# Locally Lorentz

by JDoolin
Tags: locally lorentz
P: 1,555
JDoolin you are misinformed, global Lorentz invariance only occurs in a flat spacetime.

 Quote by JDoolin For instance, I would recommend talking about how, within the gravitational pull of a planet, the rate of proper time is a function of the distance from the planet. My suggestion would be to say that in this region, we have a situation where somehow, the geometry seems to differ from Lorentz in some fashion, for it is in these local regions where we find space-time to be curved. The particle, traveling on a straight path in its own coordinates, ends up traveling on a curved path in another body's coordinates. I think that the theory behind General Relativity is strong enough, that it does not need to rely on ambiguously defined terms and attacking the fundamentals of Special Relativity. It should stand constructively on its foundations--not try to dismiss them as "only valid locally." On the global level, such slowing of proper time (implicit in general relativity) won't make a difference, because the end result, from a gloabl perspective, is just the local slowing of the speed of light. It's no more paradoxical than having glass, with its index of refraction slowing the speed of light.
Sorry but this is wrong.

 Quote by JDoolin But claiming that physics is somehow "locally lorentz" implies that physics is not "globally lorentz."
Which is the case.
In curved spacetime there is no global Lorentz invariance, there is only Lorentz invariance at the local level.
PF Gold
P: 706
 Quote by PeterDonis I don't think "direct interaction" is an apt description, because it implies that there's no difference in the physics along a photon worldline regardless of which pair of events along it I pick. The fact that the spacetime interval between any two events on a photon's worldline is zero does *not* imply that all events on that worldline are exactly the same in every physically relevant respect. So if I have two pairs of events, (A, B) and (A, C), that all lie on the same photon worldline, that does *not* imply that all the physics between A and B is exactly the same as all the physics between A and C. Simple example: a source at the origin that emits spherical wavefronts of light, and two detectors, both lying along the same radial line from the origin, one at radius R and the other at radius 2R. At time t = 0 in the frame in which all three objects (the source and both detectors) are at rest (I'm assuming flat spacetime, no gravity or other forces involved), the source emits a spherical wavefront. It arrives at detector #1 at time t = R and at detector #2 at time t = 2R. So we have three events: emission (t = 0, r = 0), detection #1 (t = R, r = R), and detection #2 (t = 2R, r = 2R). The spacetime interval between emission and detection #1 is the same as between emission and detection #2 (both are zero); however, the intensity of light measured at detection #1 is four times that measured at detection #2 (inverse square law). (In quantum terms, we would say that the amplitude for detection of a photon at detection #1 is twice the amplitude for detection of a photon at detection #2; the intensity goes as the square of the amplitude.) This difference, to me, means that saying "the locality is zero" for both pairs of events, or "direct interaction" between them, is not a good way of describing what's going on, because it gives no way of accounting for the difference in what's observed.
Proper time and proper distance are distinct from what I would usually mean by distance. Obviously, it is only in the lab frame where the distances between these events is R and 2R, and the time between these events are R/c, and 2R/c.

You are using what I would call the common definition of distance and time, which are observer dependent. I am perfectly fine with that, because it is what any ordinary person thinks of when he hears the word "distance."

By contrast, the metric distance between the source and destination events here is zero. This is the quantity that is unchanged by Lorentz Transformation. What the "metric distance" represents is an invariant quantity relating the distance and time between two events. As far as semantics are concerned, I actually object to even calling this quantity a "distance", preferring the more abstract term, "space-time-interval" and your objection makes it clearer why.

Because there is a difference between these two situations, where the receiver is 2R away, and where the receiver is R away. Yet the space-time-interval for the photons traveling the different distances is the same in both cases.

(Edit: By the way, I don't think there is any photon that is detected by both receivers. If it is detected by the first receiver, it's energy is absrobed by the first receiver. That's part of the reason why I think it is appropriate to say that a photon (a single quantum photon) is a direct interaction between particles at the source, and particles at the destination. I wasn't sure if this was a subtlety or just obvious. The subtlety comes in when you have interference, and though the light goes through both slits and there is self-interference, you still don't actually have an event. The photon is still a direct interaction with the source and destination, but somehow modified by the interference device in a geometric way, but not an "event-"ful way.)
PF Gold
P: 706
 Quote by Passionflower JDoolin you are misinformed, global Lorentz invariance only occurs in a flat spacetime. Sorry but this is wrong. Which is the case. In curved spacetime there is no global Lorentz invariance, there is only Lorentz invariance at the local level.
But I disagree. The speed of light slows down in the region of gravitational fields. This is a locally non-lorentz behavior. However, the lorentz transformation equations operate on every event in space-time. They cannot be contained to "locally lorentz."

On the other hand, it is easy to deal with a local slowing speed of light within the context of a globally lorentz spacetime.

I think the trouble occurs when distance and time are confused with the proper time and proper distance (what you call the "metric distance.") You'll notice that within the lorentz transformation equations, the metric distance is not to be found. Neither proper time nor proper distance of events appears in the operator or the inputs or the outputs of the equation. These notions are irrelevant to the geometry of spacetime at large.

However, in MTW's description of Locally Lorentz, the defintion of proper space and proper time have a central relevance. They are following geodesics where distance is used from the planetary frame, while time is used from the falling object's frame, as modified by the planetary gravity. And it is within these coordinates that somehow some principal of least action is calculated, and in some sense, if you use the right variables, the particle's trajectory is straight.

Is this straightness an application of some "local lorentzian" property? I'm not entirely sure, but I'd rather see the variables defined, and the reasoning clearly outlined.
P: 665
 Quote by Passionflower JDoolin you are misinformed, global Lorentz invariance only occurs in a flat spacetime. Sorry but this is wrong. Which is the case. In curved spacetime there is no global Lorentz invariance, there is only Lorentz invariance at the local level.
That is it!

AB
Physics
PF Gold
P: 6,132
 Quote by JDoolin As far as semantics are concerned, I actually object to even calling this quantity a "distance", preferring the more abstract term, "space-time-interval" and your objection makes it clearer why.
I agree, the term "spacetime interval" is clearer since it is unambiguous.

 Quote by JDoolin Because there is a difference between these two situations, where the receiver is 2R away, and where the receiver is R away. Yet the space-time-interval for the photons traveling the different distances is the same in both cases.
Yes, it is. But there is clearly *some* invariant, frame-independent difference between the cases, since the physical observable (light intensity, or amplitude for photon detection in the quantum case) differs. What spacetime invariant corresponds to that physically observable difference? It can't be the distance, because that is, as you point out, frame-dependent. It can't be the interval because that's the same in both cases. So whatever invariant *does* correspond to the physical difference, focusing on the spacetime interval (and the fact that it's zero for a lightlike interval) does not help in identifying what it is.
 PF Gold P: 706 Alright, I'm coming around to a legitimate meaning of "locally lorentz" but it involves defining some extra variables, and it preserves my meaning of "globally lorentz." In fact, I think, if we clearly define our variables, both are true in different contexts, and both are false in the opposite context! We can define tlocal by imagining a clock at a particular position in a gravitational field, suspended by a wall or a pole. This local time is a function of the clock's position in the gravitational field, but it is also a function of some external global observer-dependent time tglobal. The main point is we have three different sets of variables. The third set of variables is, of course, related to the geodesic of the object falling through the space. The "locally lorentz" behavior can then be described as $$\Delta \tau^2=\Delta t_{local}^2-\Delta x_{local}^2$$ While the "globally lorentz" behavior applies to the global coordinates $$\begin{pmatrix} c t_{global}' \\ x_{global}' \end{pmatrix}= \begin{pmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{pmatrix} \begin{pmatrix} c t_{global} \\ x_{global} \end{pmatrix}$$
PF Gold
P: 706
 Quote by PeterDonis I agree, the term "spacetime interval" is clearer since it is unambiguous. Yes, it is. But there is clearly *some* invariant, frame-independent difference between the cases, since the physical observable (light intensity, or amplitude for photon detection in the quantum case) differs. What spacetime invariant corresponds to that physically observable difference? It can't be the distance, because that is, as you point out, frame-dependent. It can't be the interval because that's the same in both cases. So whatever invariant *does* correspond to the physical difference, focusing on the spacetime interval (and the fact that it's zero for a lightlike interval) does not help in identifying what it is.
The main invariant of importance in Lorentz Transformations is the preservation of Maxwell's Laws, which in turn preserves the observer-dependent speed of light.

Consider that the receiver is either chasing the emitter or vice versa. Whatever effects of relativistic doppler happens to the emitter, you are assured that exactly the inverse effect will apply to the receivers. Even though the doppler effect itself is frame dependent, the coupling of two inverse effects means there is a sort of invariance in the events themselves.

Naturally, since the same events happen, regardless of what reference frame, the digital read-outs on the intensity probes will read the same.
Physics
PF Gold
P: 6,132
 Quote by JDoolin While the "globally lorentz" behavior applies to the global coordinates $$\begin{pmatrix} c t_{global}' \\ x_{global}' \end{pmatrix}= \begin{pmatrix} \gamma & -\beta \gamma \\ -\beta \gamma & \gamma \end{pmatrix} \begin{pmatrix} c t_{global} \\ x_{global} \end{pmatrix}$$
But this transformation, if it is to be valid globally in a curved spacetime, cannot be a Lorentz transformation. If you insist on writing it in the form you gave, then the coefficients $\beta$ and $\gamma$ will vary from event to event in a curved spacetime (i.e., in the presence of gravity). A global Lorentz transformation must have constant coefficients.
Physics
PF Gold
P: 6,132
 Quote by JDoolin (Edit: By the way, I don't think there is any photon that is detected by both receivers. If it is detected by the first receiver, it's energy is absrobed by the first receiver. That's part of the reason why I think it is appropriate to say that a photon (a single quantum photon) is a direct interaction between particles at the source, and particles at the destination.
The reason I object to the term "direct interaction" is that, on the one hand, you use the fact that the spacetime interval is zero to justify the term, but on the other hand, different pairs of events separated by a null spacetime interval can give rise to different physical observables. That doesn't mean the same photon gets detected by both receivers; it means that the amplitude (measured over many photon emission and detection events) for detection at the first receiver is different than the amplitude for detection at the second receiver, even though the spacetime interval between *all* the corresponding pairs of emission and detection events, for both receivers, is zero.

In other words, there must be *some* physical difference between the cases, and it can't be the spacetime interval since that's zero in both cases. Therefore, I don't think you can use the zero spacetime interval to justify the term "direct interaction", since the interval can't be the primary causal factor involved; there must be something else involved that is *not* accounted for by the interval, and which differs in the two cases, and which therefore makes them something different than just a "direct interaction".

 Quote by JDoolin I wasn't sure if this was a subtlety or just obvious. The subtlety comes in when you have interference, and though the light goes through both slits and there is self-interference, you still don't actually have an event. The photon is still a direct interaction with the source and destination, but somehow modified by the interference device in a geometric way, but not an "event-"ful way.)
I wasn't relying on interference for the point I was making, and I didn't assume it had to be present in the scenario I described. I agree that interference phenomena don't constitute "events" in the sense the term is used in SR, so they don't pose any additional issue for the topic we're discussing.
PF Gold
P: 706
 Quote by PeterDonis The reason I object to the term "direct interaction" is that, on the one hand, you use the fact that the spacetime interval is zero to justify the term, but on the other hand, different pairs of events separated by a null spacetime interval can give rise to different physical observables. That doesn't mean the same photon gets detected by both receivers; it means that the amplitude (measured over many photon emission and detection events) for detection at the first receiver is different than the amplitude for detection at the second receiver, even though the spacetime interval between *all* the corresponding pairs of emission and detection events, for both receivers, is zero. In other words, there must be *some* physical difference between the cases, and it can't be the spacetime interval since that's zero in both cases. Therefore, I don't think you can use the zero spacetime interval to justify the term "direct interaction", since the interval can't be the primary causal factor involved; there must be something else involved that is *not* accounted for by the interval, and which differs in the two cases, and which therefore makes them something different than just a "direct interaction". I wasn't relying on interference for the point I was making, and I didn't assume it had to be present in the scenario I described. I agree that interference phenomena don't constitute "events" in the sense the term is used in SR, so they don't pose any additional issue for the topic we're discussing.
Does the emission/reception of a photon result in different physical observables? On the one end, an electron drops into a lower shell. On the other end, a photon is absorbed by a photometer. Certainly, the net effect of oodles of photons is different; the closer observer receives a larger number of photons in the same area.

But I think, yes, now that I think of it, there is another invariant in the two cases. The number of wavelengths in the photon between the two events must be the same, regardless of what reference frame you are in. There are all kinds of differences that are variant, but the count of wavelengths would be an invariant.

Emitter \ \ Reciever
\ \
\ \
\ \.   (Reciever chasing emitter.)
\/\  (Nearing a zero distance in space and in time)
\ \  (the light and receiver are going in opposite directions)
\ \

.
Reciever \\\ Emitter
\\\
\\\
\\\  (Nearing an infinite distance in space and time.)
\\\  (The light has to go a long way to catch up.)
\\\
\\\

Physics
PF Gold
P: 6,132
 Quote by JDoolin Does the emission/reception of a photon result in different physical observables? On the one end, an electron drops into a lower shell. On the other end, a photon is absorbed by a photometer. Certainly, the net effect of oodles of photons is different; the closer observer receives a larger number of photons in the same area.
And the "amplitude for photon" detection I've been referring to is simply the average number of photons detected at a given detector, as a percentage of total photons emitted from the source (actually, this is the probability, and the amplitude is a "complex square root" of this since it includes phase information as well). I'm not intending even to get into all the subtleties of quantum mechanics for this discussion; I was just trying to get across the point that the intensities are different.

That said, the amplitude that is determined experimentally as I just described is *also* what we would use to determine the probability of detecting a single photon emitted from the source, at a given detector (the probability is just the square of the amplitude--more precisely the squared modulus of the amplitude, since the latter is a complex number). I don't know if the precise experiment I described has been done at the single photon level, but plenty of other single photon experiments have been done (i.e., experiments where, with very high probability, there is never more than one photon in the experiment at a time), and their results are exactly what is predicted by calculating the probability from the amplitude as I've described.

 Quote by JDoolin But I think, yes, now that I think of it, there is another invariant in the two cases. The number of wavelengths in the photon between the two events must be the same, regardless of what reference frame you are in. There are all kinds of differences that are variant, but the count of wavelengths would be an invariant.
Yes, that would work. To be a bit more precise, you could measure the frequency of the wave train as it passed the detector, and then divide by c to get a spatial "wave number" k. This number, times the distance R from the source, would give a number which should be the "number of wave crests between source and detector", and which should be (a) frame-invariant (i.e., the Doppler shift in frequency/wavenumber exactly cancels the length contraction of R, so the final result will be the same in every frame for a given source-detector pair) and (b) different for detector #1 at R and detector #2 at 2R.
PF Gold
P: 706
 Quote by PeterDonis And the "amplitude for photon" detection I've been referring to is simply the average number of photons detected at a given detector, as a percentage of total photons emitted from the source (actually, this is the probability, and the amplitude is a "complex square root" of this since it includes phase information as well). I'm not intending even to get into all the subtleties of quantum mechanics for this discussion; I was just trying to get across the point that the intensities are different. That said, the amplitude that is determined experimentally as I just described is *also* what we would use to determine the probability of detecting a single photon emitted from the source, at a given detector (the probability is just the square of the amplitude--more precisely the squared modulus of the amplitude, since the latter is a complex number). I don't know if the precise experiment I described has been done at the single photon level, but plenty of other single photon experiments have been done (i.e., experiments where, with very high probability, there is never more than one photon in the experiment at a time), and their results are exactly what is predicted by calculating the probability from the amplitude as I've described. Yes, that would work. To be a bit more precise, you could measure the frequency of the wave train as it passed the detector, and then divide by c to get a spatial "wave number" k. This number, times the distance R from the source, would give a number which should be the "number of wave crests between source and detector", and which should be (a) frame-invariant (i.e., the Doppler shift in frequency/wavenumber exactly cancels the length contraction of R, so the final result will be the same in every frame for a given source-detector pair) and (b) different for detector #1 at R and detector #2 at 2R.
I was turning this over and over in my head the other day, and realized there was a problem. If you have the emitter approaching you, you have a higher doppler frequency, and a longer distance from the event. If you have the emitter receding, you have a lower doppler frequency and a shorter distance from the event!

Approaching emitter -- higher frequency (inceases #wavelengths), more distant source event (increases #wavelengths)

Receding emitter -- lower frequency (decreases # wavelengths), nearer source event (decreases # wavelengths)
This means that the number of wavelengths between the emitter and receiver cannot be an invariant after all... or we have to throw away special relativity. Now that I look back at it, I believe it was this very issue that made me begin to say "The photon represents a direct interaction between the source and destination." (Because there is an observer dependent variation in the number of wavelengths between the events. (sometimes the photon is represented as a "packet" often a sinc wave: sin(x)/x ; it moves at the speed of light but doesn't necessarily exactly oscillate in space.))

The attached diagrams helped me to see my mistake. There are two emitters in the diagram, and one receiver between them. In the first diagram the two emitters are stationary and the number of waves is the same. In the second diagram the two emitters are traveling to the left. Because of the relativity of simultaneity, the emitter on the right began broadcasting long before the emitter on the left. This allows there to be a larger number of wavelengths present in the approaching emitter than there are in the receding emitter.

Edit: With the receding emitter and the chasing receiver, eventually the wavelength of the photon is going to be longer than the lorentz contracted distance between the emitter and receiver. This is another reason that I say "the photon is a direct interaction between the emitter and receiver."
Attached Thumbnails

Physics
PF Gold
P: 6,132
 Quote by JDoolin The attached diagrams helped me to see my mistake. There are two emitters in the diagram, and one receiver between them.
This is a different situation from the one I was discussing. You have two emitters and one receiver, instead of two receivers and one emitter; and in the "moving" frame you have the emitters moving but the receiver still at rest (if I'm reading your diagram right), whereas I was discussing the case where emitter and receivers are all mutually at rest (but where the "observer" that's watching all three might be moving relative to them). Obviously if you change the relative motion of emitter and receiver, you'll change the physics.

Try it with one emitter, at x = 0, and two receivers, one at x = R and one at x = 2R (i.e., both lying in the same direction from the emitter, but one twice as far away as the other), in the frame in which they are all mutually at rest. Then transform to a frame in which all three are moving with the same velocity v. That's the case I was talking about.
Physics
PF Gold
P: 6,132
 Quote by JDoolin Approaching emitter -- higher frequency (inceases #wavelengths), more distant source event (increases #wavelengths) Receding emitter -- lower frequency (decreases # wavelengths), nearer source event (decreases # wavelengths)
All of this is consistent with the product (frequency / c) * (distance to source) = invariant. As the frequency goes up, the distance goes down, and vice versa.

 Quote by JDoolin This means that the number of wavelengths between the emitter and receiver cannot be an invariant after all... or we have to throw away special relativity.
We would have to throw away special relativity if the number of wave crests crossing a given worldline in a given amount of proper time along that worldline--or, equivalently, the number of wave crests counted along a given line of simultaneity, in a given proper distance along that line--was *not* an invariant. If the emitter and the receivers are all at rest relative to each other, their worldlines are all parallel and they share the same lines of simultaneity, so the "number of wave crests" is a perfectly good invariant. (It's harder to calculate in a frame in which the emitter and receivers are moving, but it's still an invariant.)
PF Gold
P: 706
 Quote by PeterDonis All of this is consistent with the product (frequency / c) * (distance to source) = invariant. As the frequency goes up, the distance goes down, and vice versa.

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'll try to make this argument as simple as I can, but I think it comes down to two very important key ideas, that so far, you have not acknowledged.

The first is that the distance between the emission event and absorption event is NOT an invariant. I'm not sure how to explain it any better, except tell you to go back and look at the diagrams that I've given you and admit it. Put it in your own words, explaining why the distance is larger when the emitter is chasing, and the distance is smaller when the receiver is chasing. It's not really difficult enough to show you a bunch of intricate math. Seeing is believeing. (the ascii diagram may have failed you? Do you need me to draw six lines in a jpeg to show you?)

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. Again, this isn't something I can explain with a bunch of intricate math. You just have to think about it, and admit that it's true.

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
 We would have to throw away special relativity if the number of wave crests crossing a given worldline in a given amount of proper time along that worldline--or, equivalently, the number of wave crests counted along a given line of simultaneity, in a given proper distance along that line--was *not* an invariant. If the emitter and the receivers are all at rest relative to each other, their worldlines are all parallel and they share the same lines of simultaneity, so the "number of wave crests" is a perfectly good invariant. (It's harder to calculate in a frame in which the emitter and receivers are moving, but it's still an invariant.)
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.

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.
Physics
PF Gold
P: 6,132
 Quote by JDoolin 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.

 Quote by JDoolin 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.

 Quote by JDoolin 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.

 Quote by JDoolin 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.

 Quote by JDoolin 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."

 Quote by JDoolin 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.
Physics
PF Gold
P: 6,132
 Quote by PeterDonis 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.
I should clarify this some more to make sure I'm clear: I'm not actually saying there have to be physical wave crests *all the way* along the line of simultaneity I've described. That's not necessary (although it makes it easier to interpret the relevant spacetime diagrams if you think about it that way). All that's necessary is that we can calculate a product (frequency / c) * (distance between emitter and receiver), where both are measured along the particular line of simultaneity I've defined. If "measuring frequency along a line of simultaneity" bothers you, just redefine (frequency / c) as (spatial wavenumber), which is perfectly well-defined along the given line of simultaneity--even if the wave crests don't reach all the way from emitter to receiver along that line (because the light pulse is too short), the concept "number of wave crests per unit distance along that line" is still well-defined.
PF Gold
P: 706

You should distinguish between "distance between objects" and "distance between events."

The "unqualified distance" term you are using, is the distance between the emitter and receiver objects. This is the more familiar distance discussed under the heading of "length contraction" or "lorentz contraction", where solid bodies appear to contract in the direction of motion.

But distances can and should also be measured between nonsimultaneous events. For instance in this emission and absorption example, you shouldn't pretend that the distance traveled by the photon is equal to the distance between the emitter and receiver. Why? Because while the photon is under way, the receiver either moves toward or away from the emitter. This means that the distance between the receiver and emitter is irrelevant. It is the distance between the emission EVENT and absorption EVENT which is relevant.

You claimed the "distance between events" is not well-defined, but I disagree. Whenever you do a lorentz transformation, the x, y, and z terms refer to the coordinates of events; not objects. The distance between the emission and absorption events in any given frame is another distance $$\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}$$. The LT equations themselves do not distinguish to whom or to what the events occurred. All that is of concern is where and when, relative to a stationary origin.

Now for the snowflake example:

Let's note that this distinction (distance between events vs. distance between objects) should show up in basic "Galilean" relativity as well. We could talk about snowflakes coming down with a driver passing through them. The question is, are the snowflakes really coming straight down, or are they going almost 55 miles per hour horizontally?

The fact is, you could arrange two rulers, one long, nearly horizontal one traveling along with the car, and another short vertical one stationary with the ground, and the snowflake follows the path of both rulers. The top end of the two rulers meet when the snowflake passes the top, and the bottom end of the two rulers meet when the snowflake passes the bottom. (but at every moment, the two rulers and the snowflake align.)

My premise is that both the driver of the car, and the person on the ground use their own rulers and give correct (but different) answers for the distance between those two events. That is the premise behind the galilean transformation, and the lorentz trasformation.

First let me point out my snowflake example above. Who is correct? The guy in the car who thinks the snowflakes are coming right at him, or the person on the ground who thinks the snowflakes are coming straight down? You can pick out a frame of reference that meets some arbitrary criteria, but I think the most sensible criteria is "observer dependence." Use the distance as measured by a ruler in that reference frame, where your observer is. The driver and the person on the ground are equally correct, so long as they make it clear who and where they are.

You may still feel like one of the drivers is incorrect, but remove the planet, and just have the car and the snowflake flying through space. The snowflake is coming at the car at 55 miles per hour and travels a whole second at that speed. That distance is well-defined, even though the events did not happen at the same time.

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