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

Speed of light or speed of time?

  1. Apr 3, 2010 #1
    As far as I can tell, the equations and theory of special and general relativity can all still work if you have light travelling in a vacuum arriving instantaneously, and take the constant c to be the constant that describes the translation of distance into time.

    In this formulation, the Hubble telescope isn't looking at the state of the early universe. If you were to compare clocks at the star being observed and at the telescope, they would agree that the light is emitted and received in year 13billion or whatever the age of the universe may be. But in relation to each other, x billion of those years have been translated into distance.

    Could you explain exactly how this might be/must be wrong?

    I beg your indulgence if you think the answer is obvious.
  2. jcsd
  3. Apr 3, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You're right that c is fundamentally a property of spacetime, not a property of light.

    There are various problems with the scenario you suggest, however.

    (1) It isn't consistent with experiment, which shows that the speed of light equals the property of spacetime called c.

    (2) It would violate causality. An electromagnetic signal transmitted from A to B would be received before it was sent in certain frames of reference.

    (3) There are fundamental reasons why particles with zero rest mass must travel at a speed equal to the property of spacetime called c: http://www.lightandmatter.com/html_books/genrel/ch04/ch04.html#Section4.2 [Broken] (see subsection 4.2.2)
    Last edited by a moderator: May 4, 2017
  4. Apr 3, 2010 #3
    From the point of view of a photon (or any other massless particle for that matter), light does leave a source and arrive at its destination instantaneously. This is because for things traveling at the speed of light (thus only massless objects) the line element for measuring distances becomes null, i.e.(ds)^2 = 0. Because ds represents the distance between two objects in spacetime, this would imply that everything is right next to each other from the perspective of a photon, so it would take no time for it to travel from on spacetime location to another.
  5. Apr 3, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    FAQ: What does the world look like in a frame of reference moving at the speed of light?

    This question has a long and honorable history. As a young student, Einstein tried to imagine what an electromagnetic wave would look like from the point of view of a motorcyclist riding alongside it. But we now know, thanks to Einstein himself, that it really doesn't make sense to talk about such observers.

    The most straightforward argument is based on the positivist idea that concepts only mean something if you can define how to measure them operationally. If we accept this philosophical stance (which is by no means compatible with every concept we ever discuss in physics), then we need to be able to physically realize this frame in terms of an observer and measuring devices. But we can't. It would take an infinite amount of energy to accelerate Einstein and his motorcycle to the speed of light.

    Since arguments from positivism can often kill off perfectly interesting and reasonable concepts, we might ask whether there are other reasons not to allow such frames. There are. One of the most basic geometrical ideas is intersection. In relativity, we expect that even if different observers disagree about many things, they agree about intersections of world-lines. Either the particles collided or they didn't. The arrow either hit the bull's-eye or it didn't. So although general relativity is far more permissive than Newtonian mechanics about changes of coordinates, there is a restriction that they should be smooth, one-to-one functions. If there was something like a Lorentz transformation for v=c, it wouldn't be one-to-one, so it wouldn't be mathematically compatible with the structure of relativity. (An easy way to see that it can't be one-to-one is that the length contraction would reduce a finite distance to a point.)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook