Deriving the Lorentz formula from my own example

[JohnnyGui]

Actually, it was an issue with the browser on the particular machine I was using. On a different machine I can see the formulas, and that showed me that there was a formatting error in one of them. Fixed now.

Ah, thanks a lot. It now makes sense to me.

I have a question though that could sound a bit strange. I noticed that the factor by which the time time dilates is based on the diagonal path of the light signal and the factor by which the length contracts is based on the parallel path. Why couldn't it be so that the factor by which the time dilates is based on the parallel light clock and that the length of the diagonal path get influenced instead? In that case, I'd assume that the length of the diagonal path would have to get longer (since the path of the light signal in the parallel path is longer according to C) so that the light signal would need an equal time traveling the diagonal path as the parallel one. But lengthening the diagonal path shouldn't be a problem since in the real case there's a spatial disagreement as well, among A and C about the length between A en E.

Why isn't this possible?

 

[PeterDonis]

Why isn't this possible?

Because the length of the diagonal path is not a free variable. It's completely determined by two conditions of the problem:

(1) In the clock's rest frame, the light beam that goes from A to B and back moves in a direction exactly perpendicular to the direction of motion of observer C in that frame.

(2) In C's rest frame, A and B are moving to the right at velocity $v$, so in a time $t$ they cover a distance $vt$.

Putting those two facts together, plus the fact that distances perpendicular to the direction of relative motion are unaffected by a change of inertial frame, fixes the diagonal distance $D$; it must be equal to $L^2 + (vt)^2$. Nothing else is consistent with the above.

lengthening the diagonal path shouldn't be a problem since in the real case there's a spatial disagreement as well, among A and C about the length between A en E.

No; this "spacial disagreement" is not about the distance A covers in a time $t$ in C's frame. It's about the distance between A and E at a given instant of time in C's frame. Those two things are not the same, and length contraction only affects the second, not the first. The first is fixed by the relative velocity $v$. (Note that in C's frame, even though the distance from A to E is length contracted, both A and E still move a distance $vt$ to the right in a time $t$.)

 

[JohnnyGui]

No; this "spacial disagreement" is not about the distance A covers in a time $t$ in C's frame. It's about the distance between A and E at a given instant of time in C's frame. Those two things are not the same, and length contraction only affects the second, not the first. The first is fixed by the relative velocity $v$. (Note that in C's frame, even though the distance from A to E is length contracted, both A and E still move a distance $vt$ to the right in a time $t$.)

Ok, I think I get it. Since D is fixed by vt, if you're expecting D to change length you must say that vt will change as well but we have already described D by vt in the first place which leads to a vicious circle.

Something very random, but I think this discussion made me understand something about determining the age of galaxies or the age of events like supernovae in these galaxies when we receive their light. It always boggled me when they say that if a light of a supernova reaches our eyes, then it happened by its distance / c time ago. I always thought, because the universie is expanding, that while the light of the supernova is traveling towards us, the distance between us and the supernova is expanding and thus, light would take a lot longer to reach us than when the supernova actually happened. But apparently, that's the perspective of someone looking from "outside" the universe and seeing us and the supernova increase in distance apart. But since c is always constant, we would see that light is still coming towards us at c and thus say the supernova is distance / c time old. We're just measuring its age by our perspective.

Am I making any sense here?

 

[PeterDonis]

It always boggled me when they say that if a light of a supernova reaches our eyes, then it happened by its distance / c time ago.

This is really a sloppy way of talking, so it's actually good that it seemed fishy to you. But the way to fix it is not to pretend the universe isn't expanding and that somehow makes it ok to say that the light was emitted a distance / c time ago. Also, the issues involved here are not the same as the issues we've been discussing in this thread; everything we've discussed in this thread has assumed flat spacetime, but the spacetime of our actual universe is not flat.

There are actually several issues involved here. One is that the "distance" itself is not a direct measurement; it's inferred from other data. Some types of supernovas, as far as we can tell, have an absolute brightness (meaning, brightness as viewed from some standard distance away in flat spacetime) that is very uniform, so we can use them as "standard candles", and infer their distance from us by comparing their apparent brightness with their known absolute brightness. But that's still an inference, and there is unavoidable uncertainty associated with it.

Another issue is that, as you say, the universe is expanding, and that causes light traveling in the universe to redshift, so the apparent brightness of a supernova has to be adjusted to take into account the effect of the redshift, before we can infer a distance from it. Fortunately, we can often get a spectrum from the supernova, which gives us an independent measurement of its redshift, which makes it possible to do the adjustment I just described with reasonable confidence.

And yet another issue is that, since the universe is expanding, the distance to the supernova is not constant, so when doing all these calculations and adjustments you have to decide at what time you want the distance--the time of emission of the light, or the time of reception. The usual convention is to quote distances at the time of reception, i.e., the distance "now". But that distance is not the same as the actual travel time of the light times c, which means, conversely, that if you take the distance "now" that is usually quoted, and divide it by c, you will not get the actual travel time of the light. You will get a time that is longer, because you are basically assuming that the light had to travel all the distance "now", when in fact it only had to travel a shorter distance; the difference is the effect of the universe's expansion.

(There is also the question of what coordinates are being used in all this; I have assumed in the above that we are using the standard "comoving" coordinates that are used in cosmology, and that the supernova and we are both at rest in those coordinates, i.e., that we are both "comoving" objects. But that isn't actually true of the Earth, and it isn't exactly true of most supernovas either. So there are actually further corrections that have to be made to take that into account. I won't clutter up the discussion here with the details of those, but you should be aware of them.)

 

[JohnnyGui]

snip

I am indeed aware of the factors and influences such as the luminosity, redshift etc. I have simplified my description too much, sorry for that.

Let's say that the distance of the supernova at which the light was emitted is D1 and the distance of the supernova at which we receive the light is D2.
What I meant was calculating D1 which is done using its absolute and apparent luminosity and correcting for the redshift. Usually, we on Earth would say that the light has traveled for D1 / c time since the light was emitted at D1 distance from us. Now, suppose there's someone in space exactly in the middle between us and the supernova (at 0.5D1) on the track of the traveling light of the supernova towards the Earth. Thus, the observer sees the light emitted of the supernova passing him and going towards the Earth. The light would pass him in 0.5D1/c time (according to himself) but because of the expansion of the universe, that observer would see both the supernova and the Earth going away from him. So for that observer, he would see that, while the emitted light is traveling towards the Earth, the Earth has moved a bit further from him to a distance larger than D1 at which the light was emitted. And since c is constant for each observer, the observer would say that the emitted light of the supernova would take more than D1 / c time to reach the Earth.
So, from Earth's perspective that time would be just D1 / c but from the observer's perspective that time would be (D1 + (v_EarthD1 / c)) / c.

Again, this is all assuming having calculated D1 from luminosity after correcting for redshift and that the Earth, observer and the emitted supernova light at D1 are all on 1 straight line (flat universe). If using this scenario, wouldn't the above conclusion be correct?

 

[PeterDonis]

Usually, we on Earth would say that the light has traveled for D1 / c time since the light was emitted at D1 distance from us.

If by "usually" you mean "if the universe weren't expanding and spacetime were flat", then yes. But the universe is expanding and spacetime is flat. Just as the time D2 / c is longer than the actual time (in comoving coordinates) that the light travels, the time D1 / c is shorter than the actual time that the light travels.

for that observer, he would see that, while the emitted light is traveling towards the Earth, the Earth has moved a bit further from him to a distance larger than D1 at which the light was emitted. And since c is constant for each observer, the observer would say that the emitted light of the supernova would take more than D1 / c time to reach the Earth.

This logic isn't quite correct. But even if it were, it wouldn't show what you think it does. The exact same logic applies to an observer at the supernova itself. To this observer, the Earth is moving away, so if the Earth is a distance D1 away when the light is emitted, it will take the light longer than a time D1 / c to reach the Earth.

However, as I said, the logic isn't quite correct, because it isn't symmetric. From the viewpoint of the observer who is halfway between Earth and the supernova, according to the logic you have given, it should take a time 0.5 D1 / c for the supernova's light to reach him, because, while the supernova is moving away, that doesn't affect the speed of the light. Similarly, from the viewpoint of the observer on Earth, it should take the light a time D1 / c to reach him from the supernova, even though it takes longer from the viewpoint of an observer at the supernova. So your logic leads to the conclusion that the time the light travels, in comoving coordinates, is different for the receiver than for the emitter.

But that conclusion is wrong. It takes the light from the supernova longer than the time 0.5 D1 / c to reach the observer, just as it takes it longer than the time D1 / c to reach Earth, even from the viewpoint of an observer on Earth. Everything is homogeneous and isotropic in comoving coordinates.

 

[JohnnyGui]

But that conclusion is wrong. It takes the light from the supernova longer than the time 0.5 D1 / c to reach the observer, just as it takes it longer than the time D1 / c to reach Earth, even from the viewpoint of an observer on Earth. Everything is homogeneous and isotropic in comoving coordinates.

Ok, I thought that the difference in time duration among the receiver and the emitter would cause time dilation since it's one event (the light reaching the receiver).

I have a few questions that I think would make me understand this scenario better:
1. Are you saying that there isn't a disagreement about time duration among the receiver and emitter? (both say it would take longer than D1 / c time in the same amount?)
2. Would an observer at the supernova measure the speed of the light going towards Earth at c or at a speed greater than c since he's moving away in the opposite direction of the light?
3. At what speed would an observer in the middle between the Earth and the supernova measure the light of the supernova approaching him? If he measures it at c then that means that the Earth is also measuring the light approaching at c. If this is so, then I don't really get how the light should take more than D1 / c time to reach Earth. Light speed shouldn't be influenced by the expansion so how is this possible? Or do you mean that it's because distance D1 itself is intrinsically expanding as well?

 

[PeterDonis]

I thought that the difference in time duration among the receiver and the emitter would cause time dilation since it's one event (the light reaching the receiver).

The concept of "time dilation" that you're used to in SR doesn't apply in a curved spacetime.
"J. Random Hacker" <jrh@foobar.com>

1. Are you saying that there isn't a disagreement about time duration among the receiver and emitter? (both say it would take longer than D1 / c time in the same amount?)

If they are both using comoving coordinates, and they are both comoving observers, then yes, they will both agree on the time the light takes to travel, and both will say it is longer than D1 / c (and shorter than D2 / c).

2. Would an observer at the supernova measure the speed of the light going towards Earth at c or at a speed greater than c since he's moving away in the opposite direction of the light?

The speed of light when measured locally (i.e., measuring the speed of a light beam at your location) is always c. And it is only possible to measure speed locally in a curved spacetime; the question "what speed relative to me is a supernova moving that is a billion light years a way) does not have a well-defined answer.

3. At what speed would an observer in the middle between the Earth and the supernova measure the light of the supernova approaching him?

See above.

If he measures it at c then that means that the Earth is also measuring the light approaching at c.

Yes. See above.

If this is so, then I don't really get how the light should take more than D1 / c time to reach Earth.

Because, as I've said, you're trying to apply the rules of SR in a curved spacetime, and those rules don't work in a curved spacetime.

Light speed shouldn't be influenced by the expansion

Locally measured light speed is not influenced by the expansion. But when you try to interpret "light travels at c" as meaning "it should take light a time D1 / c to reach me from a supernova that is a distance D1 away when it emits the light", you are not talking about a locally measured light speed, so trying to reason as if it were a locally measured light speed won't work.

Or do you mean that it's because distance D1 itself is intrinsically expanding as well?

This is one way of thinking about it, heuristically, yes. But only heuristically. And it's still limited, because you are now trying to assign a well-defined meaning to the "distance" such that it works the way distances work in SR. That won't work.

 

[JohnnyGui]

Answers

Thanks a lot for the answers. I think I'm getting a grip on this. If I'm correct, can I say alternatively that the problem that I'm causing here is that I'm mixing the peculiar motion speed of light with the influences of universe expansion that causes to change that speed in a non-peculiar way?

And next to that, if an observer tries to measure the light speed from somewhere very far away where the light travels in a different factor of space-time curvature, that he would measure a different speed because there's a difference in the degree of curvature between the spacetime in which the observer stands and where the light comes from?

I'm sorry if I'm repeating this over and over again but I really want to understand this in a way that I can comprehend.

 

[PeterDonis]

if an observer tries to measure the light speed from somewhere very far away where the light travels in a different factor of space-time curvature

How is the observer going to make that measurement?

 

[JohnnyGui]

How is the observer going to make that measurement?

Like this for example :

In which the curvature of where the supernova and the galaxy lie is of a different factor than the curvature of where the observer is. I'd expect that because of the difference in curvature, and because of the expansion of the spacetime fabric the observer would measure a different speed of light because it gets influenced by the expansion and curvature in a non-peculiar motion way?

Also, is my first statement in my post #39 correct?

 

[PeterDonis]

Like this for example

That doesn't tell me how the observer makes the measurement.

I'd expect that because of the difference in curvature, and because of the expansion of the spacetime fabric the observer would measure a different speed of light because it gets influenced by the expansion and curvature in a non-peculiar motion way?

You aren't thinking about how the observer is going to actually make the measurement. Think carefully, and remember that I said the locally measured speed of light is always c, and that the "speed" of a distant object relative to an observer is not well-defined in curved spacetime.

 

[PeterDonis]

is my first statement in my post #39 correct?

I think all of your post #39 needs to be reconsidered in the light of the questions I'm asking about how you would measure the "speed" of something (light or anything else) that's spatially distant from you.

 

[JohnnyGui]

You aren't thinking about how the observer is going to actually make the measurement. Think carefully, and remember that I said the locally measured speed of light is always c, and that the "speed" of a distant object relative to an observer is not well-defined in curved spacetime.

Ok, forget about the example I gave i post #41 which isn't really effective. I'd have thought that one could consider the correctness of my first conclusion in my post #39 (mixing the peculiar motion speed of light with the non-peculiar one by the expansion) independently from my second conclusion about curvature.

Regarding measuring the speed of an object/light distant away from you, wouldn't that be the redshift? I'm trying to figure out if there would ever be a situation in which someone would measure a different speed of light because of the expansion. Is the factor by which the emitted wavelength get lengthened (to get the observed wavelength) the same factor by which the speed of light gets slowed down by the expansion when seen from the perspective of an observer looking perpendicularly at the direction in which the light travels? (since the ratio of the observed and emitted wavelength is the same as the ratio of the difference in distances between emitting and receiving the light?)

Also, doesn't a curvature merely mean an increase in distance compared to a flat spacetime? How would that make the speed of a distant object relative to the observer not well-defined? Is it because the expansion would expand a curved spacetime at a more rapid rate than a flat one?

I'm probably overthinking this.

 

[PeterDonis]

Regarding measuring the speed of an object/light distant away from you, wouldn't that be the redshift?

Not in curved spacetime, no. In curved spacetime, there is no well-defined meaning to the "speed" of a distant object.

I'm trying to figure out if there would ever be a situation in which someone would measure a different speed of light because of the expansion.

The locally measured speed of light is always c. There is no other concept of "speed" that has a well-defined meaning.

Is the factor by which the emitted wavelength get lengthened (to get the observed wavelength) the same factor by which the speed of light gets slowed down by the expansion when seen from the perspective of an observer looking perpendicularly at the direction in which the light travels?

The speed of light doesn't get slowed down. See above.

the ratio of the observed and emitted wavelength is the same as the ratio of the difference in distances between emitting and receiving the light?

I don't understand what this means.

I'm probably overthinking this.

Yes, I think you are.

 

[JohnnyGui]

Not in curved spacetime, no. In curved spacetime, there is no well-defined meaning to the "speed" of a distant object.

Is it because a curved spacetime causes an expansion redshift as well as a gravitational redshift?
If so, aren't there any ways to distinguish them?

Apart from that I read that there is no such thing as relative velocity in distant GR. I have still yet to comprehend this. Is it because everything is "moving" because of the expansion and therefore there can't be an absolute inertial frame reference that can say that you're standing still with respect to anything?

 

[PeterDonis]

Is it because a curved spacetime causes an expansion redshift as well as a gravitational redshift?

No, because in a general curved spacetime, there is no invariant way to divide up an observed redshift into "expansion" and "gravitational" parts. (Or a "Doppler" part due to "relative speed" in the SR sense.) There is just the observed redshift.

I read that there is no such thing as relative velocity in distant GR. I have still yet to comprehend this. Is it because everything is "moving" because of the expansion and therefore there can't be an absolute inertial frame reference that can say that you're standing still with respect to anything?

Not quite; it's even stronger than that. The statement that there is no such thing as "relative velocity" between distant objects applies in any curved spacetime, not just the "expanding" one that describes our universe. It is because any inertial frame in any curved spacetime can only be local--covering a small patch of spacetime ("small" in both space and time) around a chosen event. And the concept of "relative velocity" only applies within a single local inertial frame.

 

[JohnnyGui]

Not quite; it's even stronger than that. The statement that there is no such thing as "relative velocity" between distant objects applies in any curved spacetime, not just the "expanding" one that describes our universe. It is because any inertial frame in any curved spacetime can only be local--covering a small patch of spacetime ("small" in both space and time) around a chosen event. And the concept of "relative velocity" only applies within a single local inertial frame.

Sorry if I'm deviating from this topic but I'm noticing I really need to read more on this. Particularly about why curved spacetime only allows local inertial frames.
I have a feeling this has to do something with the curvature of spacetime causing a unit of the spacetime metric being stretched. A moving object in that stretched metric would seem to move slower from the perspective of someone who is sitting in a less stretched spacetime?

 

[PeterDonis]

I have a feeling this has to do something with the curvature of spacetime causing a unit of the spacetime metric being stretched.

I'm not sure what this would mean; the spacetime metric is what tells you the actual "unit" of distance (and time), so thinking of that unit as being "stretched" doesn't make sense. I suspect you are trying to think of coordinates as having some physical meaning. They don't.

 

[JohnnyGui]

I'm not sure what this would mean; the spacetime metric is what tells you the actual "unit" of distance (and time), so thinking of that unit as being "stretched" doesn't make sense. I suspect you are trying to think of coordinates as having some physical meaning. They don't.

But isn't that unit of distance (and time) larger in a curved spacetime than in a flat one?

 

[PeterDonis]

isn't that unit of distance (and time) larger in a curved spacetime than in a flat one?

The concept of the unit being "larger" or "smaller" doesn't make sense. The unit is the unit.

 

[JohnnyGui]

The concept of the unit being "larger" or "smaller" doesn't make sense. The unit is the unit.

Ah of course. Can one then say that a curved spacetime metric consists of more units than a non- or less-curved one? If not, then what is it that makes an object that is moving in a curved spacetime, differ in velocity from when it's moving in a flat spacetime, to the extent that one can't talk about relative velocity in a curved spacetime?

In the mean time, I'll try and search for some info on this since I think that I'm asking too many questions about this here.

 

[PeterDonis]

Can one then say that a curved spacetime metric consists of more units than a non- or less-curved one?

No, because there is no way of making the comparison. You can't use coordinates to do it, because the fact that two events in different spacetimes happen to have the same 4-tuple of coordinate numbers assigned to them has no physical meaning. And there is no other way to do it.

what is it that makes an object that is moving in a curved spacetime, differ in velocity from when it's moving in a flat spacetime

There's no way of making this comparison either, so the question you're asking doesn't make sense.

to the extent that one can't talk about relative velocity in a curved spacetime?

The reason one can't talk about relative velocity in curved spacetime is that the spacetime is curved. That is the difference between curved and flat spacetime that makes the concept of "relative velocity" inapplicable except locally in curved spacetime. There is no other possible comparison.

As for why curvature of the spacetime is the key property here, that's probably getting too involved for a PF thread, but I'll try. Consider how we actually compare velocities between distant objects in flat spacetime: we use an inertial frame that covers the entire spacetime. But why does the inertial frame cover the entire spacetime? Because we can take a whole fleet of observers, start them out all at rest relative to each other and moving inertially, and they will stay at rest relative to each other forever. So if observer A, over here, wants to know how fast some object is moving that is just passing observer B, he can just ask observer B how fast the object is moving relative to him, and assume that the object's velocity relative to observer A himself will be the same.

But in curved spacetime, if we take two observers, start them out at rest relative to each other, and have them move inertially, they will not stay at rest relative to each other. That is because spacetime curvature is the same thing as tidal gravity, and tidal gravity causes inertially moving objects that start out at rest relative to each other to not stay at rest relative to each other. So there is no longer any invariant way, in a curved spacetime, to relate the speed that something is moving relative to observer B, to a speed relative to observer A, because observers A and B themselves can no longer form a global inertial frame the way they could in flat spacetime.

 

[JohnnyGui]

As for why curvature of the spacetime is the key property here, that's probably getting too involved for a PF thread, but I'll try. Consider how we actually compare velocities between distant objects in flat spacetime: we use an inertial frame that covers the entire spacetime. But why does the inertial frame cover the entire spacetime? Because we can take a whole fleet of observers, start them out all at rest relative to each other and moving inertially, and they will stay at rest relative to each other forever. So if observer A, over here, wants to know how fast some object is moving that is just passing observer B, he can just ask observer B how fast the object is moving relative to him, and assume that the object's velocity relative to observer A himself will be the same.

But in curved spacetime, if we take two observers, start them out at rest relative to each other, and have them move inertially, they will not stay at rest relative to each other. That is because spacetime curvature is the same thing as tidal gravity, and tidal gravity causes inertially moving objects that start out at rest relative to each other to not stay at rest relative to each other. So there is no longer any invariant way, in a curved spacetime, to relate the speed that something is moving relative to observer B, to a speed relative to observer A, because observers A and B themselves can no longer form a global inertial frame the way they could in flat spacetime.

Ah, that made me understand it. Great example. And even if B wanted to relate the speed of something moving over a long distance just for himself (not for telling A), he wouldn't be able to do this because he knows that he won't be in an inertial frame over a long distance.

Can I say that, since inertial frames over long distances isn't maintainable, that curvature is causing acceleration for any observer? (non-inertial frames is caused by acceleration)

 

[PeterDonis]

Can I say that, since inertial frames over long distances isn't maintainable, that curvature is causing acceleration for any observer?

Not as you state it, because "acceleration" is not a precise term. And if we use the standard GR definition of "acceleration", which is proper acceleration (i.e., acceleration that you feel), the statement is false; tidal gravity by itself does not cause objects to feel any acceleration, any more than the Newtonian "force" of gravity does. Any acceleration that is felt is always due to some non-gravitational interaction.

 

[JohnnyGui]

Not as you state it, because "acceleration" is not a precise term. And if we use the standard GR definition of "acceleration", which is proper acceleration (i.e., acceleration that you feel), the statement is false; tidal gravity by itself does not cause objects to feel any acceleration, any more than the Newtonian "force" of gravity does. Any acceleration that is felt is always due to some non-gravitational interaction.

Oh, I thought the curvature of the universe is caused by gravity, i.e. acceleration, and thus we should feel gravity/acceleration wherever there is curvature.

Does this mean that gravity curving spacetime is something else than the universe being curved?

 

[PeterDonis]

I thought the curvature of the universe is caused by gravity

Spacetime curvature is the same thing as tidal gravity. But "gravity" is a broader term than just tidal gravity. Also, spacetime curvature is caused by the presence of stress-energy, through the Einstein Field Equation; it is not caused by gravity in any sense.

i.e. acceleration

Gravity is not the same thing as acceleration.

we should feel gravity/acceleration wherever there is curvature.

Why would you think that? Even in the Newtonian approximation it's obviously false: objects can fall in the neighborhood of the Earth and exhibit the effects of tidal gravity without feeling any acceleration at all.

Does this mean that gravity curving spacetime is something else than the universe being curved?

See above.

I'm getting very curious as to where you are getting your ideas about relativity. Have you actually studied any textbooks, such as Taylor & Wheeler's Spacetime Physics, or Carroll's online lecture notes? Or have you only read pop science books or articles?

 

[JohnnyGui]

I'm getting very curious as to where you are getting your ideas about relativity. Have you actually studied any textbooks, such as Taylor & Wheeler's Spacetime Physics, or Carroll's online lecture notes? Or have you only read pop science books or articles?

You could see me as someone who's just interested in cosmology and relativity and who is watching lectures (Yale) and reading some introductory articles here and there along with general books. You could expect that people like me (just having an interest on the subject) are a bit prone to having some too basic (or even false) knowledge on this. For example, I've always heard from multiple sources that gravity in general is the cause of bending spacetime and that it is the same as acceleration to explain time dilation in its presence. You're the first one who denied these for me, surely because you have studied its causes in much more detail.

I see you have named a few reliable sources on this. Now that I know them I can dive into them