Is 'charged black hole' an oxymoron?

[jartsa]

The statement was wrong*, regardless of what else you may have said wrong or right previously.

*as was the statement about a photon in a box being weightless

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.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.

 

[Austin0]

A photon in a box, bouncing up and down, in a perfectly vertical direction, in a gravity field, is weightless. That means, there is no momentum exchange with the gravitating body.

That's because, as I have explained, a falling photon gains no energy, so it does not gain any momentum either.

Magnetic field never does any work on moving charges. Gravity field never does any work on photons.

What did I write that you could interpret as my saying a falling photon gains energy or momentum?

 

[jartsa]

What did I write that you could interpret as my saying a falling photon gains energy or momentum?

Well actually, I don't know what you said. Something about photons in a box ...

So I added my two cents about photons in a box.

 

[Q-reeus]

If you are writing a covariant equation, you can't use 3-momentum; 3-vectors are never covariant in 4-D spacetime. So all the covariant quantities and equations I've written, including dp/dtau, use 4-momentum.
The Lorentz force equation you wrote down does use p to mean 3-momentum, I agree; that's because it's written in a particular frame, where we split 4-momentum into energy (a scalar) and momentum (a 3-vector).

Fine with that.

True, the fact of static equilibrium by itself doesn't tell us how the magnitudes will be represented in different frames. But it does tell us that the magnitudes must be *equal and opposite*. That, by itself, is enough to show that the two observables, O and O', are equal. I gave the chain of equal and opposite magnitudes that shows it. That chain of reasoning works regardless of which frame you choose to evaluate the magnitudes...

You were OK until you added the qualifier "in a given frame". Once again, the two observables, O and O', are actual observed numbers; they are readings on a scale. The readings on a scale cannot depend on what frame you use to calculate them. Nor can they depend on the state of motion of the "observer" relative to the scale. So they are not numbers "in a given frame"; I must be able to calculate the numbers using quantities represented in *any* frame, and come out with the same answer. You've agreed to this before, but you keep on using language that just begs to be interpreted as contradicting it.

Trouble is you have here that O and O' are readings on one and the self-same scale - which moves with the charge q. In which case I obviously agree that when someone holds up a placard reading '2', an observer moving at u relative to the placard holder still sees '2'. But that is not the definition of O and O' I gave back in first para of #102 and to which seemingly we both agreed to work from. That definition had O the reading on one scale static wrt capacitor (call it lab frame S), whereas O' was the reading on a different scale - that moving with the charge q in frame S' having relative speed u wrt S. Both O and O' are perfectly legitimate physical observables. Both detecting the same phenomenon - force on a given charge q. And one reads differently to the other! Now if working definitions have transmogrified along the way, little wonder there is continued disagreement. it impacts on the rest of your #132 the same way.

 

[Q-reeus]

Q-reeus: "or how they co-vary as measured in another frame."
False: static equilibrium is static equilibrium, regardless of relative motion in directions orthogonal to the equilibrium. That means the magnitudes that are equal and opposite stay equal and opposite regardless of that relative motion.

Quite so. Opened myself to attack on that one but it was just down to an inapt use of terminology. By "co-vary as measured in another frame" it was meant as "both varying together equally as measured in another frame". No real disagreement there.

So O and O' have to remain equal when the charge is moving relative to the capacitor, because the motion is orthogonal to the direction of the equilibrium. Their actual magnitude changes with relative motion; it goes up, by the first point above. But they are still equal and opposite; both magnitudes change in sync.

Last bit true but the confusion and disagreement is over your current definition of O and O', as discussed last post. But then it changes below...

Consider another scenario, not involving electromagnetism. I put weights on a scale and set the scale on a trolley which sits at rest on a track. There is another scale under the track, measuring the weight of the track and whatever is on it. The scales, trolley, and track all have negligible mass compared to the weights. I define two observables: O, the weight shown on the scale under the track, and O', the weight shown on the scale on the trolley.

Great so far - you have moved back to original definition in #102 of what O and O' refer to.

To the best of my knowledge, all of the following are true:
(1) When the trolley sits at rest on the track, O = O'. Call this reading A.

There are two readings involved here so I would prefer to put it as situation (1): O_A = O'_A

(2) When the trolley is rolling along the track at a constant velocity v, O = O'. (We assume the track is perfectly horizontal, so the trolley's motion is perfectly orthogonal to the direction of gravity.) Call this reading B.

My rewording; situation (2) has O_B = O'_B.

(3) In the non-relativistic approximation (v << c), reading A will be the same as reading B.

However, if we allow v to be relativistic, then the gravitational "force" on the weights is no longer just the standard Newtonian force; there is an additional velocity-dependent force which can be viewed as due to "gravitomagnetism". (In the limit as v -> c, the total force goes to twice the Newtonian value; another way of saying this is that "acceleration due to gravity" is higher on a relativistically moving object. This is why light grazing the Sun, for example, bends by twice the amount you would predict by just doing a Newtonian calculation for a mass "falling" in the Sun's field, where the mass is the light's energy divided by c^2.) So in the case of relativistic v, reading B will be larger than reading A. (I don't think anyone has actually run this experiment, of course; a real experiment like this done on Earth could only last a small fraction of a second at relativistic velocities without violating the "perfectly horizontal" requirement.) But O = O' will still hold even in this case.

I have a different take. Treating above scenario (2) as straight-line equivalent of particles circulating in a synchrotron storage ring, we assume that transferring energy from some source (e.g. bank of batteries) to the circulating particles of relativistic energy γm (m the rest energy), no overall change in gravitating mass has occurred.

[My turn to clarify better here. 'Overall' is referring to that of source+circulating particles taken together - not that the circulating particles show no net change in gravitating mass - they clearly do do! Hence if a quantity W of battery PE energy is converted into particle KE, gravitating mass lost in batteries is precisely added to particles.]

Which requires by analogy that in (2), (3),:
O_B = γO_A, O'_B = γ²O'_A, so O_B = O'_B/γ.

The matter of that doubling the Newtonian deflection has intrigued me when it comes to applying it to a spinning flywheel with spin axis oriented somewhere between radial and transverse to say Earth's gravitational field. I had at one time thought radial orientation of axis might lead to flywheel weighing more than with transverse orientation. Suggesting oblique orientation yielding a transverse net acceleration. Which no doubt is equivalence principle violating - and I presume coincidentally checked along with the precessional stuff re gravity probe B experiment? Or there have been Eotvos style experiments specifically checking that spin/orbital motion, either macroscopic (gyro) or microscopic (magnetized material) has no orientational gravitational/inertial effect re 'weight'?
And, umm, we are sort of getting pretty sidetracked from OP topic I believe...HINT!

 

[PeterDonis]

Fine with that.

Trouble is you have here that O and O' are readings on one and the self-same scale - which moves with the charge q.

No, they're not. My definition matches yours, as given just a bit later in your post:

That definition had O the reading on one scale static wrt capacitor (call it lab frame S), whereas O' was the reading on a different scale - that moving with the charge q in frame S' having relative speed u wrt S.

This is my definition too, and that's what I was using when I made my post. I see now that I worded it as "a reading on a scale", which doesn't make it clear that O and O' were supposed to refer to the readings on two *different* scales, as you describe. Sorry if that wasn't clear.

Both O and O' are perfectly legitimate physical observables. Both detecting the same phenomenon - force on a given charge q.

I would say that they are both *determined* by the same phenomenon, the Lorentz force on the charge q. One, O', is determined by that force more directly than the other (fewer pieces in between), but they both ultimately are determined by the same force. But that means that this:

And one reads differently to the other!

Makes no sense! The two scales are in force balance in the x direction; the relative motion in the y direction does not change that. They have to read the same, or they won't be motionless in the x direction; the forces in that direction won't be in balance and something will move. As you have it, with O < O' (force measured at the arm greater than force measured at the track), the arm will push the track in the negative x direction.

 

[PeterDonis]

I would say that they are both *determined* by the same phenomenon, the Lorentz force on the charge q. One, O', is determined by that force more directly than the other (fewer pieces in between), but they both ultimately are determined by the same force.

Maybe it's worth expanding on this a bit to see if it clears up an issue. Both O and O' are *determined*, ultimately, by the Lorentz force on the charge q, but neither one "detects" that force directly. That is, if I write down the most fundamental local equation that determines O and O', neither one will have the Lorentz force in it! The only fundamental local equation that has the Lorentz force in it is the force balance directly on the charge q: that balances the Lorentz force on q, q E' = q gamma E (E' is in the rest frame of q, E in the rest frame of the capacitor), against the force exerted on q by the arm, which must be equal and opposite for q to be motionless.

But the force exerted by q on the arm is not what the scale attached to the arm measures; it measures the force exerted on the arm by q. That force is determined by Newton's Third Law, not the Lorentz force law. So observable O' is not directly "detecting" the force on the charge q; it's only "detecting" it indirectly, through the force exerted on the arm by q.

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.

 

[PeterDonis]

Great so far - you have moved back to original definition in #102 of what O and O' refer to.

See my previous two posts; I never changed the definition, but my wording may not have made that clear. Hopefully it's clarified now.

There are two readings involved here so I would prefer to put it as situation (1): O_A = O'_A

My rewording; situation (2) has O_B = O'_B.

I have no real problem with this notation; I'm just too lazy to type a lot of tags, so I tend to prefer notation that doesn't require me to do that.

I have a different take. Treating above scenario (2) as straight-line equivalent of particles circulating in a synchrotron storage ring, we assume that transferring energy from some source (e.g. bank of batteries) to the circulating particles of relativistic energy γm (m the rest energy), no overall change in gravitating mass has occurred. Which requires by analogy that in (2), (3),:
O_B = γO_A, O'_B = γ²O'_A, so O_B = O'_B/γ.

I'm not sure this analogy holds; I'll have to take a bit of time to consider the rotating scenario. My only comment right now is that I would not use "gravitating mass" here; I would instead say that, if the system as a whole is isolated, then its total 4-momentum and total angular momentum must be conserved. That makes it clear that we are only considering flat spacetime; we are not considering the self-gravity of the system, only the flat spacetime effects of rotation, internal energy/momentum exchange, etc.

The matter of that doubling the Newtonian deflection has intrigued me...we are sort of getting pretty sidetracked from OP topic I believe...HINT!

It intrigues me too, but I agree it's getting too far off topic for this thread. There was at least one other recent thread on the spinning flywheel in a gravity field scenario; I don't remember offhand where that one ended up.

 

[Q-reeus]

This is my definition too, and that's what I was using when I made my post. I see now that I worded it as "a reading on a scale", which doesn't make it clear that O and O' were supposed to refer to the readings on two *different* scales, as you describe. Sorry if that wasn't clear.

OK thanks for clarifying, but wish my earlier misinterpretation on that was right, as it's back to disagreeing on reading O, which you maintain is identical to O'.

Makes no sense! The two scales are in force balance in the x direction; the relative motion in the y direction does not change that. They have to read the same, or they won't be motionless in the x direction; the forces in that direction won't be in balance and something will move. As you have it, with O < O' (force measured at the arm greater than force measured at the track), the arm will push the track in the negative x direction.

No it won't. 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 see no greater paradox here than the more familiar one involving clocks and rulers in different frames - in particular with one frame being a rotating one so the differing readings are not reciprocally observed. None of these imperfect analogies may impress you and it's getting to be a last ditch effort at seeking accord.

So I ask you to just clarify please your take on the mechanical example of spinning hollow cylinder given in first part of last paragraph in #102. What do you believe will be observed, and what explains it as seen in the two frames. I completely skip your #147 as answer to this problem may help change one of our outlooks fundamentally - cross fingers!

 

[Q-reeus]

I'm not sure this analogy holds; I'll have to take a bit of time to consider the rotating scenario. My only comment right now is that I would not use "gravitating mass" here; I would instead say that, if the system as a whole is isolated, then its total 4-momentum and total angular momentum must be conserved. That makes it clear that we are only considering flat spacetime; we are not considering the self-gravity of the system, only the flat spacetime effects of rotation, internal energy/momentum exchange, etc.

Sure I agree it's all analyzable using SR principles. Please note my edit in #145 - can't afford any more misunderstandings!

It intrigues me too, but I agree it's getting too far off topic for this thread. There was at least one other recent thread on the spinning flywheel in a gravity field scenario; I don't remember offhand where that one ended up.

OK - and best move back to center as we're not out to break the record for longest running thread. :zzz:

 

[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.