Inertial and non inertial frames

[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}<br /> <br /> A^\alpha &amp;= \frac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma<br /> \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}<br /> <br /> A^\alpha &amp;= \frac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma<br /> \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}<br /> <br /> A^\alpha &amp;= \frac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma<br /> \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)?

 

[TrickyDicky]

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)?

Well certainly the latter has been a major factor, but I think we have it moreless under control.
Let's give a final look to the point you mention first. You have just conceded that at least in the more mathematical sense, there is definition of covariant derivative as generalization of the directional derivative when a connection is needed which was the definition I was using, and this generalization always increases the order of a tensor by one. So why would be my belief mistaken?
When the outcome is the same order, it is not the generalization, it is a directional derivative proper, there is nothing covariant about it, no special strategy for parallel transport is needed to keep the vectors parallel, no connection is used, otherwise the outcome wouldn't be just another vector field.
Consider this paragraph from wikipedia:

" In the case of Euclidean space, one tends to define the derivative of a vector field in terms of the difference between two vectors at two nearby points. In such a system one translates one of the vectors to the origin of the other, keeping it parallel. With a Cartesian (fixed orthonormal) coordinate system we thus obtain the simplest example: covariant derivative which is obtained by taking the derivative of the components. In the general case, however, one must take into account the change of the coordinate system. For example, if the same covariant derivative is written in polar coordinates in a two dimensional Euclidean plane, then it contains extra terms that describe how the coordinate grid itself "rotates". In other cases the extra terms describe how the coordinate grid expands, contracts, twists, interweaves, etc."
It makes clear the Cartesian coord. flat case is a special case where the generalization is not needed and therefore

 

[TrickyDicky]

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.

What does it wrt the rank of the tensor according to you then?

 

[Muphrid]

Let's give a final look to the point you mention first. You have just conceded that at least in the more mathematical sense, there is definition of covariant derivative as generalization of the directional derivative when a connection is needed which was the definition I was using, and this generalization always increases the order of a tensor by one.

No, you're still confused. I don't blame you for this; even the wiki article is confused. It starts with the idea of covariant derivative as a generalization of directional derivatives, but everything in its examples section is more consistent with physicists' usage, which is not a generalization of directional derivatives.

At this point, I think it will be most productive if you point out the definition of the covariant derivative that you want to use, so that we may both talk about the same thing. I will then be happy to point out to how how that definition affects the rank of any rank-N tensor field it acts on.

 

[Dale]

What does it wrt the rank of the tensor according to you then?

The covariant derivative evaluated along a path does nothing to the rank. The other covariant derivative adds one index "downstairs" thereby increasing the rank by 1. They each do the same thing in curved manifolds as they do in flat ones.

 

[TrickyDicky]

The covariant derivative evaluated along a path does nothing to the rank. The other covariant derivative adds one index "downstairs" thereby increasing the rank by 1. They each do the same thing in curved manifolds as they do in flat ones.

Ok if "both covariant derivatives" do the same in flat and curved manifolds, why do the wikipedia four-acceleration entry only talks about covariant derivative for the GR case?

 

[Muphrid]

If you're referring to this line:

In general relativity the elements of the acceleration four-vector are related to the elements of the four-velocity through a covariant derivative with respect to proper time.

This should not be read as meaning that the covariant derivative is not necessary to compute four-acceleration in special relativity. It is, provided that coordinates other than Cartesian (i.e. with nonvanishing Christoffel symbols) are used.

Of course, the wiki article later clarifies this:

In special relativity the coordinates are those of a rectilinear inertial frame, so the Christoffel symbols term vanishes, but sometimes when authors uses curved coordinates in order to describe an accelerated frame, the frame of reference isn't inertial, they will still describe the physics as special relativistic because the metric is just a frame transformation of the Minkowski space metric. In that case this is the expression that must be used because the Christoffel symbols are no longer all zero.

Are these quotes the ones you're basing your interpretation on? Or something else?

 

[stevendaryl]

A^\alpha = \dfrac{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

It's hard to know how to discuss this. Why in the world do you think the right-hand side is a rank-2 tensor? The whole point of index gymnastics is that it makes it possible to figure out the rank of tensor by counting the number of free raised indices and free lowered indices. (Summed over indices, with one raised and one lowered, have no effect on the rank).

So A^\alpha is a component of a [1,0] tensor (or vector)
B_\beta is a component of a [0,1] tensor (or co-vector, or 1-form, or whatever)
g_{\alpha, \beta} is a component of a [0,2] tensor.
\eta^\alpha_\beta is a component of a [1,1] tensor, etc.

In the following equation:
A^\alpha = \dfrac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma

the right-hand side, like the left-hand side, has only one raised index, so it's a [1,0] tensor, or vector.

On the other hand, the expression

\nabla_\beta U^\alpha = \partial_\beta U^\alpha + \Gamma^\alpha_{\beta, \gamma} U^\gamma

clearly has one raised index and one lowered index, and so is a [1,1] tensor. If you contract a [1,1] tensor with a vector, you get a vector, so the following expression is a vector:

U^\beta \nabla_\beta U^\alpha = U^\beta \partial_\beta U^\alpha + \Gamma^\alpha_{\beta, \gamma} U^\beta U^\gamma

This is the same as the expression for A^\alpha if you make the identification:

\dfrac{d}{d \tau} U^\alpha = U^\beta \partial_\beta U^\alpha

 

[Dale]

Ok if "both covariant derivatives" do the same in flat and curved manifolds, why do the wikipedia four-acceleration entry only talks about covariant derivative for the GR case?

I don't know, I didn't write any part of the entry. Are you trying to suggest some specific conclusion from that? That seems weak.

 

[TrickyDicky]

It's hard to know how to discuss this. Why in the world do you think the right-hand side is a rank-2 tensor? The whole point of index gymnastics is that it makes it possible to figure out the rank of tensor by counting the number of free raised indices and free lowered indices. (Summed over indices, with one raised and one lowered, have no effect on the rank).

So A^\alpha is a component of a [1,0] tensor (or vector)
B_\beta is a component of a [0,1] tensor (or co-vector, or 1-form, or whatever)
g_{\alpha, \beta} is a component of a [0,2] tensor.
\eta^\alpha_\beta is a component of a [1,1] tensor, etc.

In the following equation:
A^\alpha = \dfrac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma

the right-hand side, like the left-hand side, has only one raised index, so it's a [1,0] tensor, or vector.

On the other hand, the expression

\nabla_\beta U^\alpha = \partial_\beta U^\alpha + \Gamma^\alpha_{\beta, \gamma} U^\gamma

clearly has one raised index and one lowered index, and so is a [1,1] tensor. If you contract a [1,1] tensor with a vector, you get a vector, so the following expression is a vector:

U^\beta \nabla_\beta U^\alpha = U^\beta \partial_\beta U^\alpha + \Gamma^\alpha_{\beta, \gamma} U^\beta U^\gamma

This is the same as the expression for A^\alpha if you make the identification:

\dfrac{d}{d \tau} U^\alpha = U^\beta \partial_\beta U^\alpha

We went over this already in the previous exchange with Muphrid, I'm aware of what you write above;yes I wrote "is" when I meant "should be", because to me the expression that has only one free index is not a covariant derivative. I can see now most physicists, and the wikipedia in certain articles, call the two expressions covariant derivatives which I find highly confusing.

 

[stevendaryl]

We went over this already in the previous exchange with Muphrid, I'm aware of what you write above;yes I wrote "is" when I meant "should be", because to me the expression that has only one free index is not a covariant derivative. I can see now most physicists, and the wikipedia in certain articles, call the two expressions covariant derivatives which I find highly confusing.

I don't know what "should" means here. It really is a vector, but it has a moral obligation to be a rank 2 tensor?

I understand there are different definitions for a "covariant derivative", but the issue is about a very specific expression:

\dfrac{d}{d \tau} U^\alpha + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma

Whether you want to call that a "covariant derivative" or not, there is no question that it is a vector, and not a rank 2 tensor.

 

[TrickyDicky]

If you're referring to this line:



This should not be read as meaning that the covariant derivative is not necessary to compute four-acceleration in special relativity. It is, provided that coordinates other than Cartesian (i.e. with nonvanishing Christoffel symbols) are used.

It is not necessary because in SR with Cartesian coordinates there's no 4-acceleration to begin with (constant 4-velocity case).

Of course, the wiki article later clarifies this:



Are these quotes the ones you're basing your interpretation on? Or something else?

Those contributed, but the main problem is still that covariant derivatives, at least in differential geometry texts are applied to fields, rather than to tangent vectors at a point(wich is simply a vector derivative along a curve, a.k.a. absolute derivative), that seems to be only valid in SR for the case when curved coordinates are used, but even if according to people in this forum that seems standard terminology among physicists it is hard to see why would anyone call the latter a covariant derivative other than to confuse.

 

[stevendaryl]

It is not necessary because in SR with Cartesian coordinates there's no 4-acceleration to begin with (constant 4-velocity case).

It seems that with every post you make, there is another misconception to address. SR does not imply that 4-acceleration is zero! The theory of classical electromagnetism implies that a charged particle of mass m and charge q has a 4-acceleration A^\alpha given by:

m A^\alpha = q F^{\alpha \beta} U_\beta

where F^{\alpha \beta} is the electromagnetic field strength tensor.

SR perfectly well describes accelerating particles, it just doesn't describe acceleration due to gravity.

 

[TrickyDicky]

I don't know what "should" means here. It really is a vector, but it has a moral obligation to be a rank 2 tensor?

No, it should be a tensor in case it really was a covariant derivative of a vector field, but it isn't so that's it.

I understand there are different definitions for a "covariant derivative", but the issue is about a very specific expression:

\dfrac{d}{d \tau} U^\alpha + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma

Whether you want to call that a "covariant derivative" or not, there is no question that it is a vector, and not a rank 2 tensor.

It is simply the absolute derivative at a point yes.

 

[TrickyDicky]

It seems that with every post you make, there is another misconception to address. SR does not imply that 4-acceleration is zero! The theory of classical electromagnetism implies that a charged particle of mass m and charge q has a 4-acceleration A^\alpha given by:

m A^\alpha = q F^{\alpha \beta} U_\beta

where F^{\alpha \beta} is the electromagnetic field strength tensor.

SR perfectly well describes accelerating particles, it just doesn't describe acceleration due to gravity.

You missed the bit about cartesian coordinates.

 

[Muphrid]

We went over this already in the previous exchange with Muphrid, I'm aware of what you write above;yes I wrote "is" when I meant "should be", because to me the expression that has only one free index is not a covariant derivative. I can see now most physicists, and the wikipedia in certain articles, call the two expressions covariant derivatives which I find highly confusing.

Well, be that as it may, the "expression with only one free index" is the generalization of the directional derivative that you've been talking about.

 

[stevendaryl]

You missed the bit about cartesian coordinates.

No, I didn't miss that part. An electron will accelerate due to an electric field. It doesn't matter whether you are describing its motion using Cartesian coordinates, or not.

 

[TrickyDicky]

An electron will accelerate alright whatever we say about it, but if you want to describe that acceleration in a flat Minkowski space-time chart you'll have to use curvilinear coordinates at some point, that or directly use the velocity space which is itself curved hyperbolic space.

 

[harrylin]

An electron will accelerate alright whatever we say about it, but if you want to describe that acceleration in a flat Minkowski space-time chart you'll have to use curvilinear coordinates at some point, that or directly use the velocity space which is itself curved hyperbolic space.

I'm sorry if this comment is off-topic in your discussion which I did not follow (my excuses again), but the accelerating electron has been part of SR from the very start and even before, without any need for use of curvilinear coordinates or hyperbolic space. Just as no curved spaces are needed for describing acceleration in classical mechanics.

 

[TrickyDicky]

I'm sorry if this comment is off-topic in your discussion which I did not follow (my excuses again), but the accelerating electron has been part of SR from the very start and even before, without any need for use of curvilinear coordinates or hyperbolic space. Just as no curved spaces are needed for describing acceleration in classical mechanics.

Note that I was referring to Minkoski spacetime, in which one of the coordinate dimensions is time. That is not so neither in clssical mechanics nor in the 1905-1907 Einstein SR pre-Minkowski, they both used time as parameter in Euclidean space, absolute in one case and proper time in the other.

 

[Dale]

An electron will accelerate alright whatever we say about it, but if you want to describe that acceleration in a flat Minkowski space-time chart you'll have to use curvilinear coordinates at some point, that or directly use the velocity space which is itself curved hyperbolic space.

Huh? What makes you think that? You can use curvilinear coordinates to describe an accelerating electron in flat Minkowski spacetime if you like, but you certainly don't have to.

 

[stevendaryl]

An electron will accelerate alright whatever we say about it, but if you want to describe that acceleration in a flat Minkowski space-time chart you'll have to use curvilinear coordinates at some point, that or directly use the velocity space which is itself curved hyperbolic space.

I don't understand what you mean by that. The equation

m \dfrac{d}{d \tau} U^\alpha = q F^{\alpha \beta} U_\beta

describes the acceleration of a spinless particle of mass m and charge q in an electromagnetic field using Cartesian coordinates.

 

[TrickyDicky]

I don't understand what you mean by that. The equation

m \dfrac{d}{d \tau} U^\alpha = q F^{\alpha \beta} U_\beta

describes the acceleration of a spinless particle of mass m and charge q in an electromagnetic field using Cartesian coordinates.

That equation describes force rather than acceleration and doesn't contradict what I'm saying, when non-inertial frames are introduced the force in the f=ma eq. includes inertial forces.
When I say curved coordinates in the context of Minkowski spacetime I'm simply alluding to the fact that if one wants to describe a non-inertial object like an accelerating particle, one is introducing a non-inertial frame of reference and the coordinate system attached to it.
I specified I was speaking about Minkowskian 4-spacetime reference frames, even if using the term curvilinear coordinates systems that are classically for Euclidean spaces and might have been a bad choice of term to refer to an obseevational non-inertial reference frame but there's certain overlap between coordinate systems and reference frames in physics.
So I should have said inertial frames instead of Cartesian coordinates.

 

[Dale]

When I say curved coordinates in the context of Minkowski spacetime I'm simply alluding to the fact that if one wants to describe a non-inertial object like an accelerating particle, one is introducing a non-inertial frame of reference and the coordinate system attached to it.

You certainly can use curvilinear coordinates if you wish, but there is no reason that you must. In flat Minkowski spacetime you can analyze a non inertial object entirely using standard straight orthonormal coordinates. Your statement that "you'll have to use curvilinear coordinates" is incorrect.

 

[stevendaryl]

That equation describes force rather than acceleration and doesn't contradict what I'm saying, when non-inertial frames are introduced the force in the f=ma eq. includes inertial forces.

The term \dfrac{d}{d \tau} U^\alpha is acceleration (proper acceleration).

What you said (I thought) was that acceleration can't be described using Cartesian coordinates in SR. That's not true.

When I say curved coordinates in the context of Minkowski spacetime I'm simply alluding to the fact that if one wants to describe a non-inertial object like an accelerating particle, one is introducing a non-inertial frame of reference and the coordinate system attached to it.

No, you don't have to use the frame of the particle in order to describe the particle's motion. Just use any inertial frame whatsoever.

 

[TrickyDicky]

The term \dfrac{d}{d \tau} U^\alpha is acceleration (proper acceleration).

What you said (I thought) was that acceleration can't be described using Cartesian coordinates in SR. That's not true.



No, you don't have to use the frame of the particle in order to describe the particle's motion. Just use any inertial frame whatsoever.

Man, you seem not to have a clue about this, if there are only inertial frames there is no acceleration.
If you don't understand the rather basic concepts I'm talking about try this excerpt from Wikipedia.
"Curvilinear coordinates and non-inertial frames

Equivalent to the original ? Curvilinear is a generalization, but the original SR can be applied locally. There can be misunderstandings over the sense in which SR can be applied to accelerating frames. The confusion here results from trying to describe three different things with just two labels. The three things are: A description of physics without gravity using just "inertial frames", i.e. non-accelerating Cartesian coordinate systems. These coordinate systems are all related to each other by the linear Lorentz transformations. The physical laws may be described more simply in these frames than in the others. This is "special relativity" as usually understood. A description of physics without gravity using arbitrary curvilinear coordinates. This is non-gravitational physics plus general covariance. Here one sets the Riemann-Christoffel tensor to zero instead of using the Einstein field equations. This is the sense in which "special relativity" can handle accelerated frames. A description of physics including gravity governed by the Einstein field equations, i.e. full general relativity. Special relativity cannot be used to describe a global frame for non-inertial i.e. accelerating frames. However general relativity implies that special relativity can be applied locally where the observer is confined to making local measurements. For example an analysis of Bremsstrahlung does not require general relativity, SR is sufficient.
The key point is that you can use special relativity to describe all kinds of accelerated phenomena, and also to predict the measurements made by an accelerated observer who's confined to making measurements at one specific location only. If you try to build a complete frame for such an observer, one that is meant to cover all of spacetime, you'll run into difficulties (there'll be a horizon, for one). The problem is that you cannot derive from the postulates of special relativity that an acceleration will not have a non-trivial effect."