Is acceleration absolute or relative - follow up

In summary: Minkowski spacetime are moving inertially, but the frame as a whole is not inertial), and the pseudorandom KVFs (corresponding to objects that are moving but have a definite velocity, but the velocity is not constant).Any static congruence is hypersurface orthogonal. That's part of the definition of "static".Most static spacetimes have only one timelike KVF (as, for example, Schwarzschild spacetime), but not all. Minkowski spacetime has two infinite families of them: the inertial KVFs (corresponding to... objects at rest in Minkowski spacetime are moving inertially, but the frame
  • #1
cianfa72
1,962
216
TL;DR Summary
Is acceleration absolute or relative - follow up: some doubts about static congruence and global coordinate chart for spacetime
Hello,

Some doubt arose me reading this thread https://www.physicsforums.com/threa...ute-or-relative-revisited.999420/post-6454462 currently closed. Sorry, I have not be able to quote directly from it :frown:

Your claim is not , however, asserting that the spacetime geometry changes when you change coordinates. And the fact that the universe is "static" and the bucket is not can be expressed as invariant properties of the spacetime geometry and particular families of worldlines within it. For example (since this is an "I" level thread, some technical jargon is not inappropriate), the family of worldlines describing the motion of objects "at rest relative to the universe" will be integral curves of a timelike Killing vector field that is hypersurface orthogonal (which is what "static" translates to in more technical GR language); whereas the family of worldlines describing the motion of the bucket will be integral curves of a timelike Killing vector field (assuming the bucket's angular velocity of rotation relative to the universe is constant) that is not hypersurface orthogonal (in more technical jargon, the bucket's motion will be stationary but not static).

From my understanding in a static spacetime there is only one 'timelike Killing Vector Field (KVF) congruence' that is hypersurface orthogonal. There can be multiple "static" congruence of KVFs however they are actually not hypersurface orthogonal.In a curved spacetime, such as the spacetime of our actual visible universe, there are no global inertial frames. The global frame we use to describe our universe is not inertial. Objects at rest in this frame are moving inertially (zero proper acceleration), but the frame as a whole is not inertial.

That does mean -- in the chosen global coordinate chart (frame) -- objects at rest (i.e. objects having worldlines described by fixed coordinate time) that move inertially actually have a not null geodesic deviation though
 
Last edited:
Physics news on Phys.org
  • #2
cianfa72 said:
In a curved spacetime, such as the spacetime of our actual visible universe, there are no global inertial frames. The global frame we use to describe our universe is not inertial. Objects at rest in this frame are moving inertially (zero proper acceleration), but the frame as a whole is not inertial.

That does mean -- in the chosen global coordinate chart (frame) -- objects at rest (i.e. objects having worldlines described by fixed coordinate time) that move inertially actually have a not null geodesic deviation though

I don't quite understand your terminology, but two properties of a global inertial coordinate system are:

  1. If an object is traveling inertially, and it is initially at rest, then it will remain at rest.
  2. The distance between objects at rest remains constant.

where "at rest" means that the spatial coordinates are not changing with time.

The universe as a whole has property 1 if you choose a coordinate system in which the expansion of the universe looks isotropic. But it doesn't have property 2. The distance between stars that are both "at rest" according to that coordinate system will not remain constant.
 
  • Like
Likes vanhees71
  • #3
stevendaryl said:
I don't quite understand your terminology, but two properties of a global inertial coordinate system are:
  1. If an object is traveling inertially, and it is initially at rest, then it will remain at rest.
  2. The distance between objects at rest remains constant.
where "at rest" means that the spatial coordinates are not changing with time.

The universe as a whole has property 1 if you choose a coordinate system in which the expansion of the universe looks isotropic. But it doesn't have property 2. The distance between stars that are both "at rest" according to that coordinate system will not remain constant.
Surely, the point I was trying to make is that if we chose a global coordinate system (chart) for which property 1. holds -- namely objects at rest in it are moving inertially -- then property 2. cannot hold because of geodesic deviation (i.e. spacetime is actually curved)
 
Last edited:
  • #4
cianfa72 said:
Surely, the point I was to make clear is that if we chose a global coordinate system (chart) for which property 1 holds -- namely objects at rest in it are moving inertially -- then property 2. cannot be true because of geodesic deviation (i.e. spacetime is actually curved)

Yes, that's true.

In any case, I don't see how this relates to the thread question of whether acceleration is absolute or relative. If everybody computes acceleration relative to geodesics, then everybody will agree about whether an object is accelerating (although the components of acceleration will depend on the coordinate system).
 
  • #5
cianfa72 said:
From my understanding in a static spacetime there is only one 'timelike Killing Vector Field (KVF) congruence' that is hypersurface orthogonal. There can be multiple "static" congruence of KVFs however they are actually not hypersurface orthogonal.

Any static congruence is hypersurface orthogonal. That's part of the definition of "static".

Most static spacetimes have only one timelike KVF (as, for example, Schwarzschild spacetime), but not all. Minkowski spacetime has two infinite families of them: the inertial KVFs (corresponding to the infinite number of possible global inertial frames) and the Rindler KVFs (corresponding to the infinite number of possible Rindler congruences). All of the KVFs in both of these infinite families are hypersurface orthogonal.

de Sitter spacetime also has one (AFAIK) infinite family of static KVFs.

As I noted in what you quoted, the "rotating bucket" congruence in Minkowski spacetime, the Langevin congruence (there are actually an infinite family of these as well), is stationary but not static. It is not hypersurface orthogonal, but that does not mean there are static congruences that are not hypersurface orthogonal. It just means that congruence is not static.

cianfa72 said:
That does mean -- in the chosen global coordinate chart (frame) -- objects at rest (i.e. objects having worldlines described by fixed coordinate time) that move inertially actually have a not null geodesic deviation though

Yes.
 
  • Like
Likes vanhees71
  • #6
stevendaryl said:
In any case, I don't see how this relates to the thread question of whether acceleration is absolute or relative. If everybody computes acceleration relative to geodesics, then everybody will agree about whether an object is accelerating (although the components of acceleration will depend on the coordinate system).
ok, surely.

PeterDonis said:
Most static spacetimes have only one timelike KVF (as, for example, Schwarzschild spacetime), but not all. Minkowski spacetime has two infinite families of them: the inertial KVFs (corresponding to the infinite number of possible global inertial frames) and the Rindler KVFs (corresponding to the infinite number of possible Rindler congruences). All of the KVFs in both of these infinite families are hypersurface orthogonal.
Does the existence of (at least one) timelike KVF 'static' congruence basically define a spacetime as 'static' ?
 
  • #7
cianfa72 said:
Does the existence of (at least one) timelike KVF 'static' congruence define a spacetime as 'static' ?

Yes. In the literature the fact that there are some spacetimes that have multiple timelike KVFs, some of which are hypersurface orthogonal and some of which are not, is not really discussed, because for any particular problem, if one is interested in a particular timelike KVF, the problem will make it clear which one and that's the only one anyone is concerned about.

Also, note that the Langevin congruence in Minkowski spacetime, which is, by the terminology we have been using, stationary but not static, is only timelike in a sufficiently small neighborhood of the axis of rotation; how small a neighborhood depends on the angular velocity ##\omega## of the congruence. At some radius ##r## from the axis, ##\omega r = c## (the speed of light), and at that radius the KVF defined by the congruence is null; outside that radius it is spacelike. By contrast, the inertial KVFs are timelike everywhere, and the Rindler KVFs are timelike everywhere inside the "wedge" in which they are standardly defined, i.e., they remain timelike all the way out to infinity. Usually, for a KVF to be recognized as "stationary" or "static" in the literature, it needs to be timelike at infinity.
 
  • Like
Likes vanhees71
  • #8
PeterDonis said:
Also, note that the Langevin congruence in Minkowski spacetime, which is, by the terminology we have been using, stationary but not static, is only timelike in a sufficiently small neighborhood of the axis of rotation; how small a neighborhood depends on the angular velocity ##\omega## of the congruence. At some radius ##r## from the axis, ##\omega r = c## (the speed of light), and at that radius the KVF defined by the congruence is null; outside that radius it is spacelike. By contrast, the inertial KVFs are timelike everywhere, and the Rindler KVFs are timelike everywhere inside the "wedge" in which they are standardly defined, i.e., they remain timelike all the way out to infinity. Usually, for a KVF to be recognized as "stationary" or "static" in the literature, it needs to be timelike at infinity.
Thus, a spacetime is defined as 'stationary' by the existence of (at least one) timelike KVF congruence; then if (at least one of) those timelike KVFs congruence is also hypersurface orthogonal (i.e. it is actually a 'static' congruence) then the spacetime is defined also to be 'static'.
 
  • #9
cianfa72 said:
Thus, a spacetime is defined as 'stationary' by the existence of (at least one) timelike KVF congruence; then if (at least one of) those timelike KVFs congruence is also hypersurface orthogonal (i.e. it is actually a 'static' congruence) then the spacetime is defined also to be 'static'.

Yes.
 
  • #10
cianfa72 said:
Summary:: Is acceleration absolute or relative - follow up: some doubts about static congruence and global coordinate chart for spacetimeFrom my understanding in a static spacetime there is only one 'timelike Killing Vector Field (KVF) congruence' that is hypersurface orthogonal. There can be multiple "static" congruence of KVFs however they are actually not hypersurface orthogonal.

I don't think this is right at all.

Consider Minkowskii space-time. There are an infinite number of inertial frames of reference, each of which will have it's own, different, timelike KVF that spans the entire space-time and is timelike everywhere. And the integral curves of each of these KVF's are different congruences.

Accelerated frames of reference, for specificity consider the Rindler line element $$-z^2 dt^2 + dx^2 + dy^2 + dz^2$$ will also have a _different_ Killing vector field that's timelike over some region of space-time, but is not globally timelike. For instance, for the above metric, there is a KVF , ##\frac {\partial}{\partial t} ## that's timelike for z>0, becomes null at z=0.

Rotating reference frames (I'm not going to bother with the line element unless there is some interest, I'd have to look it up) similarly have KVF's that are different from the others I"ve mentioned, but also are not time-like everywhere.

[[Note: corrected omitted minus sign in the Rindler metric from original post and made a few other minor changes]]
 
Last edited:
  • Like
Likes vanhees71
  • #11
pervect said:
I don't think this is right at all.

You're right, it isn't. See my posts #5 and #7.
 
  • #12
I would say that a region of space-time is defined as static if that region has at least one timelike Killing vector field.

Consider the Schwarzschild space-time, for instance. The exterior region (outside the event horizon) is clearly static. But I don't think the interior region can be regarded as static. It's possible there is some timelike KVF in the interior region, but I'm not aware of it. Certainly, MTW describes the interior region of the Schwarzschild space-time as a dynamic non-traversible wormhole. I Unfortunately, I don't have a definitive reference or mathematical proof of the absence of a timelike KVF in the interior region.
 
  • #13
pervect said:
I would say that a region of space-time is defined as static if that region has at least one timelike Killing vector field.

This is the definition of "stationary". For "static", the timelike KVF must be hypersurface orthogonal.

pervect said:
I don't think the interior region can be regarded as static.

That is the standard usage, to the best of my understanding, yes; only the region outside the horizon is considered static.

pervect said:
It's possible there is some timelike KVF in the interior region

No, there isn't. The 4th KVF (other than the 3 spacelike ones from spherical symmetry) is still there, but it's spacelike in the interior (and null on the horizon).

pervect said:
I don't have a definitive reference or mathematical proof of the absence of a timelike KVF in the interior region.

Birkhoff's Theorem is sufficient, since it proves that the geometry of any vacuum spherically symmetric region of spacetime must be the Schwarzschild geometry, and all of the KVFs of that geometry are known.
 
  • Like
Likes vanhees71
  • #14
pervect said:
Rotating reference frames (I'm not going to bother with the line element unless there is some interest, I'd have to look it up) similarly have KVF's

A KVF is not a property of a reference frame; it's a property of a spacetime geometry. The "rotating" KVFs in Minkowski spacetime (whose integral curves are the infinite family of Langevin congruences) are there in any frame; it's just easiest to show that they are KVFs in the appropriate Born chart. (And, as both you and I have noted, these KVFs are only timelike in a region sufficiently close to the axis of rotation.) However, even in standard inertial coordinates on Minkowski spacetime, one can show that a "rotating" KVF is in fact a KVF by observing that it is a linear combination, with constant coefficients, of an "inertial" KVF and one of the 3 KVFs that arise from spherical symmetry, and applying the theorem that any linear combination of KVFs with constant coefficients is also a KVF.
 
  • #15
PeterDonis said:
A KVF is not a property of a reference frame; it's a property of a spacetime geometry. The "rotating" KVFs in Minkowski spacetime (whose integral curves are the infinite family of Langevin congruences) are there in any frame; it's just easiest to show that they are KVFs in the appropriate Born chart. (And, as both you and I have noted, these KVFs are only timelike in a region sufficiently close to the axis of rotation.) However, even in standard inertial coordinates on Minkowski spacetime, one can show that a "rotating" KVF is in fact a KVF by observing that it is a linear combination, with constant coefficients, of an "inertial" KVF and one of the 3 KVFs that arise from spherical symmetry, and applying the theorem that any linear combination of KVFs with constant coefficients is also a KVF.

I didn't mean to imply that Killing vectors dependent on such things as coordinate choices or frames of reference - they do not. . I'd point to Killings equation as the formal definition.

The vector ##x^i## is a Killing vector if and only if ##\nabla_a x_b + \nabla_b x_a = 0##, where ##\nabla_a## is the covariant derivative.

Killing's equation can also be written in terms of the Lie derivative of the metric tensor being zero, though I don't usually think of it that way myself. That formulation may have some advantages, the covariant derivative needs a connection to define, while I believe the Lie derviative definition does not.

The above equation tells us if a vector field is a Killing vector field. The sign of x⋅x= ##g_{ab} x^a x^b## determines if the vector field at some particular point is timelike, spacelike, or null.

The only point I wanted to make is that the Schwazschild space-time is static in the exterior region, but not necessarily in the interior region. The vector field defined by ##\partial / \partial t## is a Killing vector field, that is defined everywhere in the Schwarzschild space-time, but this vector field is time-like only in the exterior region.

I don't believe there are any time-like Killing vector field in the interior region of the Schwarzschild metric, but I could be mistakedn - the fact that ##\partial / \partial_t## is not timelike doens't necessarily mean that there isn't some other Killiing vector field which is. But I don't believe there is.
 
Last edited:
  • #16
pervect said:
I don't believe there are any time-like Killing vector field in the interior region of the Schwarzschild metric, but I could be mistaken

I've already addressed this in post #13, but I'll run through the proof in a bit more detail. The Lie derivative formulation actually makes it easy: the Lie derivative of the metric along integral curves of a KVF is zero. So all we need to do is look at what the metric depends on.

The first thing to note is that, by spherical symmetry, we can factor out the angular part of the metric, since the 3 KVFs in that part are already known--the 3 KVFs associated with spherical symmetry. (More precisely, this is a 3-parameter group of KVFs.) So all we need to look at is the ##t##, ##r## part of the metric for additional KVFs.

Since the metric coefficients are different for every distinct value of ##r##, it is impossible for the Lie derivative of the metric to be zero along any curve that does not have constant ##r##. (Actually, there is one exception to this--can you see what it is? And why it does not affect the conclusion I am about to state?) So the only possible integral curves of a KVF are the curves of constant ##r##. And that just gives us the KVF we already know; there are no other possible ones.

(Again, to be precise, since any linear combination of KVFs with constant coefficients is also a KVF, what we have, putting back the angular part of the metric, is a 4-parameter group of KVFs. But we have shown that there cannot be any other KVFs that are linearly independent from the ones in this group, which is all we need to know.)
 
  • Like
Likes cianfa72 and vanhees71
  • #17
stevendaryl said:
I don't quite understand your terminology, but two properties of a global inertial coordinate system are:
  1. If an object is traveling inertially, and it is initially at rest, then it will remain at rest.
  2. The distance between objects at rest remains constant.
where "at rest" means that the spatial coordinates are not changing with time.

The universe as a whole has property 1 if you choose a coordinate system in which the expansion of the universe looks isotropic. But it doesn't have property 2. The distance between stars that are both "at rest" according to that coordinate system will not remain constant.
Sorry to resume this thread, I've a doubt about the definition of "distance" to employ. We said that objects "at rest" in a given arbitrary coordinate chart (a.k.a. reference frame) mean that their spatial coordinates are not changing with coordinate time.

If we take two arbitrary objects traveling inertially (i.e. their worldlines are timelike geodesics) the difference between their spatial coordinates of course stay constant with coordinate time too. Hence which kind of "distance" makes sense to employ to define the geodesic deviation between them ?
 
  • #18
cianfa72 said:
If we take two arbitrary objects traveling inertially (i.e. their worldlines are timelike geodesics) the difference between their spatial coordinates of course stay constant with coordinate time too
This statement sounds false. Surely the difference between spatial coordinates depends on the simultaneity convention that is chosen. The fact that two objects are both following timelike geodesics says nothing about the coordinate system that will be used to determine their separation at any given coordinate time.

We can be just about as pathological about choosing simultaneity conventions as we please.
 
  • #19
cianfa72 said:
If we take two arbitrary objects traveling inertially (i.e. their worldlines are timelike geodesics) the difference between their spatial coordinates of course stay constant with coordinate time too. Hence which kind of "distance" makes sense to employ to define the geodesic deviation between them ?
You mean that a comet and the Earth, both traveling on geodescics of the solar system, should have a constant difference in spatial coordinates?
 
  • #20
jbriggs444 said:
This statement sounds false. Surely the difference between spatial coordinates depends on the simultaneity convention that is chosen. The fact that two objects are both following timelike geodesics says nothing about the coordinate system that will be used to determine their separation at any given coordinate time.

We can be just about as pathological about choosing simultaneity conventions as we please.
Not sure to grasp it, fix a coordinate chart that span the spacetime region of interest. Take a timelike geodesic "at rest" in the given coordinate chart (i.e. described by a set of constant spatial coordinates).

Do you mean that another timelike geodesic "at rest" in the given coordiante chart (described by a different set of constant spatial coordinates) cannot exist in a curved spacetime ?
 
  • #21
PeroK said:
You mean that a comet and the Earth, both traveling on geodescics of the solar system, should have a constant difference in spatial coordinates?
No, that actually was the point: in a curved spacetime that is impossible. So coming back to the question...the "distance" to employ could be simple as the difference between spatial coordiantes taken at the same coordinate time.

Property 1. and 2. of post #2 of @stevendaryl qualify a coordiante chart as a global inertial coordinate chart (global inertial reference frame). Actually It does exist just in case of flat spacetime.
 
  • Skeptical
Likes PeroK
  • #22
cianfa72 said:
No, that actually was the point: in a curved spacetime that is impossible. So coming back to the question...the "distance" to employ could be simple as the difference between spatial coordiantes taken at the same coordinate time.

Property 1. and 2. of post #2 of @stevendaryl qualify a coordiante chart as a global inertial coordinate chart (global inertial reference frame). Actually It does exist just in case of flat spacetime.
You've lost me. I have no idea what you are talking about.
 
  • #23
cianfa72 said:
I've a doubt about the definition of "distance" to employ
The best invariant definition of "constant distance" (which is all that is required for the definition @stevendaryl gave) is constant round-trip light travel time. Note that this is not the same as a complete definition of "distance"--it's just a definition of constant distance.
 
  • Like
Likes cianfa72
  • #24
PeterDonis said:
The best invariant definition of "constant distance" (which is all that is required for the definition @stevendaryl gave) is constant round-trip light travel time. Note that this is not the same as a complete definition of "distance"--it's just a definition of constant distance.
So in an arbitrary curved spacetime we're always able to find a global coordiante chart for which property 1. holds for each timelike geodesic (simply assigning fixed different spatial coordinates to bodies traveling inertially).

However employing the definition of "constant distance" as pointed out by @PeterDonis we have that property 2. may only hold in flat spacetime.
 
  • #25
cianfa72 said:
So in an arbitrary curved spacetime we're always able to find a global coordiante chart for which property 1. holds for each timelike geodesic (simply assigning fixed different spatial coordinates to bodies traveling inertially).
No, even this is not always the case, because timelike geodesics can intersect if we extend them far enough, and if they intersect at an event, they can't have different spatial coordinates at that event--which means that any coordinate chart that assigns them different spatial coordinates will no longer be valid at that event. In general, it is not guaranteed to be able to find any single coordinate chart that globally covers a curved spacetime, let alone a chart that satisfies property 1.
 
  • Like
Likes cianfa72
  • #26
PeterDonis said:
In general, it is not guaranteed to be able to find any single coordinate chart that globally covers a curved spacetime, let alone a chart that satisfies property 1.
ok, thus the whole point is that in an arbitrary curved spacetime there is not guarantee to be able to "disseminate" the entire space with free bodies moving inertially that do not intersect at any event (i.e. find out a geodesic timelike congruence filling the spacetime).
 
Last edited:
  • #27
cianfa72 said:
the whole point is that in an arbitrary curved spacetime there is not guarantee to be able to "disseminate" the entire space with free bodies moving inertially that do not intersect at any event (i.e. find out a geodesic timelike congruence filling the spacetime).
Yes.
 
  • #28
cianfa72 said:
Not sure to grasp it, fix a coordinate chart that span the spacetime region of interest. Take a timelike geodesic "at rest" in the given coordinate chart (i.e. described by a set of constant spatial coordinates).

Do you mean that another timelike geodesic "at rest" in the given coordiante chart (described by a different set of constant spatial coordinates) cannot exist in a curved spacetime ?
Nothing so complicated.

You speak of a timelike geodesic that is "at rest" in a given coordinate chart. What makes you think that such a thing exists? One can pick a coordinate chart where objects following timelike geodesics have non-zero coordinate accelerations.
 
  • Like
Likes cianfa72
  • #29
jbriggs444 said:
You speak of a timelike geodesic that is "at rest" in a given coordinate chart. What makes you think that such a thing exists? One can pick a coordinate chart where objects following timelike geodesics have non-zero coordinate accelerations.
You're right. As said by @PeterDonis in post #25 in an arbitrary spacetime there is no guarantee to be able to find a global coordinate chart in which any timelike geodesic is actually "at rest" in it.
 
  • #30
jbriggs444 said:
You speak of a timelike geodesic that is "at rest" in a given coordinate chart. What makes you think that such a thing exists?
You can always find a coordinate chart in which a single timelike geodesic has constant spatial coordinates. Just construct Fermi normal coordinates centered on that geodesic. The chart might not cover the entire spacetime, but it will always cover the entire range of that single geodesic.

What you cannot always do is find a coordinate chart in which an entire family of timelike geodesics has constant spatial coordinates, since the geodesics might cross.

cianfa72 said:
As said by @PeterDonis in post #25 in an arbitrary spacetime there is no guarantee to be able to find a global coordinate chart in which any timelike geodesic is actually "at rest" in it.
No, that's not what I said in post #25. See above.
 
  • Like
Likes vanhees71
  • #31
PeterDonis said:
You can always find a coordinate chart in which a single timelike geodesic has constant spatial coordinates. Just construct Fermi normal coordinates centered on that geodesic. The chart might not cover the entire spacetime, but it will always cover the entire range of that single geodesic.
Right. If you fix the geodesic you can pick a chart that gives it constant spatial coordinates.

But if you fix a chart, you may find that there is no timelike geodesic which has constant spatial coordinates in that chart. (for instance Rindler coordinates).
 
  • #32
jbriggs444 said:
if you fix a chart
My understanding is that the OP is allowing free choice of coordinate charts, not trying to fix any particular chart.
 
  • Like
Likes SiennaTheGr8 and cianfa72
  • #33
PeterDonis said:
My understanding is that the OP is allowing free choice of coordinate charts, not trying to fix any particular chart.
Right. If he allows free choice of charts, I am allowed to fix a chart that violates his claims about it.
 
  • #34
jbriggs444 said:
If he allows free choice of charts, I am allowed to fix a chart that violates his claims about it.
I don't think he's claiming that every chart must have certain properties. I think he's just trying to figure out whether any chart can have certain proprties.
 
  • Like
Likes cianfa72
  • #35
PeterDonis said:
What you cannot always do is find a coordinate chart in which an entire family of timelike geodesics has constant spatial coordinates, since the geodesics might cross.

No, that's not what I said in post #25. See above.
My point was to say that there is no guarantee to be able to find a global coordinate chart in which any (an entire family of) timelike geodesic is actually "at rest" in it.
 

Similar threads

  • Special and General Relativity
Replies
16
Views
2K
  • Special and General Relativity
2
Replies
57
Views
2K
  • Special and General Relativity
Replies
14
Views
687
  • Special and General Relativity
Replies
31
Views
2K
  • Special and General Relativity
3
Replies
78
Views
5K
  • Special and General Relativity
Replies
12
Views
1K
  • Special and General Relativity
Replies
8
Views
969
  • Special and General Relativity
Replies
7
Views
1K
  • Special and General Relativity
Replies
23
Views
3K
  • Special and General Relativity
Replies
16
Views
2K
Back
Top