JDoolin said:
You mischaracterized what I said. The two possibilities are (a) Both the frequency and the distance go up, or (b) both the frequency and the distance go down. It may go against your intuition (I know it goes against mine), but you need to think about it some more...
I think I wasn't clear enough about what I was using the term "distance" to refer to. I should have worded my previous posts more carefully, both for that reason and because there are subtleties involved that I didn't fully consider at first.
JDoolin said:
The first is that the distance between the emission event and absorption event is NOT an invariant.
Nor is it even well-defined, as you've stated it; the interval between the emission event and the absorption event is null, so there is no "distance" between these events ("distance", in the spatial sense, only applies between spacelike separated events).
If what you meant to say is that the distance, meaning spatial distance, between the *emitter* and the *receiver* can be different in different frames (so it is not an invariant), then I agree, but there is no frame where the events on the emitter and receiver worldlines between which "distance" is measured (i.e., which are crossed by a single line of simultaneity in the frame) will be the emission and absorption *events* (again, because those events are null-separated, not spacelike-separated).
This meaning of "distance"--i.e., distance between emitter and receiver--is what I was using the term "distance," without qualification, to refer to. However, I may not have been clear enough about that, since I was also talking about an observer that sees both emitter and receiver as moving, and "distance" could have been taken to mean the distance to that observer, at some particular time, which is *not* the way I meant it. See below.
JDoolin said:
The second issue is whether the frequency is going up or down. With this, you have to put yourself in the position where the receiver passes you. If the emitter is chasing the receiver, then the frequency goes up. If the emitter is running away, then the frequency will be lower.
Agreed.
JDoolin said:
The two possibilities are (a) the frequency goes up and the distance goes up. or (b) the frequency goes down and the distance goes down. I'm sorry, but there is no option, as you claimed, for the frequency to go up while the distance goes down, or vice versa
Actually, the distance as I was using the term (i.e., distance between emitter and receiver--see above) always goes down; more precisely, if we are in any frame in which emitter and receiver are both moving with some nonzero velocity v, the spatial distance between them, as seen in that frame, will be *less* than the "proper distance" between them in the frame in which they are both at rest. This should be obvious from the fact that Lorentz contraction is the same regardless of which direction the objects are moving.
The above is what I've been talking about when I talk about "distance", and that's why I said that it was possible for the frequency to go up while the distance was going down: if the emitter is moving towards the observer, then the frequency goes up, but the distance between emitter and receiver goes down by Lorentz contraction. My quick back of the envelope check seems to indicate that in this case, the Doppler shift in frequency exactly compensates for the Lorentz contraction, so the product (frequency / c) * (distance between emitter and receiver) does in fact stay constant. (Note the way I stated the product this time--this is what I meant last time, but the term "distance to source" was ambiguous, and that may have caused confusion. Sorry about that.)
The subtlety, of course, is that the above nice compensation only works for a Doppler *blue* shift! For the case where the emitter (and hence also the receiver) is moving *away* from the observer, so there is a Doppler red shift, the nice compensation doesn't work; distance between emitter and receiver goes down (by Lorentz contraction), but frequency *also* goes down! So this product as it stands won't work as an invariant. However, there is something that will; see below.
JDoolin said:
You're argument is pretty good here, but it doesn't quite apply in this case, because technically, we're never using the same line of simultaneity... we're using an orthogonal projection of the photon's path from emitter to receiver onto different planes of simultaneity. And those different planes of simultaneity yield calculations of different numbers of wavelengths.
This is true, but it's not what I was proposing. I was simply saying the following: there is one line of simultaneity that *is* picked out by the problem as different from all the others, namely, the one in the frame in which the emitter and receiver are both at rest. The number of wave crests along that line of simultaneity between emitter and receiver *is* an invariant, because we've specified a particular spacelike line along which to measure it; anyone in any frame can calculate the number and come up with the same answer. The calculation is harder in a frame in which emitter and receiver are moving, because it has to calculate the number of wave crests along an "unnatural" spacelike line for that frame--i.e., one which is *not* a line of simultaneity in that frame. But it can still be done.
You'll note that, with reference to this particular point, I did not just use the term "distance," unqualified; I used the phrase "the number of wave crests counted along a given line of simultaneity, in a given proper distance along that line."
JDoolin said:
Edit: Do I need to mention that the number of wavelengths along the actual null-path of the photon is zero? It is akin to watching a surfer on a wave; who doesn't move forward or backward or up or down so long as the camera keeps up with the wave. So yes, in fact that is an invariant.
Yes, but one that is the same for all possible (emitter, receiver) pairs, so it does nothing to distinguish them.