Inertial and non inertial frames

[Nugatory]

When the rank is not specified it is universally understood that the word tensor refers to rank >1 tensors.

I'm sorry, but I that answer still isn't helping me make any sense of your post #85. In what way is the GR definition of proper acceleration different than the SR definition, so that "mixing the 4-acceleration in SR with its definition in GR" will lead to problems/errors?

 

[Jeronimus]

Accelerometers do not measure acceleration. They only measure different parts of the accelerometer being accelerated differently. Or otherwise said, different parts of the accelerometer being in different inertial frames of references. They measure the resulting space shift of parts of the accelerometer.

But what about a point on a disc rotating? The point(object) on the disc keeps switching the inertial reference frame it is at rest in constantly.
edit: An infinitely small accelerometer would measure nothing even on a switch of the inertial frame of reference, but such accelerometers unfortunately do not exist. The accelerometer, even the smallest, could never have all it's parts within the same inertial reference frame at all times when on a rotating disc.

An accelerometer in free fall in the right force field measures nothing, even though it is accelerating. This is because all of it's parts are in the same inertial frame at all times(ideally)

Acceleration is relative.

 

[Mentz114]

Accelerometers do not measure acceleration. They only measure different parts of the accelerometer being accelerated differently. Or otherwise said, different parts of the accelerometer being in different inertial frames of references. They measure the resulting space shift of parts of the accelerometer.

But what about a point on a disc rotating? The point(object) on the disc keeps switching the inertial reference frame it is at rest in constantly.

An accelerometer in free fall in the right force field measures nothing, even though it is accelerating. This is because all of it's parts are in the same inertial frame at all times(ideally)

Acceleration is relative.

Relative to what ? Have you read the other posts in this thread ?

 

[Dale]

Accelerometers do not measure acceleration.

The unqualified word "acceleration" is ambiguous in this context. Accelerometers do measure proper acceleration, they do not measure coordinate acceleration.

They only measure different parts of the accelerometer being accelerated differently.

This is simply false.

But what about a point on a disc rotating? The point(object) on the disc keeps switching the inertial reference frame it is at rest in constantly.
edit: An infinitely small accelerometer would measure nothing even on a switch of the inertial frame of reference, but such accelerometers unfortunately do not exist. The accelerometer, even the smallest, could never have all it's parts within the same inertial reference frame at all times when on a rotating disc.

Usually when we refer to accelerometers in theoretical discussions or thought experiments we are referring to the 6 degree of freedom kind that measures linear accelerations on three axes and rotations on three axes. Such an accelerometer would measure both the rotation and the centripetal acceleration on a rotating disk.

 

[PAllen]

Accelerometers do not measure acceleration. They only measure different parts of the accelerometer being accelerated differently. Or otherwise said, different parts of the accelerometer being in different inertial frames of references. They measure the resulting space shift of parts of the accelerometer.

It is not helpful to define your own physics. We define what an accelerometer measures as acceleration of the device as whole, given that the acceleration is sufficiently uniform (over space) and slowly changing over time.

But what about a point on a disc rotating? The point(object) on the disc keeps switching the inertial reference frame it is at rest in constantly.
edit: An infinitely small accelerometer would measure nothing even on a switch of the inertial frame of reference, but such accelerometers unfortunately do not exist. The accelerometer, even the smallest, could never have all it's parts within the same inertial reference frame at all times when on a rotating disc.

An accelerometer on the edge of rotating disk would measure that the acceleration is radial and constant.

An accelerometer in free fall in the right force field measures nothing, even though it is accelerating. This is because all of it's parts are in the same inertial frame at all times(ideally)

Free fall is defined as inertial motion in (general) relativity. Doing anything with gravity in SR is an abuse of SR. If you mean some other force, an accelerometer must be built out of matter that is non-responsive all known forces (except gravity, if you consider that a force - because nothing be neutral to gravity); else you can't call it an accelerometer.

Acceleration is relative.

Acceleration as measured by an accelerometer is absolute. All inertial frames will compute the response of a given accelerometer (in some state of motion) to be the same - even though the coordinate description of the motion will be different for each frame. If you are willing to use laws expressed in terms of tensors, with connection coefficients, then an accelerating observer's coordinate calculations will also yield the same result.

I must warn that this is not the forum to propose your own version of physics. You are welcome to pursue such an endeavor - but not on physicsforums.

 

[Jeronimus]

Maybe someone could define acceleration. To me, acceleration is the context of SR, is that which is responsible for changing which inertial frame of reference an object becomes at rest in.

I could not possibly define it to be the change in velocity over time (seen from the observer's point of view who is supposed to be accelerating), since in SR an observer considers himself at rest.

So here i am, waiting for how you define acceleration in the context of SR seen from an accelerating observer's point of view.

 

[Mentz114]

Maybe someone could define acceleration. To me, acceleration is the context of SR, is that which is responsible for changing which inertial frame of reference an object becomes at rest in.

I could not possibly define it to be the change in velocity over time (seen from the observer's point of view who is supposed to be accelerating), since in SR an observer considers himself at rest. So here i am, waiting for how you define acceleration in the context of SR seen from an accelerating observer's point of view.

Try the Wiki article. There's a definition there.

http://en.wikipedia.org/wiki/Four-vector

Imagine you're sitting in a small boat on a calm lake so the boat is at rest wrt the water. If you then throw a brick towards aft so it leaves the boat, what happens ? The brick was accelerated by your arm and the reaction ( Newton's third) pushes the boat and you. You are still at rest in the boat even if it accelerates, as long as you go with it.

 

[Jeronimus]

Imagine you're sitting in a small boat on a calm lake so the boat is at rest wrt the water. If you then throw a brick towards aft so it leaves the boat, what happens ? The brick was accelerated by your arm and the reaction ( Newton's third) pushes the boat and you. You are still at rest in the boat even if it accelerates, as long as you go with it.

Nice example, except that you do not feel acceleration when you push the brick away of you. What you feel/sense is different parts of your body being accelerated differently, and the resulting change of your body structure. This is what you sense, and not acceleration itself.
At the point your fingers touch the brick, your fingers get compressed because the brick resists being accelerated (why it resists and why inertial mass is connected to gravitational mass i do not know yet exactly). This compression moves through your body, and changes the structure which then can be sensed by you. But that is not acceleration you sense.

 

[Mentz114]

Nice example, except that you do not feel acceleration when you push the brick away of you. What you feel/sense is different parts of your body being accelerated differently, and the resulting change of your body structure. This is what you sense, and not acceleration itself.
At the point your fingers touch the brick, your fingers get compressed because the brick resists being accelerated (why it resists and why inertial mass is connected to gravitational mass i do not know yet exactly). This compression moves through your body, and changes the structure which then can be sensed by you. But that is not acceleration you sense.

Suppose you had a friend on the boat and she threw the brick ? Would you feel anything ?
Whether you did or not, an accelerometer would register the push on the boat.

You can make an accelerometer by standing a brick an its small face on a skateboard. Give the skateboard a good push and the brick falls over !

 

[Jeronimus]

Suppose you had a friend on the boat and she threw the brick ? Would you feel anything ?
Whether you did or not, an accelerometer would register the push on the boat.

You can make an accelerometer by standing a brick an its small face on a skateboard. Give the skateboard a good push and the brick falls over !

Of course it falls over, since you accelerated the bottom part of the brick touching the skateboard first. You accelerated different parts of the brick differently with the resulting change in position you measured. But you did not measure acceleration itself.

 

[PAllen]

Maybe someone could define acceleration. To me, acceleration is the context of SR, is that which is responsible for changing which inertial frame of reference an object becomes at rest in.

Where did you get this idea from? Acceleration of a body in an inertial frame in SR is taken as derivative of 4-velocity by proper time.

I could not possibly define it to be the change in velocity over time (seen from the observer's point of view who is supposed to be accelerating), since in SR an observer considers himself at rest.

Again, where do you get this from. SR says only inertial observers consider themselves at rest. The standard SR equations only apply to inertial frames; observers at rest in these frames detect no acceleration (e.g. using an accelerometer).

So here i am, waiting for how you define acceleration in the context of SR seen from an accelerating observer's point of view.

SR typically does not deal with accelerated observers; certainly Einstein never defined accelerated observers for SR. It readily deals with accelerated motion in inertial frames.

It is possible to define coordinates representing an accelerated observer, in SR. If you do, acceleration is now absolute derivative of 4-velocity by proper time. This type of derivative involves what are called connection coefficients (which are zero in an inertial frame in SR). In these coordinates, the coordinate 3-velocity of the observer will be zero (its 4-velocity will not be zero - it will have a time component). However, because of the nature of the absolute derivative, the 4-acceleration will be nonzero, and the proper acceleration (given by the norm of the 4-acceleration) will be what is measured by an accelerometer.

 

[Jeronimus]

Where did you get this idea from?

It's a pretty simple idea. You accelerate(instantaneous to keep it simple) and as a result, someone who observes you accelerating will conclude that you are no longer at rest in your former inertial frame of reference, but you are at rest in another frame.

Switching the frame an object is at rest in, is a direct result of acceleration. (seen by an observer in another inertial frame of reference)

I could continue with what acceleration is a result of, but i'd rather not.

Again, where do you get this from. SR says only inertial observers consider themselves at rest. The standard SR equations only apply to inertial frames; observers at rest in these frames detect no acceleration (e.g. using an accelerometer).

But isn't accelerating an object, seen from an observer in another IFR at rest, a constant switch of IFRs the object is at rest in? I think it is.

 

[TrickyDicky]

For the record, if the worldline of a particle is parametrised by proper time as x^\alpha(\tau), in any coordinate system in GR or SR, inertial or non-inertial, then the 4-velocity U and the 4-acceleration A are defined by
\begin{align}<br /> U^\alpha &amp;= \frac{dx^\alpha}{d\tau} \\<br /> A^\alpha &amp;= \frac{dU^\alpha}{d\tau} + \Gamma^\alpha_{\beta \gamma} U^\beta U^\gamma<br /> \end{align}
Both U and 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. Of course in Minkowski coordinates in SR, all the \Gamma^\alpha_{\beta \gamma}s are zero so the expression for 4-acceleration can be simplified.

I thought this was well-known stuff covered in every GR textbook but apparently not from the evidence of this thread.

Nope, you are wrong, I hoped someone would come up to correct it but apparently, and incredibly, no one did.
Have you tried reading Carroll notes for instance? Then you should realize that in GR the expression with christoffel corrections transforms as a tensor, not as a contravariant vector.

 

[Dale]

Maybe someone could define acceleration.

Again, your question is ambiguous in this context. You need to specify if you want a definition for coordinate acceleration or proper acceleration. Both are easy to define. Coordinate acceleration is the second time derivative of an object's position. Proper acceleration is the first covariant derivative of the object's tangent vector.

So here i am, waiting for how you define acceleration in the context of SR seen from an accelerating observer's point of view.

The above definitions hold for any point of view, inertial or not.

 

[TrickyDicky]

I'm sorry, but I that answer still isn't helping me make any sense of your post #85. In what way is the GR definition of proper acceleration different than the SR definition, so that "mixing the 4-acceleration in SR with its definition in GR" will lead to problems/errors?

See my answer to DrGreg.

 

[TrickyDicky]

When the rank isn't specified then it should be understood that the word "tensor" refers to a tensor of an arbitrary unspecified rank, including rank 0 and rank 1.

Then I guess for you a book entitled "Tensor calculus" should deal with real functions differential calculus and with vector calculus, right?

 

[stevendaryl]

Nope, you are wrong, I hoped someone would come up to correct it but apparently, and incredibly, no one did.
Have you tried reading Carroll notes for instance? Then you should realize that in GR the expression with christoffel corrections transforms as a tensor, not as a contravariant vector.

What are you talking about? Are you saying that Dr. Greg wrote down the wrong expressions for 4-velocity and 4-acceleration? Are you saying that he's wrong to call them 4-vectors? Are you saying that he's wrong to say that a 4-vector is a rank-1 tensor? Are you saying that he's wrong that this is "well-known stuff"? What do you mean wrong?

 

[PAllen]

It's a pretty simple idea. You accelerate(instantaneous to keep it simple) and as a result, someone who observes you accelerating will conclude that you are no longer at rest in your former inertial frame of reference, but you are at rest in another frame.

Switching the frame an object is at rest in, is a direct result of acceleration. (seen by an observer in another inertial frame of reference)

I could continue with what acceleration is a result of, but i'd rather not.

You made a statement about SR. As has been clarified multiple times, SR distinguishes inertial and non-inertial frames, and, as formulated by Einstein, dealt only with inertial frames. Also, FYI, pre-relativistic physics made the same distinction. Newton in Principia had extensive discussions of the absolute character of rotation and acceleration. SR did not change this (and GR, not really either).

But isn't accelerating an object, seen from an observer in another IFR at rest, a constant switch of IFRs the object is at rest in? I think it is.

There is a limited sense in which you can treat it this way. However, a derivative is has a value at a single point. If the comoving inertial frame changes, there is accelerated motion in each and every co-moving inertial frame involved.

 

[TrickyDicky]

What are you talking about? Are you saying that Dr. Greg wrote down the wrong expressions for 4-velocity and 4-acceleration? Are you saying that he's wrong to call them 4-vectors? Are you saying that he's wrong to say that a 4-vector is a rank-1 tensor? Are you saying that he's wrong that this is "well-known stuff"? What do you mean wrong?

Try reading what you quote.

 

[stevendaryl]

Try reading what you quote.

What you said made no sense, which is why I asked for clarification. Everything that Dr. Greg wrote was correct, so I have no idea what you are talking about when you call it wrong.

 

[Nugatory]

Nope, you are wrong, I hoped someone would come up to correct it but apparently, and incredibly, no one did.
Have you tried reading Carroll notes for instance? Then you should realize that in GR the expression with christoffel corrections transforms as a tensor, not as a contravariant vector.

I'm still having trouble understanding what you mean when you say this... How is "transforming as a tensor" different than "transforming as a contravariant vector"?

 

[stevendaryl]

Try reading what you quote.

Specifically, you wrote:

...in GR the expression with christoffel corrections transforms as a tensor, not as a contravariant vector.

That makes no sense, for several reasons. First, there is no such thing as the "GR expression". The expression is the same whether you are in GR or SR. The difference between the two is that the metric tensor is fixed in SR, but not in GR. But in both cases, you can define the connection coefficients and define 4-velocity and 4-acceleration, and the expressions are the same. As Dr. Greg points out, in SR it's possible to choose coordinates so that the connection coefficients are everywhere zero, but you don't have to use those coordinates, they're just simpler to work with.

The second reason what you wrote made no sense is that you said "the expression transforms as a tensor, not as a contravariant vector". That makes no sense because a vector is a special case of a tensor. If you want to say that a vector can't be a tensor, then 4-velocity is not a tensor, by that definition.

Don't tell me to reread your original post. If what you had written made sense to me, then I wouldn't have asked for clarification.

 

[stevendaryl]

I'm still having trouble understanding what you mean when you say this... How is "transforming as a tensor" different than "transforming as a contravariant vector"?

The answer is: there is no difference, TrickyDicky is confused on this point.

 

[stevendaryl]

The answer is: there is no difference, TrickyDicky is confused on this point.

There is a difference of opinion about what a vector or tensor, etc. is. Some people define these things in terms of matrices that transform in such and such a way under coordinate transformations. I don't like that definition. I prefer to know what a vector "is", rather than how it transforms. You can derive how it's components transform under coordinate changes if you know what a vector is.

The geometric way of defining things is this:

1. A parametrized path is a (continuous, differentiable, blah, blah) function P(s) mapping real numbers to points in space (or spacetime).

2. A real scalar field is the reverse: a function \Phi(p) mapping points in space (or spacetime) to real numbers.

3. A tangent vector is a linear approximation to a parametrized path. It characterizes the path locally. Technically, if P(s) is a real scalar field, then we can identify the tangent vector d/ds P(s) with the operator v that acts on scalar fields as follows: v(\Phi) = d/ds (\Phi(P(s))

4. A cotangent vector is a linear approximation to a real scalar field. It characterizes the field locally. Technically, if \Phi(X) is a real scalar field, then we can define a cotangent vector \nabla \Phi to be that operator w that acts on tangent vectors as follows: w(v) = v(\Phi)

5. Higher-level tensors are multilinear functions of tangent and cotangent vectors.

This way of defining things doesn't even mention coordinates or coordinate transformations, but is sufficient to deduce how vectors and tensors transform.

 

[harrylin]

[..] So here i am, waiting for how you define acceleration in the context of SR seen from an accelerating observer's point of view.

Hi Jeronimus,
In the context of SR, acceleration is defined relative to a to-be-specified inertial reference system (or, with "proper" acceleration: relative to a class of instantaneously co-moving inertial reference systems).
In equation: a=dv/dt. If an object accelerates relative to one inertial system, then it accelerates relative to all inertial systems.

But isn't accelerating an object, seen from an observer in another IFR at rest, a constant switch of IFRs the object is at rest in? I think it is.

Yes - however that is a complicated way of looking at it.

 

[stevendaryl]

Nope, you are wrong, I hoped someone would come up to correct it but apparently, and incredibly, no one did.
Have you tried reading Carroll notes for instance? Then you should realize that in GR the expression with christoffel corrections transforms as a tensor, not as a contravariant vector.

Here's what Sean Carroll says:

A straightforward generalization of vectors and dual vectors is the notion of a tensor...
From this point of view, a scalar is a type (0, 0) tensor, a vector is a type (1, 0) tensor, and
a dual vector is a type (0, 1) tensor.

 

[Dale]

Then I guess for you a book entitled "Tensor calculus" should deal with real functions differential calculus and with vector calculus, right?

Do you have an example of a book entitled "tensor calculus" that does not deal with rank-0 tensors (scalars) and rank-1 tensors (vectors)?

 

[TrickyDicky]

Do you have an example of a book entitled "tensor calculus" that does not deal with rank-0 tensors (scalars) and rank-1 tensors (vectors)?

Don't change what I said, I talked about calculus. They always use vectors(that always depend on a basis and therefore vector equations are depedent on the coordinate system used) in the first pages to build the concept of tensors (with the property that they are not basis-dependent and therefore tensor equations are valid on any coordinates, do you see now the difference?)
The first one I found in google books, by Synge and Child is a good example.

 

[TrickyDicky]

Here's what Sean Carroll says:

This is not about terminology, I have no problem with vectors being 1-tensors, this is a conceptual issue, instead of getting mad you could step back and try understanding what I'm saying.
Since you have Carroll notes, go to the part 3 about curvature, and see how he gets a (1,1) tensor from the covariant derivative of a contravariant vector: eq. 3.1 and following.

 

[PAllen]

Don't change what I said, I talked about calculus. They always use vectors(that always depend on a basis and therefore vector equations are depedent on the coordinate system used) in the first pages to build the concept of tensors (with the property that they are not basis-dependent and therefore tensor equations are valid on any coordinates, do you see now the difference?)
The first one I found in google books, by Synge and Child is a good example.

I have this book (the first one I used, long ago). They certainly say vectors and scalars are rank 1 and zero tensors. They use the same letter T for all cases:

T no super scripts - scalar; rank 0 tensor
T one superscript - vector (contravariant); rank 1 tensor
T more sub/superscripts - higher rank tensor.

It is instructive that they don't use, say V for vector and T for tensor.

But this is all silly. Everyone knows tensor is the general term, vector the special case, and there should be no argument unless people go looking for one.

 

[stevendaryl]

Don't change what I said, I talked about calculus. They always use vectors(that always depend on a basis and therefore vector equations are depedent on the coordinate system used) in the first pages to build the concept of tensors (with the property that they are not basis-dependent and therefore tensor equations are valid on any coordinates, do you see now the difference?)
The first one I found in google books, by Synge and Child is a good example.

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.

A 4-vector V can be written as a linear combination of basis vectors:

V = \sum V^\alpha e_\alpha

where e_\alpha is the set of basis vectors.

The components V^\alpha change when you change coordinates, and the basis vectors e_\alpha change when you change coordinates, but the combination V = \sum V^\alpha e_\alpha is the same in any coordinate system.

 

[stevendaryl]

This is not about terminology, I have no problem with vectors being 1-tensors, this is a conceptual issue, instead of getting mad you could step back and try understanding what I'm saying.

I need help from you doing that, because what you wrote makes no sense to me.

Since you have Carroll notes, go to the part 3 about curvature, and see how he gets a (1,1) tensor from the covariant derivative of a contravariant vector: eq. 3.1 and following.

I have equation 3.1 in front of me, and it doesn't help at all to clarify what you're talking about. He writes:

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

What does that have to do with your statement

Then you should realize that in GR the expression with christoffel corrections transforms as a tensor, not as a contravariant vector.

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.

 

[Dale]

Don't change what I said

Take your own advice. Your tailspins like this are rather amusing to watch.

I will look at the book you mentioned, but I bet that it uses rank-1 and -0 tensors and probably even calls them vectors and scalars.

 

[stevendaryl]

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.

The relation is kind of subtle. If you have a 4-vector V and you extend it to a vector field, then the relationship between the two is

D_\tau V^\nu = (\nabla_\mu V^\nu) V^\mu

where D_\tau V^\nu = d/d\tau V^\nu + \Gamma^\nu_{\mu \lambda} V^\mu V^\lambda

I don't know what the official term is for the operator D_\tau.

 

[PAllen]

The relation is kind of subtle. If you have a 4-vector V and you extend it to a vector field, then the relationship between the two is

D_\tau V^\nu = (\nabla_\mu V^\nu) V^\mu

where D_\tau V^\nu = d/d\tau V^\nu + \Gamma^\nu_{\mu \lambda} V^\mu V^\lambda

I don't know what the official term is for the operator D_\tau.

Absolute derivative is what I've used.

 

[TrickyDicky]

But this is all silly. Everyone knows tensor is the general term, vector the special case, and there should be no argument unless people go looking for one.

Couldn't agree more, this is the silliest discussion, even more knowing I've never argued against this and has nothing to do with the real discussion involving DrGreg's post, stevendaryl and me.
But I have it well deserved for following certain person's favourite sport.

 

[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.

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.
As you see this has nothing to do with calling vectors 1-tensors as some have wanted to pollute the discussion with.
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.

 

[Muphrid]

A directional covariant derivative of a vector field is still a vector field in GR, just as the directional derivative of a vector field is a vector field in SR.

What is a directional derivative in SR? Pick a vector $a^\mu$. The directional derivative of the vector field $V^\nu$ is $a^\mu \partial_\mu V^\nu$. Change $\partial$ to $\nabla$, and you have the correct analogue in GR. Both are still vector fields.

If instead you don't wish to consider directional derivatives, you can just look at $\partial_\mu V^\nu$ (in SR) and $\nabla_\mu V^\nu$ (in GR). Both are 2-index objects that marry the divergence and curl.

 

[PAllen]

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.
As you see this has nothing to do with calling vectors 1-tensors as some have wanted to pollute the discussion with.
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.

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 there.

 

[TrickyDicky]

A directional covariant derivative of a vector field is still a vector field in GR, just as the directional derivative of a vector field is a vector field in SR.

What is a directional derivative in SR? Pick a vector $a^\mu$. The directional derivative of the vector field $V^\nu$ is $a^\mu \partial_\mu V^\nu$. Change $\partial$ to $\nabla$, and you have the correct analogue in GR. Both are still vector fields.

If instead you don't wish to consider directional derivatives, you can just look at $\partial_\mu V^\nu$ (in SR) and $\nabla_\mu V^\nu$ (in GR). Both are 2-index objects that marry the divergence and curl.

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.

 

[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).