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Basic speed of light question

  1. Mar 12, 2015 #1
    I need to teach special and general relativity to IB Physics students in high school, and this is a question I've been wondering for a while now and want to understand before teaching it. If the speed of light is constant no matter what the speed of the source or observer, does that mean that someone who is traveling at 0.99c and someone who is “stationary” will both see light moving at the same speed (the speed of light)? If the light is shined directly at the person moving at 0.99c, with the light moving in the same direction as the motion of the observer, does the light get to the observer at a speed of c or speed of 0.01c? Is the answer due to the malleability of time?
     
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  3. Mar 12, 2015 #2

    PeterDonis

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    Yes.

    No. Velocities don't add in SR the way they do in Newtonian physics. See here:

    http://en.wikipedia.org/wiki/Velocity-addition_formula#Special_theory_of_relativity

    Note that if ##v = c##, then ##s = c## as well no matter what ##u## is.

    Sort of. Length contraction and time dilation are really best viewed as consequences of the constant speed of light, rather than the other way around.
     
  4. Mar 12, 2015 #3

    Ibix

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    As Peter notes, velocities do not add that way. If two cars are doing 60mph in opposite directions along a street, the velocity of one car seen from the other is not 120mph - it is 119.9999999999990mph (according to the Windows calculator, anyway). The effect gets more pronounced as you get to faster and faster speeds and. again as Peter notes, the speed of light is the limit and never changes.

    Einstein simply postulated that the speed of light was constant for all observers. He didn't do it just for fun - Maxwell's equations produced a speed of light that had no parameter for how fast the measurement apparatus was moving, and there were several experiments consistent with the idea (notably Michelson and Morley). Einstein was just the first to realise that you needed to take such an apparently daft idea at face value. From that postulate, you can derive the Lorentz transforms, which relate what is seen by (or, more precisely, mapped out after the fact by) one observer to what is seen by (or mapped out by) another observer moving relative to the first. Using the light-clock thought experiment it's only about half a page of high-school algebra to do so.

    The Lorentz transforms give you three things: length contraction (I will measure an object moving relative to me to be shorter than if it is stationary), time dilation (clocks moving relative to me tick slowly) and the relativity of simultaneity (clocks that show the same time to a moving observer show different times to me). These three effects do always conspire in such a way that the speed of light is constant, but that's because they were derived from that assumption. Again, as Peter says, most people regard the invariance of the speed of light to be the fundamental concept from which the rest follows.

    It also explains in one line why you can't reach light speed - that speed in constant in every frame. If you travelled at light speed, a light pulse would be stationary next to you. How can it be stationary and moving at 3×108m/s at the same time? It can't - so you can't travel at the speed of light.
     
  5. Mar 18, 2015 #4

    Meir Achuz

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    The relativistic formula for the addition of parallel velocities is simple enough for high school students to understand. The final velocity V for a bullet fired from a gun with a muzzle velocity v from a plane traveling with velocity v' is given by
    [tex]V=\frac{v+v'}{1+vv'/c^2}.[/tex]. If v'=c, then V = c.
     
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