Inertial and non inertial frames

[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."

 

[stevendaryl]

Man, you seem not to have a clue about this, if there are only inertial frames there is no acceleration.

Well, that's a tautology, because if a particle accelerates, then the frame of that particle is noninertial. But the question was: Can you describe acceleration using only inertial Cartesian coordinates? Of course, you can!

Look, you said something that is blatantly false. You said this:

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.

That's just not true. You do not "have to use curvilinear coordinates at some point" in order to describe acceleration in flat Minkowski space-time. In SR, you don't ever have to use curvilinear coordinates, in the same way that in Newtonian physics you don't ever have to use curvilinear coordinates. For both SR and Newtonian physics, there are situations where curvilinear or noninertial coordinates are more convenient than inertial Cartesian coordinates, but as long as space or spacetime is flat, you are never forced to use curvilinear or noninertial coordinates.

 

[TrickyDicky]

Well, that's a tautology, because if a particle accelerates, then the frame of that particle is noninertial. But the question was: Can you describe acceleration using only inertial Cartesian coordinates? Of course, you can!

I explained what I meant in a subsequent post, but you prefer to ignore it.
It all comes down to what one means by "describe acceleration", if one restricts the description to a point, you can of course describe it in SR regardless of the coordinates, but what I understand by "describing acceleration" is something more global that includes the accelerating (non-inertial) frame-more so if you look at the title of this thread-and for that you use noninertial coordinates, like Rindler, and you encounter horizons.

Look, you said something that is blatantly false. You said this:



That's just not true. You do not "have to use curvilinear coordinates at some point" in order to describe acceleration in flat Minkowski space-time. In SR, you don't ever have to use curvilinear coordinates, in the same way that in Newtonian physics you don't ever have to use curvilinear coordinates. For both SR and Newtonian physics, there are situations where curvilinear or noninertial coordinates are more convenient than inertial Cartesian coordinates, but as long as space or spacetime is flat, you are never forced to use curvilinear or noninertial coordinates.

See above. You keep ignoring the part of my posts where I qualified that statement. I guess for you this is not about learning but about telling people they must be wrong.

 

[stevendaryl]

I explained what I meant in a subsequent post, but you prefer to ignore it.

What you said was perfectly clear, but was wrong.

It all comes down to what one means by "describe acceleration", if one restricts the description to a point, you can of course describe it in SR regardless of the coordinates, but what I understand by "describing acceleration" is something more global that includes the accelerating (non-inertial) frame-more so if you look at the title of this thread-and for that you use noninertial coordinates, like Rindler, and you encounter horizons.

If you use inertial Cartesian coordinates, you can describe accelerated motion globally in SR, with no horizons.

See above. You keep ignoring the part of my posts where I qualified that statement.

With the qualification, it's still wrong.

 

[TrickyDicky]

Also if you read the quote from wikipedia you'll see that for doing the local computation of 4-acceleration in SR with Cartesian coordinates one is relying on implicit knowledge from GR.

 

[TrickyDicky]

I guess for you this is not about learning but about telling people they must be wrong.

Thanks for confirming this.

 

[TrickyDicky]

If you use inertial Cartesian coordinates, you can describe accelerated motion globally in SR, with no horizons.

Nope, but I won't bother explaining it to you anymore, bye.

 

[stevendaryl]

Thanks for confirming this.

Look, you make an incorrect statement, and then I tell you that it is incorrect. You make a qualification, but that qualification doesn't make it correct. You are deeply confused on this point. I guess you can blame me for not working hard enough to explain why you are wrong, so that you can learn something from it.

You are mixing up (1) describing the path of an accelerated particle using Cartesian inertial coordinates, and (2) describing physics using inertial Cartesian coordinates in which the accelerated particle is at rest.

I thought you were saying that (1) is impossible, globally. That's just not true.

Now, it is true that (2) is impossible, except locally. But that doesn't at all imply that (1) is impossible.

 

[stevendaryl]

Nope, but I won't bother explaining it to you anymore, bye.

You are just wrong on this point. There is no "explaining" involved. Here's a quote from Wikipedia:

In flat spacetime, the use of non-inertial frames can be avoided if desired. Measurements with respect to non-inertial reference frames can always be transformed to an inertial frame, incorporating directly the acceleration of the non-inertial frame as that acceleration is seen from the inertial frame.

The combination of Maxwell's equations and the Lorentz force law gives a description of physics in flat spacetime that uses only inertial, Cartesian coordinates. If you think otherwise, you are mistaken. I'm sorry that you are confused, but I don't really think it is my fault.

 

[stevendaryl]

Nope, but I won't bother explaining it to you anymore, bye.

Look, everyone has limits to their understanding. Your problem is that you believe that you understand things much better than you actually do. I don't believe that you will find a single competent physicist who will agree that one must use curvilinear coordinates to globally describe accelerated motion.

This is almost true by definition. Locally, you can use inertial Cartesian coordinates even in curved spacetime. You can use coordinates that are defined within a local chart, that are only good within a small region. Pretty much the definition of "flat spacetime" is the existence of a global chart, a single chart covering all of spacetime within which one can use inertial Cartesian coordinates.

 

[TrickyDicky]

Look, you make an incorrect statement, and then I tell you that it is incorrect. You make a qualification, but that qualification doesn't make it correct. You are deeply confused on this point. I guess you can blame me for not working hard enough to explain why you are wrong, so that you can learn something from it.

You are mixing up (1) describing the path of an accelerated particle using Cartesian inertial coordinates, and (2) describing physics using inertial Cartesian coordinates in which the accelerated particle is at rest.

I thought you were saying that (1) is impossible, globally. That's just not true.

Now, it is true that (2) is impossible, except locally. But that doesn't at all imply that (1) is impossible.

Wow, you finally got it .
You're welcome.