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How is Constant speed of light referenced

  1. Oct 14, 2009 #1
    The speed of light is constant but with reference to what?
    Distances are not absolute and neither is time.

    Is it that all inertial observers agree upon the ratio of the distance to the time that it takes for light to go between the two ends of some object even though they don’t necessarily agree upon the length of the object or the time it takes light to traverse that object?

    One of the fundamental postulates of Einstein's special theory of relativity is that all inertial observers will measure the same speed of light in vacuum regardless of their relative motion with respect to each other or the source.

    I am sure that this statement is definitive but I find it confusing.

    I find it confusing because defining speed means defining it with reference to a distance as well as a ‘time span’ and saying that the distance divided by the ‘time span’ is the same for all inertial observers. But distances in SR are not absolute. Different inertial observers will measure different distances for a given object.
     
  2. jcsd
  3. Oct 14, 2009 #2
    You just take the distances and time intervals that are in your equations. The value of c is invariant of the reference frame change.
     
  4. Oct 14, 2009 #3

    A.T.

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    With reference to every observer measured by rulers and clocks at rest to the observer.
     
  5. Oct 14, 2009 #4
    Given that c is an invariant in equations I am trying to interpret it in a physical sense.
    For instance, in order to be invariant in different time flows does the speed of light vary with the time flow?
     
  6. Oct 14, 2009 #5

    HallsofIvy

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    The speed of light is constant with respect to any reference frame. If x beam of light passes both of observors A and B, who are moving a very high speed with respect to each other, each will observe the speed of light to be c in their own reference frame. This in spite of the fact that each observes the other's time to be slowed up.

    Another point. If a person standing in the back of a truck, with speed 30 mph relative to you, standing on the side of the road, throws a ball to you at, relative to him, 70 mph, it will hit you with speed, relative to you of (approximately) 30+ 70= 100 mph.

    If that same person directs a flashlight at you the light will move away from him at c, in his reference frame, and will move toward you at c, in your reference frame.

    That, and the reason I said "(approximately)" before is because the "addition of velocities" formula in relativity is
    [tex]\frac{v_1+ v_2}{1+ \frac{v_1v_2}{c^2}}[/tex]

    If [itex]v_1[/itex] and [itex]v_2[/itex] are small compared to c, that [itex]v_1v_2/c^2[/itex] will be very small compared to 1 and difference between that and [itex]v_1+ v_2[/itex] will be unmeasurably small. But if [itex]v_2= c[/itex] the formula becomes
    [tex]\frac{v_1+ c}{1+ \frac{v_1c}{c^2}}= \frac{v_1+ c}{1+ \frac{v_1}{c}}[/tex]
    and, multiplying both numerator and denominator by c,
    [tex]= \frac{(v_1+ c)c}{c+ v_1}= c[/tex]
     
  7. Oct 14, 2009 #6
    This is basically the right answer. The entire reason observers disagree on distances and times is because they must agree on c. So yes, these differences conspire to keep the observed speed of light c, for all observers.
     
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