For Brent ...

[Alan Grayson]

You seem to know a lot about relativity, so please explain how the metric tensor can be defined unambiguously at some point P on the underlying manifold, spacetime, if there is an uncountable set of pairs on a vector space on the tangent space at some point P on which the metric tensor is defined (as a bilinear function which maps to the real numbers.) TY, AG

[Alan Grayson]

Brent; do me a small favor. If you are unable to answer my question, just say so. OK? AG

[Brent Meeker]

I should think it was obvious that the metric tensor, M, at a point g*Mg gives the length of g.

I generally ignore these questions because you're asking questions that take some exposition which is found in any text book and many online sites.  I also recommend "Relativity Demystified" by David McMahon.

Bretn

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[Alan Grayson]

On Monday, September 2, 2024 at 7:58:04 PM UTC-6 Brent Meeker wrote:

I should think it was obvious that the metric tensor, M, at a point g*Mg gives the length of g.

It's not obvious and I don't see how it's relevant. At each point the metric tensor can be evaluated on an uncountable set of pairs of vectors. So which pair does one choose? AG 

I generally ignore these questions because you're asking questions that take some exposition which is found in any text book and many online sites.  I also recommend "Relativity Demystified" by David McMahon.

TY for the reference. I'll check if it has the issue I need addressing. It's not on any of the sites I checked, so far. AG 

[Alan Grayson]

I downloaded your reference. It has a wealth of information and it's not at all obvious where I should look to answer my question. Typical situation!  If you could tell me where exactly my question is dealt with, I won't bother you again. TY, AG

[Brent Meeker]

On 9/2/2024 9:06 PM, Alan Grayson wrote:

On Monday, September 2, 2024 at 7:58:04 PM UTC-6 Brent Meeker wrote:

I should think it was obvious that the metric tensor, M, at a point g*Mg gives the length of g.

It's not obvious and I don't see how it's relevant. At each point the metric tensor can be evaluated on an uncountable set of pairs of vectors. So which pair does one choose? AG

Don't pick a pair.  Pick one and it's conjugate.

Brent

I generally ignore these questions because you're asking questions that take some exposition which is found in any text book and many online sites.  I also recommend "Relativity Demystified" by David McMahon.

TY for the reference. I'll check if it has the issue I need addressing. It's not on any of the sites I checked, so far. AG 

 

Bretn

On 9/2/2024 5:25 PM, Alan Grayson wrote:

Brent; do me a small favor. If you are unable to answer my question, just say so. OK? AG

On Saturday, August 31, 2024 at 8:48:13 PM UTC-6 Alan Grayson wrote:

You seem to know a lot about relativity, so please explain how the metric tensor can be defined unambiguously at some point P on the underlying manifold, spacetime, if there is an uncountable set of pairs on a vector space on the tangent space at some point P on which the metric tensor is defined (as a bilinear function which maps to the real numbers.) TY, AG

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[John Clark]

On Sat, Aug 31, 2024 at 10:48 PM Alan Grayson <agrays...@gmail.com> wrote:

 > please explain how the metric tensor can be defined unambiguously at some point P on the underlying manifold, spacetime, if there is an uncountable set of pairs on a vector space on the tangent space at some point P on which the metric tensor is defined

If, as I suspect, your interest is physics and not pure mathematics then it's a non-issue. The fact is nobody is even sure that 4D space-time contains an infinite number of points, for all we know it may only contain an astronomical number to an astronomical power number of points. That's undoubtedly a very big number but it's no closer to being infinite than the number one is.  

And even if 4D space-time does contain an uncountabley infinite number of points, if you simplify your physical theory by assuming there is only a countably infinite number of points it will have a negligible effect on your theory; that is to say you could make the discrepancy between what your theory predicts will happen and what you actually observed to happen in experiments to be arbitrarily small. I am not aware of any physical theory in which the difference between countable infinity and uncountable infinity leads to different experimentally testable predictions. 

 John K Clark    See what's on my new list at  Extropolis

n4x

[Alan Grayson]

We're dealing with the tangent space defined by a vector space of velocities on the spacetime manifold, so there are at least a countable infinity of such vectors. And the issue is whether the metric tensor is well defined since it's a bilinear function of a pair of velocity vectors, which maps to the real numbers. But for the metric tensor FIELD to be well-defined, we need a unique real number as the result, but this seems impossible since there are many possible pairs which map to different real numbers. I don't understand Brent's suggestion. AG 

[Alan Grayson]

While spacetime might not have an infinite set of events, countable or uncountable, the tangent space is constructed via a vector space with at least a countable number of elements. To see this, consider velocity vectors with rational velocities AG

On Tuesday, September 3, 2024 at 6:56:27 AM UTC-6 John Clark wrote:

[John Clark]

On Tue, Sep 3, 2024 at 10:07 AM Alan Grayson <agrays...@gmail.com> wrote:

> While spacetime might not have an infinite set of events, countable or uncountable, the tangent space is constructed via a vector space with at least a countable number of elements. 

Even though there are an uncountably infinite number of real numbers but only a countable number of rational numbers, you can always find a rational number that is arbitrarily close to any real number, provided that the real number in question is computable. Thus the thing that you're so worried about is not a concern for a physicist because, at least so far as anybody knows, there is no physical experiment you can perform that can reveal the difference between the countable infinite and the uncountable infinite.  

Mathematics is the language of physics but mathematics is not physics. And any language can be used for both fiction and nonfiction, so it could be that the hierarchies of infinity and even the very concept of infinity is the mathematical equivalent of a Harry Potter novel. Mathematical consistency may not be enough to ensure physical reality.

 John K Clark    See what's on my new list at  Extropolis

acr

[Alan Grayson]

I don't think you understand the issue. Velocities greater than c in the underlying spacetime manifold are allowed in the construction of the tangent plane, so the issue I raise has nothing to do with countable vs. uncountable, or computability. AG

[John Clark]

On Tue, Sep 3, 2024 at 3:58 PM Alan Grayson <agrays...@gmail.com> wrote:

> I don't think you understand the issue. Velocities greater than c in the underlying spacetime manifold are allowed in the construction of the tangent plane,

I don't think you understand that the things that mathematics allows and the things that the laws of physics allow are not necessarily the same thing. Mathematics allows for the existence of Newtonian physics, and Newton allows objects to travel much faster than the speed of light, but Einstein taught us that the laws of physics forbid it.  

Nearly all new theories in physics are mathematically consistent, but nearly all new theories in physics are also eventually proven to be wrong.  

 John K Clark    See what's on my new list at  Extropolis

wpb

 

[Alan Grayson]

The tangent space and the metric tensor field are used in GR. I fail to see how your comments relate to the possibly ambiguous concept of the latter. The metric tensor field seems ambiguously defined. AG

[Alan Grayson]

IOW, without using vectors representing velocities greater than c, we can get different values for the metric tensor field. Brent says don't choose a pair to do the calculation, choose the second as conjugate of the first. How does this help? AG
G

[John Clark]

On Tue, Sep 3, 2024 at 6:44 PM Alan Grayson <agrays...@gmail.com> wrote:

> I fail to see how your comments relate to the possibly ambiguous concept of the latter. The metric tensor field seems ambiguously defined.

A N dimensional space is composed of an uncountable number of real numbers but it can be unambiguously defined by just N countable rational numbers, you can pair them up one to one. This is possible because there is only a countably infinite number of COMPUTABLE real numbers, the same rank of infinity as the rational numbers. So you can in effect give a rational number name to every real number you are able to find on the number line. You can do this even for a number such as π which is not only irrational, it's transcendental, because it is also computable. You can use an infinite series to get arbitrarily close to π.  

The vast majority of numbers on the number line are NOT computable (and have no name) but that's not really a problem despite the fact that the vast majority of numbers on the number line are NOT computable because, except for Chaitin's Omega Number, every number that a mathematician has ever heard of is a computable number. Computable numbers can have names, uncomputable numbers can not.

  John K Clark    See what's on my new list at  Extropolis

und

[Alan Grayson]

What you write seems correct but doesn't address the issue I raised; namely, that the metric tensor is defined on pairs of vectors in the vector space in the tangent plane of the spacetime manifold, and yields different real values for most different pairs. So, it seems that the metric tensor FIELD is NOT well defined. AG

[John Clark]

On Wed, Sep 4, 2024 at 8:08 AM Alan Grayson <agrays...@gmail.com> wrote:

> it seems that the metric tensor FIELD is NOT well defined. AG

 

The metric tensor encodes spacetime curvature, and for every point in spacetime that you can name I can give you a 4x4 matrix of unique computable numbers that defines the curvature at that point. What's ambiguous about that? It's true that I can't do that for points in spacetime that you cannot name, but that is not a problem because you cannot get experimental results from points that you cannot name.

John K Clark    See what's on my new list at  Extropolis

wai

[Alan Grayson]

Firstly, the metric tensor does NOT directly give the curvature. Rather, it's a bilinear function whose range is a real number, but which one? That depends on the two domain vectors in the vector space on the tangent plane at some point P where the tangent plane touches the spacetime manifold. The 4x4 matrix you refer to is a representation of its mapping, not the curvature at some point P on the spacetime manifold. To get the real number it maps to, you need 2 vectors, a column vector followed by a row vector, but which give the real number we're seeking? This is where the ambiguity that bothers me is manifested. In GR, to get the curvature, we need to solve for the metric tensor field, and then solve for the curvature scalar and the Riemann curvature tensor. Recall, that in Einstein's Field Equation, there are these 3 tensors on the LHS, if  we place the Energy-Momentum tensor is on the RHS. As for the uncountable set of real numbers, since we can't write them down, they can't be put in list (which doesn't omit any), but which can be done for the rational numbers as proven by Cantor using his diagonal proof.  AG

[Brent Meeker]

Most different vectors are different lengths.

Brent

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[Alan Grayson]

Of course. How does this help? AG 

[Brent Meeker]

It explains this "...and yields different real values for most different pairs."

Brent

[Alan Grayson]

But my point is therefore that the metric tensor field is ambiguous !

[Brent Meeker]

The metric field is the set of metric tensors, one at each point.  It's not some vector lengths.

When are you gonna read "Relativity DeMystifie".  I told you the page numbers and you can look up more in the index.

Brent

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[Alan Grayson]

No; you never posted any page numbers in your reference. Had you done that, I would have immediately studied your specific reference. But I was about to post something like what you wrote; namely, that solving for the metric tensor MEANS solving for the 16 components defining its function, given a stress-energy tensor and a coordinate system. The form of the components must depend on the coordinates. And given a coordinate system, the components will vary depending on location in spacetime, and this is what's meant by the "metric tensor field". I thought the "field" refers to a unique real number at each point in spacetime, but it must refer to the components of the matrix representing the tensor.  Wiki's definition seems misleading since it states that the metric tensor is a bilinear function of two vectors on the tangent plane. Those vectors are its arguments, but sort-of misleading. Is there anything in the foregoing that I got wrong? AG

[Alan Grayson]

How is the foregoing consistent with the statement that tensors are independent of coordinate systems? TY, AG

[Brent Meeker]

On 9/4/2024 9:39 PM, Alan Grayson wrote:

No; you never posted any page numbers in your reference. Had you done that, I would have immediately studied your specific reference.

If you have the book you can look up "metric" in the index.  There are multiple entries: 23, 32-45

But I was about to post something like what you wrote; namely, that solving for the metric tensor MEANS solving for the 16 components defining its function, given a stress-energy tensor and a coordinate system. The form of the components must depend on the coordinates. And given a coordinate system, the components will vary depending on location in spacetime, and this is what's meant by the "metric tensor field". I thought the "field" refers to a unique real number at each point in spacetime, but it must refer to the components of the matrix representing the tensor.  Wiki's definition seems misleading since it states that the metric tensor is a bilinear function of two vectors on the tangent plane. Those vectors are its arguments, but sort-of misleading. Is there anything in the foregoing that I got wrong? AG

Nope that's it. 

On Wednesday, September 4, 2024 at 8:09:07 PM UTC-6 Brent Meeker wrote:

The metric field is the set of metric tensors, one at each point.  It's not some vector lengths.

When are you gonna read "Relativity DeMystifie".  I told you the page numbers and you can look up more in the index.

Brent

On 9/4/2024 5:59 PM, Alan Grayson wrote:

But my point is therefore that the metric tensor field is ambiguous !

On Wednesday, September 4, 2024 at 6:40:38 PM UTC-6 Brent Meeker wrote:

It explains this "...and yields different real values for most different pairs."

Brent

On 9/4/2024 1:03 PM, Alan Grayson wrote:

and yields different real values for most different pairs.

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[Brent Meeker]

If  you change coordinate systems the vectors and tensors transform in such a way that the physics is unchanged.  That's the defining property of vectors and tensors and why they are not just the arrays used to represent them.

Brent

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[Alan Grayson]

Since the metric tensor is defined on the flat tangent space, will its matrix representation always be diagonal? TY, AG

[Brent Meeker]

Ever use cylindrical coordinates?

Brent

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[Alan Grayson]

I know how they're defined but I never used them. AG

[Alan Grayson]

If you're interested in the Tao instead of laughing at it, google "chakra'. AG

[John Clark]

On Mon, Sep 9, 2024 at 1:19 PM Alan Grayson <agrays...@gmail.com> wrote:

> If you're interested in the Tao instead of laughing at it, google "chakra'. AG

If you're really interested in The Tao then forget chakra and read "The Tao Is Silent" by Raymond Smullyan. It's one of the most brilliant books I've ever read.

The Tao Is Silent

John K Clark    See what's on my new list at  Extropolis

rsb