Inertial and non inertial frames

[TrickyDicky]

No, there are two separate operations:

1) absolute derivative along a curve; this is equally valid in curved and flat spacetime; in flat the connection components vanish. Absolute derivative of a contravariant vector along a curve is a contravariant vector (along the same curve). As I see it, this is defined and motivated without respect to (2) - or can be.

2) Covariant derivative. This makes a tensor of one higher rank than its operand. Covariant derivative of contravariant vector field produces rank two mixed tensor.

3) A derived fact is that if you extend a vector along a path to be a field (doesn't matter how), apply (2), then contract by the original vector field and restrict to the curve, you get the same result as (1).

I am of the school to treat (3) as an interesting consequence, but understand (1) as a separate operation, with no need to go through (2) and (3) to get their.

I spoke at all times of the covariant derivative, you can check it, so I'm not sure what you mean by No.

 

[PAllen]

I spoke at all times of the covariant derivative, you can check it, so I'm not sure what you mean by No.

My misunderstanding.

 

[TrickyDicky]

My misunderstanding.

No problem, you are one of the very few regular people in this subforum (peter might be another) , that even when disagreeing with me (and we disagree a lot which is Ok) I feel that is honestly concentrating on the physics rather than in who is posting or in silly ego games that only contribute to confusion.

 

[Dale]

They always use vectors(that always depend on a basis and therefore vector equations are depedent on the coordinate system used)

I think you have a fundamental misunderstanding here. Vectors cannot depend on a basis since the basis is a set of vectors. That would be circular.

The components of a vector certainly depend on the basis, but not the vector itself. Also, although you can construct a basis from coordinates they are not the same thing and you can change coordinates without changing basis and vice versa.

So, basically vector equations do not depend on coordinates, as you suggest. Vectors are tensors and have the same coordinate independence.

 

[TrickyDicky]

We're actually talking about two different, though related, things. If $V$ is the velocity 4-vector, then you can define an operation producing another 4-vector, $A$ via:

$A^\nu = d/d\tau V^\nu + \Gamma^\nu_{\mu \lambda} V^\mu V^\lambda$

That produces a 4-vector. In contrast, if $V$ is a vector field then you can define an operation producing a [1,1] tensor via:

$\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{\mu \lambda} V^\lambda$

Those are different, but related operations.

As I told PAllen I always referred to the covariant derivative, not the absolute derivative.

Some people might have been misled by the wikipage on 4-acceleration where they call covariant derivative the formula for the absolute derivative that you use in your first formula above.
The bottom line is that in a general manifold using the Levi-Civita connection like in GR's curved manifold, the two formulas amount to the same thing. The vector at a point in the tangent vector space is always the linearized 4-vector in which the Christoffel coefficients vanish. That is common for SR and GR, but since I was inciding in the difference it was clear I was referring to the field extension. Well, or so I thought.

 

[Muphrid]

Directional covariant derivative?? That's a first, I only know the covariant derivative of a vector field, and in general manifolds it is a 2-index tensor. When the Christoffel coefficients can be made to vanish like in the flat case the covariant derivative reduces to directional derivative and gives another vector field.

I'm trying to show you that if you applied that logic to flat space the same way, you would not get a directional derivative.

You're starting with $\nabla_\mu V^\nu$, which we both agree is a two-index tensor. In flat space, cartesian coordiantes, $\nabla_\mu = \partial_\mu$, and you get $\partial_\mu V^\nu$. This still has two free indices, man. It's not a directional derivative.

 

[pervect]

See my answer to DrGreg.

I don't see anything in Caroll on the issue - except that it seems to support Dr. Greg's obsrevation that this point could use a fuller discussion in textbooks.

I believe that Dr. Greg's expression is right, as is his comment that this isn't well-enough discussed in most textbooks. MTW, Caroll's online notes, and even Wald come up short in this department from what I could tell, with respect to the 4-accleration. (I did have to rely on the index in my search, it's possible I missed something in the texts that wasn't well indexed.)

Wald does mention in a few places, mostly in exercises, that $a^a = u^a \nabla_a u^a$ however. This is pretty close - it's guaranteed to be right if it exists, but as Bill K pointed out a while ago using this as a defintion is a bit sloppy. It requires the existence of a vector field to apply Wald's defintion, and this is not guaranteed, all that is guaranteed is that we have one curve with a tangent vector on it, we don't necessarily have a congruence of curves defining a vector field.Given that Wald's treatment was sloppy, and that I couldn't find anything really clear in MTW or in Caroll's online lecture note, I'd have to agree that this basic point could use further clarification. And as far as I'm concerned Bill K and Dr Greg have provided a useful service in clarifying it.

I'm at a loss as to what TrickyDicky's intended point was even after reading it several times and perusing Caroll. I might be having an unusally dense day, but I rather think TrickyDicky is wrong on this one.

 

[TrickyDicky]

I agree that I should have qualified better my critique to DrGreg's post instead of just saying it was wrong, since most of what it said is right. But I didn't understand his insistence on saying the GR and SR case were equal even when it was specifically mentioned in the post that in the SR case the Christoffel coefficients vanished.
I think this is too important a difference to say that both objects are the same regardless of whether it is the flat special case or the curved manifold case with Levi-Civita connection.
My understanding is that in the latter case the absolute
derivative of a vector field equals the covariant derivative and it is also a (1,1) tensor.

 

[TrickyDicky]

I'm trying to show you that if you applied that logic to flat space the same way, you would not get a directional derivative.

You're starting with $\nabla_\mu V^\nu$, which we both agree is a two-index tensor. In flat space, cartesian coordiantes, $\nabla_\mu = \partial_\mu$, and you get $\partial_\mu V^\nu$. This still has two free indices, man. It's not a directional derivative.

If what you mean is that you get a matrix of partial derivatives, that is right, but I believe that is equivalent to a directional derivative, you don't need any connection for that(this is flat space).

 

[PeterDonis]

I am of the school to treat (3) as an interesting consequence, but understand (1) as a separate operation, with no need to go through (2) and (3) to get there.

I think that the route through (2) and (3) is easier for people (such as me) who are more used to working with the machinery of coordinate charts than with the machinery of parametrized curves. If you're working with a coordinate chart you have no choice but to go through (2) and (3) to get to (1) since your description of the curve is in terms of functions of the coordinates, not functions of an affine parameter. But I think that's what programmers call an "implementation detail"; I agree that logically speaking, (1) can be formulated in a way that is independent of (2) and (3).

 

[Muphrid]

If what you mean is that you get a matrix of partial derivatives, that is right, but I believe that is equivalent to a directional derivative, you don't need any connection for that(this is flat space).

You're missing my point still. I'm trying to tell you that a directional derivative, while true it doesn't need a connection, is still entirely different from what you think it is. Or at the least, you're applying it inconsistently.

Let's go back to 3d. What is the directional derivative in 3d? $a \cdot \partial V$, for some field $V$ and some direction $a$. In index notation, this is $a^i \partial_i V^j$.

But there are two derivatives that aren't directional. $\partial \cdot V$ and $\partial \times V$. They are effectively treated with one expression, one 2-index tensor: $\partial_i V^j$.

You keep effectively saying the latter are directional. They're not. When you introduce a connection and $\partial$ is no longer equal to $\nabla$, that doesn't change. Only the first expression, with some explicit direction $a$, is directional. The second expression is not. Both useful things, but not the same.

 

[Dale]

That's a first, I only know the covariant derivative of a vector field, and in general manifolds it is a 2-index tensor.

See Carroll's 3.30 and the nearby material. There he defines the covariant derivative of an arbitrary tensor along a path, which is a tensor of the same indices as the tensor for which you are taking the derivative. Even if you were not previously aware of it, it should seem familiar since it is closely tied to parallel transport.

Some people might have been misled by the wikipage on 4-acceleration where they call covariant derivative the formula for the absolute derivative that you use in your first formula above.

Carroll calls it a covariant derivative also, so the term "absolute derivative" is by no means universal. Since the Wikipedia page expanded the formula there should have been no confusion about what was being discussed and what rank tensor it was.

 

[stevendaryl]

Man, finally it dawns on you I was referring to a vector field, what else would you take the covariant derivative of in curved space (like GR's)?, that is the difference with SR, in flat spacetime the covariant derivative reduces to a simple directional derivative and you get another 4-vector.

I don't know why you say that. In SR (using Minkowsky coordinates), the formula for covariant derivative reduces to

$\nabla_\mu V^\nu = \partial_\mu V^\nu$

That's not a 4-vector, it's a [1,1] tensor. There is no distinction between SR and GR on this point.

I just can't make out how so many people can misunderstand this when I was specifically speaking about the difference SR/GR, and specifically talking about the "covariant derivative" of a vector field, we weren't at any moment talking about the tangent space of vector uniquely because at that space it makes no sense to talk about the difference SR/GR, and besides as I said covariant derivatives are always of vector fields, with the only caveat that in the flat special case it happens to be equivalent to a directional derivative (the Christoffel symbols vanish), and the result is another vector. But in the general manifold case the result is a order 2 tensor.

Well, my participation started when Dr. Somebody wrote down the usual equations for 4-velocity and 4-acceleration, and you said he was wrong.

I still don't agree with what you're saying. You talk about the difference between SR and GR, but everything mentioned is pretty much the same in GR and SR. The difference is that in SR it is always possible to choose coordinates so that $\Gamma^\nu_{\mu \lambda}$ is always zero. But it's still the case that if you choose different coordinates (for instance, accelerated coordinates or spherical polar coordinates), the $\Gamma$ coefficients will be nonzero.

 

[TrickyDicky]

I don't know why you say that. In SR (using Minkowsky coordinates), the formula for covariant derivative reduces to

$\nabla_\mu V^\nu = \partial_\mu V^\nu$

That's not a 4-vector, it's a [1,1] tensor. There is no distinction between SR and GR on this point.
Well, my participation started when Dr. Somebody wrote down the usual equations for 4-velocity and 4-acceleration, and you said he was wrong.

I still don't agree with what you're saying. You talk about the difference between SR and GR, but everything mentioned is pretty much the same in GR and SR. The difference is that in SR it is always possible to choose coordinates so that $\Gamma^\nu_{\mu \lambda}$ is always zero. But it's still the case that if you choose different coordinates (for instance, accelerated coordinates or spherical polar coordinates), the $\Gamma$ coefficients will be nonzero.

My understanding is that the covariant derivative of a vector field generalizes the directional derivative, and it is always in the direction of a certain vector so it can't be simply the formula for the partials derivatives of the components of a vector, that has
always the coordinate directions.

 

[TrickyDicky]

You're missing my point still. I'm trying to tell you that a directional derivative, while true it doesn't need a connection, is still entirely different from what you think it is. Or at the least, you're applying it inconsistently.

Let's go back to 3d. What is the directional derivative in 3d? $a \cdot \partial V$, for some field $V$ and some direction $a$. In index notation, this is $a^i \partial_i V^j$.

But there are two derivatives that aren't directional. $\partial \cdot V$ and $\partial \times V$. They are effectively treated with one expression, one 2-index tensor: $\partial_i V^j$.

You keep effectively saying the latter are directional. They're not. When you introduce a connection and $\partial$ is no longer equal to $\nabla$, that doesn't change. Only the first expression, with some explicit direction $a$, is directional. The second expression is not. Both useful things, but not the same.

In fact I don't think the latter is directional, what you seem to be missing is that it is not actually a covariant derivative, that's simply the matrix of partial derivatives of a vector, not of a vector field, in flat space, a cartesian tensor.

 

[TrickyDicky]

You talk about the difference between SR and GR, but everything mentioned is pretty much the same in GR and SR. The difference is that in SR it is always possible to choose coordinates so that $\Gamma^\nu_{\mu \lambda}$ is always zero. But it's still the case that if you choose different coordinates (for instance, accelerated coordinates or spherical polar coordinates), the $\Gamma$ coefficients will be nonzero.

Sure , that's because $\Gamma^\nu_{\mu \lambda}$ is not a tensor and therefore it can be zero in certain coordinates and not in others, and that is the same in SR and in GR. In fact I was only considering Cartesian coordinates for the SR case, so that might have added misunderstandings.

The fact is that DrGreg's post seemed to be referring to vectors at a point rather than vector fields, and I mistakenly thought he was referring to vector fields, that should close the case wrt to that post.

 

[Muphrid]

In fact I don't think the latter is directional, what you seem to be missing is that it is not actually a covariant derivative, that's simply the matrix of partial derivatives of a vector, not of a vector field, in flat space, a cartesian tensor.

First, in $\partial_i V^j$, $V$ is still a vector field.

Second, just what did you mean when you said this?

I just can't make out how so many people can misunderstand this when I was specifically speaking abou the difference SR/GR, and specifically talking about the "covariant derivative" of a vector field, we weren't at any moment talking about the tangent space of vector uniquely because at that space it makes no sense to talk about the difference SR/GR, and besides as I said covariant derivatives are always of vector fields, with the only caveat that in the flat special case it happens to be equivalent to a directional derivative (the Christoffel symbols vanish), and the result is another vector. But in the general manifold case the result is a order 2 tensor.

Quite frankly, it's up to you to tell us what you mean. Either you mean $a^\mu \nabla_\mu V^\nu$ or you mean $\nabla_\mu V^\nu$. But my statement still stands: going from flat space, Cartesian coordinates is not going to make the former become the latter, is not going to make the analogue of a directional derivative (which still results in a vector field) become this rank-2 tensor.

 

[TrickyDicky]

I think you are confused. A vector is a special case of a tensor. A vector equation is "coordinate free" in exactly the same sense that a tensor equation is (since it is) a tensor equation.

An important distinction is between a vector and the n-tuple of components of the vector. The former is coordinate-independent, while the latter is coordinate-dependent.

You are of course right, I was thinking about the basis dependency of the components and I ended up writing a misleading statement about vector equations which are of course invariant under linear transformations.
The distinction I was trying to make between 1-tensors and higher order tensors was because I think that to deal with curvature(nonlinearity) like in GR and extend the invariance from linear to nonlinear transformations one needs the higher than one-index tensors that are explicitly multilinear mappings.

 

[TrickyDicky]

I managed to track back the initial disagreement:

Quote by TrickyDicky

you specifically linked to the wikipage where it says that in GR 4-acceleration is the covariant derivative of 4-velocity wrt proper time, that is a rank 2- tensor.

No, look more carefully. It is given as a rank-1 tensor, $A^\lambda$.

And then

Carroll calls it a covariant derivative also, so the term "absolute derivative" is by no means universal. Since the Wikipedia page expanded the formula there should have been no confusion about what was being discussed and what rank tensor it was.

Apparently everyone has agreed that the covariant derivative of the 4-velocity vector field must be a higher rank tensor field, in this case a (1,1) or rank-2 tensor (except in the commented special case of flat space when not using curved coordinates in which case is a purely vector derivative).
So I'm still puzzled as to why it was denied in the above quote, even if in the next quote it is apparently acknowledged that the formula used in the wikipedia page is equivalent to that of a covariant derivative according to Carroll.
I'm also still waiting for someone to explain to me why DrGreg's post claims that "in any coordinate system in GR or SR, inertial or non-inertial, ... the 4-acceleration A are defined by
$$\begin{align} A^\alpha &= \frac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma \end{align}$$
... A are 4-vectors, i.e. rank-1 tensors.

All of the above is true in both SR and GR, in inertial or non-inertial coordinates."

when it has been apparently agreed by many that in the flat case with curvilinear coordinates and in the curved manifold case the Chistoffel symbols can't be set to zero so the RHS of the formula is clearly a 2-rank tensor.
Thanks

 

[Muphrid]

Apparently everyone has agreed that the covariant derivative of the 4-velocity vector field must be a higher rank tensor field, in this case a (1,1) or rank-2 tensor (except in the commented special case of flat space when not using curved coordinates in which case is a purely vector derivative).

As I've repeatedly tried to point out to you, this is not true. Why do you keep saying this? You keep failing to appreciate the difference between a directional derivative (which preserves the rank of the tensor it acts on) and the "ordinary" covariant derivative (which increases the rank). This is true in both curved space and in flat space. The only difference is that in flat space, cartesian coordinates, the covariant derivative $\nabla$ is also equal to $\partial$, the vector partial derivative.

So I'm still puzzled as to why it was denied in the above quote, even if in the next quote it is apparently acknowledged that the formula used in the wikipedia page is equivalent to that of a covariant derivative according to Carroll.
I'm also still waiting for someone to explain to me why DrGreg's post claims that "in any coordinate system in GR or SR, inertial or non-inertial, ... the 4-acceleration A are defined by
$$\begin{align} A^\alpha &= \frac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma \end{align}$$
... A are 4-vectors, i.e. rank-1 tensors.

All of the above is true in both SR and GR, in inertial or non-inertial coordinates."

when it has been apparently agreed by many that in the flat case with curvilinear coordinates and in the curved manifold case the Chistoffel symbols can't be set to zero so the RHS of the formula is clearly a 2-rank tensor.
Thanks

Because what's written is $a^\alpha = u^\beta \nabla_\beta u^\alpha$, a directional covariant derivative, and the RHS is clearly rank-1. There's only one free index! Setting the Christoffel symbols to zero doesn't change the rank of the result. You seem very confused on this point.

The notation may be misleading when talking about the velocity and acceleration vectors of a trajectory, but if they happen to be fields, then this notation is perfectly valid.

 

[TrickyDicky]

As I've repeatedly tried to point out to you, this is not true. Why do you keep saying this? You keep failing to appreciate the difference between a directional derivative (which preserves the rank of the tensor it acts on) and the "ordinary" covariant derivative (which increases the rank). This is true in both curved space and in flat space. The only difference is that in flat space, cartesian coordinates, the covariant derivative $\nabla$ is also equal to $\partial$, the vector partial derivative.

This is quite amazing, you have drscribed what I wrote in slightly different words and yet you claim it is not true.

Because what's written is $a^\alpha = u^\beta \nabla_\beta u^\alpha$, a directional covariant derivative, and the RHS is clearly rank-1. There's only one free index! Setting the Christoffel symbols to zero doesn't change the rank of the result. You seem very confused on this point.

Your expression is different to the one DrGreg wrote.
Also, would you mind giving the more widely used name of what you are calling "directional covariant derivative"?

 

[TrickyDicky]

Can you apply the directional covariant derivative to the 4-velocity vector field in GR?

 

[Muphrid]

This is quite amazing, you have drscribed what I wrote in slightly different words and yet you claim it is not true.

Then let me highlight, in no uncertain terms, what I disagree with:

Apparently everyone has agreed that the covariant derivative of the 4-velocity vector field must be a higher rank tensor field, in this case a (1,1) or rank-2 tensor (except in the commented special case of flat space when not using curved coordinates in which case is a purely vector derivative).

I've tried to show over several posts that the distinction you think exists here does not exist. Even in the flat space, cartesian coordinates case, the result does not require a different number of indices to express. Perhaps that is not what you're saying here, but you're still implying that flat space, cartesian coordinates, is somehow different from the more general case when it isn't.

Your expression is different to the one DrGreg wrote.
Also, would you mind giving the more widely used name of what you are calling "directional covariant derivative"?

The use of the connection is in common, and that is at issue when you said,

$$A^\alpha = \frac{dU^\alpha}{d\tau} + \Gamma^{\alpha}_{\beta \gamma} U^\beta U^\gamma$$

when it has been apparently agreed by many that in the flat case with curvilinear coordinates and in the curved manifold case the Chistoffel symbols can't be set to zero so the RHS of the formula is clearly a 2-rank tensor.
Thanks

Under no circumstances is the right-hand side of this expression a rank-2 tensor. This is obvious for the $d/d\tau$ term, and two indices are contracted for the Christoffel symbol term.

I'm merely using "directional" to be descriptive. If there's a more conventional name for it, I'm all for using it. Nevertheless, you still don't seem to appreciate that $u^\alpha \nabla_\alpha$ is a scalar differential operator (it always preserves rank) while $\nabla_\alpha$ is not and that this behavior doesn't change based on flat vs. curved, cartesian vs. curvillinear coordinates.

 

[Dale]

Apparently everyone has agreed that the covariant derivative of the 4-velocity vector field must be a higher rank tensor field, in this case a (1,1) or rank-2 tensor

There are at least two related but different operations referred to by the term "covariant derivative". One is the one you mention here, and the other is the one Carroll describes in eq 3.30 and others call an absolute derivative. This other one is the one that applies to a covariant derivative taken over a specific path, such as when describing the value measured by an accelerometer.

So I'm still puzzled as to why it was denied in the above quote

It was denied because what you said was wrong. You said, "you specifically linked to the wikipage where it says that in GR 4-acceleration is the covariant derivative of 4-velocity wrt proper time, that is a rank 2- tensor". This is clearly wrong since the wikipage I linked to said that the 4-acceleration was a rank-1 tensor, and unambiguously gave the correct expression that it was referring to by the term "covariant derivative".

I'm also still waiting for someone to explain to me why DrGreg's post claims that "in any coordinate system in GR or SR, inertial or non-inertial, ... the 4-acceleration A are defined by
$$\begin{align} A^\alpha &= \frac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma \end{align}$$
... A are 4-vectors, i.e. rank-1 tensors.

All of the above is true in both SR and GR, in inertial or non-inertial coordinates."

when it has been apparently agreed by many that in the flat case with curvilinear coordinates and in the curved manifold case the Chistoffel symbols can't be set to zero so the RHS of the formula is clearly a 2-rank tensor.

Who agreed with that? The RHS is clearly never a rank-2 tensor. It does not matter if the coordinates are curvilinear or straight or if the manifold is flat or curved. The RHS is always a rank-1 tensor, aka a vector.

Perhaps you can explain why you think otherwise, noting the number and placement of indices and recalling the Einstein summation convention when covariant and contravariant indices have the same variable.

 

[TrickyDicky]

There are at least two related but different operations referred to by the term "covariant derivative". One is the one you mention here, and the other is the one Carroll describes in eq 3.30 and others call an absolute derivative. This other one is the one that applies to a covariant derivative taken over a specific path, such as when describing the value measured by an accelerometer.

It was denied because what you said was wrong. You said, "you specifically linked to the wikipage where it says that in GR 4-acceleration is the covariant derivative of 4-velocity wrt proper time, that is a rank 2- tensor". This is clearly wrong since the wikipage I linked to said that the 4-acceleration was a rank-1 tensor, and unambiguously gave the correct expression that it was referring to by the term "covariant derivative".

Who agreed with that? The RHS is clearly never a rank-2 tensor. It does not matter if the coordinates are curvilinear or straight or if the manifold is flat or curved. The RHS is always a rank-1 tensor, aka a vector.

Perhaps you can explain why you think otherwise, noting the number and placement of indices and recalling the Einstein summation convention when covariant and contravariant indices have the same variable.

I always thought "covariant derivative" was only related to the operation that produced one higher order tensors. That's why, and so I was supposing that like it is often the case the indices were wrong in the wikipedia page.
In any case what I am saying is that in the curved manifold GR case, the covariant derivative of a vector field is a 2-index tensor, so the expression in the 4-acceleration page in which the indices cancel cannot be a covariant derivative.

 

[Muphrid]

I always thought "covariant derivative" was only related to the operation that produced one higher order tensors. That's why, and so I was supposing that like it is often the case the indices were wrong in the wikipedia page.
In any case what I am saying is that in the curved manifold GR case, the covariant derivative of a vector field is a 2-index tensor, so the expression in the 4-acceleration page in which the indices cancel cannot be a covariant derivative.

My interpretation is that the covariant derivative, as a differential operator, will increase the rank of whatever it acts on regardless of whether the space itself is curved or flat, in cartesian coordinates or other coordinates. Do you understand this is one aspect of what I'm saying? Do you agree that this is the case?

 

[TrickyDicky]

Under no circumstances is the right-hand side of this expression a rank-2 tensor. This is obvious for the $d/d\tau$ term, and two indices are contracted for the Christoffel symbol term.

I'm merely using "directional" to be descriptive. If there's a more conventional name for it, I'm all for using it. Nevertheless, you still don't seem to appreciate that $u^\alpha \nabla_\alpha$ is a scalar differential operator (it always preserves rank) while $\nabla_\alpha$ is not and that this behavior doesn't change based on flat vs. curved, cartesian vs. curvillinear coordinates.

I see that, perhaps I should have said "the RHS should be" rather than is, to avoid problems, but what I don't know is why you keep mixing what you call scalar diff. op. which is a directional derivative, with its generalization for spaces that require a connection, the covariant derivative.

 

[TrickyDicky]

My interpretation is that the covariant derivative, as a differential operator, will increase the rank of whatever it acts on regardless of whether the space itself is curved or flat, in cartesian coordinates or other coordinates. Do you understand this is one aspect of what I'm saying? Do you agree that this is the case?

No, because at least in differential geometry the covariant derivative only justification is to operate when vector spaces in different points of a manifold require a connection. You don't need a connection in flat space, where the tangent vector space coincides with the manifold.

 

[Matterwave]

The terminology used varies between sources and conventions. Sometimes, Covariant Derivative can also mean directional covariant derivative...sometimes one uses the term "total covariant derivative" (usually denoted with a big D) to mean the same thing.

 

[Muphrid]

No, because at least in differential geometry the covariant derivative only justification is to operate when vector spaces in different points of a manifold require a connection. You don't need a connection in flat space, where the tangent vector space coincides with the manifold.

The covariant derivative exists even in spaces without a connection. It just coincides with $\partial$, the vector derivative.

I see that, perhaps I should have said "the RHS should be" rather than is, to avoid problems, but what I don't know is why you keep mixing what you call scalar diff. op. which is a directional derivative, with its generalization for spaces that require a connection, the covariant derivative.

See above. The directional derivative of a space without a connection does not generalize to the covariant derivative in spaces with a connection.

Space without a connection has the following two differential operators:
1a) Vector (increases rank by 1): $\partial_i X^j$
1b) Scalar (does not change rank): $Y^i \partial_i X^j$, for some vector $Y^i$.

Space with a connection has the following two differential operators:
2a) Vector (increases rank by 1): $\nabla_\alpha X^\beta$
2b) Scalar (does not change rank): $Y^\alpha \nabla_\alpha X^\beta$, for some vector $Y^\alpha$.

Operator (2a) is the generalization of operator (1a). Operator (2b) is the generalization of operator (1b). Do you agree or disagree?

Do you agree that the scalar operators could both equally be called "directional"?

Do you agree that (2a) is the covariant derivative, which is not the generalization of the directional derivative (1b)?

 

[TrickyDicky]

Then let me highlight, in no uncertain terms, what I disagree with:



I've tried to show over several posts that the distinction you think exists here does not exist. Even in the flat space, cartesian coordinates case, the result does not require a different number of indices to express. Perhaps that is not what you're saying here, but you're still implying that flat space, cartesian coordinates, is somehow different from the more general case when it isn't.

Well here I think you are referring to the Jacobian matrix of a vector valued function which has order two obviously, but I have not seen it called a covariant derivative.

 

[TrickyDicky]

The covariant derivative exists even in spaces without a connection. It just coincides with $\partial$, the vector derivative.



See above. The directional derivative of a space without a connection does not generalize to the covariant derivative in spaces with a connection.

Space without a connection has the following two differential operators:
1a) Vector (increases rank by 1): $\partial_i X^j$
1b) Scalar (does not change rank): $Y^i \partial_i X^j$, for some vector $Y^i$.

Space with a connection has the following two differential operators:
2a) Vector (increases rank by 1): $\nabla_\alpha X^\beta$
2b) Scalar (does not change rank): $Y^\alpha \nabla_\alpha X^\beta$, for some vector $Y^\alpha$.

Operator (2a) is the generalization of operator (1a). Operator (2b) is the generalization of operator (1b). Do you agree or disagree?

Do you agree that the scalar operators could both equally be called "directional"?

Do you agree that (2a) is the covariant derivative, which is not the generalization of the directional derivative (1b)?

No, I don't , I find your terminology very confusing, and too far from the standard definitions in differential geometry or even in the wikipedia. Take a look at the wikipedia page on covariant derivative, it explicitly says that the covariant derivative is a generalization of the directional derivative for spaces that require a connection.

In any case we should agree about the underlying concepts, if not about the notation, but I'm finding it frankly tough in this forum context.

 

[Dale]

You don't need a connection in flat space, where the tangent vector space coincides with the manifold.

It is true that you don't NEED a connection in flat space, but you can still have one, a trivial connection with all the Christoffel symbols equal to 0. So a covariant derivative is perfectly well defined in flat space just as in curved spaces. It does the same thing, wrt the rank of the tensor, in both curved and flat space.

 

[WannabeNewton]

Take a look at the wikipedia page on covariant derivative, it explicitly says that the covariant derivative is a generalization of the directional derivative for spaces that require a connection.

This is indeed, more or less, how you will see it defined in proper riemannian geometry texts.

 

[Muphrid]

No, I don't , I find your terminology very confusing, and too far from the standard definitions in differential geometry or even in the wikipedia. Take a look at the wikipedia page on covariant derivative, it explicitly says that the covariant derivative is a generalization of the directional derivative for spaces that require a connection.

In any case we should agree about the underlying concepts, if not about the notation, but I'm finding it frankly tough in this forum context.

Okay, I see wikipedia explicitly uses a dummy vector (they call it $u$, I've called it $Y$ a couple posts ago) and does indeed define the covariant derivative as the generalization of the directional derivative. And that's fine. It's by no means universal. I have Alcubierre's book on 3+1 relativity in front of me, and it defines $\nabla_\alpha$ as the covariant derivative, a definition which is decidedly not the generalization of the directional derivative. Honestly, it doesn't bother me to say they're both so inextricably related that people use the same term for one or the other almost at will.

Nevertheless, all this confusion about what "covariant derivative" really means could've been avoided. As has been pointed out, the generalization of the directional derivative of a vector field is still going to be a vector field (a rank-1 tensor) in curved spaces. Ultimately, is that not what started this disagreement--the mistaken belief that it would somehow be rank-2? Or did that all stem from the unclear definition of "covariant derivative" (which, I concede, it may be that physicists have broken with strict mathematical definitions and caused the term to be used more broadly than it was historically defined, in an abuse of the term)?