Black holes

[xtifr]

It's the measured speed of light that's invariant.

Ok, as others said, that makes a certain amount of sense. We'll come back to it.

Oh here we go, dismissal of evidence, total denial, and absolute fiction. The typical traits of conviction. Forget it xtifr, I'll give you the hard scientific evidence, I'll give you the Einstein quotes, and I'll give you the reason why people like you believe what you do. But you won't believe any of it. It is exactly like trying to talk to a creationist.

Once again, you're the one that wants me to believe your theory. I am expressing a willingness to consider your idea if you can explain a logical inconsistency in the historical facts. This does not equate to "dismissal of evidence, total denial, and absolute fiction". I simply want to know how it is possible that Einstein never corrected anyone's misinterpretation of his ideas. GR was known and studied by numerous physicists throughout Einstein's life, even if it was, as you claim, a relatively (pun not intended) obscure theory until after his death.

The quotes you gave me don't address this point, as they only go up to 1915.

Bottom line: I'm not trying to use this as an excuse to dismiss your ideas, but it seems like a discrepancy that's very hard to explain, and makes me much more dubious about what you say, and I'm wondering if you have a reasonable explanation that would make those doubts go away. Even if you say no, I'm not going to claim that that proves you wrong, because it doesn't, but if you can provide an answer, it would make it that much easier for me to think about taking you seriously.

 

[Farsight]

I don't want you to believe my theory, xtifr, I want you to believe Einstein's theory. The "logical inconsistency in historical facts" just isn't there. You only think it's there because the history you've been spoon-fed bears little resemblance to the real thing. Because people with some vested interest or reputation to uphold have been airbrushing over the real history and the real facts. If you're sharp you can spot this. Check out Graham Farmelo's Dirac biography The Strangest Man where on page 53 you can read this:

"At that time, Cunningham and Eddington were streets ahead of the majority of their Cambridge colleagues, who dismissed Einstein's work, ignored it, or denied its significance".

That's talking about 1923. Eight years after Einstein did GR, and three years after his media fame. Nothing much changed until the Golden Age forty years later, and then GR changed. Oh, and see this paper called The Confrontation between General Relativity and Experiment by Clifford M. Will? Take a look at what it says on page 6:

"Although special relativity itself never benefited from the kind of “crucial” experiments, such as the perihelion advance of Mercury and the deflection of light, that contributed so much to the initial acceptance of GR and to the fame of Einstein, the steady accumulation of experimental support, together with the successful merger of special relativity with quantum mechanics, led to its being accepted by mainstream physicists by the late 1920s..."

Did you catch that? SR wasn't mainstream until the late twenties! What you'd call "the mainstream physics community" was sceptical of Einstein for decades. Then when they stopped being sceptical, whoosh, out comes the airbrush. Now when you read some typical magazine or book or website, there's hardly a trace of it. People like editors do this out of loyalty to physics. They don't want to say that "the mainstream" was talking bollocks for decades in case you it occurs to you, and the rest of the paying public, that about some things, it still is. The road to hell is paved with good intentions, and this is the true Trouble with Physics. In some respects the end product bears some resemblance to a one-party state that brooks no dissent and admits no wrong and censors the truth. And gets nothing done. And is increasingly irrelevant to modern life, etc etc etc.

 

[DeiRenDopa]

I don't want you to believe my theory, xtifr, I want you to believe Einstein's theory. The "logical inconsistency in historical facts" just isn't there. You only think it's there because the history you've been spoon-fed bears little resemblance to the real thing. Because people with some vested interest or reputation to uphold have been airbrushing over the real history and the real facts. If you're sharp you can spot this. Check out Graham Farmelo's Dirac biography The Strangest Man where on page 53 you can read this:

"At that time, Cunningham and Eddington were streets ahead of the majority of their Cambridge colleagues, who dismissed Einstein's work, ignored it, or denied its significance".

That's talking about 1923. Eight years after Einstein did GR, and three years after his media fame. Nothing much changed until the Golden Age forty years later, and then GR changed. Oh, and see this paper called The Confrontation between General Relativity and Experiment by Clifford M. Will? Take a look at what it says on page 6:

"Although special relativity itself never benefited from the kind of “crucial” experiments, such as the perihelion advance of Mercury and the deflection of light, that contributed so much to the initial acceptance of GR and to the fame of Einstein, the steady accumulation of experimental support, together with the successful merger of special relativity with quantum mechanics, led to its being accepted by mainstream physicists by the late 1920s..."

Did you catch that? SR wasn't mainstream until the late twenties! What you'd call "the mainstream physics community" was sceptical of Einstein for decades. Then when they stopped being sceptical, whoosh, out comes the airbrush. Now when you read some typical magazine or book or website, there's hardly a trace of it. People like editors do this out of loyalty to physics. They don't want to say that "the mainstream" was talking bollocks for decades in case you it occurs to you, and the rest of the paying public, that about some things, it still is. The road to hell is paved with good intentions, and this is the true Trouble with Physics. In some respects the end product bears some resemblance to a one-party state that brooks no dissent and admits no wrong and censors the truth. And gets nothing done. And is increasingly irrelevant to modern life, etc etc etc.

Thanks Farsight.

No, really.

I've often wondered what physics - especially physics of the last century or so - looks like to someone who wants to be a physicist but can't do the math. Now I know.

"Einstein's theory"? Isn't that the one he wrote up, in some paper published around 1916? I seem to recall that someone - or maybe more than one - here has written a post of two on the content of that paper, commenting on the fact that it's essentially meaningless unless you can follow the math in it. Do I understand you to be saying that you are now - after many years of saying otherwise - that you understand the math in that paper by Einstein?

 

[Farsight]

Up to there you have it sort of right but then you go off the rails yet again:

There is no "tilt" because there is no external reference against which to measure such a "tilt". The curvature of spacetime is intrinsic.

There is a "tilt". If there wasn't a tilt things wouldn't fall down. Fall off a ladder and bang, the ground gives you a reference.

FYI: The Riemann curvature tensor is zero in any Euclidean space.

Correct.

AFAIK: It is the manifold's global intrinsic curvature which means that "things still fall down and light still curves"

No, it's the tilt. The Reimann curvature changes it from flat-horizontal to flat-tilted. Go and look at pictures of "curved spacetime". The curvature you can see is Riemann curvature. Look at say this picture. Follow one of the grid lines from the red dot up towards the top right. The line is pretty straight near the edge of the picture. There's not much Riemann curvature there, but the grid square is tilted. If you were in that square you wouldn't measure any detectable tidal force, but a light-clock on the floor still runs slower than a light-clock at the ceiling, and things still fall down and light still curves.

but the property of having the manifold locally Euclidean means that there is no way to tell the effects of this curvature (gravity!) from an acceleration.

Sure. But it isn't really Euclidean. It's just nearly Euclidean. If it was absolutely Euclidean then the gravitational field wouldn't diminish with distance. Or if you start off at a great distance where the gravitational field is undetectable, then you could approach the red spot and the gravitational field would still be undetectable.

This statement is especially wrong:

If you're walking across a plane and the gradient doesn't alter, you've got a flat hill. You've got no hill!

It isn't wrong at all. It's bang-on right. Take away the Rieman curvature and what you're left with is this. It's as flat as a board. There's no curvature to give you the tilt you need for light to curve and for things to fall down. Sure you could pick up the whole board and tilt it, but real gravitational fields just aren't like that. NB: Einstein used the word "special" instead of "real", but I think the latter is the better word here.

Think about the geodesic dome analogy yet again: If you walk along the surface of the dome then the gradient does alter and you are not walking on flat ground !

Imagine you're standing on top of it. I wave my magic wand to take away the Riemann curvature, and hey presto you're suddenly standing on a flat plane of glass panels.

One more time for you, Farsight: A manifold that is locally Euclidean does not mean that the manifold is globally or intrinsically Euclidean.

It's locally Euclidean only in an infinitestimal region. Replace your geodesic dome with a spherical glass dome and make yourself the size of an ant. You're sitting on top of it with your feelers and your instruments. It looks locally Euclidean. It's so close to Euclidean that you can't detect any curvature at all. But if it really was absolutely Euclidean you'd be sitting on a glass plane, not a glass dome. And when you went for a walk you wouldn't fall off.

I'm not wrong about this RC. Note the ominous silence and the evasion from your "friends". They don't have the honesty to tell you that Actually, RC, Farsight is right about this.

 

[Roboramma]

They don't have the honesty to tell you that Actually, RC, Farsight is right about this.

You seem to be ignoring the possibility (again) that doing so would be dishonest, because they believe that you are wrong.

 

[Vorpal]

It's useless. Farsight is completely incapable of telling apart the gravitational field and the gravitational tidal forces. If we were back in Newtonian potential, this would be like confusing the field components -∂Φ/∂x (etc.) with the tidal force components ∂²Φ/∂x∂y (etc.).

That's exactly what's going on here. In GTR, metric g_μν plays the role of the old potential Φ, the field components Γ^α_μν are still found from the first derivatives, just in a substantially more complicated way. The whole hubbub is then about the claim that in a freefalling frame, the field vanishes (locally!). In other words, in the corresponding coordinates the first derivatives of the 'potential' (metric) vanish.
So it's no wonder that everyone just shakes their at Farsight's yet another repetition of the the impossibility of making Riemann curvature vanish. No one ever said it was possible. Because the Riemann curvature describes the deviation of nearby geodesics, i.e., tidal forces, and in a freefalling frame, it's completely determined by the second derivatives of the 'potential' (metric).

First derivatives and second derivatives are different things. One is different from two. I have trouble apprehending the total confusion of ideas that it takes to conflate them, and yet, he manages it.

 

[ctamblyn]

Now you may feel I'm being too harsh; after all, hasn't Farsight written at least something quantitative? Hasn't he tried - at least once - to put hard numbers (with the appropriate units) onto this signature diagrams, to at least allow the possibility that FFGR could be tested, using hard scientific evidence? Well, no, I don't think he's ever done that.

Don't believe me? A single counter-example would demolish my hypothesis.

I'd greatly welcome hard evidence that such a counter-example exists.

To my surprise, I recently came across some predictions of FGR/"relativity+" which are experimentally falsifiable, if not quantitative. I don't recall seeing them mentioned in this thread, though my memory's not what it used to be and all these threads look the same after about 10 pages . They were essentially claims that quantising gravity is impossible, and Higgs bosons and strangelets (IIRC) don't exist.

 

[Reality Check]

There is a "tilt". If there wasn't a tilt things wouldn't fall down. Fall off a ladder and bang, the ground gives you a reference.

You cannot tell the difference between the geodesic dome analogy and spacetime!
There is no ladder or ground in spacetime. I will make this explicit for you:

Originally Posted by Reality Check
Up to there you have it sort of right but then you go off the rails yet again:

There is no "tilt" in spacetime because there is no external reference in spacetime against which to measure such a "tilt"in spacetime . The curvature of spacetime is intrinsic.

No, it's the tilt. The Reimann curvature changes it from flat-horizontal to flat-tilted.
...description of a cartoon....

Still ignorant of GR and dependent on the "pretty picture" physics used to teach children and undergraduates, Farsight.
Spacetime is intrinsically curved. It does not have a magic plane to with to define an "tilt".
The Riemann curvature tensor is about the intrinsic curvature of spacetime.

But it isn't really Euclidean. It's just nearly Euclidean.

The manifold is constructed to be locally Minkowskian (exactly). This does not effect gravity because it is caused by the global properties of the manifold, e.g. its curvature.

Take away the Rieman curvature and what you're left with is this. It's as flat as a board.

Take away the Riemann curvature and you have a globally Minkowskian spacetime. We are talking about a curved space time where the Riemann curvature tensor is non-zero.
FYI (not that you will take any notice or understand)
Riemann curvature tensor

In the mathematical field of differential geometry, the Riemann curvature tensor, or Riemann–Christoffel tensor after Bernhard Riemann and Elwin Bruno Christoffel, is the most standard way to express curvature of Riemannian manifolds. It associates a tensor to each point of a Riemannian manifold (i.e., it is a tensor field), that measures the extent to which the metric tensor is not locally isometric to a Euclidean space. The curvature tensor can also be defined for any pseudo-Riemannian manifold, or indeed any manifold equipped with an affine connection. It is a central mathematical tool in the theory of general relativity, the modern theory of gravity, and the curvature of spacetime is in principle observable via the geodesic deviation equation. The curvature tensor represents the tidal force experienced by a rigid body moving along a geodesic in a sense made precise by the Jacobi equation.

So in Euclidean (or Minkowskian) space, the Riemann curvature tensor is zero and the tidal experienced by a rigid body moving along a geodesic is zero.

This is of course what Einstein explicitly stated in the equivalence principle. Locally spacetime is observed to be Minkowskian so GR is based on a manifold which is locally Minkowskian. The Riemann curvature tensor is zero in this local Minkowskian space and so there are no tidal forces which you could use to distinguish between being in a uniform gravitational field and being under constant acceleration.

Imagine you're standing on top of it. I wave my magic wand to take away the Riemann curvature, and hey presto you're suddenly standing on a flat plane of glass panels.

Wave a magic wand to make the Riemann curvature tensor zero and the "glass panels" no longer exist. There is no geodesic dome existing anymore in the analogy. It is the Riemann curvature tensor that decribes the dome in the analogy.

I'm not wrong about this RC. Note the ominous silence and the evasion from your "friends". They don't have the honesty to tell you that Actually, RC, Farsight is right about this.

You are wrong about this, Farsight. Note the stunned silence from the other posters in this thread as they cannot believe the ignorance of GR that you are displaying. They have the knowledge that you display so much ignorance of GR and so unwilling to learn about GR (how many years have you bee obsessing about Einstein quotes without actually learning GR, Farsight?) that they know that it is useless to state Actually, Farsight, RC is right about this.

P.S. I am not a GR expert so it is the the general gist of what I am stating is correct. The language that I am using is probably wrong (actually wrong in the case of the locally Euclidean Minkowskian mistake).

 

[ben m]

Actually, Farsight, RC is right about this.

 

[edd]

Note the stunned silence from the other posters in this thread as they cannot believe the ignorance of GR that you are displaying.

This too.

 

[Farsight]

It's useless. Farsight is completely incapable of telling apart the gravitational field and the gravitational tidal forces. If we were back in Newtonian potential, this would be like confusing the field components -∂Φ/∂x (etc.) with the tidal force components ∂²Φ/∂x∂y (etc.).

I'm not confusing anything.

That's exactly what's going on here. In GTR, metric g_μν plays the role of the old potential Φ, the field components Γ^α_μν are still found from the first derivatives, just in a substantially more complicated way. The whole hubbub is then about the claim that in a freefalling frame, the field vanishes (locally!). In other words, in the corresponding coordinates the first derivatives of the 'potential' (metric) vanish.

No, we're talking about the room you're in. Things falls down because the potential at the floor is lower than at the ceiling. Because there's a gradient in the potential. That's the first derivative. The change in the gradient is the second derivative, and we need it because when we start off flat and level, with no change in gradient we stay flat and level.

So it's no wonder that everyone just shakes their at Farsight's yet another repetition of the the impossibility of making Riemann curvature vanish. No one ever said it was possible. Because the Riemann curvature describes the deviation of nearby geodesics, i.e., tidal forces, and in a freefalling frame, it's completely determined by the second derivatives of the 'potential' (metric). First derivatives and second derivatives are different things. One is different from two. I have trouble apprehending the total confusion of ideas that it takes to conflate them, and yet, he manages it.

I'm not conflating anything. It's all perfectly simple.

 

[Farsight]

You cannot tell the difference between the geodesic dome analogy and spacetime! There is no ladder or ground in spacetime. I will make this explicit for you:

You introduced the geodesic dome analogy. I was just going with the flow.

Still ignorant of GR and dependent on the "pretty picture" physics used to teach children and undergraduates, Farsight. Spacetime is intrinsically curved. It does not have a magic plane to with to define an "tilt". The Riemann curvature tensor is about the intrinsic curvature of spacetime.

I need the pretty pictures to explain it to you, because you don't understand it. The word "intrinsic" is a label that covers up a lack of understanding.

The manifold is constructed to be locally Minkowskian (exactly). This does not effect gravity because it is caused by the global properties of the manifold, e.g. its curvature.

If your manifold is exactly flat locally in a region that is other than infinitesimal, it's exactly flat globally. It's like saying this part of the surface of the earth is exactly flat. Absolutely utterly flat. And the next part, and the next, and all other parts. Then the surface of the earth isn't a sphere any more, it's a flat plane. It isn't curved at all.

Take away the Riemann curvature and you have a globally Minkowskian spacetime. We are talking about a curved space time where the Riemann curvature tensor is non-zero.

Go read my response to Vorpal about first and second derivatives. The first derivative is the gradient in the potential. The second derivative is the change in the gradient. That's the Reimann curvature. Take that away and your initially flat-horizontal spacetime with a zero gradient has no change in gradient. So it's a flat plane. It isn't curved at all.

FYI (not that you will take any notice or understand) Riemann curvature tensor

I understand it. It "measures the extent to which the metric tensor is not locally isometric to a Euclidean space".

So in Euclidean (or Minkowskian) space, the Riemann curvature tensor is zero and the tidal [force] experienced by a rigid body moving along a geodesic is zero.

Sure thing. Only take care that you don't mix up space and spacetime, and understand the static situation first. Look at the pictures of curved spacetime again, imagine you're in a room at some point on the grid, and that you aren't moving. If things fall down there's a "tilt". If you can measure some tidal force, there's a "bend" too. After that, put yourself on horizontal flat grid, and accelerate. The grid appears to tilt and in your rocketship things appear to fall down. It's really simple.

This is of course what Einstein explicitly stated in the equivalence principle. Locally spacetime is observed to be Minkowskian so GR is based on a manifold which is locally Minkowskian. The Riemann curvature tensor is zero in this local Minkowskian space and so there are no tidal forces which you could use to distinguish between being in a uniform gravitational field and being under constant acceleration.

Sure. But it's only exactly Minkowskian in an infinitesimal region. A region of zero extent. In a bigger region like the room you're in now, the tidal force is so slight as to be immeasurable, but it isn't zero. If it was, the same applies to the room above that, ad infinitum. Then the gradient in gravitational potential wouldn't flatten out and the force of gravity wouldn't diminish with distance.

Wave a magic wand to make the Riemann curvature tensor zero and the "glass panels" no longer exist. There is no geodesic dome existing anymore in the analogy. It is the Riemann curvature tensor that decribes the dome in the analogy.

Don't scrabble to escape what I said. Face up to it.

You are wrong about this, Farsight. Note the stunned silence from the other posters in this thread as they cannot believe the ignorance of GR that you are displaying. They have the knowledge that you display so much ignorance of GR and so unwilling to learn about GR (how many years have you bee obsessing about Einstein quotes without actually learning GR, Farsight?) that they know that it is useless to state Actually, Farsight, RC is right about this. P.S. I am not a GR expert so it is the the general gist of what I am stating is correct. The language that I am using is probably wrong (actually wrong in the case of the locally Euclidean Minkowskian mistake).

Your language is clear enough. And I'm not wrong. I would urge you to ask elsewhere to check what I'm saying. Then you'll understand something important about some of the other posters here.

 

[Ziggurat]

If your manifold is exactly flat locally in a region that is other than infinitesimal, it's exactly flat globally.

This is simply false. I have no idea what made you think it was true, but it has no connection to reality or math. It is trivial to come up with such a manifold. And one can even find simple GR solutions which have this same property of regions with finite but nonzero extent which are flat, with curvature outside that region.

It's like saying this part of the surface of the earth is exactly flat. Absolutely utterly flat. And the next part, and the next, and all other parts. Then the surface of the earth isn't a sphere any more, it's a flat plane. It isn't curved at all.

Utter nonsense. If part of the surface is flat, then obviously it's not a sphere. But that doesn't mean the whole thing is flat. Hell, according to your logic, there can't even be flat surfaces for 3D objects,

Go read my response to Vorpal about first and second derivatives. The first derivative is the gradient in the potential. The second derivative is the change in the gradient. That's the Reimann curvature. Take that away and your initially flat-horizontal spacetime with a zero gradient has no change in gradient. So it's a flat plane. It isn't curved at all.

A second derivative being zero over a range doesn't mean it's zero everywhere. How could you even think otherwise?

 

[Farsight]

This is simply false. I have no idea what made you think it was true, but it has no connection to reality or math. It is trivial to come up with such a manifold. And one can even find simple GR solutions which have this same property of regions with finite but nonzero extent which are flat, with curvature outside that region.

Are you for real? It isn't false, and it is connected to reality and math. Go and read up on a manifold. Pay attention to the sphere differs from the plane "in the large".

Utter nonsense. If part of the surface is flat, then obviously it's not a sphere. But that doesn't mean the whole thing is flat. Hell, according to your logic, there can't even be flat surfaces for 3D objects

Don't be specious. A cube has six flat surfaces. A sphere has one surface, and it isn't flat. It's curved. And no finite region of that sphere is locally flat. Only an infinitesimal region is locally flat. A region of zero extent.

A second derivative being zero over a range doesn't mean it's zero everywhere. How could you even think otherwise?

I didn't say that. Let me reiterate. Here's a picture of gravitational potential. If you take a small region, like at the bottom in the middle and say it's absolutely flat, and then shuffle over a little and say the next small region is absolutely flat, and then repeat ad infinitum, what you do is throw away the gravitational field. You end up with this instead of this. How much simpler do I have to make it before it sinks in?

 

[sol invictus]

Are you for real? It isn't false, and it is connected to reality and math.

It's false, Farsight. You're wrong. Absolutely nothing in either reality or mathematics or the theory of manifolds forbids finite sized flat regions. Admit it, learn from it, and move on. Can you do that?

In physics, a good example is the spacetime inside a spherically symmetric shell of mass. In mathematics, you gave an example yourself:

A cube has six flat surfaces.

Indeed. And a cube obviously isn't flat globally, so you've given a counterexample to your own assertion.

In case that somehow isn't obvious to you - who knows? - the surface of a cube is a 2D surface and therefore satisfies the Gauss-Bonnet theorem. Since its Euler character is 2, it cannot possibly have zero curvature everywhere (the theorem relates the integral of the curvature to the Euler character). Indeed, the curvature of a cube is concentrated in its eight corners, each of which contributes pi/2 to the total in the appropriate units.

How much simpler do I have to make it before it sinks in?

The only thing that is sinking in is your complete lack of understanding of the basics of differential calculus.

 

[Farsight]

It's false, Farsight. You're wrong. Absolutely nothing in either reality or mathematics or the theory of manifolds forbids finite sized flat regions. Admit it, learn from it, and move on. Can you do that?

Don't be facile. The reality of gravitational fields forbids all finite regions from being flat.

In physics, a good example is the spacetime inside a spherically symmetric shell of mass.

It's space inside the spherical shell, not spacetime. There's a big difference between a finite region of space and an infinitesimal region of spacetime - the latter offers neither space nor time for motion. But that apart, the small region at the bottom centre of the upturned-hat depiction of gravitational potential appears locally flat. However if you insist that it's absolutely flat and shuffle sideways, and then insist that it's still absolutely flat, and repeat ad infinitum, it isn't a hat any more. It's a flat disc.

In mathematics, you gave an example yourself... Indeed. And a cube obviously isn't flat globally, so you've given a counterexample to your own assertion. In case that somehow isn't obvious to you - who knows? - the surface of a cube is a 2D surface and therefore satisfies the Gauss-Bonnet theorem. Since its Euler character is 2, it cannot possibly have zero curvature everywhere (the theorem relates the integral of the curvature to the Euler character). Indeed, the curvature of a cube is concentrated in its eight corners, each of which contributes pi/2 to the total in the appropriate units. The only thing that is sinking in is your complete lack of understanding of the basics of differential calculus.

The cube isn't relevant. We're talking about a gravitational field. We're talking about this. The curvature you can see in the depiction is Riemann curvature. Without it the gravitational potential doesn't get off the ground, and nor does the gradient. So the force of gravity is zero everywhere, you've got a flat level plane, and the gravitational field has vanished. It's gone.

 

[sol invictus]

Don't be facile. The reality of gravitational fields forbids all finite regions from being flat.

Utter and complete nonsense. Birkhoff proved in 1923 that the spacetime inside a shperically symmetric shell is flat. Fail.

It's space inside the spherical shell, not spacetime.

What?? I'm not talking about a shell that exists for an instant and then vanishes, Farsight. I'm talking about a rigid shell that can exist for an arbitrarily long time, so there's a spacetime region inside it.

The cube isn't relevant.

You were wrong. Admit it, learn from it, move on.

So the force of gravity is zero everywhere, you've got a flat level plane, and the gravitational field has vanished. It's gone.

Wrong yet again. The Riemann curvature is not the gravitational field - the connection is.

 

[Dave_46]

<snip> I'm talking about a rigid shell that can exist for an arbitrarily long time, ...<snip>

Would that rigid shell be the surface of the sun?

Sorry, I couldn't resist it.

Dave

 

[dafydd]

Farsight, why can you never admit that you were wrong about something? A person can't be right all the time.

 

[DeiRenDopa]

The following expression springs to mind: when you're in a hole, stop digging.