Black holes

[ben m]

You lost me, and everybody else. What does sol's metric represent again? All I've managed to squeeze out of him is "flat spacetime", and more recently "constant proper acceleration".

Nope. Clinger has applied the standard differential geometry to Sol's standard presentation of a spacetime metric. Just like I did a few pages ago, except that I used Mathematica to brute-force the indices instead of thinking them through one by one.

I repeat, this is bog-standard geometry, and if you don't understand this it means you don't know what a metric is. I also repeat my quote from Misner, Thorne, and Wheeler, a textbook accessible to 20-year-old undergrads but (apparently) opaque to you.

Box 14.2: Straightforward curvature computation. The elementary and universally applicable method for computing the components R^mu_nu_alpha_beta of the Riemann curvature tensor starts from the metric components g_mu_nu in a coordinate basis ... to compute the curvature by the standard method use the formula for ds^2

Clinger's calculation simply "computes the curvature by the standard method", which is "universally applicable", using only "the formula for ds^2" which is what Sol gave you. This isn't specific to MTW, or to Einstein. This is the basic mathematical work that Einstein used to build GR.

The mathematical apparatus useful for the general relativity theory, lay already complete in the "Absolute Differential Calculus", which were based on the researches of Gauss, Riemann and Christoffel on the non-Euclidean manifold

 

[Complexity]

And I sense a troll. Now start making a sincere contribution to this discussion, or sling your hook.

Nonsense.

He is pointing out that you are consistently wrong, deluded, and pigheaded.

The rest of us have been making the same observations, to no effect.

I don't think any of us has any real expectation of educating you or changing your mind, because I don't think you are capable of those things.

We, especially the real phycisists here, are trying to explain to current and future readers why you are wrong.

You merely provide the excuse for the lesson.

 

[sol invictus]

I'm not lying about Einstein, I'm quoting him.

You're lying, as Clinger already pointed out:

Farsight: "When spacetime is flat, no gravitational field is present."
Einstein: "the Γ^τ_μν...are the components of the gravitational field."

Γ^τ_μν is non-zero even in flat spacetime in some (actually, almost all) coordinate systems - for instance, the coordinates I've been asking you about all this time.

The change in spin orientation is a movement

That's not true - but even if it was, you'd still be wrong. The spin can stay exactly constant even in the presence of a magnetic field.

You were wrong. Admit it, learn from it, move on. Can you do that?

 

[sol invictus]

You lost me, and everybody else.

No, Farsight. It's just you.

 

[Ziggurat]

You lost me, and everybody else.

You keep presuming to speak for everybody else. And yet, your pronouncements about what everyone else thinks keep being wrong. He might have lost some people, but he most certainly didn't lose everybody. And if he lost you, it's because you don't know GR. But then, that has been obvious for a long time.

What does sol's metric represent again? All I've managed to squeeze out of him is "flat spacetime"

That's exactly what it represents. And W.D. just got done proving it was flat spacetime in that very post.

and more recently "constant proper acceleration".

That's more properly a description of how the coordinates Sol used relate to the more standard coordinates for flat spacetime. But it's still the same spacetime.

You can't make a gravitational field by zooming through space, Clinger.

Call it what you will, the metric Sol gave is a correct metric for flat spacetime. And it has exactly the features you object to for a black hole.

Things don't fall down as you pass by.

What makes you say that?

Look, go back to what Einstein said and think it through then contribute sensibly to the discussion instead of hiding behind mathematics.

The words Einstein used were never as important as the math he used. W.D. isn't hiding behind math, you are hiding behind words. But then, you still haven't gotten this whole quantifiability thing down.

Do you get it yet? Space isn't nothing. And you don't change space by moving through it.

Who said you did? The metric is just a representation. Space is the same regardless of what metric you use. Nobody is contesting that. But that is rather the point: you have mistaken features of the metric for features of the space. And so when Sol presents you with a metric for flat spacetime which has these strange features, you can't grasp that they are only features of the metric, and of the space itself. So you object to the metric, because you can't accept that there would be such a difference, since doing so would invalidate your entire thesis about black holes. In essence you, not Sol, have taken the position that space is changed by the choice of metric. But because you can't quantify anything, because you can't understand any of the math, you can't even grasp that this is what you have done.

 

[Hellbound]

Zig,

Have I told you lately that you're my hero?

Excellent post.

 

[Reality Check]

You lost me, and everybody else.

Wrong again, Farsight - you are the only one lost.
As soon as I saw the metric I suspected that it was flat spacetime and W.D.Clinger confirmed it.

 

[Dancing David]

And I sense a troll. Now start making a sincere contribution to this discussion, or sling your hook.

Rudeness might be part of the pattern as well.

 

[Farsight]

You keep presuming to speak for everybody else. And yet, your pronouncements about what everyone else thinks keep being wrong. He might have lost some people, but he most certainly didn't lose everybody. And if he lost you, it's because you don't know GR. But then, that has been obvious for a long time.

I'm not wrong about the speed of light varying with gravitational potential, which is what Einstein said repeatedly whilst formulating GR. We can see it. And I'm not wrong about a gravitational field being inhomogeneous space, which Einstein said in 1920. Clinger's dollop of Emperor's New Clothes doesn't change that one bit. Care to list the terms, explain what they mean, and talk us through that maths? Never seems to happen does it?

That's exactly what it represents. And W.D. just got done proving it was flat spacetime in that very post.

And yet he presented it as a fait accompli without explanation, and there's still no explanation of what [latex]$ds^2 = -(r-r_0) dt^2 + dr^2/(r-r_0) + dy^2 + dz^2$[/latex] represents. This is getting laughable. It's like priests in a corner desperately spouting Latin.

That's more properly a description of how the coordinates Sol used relate to the more standard coordinates for flat spacetime. But it's still the same spacetime.

I related the standard coordinates of flat space time to sol's. And sol is still fighting shy of explaining why his expression relates to [latex]$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$[/latex]. Talk about smoke and mirrors. Do you really think you can ignore the hard scientific evidence and what Einstein said, and get away with it by use of a mathematical smokescreen?

You can't make a gravitational field by zooming through space, Clinger.

Call it what you will, the metric Sol gave is a correct metric for flat spacetime. And it has exactly the features you object to for a black hole.

And it is academic, because light doesn't curve in flat spacetime. A metric is a description of what you measure, and your motion through space changes your measurements, but it doesn't change the other things in space. It won't make a light beam curve, just as running through the rain doesn't make it slant. That's just how it looks to you.

Things don't fall down as you pass by.

What makes you say that?

You can't create a gravitational field in space by moving through that space. If it isn't there, it isn't there.

The words Einstein used were never as important as the math he used. W.D. isn't hiding behind math, you are hiding behind words. But then, you still haven't gotten this whole quantifiability thing down.

Yeah yeah, dismiss Einstein.

Who said you did? The metric is just a representation. Space is the same regardless of what metric you use. Nobody is contesting that. But that is rather the point: you have mistaken features of the metric for features of the space.

No, I haven't. You have. See below.

And so when Sol presents you with a metric for flat spacetime which has these strange features, you can't grasp that they are only features of the metric, and of the space itself.

That's my bold. You're the one confusing the metric with space Zig. Just as you confused distance with displacement.

So you object to the metric, because you can't accept that there would be such a difference, since doing so would invalidate your entire thesis about black holes. In essence you, not Sol, have taken the position that space is changed by the choice of metric. But because you can't quantify anything, because you can't understand any of the math, you can't even grasp that this is what you have done.

Garbage. I'm objecting to the lack of explanation of terms and scenario. And I'm the one who said you don't change space by zooming through it. Those light beams don't curve. Look, there's Zig zooming through space, riding his negative carpet. And oh wait, he's zooming past a light beam. See it curve? Er, no.

 

[edd]

That's my bold. You're the one confusing the metric with space Zig. Just as you confused distance with displacement.

Sounds like he typo'd. We've all done it, I know I have.

While I'm posting light rays moving radially to a black hole don't curve either (well, depending on one's point of view, but I think it's clear enough what I mean). They might red or blueshift of course, but light rays can do that in flat space too.

 

[sol invictus]

And I'm not wrong about a gravitational field being inhomogeneous space, which Einstein said in 1920.

You're completely wrong, not just about the physics, but about what Einstein said (in 1920 and at other times). For example:

WE imagine a large portion of empty space, so far removed from stars and other appreciable masses that we have before us approximately the conditions required by the fundamental law of Galilei....As reference-body let us imagine a spacious chest resembling a room with an observer inside who is equipped with apparatus. Gravitation naturally does not exist for this observer....To the middle of the lid of the chest is fixed externally a hook with rope attached, and now a “being” (what kind of a being is immaterial to us) begins pulling at this with a constant force. The chest together with the observer then begin to move “upwards” with a uniformly accelerated motion. ...how does the man in the chest regard the process? The acceleration of the chest will be transmitted to him by the reaction of the floor of the chest. He must therefore take up this pressure by means of his legs if he does not wish to be laid out full length on the floor. He is then standing in the chest in exactly the same way as anyone stands in a room of a house on our earth. If he release a body which he previously had in his hand, the acceleration of the chest will no longer be transmitted to this body, and for this reason the body will approach the floor of the chest with an accelerated relative motion. The observer will further convince himself that the acceleration of the body towards the floor of the chest is always of the same magnitude, whatever kind of body he may happen to use for the experiment.
Relying on his knowledge of the gravitational field (as it was discussed in the preceding section), the man in the chest will thus come to the conclusion that he and the chest are in a gravitational field which is constant with regard to time. Of course he will be puzzled for a moment as to why the chest does not fall in this gravitational field. Just then, however, he discovers the hook in the middle of the lid of the chest and the rope which is attached to it, and he consequently comes to the conclusion that the chest is suspended at rest in the gravitational field.
Ought we to smile at the man and say that he errs in his conclusion? I do not believe we ought if we wish to remain consistent; we must rather admit that his mode of grasping the situation violates neither reason nor known mechanical laws. Even though it is being accelerated with respect to the “Galileian space” first considered, we can nevertheless regard the chest as being at rest. We have thus good grounds for extending the principle of relativity to include bodies of reference which are accelerated with respect to each other, and as a result we have gained a powerful argument for a generalised postulate of relativity.

According to Einstein gravity can exist even in perfectly flat, homogeneous spacetime. You're directly contradicting what Einstein said in 1920, Farsight. Moving on:

And it is academic, because light doesn't curve in flat spacetime.
....
And I'm the one who said you don't change space by zooming through it. Those light beams don't curve. Look, there's Zig zooming through space, riding his negative carpet. And oh wait, he's zooming past a light beam. See it curve? Er, no.

we obtain a new result of fundamental importance when we carry out the analogous consideration for a ray of light. With respect to the Galileian reference-body K, such a ray of light is transmitted rectilinearly with the velocity c. It can easily be shown that the path of the same ray of light is no longer a straight line when we consider it with reference to the accelerated chest (reference-body K'). From this we conclude, that, in general, rays of light are propagated curvilinearly in gravitational fields.

According to Einstein, rays of light curve in perfectly flat spacetime when viewed from accelerated reference frames, because acceleration is gravity and light curves in response to gravity. Wrong again, Farsight.

 

[ben m]

Care to list the terms, explain what they mean, and talk us through that maths? Never seems to happen does it?

Are you deaf? ds is an interval. t,r,x,y are coordinates, labeling a network of points in some spacetime. r0 is a constant with the same units as r. The equation relating them includes BOTH information about the shape of the spacetime, and about how the coordinate-labels are laid down.

I repeat, for the third time: what Clinger (and Sol and I and apparently others) have done is (in the words of MTW) an "elementary and universally applicable" bit of differential geometry.

And yet he presented it as a fait accompli without explanation, and there's still no explanation of what [latex]$ds^2 = -(r-r_0) dt^2 + dr^2/(r-r_0) + dy^2 + dz^2$[/latex] represents. This is getting laughable. It's like priests in a corner desperately spouting Latin.

"Hey, you claim to be doing calculus, but you wrote down this big-squiggly-S thing and you stick the letter "d" in your equations without explaining it."

ETA: Here is Einstein telling you to learn the math:

The mathematical apparatus useful for the general relativity theory, lay already complete in the "Absolute Differential Calculus", which were based on the researches of Gauss, Riemann and Christoffel on the non-Euclidean manifold, and which have been shaped into a system by Ricci and Levi-Civita, and already applied to the problems of theoretical physics. I have in part B of this communication developed in the simplest and clearest manner, all the supposed mathematical auxiliaries, not known to Physicists, which will be useful for our purpose ...

 

[Farsight]

Nope. Clinger has applied the standard differential geometry to Sol's standard presentation of a spacetime metric. Just like I did a few pages ago, except that I used Mathematica to brute-force the indices instead of thinking them through one by one.

A spacetime metric that represents what, exactly? Here it is, it's like Pythagoras's theorem, we've got a spacetime interval and the usual y and z terms, but we've now got r terms. What do they represent exactly, and what real-world scenario does the overall expression describe?

[latex]$ds^2 = -(r-r_0) dt^2 + dr^2/(r-r_0) + dy^2 + dz^2$[/latex]

I repeat, this is bog-standard geometry, and if you don't understand this it means you don't know what a metric is. I also repeat my quote from Misner, Thorne, and Wheeler, a textbook accessible to 20-year-old undergrads but (apparently) opaque to you.

I know what a metric is.

Clinger's calculation simply "computes the curvature by the standard method", which is "universally applicable", using only "the formula for ds^2" which is what Sol gave you. This isn't specific to MTW, or to Einstein. This is the basic mathematical work that Einstein used to build GR.

But we still don't know what Clinger is computing the curvature of now do we? It's been weeks now, you've got nothing, and you're still trying to dismiss Einstein and scientific evidence and defend your waterfall neverland fantasies with mathematics that you will not relate to a real-world scenario.

 

[Farsight]

Sounds like he typo'd. We've all done it, I know I have.

I don't think so Edd. One of the big problems here is that people confuse spacetime with space and think of "the metric" as space.

While I'm posting light rays moving radially to a black hole don't curve either (well, depending on one's point of view, but I think it's clear enough what I mean).

Yes, no problem. I've talked about the vertical light beam emitted from the event horizon that doesn't curve. I maybe should have said orthogonal light beams or something to clarify matters.

They might red or blueshift of course, but light rays can do that in flat space too.

That vertical light beam doesn't actually change frequency. If I'm up at some elevation and shine a light beam down to you on the ground, we don't agree on the frequency because our clocks are going at different rates. Your clock is going slower than mine so you say the frequency is higher, but it hasn't actually changed. Conservation of energy applies. In similar vein if you move towards a light beam, you don't change its frequency, you just measure it differently.

 

[ben m]

A spacetime metric that represents what, exactly? Here it is, it's like Pythagoras's theorem, we've got a spacetime interval and the usual y and z terms, but we've now got r terms. What do they represent exactly, and what real-world scenario does the overall expression describe?

[latex]$ds^2 = -(r-r_0) dt^2 + dr^2/(r-r_0) + dy^2 + dz^2$[/latex]

I know what a metric is.

I don't think you know what a metric is.

You are in some spacetime. There's a 4D measurement grid laid out in that spacetime, and at each grid-line-intersection you'll find a little label like "r = 4.5, z = 0.00, y = 1.39, t = 86400.2". You have no other measuring devices, so any time you do an experiment you'll label the experimentally relevant events by consulting the nearest grid-label. ("I'm bouncing light between two mirrors. Light hit the first mirror at the grid point labeled {4.5,0.00,1.39,86400.2}, later it hit another mirror near the grid point labeled {6.8,1.9,1.39,86401.2}") The expression that Sol gave you is the equivalent of the Pythagorean Theorem, in that it tells you how to calculate a hypotenuse from these {r,x,y,t} grid-point-labels.

The point (or one of the points) that you're missing is that Sol's hypotenuse-calculation-prescription, IF YOU KNEW ENOUGH MATH TO INTERPRET IT would tell you how the gridlines work, and would tells you whether (and where, and how, and how much) the spacetime is curved. Clinger did the algebra for you using a straightforward plug-and-chug technique.

If you don't like Sol's grid, you can use his hypotenuse-prescription to lay down an alternative grid, and to convert Sol's {r,z,y,t} to your own arbitrary {u,v,w,t'}.

 

[sol invictus]

A spacetime metric that represents what, exactly? Here it is, it's like Pythagoras's theorem, we've got a spacetime interval and the usual y and z terms, but we've now got r terms. What do they represent exactly, and what real-world scenario does the overall expression describe?

[latex]$ds^2 = -(r-r_0) dt^2 + dr^2/(r-r_0) + dy^2 + dz^2$[/latex]

I know what a metric is.

But we still don't know what Clinger is computing the curvature of now do we? It's been weeks now, you've got nothing, and you're still trying to dismiss Einstein and scientific evidence and defend your waterfall neverland fantasies with mathematics that you will not relate to a real-world scenario.

Your questions are nonsensical, and simply make it obvious that you don't know what a metric is. There's no shame in that - the vast majority of the world's population has even less of a clue than you do. But if you want to discuss it, the appropriate course of action is to try to learn what you don't know.

Stop pretending to understand when you don't, drop the attitude, start over, and ask some basic questions, nicely. We'll help you.

 

[DeiRenDopa]

That vertical light beam doesn't actually change frequency. If I'm up at some elevation and shine a light beam down to you on the ground, we don't agree on the frequency because our clocks are going at different rates. Your clock is going slower than mine so you say the frequency is higher, but it hasn't actually changed. Conservation of energy applies. In similar vein if you move towards a light beam, you don't change its frequency, you just measure it differently.

(bold added)

Ladies and gentlemen, we are now back in a universe where stuff really is one thing or another, where observers and observations don't really matter.

What's that? You thought Einstein (and others) rather convincingly showed that absolutes are rather uninteresting because the only thing anyone can tell about the universe is what one observes? That relativity rules?

Well, yes, I thought so too, once. But then Farsight came along and edukuted edacuted edumcated me; now I realize that relative is, well, relative. That beneath all the smoke and mirrors Einstein said, there's another, hidden Einstein. This secret Einstein believed that observers really don't matter, that there is a way to get at what's really happening.

What's that? You say that this isn't really what relativity is about? That this is just a thinly disguised version of the absolute time and space of Newton the Greeks?

No, you're wrong! just read Farsight's insightful gloss of a small handful of carefully chosen words (out of the millions Einstein wrote), then you'll see.

How? You're asking how you can get to understand what really happens? And not be stuck merely with what you can actually observe?

Well look, Farsight explained how; just think of a pair of trains with bombs aboard, got that? Good. Now, which train will explode first?

 

[Ziggurat]

I'm not wrong about the speed of light varying with gravitational potential, which is what Einstein said repeatedly whilst formulating GR. We can see it.

We can indeed, given a specific set of coordinates and performing our experiments in specific ways. Nobody ever said otherwise.

And yet he presented it as a fait accompli without explanation

Not at all. He gave a thorough explanation. You just couldn't follow it, because you can't do the math. You seem to think that with the right conceptual framework, the math becomes irrelevant. But that's simply not true. You can't construct the right conceptual framework without the math.

and there's still no explanation of what [latex]$ds^2 = -(r-r_0) dt^2 + dr^2/(r-r_0) + dy^2 + dz^2$[/latex] represents.

Also not true. It's been explained to you multiple times. It's flat space-time, the same flat space-time that the canonical Minkowski metric describes. We can pick different coordinates to describe flat space-time, and this is just one set of coordinates. What else do you want to know? Do you need an equation to describe how to get from the more conventional coordinates to these ones? I don't recall you asking, but if you had a clue about GR you could find the transforms yourself, and even if we gave them to you, you probably wouldn't know what to do with them.

This is getting laughable. It's like priests in a corner desperately spouting Latin.

That's how you think of math, isn't it? It's just Latin: incomprehensible and irrelevant gibberish, as far as you're concerned.

And it is academic, because light doesn't curve in flat spacetime.

How do you define "curve"? If you mean that light always follows a geodesic (in a vacuum), then that's true no matter what spacetime you're in. If you define "curve" in terms of how it relates to your coordinate system, then whether or not light curves depends on what you choose for that coordinate system. But of course, this is a mathematical question, and you're allergic to those.

You can't create a gravitational field in space by moving through that space. If it isn't there, it isn't there.

That wasn't what you said. You said, "Things don't fall down as you pass by." And you said this in regards to the use of an accelerating reference frame. But you're rather obviously wrong. If you're accelerating, then everything in the universe is falling down in your reference frame, where down is the direction opposite to your acceleration. You don't want to call it gravity. Fine, that's a semantic issue. The math, however, is not. But of course, math is what you're allergic to.

Yeah yeah, dismiss Einstein.

I haven't dismissed anything Einstein has said. But what he says with words is often ambiguous, because that's a fundamental problem with human language. It is imprecise. But his math does not suffer this weakness. Furthermore, math is what makes his theory quantifiable and testable. The words can help explain the math, but at the end of the day, it is the math which actually matters.

And you have done nothing but dismiss Einstein's math. Because it's so far beyond your capacity that it's like Latin to you. And so you lash out at people who understand the math, and can use the math, because they don't agree with you. But the people who can do the math, and who use it correctly, are in far better agreement with Einstein than the person who doesn't have a clue how any of that math works.

I'm objecting to the lack of explanation of terms and scenario.

Meaning you want someone to hold your hand and present you with a coordinate transform. Because you can't figure it out yourself.

And I'm the one who said you don't change space by zooming through it. Those light beams don't curve. Look, there's Zig zooming through space, riding his negative carpet. And oh wait, he's zooming past a light beam. See it curve? Er, no.

Same thing happens if you bound a light beam around while falling in a gravitational field. You will perceive the light beam as being straight. Someone stationary in the field will perceive it being curved. So what is it really?

Well, let's start by defining what it means to be curved.

 

[Farsight]

You're completely wrong, not just about the physics, but about what Einstein said (in 1920 and at other times). For example:

You're wrong again. Einstein said this in 1916. The German book was entitled Über die spezielle und die allgemeine Relativitätstheorie. It was translated into English in 1920, and we know it as Relativity: The Special and General Theory. Your excerpt is where Einstein is talking about mass and giving us the principle of equivalence. Yes, we all know that if you're accelerating it's like you're in a gravitational field. Things fall to the floor and light curves. But look at what he says a bit later on:

"Now we might easily suppose that the existence of a gravitational field is always only an apparent one. We might also think that, regardless of the kind of gravitational field which may be present, we could always choose another reference-body such that no gravitational field exists with reference to it. This is by no means true for all gravitational fields, but only for those of quite special form. It is, for instance, impossible to choose a body of reference
such that, as judged from it, the gravitational field of the earth (in its entirety) vanishes".

This is why your flat spacetime isn't the same as a real gravitational field.

According to Einstein gravity can exist even in perfectly flat, homogeneous spacetime. You're directly contradicting what Einstein said in 1920, Farsight. Moving on:

I'm not contradicting him. You're conveniently ignoring what he actually said. He also said this remember?

"In the second place our result shows that, according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity and to which we have already frequently referred, cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Now we might think that as a consequence of this, the special theory of relativity and with it the whole theory of relativity would be laid in the dust. But in reality this is not the case. We can only conclude that the special theory of relativity cannot claim an unlinlited domain of validity; its results hold only so long as we are able to disregard the influences of gravitational fields on the phenomena (e.g. of light)."

Only we know that velocity is the common usage, as in "high velocity bullet". Don't we? Because he's referring to the postulate of special relativity. And he's previously talked about c varying with gravitational potential. On numerous occasions.

According to Einstein, rays of light curve in perfectly flat spacetime when viewed from accelerated reference frames, because acceleration is gravity and light curves in response to gravity. Wrong again, Farsight.

We've been through all this. You see a light beam as curved when you accelerate past it. And acceleration is acceleration. It isn't what gravity is. It's only a principle of equivalence. You are free to stop, whereupon that curvature has gone away. But it is impossible to choose a body of reference such that, as judged from it, the gravitational field of the earth (in its entirety) vanishes.

 

[sol invictus]

But look at what he says a bit later on:

"Now we might easily suppose that the existence of a gravitational field is always only an apparent one. We might also think that, regardless of the kind of gravitational field which may be present, we could always choose another reference-body such that no gravitational field exists with reference to it. This is by no means true for all gravitational fields, but only for those of quite special form. It is, for instance, impossible to choose a body of reference
such that, as judged from it, the gravitational field of the earth (in its entirety) vanishes".

This is why your flat spacetime isn't the same as a real gravitational field.

Saying that it isn't a "real" gravitational field directly and explicitly contradicts Einstein, Farsight. Yes, it's true that in flat spacetime one can choose coordinates in which there is no gravitational field. But it's also true that one can choose coordinates in which there is one, just as he says:

...come to the conclusion that he and the chest are in a gravitational field...Ought we to smile at the man and say that he errs in his conclusion? I do not believe we ought if we wish to remain consistent....the Γ^τ_μν...are the components of the gravitational field.

You're wrong. Accept it, learn from it, move on.

We've been through all this. You see a light beam as curved when you accelerate past it. And acceleration is acceleration. It isn't what gravity is.

You're contradicting Einstein again:

It can easily be shown that the path of the same ray of light is no longer a straight line when we consider it with reference to the accelerated chest (reference-body K')...rays of light are propagated curvilinearly in gravitational fields...the Γ^τ_μν...are the components of the gravitational field.