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Speed of light

  1. Nov 12, 2007 #1
    A rocket can't travel faster than the speed of light.

    What would happen if you sent off a rocket going at 0.5c in one direction, and then sent off another rocket going at 0.5c in the opposite direction?

    Wouldn't one be travelling at the speed of light, relative to the other?
     
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  3. Nov 12, 2007 #2

    ZapperZ

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  4. Nov 12, 2007 #3

    Chris Hillman

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    Comparing three trigonometries

    No; ditto Zapper Z, plus this:

    One way to understand this is to learn and compare hyperbolic, parabolic, and elliptic trigonometry. Here, elliptic trig is better known as "high school trig", parabolic trig corresponds to the transformation laws of Galilean relativity, and hyperbolic trig corresponds to the transformation laws of special relativity. Then, your confusion comes down to expecting that velocities will be additive, or more geometrically, that slopes will be additive. But this is not true in either hyperbolic or elliptic trig! The correct addition laws for slopes are respectively
    [tex]
    v_{12} = \frac{v_1 + v_2}{1+v_1 \, v_2}, \;
    v_{12} = v_1 + v_2, \;
    v_{12} = \frac{v_1 + v_2}{1-v_1 \, v_2}
    [/tex]
    for hyperbolic, parabolic, elliptic trig. In the last case, you should recognize the "addition law for tangents" in high school trig; recall that [itex]v=\tan(\theta)[/tex] is the slope of a line making angle [itex]\theta[/itex] with the appropriate coordinate axis.
     
  5. Nov 12, 2007 #4

    robphy

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    Indeed.. it is the "angle" (arc length on a unit "circle") not the "slopes" (the "tangent"-function of the "angle") that are additive. [The quotes mean that these structures are defined analogously with respect to the three geometries.]
    In Galilean geometry, it turns out that the "tangent"("angle") is equal to the "angle"
    ... hence in the special case of Galilean kinematics, "velocities are additive".
     
  6. Nov 12, 2007 #5

    Chris Hillman

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    Right.

    I thought I had written a WP article with a small table comparing the trig functions for hyperbolic, parabolic, elliptic trig but right now I can't seem to locate that. However, the OP can try this poster
    www.aapt-doorway.org/Posters/SalgadoPoster/SalgadoPoster.htm; page 6 contains diagrams similar to the ones I drew when I was first learning this stuff (inspired by the first edition of Spacetime Physics by Taylor and Wheeler). The general idea is of course to interpret a straight line as the "world line" of an inertial observer and to interpret the slope of a straight line as the (constant) velocity of the observer.

    For more advanced readers: as the poster hints, there are beautiful connections here with two dimensional real Cayley-Dickson algebras (which were employed by Dirac to "factor" the wave equation) and with the nine Cayley-Klein plane geometries (which are connected with Cayley's discovery of projective metrics and were instrumental in the formulation by Klein and Lie of the Erlangen Program, which led both to Lie theory and to the notion of symmetry groups, both of which of course play crucial roles in modern physics!). A very readable book is I. M. Yaglom, A simple non-Euclidean geometry and its physical basis: an elementary account of Galilean geometry and the Galilean principle of relativity, Springer, 1979. See also I. M. Yaglon, Felix Klein and Sophus Lie: evolution of the idea of symmetry in the Nineteenth Century, Birkhauser, 1988.
     
    Last edited: Nov 12, 2007
  7. Nov 12, 2007 #6

    rbj

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    Chris, probably some math-wannabe at WP took it out long ago, and if you tried to defend it's inclusion this wannabe would find an admin who also doesn't have any expertise in the topic to side with him (or her). then you would get blocked for a revert war and if you protested that, no one (with any authority) would listen. and if you made a big stink about it, they would ban you. but i know you got out before all this happened.
     
  8. Nov 12, 2007 #7

    Chris Hillman

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    Blechch, rampant POV-pushing at WP

    It seems that I am not the only one here who is wont to rant about the failings of Wikipedia :wink:

    Did you hear that Jimbo's underlings are about to roll back the tiny prohibition on IP anons creating new articles?

    As you know, I have not been active at WP since the fall of 2006, but from time to time I still check for wikishilling and continue to find evidence that this is common in the math and physics pages :yuck:
     
    Last edited: Nov 13, 2007
  9. Nov 14, 2007 #8
    Thanks for your answers.

    I think I misphrased my question. I have a more oblique question:

    Would one be travelling at the speed of light, relative to the other, from the perspective of an observer on Earth?

    No information would be able to travel faster than c, but the separation between the two objects would appear to increase at c or faster.

    So is the answer "Yes", or is it that the phrase "relative to the other, from the perspective of an observer on Earth" only made sense before Einstein, and now it is nonsense?
     
  10. Nov 14, 2007 #9

    HallsofIvy

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    I'm not sure that the phrase "relative to the other, from the perspective of an observer on Earth" ever made sense! Suppose, ignoring Einstein, I am standing on the side of a road observing a car moving at 90 kph due east and another moving at 100 kph due west. How would I determine "the speed of each relative to the other, from my perspective"? The speed of one vehicle "relative to the other" has to be independent of "my perspective". Still ignoring Einstein, I could calculate that the speed of one relative to the other is 90- (-100)= 190 kph, but whether that meant anything physical or not would depend upon my reason for doing that calculation. Experiment (the "court of last resort" for any scientific question) shows that, as long as velocities are low relative to the speed of light, subtracting velocities like that (or adding speeds) give a very good correspondence to experiment.

    However, for speeds that are a fair fraction of the speed of light, the formula given before, in this case
    [tex]\frac{0.5c+ 0.5c}{1+ \frac{(0.5c)(0.5c)}{c^2}}= \frac{c}{1.25}= 0.8 c[/tex]
    gives a much better correspondence to experiment.
     
  11. Nov 14, 2007 #10

    robphy

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    Lord Ping,
    You might find this thread helpful
    https://www.physicsforums.com/showthread.php?t=184920
    especially #7 and beyond.

    The best strategy for answering [hopefully for one's self] these relativistic-kinematic problems is to draw a spacetime diagram and [learn to] interpret the physics read off from it.
     
    Last edited: Nov 14, 2007
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