# Locally Lorentz

PeterDonis
Mentor
Re: "Locally Lorentz"

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.

PeterDonis
Mentor
Re: "Locally Lorentz"

(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".

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.

Gold Member
Re: "Locally Lorentz"

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.

Code:
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.)
\\\
\\\


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PeterDonis
Mentor
Re: "Locally Lorentz"

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.

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.

Gold Member
Re: "Locally Lorentz"

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."

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PeterDonis
Mentor
Re: "Locally Lorentz"

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.

PeterDonis
Mentor
Re: "Locally Lorentz"

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.

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.)

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Gold Member
Re: "Locally Lorentz"

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.

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PeterDonis
Mentor
Re: "Locally Lorentz"

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.

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.

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.

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.

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."

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.

PeterDonis
Mentor
Re: "Locally Lorentz"

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.

Gold Member
Re: "Locally Lorentz"

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.

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.

Agreed.

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.

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."

Yes, but one that is the same for all possible (emitter, receiver) pairs, so it does nothing to distinguish them.

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.

Gold Member
Re: "Locally Lorentz"

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.
The problem with this is that your "given" line of simultaneity, while it sounds like "THE ONE" is actually pretty arbitrarily chosen. You are saying there is something special about the reference frame where both emitter and receiver is at rest. And, I agree that there is something special about that frame. Namely, you've got built in observers (the emitter and receiver) that are in that frame.

You're making kind of an odd claim, though, regarding this reference frame. You say that the number of wavelengths is an invariant as long as we always use the same line of simultaneity, which is of course true, but it's rather like saying, "the radio will always be at the same volume, so long as we never turn the volume knob."

Another issue I see with your "given line of simultaneity" is that it can only apply when you have the emitter and receiver moving on parallel world-lines. If the emitter and receiver were moving toward or away from each other, there is no built-in line of simultaneity which asserts itself as "THE ONE"

PeterDonis
Mentor
Re: "Locally Lorentz"

You should distinguish between "distance between objects" and "distance between events."
I disagree; any use of the term "distance" implies a distance between particular events. What you are calling "distance between objects" really means "distance between two particular events, one on each object's worldline, that are somehow picked out as seeming natural to measure the distance between." However, that notion in itself only makes sense if the interval between the two events is spacelike; see next comment.

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.
I think your terminology is confused. What I think you are really talking about here is *not* the distance between the emission event and the absorption event; as I said before, those events are separated by a null interval so the term "distance" between them makes no sense. However, there is a distance you could be referring to. Suppose I draw a line of simultaneity in the emitter's frame that passes through the absorption event. This line will cross the emitter's worldline at some event, and the distance between *that* event and the absorption event is well-defined, since the interval between them is spacelike. Also, if the receiver is moving relative to the emitter, that distance will indeed be different than the corresponding "distance" I would measure if I drew a line of simultaneity in the emitter's frame passing through the *emission* event, and measured the distance from that event to the event on the receiver's worldline where the "emission" line of simultaneity intersects it.

Of course, the "distance" I've just defined is observer-dependent; if I go through the same process in the receiver's frame, I will come up with different answers if the receiver is moving relative to the emitter. See next comment.

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.
Yes, they are. So what? I've never denied that distances, defined this way (more precisely, defined the way I did above) are observer-dependent. I'm talking about a different issue; the issue of what quantities can figure in the physical laws involved. Those quantities have to be invariants, which means you can't use observer-dependent quantities like the distance I defined above, *unless* there's a particular symmetry in the problem that picks out a state of motion. See my next post.

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.
I assume you mean a second in the car's frame, since in the snowflake's frame the snowflake's speed is zero. Yes, if the observer in the car wants to know the "distance" the snowflake was, in his frame, away from the car one second before it hit the car, he can just multiply speed by time as you say. But that will *not* give him any sort of "length along the snowflake's worldline", because the snowflake's worldline is timelike. What it will give him is a distance defined as I defined it above, between the event on the snowflake's worldline one second (in the car's frame) before it hit the car, and the event on the car's worldline that intersects the car line of simultaneity through the snowflake event I just described.

PeterDonis
Mentor
Re: "Locally Lorentz"

The problem with this is that your "given" line of simultaneity, while it sounds like "THE ONE" is actually pretty arbitrarily chosen. You are saying there is something special about the reference frame where both emitter and receiver is at rest. And, I agree that there is something special about that frame. Namely, you've got built in observers (the emitter and receiver) that are in that frame.
More precisely, it's the very fact that there *are* two observers who are at rest in the same frame (i.e., at rest relative to each other) that picks out that particular frame as special *for this problem*. Another way of stating the specialness is that this problem has a symmetry that the general problem lacks: both observers have the same lines of simultaneity (or, equivalently, their worldlines are parallel, as you put it later in your post). Given that symmetry in the problem, the particular frame that respects that symmetry is not "arbitrarily" chosen; it's naturally picked out by the symmetry.

You're making kind of an odd claim, though, regarding this reference frame. You say that the number of wavelengths is an invariant as long as we always use the same line of simultaneity, which is of course true, but it's rather like saying, "the radio will always be at the same volume, so long as we never turn the volume knob."
No, it's like saying that if there's a particular radio station I want to listen to, there's only one place I can put the tuning knob. Given the particular symmetry in the problem, the line of simultaneity is not arbitrarily chosen; there is a particular property that naturally picks it out.

Another issue I see with your "given line of simultaneity" is that it can only apply when you have the emitter and receiver moving on parallel world-lines. If the emitter and receiver were moving toward or away from each other, there is no built-in line of simultaneity which asserts itself as "THE ONE"
You're correct; the case where emitter and receiver are in relative motion is more complicated. I think the idea of "number of wave crests counted along some particular line" can be generalized to cover that case, but I haven't completely worked it through yet (though I suspect that the fact that I've only stated it so far in terms of space instead of spacetime is an issue--instead of counting wave crests along a line, the answer may be to count them in a spacetime area, which would translate into a 4-volume if we put back in the other 2 spatial dimensions we've been suppressing). I am pretty sure, though, that no notion of "distance" by itself, such as the one discussed in my last post, will work, because of the observer-dependence I mentioned there.

Gold Member
Re: "Locally Lorentz"

Code:
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.)
\\\
\\\

Peter,

From your description of what you think I mean, I can tell that this ascii picture was woefully insufficient to get my idea across, so I took the time to attempt to remedy the situation with the attached diagram.

As you can see, in the diagram on the left, the emitter is chasing the receiver, and vice-versa on the right. I'm not saying something about their motion relative to each other but about their motion relative to a given observer.

I don't want to go through and respond to your last response point-by-point, because if you've missed this idea, you won't make sense out of anything else I've said.

However, I do want to clarify a couple of ideas.

One, the difference between a distance between events, and a distance between objects.

If you draw a line of simultaneity in some observer's reference frame, intersecting two object's worldlines, this represents the distance between objects. This is an observer dependent quantity. If those objects happen to have parallel worldlines, this distance is subject to the lorentz contraction.

If you have the space-time coordinates of two events, (regardless of whether their separation is null, timelike, or space-like) you can calculate the distance between events by using the distance formula, (sqrt(x^2+y^2+z^2). This is also an observer dependent quantity.The distance between two events, is to be distinguished, also, from the space-time INTERVAL between the two events, which of course is an invariant sqrt(t^2-x^2-y^2-z^2).

There is one other aspect of your argument where I think you should take a bit more care, because at one point, I think you have suggested that the "distance between events" is meaningless unless those events have a space-like separation. Yet, you were able to fairly well describe the distance between time-like separated events from the frame of the car, and from the frame of the ground, and the frame of the snowflake, so I know you cannot really believe these ideas to be meaningless.

So what? I've never denied that distances, defined this way (more precisely, defined the way I did above) are observer-dependent. I'm talking about a different issue; the issue of what quantities can figure in the physical laws involved. Those quantities have to be invariants, which means you can't use observer-dependent quantities like the distance I defined above, *unless* there's a particular symmetry in the problem that picks out a state of motion. See my next post.
Here is where we may have a central disagreement, which is the question, "Do observer dependent quantities have any place in physics?" I would say, emphatically YES. In any actual description of phenomena, one invokes a hypothetical observer,from whose viewpoint one measures all of the phenomena. Often called "the lab frame." It's not a symmetry of the problem that picks out the state of motion. It's the velocity of the rulers and clocks and apparatus that you're using that picks out the state of motion.

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PeterDonis
Mentor
Re: "Locally Lorentz"

I took the time to attempt to remedy the situation with the attached diagram.
Thanks, the diagram and your clarifications make it clearer what you're using your terms to refer to. I don't think we have any disagreement about the physics of this particular scenario, only about what terminology is "appropriate" to describe it.

As you can see, in the diagram on the left, the emitter is chasing the receiver, and vice-versa on the right. I'm not saying something about their motion relative to each other but about their motion relative to a given observer.
Got it. No disagreement here.

If you draw a line of simultaneity in some observer's reference frame, intersecting two object's worldlines, this represents the distance between objects. This is an observer dependent quantity. If those objects happen to have parallel worldlines, this distance is subject to the lorentz contraction.
No disagreement here.

If you have the space-time coordinates of two events, (regardless of whether their separation is null, timelike, or space-like) you can calculate the distance between events by using the distance formula, (sqrt(x^2+y^2+z^2). This is also an observer dependent quantity.The distance between two events, is to be distinguished, also, from the space-time INTERVAL between the two events, which of course is an invariant sqrt(t^2-x^2-y^2-z^2).
I understand now what you meant by "distance between events" (and how it's distinct from the interval, but I was never in any doubt that you meant something different than the interval by your "distance" terms). I would point out, however, that there is one key difference between this "distance between events" and the "distance between observers" as you defined the latter term above: the "distance between events" does not Lorentz transform like you expect a distance to transform. It can't possibly, because it's longer if the emitter and receiver are moving in one direction, and shorter if they're moving in the other direction. Lorentz contraction doesn't work like that, as I pointed out in an earlier post.

There is one other aspect of your argument where I think you should take a bit more care, because at one point, I think you have suggested that the "distance between events" is meaningless unless those events have a space-like separation. Yet, you were able to fairly well describe the distance between time-like separated events from the frame of the car, and from the frame of the ground, and the frame of the snowflake, so I know you cannot really believe these ideas to be meaningless.
But you'll note that I was careful to define "distance" in such a way that it was always measured along a spacelike line; I just obscured one aspect of that by defining the events such that one of them was on the worldline of the observer whose lines of simultaneity were being used (so the "spacelike distance" from the observer to that event was zero). The scenario in your diagrams above doesn't have that property, so one has to be more explicit in defining exactly what "distances" are being used. In your scenario, I would say that the "distance between events" that you've defined is actually the difference between two distances that are each measured along spacelike lines: the distance between the observer and the emission event (measured along the observer's line of simultaneity that passes through that event) and the distance between the observer and the absorption event (measured along the observer's line of simultaneity that passes through that event). Your formulation in terms of the differences in x, y, and z coordinates implicitly uses this definition, because the coordinate values at each event are defined as the components along each spatial coordinate axis of the distance from the observer's worldline to the event along the observer's line of simultaneity through that event.

Here is where we may have a central disagreement, which is the question, "Do observer dependent quantities have any place in physics?" I would say, emphatically YES. In any actual description of phenomena, one invokes a hypothetical observer,from whose viewpoint one measures all of the phenomena. Often called "the lab frame." It's not a symmetry of the problem that picks out the state of motion. It's the velocity of the rulers and clocks and apparatus that you're using that picks out the state of motion.
You may be misunderstanding the point I was making. I wasn't saying that observer-dependent quantities don't have any place in physics; I was saying only that observer-dependent quantities can't appear in the *laws* of physics. The law that determines what light intensity a particular observer will measure from a particular source is one of the laws in which observer-dependent quantities can't appear. So the correct statement of that law can't use "distance" in general (but see below for the special case where the emitter and receiver are at mutual rest), since that is observer-dependent (regardless of which of the definitions discussed above we use). However, it also can't use the spacetime interval between the emission and absorption events, because that's always zero even though different (source, receiver) pairs produce different measured intensities. So there must be some *other* invariant, observer-independent quantity that appears in the law.

I think you're also somewhat mis-stating the role of symmetries in a physical situation. For the special case we've been discussing, where the emitter and receiver are at mutual rest, there is a special state of motion in which the problem looks simplest: the state of motion which corresponds to the mutual rest frame of emitter and receiver. This is what I meant by saying that this particular problem (*not* the general problem, which allows emitter and receiver to be moving relative to one another) has a "symmetry"--it looks simpler if you pick a particular reference frame in which to describe it. In that frame, all the different kinds of distances we've been discussing are equal: the "distance between observers" is the *same* as the "distance between emission and absorption events". This distance is also the "proper distance" between the emitter and the receiver, and "proper distance" has the same kind of special status in relativity as "proper time" does--it qualifies as an "invariant", because it can be defined in a coordinate-independent way, just as proper time can. (It's true that it can't be the general invariant that appears in the light intensity law I referred to above, because it only works for this special case, as I noted in a previous post. The general invariant, I think, is a generalization of this one, but I haven't completely worked it through yet.)