Space time curvature caused by fast electron

[PeterDonis]

Asymptotically flat space time can be described using flat geometry.

Really? Then please exhibit, explicitly, such a description. Asymptotically flat is not the same as flat.

You can't do that in GR because c is postulated to be always the same and valid geometries are restricted by this postulate.

Really? Why do you think so? It's true that *SR* postulates that c is always the same, but that's because SR only applies globally in a globally flat spacetime, including the postulate that "c is always the same". The rule that "nothing can go faster than light" in a curved spacetime must be generalized to "nothing can move outside the light cones", and the light cones in a curved spacetime can "tilt" from event to event, which is what gives the appearance of "c" changing. But all this can be perfectly well described by a geometry; it's just a curved geometry.

Yes, that's the point. And because of that we can't say that GR satisfies relativity principle.

No, we can't say that a generic spacetime in GR is globally flat, which means a global Lorentz transformation can't possibly apply, since those only apply in globally flat spacetimes. An asymptotically flat spacetime, such as that around a single gravitating body, is not globally flat.

But we *can* describe a generic spacetime in GR using a curved geometry (*which* curved geometry depends on the specific spacetime), and we *can* describe any given curved geometry using various coordinate charts, and transform between them. We can also show that any physical observable in GR (such as the spacetime curvature observed around a given object by an observer traveling on a given worldline) is described by an invariant, something that is the same in all coordinate charts.

If you want to say that all this does not satisfy the "relativity principle", that's your choice of words, I guess, but it doesn't affect the physics. Basically you are trying to apply the rules of SR outside their domain of applicability; those rules only apply globally in a globally flat spacetime, and you are trying to apply them to a curved spacetime. That doesn't work.

 

[Passionflower]

You can't do that in GR because c is postulated to be always the same and valid geometries are restricted by this postulate.

That is incorrect the speed of light varies in a gravitational field.

However I blame your mistake on educators who intentionally only talk about the local speed of light as if a speed between two objects some distance removed is not important at all.

 

[PeterDonis]

However I blame your mistake on educators who intentionally only talk about the local speed of light as if a speed between two objects some distance removed is not important at all.

Well, the educators do have something of a dilemma, since talking about the ways that you *can* compare "speeds" between distant objects in a curved spacetime also causes a lot of confusion, as plenty of threads on these forums bear witness.

 

[Passionflower]

Well, the educators do have something of a dilemma, since talking about the ways that you *can* compare "speeds" between distant objects in a curved spacetime also causes a lot of confusion, as plenty of threads on these forums bear witness.

True, but to reply to every single question with straight lips that the local speed of light is constant (period), is not helpful in my opinion.

For instance we all know that a test observer falling in at escape velocity in a Schwarzschild solution does that in a finite proper time. The reason far away observers do not see that is that the light at the event horizon has not enough speed to escape. In fact in a Schwarzschild solution we can use the factor 1-rs/r to determine the speed of light and this is directly related how we see the velocity at a distance:

[PLAIN]http://img254.imageshack.us/img254/5750/velocitycompare.png
In this plot the Schwarzschild radius is 1 and for a distant observer free falling observers come to a halt at the event horizon while locally (wrt to a stationary observer) they approach c.

 

[pervect]

So and how we can go from transformations defined in tangent space to transformations that are defined in manifold itself?
Say we take 2-d surface of a sphere and at some point we want to make x'=2x transformation. Obviously this will not work as global transformation.

If you have a local transformation, there's no logical necessity for it to define a global transformation. It's only part of a bigger entity.

However, it is logically necessary for a global transformation to define a local transformation.

If you use a coordinate basis for your local tangent space, the local coordinate labels you put on your tangent space are the same as your global labels, and the issue of going from global to local is trivial.

Example: suppose you use polar coordinates (r,theta) on a 2 dimensional manifold.

Then it has a cotangent space, the dual of a tangent space, defined by (dr, dtheta). There's a little bit of a trick here. You can think of dr as just being a number. But you can also think of dr as being a map from a vector in the tangent space to a number which gives the value of dr. It's in the later sense that dr is in the cotangent space. Strictly as a number, it would just be a scalar.

The dual of a vector in the cotangent space is a vector in the tangent space. So the dual of dr would be a vector in the tangent space, and that vector is called d/dr. Thus one can (and does) identify the vectors in the tangent space with partial derivatives.

If you have some global mapping that maps, say, (r,theta) -> (r', theta') the mapping from d/dr to d/dr' is just defined by the chain rule for partial derivatives.

It's often convenient to work in the cotangent space, and then you can just use algebra and partial derivatives and the chain rule to work out the mapping.

i.e if r' = g(r, theta)

$$dr' = \frac{\partial g}{ \partial r } \, dr + \frac{\partial g}{ \partial \theta} \, d \theta$$

But if you have some striclty local mapping, say a Lorentz Boost, that's defined in the tangent or cotangent space, there isn't necessarily any way to define it over the whole manifold. You might be able to come up with some mapping that has some global properties that you want, but you might not.

SO there isn't necessarily any global equivalent to a "Lorentz boost", i.e. a global transformation that acts as a Loretnz boost in ALL tangent spaces, but if you pick one particular tangent space, you can find a global transformation that acts like a Lorentz boost in that particular choice of tangent space.

 

[kmarinas86]

Hi everybody!

what happens if an electron passes by with a speed of, say, 99.999999999...% of the speed of light (relative to me). Its mass will then be enormous. Will this electron cause a relevant curvature of spacetime?

Consider a situation where an electron is accelerated by photons produced via a fusion event.

If you fuse two atoms of hydrogen together, the mass that results, the helium atom, is less. $m_0$ decreased. This change is frame-invariant ($m_0$ is decreased for all frames). However, the photon produced is also able produce curvature of space time. Yet, the photon does not have any $m_0$ because its $E$ equals its $|\mathbf{p}c|$. The energy of a photon is frame-dependent, but so are the $\gamma$'s, and thus also the relativistic kinetic energies, of the original two hydrogen atoms, in addition to that of the subsequent helium atom post-fusion. This frame-dependent energy is not what matters. What matters is the amount of rest mass that was lost to the photon. That is what determines the curvature caused by the existence of that photon. There is a reference frame when the corresponding energy of the rest mass that was lost (i.e. $-\Delta m_0c^2$) and the observed energy of the photon are equal. Let's now say that you accelerate an electron to light speed with bombardment of numerous photons onto that electron. Conclusion: The rest mass that was lost by masses to generate those photons which subsequently accelerate the electron determines how much additional curvature the electron produces (in a non-linear way of course).

 

[PeterDonis]

True, but to reply to every single question with straight lips that the local speed of light is constant (period), is not helpful in my opinion.

Just with a period, I don't disagree. But if the "local speed of light" is clarified to mean "the speed of light as seen by a local observer, i.e., an observer whom the light ray is just passing at the time the speed is measured", then the "local speed of light" *is* constant, and is always c.

For instance we all know that a test observer falling in at escape velocity in a Schwarzschild solution does that in a finite proper time. The reason far away observers do not see that is that the light at the event horizon has not enough speed to escape.

As seen by the faraway observers, yes (if one is willing to interpret the coordinate speed of the light as "the speed seen by faraway observers", which can be justified but is not the only possible interpretation). But as seen by local observers (at the same radial coordinate r as the light), the light is moving radially outward at c.

 

[PeterDonis]

There is a reference frame when the corresponding energy of the rest mass that was lost (i.e. $-\Delta m_0c^2$) and the observed energy of the photon are equal. Let's now say that you accelerate an electron to light speed with bombardment of numerous photons onto that electron. Conclusion: The rest mass that was lost by masses to generate those photons which subsequently accelerate the electron determines how much additional curvature the electron produces (in a non-linear way of course).

I'm not sure about this because you haven't expressed it in a frame-invariant way. The frame-invariant way to express it would be to find a suitable stress-energy tensor describing the "source" of the spacetime curvature. For an electron in isolation, the stress-energy tensor is such that in the electron's rest frame, T_00 is the electron's rest energy density, and all other components are zero.

However, if you're considering the entire system you describe, you have to find a stress-energy tensor for the whole system. The simplest way to do that is to find that tensor's components in the center of momentum frame of the system. Let's do that first for a single fusion reaction in isolation, and then for the ensuing photon-electron reaction from that single fusion reaction.

For a single fusion reaction, in the center of momentum frame, the momenta of the two hydrogen atoms before the reaction are equal and opposite, and so are the momenta of the helium atom and the photon after the reaction. In this frame, T_00 is the total energy of the objects present (including both rest energy and kinetic energy), and all other components are zero (because the net momentum is zero). Therefore, by conservation of energy, T_00 remains constant through this reaction, and the sum of all energies before equals the sum of all energies after. So we have (in units where c = 1)

$$T_{00} = 2 m_{D} + 2 K_{D} = m_{He} + K_{He} + E$$

where $m_{D}$ is the rest mass of each hydrogen (deuterium) atom that fuses, $m_{He}$ is the rest mass of the helium atom, $K_{D}$ is the kinetic energy of each hydrogen atom before the reaction (we assume they are equal), $K_{He}$ is the kinetic energy of the helium atom after the reaction, and $E$ is the energy of the photon.

Now, suppose we have an electron sitting at rest in this same frame. It contributes an additional $m_{e}$ to $T_{00}$ at the start; and if it absorbs the energy of the photon, that energy simply becomes its kinetic energy in the same frame, so we can just add another term to our equation above:

$$T_{00} = 2 m_{D} + 2 K_{D} + m_{e} = m_{He} + K_{He} + m_{e} + E$$

You can see that the RHS of this equation remains *exactly the same* when the electron absorbs the photon; $E$ remains the same, it's just now attached to the electron (as its kinetic energy) rather than the photon. So in this frame, there is an additional contribution to the stress-energy tensor, and therefore to the source of spacetime curvature, equal to $E$, which can be thought of as "attached" to the electron.

But that "attachment" is frame-dependent; if we transform to a frame in which the electron is at rest, we will find that $T_{00}$ has become larger, because although the electron has lost kinetic energy E, the helium atom has gained *more* kinetic energy (because its velocity change is the same magnitude as the electron's and it has more rest mass). But also, there are now other nonzero components of the stress-energy tensor due to the net momentum of the system as a whole in the electron's rest frame. These two effects offset each other in such a way that frame-invariant quantities, such as the tidal gravity experienced by an observer traveling on a particular worldline that passes close to this system, are unchanged. So thinking of E as causing additional curvature "due to the electron" is not really a good way of thinking of it; you have to look at the entire system.

 

[kmarinas86]

But that "attachment" is frame-dependent; if we transform to a frame in which the electron is at rest, we will find that $T_{00}$ has become larger, because although the electron has lost kinetic energy E, the helium atom has gained *more* kinetic energy (because its velocity change is the same magnitude as the electron's and it has more rest mass). But also, there are now other nonzero components of the stress-energy tensor due to the net momentum of the system as a whole in the electron's rest frame. These two effects offset each other in such a way that frame-invariant quantities, such as the tidal gravity experienced by an observer traveling on a particular worldline that passes close to this system, are unchanged. So thinking of E as causing additional curvature "due to the electron" is not really a good way of thinking of it; you have to look at the entire system.

This is all rather strange. The photon does indeed carry energy with it from the helium atom to the electron, and this is true even in the new frame you are describing. The helium emits energy and the electron absorbs it. If the helium atom gains kinetic energy while the electron loses kinetic energy, then it must be saying something about the amount of non-kinetic energy. Presumably, that can only be as a result of potential energy which is lost by the helium atom and gained by the electron. However, there seems to be no account in SR for "relativistic potential energy" in the same sense that there is for relativistic kinetic energy.

 

[PAllen]

This is all rather strange. The photon does indeed carry energy with it from the helium atom to the electron, and this is true even in the new frame you are describing. The helium emits energy and the electron absorbs it. If the helium atom gains kinetic energy while the electron loses kinetic energy, then it must be saying something about the amount of non-kinetic energy. Presumably, that can only be as a result of potential energy which is lost by the helium atom and gained by the electron. However, there seems to be no account in SR for "relativistic potential energy" in the same sense that there is for relativistic kinetic energy.

If you look at this from the frame where the electron will be at rest after absorbing a photon, you have an electron with initial significant KE and momentum. As a series of photons interact with it, all of its momentum and KE are carried away by photons (that have increased energy and momentum). So, in this frame, the electron loses energy and momentum, ending up at rest, and contributes the minimal amount possible (for an electron) to T. Of course, the photons, as well as the Helium atom contribute more in this frame.

All of your complications don't change the fact that the speed and KE of an electron are frame dependent. They also are not very relevant to the original question about an isolated 'high speed' electron. Whether that electron is high speed relative to a detector because the detector is in a rocket going at high speed, or because the electron has been bombarded by photons somewhere else, cannot be relevant to what the detector measures about properties related to its curvature.

 

[kmarinas86]

If you look at this from the frame where the electron will be at rest after absorbing a photon, you have an electron with initial significant KE and momentum. As a series of photons interact with it, all of its momentum and KE are carried away by photons (that have increased energy and momentum). So, in this frame, the electron loses energy and momentum, ending up at rest, and contributes the minimal amount possible (for an electron) to T. Of course, the photons, as well as the Helium atom contribute more in this frame.

All of your complications don't change the fact that the speed and KE of an electron are frame dependent. They also are not very relevant to the original question about an isolated 'high speed' electron. Whether that electron is high speed relative to a detector because the detector is in a rocket going at high speed, or because the electron has been bombarded by photons somewhere else, cannot be relevant to what the detector measures about properties related to its curvature.

I don't believe that the photon can carry negative energy. If a photon moves from region A to region B, and if it carries $x$ amount of energy according to observer $o$, that amount has to be positive. Region A must lose $x$ amount of energy and region B must gain $x$ amount of energy, as far as observer $o$ is concerned. The energy, momentum, and pressure should increase for region B and decrease for region A for all observers because this energy cannot be negative in any frame. Also, the direction of energy flow at the speed of light cannot be so dependent as to appear to move in diametrically opposite directions in different inertial frames of reference. Such is saying that a photon in one reference frame moves from the sun to the Earth in one frame but can, in another, move from the Earth to the sun, and that's simply an untenable notion, even in SR.

 

[PAllen]

I don't believe that the photon can carry negative energy. If a photon moves from region A to region B, and if it carries $x$ amount of energy according to observer $o$, that amount has to be positive. Region A must lose $x$ amount of energy and region B must gain $x$ amount of energy, as far as observer $o$ is concerned. The energy, momentum, and pressure should increase for region B and decrease for region A for all observers because this energy cannot be negative in any frame. Also, the direction of energy flow at the speed of light cannot be so dependent as to appear to move in diametrically opposite directions in different inertial frames of reference. Such is saying that a photon in one reference frame moves from the sun to the Earth in one frame but can, in another, move from the Earth to the sun, and that's simply an untenable notion, even in SR.

You're still not getting the idea that all of this is frame dependent. Simplify to a single electron. In one frame it carries a large amount of KE from A to B. In another frame, A and B are the 'same point', and the amount of energy in A=B stays the same (the energy equivalent of the electron rest mass). None of your confusion here has anything to do with special or general relativity. These aspects are all the same in Newtonian/Galilean mechanics.

 

[kmarinas86]

You're still not getting the idea that all of this is frame dependent. Simplify to a single electron. In one frame it carries a large amount of KE from A to B. In another frame, A and B are the 'same point', and the amount of energy in A=B stays the same (the energy equivalent of the electron rest mass). None of your confusion here has anything to do with special or general relativity. These aspects are all the same in Newtonian/Galilean mechanics.

If the electron is bounded to an atom, and if we increase the electron's energy level, its kinetic energy decreases (due to being further from the nucleus on average) but its potential energy increases by a greater amount, allowing the energy to increase. I don't see how that overall increase could become a overall decrease in a different frame. It would be as if there would be a frame of reference where accumulation of asteroid material would decrease the mass of the earth, instead of increasing it, which makes no sense.

Now, if you have free electron and it reflects (not absorbs - a word with a different meaning by the way) the energy of a photon, then I can see your point. If you consider only those kinds of interactions, no change on the impact to spacetime curvature by the electron should be observed. The electron loses energy to the photon in that case in the same sense that a train would slow down and give energy to whatever object it collides with head-on.

 

[PAllen]

If the electron is bounded to an atom, and if we increase the electron's energy level, its kinetic energy decreases (due to being further from the nucleus on average) but it's potential energy increases by a greater amount, allowing the energy to increase. I don't see how that overall increase could become a overall decrease in a different frame. It would be as if there would be a frame of reference where accumulation of asteroid material would decrease the mass of the earth, instead of increasing it, which makes no sense.

Now, if you have free electron and it reflects (not absorbs - a word with a different meaning by the way) the energy of a photon, then I can see your point. If you consider only those kinds of interactions, no change on the impact to spacetime curvature by the electron should be observed.

(absorbs was a typo, corrected in the next sentence).

If you're talking about quantum phenomena like energy levels in an atom, we can't really talk about anything like the frame of an electron. The OP and all other discussion in this thread deals with a domain where a classical (including SR/GR as classical, of course) approximation is adequate.

However, the key distinction I think you are wrestling with can arise classically, and Peter Donis already explained it in some detail (please read his #43 carefully). If you consider an electron as part of a system of matter and energy, the contribution of its KE to curvature cannot be removed by a coordinate transform, however it can be relocated. A system of particles can be described intrinsically as absorbing energy contributing to gravity. However, the distribution of energy within the system is coordinate dependent, and nothing changes about something fundamental like an electron. There are no 'excited states' of electrons, where the electron intrinsically has more energy.

 

[kmarinas86]

(absorbs was a typo, corrected in the next sentence).

If you're talking about quantum phenomena like energy levels in an atom, we can't really talk about anything like the frame of an electron. The OP and all other discussion in this thread deals with a domain where a classical (including SR/GR as classical, of course) approximation is adequate.

However, the key distinction I think you are wrestling with can arise classically, and Peter Donis already explained it in some detail (please read his #43 carefully). If you consider an electron as part of a system of matter and energy, the contribution of its KE to curvature cannot be removed by a coordinate transform, however it can be relocated. A system of particles can be described intrinsically as absorbing energy contributing to gravity. However, the distribution of energy within the system is coordinate dependent, and nothing changes about something fundamental like an electron. There are no 'excited states' of electrons, where the electron intrinsically has more energy.

What happens if you have an electron moving non-inertially in a curved path? Surely such an electron would have more energy than one that is not. The only way for this non-inertial motion to disappear is by assuming that a non-inertial frame matching that of the electron is somehow inertial, which is a contradiction of course, so there isn't really a way to get rid of the fact that such a case involves additional energy. Of course, one could argue that the energy associated with the field causing this gyration is not intrinsic to the electron, but what is clear is that the energy associated with the electron (or its coupling with the surroundings rather) can have a component that is (not so) dependent on the inertial frame of the observer. However, it is not clear to me how such energy, dependent on non-inertial motion, would be re-distributed between the electron and the field source depending on the inertial observer. In such a situation, could we say that energy arising from non-inertial motion is not subject to redistribution with respect to the inertial observer chosen?

 

[PAllen]

What happens if you have an electron moving non-inertially in a curved path? Surely such an electron would have more energy than one that is not. The only way for this non-inertial motion to disappear is by assuming that a non-inertial frame matching that of the electron is somehow inertial, which is a contradiction of course, so there isn't really a way to get rid of the fact that such a case involves additional energy. Of course, one could argue that the energy associated with the field causing this gyration is not intrinsic to the electron, but what is clear is that the energy associated with the electron (or its coupling with the surroundings rather) can have a component that is (not so) dependent on the inertial frame of the observer. However, it is not clear to me how such energy, dependent on non-inertial motion, would be re-distributed between the electron and the field source depending on the inertial observer. In such a situation, could we say that such energy is not subject to redistribution with respect to the inertial observer chosen?

Let's say you have an electron and positron co-orbiting under EM force, and let's posit that QM prevents radiation from carrying away the energy (while still somehow allowing a classical world view), and further, let's ignore field energy. The distribution of KE between the electron and positron over time is coordinate and frame dependent. However, all frames will have a total system energy at least as great as the COM frame, at all times. Thus, wherever it is located (in a given moment in a given frame), the KE will contribute to gravity.

I think you should focus on understanding a simpler case, that doesn't require a bunch of absurd assumptions to discuss classically. Just consider a ball bouncing around in a box with all collisions elastic, versus a ball in isolation. In the former, the distribution of KE between the ball and the box is time and frame dependent. However, at all times, in all frames, the total KE of the box plus particle is at least the COM KE. Meanwhile, the KE of the isolated ball can trivially be made zero in a suitable frame. The ball+box system intrinsically contains a KE component to its total energy, but it is absurd to try to localize this to 'the ball is different from the isolated ball' or 'the box is different from an isolated box'. The system is different, but you just can't localize the difference.

Again, all of this can be discussed without reference to relativity. The issues you raise are completely pre-relativity classical physics (except, insofar as energy contributes to gravity).

 

[kmarinas86]

Let's say you have an electron and positron co-orbiting under EM force, and let's posit that QM prevents radiation from carrying away the energy (while still somehow allowing a classical world view), and further, let's ignore field energy. The distribution of KE between the electron and positron over time is coordinate and frame dependent.

However, the energy between the electron and the positron is the field energy. The KE of the electron and positron are frame-dependent. However, due to the electrical neutrality of the combined electron and positron, the magnetic field should be unaffected, at large distances, by the choice of inertial frame. Unlike inertial motion, the non-inertial motion, which is tied to acceleration of the orbits, does not depend on the choice of the inertial frame. Wouldn't an energy field tied to non-inertial motion be invariant under Lorentz transformations? This would mean that not all the energy would subjected to redistribution, when we consider a more general case where the motion is not fully inertial.

 

[PAllen]

However, the energy between the electron and the positron is the field energy. The KE of the electron and positron are frame-dependent. However, due to the electrical neutrality of the combined electron and positron, the magnetic field should be unaffected, at large distances, by the choice of inertial frame. Unlike inertial motion, the non-inertial motion, which is tied to acceleration of the orbits, does not depend on the choice of the inertial frame. Wouldn't an energy field tied to non-inertial motion be invariant under Lorentz transformations? This would mean that not all the energy would subjected to redistribution, when we consider a more general case where the motion is not fully inertial.

To posit localization of energy to an electron, you need to posit internal states for it, period. Similarly, to the extent you treat a rigid ball as an ideal object with no internal structure (thus temperature and entropy undefined for it), nothing about its motion, inertial or otherwise, adds to its rest mass (we have defined it so by idealization). Similarly, to the extent you consider an electron fundamental, with no internal degrees of freedom, you cannot treat any electron as having internal energy of any type.

 

[kmarinas86]

To posit localization of energy to an electron, you need to posit internal states for it, period. Similarly, to the extent you treat a rigid ball as an ideal object with no internal structure (thus temperature and entropy undefined for it), nothing about its motion, inertial or otherwise, adds to its rest mass (we have defined it so by idealization). Similarly, to the extent you consider an electron fundamental, with no internal degrees of freedom, you cannot treat any electron as having internal energy of any type.

In line with what I was saying though, couldn't energy of the field components which are generated by non-inertial motion of the particles themselves (to not say anything about non-existent internal structure, of course), be localized as a result of being non-inertial, and consequently have a localization of the energy (in the field itself) that does not in anyway have a distribution dependent on the choice of an inertial frame?

 

[PAllen]

In line with what I was saying though, couldn't energy of the field components which are generated by non-inertial motion of the particles themselves (to not say anything about non-existent internal structure, of course), be localized as a result of being non-inertial, and consequently have a localization of the energy (in the field itself) that does not in anyway have a distribution dependent on the choice of an inertial frame?

Yes, that's fine. The EM field (viewed classically or as QFT) certainly contributes to the stress energy tensor. And an oscillating EM field carries energy. Thus I could agree, in general, with this statement.

 

[pervect]

Just with a period, I don't disagree. But if the "local speed of light" is clarified to mean "the speed of light as seen by a local observer, i.e., an observer whom the light ray is just passing at the time the speed is measured", then the "local speed of light" *is* constant, and is always c.

As seen by the faraway observers, yes (if one is willing to interpret the coordinate speed of the light as "the speed seen by faraway observers", which can be justified but is not the only possible interpretation). But as seen by local observers (at the same radial coordinate r as the light), the light is moving radially outward at c.

I would have to say that saying "the local speed of light" is always "c" is one of the most helpful things one can say about general relativity. Given the well-known problems defining relative velocities at a distance (see , for instance, Baez http://math.ucr.edu/home/baez/einstein/node2.html), I'd much rather hear someone say "the velocity of light measured locally is always c" than the alternatives.

In general relativity, we cannot even talk about relative velocities, except for two particles at the same point of spacetime -- that is, at the same place at the same instant. The reason is that in general relativity, we take very seriously the notion that a vector is a little arrow sitting at a particular point in spacetime. To compare vectors at different points of spacetime, we must carry one over to the other. The process of carrying a vector along a path without turning or stretching it is called `parallel transport'. When spacetime is curved, the result of parallel transport from one point to another depends on the path taken! In fact, this is the very definition of what it means for spacetime to be curved. Thus it is ambiguous to ask whether two particles have the same velocity vector unless they are at the same point of spacetime.

 

[TrickyDicky]

I agree.
And yet you'll find that in cosmology high redshifts are still discussed in terms of velocities, (including values much higher than c for far enough objects). This will always lead to confusion for lay people IMO, even if every once in a while it is explained that those velocities are really coordinate artifacts.

 

[Passionflower]

I agree.
And yet you'll find that in cosmology high redshifts are still discussed in terms of velocities, (including values much higher than c for far enough objects). This will always lead to confusion for lay people IMO, even if every once in a while it is explained that those velocities are really coordinate artifacts.

So what we actually measure is a coordinate artifact to you?

 

[PAllen]

So what we actually measure is a coordinate artifact to you?

Well, what we measure is redshift. We interpret it as recession velocity according to a model with lots of corroboration. However, this recession velocity is in the CMB frame, and is similar to a separation speed in SR. Other reasonable ways of interpreting the same measured redshift lead to different velocities (e.g. parallel transporting the remote object's 4-velocity on the null path to the detector leads to a relative velocity always less than c, and further, one such that the red shift is identical to a local kinematic red shift produced by a local object whose 4-velocity matches the parallel transported 4-velocity).

 

[Naty1]

Pervect, Peter Donis, PAllen:

Lots of great explanations posted ... thank you!