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Doppler effect and invariability of C

  1. Jul 8, 2012 #1
    If the doppler effect is the changing if wavelengths due to movement of a light source (In reference to light of course), does this mean the distance between the wavefront and the source changes at a speed DIFFERENT from C? doesn't this mean that if the observer is in the plane of the source the speed of light for that observer is NOT C? I'm sure I'm missing something, because there's no way the rule of invariable C is shown to be wrong that easily! Especially not by me! haha! Not sure if I made it clear but the source would be moving in the direction of the direction of travel of the wave at any velocity with regards to the plane of the wavefront.
  2. jcsd
  3. Jul 8, 2012 #2


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    If I understand you correctly, you are saying that you have an observer (who we'll take to be stationary) and a light source that is moving towards him with speed v. The problem is, you are then reasoning that if the light moves with speed c towards the observer, and the source moves with speed v towards the observer, then light must move with speed c - v away from the source. The problem with this is that, you have implicitly, without realizing it, changed frames of reference from the observer's frame to the source's frame, and you used a velocity transformation rule to transform the velocity of the light as measured in the first frame to the velocity of light as measured in the second frame. Unfortunately, you used the wrong velocity transformation law. You used the classical or "Galilean" velocity transformation rule, which says that to transform from the velocity of an object u as measured in one reference frame, to the velocity of that same object (u') as measured in another reference frame that is moving with speed v relative to the first, you simply add the two relative motions together, so that u = u' + v. However, velocities do not add together in this simple way in special relativity.

    Instead, the correct velocity transformation rule is u = (u' + v) / (1 - u'v/c^2). Notice that if the "object" whose velocity u we are measuring in either frame is a photon, i.e. u' = c, then we end up with:

    u = (c + v)/(1 +cv/c^2) = (c + v)/(1 + v/c)

    Multiply the numerator and the denominator by c:

    u = c(c+v) / (c + v) = c

    u = c

    If the velocity is c in one reference frame, it will be c in any other reference frame. EDIT: I am talking about inertial reference frames here. This makes sense, because the velocity transformation rule above was derived by assuming (i.e. taking it to be a postulate of special relativity) that light always moves at c in any frame of reference.

    At first it might seem preposterous that if a moving source emits light at speed c relative to it, then you will measure that light to be coming toward you at speed c as well (as opposed to faster than c). However, it soon makes sense if you realize that the receiver and the sender do not agree on the amount of time that it takes for the light to travel between them, or the amount of distance covered by the light. This is the unique result of special relativity: space and time are relative.
    Last edited: Jul 8, 2012
  4. Jul 8, 2012 #3
    I have come across this problem. People disagree about the time at which the light hits a point, but, say I have a detector wired up to the same clock as the source, to prevent this misalignment that seems to happen between frames of reference otherwise. What will they tell me about where and when the light hits its target?
  5. Jul 8, 2012 #4


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    Umm, okay, first of all: did you read the statement that I put in boldface and italics? That, fundamentally, is the answer to your question. Your reasoning that the velocity of the light relative to the source should be the velocity of the light relative to the observer minus the velocity of the source relative to the observer is simply wrong. Relative velocities DO NOT add up in this way according to special relativity. I showed you how they do add together, and that, if you use the correct velocity addition formula, you arrive at the conclusion that the speed of the light wave is c both relative to the source and relative to the observer.

    Secondly, the "disagreement" that I referred to is NOT merely a result of logistical or measurement difficulties. We have two events in spacetime. Event 1: light leaves source. Event 2: light arrives at receiver. The time intervals and distance intervals between these two events are actually different for the two observers.
  6. Jul 8, 2012 #5
    laura: Cephid has the correct explanation but it is tough to overcome the perception everyone has as do you when initially considering lightspeed.

    Maybe a simple minded explanation will help: consider water waves coming towards you. The waves have a certain spacing between, say, crests.....that's the wavelength. If you move towards the incoming waves those crests come faster, the wavelength appears shortened, but the speed of the waves themselves has not changed. Just the wavelength relative to you has changed. [Don't take this analogy too far ,however, as water waves travel via a medium, light needs no such medium.

    Whatever trick you decide to use to have it make sense to you, if you remember that observed locally, where you are, the speed of light in a vacuum is always 'c', you'll be able to begin to make sense of other aspects of relativity. Another trick is to always ask youself "What frame of reference is being used in an explanation?" This is a frequent source of confusion in some discussions here. Remember it took an "Einstein" to figure it out so it's ok to fuss about this for a while!
  7. Aug 19, 2012 #6
    I agree that it may be very helpful to use the example of water or sound waves for understanding the Doppler effect. However, the analogy can be perfect and the above explanation may benefit from some expansion - notably the part "If you move towards the incoming waves" can be confusing, for which reference system is implied? Clearly another one! So Laura, here's my try at this. :tongue2:

    Assume that your reference system is at rest in a "light medium". Effectively that is the model that SR uses, and which works (SR is based on Maxwell's electrodynamics which includes light). If a source is moving towards a detector that is in rest (that's the simplest case, but it happens to be the one that you brought up!) then the different vibrations will pile up in front of the source, so that the detected frequency will be higher. This is exactly the same for light as for sound (as long as the air in-between is undisturbed). And of course, the speed of sound in this example is simply c (the speed of sound in air). Thus, if you understand the Doppler effect for sound, that can be a great help. :smile:

    Note also that as long as you stick to the same reference frame then velocities do "add up", as they should, by mathematical necessity. It's only when you switch reference frames that the classical law of addition of velocities doesn't work with SR.
    Last edited: Aug 19, 2012
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