Is 'charged black hole' an oxymoron?

[PeterDonis]

I'm not sure this analogy holds; I'll have to take a bit of time to consider the rotating scenario.

I've thought about this case some more, and I'm still not sure the analogy holds.

To simplify things somewhat from the "beams inside a synchrotron" scenario, suppose we have a single object of charge q inside a sort of circular capacitor, which I'll call the "ring" for brevity: a device that can create an electric field directed radially. We put the charged object at the end of a massless arm that runs on a track on the inside surface of the ring. The arm holds the charged object at a particular radius r, which is between the inner radius r_i and the outer radius r_o of the ring. The whole thing is floating in free space (flat spacetime).

Our two observables are, again, a scale under the track (O) and a strain gauge on the arm (O').

First, we consider the case where the q+arm assembly is at rest relative to the ring. We turn on the capacitor and find that the scale and strain gauge show nonzero readings. Call this set of readings A. We expect the readings to be equal: O_A = O'_A.

Now we set the q+arm assembly to rotating around the ring with constant angular velocity omega, and therefore constant relative speed v relative to the ring (but with changing direction). At a given instant, we can set up two local inertial frames (MCIFs): one in which the ring+track is at rest and the q+arm assembly is moving at speed v in a direction orthogonal to the E field generated by the ring; and one in which the q+arm assembly is at rest and the ring+track is moving at -v in a direction orthogonal to the E field generated by the ring.

It seems to me that, locally, these two inertial frames are the same in almost all respects (I'll elucidate that "almost all" in a moment) as the two frames in the straight line case: the rest frame of the charge q, and the rest frame of the capacitor. There is one key complication, though; in the rotating case, there is no longer an exact "force balance" in both frames, because the motions are no longer unaccelerated; the instantaneous relative velocity of the two is orthogonal, but the instantaneous acceleration of q+arm is not zero, it has an inward radial component.

This, to me, means we can't draw an exact analogy between the two scenarios, because we no longer have an exact force balance in the direction of the E field. The reasoning that led to O = O' depended on there being an exact force balance. So I don't see the two cases as analogous.

 

[Dale]

If you know how photons in a box behave, then educate us, because we have a small problem here. We all know about gravitational redshift of photons already.

It sure doesn't seem like you do. Photons in a box in gravity or in an accelerating reference frame gain energy on the way down (gravitational blueshift) and lose energy on the way up (gravitational redshift).

EDIT: Here's a short course of Jartsa's photon redshift theory:

Electric charges are deposited on the clock hands of a clock. Now the clock produces an EM-wave. The clock is dropped into a gravity well. Now clock hands move more slowly, and the frequency of the EM-wave is lower.

And how do you think this justifies your experimentally-contradicted claim that "a falling photon gains no energy"?

 

[Austin0]

It sure doesn't seem like you do. Photons in a box in gravity or in an accelerating reference frame gain energy on the way down (gravitational blueshift) and lose energy on the way up (gravitational redshift).

And how do you think this justifies your experimentally-contradicted claim that "a falling photon gains no energy"?

Pound-Rebka and GPS clock dilation are well known but have you encountered any results regarding direct EM shift to and from geosynchronous satellites?

Also when you say gains energy, do you mean as a change in transit through kinetic gain in momentum or simply an intrinsic difference at emission that is equivalent to gaining energy through relative potential differences?

This seems to be somewhat of an on going question without consensus ;-)

 

[jartsa]

It sure doesn't seem like you do. Photons in a box in gravity or in an accelerating reference frame gain energy on the way down (gravitational blueshift) and lose energy on the way up (gravitational redshift).

And how do you think this justifies your experimentally-contradicted claim that "a falling photon gains no energy"?

My experiment and the Pound-Rebka experiment are both
"a gravitational redshift experiment, which measures the redshift of light moving in a gravitational field, or, equivalently, a test of the general relativity prediction that clocks should run at different rates at different places in a non-uniform gravitational field."

I copied the quoted part from the Wikipedia page.

Now let's compare a clock that was dropped into a gravity well, to clocks outside the gravity well, after fetching the clock back from the well. Now we notice that clocks run at different rates at different places in a non-uniform (or uniform) gravitational field.

And here is a story how this latter experiment was done:
http://en.wikipedia.org/wiki/Hafele–Keating_experimentEDIT: A shorter answer: In my experiment a climbing EM-wave did not lose frequency, so redshift is disproved.

 

[Austin0]

But the "energy" could come from static equilibrium; the boxes could have started out sitting on a platform at the higher height, which is held up by columns, for example. The scenario does not specify how any "static" objects--objects at a constant height--are held there, since it's not relevant to the question the scenario was intended to answer. If it matters, assume all objects are held in static equilibrium; no rocket engines or other "hovering" methods that require energy expenditure.

If we assume that the box is suspended by thrust and then the whole system moves to a lower potential under thrust do you think this changes conditions in the box .i.e. producing a different result from this scenario?

During the time they are not reflecting, yes, that's true. But the reflection events are *not* the same. See below.



It's not just the velocity of the mirrors that's different; it's the *acceleration* of the mirrors, as in proper acceleration. The free-falling box's mirrors feel no force; they are weightless. The lowered box's mirrors feel a force, from the rope, so they are accelerated. That makes a difference when the photons reflect off the mirrors. See below.



No, it's not. Look at it in the momentarily comoving inertial frame (MCIF) of the box. The freely falling box will be at rest in this frame, and will remain at rest for the time of flight of a photon from one wall to the other. (Strictly speaking, we need to assume that the length of the box is small enough that light can cross it before tidal effects become measurable; but that is a pretty easy condition to meet.) So the walls won't be moving relative to the photon, and the two impacts will just cancel each other out, as you say.

The lowered box, however, is accelerated upward; so if the box is momentarily at rest when the photon leaves one wall, it will be moving when the photon reaches the other wall. If the photon is going from the upper to the lower wall, the wall will actually be moving *towards* the photon in the MCIF when the photon reaches the lower wall. The wall will be moving away from the photon if the photon is moving from the lower towards the upper wall. So here the two impacts do *not* cancel out; the downward impact, on the lower wall, exerts more force than the upward impact, on the upper wall, does. So on net the photons exert a downward force on the box; this force does work that is transmitted up the rope.

(Note that we can ignore the constant downward velocity of the box; we simply adjust the MCIF so the box is at rest in it at the instant a photon leaves one wall. That means the MCIF will have a small downward boost relative to the MCIF of a static observer at the same altitude. That has no effect on the analysis I just gave, but if you're worried about it having some other effect, we can make it negligibly small by lowering the box slowly enough.)

it is this downward boost of the MCIF that gives you the relative up motion of the floor of the box.

If we have a suspended mirror , horizontal and facing up at ground level: If we locally reflect a photon vertically off the mirror will it be redshifted as measured locally??

 

[PeterDonis]

Also when you say gains energy, do you mean as a change in transit through kinetic gain in momentum or simply an intrinsic difference at emission that is equivalent to gaining energy through relative potential differences?

This seems to be somewhat of an on going question without consensus ;-)

Part of the reason there appears to be confusion is that energy is frame-dependent; put another way, the word "energy" can correspond to a number of different things. If you don't keep careful track of which one you're talking about, it's easy to get confused.

When DaleSpam said that photons gain energy in falling, and lose energy in rising, due to the gravitational blueshift/redshift, he was talking about energy relative to static observers at different altitudes. The Pound-Rebka experiment tested this directly. Whether you interpret this result as "kinetic gain in momentum" or as "relative potential differences" is a matter of which viewpoint you want to adopt: a static observer would say it is relative potential differences, but a freely falling observer would look at the changing proper accelerations of static observers with altitude and say that the gravitational redshift/blueshift is the result of "kinetic gain in momentum" (or loss), similar to the way an inertial observer in flat spacetime would interpret the redshift/blueshift of light signals between different Rindler observers.

When I said that the photon temperature (i.e., average energy) inside a freely falling box of photons is unchanged, I was talking about energy relative to the box. At another point, I believe I said that the "energy at infinity" of the freely falling box is unchanged; as it falls, it gains kinetic energy but loses potential energy, so the total energy remains constant.

When I said that the photon temperature (i.e., average energy) inside a box of photons being slowly lowered, with work being extracted, decreases, I was again talking about energy relative to the box. I could also have framed my description of that scenario in terms of "energy at infinity", and in fact I may have, implicitly, since I think I said that the box is losing potential energy and its kinetic energy is unchanged.

AFAIK, nobody has run actual experiments to test the latter two scenarios, but I gave several theoretical arguments for why the results would have to be as I said, given the known result of the Pound-Rebka experiment.

 

[PeterDonis]

it is this downward boost of the MCIF that gives you the relative up motion of the floor of the box.

No, it isn't; it's the acceleration of the floor of the box. The boost only gives a constant relative velocity; but a constant relative velocity won't cause a *difference* in impacts on the upper and lower surfaces of the box. (If you doubt that, try analyzing the freely falling case, not in the MCIF as I did, but in a frame that is Lorentz boosted relative to the MCIF. In that frame, there will be a Doppler redshift at one end and a blueshift at the other, but they will cancel because the relative velocity is constant.) A difference in impacts, producing a net force, requires a *change in velocity* during the flight time of the photons, i.e., acceleration.

If we have a suspended mirror , horizontal and facing up at ground level: If we locally reflect a photon vertically off the mirror will it be redshifted as measured locally??

No, because there's only a single surface, and "locally" means I can get as close to the surface as I like to make the measurement. There has to be a change in velocity of the "box" *during the flight time of the photon*. If the measurement is purely local, there is no "flight time". The "box" has to have a finite distance between the upper and lower surfaces, to give the acceleration time to produce a measurable effect.

This does raise a point: I said previously that the box has to be small enough that tidal effects are negligible; but now I'm saying that the box can't be too small or acceleration effects are negligible! That's true; there has to be some finite range of "box size" where we can see the effects of acceleration, but can't yet see tidal effects. The reason there is in fact such a range of box size is that acceleration is a first-order effect but tidal effects are second order. More precisely, acceleration is due to the first derivative of the metric coefficients--the connection--while tidal effects are due to the second derivatives--curvature.

 

[Dale]

EDIT: A shorter answer: In my experiment a climbing EM-wave did not lose frequency, so redshift is disproved.

Nonsense. It was emitted at one frequency and detected at another with emitter and detector both at rest. Pound Rebka unambiguously proved gravitational blueshift, it is an experimental fact. You can attempt another explanation for the evidence, but you don't get to ignore the evidence.

 

[Dale]

Pound-Rebka and GPS clock dilation are well known but have you encountered any results regarding direct EM shift to and from geosynchronous satellites?

I have never looked for such evidence. I suspect that some communications company has put an atomic clock into geosynchronous orbit, but there is no conceptual difference between that and Pound Rebka experiment so there is little likelihood that such an experiment would attract funding for it's own sake.

Also when you say gains energy, do you mean as a change in transit through kinetic gain in momentum or simply an intrinsic difference at emission that is equivalent to gaining energy through relative potential differences?

I have no idea what you mean. The lights energy at the bottom is more than its energy at the top for stationary emitters and detectors.

 

[Dale]

When DaleSpam said that photons gain energy in falling, and lose energy in rising, due to the gravitational blueshift/redshift, he was talking about energy relative to static observers at different altitudes. The Pound-Rebka experiment tested this directly.

Yes.

I don't know why this is controversial. Even in Newtonian mechanics energy is frame variant. Why should we expect that inertial and non inertial observers will agree on energy when we know that not even different inertial observers agree?

 

[jartsa]

Nonsense. It was emitted at one frequency and detected at another with emitter and detector both at rest. Pound Rebka unambiguously proved gravitational blueshift, it is an experimental fact. You can attempt another explanation for the evidence, but you don't get to ignore the evidence.

Perhaps, if I tell you that it's not really my theory, it is possible for you to understand.

Now, think about a clock with charged clock hands in a really deep gravity well, the clock hands turning at ridiculouly slow pace. This clock can't emit an EM-wave with anything but a very low frequency.

 

[Dale]

Now, think about a clock with charged clock hands in a really deep gravity well, the clock hands turning at ridiculouly slow pace. This clock can't emit an EM-wave with anything but a very low frequency.

It can emit a EM wave with any frequency it likes. We will only receive the low frequency after it loses energy.

Look, this is very basic stuff. It happens to any massive object in an accelerating frame or in gravity even in Newtonian mechanics. Do you disagree? If not, then why is it a surprise that it happens to light?

 

[Austin0]

Pound-Rebka and GPS clock dilation are well known but have you encountered any results regarding direct EM shift to and from geosynchronous satellites?

I have never looked for such evidence. I suspect that some communications company has put an atomic clock into geosynchronous orbit, but there is no conceptual difference between that and Pound Rebka experiment so there is little likelihood that such an experiment would attract funding for it's own sake.

I thought that with satellite altitude that potential dilation might possibly be detected not through clocks but directly through shift in EM transmissions. Radio or light signals between satellite and ground

Also when you say gains energy, do you mean as a change in transit through kinetic gain in momentum or simply an intrinsic difference at emission that is equivalent to gaining energy through relative potential differences

I have no idea what you mean. The lights energy at the bottom is more than its energy at the top for stationary emitters and detectors.

I find it hard to believe you haven't encountered any of the threads regarding this question.
In one view the difference in observed frequency is purely due to the dilation differential be tween the electrons of emission and reception.The photon retains its initial frequency throughout . There is no change in the photon during transit.
The other view is that the photon does change in transit. Gaining momentum through gravitational coordinate acceleration into the well and comparably loosing it going uphill.

some people seem to hold the view that these are just two ways of looking at the same thing but I question this. Gravitational time dilation itself, is now a fact. SO this explanation totally accounts for the results. I.e. A difference in resonant frequencies of emitting and absorbing electrons due to dilation.
so the assumption of any additional factor at work would seem to necessitate additional observed end results above the gamma factor or be rejected.
Your thought?

 

[PeterDonis]

Now, think about a clock with charged clock hands in a really deep gravity well, the clock hands turning at ridiculouly slow pace.

Not to an observer who is there next to the clock. The hands can turn at any pace as seen locally. They will only seem to turn very slowly to an observer much higher up in the gravity well.

 

[Dale]

I thought that with satellite altitude that potential dilation might possibly be detected not through clocks but directly through shift in EM transmissions. Radio or light signals between satellite and ground

The Pound Rebka experiment was directly through a shift in EM transmissions, not a clock.

I find it hard to believe you haven't encountered any of the threads regarding this question.

Certainly, but I have never encountered the terminology you used. Like "kinetic gain in momentum". I don't know what you are referring to with your different options.

Your thought?

Energy is different in different reference frames. That's all.

Pound Rebka proved unambiguously that in stationary frames (relative to the gravitating source) the energy increases as it goes down. That different frames disagree should be completely expected. There is no need for all sorts of mental gymnastics or contorted logic here. It is just a simple and clear experimental result (that light's energy is affected by gravity just like everything else's energy) and the usual understanding that energy is frame variant.

 

[jartsa]

Let us consider a clock with one light second long charged clock hands. This clock can procuce an EM-wave with maximum frequency about 1/6.28 Hz.

In a gravity well where the speed of light is let's say half of the normal, the maximum frequency is about 1/12.56 Hz

 

[Dale]

My point is that maybe improper assumptions have been made in coupling the two

I found the original paper, but it was in German, so that wasn't too helpful for me. So I looked around and found a lot of proofs that were just too brief for me to really grasp, until this one by Eric Poisson:

http://books.google.com/books?id=bk...qySBA&ved=0CD4Q6AEwAA#v=onepage&q=176&f=false

Just click any of the hyperlinks to the proof on the bottom of p 176 and all of p 177. Seems pretty solid to me.

 

[Dale]

Let us consider a clock with one light second long charged clock hands. This clock can procuce an EM-wave with maximum frequency about 1/6.28 Hz.

In a gravity well where the speed of light is let's say half of the normal, the maximum frequency is about 1/12.56 Hz

That is the frequency received by an observer "uphill" where the light has lost energy, not the frequency emitted by the clock.

In your view, what makes the uphill observer so omniscient that what his clock receives defines what the downhill clock emitted?

 

[PeterDonis]

Just spotted this on re-reading through the thread:

Relative motion of scales generates a 'lever advantage' situation of sorts. Scales in S' are closer to the lever 'fulcrum point' so to speak and record a larger applied force than scales in S - in fact necessarily so if the 'lever' (static equilibrium in x direction) is to stay in balance.

I don't see how this would work at all. First of all, there is no "fulcrum point", or at least I don't see one; that would require the track to only be supported at one spot between the scale under the track and the location of the arm. That makes no sense, and anyway, we could just change the scenario to have a support directly underneath the scale, so its distance to any "fulcrum" would be zero. Would that change your prediction of the relative forces?

Second, even if we leave out the above, any fulcrum point would have to be moving relative to either the capacitor+track or q+arm, so the amount of leverage, which would be the ratio of the two distances to the fulcrum, will be time-dependent. Are you claiming that one of the scale readings in this scenario is time dependent?

Finally, even if we leave out both of the above, for leverage to work, the two forces being measured need to be in opposite directions (push down on one end of the lever, the other end pushes up). Here both forces we are looking at, O and O', are in the same direction: the force of q on the arm, and the force of the arm on the scales (transmitted via the track). "Leverage" due to horizontal distance can't affect the ratio of those two forces. It would be like a kid sitting way out on a seesaw, moving down, and somehow making me, closer to the fulcrum on the other side, move down as well. Doesn't make sense.

 

[jartsa]

That is the frequency received by an observer "uphill" where the light has lost energy, not the frequency emitted by the clock.

In your view, what makes the uphill observer so omniscient that what his clock receives defines what the downhill clock emitted?

Maybe you think gravitational time dilation is relative. But it's absolute.

Let's consider just a shallow gravity well, like a groove on the ground. We know we don't see 10 seconds into the past when looking into a ditch. A clock will be 10 seconds gravitationally delayed after being in the ditch for a long enough time.

We can deduce that a clock in a ditch is running slow, if we see it's 30 seconds late, for example. And I don't mean a broken clock here, or some other silly thing.

 

[TrickyDicky]

Maybe you think gravitational time dilation is relative. But it's absolute.

Let's consider just a shallow gravity well, like a groove on the ground. We know we don't see 10 seconds into the past when looking into a ditch. A clock will be 10 seconds gravitationally delayed after being in the ditch for a long enough time.

We can deduce that a clock in a ditch is running slow, if we see it's 30 seconds late, for example. And I don't mean a broken clock here, or some other silly thing.

You might have some confusion about gravitational dilation here, the time dilation depends on the location on the gravitational potential, and it's relative between points located at different points along the gravitational potential, what is absolute about that?

 

[TrickyDicky]

Similarly, observable O, the force exerted on the track by the arm, is determined by Newton's Third Law; it must be equal and opposite to the force exerted on the arm by the track. As I said before, the latter force must be equal and opposite to O', for the arm to be motionless; so again observable O is determined by Newton's Third Law, not the Lorentz force law. Observable O "detects" the Lorentz force on q indirectly, just as O' does; it just detects it more indirectly since there are more intermediate steps in the chain that links them.

I'm a bit confused by this. Classical electrodynamics has this curious thing about charges and Newton's third law, they are not supposed to folllow 3rd law by themselves (not considering the field) and I think this is the only instance this happens. But there's no momentum conservation issues because according to CED textbooks what one has to take into account is not only the charges but the field and what is conserved is the sum of the charge's momentum plus the field's momentum and this is all taken into account when using Lorentz forces (actually more practical to use the full Maxwell tensor that simplifies the Lorentz forces bookkeeping a great deal). So the third law is accounted for when using the Lorentz force. And all this should be compatible with SR also.
This kind of reminds me of the recent turmoil about the paper claiming Lorentz force is not relativistic that was quickly resolved.
So I'm not sure your analysis here is totally accurate in this sense.

 

[stevendaryl]

Maybe you think gravitational time dilation is relative. But it's absolute.

Let's consider just a shallow gravity well, like a groove on the ground. We know we don't see 10 seconds into the past when looking into a ditch. A clock will be 10 seconds gravitationally delayed after being in the ditch for a long enough time.

We can deduce that a clock in a ditch is running slow, if we see it's 30 seconds late, for example. And I don't mean a broken clock here, or some other silly thing.

There are two different effects--one is coordinate-dependent, and one is independent of coordinates. To say that one clock is running slow compared with another clock is a coordinate-dependent effect. In one coordinate system, one clock is slow, and in another coordinate system, the other clock might be running slow. That's what the mutual time dilation of Special Relativity is all about--two observers with a relative speed of 90% the speed of light will each see the clock of the other run slow. The same thing is true of General Relativity. How fast a clock runs is dependent on a choice of coordinates.

Let \tau_A be elapsed time on clock A, and let \tau_B be elapsed time on clock B. Let t be time in one coordinate system, and let T be time in another. Then the rate of clock A will be
d\tau_A/dt in one coordinate system, and
d\tau_A/dT in the other. Similarly for the rate of B.

If A and B are spatially separated, it is certainly possible to have

• d\tau_A/dt > d\tau_B/dt
• d\tau_A/dT < d\tau_B/dT

So there is nothing absolute or objective about the claim that clock A runs faster or slower than clock B.

On the other hand, if A and B are initially together, they separate and then come back together again, all coordinate systems will agree that the total elapsed time on A is greater than the total elapsed time on B (or the other way around, whichever is the case). The fact that all coordinate systems agree on total elapsed time does not mean that they agree on clock rates along the way. One coordinate system will attribute the elapsed time difference to one part of the journey, while another coordinate system will attribute it to another part of the journey.

Elapsed times are absolute. Clock rates are not.

 

[jartsa]

There are two different effects--one is coordinate-dependent, and one is independent of coordinates. To say that one clock is running slow compared with another clock is a coordinate-dependent effect. In one coordinate system, one clock is slow, and in another coordinate system, the other clock might be running slow. That's what the mutual time dilation of Special Relativity is all about--two observers with a relative speed of 90% the speed of light will each see the clock of the other run slow. The same thing is true of General Relativity. How fast a clock runs is dependent on a choice of coordinates.

Let \tau_A be elapsed time on clock A, and let \tau_B be elapsed time on clock B. Let t be time in one coordinate system, and let T be time in another. Then the rate of clock A will be
d\tau_A/dt in one coordinate system, and
d\tau_A/dT in the other. Similarly for the rate of B.

If A and B are spatially separated, it is certainly possible to have

• d\tau_A/dt > d\tau_B/dt
• d\tau_A/dT < d\tau_B/dT

So there is nothing absolute or objective about the claim that clock A runs faster or slower than clock B.

On the other hand, if A and B are initially together, they separate and then come back together again, all coordinate systems will agree that the total elapsed time on A is greater than the total elapsed time on B (or the other way around, whichever is the case). The fact that all coordinate systems agree on total elapsed time does not mean that they agree on clock rates along the way. One coordinate system will attribute the elapsed time difference to one part of the journey, while another coordinate system will attribute it to another part of the journey.

Elapsed times are absolute. Clock rates are not.

Let's consider a 1 cm thick clock hand. Can we say the under side of this clock hand always runs at the same rate as the upper side?

Can this much be said?

 

[jartsa]

You might have some confusion about gravitational dilation here, the time dilation depends on the location on the gravitational potential, and it's relative between points located at different points along the gravitational potential, what is absolute about that?

Yes, and space is filled with these potential fields, and we can take any two objects in space, and say which is the upper object.

Pushing is not absolute work. Pushing upwards, also known as lifting, just absolutely is work.

 

[stevendaryl]

Let's consider a 1 cm thick clock hand. Can we say the under side of this clock hand always runs at the same rate as the upper side?
Can this much be said?

As I said, if two clocks A and B are SPATIALLY SEPARATED, then the question of which clock has a faster rate is coordinate-dependent. If they are at the same location, then the question is not coordinate-dependent.

Do you understand that for two clocks A and B that are spatially separated, and for two different coordinate systems (with time coordinates t and T), you can have

d\tau_A/dt > d\tau_B/dt

and also have

d\tau_A/dT < d\tau_B/dT

If A and B are at the same location, that is not possible.

 

[stevendaryl]

Yes, and space is filled with these potential fields, and we can take any two objects in space, and say which is the upper object.

Pushing is not absolute work. Pushing upwards, also known as lifting, just absolutely is work.

This discussion is completely full of misunderstandings about General Relativity. The big one is that there is no such thing as a "gravitational potential" in General Relativity. Gravity is not a force in GR, and it does not have a potential. It has a metric, which determines the distance between two points that are spacelike separated and also determines the proper time between points that are timelike separated.

Now, it is possible to choose a coordinate system (the Schwarzschild coordinates) such that the time component of the metric tensor, g_tt is related to the Newtonian gravitational potential:

g_tt = 1 - 2GM/(c² r)

But g is NOT a potential in GR, and that form of the time component is coordinate-dependent.

 

[Dale]

Maybe you think gravitational time dilation is relative. But it's absolute.

I disagree in general, but so what? What does the label "absolute" or "relative" have to do with the question of whether the light's energy increases as it goes down?

If you want to label the energy change "absolute" rather than "relative" it doesn't bother me. I think you are wrong to do so, but it is beside the point which is that it changes.

 

[jartsa]

As I said, if two clocks A and B are SPATIALLY SEPARATED, then the question of which clock has a faster rate is coordinate-dependent. If they are at the same location, then the question is not coordinate-dependent.

Do you understand that for two clocks A and B that are spatially separated, and for two different coordinate systems (with time coordinates t and T), you can have

d\tau_A/dt > d\tau_B/dt

and also have

d\tau_A/dT < d\tau_B/dT

If A and B are at the same location, that is not possible.

Gravitational time dilation is different to velocity time dilation. It's much more simple: Upper clock runs faster. You can not find a cordinate system were lower clock runs faster.

Actually, you can find a cordinate system where lower clock runs faster, if lower and upper clock are in relative motion to each other.

 

[PeterDonis]

Classical electrodynamics has this curious thing about charges and Newton's third law, they are not supposed to folllow 3rd law by themselves (not considering the field)

I see what you mean, but I'm not trying to apply Newton's Third Law in this way. Momentum stored in the EM field doesn't play a role in this problem, because we're talking about balancing the Lorentz force on the charge q with the mechanical force exerted on q by the arm, for a net force on q of zero. Momentum stored in the field would only come into play if q was moving freely, with the EM force the only force on it. Then, yes, you would have to keep track of momentum stored in the field for things to balance; if you only looked at the charge q itself it would seem like it was gaining momentum (and energy) "from nowhere". But in this problem q is not gaining momentum and energy at all, because it's not moving at all in the x direction and its motion in the y direction (in situation #2) is at a constant speed.

 

[Dale]

The big one is that there is no such thing as a "gravitational potential" in General Relativity. Gravity is not a force in GR, and it does not have a potential.

You are correct, in general. However, just to be clear, in a static spacetime you can define a potential, and in stationary spacetimes you can define a pair of potentials. But I agree with you that you shouldn't rely on reasoning based on potentials too heavily in GR. It will necessarily be limited in application at best.

 

[PeterDonis]

Now, it is possible to choose a coordinate system (the Schwarzschild coordinates) such that the time component of the metric tensor, g_tt is related to the Newtonian gravitational potential:

g_tt = 1 - 2GM/(c² r)

But g is NOT a potential in GR, and that form of the time component is coordinate-dependent.

It's true that g_tt is coordinate dependent, but it's not correct to state flatly that g_tt is not a potential, or that there is never a potential in GR. "Potential" is a matter of interpretation. In a generic spacetime, you are correct, there is not in general going to be anything that can be interpreted as a potential. But in this case, the spacetime is static and asymptotically flat, so you can indeed define a "potential" (it's basically defined the way the Newtonian potential is defined, but with relativistic corrections) and show that the "acceleration due to gravity" is the gradient of that potential, which is true of the GR version just as it's true of the Newtonian version in the Newtonian approximation. These statements can be formulated in a coordinate-independent way, btw, so even though g_tt itself is coordinate dependent, it's not correct that the potential itself is just a coordinate-dependent thing with no physical meaning, in this particular spacetime. You have to be careful to understand the limitations of viewing things this way, but within its limitations, it's a useful view.

 

[stevendaryl]

Gravitational time dilation is different to velocity time dilation. It's much more simple: Upper clock runs faster. You can not find a coordinate system were lower clock runs faster.

First of all, I didn't say anything about velocities. I was talking about coordinate transformations. Coordinate transformations change velocities, and clock rates.

Second, no, gravitational time dilation is not different from velocity time dilation--not in any absolute sense, anyway. What looks like gravitational time dilation in one coordinate system will look like velocity dependent time dilation in another coordinate system. "Gravitational time dilation" doesn't have anything specifically to do with gravity, it has to do with using curvilinear coordinates. If you use inertial coordinates, there is no gravitational time dilation.

Third, what you said is just not true. You can certainly find a coordinate system in which, at one particular time, the lower clock runs faster than the upper clock.

Let (x,t) be the Schwarzschild coordinates in which the two clocks are at rest. Then, according to those coordinates, Rate_{upper clock}/Rate_{lower clock} = 1+ gh/c². Now, switch coordinates to coordinates (X,T) defined by:

X = x
T = (1+Ax)t

Then in the coordinate system (X,T), the ratio of the clock rates will be given by:

Rate'_{upper clock}/Rate'_{lower clock} = (1+gh/c²)/ (1+Ah)

If A > g/c², then in this coordinate system, the lower clock runs faster than the upper clock.

Clock rates are a coordinate-dependent quantity. There is no absolute sense in which the upper clock always runs faster than the lower clock.

 

[stevendaryl]

It's true that g_tt is coordinate dependent, but it's not correct to state flatly that g_tt is not a potential, or that there is never a potential in GR.

Not according to GR, it's not. GR does not describe gravity in terms of potentials.

"Potential" is a matter of interpretation. In a generic spacetime, you are correct, there is not in general going to be anything that can be interpreted as a potential. But in this case, the spacetime is static and asymptotically flat, so you can indeed define a "potential" (it's basically defined the way the Newtonian potential is defined, but with relativistic corrections) and show that the "acceleration due to gravity" is the gradient of that potential,

But "acceleration" is a coordinate-dependent quantity. An object that is accelerating under gravity in one coordinate system is not accelerating at all in another coordinate system.

which is true of the GR version just as it's true of the Newtonian version in the Newtonian approximation. These statements can be formulated in a coordinate-independent way,

Hmm. I would have to see what you mean by that. I suppose you could define a "stationary observer" to be one for which the local spacetime curvature is unchanging as a function of time. Then you could define "the acceleration due to gravity" to be the relative acceleration between a freefalling object initially at "rest" and a nearby "stationary" observer. But I don't see how those definitions involve potentials.

btw, so even though g_tt itself is coordinate dependent, it's not correct that the potential itself is just a coordinate-dependent thing with no physical meaning, in this particular spacetime. You have to be careful to understand the limitations of viewing things this way, but within its limitations, it's a useful view.

I would say that definitely "gravitational potential" is a coordinate-dependent thing. In the case of the Schwarzschild geometry, there is a set of "preferred" coordinate systems, which are the ones for which the metric is spherically symmetric and time-independent.

 

[stevendaryl]

You are correct, in general. However, just to be clear, in a static spacetime you can define a potential, and in stationary spacetimes you can define a pair of potentials. But I agree with you that you shouldn't rely on reasoning based on potentials too heavily in GR. It will necessarily be limited in application at best.

Maybe it's a matter of terminology. In the Schwarzschild spacetime (and is that the only one?) you can pick a coordinate system in which the metric tensor is time-independent and spherically symmetric. For that particular coordinate system, you can solve for the trajectories of test particles in a way that looks a lot like solving a problem in Newtonian physics with a gravitational potential. But I would not call it a potential.

 

[George Jones]

For good or for bad, "effective potential" is a term used in most GR texts (e.g, Carroll, Wald, Taylor and Wheeler, Hartle, Schutz) in their treatments of geodesics in Schwarzschild.

 

[PeterDonis]

Maybe it's a matter of terminology. In the Schwarzschild spacetime (and is that the only one?) you can pick a coordinate system in which the metric tensor is time-independent and spherically symmetric.

Schwarzschild is not the only one; you can do it in Reissner-Nordstrom spacetime as well. But you can also express the fact that the metric is time-independent and spherically symmetric in a coordinate-free manner, in terms of the Killing vector fields of the spacetime. And you can define the "potential" in terms of the Killing vector fields, so that it has a coordinate-free definition as well.

 

[Q-reeus]

To simplify things somewhat from the "beams inside a synchrotron" scenario, suppose we have a single object of charge q inside a sort of circular capacitor, which I'll call the "ring" for brevity: a device that can create an electric field directed radially. We put the charged object at the end of a massless arm that runs on a track on the inside surface of the ring. The arm holds the charged object at a particular radius r, which is between the inner radius r_i and the outer radius r_o of the ring. The whole thing is floating in free space (flat spacetime).

Our two observables are, again, a scale under the track (O) and a strain gauge on the arm (O').

First, we consider the case where the q+arm assembly is at rest relative to the ring. We turn on the capacitor and find that the scale and strain gauge show nonzero readings. Call this set of readings A. We expect the readings to be equal: O_A = O'_A.

Now we set the q+arm assembly to rotating around the ring with constant angular velocity omega, and therefore constant relative speed v relative to the ring (but with changing direction). At a given instant, we can set up two local inertial frames (MCIFs): one in which the ring+track is at rest and the q+arm assembly is moving at speed v in a direction orthogonal to the E field generated by the ring; and one in which the q+arm assembly is at rest and the ring+track is moving at -v in a direction orthogonal to the E field generated by the ring.

It seems to me that, locally, these two inertial frames are the same in almost all respects (I'll elucidate that "almost all" in a moment) as the two frames in the straight line case: the rest frame of the charge q, and the rest frame of the capacitor. There is one key complication, though; in the rotating case, there is no longer an exact "force balance" in both frames, because the motions are no longer unaccelerated; the instantaneous relative velocity of the two is orthogonal, but the instantaneous acceleration of q+arm is not zero, it has an inward radial component.

This, to me, means we can't draw an exact analogy between the two scenarios, because we no longer have an exact force balance in the direction of the E field. The reasoning that led to O = O' depended on there being an exact force balance. So I don't see the two cases as analogous.

Centripetal acceleration introduced here is of no fundamental consequence. Covered that in the similar and better spinning carousel example back in #92; simply increase radii of your ring capacitor until u²/r acceleration is negligible, or just specify we are concerned only with EM interaction component. The carousel example was better in a number of respects. Centripetal acceleration and electric forces owing to an axially uniform applied E are there orthogonal not co-radial as above, making for a clean separation. But it's your previous response, maintaining there will be an increased axial net force on carousel charges (as measured in non-spinning frame) owing to rotation speed that has an obvious and easily checkable corollary. A necessarily equal and opposite force back on the stationary capacitor charges - assuming one accepts overall force balance must apply.

That no such speed dependent increase in carousel axial forces occurs, just note that it is experimental fact, consistent with charge invariance strictly holding in SR, there is no electrostatic field generated by a current flowing in an electrically neutral circuit. Only an exterior Biot-Savart B field - explained in the usual SR way as differential charge densities seen in a frame moving wrt the circuit. Thus an exterior charge q_e stationary wrt the circuit feels nothing (we assume for simplification here there is no screening charges on the circuit's conducting wire). In the frame of a moving conduction charge q_c, the transverse E field component of q_e is greater for q_c than experienced by a fixed lattice charge q_f, by the drift-velocity gamma factor. Your logic maintains that larger proper 3-force acting on q_c should persist unchanged as measured in the circuit rest frame, where conduction and lattice charge number-densities are equal. Integrating over all conduction and lattice charges, it follows a net force would be exerted by field of q_e on the circuit, despite no such force being exerted by the circuit on q_e. That is the logical consequence of your position. And I've said as much on previous occasions. Force balance is naturally satisfied though by applying the usual 3-force transformation rule - requiring transverse components of both forces and fields measure differently in different frames.

 

[Q-reeus]

I found the original paper, but it was in German, so that wasn't too helpful for me. So I looked around and found a lot of proofs that were just too brief for me to really grasp, until this one by Eric Poisson:
http://books.google.com/books?id=bk...qySBA&ved=0CD4Q6AEwAA#v=onepage&q=176&f=false
Just click any of the hyperlinks to the proof on the bottom of p 176 and all of p 177. Seems pretty solid to me.

Thanks for reference, but apart from p 176 being unavailable I got the impression of a derivation following formal rules without really explaining the guiding philosophy. What raises my eyebrows is a passage down on p 180:
"It therefore appears that RN metric describes more than just a single black hole. Indeed it describes an infinite lattice of asymptotically flat universes connected by black-hole tunnels. Such a fantastic spacetime structure...", 'fantastic' being the key word - derivation from 'fantasy'.

Found this article which only asks more questions imo: https://facultystaff.richmond.edu/~ebunn/ajpans/ajpans.html
I come back to asking how coupling together the implications in #109 together with case 3: in #1 can be dismissed on logical grounds.

 

[Q-reeus]

I don't see how this would work at all. First of all, there is no "fulcrum point", or at least I don't see one; that would require the track to only be supported at one spot between the scale under the track and the location of the arm. That makes no sense, and anyway, we could just change the scenario to have a support directly underneath the scale, so its distance to any "fulcrum" would be zero. Would that change your prediction of the relative forces?

You're taking my metaphors way too literally! Could you please this time answer what I asked of you in last para of #149. And just to be completely clear, in #102 - that's a thin walled right-circular cylinder subject to a uniform axially directed end-forces via say frictionless pistons. Your take on the axial strain in both non-spinning and spinning cases. And to eliminate distractions like effect of coupling centrifugal stresses to axial strain via Poisson's ratio, we specify force is only applied after spin-up has occurred in spinning case. Or if you want, that frictionless thick outer sleeve absorbs all centrigugal stresses, etc. etc. Your understanding of what happens and why, as seen and interpreted in the two frames. Please.

 

[Dale]

Thanks for reference, but apart from p 176 being unavailable I got the impression of a derivation following formal rules without really explaining the guiding philosophy.

I know that we probably disagree on this, but I fundamentally believe that the universe can be analyzed logically. So the formal rules are important.

The assumptions from the derivation, as far as I can tell are:
1) EFE holds
2) Maxwell's equations hold
3) static, spherical symmetry
4) exterior EM field has no sources
5) exterior metric has only EM field as source

I cannot see any other assumptions, and, as you say, the remainder is simply following the formal rules from those.

I will look at the other points you cited, but can you look at those 5 assumptions and see if any of them seem wrong to you? Or let me know if you think I missed one.

 

[PeterDonis]

Centripetal acceleration introduced here is of no fundamental consequence. Covered that in the similar and better spinning carousel example back in #92; simply increase radii of your ring capacitor until u²/r acceleration is negligible

Yes, I can see that you can reduce the acceleration until its effect is very small, so you are very close to having a force balance.

, or just specify we are concerned only with EM interaction component.

Not sure what good that does; force balance is force balance, it has to include all forces acting.

The carousel example was better in a number of respects. Centripetal acceleration and electric forces owing to an axially uniform applied E are there orthogonal not co-radial as above, making for a clean separation.

Yes, this is true, having the forces orthogonal changes the analysis.

But it's your previous response, maintaining there will be an increased axial net force on carousel charges (as measured in non-spinning frame) owing to rotation speed that has an obvious and easily checkable corollary.

I don't think I've even analyzed the case of axial forces at all, have I? I've only analyzed the linear relative motion case and the case of radial forces in a circular ring.

In any case, you keep on multiplying scenarios when we haven't even got the first one taken care of. See below.

You're taking my metaphors way too literally! Could you please this time answer what I asked of you in last para of #149.

Which referred back to #102...

And just to be completely clear, in #102 - that's a thin walled right-circular cylinder subject to a uniform axially directed end-forces via say frictionless pistons.

Which involves forces in two different directions, axial and radial, if the cylinder is spinning. Can't you stick to one scenario? We were talking about a simple scenario with forces in only one direction. I have asked you, repeatedly, how the arm can possibly be motionless if the forces on the two ends of it are different, which is a necessary implication of your claim that the reading on the scale under the track, O, is different than the reading on the strain gauge on the arm, O', when q+arm is moving relative to capacitor+track. Are you ever going to just answer this question, without dragging in multiple other scenarios that add other elements that aren't in this one, and then asking me to wade through all your arguments that purport to show how they really really are related after all?

Sorry if I sound frustrated, but if you really have a good answer to the question I posed (again) in the last paragraph, you ought to be able to give it in a single post that doesn't refer to any other scenarios. Until you do, I'm not going to talk about any of those other scenarios. They might well be interesting to talk about, but let's get the one simple scenario, with forces in just one direction, taken care of first.

 

[PeterDonis]

if you really have a good answer to the question I posed (again) in the last paragraph, you ought to be able to give it in a single post that doesn't refer to any other scenarios.

In fact, I'll make it even easier by simplifying the scenario to show that the fundamental question about force balance doesn't even involve EM fields at all. (That should also keep up my tradition for provocative disagreement .)

Consider an object "q" on an arm that has a strain gauge attached, which runs on a track with a scale under it. There is some unspecified force acting on q in the positive x direction. It doesn't matter what kind of force it is; the only requirement is that it only acts on q, not on any other objects we're considering. (In the version where q is a charged object, this is ensured by stipulating that the arm with its strain gauge, and the track with its scale under it, are electrically neutral.) The reading on the scale under the track is observable O; the reading on the strain gauge on the arm is O'.

(1) In the case where everything is at rest, no relative motion, the force balance in the x direction implies O = O'.

(2) In the case where "q"+arm+gauge is moving in the y direction, relative to track+scale, the force balance in the x direction still implies O = O'. The actual magnitude of O and O' may be different from what it is in #1 above, but the two will still be the same.

Okay, Q-reeus: you agree with #1 but not with #2. What's the difference? Don't use any specific facts about what kind of force acts on "q". Give a general argument that will hold regardless of what kind of force it is, given what is specified above.

(Edit: Just to be clear, though we've discussed it before, by "readings" O and O' I mean the actual numbers appearing on the scale's or strain gauge's readouts.)

 

[Q-reeus]

I know that we probably disagree on this, but I fundamentally believe that the universe can be analyzed logically. So the formal rules are important.

The assumptions from the derivation, as far as I can tell are:
1) EFE holds
2) Maxwell's equations hold
3) static, spherical symmetry
4) exterior EM field has no sources
5) exterior metric has only EM field as source

I cannot see any other assumptions, and, as you say, the remainder is simply following the formal rules from those.

I will look at the other points you cited, but can you look at those 5 assumptions and see if any of them seem wrong to you? Or let me know if you think I missed one.

That summary is helpful. As I have claimed global failure of charge invariance is implied by examples given earlier, that in turn is questioning how 1) and 2) are made to combine in getting to RN metric. The only coupling of relevance there is afaik that EM field acts as an exterior SET source - of further gravity. But that otherwise static gravity modifies static EM field not. You know now what point in particular I refer to - effective charge is not subject to any 'redshift'. Points 3) to 5) are just boundary conditions so no problem there.

 

[Q-reeus]

Which involves forces in two different directions, axial and radial, if the cylinder is spinning.

You may have noticed though, the uncut version of your quote specified how to eliminate radial stresses - and one further option in that regard is as for your radial capacitor - just increase radius arbitrarily, or limit spin-rate to small value. That way, ratio of gamma factor effecting axial strain, to centrifugul stresses is arbitrarily large. Many ways to tame the 'monster' of two forces here! But if you don't want to, ok no gun at your head.

Can't you stick to one scenario? We were talking about a simple scenario with forces in only one direction. I have asked you, repeatedly, how the arm can possibly be motionless if the forces on the two ends of it are different, which is a necessary implication of your claim that the reading on the scale under the track, O, is different than the reading on the strain gauge on the arm, O', when q+arm is moving relative to capacitor+track. Are you ever going to just answer this question, without dragging in multiple other scenarios that add other elements that aren't in this one, and then asking me to wade through all your arguments that purport to show how they really really are related after all?

Well they are related Peter, and honestly your question has been answered umpteen times but not to your liking. We are dealing with effects of motion in SR context and readings in general will distort from one frame to another. In particular for our discussions, the 3-force transformation rules say so!
That bit in #188 about exterior charge and circuit is a good example of how force measured in one frame must measure differently in another. Only real paradox would be if they didn't vary. NASA would love it to be so - "off to the stars with breakthrough propulsion physics."

Sorry if I sound frustrated, but if you really have a good answer to the question I posed (again) in the last paragraph, you ought to be able to give it in a single post that doesn't refer to any other scenarios. Until you do, I'm not going to talk about any of those other scenarios. They might well be interesting to talk about, but let's get the one simple scenario, with forces in just one direction, taken care of first.

To repeat, there is no paradoxical imbalance because, as the 3-force transformation formula insists, what is measured by strain gauge moving with charge and arm, must be greater by gamma factor from what the stationary-in-lab-frame scales under the tracks measures, in order that arm has no net dp_x/dt. We cannot directly compare readings because relative motion is screwing up an exact correspondence. As you know I have coined it RW - strain gauge (or better for our purposes - spring scales holding charge) is 'weakened' as perceived in lab frame - the higher reading is out of whack as far as lab frame folks are concerned. And vice versa from moving frame perspective.

And now that I've caught your #193, I guess the answer to your specific there is going to be obvious. For any transverse force F'_x acting on q in frame S' - in motion at relative velocity u wrt lab frame S, transverse force F_x read by stationary scales in lab frame is, by simply applying the 3-force transformation formulae, reduced according to F_x = F'_x/γ. And conversely F'_x = γF - ie reciprocity of readings applies. And there is then no actual imbalance along x - dp_x/dt (net) = 0.
In the 4D perspective you are more familiar with, isn't it just a case of 4-invariance demanding certain spatial components must therefore vary to preserve overall 4-invariance?
And yes mutual crankiness does seem to be cranking up - what!

 

[Dale]

That summary is helpful. As I have claimed global failure of charge invariance is implied by examples given earlier, that in turn is questioning how 1) and 2) are made to combine in getting to RN metric. The only coupling of relevance there is afaik that EM field acts as an exterior SET source - of further gravity. But that otherwise static gravity modifies static EM field not. You know now what point in particular I refer to - effective charge is not subject to any 'redshift'. Points 3) to 5) are just boundary conditions so no problem there.

Thanks, that really helps to, I think, pinpoint the concern.

Re: the coupling of the EFE and Maxwell's Equations (ME). It isn't just one-way. In other words, the EFE are impacted by the stress energy tensor of the EM field, but ME in curved spacetime are, in turn, impacted by the metric. So they really are a set of coupled equations, not just an input and an output.

There is a specific line in the derivation that shows this quite clearly, I will post it shortly.

EDIT: Here is the line:
"Maxwell's equations in vacuum are 0=F^{\alpha\beta}_{\;\;\; ;\beta} = |g|^{-1/2}(|g|^{1/2}F^{\alpha\beta})_{,\beta}."

I am still going through the link you posted, that won't be as "shortly".

 

[PeterDonis]

To repeat, there is no paradoxical imbalance because, as the 3-force transformation formula insists, what is measured by strain gauge moving with charge and arm, must be greater by gamma factor from what the stationary-in-lab-frame scales under the tracks measures, in order that arm has no net dp_x/dt.

I understand the argument you are making; I'm just not sure you are properly representing the actual observables in the math. However, I admit I'm having trouble myself coming up with an invariant description of both observables, O and O', so I can't exactly fault you for not coming up with one either.

We cannot directly compare readings because relative motion is screwing up an exact correspondence.

But consider an implication of this. Suppose that the reading O' does not change when we put "q"+arm in motion. (Again, we're not specifying anything about how the force on "q" is produced; just that it doesn't change when "q"+arm is in relative motion wrt scale+track.) Your argument implies that reading O must *decrease* in this scenario.

And conversely F'_x = γF - ie reciprocity of readings applies. And there is then no actual imbalance along x - dp_x/dt (net) = 0.
In the 4D perspective you are more familiar with, isn't it just a case of 4-invariance demanding certain spatial components must therefore vary to preserve overall 4-invariance?

Again, I see the argument for why the mathematical objects you are talking about should vary in the way they are varying; if we restrict attention to one frame at a time the math equations in each frame appear to describe a balance. But again, I'm not sure we agree on which particular math expressions represent the actual physical observables. However, as I said, I'm having trouble coming up with invariant expressions for the observables, so perhaps we need to table this until I can.

 

[PeterDonis]

In the 4D perspective you are more familiar with, isn't it just a case of 4-invariance demanding certain spatial components must therefore vary to preserve overall 4-invariance?

One other thought on this: in cases where what you are describing happens, the time component varies as well (for example, the obvious case of a Lorentz boost changing both the time and space components of a 4-vector).

Here the time component of dp/dt (p is the 4-momentum here, obviously) is *zero* for all the objects involved; the time component of dp/dt describes a change in energy, and that isn't happening for any of the objects. So whatever is going on with the space components, it isn't a simple case of being part of overall 4-invariance as you're describing. Another way of putting this, as I've said before, is that the "variance" in space components is not between representations of the same geometric object, the same 4-vector, in two different frames (which is where "preserving overall 4-invariance" would apply). The variance here is between the space components of two *different* geometric objects, two different 4-vectors--one describing the arm+gauge, the other describing the scale+track.

 

[TrickyDicky]

I see what you mean, but I'm not trying to apply Newton's Third Law in this way. Momentum stored in the EM field doesn't play a role in this problem, because we're talking about balancing the Lorentz force on the charge q with the mechanical force exerted on q by the arm, for a net force on q of zero. Momentum stored in the field would only come into play if q was moving freely, with the EM force the only force on it. Then, yes, you would have to keep track of momentum stored in the field for things to balance; if you only looked at the charge q itself it would seem like it was gaining momentum (and energy) "from nowhere". But in this problem q is not gaining momentum and energy at all, because it's not moving at all in the x direction and its motion in the y direction (in situation #2) is at a constant speed.

If we were restricting the discussion to SR you'd be right, but in GR there isn't such thing as a purely inertial geodesic.

 

[PeterDonis]

If we were restricting the discussion to SR you'd be right, but in GR there isn't such thing as a purely inertial geodesic.

I don't know what you mean by this. If by "purely inertial geodesic" you mean "freely falling worldline", that's obviously false; there are an infinite number of freely falling worldlines (timelike curves) in any spacetime. I can't come up with any other meaning for "purely inertial geodesic", so if the above is not what you meant, please clarify.