On Feb 10, 2013, at 9:58 AM, Alexander Poltorak <
Paul,
I am not denying that there is a tensor part in a LC connection – I use it in my papers – the only thing I am saying is, to extract it, you need the second connection.
Right, no problem there mathematically. Physically it means adding new tensor fields like torsion & non-metricity.
Your previous assertion that there is a unique decomposition of LC into tensor and non-tensor part is incorrect. Every time you subtract another connection (affine or LC) from your first LC connection, you get a tensor of affine deformation. Since you can define infinite number of various connections on the manifold, there is infinite number of ways to decompose you LC into a tensor and non-tensor, as the affine deformation tensor will be different depending on the second connection.
However, what I think you are trying to say, is that there is one way to extract a tensor out of LC connection, which contains all information about the geometry but is a true tensor. If this is what you are saying, that is certainly correct. There is very simple way to do it – just subtract from your first LC connection another affine connection with zero curvature and torsion. What you will get is a tensor of affine deformation that contains all information about the geometry defined by your original LC connection. Essentially, what you are doing through this procedure, you are stripping the information about the coordinate system from your LC connection and leaving only information about the geometry imbedded in the tensor of affine deformation (which is also the tensor of nonmetricity for the affine connection with respect to the metric associated with your LC connection). This gives you a unique tensor part of the LC connection that you are seeking. But why reinvent the wheel and call it a “tensor of metricity” when everyone in the world calls tensor of nonmetricity or tensor of affine deformation? You will only confuse people by inventing new terminology for well-known objects. So far, it’s all pretty obvious.
OK, but then the question is what is the explicit structure, the formula for, this allegedly unique affine connection A with zero curvature and zero torsion? Also, "curvature" and torsion with respect to itself A? Or with respect to the original LC connection? It seems it must be the latter. The simplest connection with zero curvature and zero torsion relative to the LC connection is A = 0. But that is obviously not a good choice. Also connections describe frames of reference as well as parallel transport in the appropriate fiber space of the physically relevant fiber bundle.
I think where we run into a philosophical argument, is where you propose to discard the second connection. I understand that mathematically speaking, you can do it. But what is the physical meaning of this? What is the meaning of your flat connection that you need to subtract from the LC connection to extract the tensor of affine deformation? You can take two approaches: first, a-la Rosen, you can say that without gravitational field the spacetime ought to be flat and gravity curves it – hence we start with the flat connection (or Rosen’s flat metric -- either way, it’s bimetrism, because LC connection always has its proper metric associated with it) and then introduce the second LC connection (and the second metric) describing the geometry change by the presence of gravitational field – the difference between the two will be your affine deformation tensor that describes the strength of gravitational field. Or you can follow my approach, where I propose that the first affine connection describes the choice of the frame of reference (in an IFR the connection has no curvature or torsion, but in a NIFR, the connection has curvature and, possibly, torsion). But this is a question of interpretation. The result will be the same – the use of the tensor of affine deformation or tensor of nonmetricity (if there is at least one metric) as the strength of the gravitational field, as I’ve done in my papers. ... But I don’t see what you are adding to what I have described more than 30 years ago.
OK, but here there is a conceptual philosophical problem. Frames of reference are only descriptions of frame-invariant geometric objects. Curvature and torsion are frame-invariant geometric objects. So this appears to be a contradiction since in your idea of "frame" geometric objects are no longer frame invariant. In terms of Plato's Allegory of the Cave, what is real are the objects what is frame dependent are the projected shadows from the objects. The shadows are the subjective frame-dependent representations of the real objects.
Best regards,
Alex
From: Paul Zielinski [mailto:
To: Alexander Poltorak
Cc: JACK SARFATTI; d14947 Gladstone; Waldyr A. Rodrigues Jr.; james Woodward; Gerry Pelligrini; Saul-Paul Sirag
Subject: Re: KISS OFF! ;-)
Alex,
Thanks for your response. Comments below.
On 2/9/2013 8:43 PM, Alexander Poltorak wrote:
Paul: see my comments below:
From: Paul Zielinski [mailto:
To: Alexander Poltorak
Cc: JACK SARFATTI; d14947 Gladstone; Waldyr A. Rodrigues Jr.; james Woodward; Gerry Pelligrini; Saul-Paul Sirag
Subject: Re: KISS OFF! ;-)
Alex,
I'm sorry but I have to say that what you wrote below is simply erroneous.
If I understand your position correctly, you are saying that it is only possible to extract a non-zero tensor from the LC connection as the
non-metricity of a second "Affine" connection, and that since no such connection is available in orthodox GR (which I think everyone
agrees with), the LC connection *has no tensor part in that theory*. In other words, the second Affine connection being unavailable, the
LC connection is "irreducibly non-tensorial" in that context.
[AP] That is correct, that’s my assertion.
But this is clearly false, since all that is required here is that it be shown that there is a quantity contained in the LC connection whose
components A^u_vw transform according to tensor rules.
[AP] Be my guest, try to prove it. I don’t think you will succeed.
OK then suppose we have the second connection, and use it your way to identify a class of (1, 2) tensors Acontained in the LC connection.
How can removing the second connection from the formalism change the coordinate transformation properties of the quantity A^u_vw once
it is identified? How can that be possible?
It's one thing to say that it is not "explicitly" present, as you did, but it's quite another to say that it's not present at all.
My position here is that once it is identified (or "extracted" in your terminology) and its transformation properties are established, the removal of the second
connection that is used to identify it only prevents us from calling it the "non-metricity" of the missing connection. It doesn't prevent us from classifying it as
a tensor.
Or is it your position that this quantity goes to zero when the second connection is removed from the theory?
Once this is established, there is no need for a second connection since then the existence of such a quantity depends only on this independent condition being satisfied. So you can use a second connection as a "construction for the sake of proof" in order to isolate the tensor part of the LC connection, and then discard it once the existence of the tensor quantity is established.
[AP] Yes, if you can show that there is a quantity contained in the LC connection whose components A^u_vw transform according to tensor rules, the second connection would not be required. However, you have not shown it and, I am afraid, you will not be able to show it.
Then what is your -Q^u_vw? This is the negative of the non-metricity of the second "Affine" connection, right? You seem to be saying that the components
-Q^u_vw no longer transform according to the (1, 2) tensor rules when the second connection is excluded from the theory. Or else that they all go to zero.
How exactly does that work, in your view?
Here is a simple illustration of the fallacy of this proposition. If you chose normal (aka Riemannian) coordinates in the vicinity of point p, Christoffel Symbols of your LC connection vanish in the vicinity of p, which could not happen if LC connection contained a tensor component.
Ah OK I see.
But in my theory of the LC connection, the Riemann coordinates make a non-tensor contribution to the LC connection that cancels the tensor geometric contribution, i.e., the matrix representations of the coordinate and geometric contributions to the LC connection sum in the Riemann CS to give a *zero matrix*.
According to my understanding the coordinate contributions to the LC connection depend only on the non-linear character of the diffeomorphic transformations on the coordinate
space R^4, and do not at all depend on the intrinsic geometry of the object manifold.
So this is a fundamental difference in our respective understandings of the nature of the LC connection and its relationship to the coordinates and the coordinate space R^n.
From my perspective your argument is circular, since according to my understanding you still get a vanishing LC connection around any given point p in a Riemann CS
even with a non-zero tensor geometric part.
Here is even simpler proof: take a flat Minkowski space. In curvilinear coordinates, Christoffel symbols will be non-zero, but in Minkowski coordinates they all vanish globally. How would that be possible if there was a tensor component there?
Easy. The tensor geometric part of the LC connection is zero everywhere on a Minkowski manifold.
Which means that on a flat manifold in R^n-curvilinear coordinates you have a non-zero pure affine connection except for a *zero*
geometric contribution (i.e., a zero tensor).
Referring to the theory of parallel transport, this is because on a flat manifold the inner product defined by the Minkowski metric
is invariant under transport of vectors along the manifold, and thus there is nothing to correct for in the partial derivatives of tensors,
except for curved-coordinate artifacts. So all that applies in this case, and all that is present, is the coordinate part of the LC connection,
which enables the LC covariant derivative to correct the coordinate artifacts. The zero geometric part does nothing.
In other words, in the *unique* decomposition
Γ^u_vw = G^u_vw + X^u_vw
on a globally flat Minkowski manifold, all G^u_vw = 0, some X^u_vw =/= 0.
This makes perfect sense to me, since here we are interested in the covariant first order *geometric* variation of the inner product under
infinitesimal displacement of vectors/tensors along the manifold, which clearly vanishes for a Minkowski manifold.
Also it is not true that defining a second connection is the *only* way to extract such a tensor, since for example as I've mentioned you can
take the difference between two LC connections (compatible with different metrics) to get a similar results, and then discard the second
metric as a "construction for the sake of proof" afterwards.
[AP] I don’t understand your argument here. By introducing the second LC connection associated with another metric you have introduced the second connection, haven’t you? Of course, the difference between two LC connection will always be a tensor.
Yes exactly.
The difference between any two Affine connections is always a tensor, called tensor of affine deformation.
Correct.
But you need the second connection, metric (i.e., LC) or not (i.e. Affine)!
Yes but because the geometric contribution G^u_vw to the LC connection is zero for the flat manifold, it does nothing,
and we can simply remove it from the definition of the resulting tensor quantity without disturbing anything. Then we have a
standalone tensor G^u_vw that only refers by definition to *one* metric. That's the trick.
That's what I meant by "kicking down the ladder behind you". It's just a mathematical construction for the sake of proof.
It's very important to understand that the resulting tensor is a standalone quantity whose transformation properties are not
affected in the slightest by removal of the zero flat space contribution.
I can show that the resulting tensor is the negative of the non-metricity of what I'm calling the "pure affine connection",
which has no geometric part and is thus irreducibly tensorial.
So all roads lead to Rome.
Thus the correct statement here would be that while the tensor that is exposed by the application of the covariant derivative associated
with your second "Affine" connection is not the *non-metricity* tensor of the Affine connection if the second Affine connection is not
defined, it is still a *tensor* quantity transforming according to tensor rules that is mathematically present in the LC connection,
regardless.
[AP] You lost me here again. A tensor obtained by replacing partial derivatives of the metric tensor in Christoffel symbols by covariant derivatives with respect to another connection is by definition the tensor of nonmetricity for that second connection.
Yes exactly. But if the second connection is removed from the theory, there can be no "non-metricity" tensor of *that* connection in the theory.
So in that case you can no longer *call* the tensor quantity extracted by that method a "non-metricity" tensor. But it still has the same tensor
transformation properties, and is therefore still a tensor, and is still present in the LC connection, regardless.
That is shown clearly by the Levi-Civita dual metric construction, which exposes the same family of tensors without reference to a second
connection; and according to the argument above, when you take one of the metrics to be flat, you can also remove all reference to the flat
metric once you have identified a *unique* 3rd rank geometric tensor inside the LC connection, without disturbing the value or the
transformation properties of the resulting quantity.
You can do the same thing in your model where you have two metrics. When you construct to LC connections based on each respective metric and then take a covariant derivative of the first metric with respect to LC connection associated with the second metric, you will get the tensor of nonmetricity of that second connection with respect to the first metric. Or vice versa. In this scenario, albeit you start with two metrics, you still have two connections.
Yes but see above. The kicking-down-the-ladder trick of removing the always-zero geometric part of the flat LC connection from the *definition* of the
resulting tensor yields a standalone quantity whose definition refers only to a *single* metric. Because removing a quantity that is *always zero* from
the definition numerically leaves the same tensor in place.
So we are talking about two things here: (1) the method used to isolate the tensor part of the LC connection; and (2) the tensor
properties of the quantity so isolated. [AP] This is a tautology. Once you isolated “the tensor part of the LC connection” in (1), obvious, “the tensor properties of the quantity so isolated,” which is a tensor by your own definition, are guaranteed. Only (1) depends on the existence of a second connection in your theory, while (2) stands quite independently of (1) in your theory since it depends only on the transformation properties of the components A^u_vw under coordinate transformations, which are not at all dependent on the existence of the second connection.
So I stand by what I said: your argument in favor of what you understood to be Jack's position on this question is not logically consistent
with your position on the extraction of a non-zero tensor from the LC connection using your second Affine connection.
[AP] I respectfully disagree.
Are you now willing to acknowledge the error? [AP] I’d be glad to acknowledge, if I knew were the error was.
See above.
There is clearly a fundamental difference in our respective understandings of the LC connection. I am saying that Riemann coordinates
are R^n-curvilinear (in the coordinate space R^n) and therefore make a non-zero contribution to the LC connection, and that this cancels
the geometric part around any point in such a CS. In other words, the respective matrix representations of the two linearly independent
contributions mutual cancel in such a CS.
This is a basic point that I think will have to be resolved before we can go any further with this.
Regards,
Paul
Regards,
Paul
On 2/7/2013 1:34 PM, Alexander Poltorak wrote:
There is no logical contradiction. To get a tensor, you must introduce the second connection. It is not present in the standard formulation of GR – hence Jack correctly states that LC connection is irreducibly nontensorial. We can get to a tensor, but for that we need the second connection (not the second metric, as you suggest, but the Affine connection), which does not exist explicitly in Einstein’s GR.
—Alex
From: Paul Zielinski [mailto:
To: Alexander Poltorak
Cc: JACK SARFATTI; d14947 Gladstone; Waldyr A. Rodrigues Jr.; james Woodward; Gerry Pelligrini; Saul-Paul Sirag
Subject: Re: KISS OFF! ;-)
Alex,
How can you say that the LC connection decomposes into a tensor and a non-tensor, and at the same time argue that Jack is
right when he says that the LC connection has no tensor part? This seems like a logical contradiction to me.
Of course the LC connection as a whole is a non-tensor, and of course the non-metricity Q^u_vw of the metric compatible LC
connection is zero *by definition*. However, it doesn't follow that there is no tensor part in the "LC connection of GR". The
LC connection of GR is the LC connection of Riemannian geometry, and the LC connection of Riemannian geometry contains
an infinite class of (generally) non-zero tensors, as you yourself have argued.
It seems to me that the correct statement here is that the LC connection of GR does contain this tensor part, but this quantity
has not previously been physically interpreted in *orthodox* GR.
On 2/7/2013 12:38 PM, Alexander Poltorak wrote:
What Jack is talking about by saying there is no tensor component in 1916 GR's LC connection is as follows: a general Affine connection, as is well known, is a sum of a metric connection (aka LC connection), which is a non-tensorial quantity, nonmetriciy and torsion, which are both tensors. The only thing Jack is saying is that in standard 1916 GR, both nonmetriciy and torsion are zero and, therefore the Affine connection is equal to a LC connection, which is non-tensor -- hence, Jack says, there is no tensorial part in GR's connection and he is right of course.
Can we define a real gravity field in the sense of Newton in Einstein's GR?
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