think it might move slower is because you are using what we call "Galilean transformation", i.e. the way we add velocities. It has already been shown that this is only valid for velocity that is very much smaller than c, i.e. an approximation.

    If you start with the postulate of SR, then there is a more generalized way to add velocities. If you are asking WHY c is a constant in all reference frames, we have no answer to that (yet), because that is how Nature behaves as we know it now.

  10. Feb 15, 2013 #9
    That site is inaccurate, but it does explain the basics. Not clear enough? Regretfully I don't understand your (mis)understanding about observing a moving object, or what that has to do with the speed of light. No object can reach the speed of light.
  11. Feb 15, 2013 #10
    Ahh I see, so no one knows why it happens but we just have to accept that it's true.

    Thank you for you help.
  12. Feb 15, 2013 #11
    Different people give different answers as to the "why". For example, assuming that Maxwell's electrodynamics is correct then it has been shown that special relativity (incl. the aspect that you mentioned) follows from the conservation laws (conservation of momentum and energy). Perhaps it can also be shown to follow from the wave nature of matter (I'm not sure about that one).
  13. Feb 15, 2013 #12
    Could you please expand on the equation, I'm not familiar with the terms "tex" and "frac".
  14. Feb 15, 2013 #13
    What you should see is something like:

    v - u
    1 - uv/c2

    If you see terms like "tex" and "frac", then there is a problem with your browser or internet reception.
  15. Feb 15, 2013 #14


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    See here:


    There's also Wikipedia, but it probably has more "extra" stuff than you really want:


    As Harrylin noted, your browser isn't decoding LaTeX code for you for some reason. If you're going to hang around here a lot, you really should get that fixed, because we use LaTeX a lot for equations. Try asking about it in "Forum Feedback and Announcements", down at the bottom of the list of forums on our home page.
  16. Mar 12, 2013 #15


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    I believe MathJax doesn't work in tapatalk.
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