Locally Lorentz

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"Locally Lorentz"

Mister Thorne Wheeler, "Gravitation" asks "What does it mean to say that the geometry of a sufficiently limited region of spacetime in the real physical world is Lorentzian?"

The follow this up with two answers, neither of which appears to have much to do with the question. Instead, they give formulas to calculate proper time and proper distance. This barely scratches the surface of what it means to be Lorentz. Certainly, a body that takes an accelerated path through spacetime ages less than a body that takes an inertial path.

But what MTW fails to do is to get across a deeper undestanding of the paradox involved, and more particularly the resolution to this paradox. In fact, I can't find a single example in MTW where they demonstrate any competent expanation or deeper understanding of Special Relativity. They seem to start and end with the notion that Special Relativity is completely summed up by one equation:

[tex]s^2=-\tau^2=x^2+y^2+z^2-t^2[/tex]

I have no problem with having one equation to sum up Special Relativity, I just think the chose the wrong one. As for me, I don't think that SR is summed up by the calculation of the space-time-interval between events. Instead, it is summed up by the Lorentz Transformation Equations.

Let me offer a single example of what the Lorentz Transformations do. Then we can ask whether that operation represents a "global" or a "local" application of Lorentz, and whether it makes sense to say that physics is only "locally" Lorentz.


Here is the example:

I have a particle observer one foot away from a wall. Roughly 1 nanosecond ago, light came from the wall which is currently being observed by our observer.

The space-time coordinates of this observer is (ct, x) = (-1, 1); that is, 1 second ago, and one foot away.

Now, our observer undergoes a tremendous acceleration toward the wall of 0.99999999959c. (This corresponds to a change in rapidity of 10. tanh(10)=0.99999999959c.

The event (-1,1) is transformed by Lorentz Transformation as

[tex]\left(
\begin{array}{cc}
\cosh (\varphi ) & -\sinh (\varphi ) \\
-\sinh (\varphi ) & \cosh (\varphi )
\end{array}
\right)=
\left(
\begin{array}{c}
-22026 \\
22026
\end{array}
\right)
[/tex]

So this event, which happened only 1 foot away, now happened 22026 feet away; that is, over four miles away.

In general with these large numbers, (velocities extremely close to c) it is a simple calculation, once you know the rapidity. If you have a rapidity of 100, then the multiple is 2*cosh(100)=2.68*1043. With a rapidity change of 1000, the multiple is 2*cosh(1000)=1.97*10434 feet (6.36*10414 light years.)


So...

Is this exampe local? The proper time between these two events was zero. The proper distance between these two events is zero. This would seem to be about as "local" as you can get.

However, by applying the Lorentz Transformation to these two events, we were able to make them as far apart as we wanted in coordinate space and time. A google google google google times as big as the universe. However, the proper time, and proper distance between these events remains zero.

So how can anyone justify even using the phrase "locally Lorentz" If any physics is Lorentz at all, it can be stretched over arbitrarily large swaths of spacetime--not local at all.
 

Answers and Replies

PAllen
Science Advisor
7,813
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Re: "Locally Lorentz"

Mister Thorne Wheeler, "Gravitation" asks "What does it mean to say that the geometry of a sufficiently limited region of spacetime in the real physical world is Lorentzian?"

The follow this up with two answers, neither of which appears to have much to do with the question. Instead, they give formulas to calculate proper time and proper distance. This barely scratches the surface of what it means to be Lorentz. Certainly, a body that takes an accelerated path through spacetime ages less than a body that takes an inertial path.

But what MTW fails to do is to get across a deeper undestanding of the paradox involved, and more particularly the resolution to this paradox. In fact, I can't find a single example in MTW where they demonstrate any competent expanation or deeper understanding of Special Relativity. They seem to start and end with the notion that Special Relativity is completely summed up by one equation:

[tex]s^2=-\tau^2=x^2+y^2+z^2-t^2[/tex]

I have no problem with having one equation to sum up Special Relativity, I just think the chose the wrong one. As for me, I don't think that SR is summed up by the calculation of the space-time-interval between events. Instead, it is summed up by the Lorentz Transformation Equations.

Let me offer a single example of what the Lorentz Transformations do. Then we can ask whether that operation represents a "global" or a "local" application of Lorentz, and whether it makes sense to say that physics is only "locally" Lorentz.


Here is the example:

I have a particle observer one foot away from a wall. Roughly 1 nanosecond ago, light came from the wall which is currently being observed by our observer.

The space-time coordinates of this observer is (ct, x) = (-1, 1); that is, 1 second ago, and one foot away.

Now, our observer undergoes a tremendous acceleration toward the wall of 0.99999999959c. (This corresponds to a change in rapidity of 10. tanh(10)=0.99999999959c.

The event (-1,1) is transformed by Lorentz Transformation as

[tex]\left(
\begin{array}{cc}
\cosh (\varphi ) & -\sinh (\varphi ) \\
-\sinh (\varphi ) & \cosh (\varphi )
\end{array}
\right)=
\left(
\begin{array}{c}
-22026 \\
22026
\end{array}
\right)
[/tex]

So this event, which happened only 1 foot away, now happened 22026 feet away; that is, over four miles away.

In general with these large numbers, (velocities extremely close to c) it is a simple calculation, once you know the rapidity. If you have a rapidity of 100, then the multiple is 2*cosh(100)=2.68*1043. With a rapidity change of 1000, the multiple is 2*cosh(1000)=1.97*10434 feet (6.36*10414 light years.)


So...

Is this exampe local? The proper time between these two events was zero. The proper distance between these two events is zero. This would seem to be about as "local" as you can get.

However, by applying the Lorentz Transformation to these two events, we were able to make them as far apart as we wanted in coordinate space and time. A google google google google times as big as the universe. However, the proper time, and proper distance between these events remains zero.

So how can anyone justify even using the phrase "locally Lorentz" If any physics is Lorentz at all, it can be stretched over arbitrarily large swaths of spacetime--not local at all.
You can say that the metric sums up everything because the complete Lorentz transform is derivable from the metric.

I think the definition of locality you want to use is one phrased in terms of invariants: proper distance and proper time; also any region over which the the metric deviation from Minkowski is insignificant. In your example, despite the huge coordinate distance, the metric remains flat over this region, so it is locally Lorentz.

(At least this is my understanding).
 
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You can say that the metric sums up everything because the complete Lorentz transform is derivable from the metric.

I think the definition of locality you want to use is one phrased in terms of invariants: proper distance and proper time; also any region over which the the metric deviation from Minkowski is insignificant. In your example, despite the huge coordinate distance, the metric remains flat over this region, so it is locally Lorentz.

(At least this is my understanding).
It is my understanding too.
 
bcrowell
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Re: "Locally Lorentz"

MTW isn't a book on SR, it's a book on GR. As PAllen points out, you can derive the Lorentz transformation from the metric or the metric from the metric from the Lorentz transformation. But in the context of GR, it's more natural to emphasize the metric, because a metric is what GR has. GR does not have frames of reference, except locally, so a transformation between frames of reference isn't a central concern.

-Ben
 
JDoolin
Gold Member
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Re: "Locally Lorentz"

MTW isn't a book on SR, it's a book on GR. As PAllen points out, you can derive the Lorentz transformation from the metric or the metric from the metric from the Lorentz transformation. But in the context of GR, it's more natural to emphasize the metric, because a metric is what GR has. GR does not have frames of reference, except locally, so a transformation between frames of reference isn't a central concern.

-Ben
I know how to go from the Lorentz Transformation to the metric, but not the other way around.

To go from the LT's to the metric, you must either select two events which (a) you could travel between; i.e. the d/c < t. or you can select two events (b) which you can't travel between, i.e. the distance/speed of light > t.

In the first case, you apply the LT so that the two events are in the same position, but at different times. The time between those two events in this new reference frame is the proper time.

In the second case, you apply the LT so that the two events are in different positions, but at the same time. The distance in this reference frame is the proper distance.

(by the way going from lorentz transformation to the metric isn't so much derivation as definition.)

There is of course, a third case, where the the distance/speed of light = 1. In which case, you cannot transform the events so they happen at the same time, or the same place. You can make them arbitrarily close in space, and arbitrarily close in time, but never zero. For this relationship, the "proper time" and "proper distance" are both zero.

I'm interested to see you go the other direction, deriving the Lorentz Transformation from the definition of proper time.
 
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Mentz114
Gold Member
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Re: "Locally Lorentz"

It is possible to find the transformation that leaves the proper length unchanged. In 2 dimensions,

[tex]
\left[ \begin{array}{c}
T' \\\
X' \end{array} \right]=
\left[ \begin{array}{cc}
a & b \\\
c & d\end{array} \right]
\left[ \begin{array}{c}
T \\\
X \end{array} \right]
[/tex]

[tex]
\begin{align*}
ds^2 &= T'^2-X'^2=(aT+bX)^2-(cT+dX)^2\\
&= T^2(a^2-c^2)-X^2(b^2-d^2)+2XT(ab-cd)
\end{align*}
[/tex]

If

[tex]
a=d=cosh(r), \ \ c=d=sinh(r)
[/tex]

then [itex]ds^2[/itex] is invariant under the transformation.
 
1,538
0
Re: "Locally Lorentz"

Here is the example:

I have a particle observer one foot away from a wall. Roughly 1 nanosecond ago, light came from the wall which is currently being observed by our observer.

The space-time coordinates of this observer is (ct, x) = (-1, 1); that is, 1 second ago, and one foot away.

Now, our observer undergoes a tremendous acceleration toward the wall of 0.99999999959c. (This corresponds to a change in rapidity of 10. tanh(10)=0.99999999959c.

The event (-1,1) is transformed by Lorentz Transformation as

[tex]\left(
\begin{array}{cc}
\cosh (\varphi ) & -\sinh (\varphi ) \\
-\sinh (\varphi ) & \cosh (\varphi )
\end{array}
\right)=
\left(
\begin{array}{c}
-22026 \\
22026
\end{array}
\right)
[/tex]

So this event, which happened only 1 foot away, now happened 22026 feet away; that is, over four miles away.

In general with these large numbers, (velocities extremely close to c) it is a simple calculation, once you know the rapidity. If you have a rapidity of 100, then the multiple is 2*cosh(100)=2.68*1043. With a rapidity change of 1000, the multiple is 2*cosh(1000)=1.97*10434 feet (6.36*10414 light years.)


So...

Is this exampe local? The proper time between these two events was zero. The proper distance between these two events is zero. This would seem to be about as "local" as you can get.
Do not mistake the location of the points on the manifold for the distance measure between those same points.
 
pervect
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Re: "Locally Lorentz"

You can think of it like this. Rotations are linear transformations that leave distances unchanged. The Lorentz transformation is a linear transformation that leaves the Lorentz interval unchanged.

There are transforms other than Lorentz transforms that leave the Lorentz interval unchanged. These, however, are combinations of the Lorentz transform with standard spatial rotations. If you restrict yourself to one space and one time dimension, there is only the Lorentz transform.

You might try reading "Space Time Physics" by Taylor & Wheeler for more backgorund, they explain this in more detail, taking more time to do it.

But if you want to have a go at it yourself, it's not terribly hard to derive the Lorentz transforms from the invariance of the Lorentz interval. It's easiest if you choose units such that the speed of light, c, is equal to 1.

You just need to look for a linear transformation

x' = ax + bt
t' = fx + gt

such that

x'^2 - t'^2 = x^2 - t^2

Expanding, you get

(a^2 - f^2) x^2 + (2ab-2fg) xt + (b^2 - g^2) t^2 = x^2 - t^2

From this you conclude a^2 - f^2 = 1 , b^2 - g^2 = -1, and ab=fg. It's easy enough to confirm that the Lorentz transforms (if they look unfamiliar, it's because they're the Lorentz transforms in geometric units where c=1, so all factors of c have been omitted) satisfy this.

a = gamma = 1/sqrt(1-v^2)
b = v*gamma = v/sqrt(1-v^2)
f = v*gamma = v/sqrt(1-v^2)
g = gamma = 1/sqrt(1-v^2)

It's a little more work (more than I care to do, and I'm not sure where to refer you to as a reference) that there aren't any other solutions that aren't equivalent to the above. (Replacing v by -v is something that falls in the category of equivalent, for instance).
 
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atyy
Science Advisor
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Re: "Locally Lorentz"

If the metric is everywhere diag(-1,1,1,1) then it is Minkowskian.

However, in GR, the metric obeys the Einstein field equations, and is specified to have signature 2. So it is not diag(-1,1,1,1) everywhere. However, there are coordinates where the metric is diag(-1,1,1,1) at any particular point, and the first derivatives of the metric also vanish (but not second derivatives), and so the metric is said to be locally Lorentz. The deviation from local Lorentzianess as one goes away from the point is specified by terms in Taylor series, so we understand exactly how locally Lorentz it is or isn't.
 
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Re: "Locally Lorentz"

I think the definition of locality you want to use is one phrased in terms of invariants: proper distance and proper time
Is there an invariant way of defining locality for lightlike directions?
 
PAllen
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Re: "Locally Lorentz"

Is there an invariant way of defining locality for lightlike directions?
Not that I can think of, but even if you pick a coordinate system with lightlike basis vectors, you can discuss the size of region in terms of proper spacelike interval and proper timelike interval.
 
JDoolin
Gold Member
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Re: "Locally Lorentz"

Do not mistake the location of the points on the manifold for the distance measure between those same points.
The only distances I know of are "proper distance," "proper time," and "distance." The last distance has no adjective, and it refers to the Euclidean, observer dependent distance. (It is observer dependent, because two observers with different rapidities will observe different distances.)

The "distance measure" as you put it, is an ambiguous term, since relativistically traveling observers would measure different distances between the same two events.

Of course, I am referring to the Euclidean, observer dependent distance, with the critera that the space is "locally lorentz," I applied a Lorentz Transform to a "local" space, and find that for all intents and purposes, local is global.
 
JDoolin
Gold Member
720
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Re: "Locally Lorentz"

You can think of it like this. Rotations are linear transformations that leave distances unchanged. The Lorentz transformation is a linear transformation that leaves the Lorentz interval unchanged.

There are transforms other than Lorentz transforms that leave the Lorentz interval unchanged. These, however, are combinations of the Lorentz transform with standard spatial rotations. If you restrict yourself to one space and one time dimension, there is only the Lorentz transform.

You might try reading "Space Time Physics" by Taylor & Wheeler for more backgorund, they explain this in more detail, taking more time to do it.

But if you want to have a go at it yourself, it's not terribly hard to derive the Lorentz transforms from the invariance of the Lorentz interval. It's easiest if you choose units such that the speed of light, c, is equal to 1.

You just need to look for a linear transformation

x' = ax + bt
t' = fx + gt

such that

x'^2 - t'^2 = x^2 - t^2

Expanding, you get

(a^2 - f^2) x^2 + (2ab-2fg) xt + (b^2 - g^2) t^2 = x^2 - t^2

From this you conclude a^2 - f^2 = 1 , b^2 - g^2 = -1, and ab=fg. It's easy enough to confirm that the Lorentz transforms (if they look unfamiliar, it's because they're the Lorentz transforms in geometric units where c=1, so all factors of c have been omitted) satisfy this.

a = gamma = 1/sqrt(1-v^2)
b = v*gamma = v/sqrt(1-v^2)
f = v*gamma = v/sqrt(1-v^2)
g = gamma = 1/sqrt(1-v^2)

It's a little more work (more than I care to do, and I'm not sure where to refer you to as a reference) that there aren't any other solutions that aren't equivalent to the above. (Replacing v by -v is something that falls in the category of equivalent, for instance).
Thanks pervect. (Thank you, too, Mentz, but pervect has a few extra equal signs so I looked at his first.)

So the lorentz transformation can be derived from the metric by asking "what set of transformations keep this quantity t^2 - x^2 constant?

Vice versa, the metric is derived from the lorentz transformation by asking "what is the minimum distance I can make these two events?" or "What would that clock measure, that goes between those two events" or "what would that ruler measure that goes between those two events."

The two things seem almost equivalent, but actually, I think there is an almost ideological difference between the two approaches. Going from the LT's to the metric, I am assuming a global geometric feature of the universe. Just as when I turn around, the universe will appear to undergo a rotational transformation, when I accelerate, the universe will undergo a Lorentz Transformation. Just as rotation applies equally to all objects at all distances, the Lorentz Transformation applies to all objects at all points in space and time. When I apply the lorentz transformation to derive the metric, I am using a universally applicable theory and finding the value of some local quantity.

On the other hand, if you go from the metric to the LT's, you are only concerned with the geometric features of nearby objects, traveling fairly slowly, along geodesics within a gravitational field. The lorentz transformations, themselves, do preserve these invariants, of the clock's time, or the ruler's length, but they are merely a mathematical curiosity, of little importance, since we are only concerned with what happens on earth.

bcrowell put it succinctly "GR does not have frames of reference, except locally, so a transformation between frames of reference isn't a central concern." I think bcrowell is correct, but I have often seen people claim that transformation between frames is not even a "valid" concern; they overstep the boundaries and say that Lorentz Transformations are only valid locallly.
 
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Re: "Locally Lorentz"

The only distances I know of are "proper distance," "proper time," and "distance." The last distance has no adjective, and it refers to the Euclidean, observer dependent distance. (It is observer dependent, because two observers with different rapidities will observe different distances.)
The metric distance between two points on the manifold is observer independent.

The "distance measure" as you put it, is an ambiguous term, since relativistically traveling observers would measure different distances between the same two events.
That is not true, the metric distance is not ambiguous and observer independent.

Of course, I am referring to the Euclidean, observer dependent distance, with the critera that the space is "locally lorentz," I applied a Lorentz Transform to a "local" space, and find that for all intents and purposes, local is global.
Then you clearly misapply the notion of 'locally Lorentzian'.
 
JDoolin
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Re: "Locally Lorentz"

Not that I can think of, but even if you pick a coordinate system with lightlike basis vectors, you can discuss the size of region in terms of proper spacelike interval and proper timelike interval.
I think I've seen somewhere, where they just take the space-time graph and rotateit it 45 degrees, and then the transformation becomes a contraction/expansion on the horizontal axis, and vice-vers on the vertical axis.

I think this constant sized region here involves four distinct events. Two photons leave a spot, hit something to turn around, and meet again. That would enclose a region of space-time that would have constant area under lorentz transformation.

But in general, no. If you have a photon going from one place to another, the best description of locality for it is zero. Effectively, there is a direct interaction between the particles at the origin, and the particles at the destination, with one caveat: The destination event definitely happens after the source event.

This is totally off-topic from the idea of local lorentz, but there is another interaction that I find interesting and related; when the photon is produced in, say, a Helium atom, this occurs when the electron falls into a lower shell, or perhaps when it hits the lower shell. Does the photon arise from the interaction of two particles, or does it arise from the acceleration of one particle? Is there a direct interaction between the proton and electron (one event?) or are they separated by a finite distance when they interact (two events?), or is the photon generated in a process that takes place in a region of time and space?

So (1) at the source, there are two events which occur at the same time, and (2) in between, there are events which definitely occur one after the other, and (3) at the destination, there are two events which occur at the same time.
 
1,384
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Re: "Locally Lorentz"

I think I've seen somewhere, where they just take the space-time graph and rotateit it 45 degrees, and then the transformation becomes a contraction/expansion on the horizontal axis, and vice-vers on the vertical axis.
Sounds like light-cone coordinates.
 
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Re: "Locally Lorentz"

If you have a photon going from one place to another, the best description of locality for it is zero. Effectively, there is a direct interaction between the particles at the origin, and the particles at the destination, with one caveat: The destination event definitely happens after the source event.
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.
 
JDoolin
Gold Member
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Re: "Locally Lorentz"

The metric distance between two points on the manifold is observer independent.


That is not true, the metric distance is not ambiguous and observer independent.


Then you clearly misapply the notion of 'locally Lorentzian'.

There's really no point in saying "locally lorentz," since the Lorentz Transformations apply to every event in spacetime, it is either globally Lorentz, or not Lorentz at all. Perhaps, MTW should use some other set of words to describe what they are talkng about.

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.

But claiming that physics is somehow "locally lorentz" implies that physics is not "globally lorentz." This is something that MTW tries to do throughout "Gravitation" is dismiss Special Relativity as being somehow incompatible with General Relativity. However, I have not yet found any logic in any of their arguments. Only weird claims, like this "locally lorentz" one.
 
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Re: "Locally Lorentz"

JDoolin you are misinformed, global Lorentz invariance only occurs in a flat spacetime.

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.

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.
 
JDoolin
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Re: "Locally Lorentz"

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.)
 
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JDoolin
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9
Re: "Locally Lorentz"

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.
 
660
0
Re: "Locally Lorentz"

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
 
27,067
7,297
Re: "Locally Lorentz"

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.

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.
 
JDoolin
Gold Member
720
9
Re: "Locally Lorentz"

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

[tex]\Delta \tau^2=\Delta t_{local}^2-\Delta x_{local}^2[/tex]

While the "globally lorentz" behavior applies to the global coordinates

[tex]\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}[/tex]​
 
JDoolin
Gold Member
720
9
Re: "Locally Lorentz"

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.
 

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