Black holes

[Brian-M]

You don't believe me? OK, try to explain what an "electron" is, without using any of the physics of the last ~500 years.

Challenge accepted...

An electron is an atom of lightning.



(That's atom as in the Ancient Greek philosophy of atomism, not atom as used by scientists today.)

 

[Mr.D]

No, it's a principle. The principle of equivalence isn't exact. You can tell the difference between being in an accelerating spaceship and being on a planet. There's a 1/r versus a 1/r² factor that distinguishes the two.

Really?

Could you please remind me what Einstein said about this?

 

[Vorpal]

Over extended region of spacetime, you will have curvature that tells them apart. But mathematically, the equivalence principle corresponds to an exact differential statement. One needs calculus to appreciate the significance.

 

[sol invictus]

Over extended region of spacetime, you will have curvature that tells them apart. But mathematically, the equivalence principle corresponds to an exact differential statement. One needs calculus to appreciate the significance.

There's a reasonably straightforward physical statement of it. In a given gravitational field near a given point, all experiments performed within a freely falling laboratory will give results that are identical to those in zero gravity, apart from tidal effects. The smaller the lab the less effect tidal forces have, and so if the lab is small enough they are undetectable and the results of experiments will be indistinguishable from those in zero gravity.

The same statement holds true for uniformly accelerated laboratories like those on the earth's surface (obviously, since one of the experiments you could perform in your freely-falling lab is to accelerate a smaller lab upwards).

In short, gravity and acceleration are locally indistinguishable.

 

[Vorpal]

Yep. Though that experimental statement is very clear, I think you might be a bit too used to treating elementary mathematics as obvious:

See above. It's the other way round - there's an issue with the "locally-STR nature of spacetime" in that it's of infinitesimal extent. I'm not happy with Einstein's usage of this, but nobody's perfect. It means it's actually of zero extent. So it isn't actually there at all.

A hypothesis that explains that kind of statement was suggested previously, and since then, great amounts of evidence have corroborated it.

 

[Brian-M]

No, it's a principle. The principle of equivalence isn't exact. You can tell the difference between being in an accelerating spaceship and being on a planet. There's a versus a 1/r² factor that distinguishes the two.

Really?

Could you please remind me what Einstein said about this?

What I'd like to know is what he means by r. I'm inclined to assume it means radius. If you draw a circle that intersects with an object, and place a planet or black-hole at the center of the circle, then r is the same as d, so gravitational acceleration would be proportional to 1/r².

But how is acceleration proportional to 1/r in an accelerating spaceship? I'd thought when he first mentioned this at the end of post 613 he'd mistakenly assumed I'd meant a rotating space-station, in which case acceleration would be proportional to 1/r (which is why I brought up the subject of a rotating space-station to begin with). But now he's clearly indicating an accelerating spaceship. That doesn't even make sense.

ETA: A thought experiment for Farsight...

You're in a laboratory that's hovering a fixed distance above the planet's surface. Maybe it's located on Sky High or Cloud City. Doesn't matter.

Instead of hovering over a normal planet, it's hovering over a vast black-hole with the combined mass of millions of galaxies.

But it's also hovering high enough that the force of gravity is only 1g inside the lab.

One day you realize that you've never actually seen the black hole that your lab is supposed to be hovering over, and you wonder how you can be sure that you really are hovering over a super-enormous black-hole and not simply accelerating through space.

Doing some quick calculations, you also realize that due to your great distance from the black hole, the difference in acceleration between the floor and ceiling of your lab would be so small that it'd be impossible to build an accelerometer precise enough to tell the difference, and simply placing atomic clocks at different heights would not result in any measurable loss of synchronization within your lifetime.

So how do tell whether the acceleration you're experiencing comes from gravity or accelerating through empty space, without looking outside?

What IS the difference?

 

[Mr.D]

There's a reasonably straightforward physical statement of it. In a given gravitational field near a given point, all experiments performed within a freely falling laboratory will give results that are identical to those in zero gravity, apart from tidal effects. The smaller the lab the less effect tidal forces have, and so if the lab is small enough they are undetectable and the results of experiments will be indistinguishable from those in zero gravity.

The same statement holds true for uniformly accelerated laboratories like those on the earth's surface (obviously, since one of the experiments you could perform in your freely-falling lab is to accelerate a smaller lab upwards).

In short, gravity and acceleration are locally indistinguishable.

It should be even a bit more general than that, though.

Even if we suppose our gedanken-lab is sufficiently tall enough and/or has instrumentation sensitive enough to measure the dg/dr of some hypothetical nearby gravity well, we still couldn't distinguish from a single experiment if our lab is subject to said gravity source or in a spaceship accelerating non-uniformly in just the right way.


ETA: There are hidden assumptions too, but I doubt farsight would notice/appreciate them anway.

 

[DeiRenDopa]

You don't believe me? OK, try to explain what an "electron" is, without using any of the physics of the last ~500 years.

Challenge accepted...

An electron is an atom of lightning.



(That's atom as in the Ancient Greek philosophy of atomism, not atom as used by scientists today.)

Very good!

What, then, is an ion (as in: a non-neutral atom, in the modern sense)? Is it, too, an atom of lightning? Or an atom of a certain kind of lightning?

And is an electron also an atom of 'static', the thing which crackles on a dry, cold day when removing certain items of clothing are removed (and sparkles too, in the dark)?

 

[DeiRenDopa]

What I'd like to know is what he means by r. I'm inclined to assume it means radius. If you draw a circle that intersects with an object, and place a planet or black-hole at the center of the circle, then r is the same as d, so gravitational acceleration would be proportional to 1/r².

But how is acceleration proportional to 1/r in an accelerating spaceship? I'd thought when he first mentioned this at the end of post 613 he'd mistakenly assumed I'd meant a rotating space-station, in which case acceleration would be proportional to 1/r (which is why I brought up the subject of a rotating space-station to begin with). But now he's clearly indicating an accelerating spaceship. That doesn't even make sense.

ETA: A thought experiment for Farsight...

You're in a laboratory that's hovering a fixed distance above the planet's surface. Maybe it's located on Sky High or Cloud City. Doesn't matter.

Instead of hovering over a normal planet, it's hovering over a vast black-hole with the combined mass of millions of galaxies.

But it's also hovering high enough that the force of gravity is only 1g inside the lab.

One day you realize that you've never actually seen the black hole that your lab is supposed to be hovering over, and you wonder how you can be sure that you really are hovering over a super-enormous black-hole and not simply accelerating through space.

Doing some quick calculations, you also realize that due to your great distance from the black hole, the difference in acceleration between the floor and ceiling of your lab would be so small that it'd be impossible to build an accelerometer precise enough to tell the difference, and simply placing atomic clocks at different heights would not result in any measurable loss of synchronization within your lifetime.

So how do tell whether the acceleration you're experiencing comes from gravity or accelerating through empty space, without looking outside?

What IS the difference?

Like sol, I don't speak Farspeak.

However, he may be referring to the fact that, at some limit, near the surface of a planet, the planet acts like an infinite horizontal block, so as you go 'up', the field goes as 1/r ...

 

[Farsight]

I can't grasp why you are allowed to talk about event horizons this way (as something physical, not just talking about them incorrectly I mean) when you don't want us to talk about light cones (as an example) in the same way.

There's a big difference. If you had a black hole in front of you, you see gravitational lensing surrounding a black spot. There's something actually there. Something physical. Think about what I've been saying about the speed of light going to zero where gravitational time dilation goes infinite, then think of the event horizon as a place which marks "the end of events".

There's other things I should probably say about your recent posts but my brain seems to have some kind of kernel panic when I attempt to formulate them.

Write things down, it helps.

 

[Farsight]

Clinger has nothing but abuse I see, and sol still isn't coming clean on the expression so moving on:

Really?

Yes. Really.

Could you please remind me what Einstein said about this?

He imagined a painter falling off a roof, and had a "happiest thought of my life" brainwave. The principle of equivalence and indeed GR came from this. But again, it's a principle rather than an absolute rule. If you found yourself in a series of windowless rooms, one above the other, you might be in a skyscraper or in a very long accelerating rocket. With the right equipment you could tell the difference.

ETA:

Over extended region of spacetime, you will have curvature that tells them apart. But mathematically, the equivalence principle corresponds to an exact differential statement. One needs calculus to appreciate the significance.

Well said Vorpal.

 

[Farsight]

Yep. Though that experimental statement is very clear, I think you might be a bit too used to treating elementary mathematics as obvious:

See above. It's the other way round - there's an issue with the "locally-STR nature of spacetime" in that it's of infinitesimal extent. I'm not happy with Einstein's usage of this, but nobody's perfect. It means it's actually of zero extent. So it isn't actually there at all.

A hypothesis that explains that kind of statement was suggested previously, and since then, great amounts of evidence have corroborated it.

The small extent means you're only considering the tangent of the curve and you've thrown away the Riemann curvature, which is the defining element of the gravitational field. I think Einstein's gravitational field by Pete Brown is a good read. Have a look at page 20 and the comment by Synge:

"I have never been able to understand this principle…Does it mean that the effects of a gravitational field are indistinguishable from the effects of an observer’s acceleration? If so, it is false. In Einstein’s theory, either there is a gravitational field or there is none, according as the Riemann tensor does not or does vanish. This is an absolute property; it has nothing to do with any observers world line … The Principle of Equivalence performed the essential office of midwife at the birth of general relativity, but, as Einstein remarked, the infant would never have gone beyond its long clothes had it not been for Minkowski’s concept [of space-time geometry]. I suggest that the midwife be buried with appropriate honours and the facts of absolute space-time faced".

 

[Vorpal]

The small extent means you're only considering the tangent of the curve and you've thrown away the Riemann curvature, which is the defining element of the gravitational field.

It isn't. The most important element would be the metric, which is analogous to a potential for the gravitational field, if you're going to have the term "gravitational field" in the first place (it's perfectly possible to do without explicit interpretations, e.g., as MTW do).

"I have never been able to understand this principle…Does it mean that the effects of a gravitational field are indistinguishable from the effects of an observer’s acceleration? If so, it is false. In Einstein’s theory, either there is a gravitational field or there is none, according as the Riemann tensor does not or does vanish. This is an absolute property; it has nothing to do with any observers world line …

Let's see how this argument works for electromagnetism in flat spacetime, say with some electromagnetic tensor F_αβ. Suppose that the field is such that F^α_β,γ = 0. I think that would have nearby test charges not accelerate away from each other. Now suppose someone said

Either there is an electromagnetic field or there is none, according as the F^α_β,γ tensor does not or does vanish.​

That makes no sense. The tidal forces vanishing does not mean the field vanishes. The Riemann tensor represents the gravitational tidal forces. Not the presence or absence of a gravitational field.

The Principle of Equivalence performed the essential office of midwife at the birth of general relativity, but, as Einstein remarked, the infant would never have gone beyond its long clothes had it not been for Minkowski’s concept [of space-time geometry]. I suggest that the midwife be buried with appropriate honours and the facts of absolute space-time faced".

Synge's argument against the equivalence principle seems to be based on only his personal non-understanding of it.

 

[Farsight]

What I'd like to know is what he means by r. I'm inclined to assume it means radius. If you draw a circle that intersects with an object, and place a planet or black-hole at the center of the circle, then r is the same as d, so gravitational acceleration would be proportional to 1/r².

Yes. The force of gravity follows the inverse square rule.

But how is acceleration proportional to 1/r in an accelerating spaceship? I'd thought when he first mentioned this at the end of post 613 he'd mistakenly assumed I'd meant a rotating space-station, in which case acceleration would be proportional to 1/r (which is why I brought up the subject of a rotating space-station to begin with). But now he's clearly indicating an accelerating spaceship. That doesn't even make sense.

It's to do with Born rigidity. If your accelerating spaceship was absolutely rigid, then as you climbed up the stairs to higher and higher decks the force of gravity wouldn't diminish at all. Sorry, I can't find a reference for this 1/r.

ETA: A thought experiment for Farsight...

You're in a laboratory that's hovering a fixed distance above the planet's surface. Maybe it's located on Sky High or Cloud City. Doesn't matter.

Instead of hovering over a normal planet, it's hovering over a vast black-hole with the combined mass of millions of galaxies.

But it's also hovering high enough that the force of gravity is only 1g inside the lab.

OK.

One day you realize that you've never actually seen the black hole that your lab is supposed to be hovering over, and you wonder how you can be sure that you really are hovering over a super-enormous black-hole and not simply accelerating through space.

Doing some quick calculations, you also realize that due to your great distance from the black hole, the difference in acceleration between the floor and ceiling of your lab would be so small that it'd be impossible to build an accelerometer precise enough to tell the difference, and simply placing atomic clocks at different heights would not result in any measurable loss of synchronization within your lifetime.

That's wrong. If the clock at the ceiling stayed synchronised with the clock at the floor, things wouldn't fall down.

So how do tell whether the acceleration you're experiencing comes from gravity or accelerating through empty space, without looking outside? What IS the difference?

There's a 1000-storey skyscraper in cloud city. You put clocks on every floor and plot their loss of synchronisation on a thousand-point graph. If the plot maps onto a section of the inverse square rule, you're in a gravitational field.

 

[Farsight]

It isn't. The most important element would be the metric, which is analogous to a potential for the gravitational field, if you're going to have the term "gravitational field" in the first place (it's perfectly possible to do without explicit interpretations, e.g., as MTW do).

It is the defining feature of a gravitational field. Yes the potential is more fundamental than the field, but you really cannot transform away that Riemann curvature. You can ignore it by taking a region of infinitesimal extent, but that's throwing out the baby with the bathwater.

Let's see how this argument works for electromagnetism in flat spacetime, say with some electromagnetic tensor F_αβ. Suppose that the field is such that F^α_β,γ = 0. I think that would have nearby test charges not accelerate away from each other. Now suppose someone said

Either there is an electromagnetic field or there is none, according as the F^α_β,γ tensor does not or does vanish.​

That makes no sense. The tidal forces vanishing does not mean the field vanishes. The Riemann tensor represents the gravitational tidal forces. Not the presence or absence of a gravitational field.

Again potential is more fundamental than the field, as per this section of the wikipedia article on the Aharonov-Bohn effect. But if the test particle that you place in this "field" doesn't move, you can't say that it's in an electromagnetic field. Just as you can't say you're in a gravitational field when you don't fall down.

Synge's argument against the equivalence principle seems to be based on only his personal non-understanding of it.

No. Yours is. You don't understand the electromagnetic field either, and use words like "eerie" instead of paying attention.

 

[Vorpal]

It is the defining feature of a gravitational field. Yes the potential is more fundamental than the field, but you really cannot transform away that Riemann curvature. You can ignore it by taking a region of infinitesimal extent, but that's throwing out the baby with the bathwater.

Nonsense. It has direct experimental interpretation, which has been covered. What finding a local Lorentz frame means mathematically is that at any event, it's possible to find a frame in which not only is the metric takes the Minkowski there, but the its first derivatives vanish identically.

No one claimed that you can transform away Riemann curvature, and it's very strange that you seem to think that anyone did. The above doesn't mean that the curvature vanishes, because that depends entirely on the second derivatives (just as analogously in Newtonian gravity, tidal forces are found by second derivatives of the potential). And it's not some triviality. For one, it directly implies that a particle at rest in a local Lorentz frame is moving along a geodesic.

Again potential is more fundamental than the field, as per this section of the wikipedia article on the Aharonov-Bohn effect.

Once again, you bring up something completely and utterly irrelevant. It has nothing to do with what I said, and I'm not interested in discussing that with you. Let's stay on topic here:
1) Synge claims that there is no gravitational field present whenever gravitational tidal forces (i.e., Riemann curvature tensor) vanishes.
2) I say that's a nonsense implication.
Your comment addresses precisely 0% of that.

But if the test particle that you place in this "field" doesn't move, you can't say that it's in an electromagnetic field. Just as you can't say you're in a gravitational field when you don't fall down.

I don't know whether you're aware of this, but things like the electric field are defined by the behavior of test charges. In fact, it is precisely when a test charge does not move in some inertial frame that you are measuring the electric field directly. As for relationship between gravitational freefall and not moving in some frame, see above.

No. Yours is. You don't understand the electromagnetic field either, and use words like "eerie" instead of paying attention.

What are you on about? Some time ago I did say that the "metric:connection coefficients :: potential:field" analogy works so well as to be "almost eerie." Do you have an argument that the identification of gravitational field and the connection coefficients is wrong?** Because that would actually begin to be relevant. My subjective sense wonder is less so.


P.S. If you had any idea of how the EM field works, you'd have realized just how utterly off the mark your comment above was. Literally, the only thing I'd have to add to meet your pedantic standards is add "oh, and F_αβ = A_β,α - A_α,β" and absolutely nothing about the point changes.
**P.P.S. And if you do: why do you hate Einstein, Farsight?

 

[Farsight]

Nonsense. It has direct experimental interpretation, which has been covered. What finding a local Lorentz frame means mathematically is that at any event, it's possible to find a frame in which not only is the metric takes the Minkowski there, but the its first derivatives vanish identically.

Big deal, you're in a falling box and everything is floating around, but look out the window and see the ground coming up fast. You'll soon be disabused of the notion that the gravitational field has vanished.

No one claimed that you can transform away Riemann curvature, and it's very strange that you seem to think that anyone did.

Then make it clear to all the guys that this is what distinguishes a real live gravitational field from sitting in an accelerating spaceship.

The above doesn't mean that the curvature vanishes, because that depends entirely on the second derivatives (just as analogously in Newtonian gravity, tidal forces are found by second derivatives of the potential). And it's not some triviality. For one, it directly implies that a particle at rest in a local Lorentz frame is moving along a geodesic.

And therefore isn't at rest. It's only at rest with respect to you because you're moving too.

Once again, you bring up something completely and utterly irrelevant. It has nothing to do with what I said, and I'm not interested in discussing that with you.

You brought up the "eerie" connection with electromagnetism, and you referred to the electromagnetic field. What I said concerned potential, and was absolutely relevant.

Let's stay on topic here:
1) Synge claims that there is no gravitational field present whenever gravitational tidal forces (i.e., Riemann curvature tensor) vanishes.
2) I say that's a nonsense implication.
Your comment addresses precisely 0% of that.

I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication.

I don't know whether you're aware of this, but things like the electric field are defined by the behavior of test charges. In fact, it is precisely when a test charge does not move in some inertial frame that you are measuring the electric field directly. As for relationship between gravitational freefall and not moving in some frame, see above.

It's the electromagnetic field, not the electric field.

What are you on about? Some time ago I did say that the "metric:connection coefficients :: potential:field" analogy works so well as to be "almost eerie." Do you have an argument that the identification of gravitational field and the connection coefficients is wrong?** Because that would actually begin to be relevant. My subjective sense wonder is less so.

P.S. If you had any idea of how the EM field works, you'd have realized just how utterly off the mark your comment above was. Literally, the only thing I'd have to add to meet your pedantic standards is add "oh, and F_αβ = A_β,α - A_α,β" and absolutely nothing about the point changes.

The point charges? There are no point charges. I know how the EM field works, and why the analogy works. You don't.

**P.P.S. And if you do: why do you hate Einstein, Farsight?

What? I don't hate Einstein.

 

[sol invictus]

Big deal, you're in a falling box and everything is floating around, but look out the window and see the ground coming up fast. You'll soon be disabused of the notion that the gravitational field has vanished.

And that is relevant... how? The precise statement of the equivalence principle concerns local experiments. Looking out the window isn't a local experiment, at least not unless you're close to the earth - but in that case, you cannot distinguish between the earth accelerating towards you and you falling towards the earth.

Then make it clear to all the guys that this is what distinguishes a real live gravitational field from sitting in an accelerating spaceship.

We all know that a non-zero Reimann tensor cannot be transformed away by a coordinate transformation, because (unlike you) we know what the term "tensor" means. The Christoffel connection is a different story (it's not a tensor), and it's a better analog of "gravitational field" than the Riemann tensor is.

What? I don't hate Einstein.

Then why do you continuously contradict and mis-represent what he said about his own theory?

 

[DeiRenDopa]

It is the defining feature of a gravitational field. Yes the potential is more fundamental than the field, but you really cannot transform away that Riemann curvature. You can ignore it by taking a region of infinitesimal extent, but that's throwing out the baby with the bathwater.

Nonsense. It has direct experimental interpretation, which has been covered. What finding a local Lorentz frame means mathematically is that at any event, it's possible to find a frame in which not only is the metric takes the Minkowski there, but the its first derivatives vanish identically.

No one claimed that you can transform away Riemann curvature, and it's very strange that you seem to think that anyone did. The above doesn't mean that the curvature vanishes, because that depends entirely on the second derivatives (just as analogously in Newtonian gravity, tidal forces are found by second derivatives of the potential). And it's not some triviality. For one, it directly implies that a particle at rest in a local Lorentz frame is moving along a geodesic.

Again potential is more fundamental than the field, as per this section of the wikipedia article on the Aharonov-Bohn effect.

Once again, you bring up something completely and utterly irrelevant. It has nothing to do with what I said, and I'm not interested in discussing that with you. Let's stay on topic here:
1) Synge claims that there is no gravitational field present whenever gravitational tidal forces (i.e., Riemann curvature tensor) vanishes.
2) I say that's a nonsense implication.
Your comment addresses precisely 0% of that.

But if the test particle that you place in this "field" doesn't move, you can't say that it's in an electromagnetic field. Just as you can't say you're in a gravitational field when you don't fall down.

I don't know whether you're aware of this, but things like the electric field are defined by the behavior of test charges. In fact, it is precisely when a test charge does not move in some inertial frame that you are measuring the electric field directly. As for relationship between gravitational freefall and not moving in some frame, see above.

No. Yours is. You don't understand the electromagnetic field either, and use words like "eerie" instead of paying attention.

What are you on about? Some time ago I did say that the "metric:connection coefficients :: potential:field" analogy works so well as to be "almost eerie." Do you have an argument that the identification of gravitational field and the connection coefficients is wrong?** Because that would actually begin to be relevant. My subjective sense wonder is less so.


P.S. If you had any idea of how the EM field works, you'd have realized just how utterly off the mark your comment above was. Literally, the only thing I'd have to add to meet your pedantic standards is add "oh, and F_αβ = A_β,α - A_α,β" and absolutely nothing about the point changes.
**P.P.S. And if you do: why do you hate Einstein, Farsight?

Perhaps, like me, you're wondering what Farsight is talking about, in this exchange.

I mean, Vorpal is fairly polite, but is basically saying "you're not making a bit of sense, Farsight!" And I expect that every reader shares that sentiment.

Well, I have a simple, straight-forward explanation that I'd like to present, for your consideration; namely, that Farsight's basic understanding of physics is so radically different from Vorpal's (and, I should add, almost everyone else's, here in this thread) that there's no communication actually taking place here.

And why is this? I mean, there's nothing particularly unusual in what Vorpal writes, is there? Nothing that you won't find in a suitable, contemporary physics textbook, is there?

Well, Farsight's physics is, fundamentally, non-quantitative. It is the physics of the time before Galileo and Newton, a physics which is done without numbers, equations, abstract symbols, and the like; it is a physics like most religious studies, the reading of sacred texts, and attempting to divine meaning from the words alone. There is no room in Farsight's physics for any mathematics.

In short, as I said earlier, Farsight doesn't have a quantitative clue.

If I am right, then the only meaningful dialog anyone can have with Farsight - on physics-related topics (excluding the history of physics and philosophy) - needs to begin with the quantitative foundations of physics. And if Farsight is unwilling to engage in such discussions, then ...

 

[Vorpal]

Big deal, you're in a falling box and everything is floating around, but look out the window and see the ground coming up fast. You'll soon be disabused of the notion that the gravitational field has vanished.

Nope, not in that frame, I wouldn't. "It's not the fall that kills you--it's the sudden stop at the end." My terminal ends is because the ground puts a tremendous force on me.

I don't think you understand what local means.

And therefore isn't at rest. It's only at rest with respect to you because you're moving too.

Of course--what is or isn't rest is frame-dependent. But the major difference is that in Newtonian physics, making a frame that comoves with a gravitationally freefalling particle gives an accelerated frame--it doesn't follow a geodesic. In GTR, it does: the acceleration four-vector along the trajectory identically vanishes. That's the point of EP.

You brought up the "eerie" connection with electromagnetism, and you referred to the electromagnetic field. What I said concerned potential, and was absolutely relevant.

Synge's argument is about the relationship between the gravitational field and the gravitational tidal forces. You're simply claiming some relevance without even a word as to why it is so.

It's the electromagnetic field, not the electric field.

Since you yourself specified the case where the charge is not moving in our local inertial frame, the electric field is what you get to measure. Remember Lorentz force and it'll be obvious.

The point charges? There are no point charges.

For the purposes of this discussion, it doesn't matter whether there are or aren't any point charges. Read what I wrote again: I'm saying that even if that particular statement of yours was 100% absolute truth, nothing about my point changes. Because what you said literally adds nothing more than a side comment of "... and the field comes from the potential." Yes, everyone knows that. I don't mind if you consider it the mostest fundamentalest thing there ever was. What I said was:
(*) the gravitational field and gravitational tidal forces are different things
and the electromagnetic field was brought in to illustrate why Synge's argument is mistaken: because we can define a tensor that describes electromagnetic tidal forces, in the sense of deviation of equally-charged test particles, and directly confirm that its vanishing does not imply that the electromagnetic field vanishes.

Or, more bluntly: Synge's implication is very clearly false for other fields. Why should we believe him about the gravitational field?

I know how the EM field works, and why the analogy works. You don't.

Evidently, you don't agree with it, because if you accepted it as valid, you'd see that Synge is wrong immediately. Hence my question about your reasons against it.

What? I don't hate Einstein.

It's just poking fun at the fact that for someone who incessantly insistent that he was "with Einstein", you spend extraordinary amounts of time arguing against him. And I'll put this last:

I've said it already. When gravitational tidal forces vanish you've got a gravitational field that doesn't diminish with distance. That's a nonsense implication.

What's wrong with that? If you believe the analogy to be valid, then the gravitational field is the connection, and it's trivial that all you need for a homogeneous but nonvanishing gravitational field is to take flat spacetime in a uniformly accelerated frame. So I ask yet again: do you have any reasons to consider the analogy wrong?