Space time curvature caused by fast electron

[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!