Metric or Kronecker delta?

[Genady]

My question is about the following step in a derivation of energy-momentum tensor:

When the ∂νL in (3.33) moves under the ∂μ in (3.34) and gets contracted, I'd expect it to become \(\delta^{\mu}_{\nu} \mathcal L\). Why is it rather gμνL ? Typo?

(In this text, gμν=ημν )

Edited February 25 by Genady

[joigus]

30 minutes ago, Genady said:

My question is about the following step in a derivation of energy-momentum tensor:

When the ∂νL in (3.33) moves under the ∂μ in (3.34) and gets contracted, I'd expect it to become δμνL . Why is it rather gμνL ? Typo?

(In this text, gμν=ημν )

No, no typo. It is actually a theorem (or lemma, etc) of tensor calculus that the gradient wrt contravariant coordinates is itself covariant. It is just a fortunate notational coincidence that the "sub" position in the derivative symbol seems to suggest that.

Proof:

Spoiler

\[ x^{\mu}=\left.\left(\varLambda^{-1}\right)^{\mu}\right._{\beta}\hat{x}^{\beta} \]

\[ \frac{\partial}{\partial\hat{x}^{\mu}}=\frac{\partial x^{\alpha}}{\partial\hat{x}^{\mu}}\frac{\partial}{\partial x^{\alpha}}=\left.\left(\varLambda^{-1}\right)^{\alpha}\right._{\mu}\frac{\partial}{\partial x^{\alpha}} \]

 

[Genady]

1 hour ago, joigus said:

No, no typo. It is actually a theorem (or lemma, etc) of tensor calculus that the gradient wrt contravariant coordinates is itself covariant. It is just a fortunate notational coincidence that the "sub" position in the derivative symbol seems to suggest that.

Proof:

  Reveal hidden contents

 

xμ=(Λ−1)μβx^β

 

 

∂∂x^μ=∂xα∂x^μ∂∂xα=(Λ−1)αμ∂∂xα

 

 

I understand this but I don't think it answers my question. This is what I mean:

I can rewrite (3.34) so: \[ \partial_{\mu} (\sum_n \frac {\partial \mathcal L} {\partial (\partial_{\mu} \phi_n)} \partial_{\nu} \phi_n) = \partial_{\mu} (g_{\mu \nu} \mathcal L)\]

Then, from this and (3.33), we get \[\partial_{\nu} \mathcal L = \partial_{\mu} (g_{\mu \nu} \mathcal L)\]

I think, it is incorrect.

It rather should be \[\partial_{\nu} \mathcal L = \partial_{\mu} (\delta^{\mu}_{\nu} \mathcal L)\]

P.S. Ignore positions of indices; Schwartz does not follow upper/lower standard. The difference is between \(g\) and \(\delta\).

[joigus]

Oh. Got you. Yes, you're right. It should be what you say. Classic books in QFT tend to be rather fast-and-loose with the indices. \( \partial_{\nu}\mathcal{L}=\partial_{\mu}\left(g_{\mu\nu}\mathcal{L}\right) \) is not a tensor equation. \( \partial_{\nu}\mathcal{L}=\partial_{\mu}\left(\left.\delta^{\mu}\right._{\nu}\mathcal{L}\right) \) is.

Although I should say there is no fundamental difference between \( g \) and \( \delta \) really. \( \delta \) is just \( g \) (viewed as just another garden-variety tensor) with an index raised (by using itself).

 

Bogoliubov is similarly cavalier with the indices if I remember correctly. Is it from the 50's?

[Genady]

Just now, joigus said:

Oh. Got you. Yes, you're right. It should be what you say. Classic books in QFT tend to be rather fast-and-loose with the indices. ∂νL=∂μ(gμνL) is not a tensor equation. ∂νL=∂μ(δμνL) is.

Although I should say there is no fundamental difference between g and δ really. δ is just g (viewed as just another garden-variety tensor) with an index raised (by using itself).

 

They are different: \[\delta=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\] \[g=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}\]

 

 

 

[joigus]

As matrices, they are. But as tensors, they aren't. They are one and the same basis-independent object, coding the same physical information. This is exactly the same as a vector in one basis looking, eg, like matrix (0 1 0 0) but looking like (0 -1 0 0) in another basis. You are confusing the tensor with its coordinates.

IOW: All metrics, no matter the dimension and signature, look exactly like the identity matrix (the Kronecker delta) when the scalar product is expressed under the convention that the first factor is written in covariant components, and the second one in contravariant ones.

A tensor is a physical object. A matrix is just a collection of numbers used to represent that object.

Let me put it this way:

\[ U_{\mu}V^{\mu}=U_{\mu}\left.\delta^{\mu}\right._{\nu}V^{\nu}=U^{\mu}\left.\delta_{\mu}\right.^{\nu}V_{\nu}=U^{\mu}g_{\mu\nu}V^{\nu}=U_{\mu}g^{\mu\nu}V_{\nu} \]

Edited February 26 by joigus
minor correction

[Genady]

31 minutes ago, joigus said:

As matrices, they are. But as tensors, they aren't. They are one and the same basis-independent object, coding the same physical information. This is exactly the same as a vector in one basis looking, eg, like matrix (0 1 0 0) but looking like (0 -1 0 0) in another basis. You are confusing the tensor with its coordinates.

IOW: All metrics, no matter the dimension and signature, look exactly like the identity matrix (the Kronecker delta) when the scalar product is expressed under the convention that the first factor is written in covariant components, and the second one in contravariant ones.

A tensor is a physical object. A matrix is just a collection of numbers used to represent that object.

Let me put it this way:

 

UμVμ=UμδμνVν=UμδμνVν=UμgμνVν=UμgμνVν

 

Please, I really, really know this. I know this index gymnastics, lowering and raising indices, tensors vs. basis representations, etc. I appreciate your time, but there is no need to teach basics here. Let's focus.

Back to my question. 

Let's take \(\nu=1\).

If \(\partial_{\nu} \mathcal L = \partial_{\mu} (g_{\mu \nu} \mathcal L)\),

then \(\partial_1 \mathcal L = \partial_{\mu} (g_{\mu 1} \mathcal L) = -\partial_1 \mathcal L \).

Where is my mistake?

Edited February 26 by Genady

[joigus]

16 minutes ago, Genady said:

Where is my mistake?

Nowhere. You made no mistake. None whatsoever. You are just realising that Schwartz made a mistake, not you. So it was a typo.

He probably meant to write something like,

\[ \partial^{\mu}\left(\sum_{n}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi_{n}\right)}\partial_{\nu}\phi_{n}-g_{\mu\nu}\mathcal{L}\right)=0 \]

which is correct, and consistent with your derivation.

That's the problem with books that don't follow the covariant/contravariant convention. Index gymnastics does that for you automatically.

Sorry, I thought I'd told you:

1 hour ago, joigus said:

Oh. Got you. Yes, you're right. It should be what you say. Classic books in QFT tend to be rather fast-and-loose with the indices.

 

[Genady]

Thank you. All's well. 

14 minutes ago, joigus said:

That's the problem with books that don't follow the covariant/contravariant convention. Index gymnastics does that for you automatically.

Yes, I like the book otherwise, but it would be so much easier to follow if the indices were where they should be.

[joigus]

No problem.

By the way, I should have written,

\[ \partial^{\mu}\left(\sum_{n}\frac{\partial\mathcal{L}}{\partial\left(\partial^{\mu}\phi_{n}\right)}\partial_{\nu}\phi_{n}-g_{\mu\nu}\mathcal{L}\right)=0 \]

That is correct.

[KJW]

3 hours ago, joigus said:

They are one and the same basis-independent object, coding the same physical information.

But they don't encode the same information. The metric tensor encodes the spacetime curvature whereas the Kronecker delta does not. But the inverse of the metric tensor does encode the same information as the metric tensor.

 

Edited February 26 by KJW

[joigus]

5 hours ago, KJW said:

But they don't encode the same information. The metric tensor encodes the spacetime curvature whereas the Kronecker delta does not. But the inverse of the metric tensor does encode the same information as the metric tensor.

 

What curvature? This is all pseudo-Euclidean metric we're talking about. This is QFT. \( g_{\mu\nu}=\eta_{\mu\nu} \)

I think you mean in GR. But even in GR, the metric tensor does not encode curvature. The Riemann tensor does. And the metric tensor is covariantly constant. Because it is. That should give us a clue. It doesn't really encode much, does it? This is a common misconception, that the metric tensor components "encode" something. In terms of tetrads it's very clear that it's nothing but the identity operator. What's perplexing is that covariantly shifting with the Christoffels obtained from it around an infinitesimally-small closed lood gives you something else, but that's a different story.

[KJW]

17 minutes ago, joigus said:

What curvature? This is all pseudo-Euclidean metric we're talking about. This is QFT. gμν=ημν

I think you mean in GR.

Of course.

 

 

17 minutes ago, joigus said:

But even in GR, the metric tensor does not encode curvature. The Riemann tensor does.

The metric tensor does encode curvature in the sense that one can mathematically obtain the Riemann tensor from it. One can't mathematically obtain the Riemann tensor from the Kronecker delta. Therefore, the Kronecker delta is not mathematically equivalent to the metric tensor.

 

 

17 minutes ago, joigus said:

And the metric tensor is covariantly constant.

By definition... it need not be covariantly constant.

 

 

17 minutes ago, joigus said:

In terms of tetrads it's very clear that it's nothing but the identity operator.

That's because it has been replaced with a different quantity to represent the anholonomy.

 

 

17 minutes ago, joigus said:

What's perplexing is that covariantly shifting with the Christoffels obtained from it around an infinitesimally-small closed lood gives you something else, but that's a different story.

But it's not a different story. It's saying that the metric tensor contains information to mathematically obtain the Riemann curvature tensor.

 

 

[joigus]

39 minutes ago, KJW said:

Of course.

That's all I wanted to hear, really. If you want to discuss GR, I suggest you open a new thread doing so. Discussing topological aspects of GR on a thread about index gymnastics in flat QFT would be highly misleading.

 

39 minutes ago, KJW said:

The metric tensor does encode curvature in the sense that one can mathematically obtain the Riemann tensor from it.

One can obtain Shakespeare's Sonnets from the alphabet, but I see no Shakespeare in ABCDEFGHIJKLMNOPQRSTUVWXYZ. Do you?

It's what you do with the alphabet that matters. I can do no nothing like what Shakespeare did. Same happens with the metric.

Yours is actually a common misconception. I'm just saying.

Edited February 26 by joigus
minor correction

[KJW]

1 hour ago, joigus said:

That's all I wanted to hear, really. If you want to discuss GR, I suggest you open a new thread doing so. Discussing topological aspects of GR on a thread about index gymnastics in flat QFT would be highly misleading.

Why would I open a new thread? You wrote something I disagreed with in this thread. It was not my intention to discuss topological aspects of GR. It was my intention to argue that the metric tensor and the Kronecker delta are not equivalent. In spite being numerically identical, not even the Euclidean metric tensor is equivalent to the Kronecker delta because the Euclidean metric tensor contains information that the space is flat, whereas the Kronecker delta does not.

 

 

1 hour ago, joigus said:

One can obtain Shakespeare's Sonnets from the alphabet, but I see no Shakespeare in ABCDEFGHIJKLMNOPQRSTUVWXYZ. Do you?

It's what you do with the alphabet that matters. I can do no nothing like what Shakespeare did. Same happens with the metric.

This is a false analogy. In the case of the Riemann tensor, it is not what you do with the metric tensor that matters because what you do with the metric tensor to form the Riemann tensor doesn't change. But if I change the metric tensor, I may change the Riemann tensor. A more correct analogy is DNA sequencing where the process of converting the DNA to a sequence of letters "A", "G", "C", and "T" is a procedure that doesn't change with the sample. The letters "A", "G", "C", and "T" are not the same as the DNA molecule, but we assume an informational connection between the two. There are many possible values of the metric tensor, leading to many possible values of the Riemann tensor. But there is only one Kronecker delta (for a particular number of dimensions).

 

 

1 hour ago, joigus said:

Yours is actually a common misconception. I'm just saying.

I don't think it is a misconception. I think you are operating under the common misconception that all the various subscripted/superscripted versions of a given tensor are necessarily equivalent. But for the same reason that the Kronecker delta is not equivalent to the metric tensor, raising or lowering indices may alter the information content of the tensor. However, the reversibility of raising or lowering indices is due to the metric tensor or the inverse of the metric tensor retaining any information that might have been lost as a result of raising or lowering indices. This does imply that the given tensor is dependent on the metric tensor or its inverse. Otherwise, raising or lowering indices can't alter the information content, and all the various subscripted/superscripted versions are equivalent.

 

 

Edited February 26 by KJW

[joigus]

7 minutes ago, KJW said:

This is a false analogy.

No, yours is. Wanna know of a better analogy? Mine. Mine is better.

It's been 50 years since I last did this.  

Edited February 26 by joigus

[KJW]

7 minutes ago, joigus said:

No, yours is. Wanna know of a better analogy? Mine. Mine is better.

It's been 50 years since I last did this.  

This is an appeal to your own authority. I think that means you lost the argument.

 

[joigus]

8 hours ago, KJW said:

This is an appeal to your own authority.

No, it was a joke. I thought the emoji had given it away. I'm sorry it didn't, and you took it seriously.

Value judgement is falacious. Like "your analogy is false", "I think you're wrong". That's what I was mocking.

8 hours ago, KJW said:

I think that means you lost the argument.

Funny (and may I say telling), that you consider it a contest. It's not. Nor should it be.

You are just wrong, or you sound to me very much like you are in what seems to be your interpretation of tensors. Take this example: flat \( \mathbb{R}^{2} \) (the plane).

All of this should be self-explanatory.

\[ ds^{2}=dx^{i}g_{ij}dx^{j}=\left(\begin{array}{cc} dr & d\theta\end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 0 & r^{2} \end{array}\right)\left(\begin{array}{c} dr\\ d\theta \end{array}\right)=dr^{2}+r^{2}d\theta^{2} \]

\[ ds^{2}=dx_{i}g^{ij}dx_{j}=\left(\begin{array}{cc} dr & r^{2}d\theta\end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 0 & r^{-2} \end{array}\right)\left(\begin{array}{c} dr\\ r^{2}d\theta \end{array}\right)=dr^{2}+r^{2}d\theta^{2} \]

\[ ds^{2}=dx_{i}\left.\delta^{i}\right._{j}dx^{j}=\left(\begin{array}{cc} dr & r^{2}d\theta\end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)\left(\begin{array}{c} dr\\ d\theta \end{array}\right)=dr^{2}+r^{2}d\theta^{2} \]

In fact, \( g_{ij} \) and \( g^{ij} \) will give you more than you bargained for: They will give you a spurious sigularity that isn't there at all. There's nothing wrong at \( r=0 \). The covariant methods tell you that.

But the form of the once-covariant once-contravariant components of the same silly little thing give you a clue, as \( \left.\delta^{i}\right._{j} \) tells you clearly that nothing funny is going on at that point. The Kronecker delta, in this case, is more honest-to-goodness than the other ones.

If you calculate the Riemann, of course, it will tell you that beyond any doubt. It's at that level that you can talk about anoholonomy and curvature.

I'm sure you know all that from what we've talked before. These are cautionary tales that are in the literature.

Another possibility is that we didn't understand each other's point. One can never be sure.

And I apologise to @Genady. For this was his thread on variational derivatives really and wasn't about curvature at all. 🤷‍♂️

[KJW]

On 2/27/2024 at 8:49 AM, joigus said:

All of this should be self-explanatory.

Actually, your example of [math]ds^2 = dx^i g_{ij} dx^j[/math], [math]ds^2 = dx_i g^{ij} dx_j[/math], and [math]ds^2 = dx_i \delta^i_j dx^j[/math] only serves to strengthen my point. Of the three metrics, only in the case of [math]ds^2 = dx^i g_{ij} dx^j[/math] are both vectors differentials of the coordinates. The differentials of the coordinates vector is naturally contravariant as a result of the chain rule. The [math]dx_i = (dr \ \ \ \ r^2 d\theta)[/math] vector is not a differentials of the coordinates vector as it is clearly seen to contain components of the metric tensor. Indeed, this vector required lowering the index of [math]dx^i = (dr \ \ \ \ d\theta)[/math] with the metric tensor. One can't lower indices with the Kronecker delta, another difference between the Kronecker delta and the metric tensor. In all three cases, the metrics [math]ds^2 = dr^2 + r^2 d\theta^2[/math] are the same. Note that the coefficients of the differentials of the coordinates quadratics are components of the metric tensor [math]g_{ij}[/math]. There is not a Kronecker delta in sight. Although the rules of index manipulation do allow the three expressions [math]ds^2 = dx^i g_{ij} dx^j[/math], [math]ds^2 = dx_i g^{ij} dx_j[/math], and [math]ds^2 = dx_i \delta^i_j dx^j[/math], [math]ds^2 = dx^i g_{ij} dx^j[/math] is the natural expression.

 

 

On 2/27/2024 at 8:49 AM, joigus said:

In fact, [math]g_{ij}[/math] and [math]g^{ij}[/math] will give you more than you bargained for: They will give you a spurious singularity that isn't there at all. There's nothing wrong at [math]r = 0[/math]. The covariant methods tell you that.

At [math]r = 0[/math], there is a (removable) singularity that is a natural part of the polar coordinate system. If you want to remove this singularity, you change the coordinate system to something like the Euclidean coordinate system.

 

 

On 2/27/2024 at 8:49 AM, joigus said:

But the form of the once-covariant once-contravariant components of the same silly little thing give you a clue, as [math]\delta^i_j[/math] tells you clearly that nothing funny is going on at that point. The Kronecker delta, in this case, is more honest-to-goodness than the other ones.

The Kronecker delta tells you nothing. Its invariance guarantees that. You chose a flat space for your example. How about choosing the spacetime of a Schwarzschild black hole, where the central singularity is not removable.

 

 

On 2/27/2024 at 8:49 AM, joigus said:

You are just wrong, or you sound to me very much like you are in what seems to be your interpretation of tensors.

I don't think that what I'm talking about is my interpretation of tensors. I see it as more about general mathematical logic.

 

 

Edited March 1 by KJW

[joigus]

On 3/1/2024 at 6:22 PM, KJW said:

The Kronecker delta tells you nothing. Its invariance guarantees that. You chose a flat space for your example. How about choosing the spacetime of a Schwarzschild black hole, where the central singularity is not removable.

This is getting farther and farther away form discussing anything substantial (let alone anything within the OP context), and more and more about you getting out of your way in order to shift the context so that I could be proven wrong in that context.

Anyway, Kronecker delta with a superindex and a subindex is an isotropic tensor. Kronecker delta with, eg, two covariant indices (like Tαβ=δαβ ) tells you much, much less, as it's a frame-dependent equality. δαβ is telling you that what you have here is just a rule to dot-multiply vectors. How much or how little does that "encode"? The dimension, and the fact that you're dealing with a scalar product? That's about all. I'll leave to you to decide how much that is telling you. There is a reason why the connection is given in terms of g−1∂g . Neither covariant indices nor contravariant ones give you the curvature, and it's an interplay between the two that does it.

There is just another isotropic tensor in every space (the ϵ tensor). It is kind of telling you about orthogonality. The Kronecker only looks standard (1's & 0's) with one index up and the other down. The epsilon tensor only looks standard (1's, 0's, and -1's) with all indices down or all up. Otherwise, they show you all kinds of misleading info, as I clearly showed you with my textbook-standard example.

Also, multilinear operators are not just "tensors" independently of a context, as you seem to imply. Multilinear operators are or are not tensors depending on the relevance of a certain group of transformations. There are such things as O(3) tensors (orthogonal tensors), U(n) tensors,... there are pseudo-Euclidean tensors (the only ones we were talking about to begin with), there are tensors under diffeomorphisms (the ones you, for some reason, want to shift the conversation to, although they have little to do with the initial discussion), etc. 

Let me point out more mistakes that you're making:

On 3/1/2024 at 6:22 PM, KJW said:

The differentials of the coordinates vector is naturally contravariant as a result of the chain rule. The dxi=(dr    r2dθ) vector is not a differentials of the coordinates vector as it is clearly seen to contain components of the metric tensor

Again, you are wrong. There's nothing special about differentials of the coordinates:

All such objects form a basis. They are called coordinate bases or holonomic bases --see below. But not all bases are made up of derivatives of coordinates. It's only when they are that they thus called. This is from Stewart, Advanced General Relativity, Cambridge 1991:

 

So no, not all bases are coordinate bases. And I wrote down a totally legit basis.

More on that:

https://www.physicsforums.com/threads/non-coordinate-basis-explained.950852/#:~:text=Some examples of non-coordinate basis vectors include polar basis,defined by traditional coordinate axes.

Under a different name:

https://en.wikipedia.org/wiki/Holonomic_basis

So one thing is a basis, and quite a very specific (and a distinct) thing is a coordinate basis. In still other words: A coordinate basis is made up of a set of exact differential forms and their duals. This is in close analogy to what happens in thermodynamics: You can use the Pfaffian forms of heat and work to define any change in the energy of a system, even though there are no "heat coordinate" or "work coordinate". But a basis they are: dU=TdS-PdV, even though TdS is not d(anything) and PdV is not d(anything).

Do you or do you not agree that the variational derivative @Genady was talking about should be written as,

\[ \partial^{\mu}\left(\sum_{n}\frac{\partial\mathcal{L}}{\partial\left(\partial^{\mu}\phi_{n}\right)}\partial_{\nu}\phi_{n}-g_{\mu\nu}\mathcal{L}\right)=0 \]

instead of,

\[ \partial_{\mu}\left(\sum_{n}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi_{n}\right)}\partial_{\nu}\phi_{n}-g_{\mu\nu}\mathcal{L}\right)=0 \]

As far as I can tell, that was the question, your detour into differential geometry has been satisfactorily answered, and you seem to have nothing further to say that's remotely on-topic.

[Genady]

2 hours ago, joigus said:

should be written as

I'm at ease with the Schwartz's notation by now. It is as simple as mentally substituting \(A_{\mu} g^{\mu \alpha} B_{\alpha}\) every time he writes \(A_{\mu} B_{\mu}\).

[KJW]

4 hours ago, joigus said:

Do you or do you not agree that the variational derivative @Genady was talking about should be written as,

[math]\partial^\mu \left( \displaystyle \sum_{n} \dfrac{\partial\mathcal{L}}{\partial(\partial^\mu \phi_n)} \partial_\nu \phi_n - g_{\mu \nu} \mathcal{L} \right) = 0[/math]     [Expression 1]

instead of,

[math]\partial_\mu \left( \displaystyle \sum_{n} \dfrac{\partial\mathcal{L}}{\partial(\partial_\mu \phi_n)} \partial_\nu \phi_n - g_{\mu \nu} \mathcal{L} \right) = 0[/math]     [Expression 2]

[Expression 2] is incorrect. [Expression 1], although not incorrect (but see below), is not how I would choose to write it. I generally write partial differential operators in covariant (subscripted) form. Thus, I would use [Expression 2] in the corrected form:

[math]\partial_\mu \left( \displaystyle \sum_{n} \dfrac{\partial\mathcal{L}}{\partial(\partial_\mu \phi_n)} \partial_\nu \phi_n - \delta^\mu_\nu \mathcal{L} \right) = 0[/math]

At this point, I must admit that I have made an error. I took something you said out of context and placed it into another context in which I said you made an error. I apologise to you and Genady for this.

However, I will ask you this: In general, does [math]\partial_\mu \partial^\mu \phi = \partial^\mu \partial_\mu \phi[/math] ?

 

 

Edited March 2 by KJW

[KJW]

4 hours ago, joigus said:

Anyway, Kronecker delta with a superindex and a subindex is an isotropic tensor. Kronecker delta with, eg, two covariant indices (like Tαβ=δαβ ) tells you much, much less, as it's a frame-dependent equality.

Hmmm, I nearly misread this. But it seems you are not giving me the credit to know that the identity matrix can only be a tensor if it has one superscript and one subscript.

But the metric tensor and its inverse are not only frame-dependent quantities, but they also depend on the intrinsic manifold. By contrast, the Kronecker delta is not only frame-independent, but also independent of the intrinsic manifold. That is, the metric tensor or its inverse tells you both what intrinsic manifold you have, and also the frame within that intrinsic manifold. The Kronecker delta tells you neither. Quantities that are frame-independent but tell you what intrinsic manifold you have would be very useful, but the Kronecker delta is not that quantity. Neither is the metric tensor, but the Riemann tensor is frame-independent on a flat manifold.

 

 

4 hours ago, joigus said:

There is just another isotropic tensor in every space (the ϵ tensor). It is kind of telling you about orthogonality.

Only the metric tensor and its inverse can tell you about orthogonality. The ϵ tensor can only tell you about the weaker condition of linear-independence. I suspect you know this, but are underestimating me.

 

 

4 hours ago, joigus said:

Multilinear operators are or are not tensors depending on the relevance of a certain group of transformations.

I am fully aware that the concept of a tensor is dependent on the group of transformations. In my own research, I explored tensors within the context of the general linear group. This requires that the definition of the partial differential operator be extended beyond the chain rule. It introduces a new quantity to deal with the anholonomy. At the time, it was unclear to me about how to deal with the oxymoronic "anholonomic coordinate systems". Only later did I realise that they were a generalisation of the vierbein.

 

 

4 hours ago, joigus said:

There's nothing special about differentials of the coordinates

What is special about the differentials of the coordinates is that they are differentials of the coordinates. What more needs to be said? I'm perfectly aware of the modern trend of getting rid of the entire notion of coordinate systems. But I reject this trend, regarding it as throwing the baby out with the bathwater. The notion of covariance is about putting all coordinate systems on equal footing by ensuring that equalities are the same in all coordinate systems. But somehow, even the notion of a coordinate system is an anathema. Even though tensors can be reformulated without coordinate systems such as I mentioned above, there is still an underlying aspect of the notion of coordinate systems within this formulation. It underlies the concept of curvature, regardless of how curvature is defined in the modern formalism.

 

 

[joigus]

10 hours ago, KJW said:

I'm perfectly aware of the modern trend of getting rid of the entire notion of coordinate systems. But I reject this trend, regarding it as throwing the baby out with the bathwater.

Ok, so you're old school. I respect that.

But mind you that coordinates could be misleading you in some respects while they're helping you in others. This observation should always be carried along.

13 hours ago, KJW said:

However, I will ask you this: In general, does ∂μ∂μϕ=∂μ∂μϕ ?

In flat coordinates, sure. In curvilinear coordinates, it's a bit more involved than that. That's called the Laplace-Beltrami operator and you have to write some metric tensors in between, and also some epsilons, if I remember correctly. I would need some time to remember all the machinery. If yours is a genuine question.

https://en.wikipedia.org/wiki/Laplace–Beltrami_operator

Also books like Gockeler-Schucker, etc. on differential-geometry methods for theoretical physics.

Are you just asking or trying to catch me again?

14 hours ago, Genady said:

I'm at ease with the Schwartz's notation by now. It is as simple as mentally substituting AμgμαBα every time he writes AμBμ .

Yes. That's what Feynman did all the time. For some reason he didn't like the g's.

[KJW]

8 hours ago, joigus said:

Are you just asking or trying to catch me again?

You could say that, but in fact I was making a point. The subscripted form of the partial differential operator is the natural form of this operator, whereas the superscripted form is obtained from the subscripted form by raising the index with the inverse of the metric tensor. The superscripted form of the partial differential operator therefore implicitly contains the inverse of the metric tensor which has to be explicitly taken into account in [math]\partial_\mu \partial^\mu \phi[/math]. Thus:

[math]\partial_\mu \partial^\mu \phi = \partial_\mu (g^{\mu\nu} \partial_\nu \phi) = \partial_\nu (g^{\nu\mu} \partial_\mu \phi) = \partial_\nu g^{\nu\mu}\, \partial_\mu \phi + g^{\nu\mu}\partial_\nu \partial_\mu \phi = \partial_\nu g^{\nu\mu}\, \partial_\mu \phi + \partial^\mu \partial_\mu \phi[/math]

A similar problem occurs with [math]dx_\mu[/math]. Is [math]dx_\mu = g_{\mu\nu} dx^\nu[/math]? Or is [math]dx_\mu = d(g_{\mu\nu} x^\nu)[/math]? I generally avoid this problem by not using [math]dx_\mu[/math]. However, in the case of:

[math]u^\nu = \dfrac{dx^\nu}{ds}[/math]

it seems unambiguous to me that:

[math]u_\mu = g_{\mu\nu} u^\nu = g_{\mu\nu} \dfrac{dx^\nu}{ds}[/math]

and not:

[math]u_\mu = \dfrac{dx_\mu}{ds}[/math]

 

The point is that the various subscripted/superscripted forms of a given quantity are not created equally and that there is often one form that is the natural form from which all the other forms are derived, and sometimes it does require one to know which is the natural form.

 

 

Edited March 3 by KJW