Question on Dirac's derivatives of the 4-velocity w.r.t. coordinates

[PeterDonis]

I insist that it is not defined.

You insist incorrectly. Apparently you have not grasped the implications of the term field which I keep using, and which Wald uses.

You are working in the opposite direction from the direction Wald does. You are starting with something that you think of as only defined on a single curve, and trying to extend it to an open region. Wald starts with things defined on open regions (tensor fields and operators on tensor fields), and from those things he constructs things that express derivatives along curves. (MTW is more explicit about this aspect than Wald is.) And since your question quoted an equation from Wald, the relevant framework to use is Wald's, not yours. And in Wald's framework, my answer is correct.

Sure, if you start only with the tangent vector of a single curve, your question about $\nabla_x T^t$ can't even be asked; there is no such thing. But Wald is not doing what you're doing. Nor am I. (Nor, as far as I can see, is Dirac, to go back to what this thread was originally about.)

 

[PeterDonis]

Says who?

You did, by defining $T^a$ such that its components are $\delta^a_t$ in your chosen frame. All I did was recognize what you failed to recognize, that $T^a$ so defined, using Wald's framework (and as I said in my last post, you're quoting Wald, so it's his framework we're using, not yours), is a vector field, defined on an open region of spacetime (in this case all of Minkowski spacetime).

 

[PAllen]

Note also, that Wald does derive, in passing, that parallel transport along a curve ends up depending only on values along the curve; and obviously, this would also be true of the geodesic equation. However, he never uses or relies on this fact. Nor does he ever define the absolute derivative, so far as I see.

 

[PeterDonis]

Here, the absolute derivative (without using that name) is introduced

Note, though, that what this absolute derivative acts on in MTW Box 10.2 is a vector field; i.e., it's assumed (implicitly) that it's defined in an open region around the curve whose tangent is used in the absolute derivative. So when we get to the geodesic equation, where the absolute derivative is acting on that same tangent, MTW is implicitly assuming a tangent field defining a family of curves in an open region, not just a single curve.

 

[JimWhoKnew]

You insist incorrectly. Apparently you have not grasped the implications of the term field which I keep using, and which Wald uses.

You are working in the opposite direction from the direction Wald does. You are starting with something that you think of as only defined on a single curve, and trying to extend it to an open region. Wald starts with things defined on open regions (tensor fields and operators on tensor fields), and from those things he constructs things that express derivatives along curves. (MTW is more explicit about this aspect than Wald is.) And since your question quoted an equation from Wald, the relevant framework to use is Wald's, not yours. And in Wald's framework, my answer is correct.

Sure, if you start only with the tangent vector of a single curve, your question about $\nabla_x T^t$ can't even be asked; there is no such thing. But Wald is not doing what you're doing. Nor am I. (Nor, as far as I can see, is Dirac, to go back to what this thread was originally about.)

There is an infinite ways in which you can chose the family of geodesics surrounding the given one. $\nabla_x T^t$ depends on this choice. Where does Wald say what this choice should be?

To make things clear: I don't say the books are wrong. I do hint that some of them are sloppy at this point.

 

[PeterDonis]

There is an infinite ways in which you can chose the family of geodesics surrounding the given one.

Perhaps, but you chose one particular one.

Where does Wald say what this choice should be?

Wald doesn't. You did. You defined $T^a$ to have components $(1, 0, 0, 0)$ in your chosen frame.

What you didn't realize was that, since you were asking about Wald's geodesic equation, you were defining a vector field that has those components in an open region, not just along one single curve. And that vector field of course defines a unique set of integral curves in that same open region, which are the geodesics I described.

 

[PeterDonis]

I don't say the books are wrong. I do hint that some of them are sloppy at this point.

I don't think the books are sloppy about this point. I think they are just doing a different thing from what you are trying to do, and you are interpreting what they do according to what you are trying to do, instead of understanding what the books are trying to do.

 

[PeterDonis]

There is an infinite ways in which you can chose the family of geodesics surrounding the given one.

Perhaps

For the actual case you gave, namely, flat Minkowski spacetime, no, there aren't. Once you pick one geodesic worldline in Minkowski spacetime, there is only one family of geodesics that it can possibly be a part of--the one I described. Any other geodesic outside that family will intersect your chosen geodesic and hence can't be a part of any family of geodesics that includes it.

 

[JimWhoKnew]

You did, by defining $T^a$ such that its components are $\delta^a_t$ in your chosen frame. All I did was recognize what you failed to recognize, that $T^a$ so defined, using Wald's framework (and as I said in my last post, you're quoting Wald, so it's his framework we're using, not yours), is a vector field, defined on an open region of spacetime (in this case all of Minkowski spacetime).

I'm having the LaTeX rendering issue, and can't preview my posts properly. It causes me to edit them after I submit them. But you reply so fast, that my corrections are lagging behind.

I apologize for that.

 

[JimWhoKnew]

Any other geodesic outside that family will intersect your chosen geodesic and hence can't be a part of any family of geodesics that includes it.

If the geodesics for $x\neq 0$ have a smoothly varying velocity (as a function of x; each geodesic has a constant velocity, of course) in the y direction, they will not intersect my chosen geodesic.

 

[PAllen]

Note, though, that what this absolute derivative acts on in MTW Box 10.2 is a vector field; i.e., it's assumed (implicitly) that it's defined in an open region around the curve whose tangent is used in the absolute derivative. So when we get to the geodesic equation, where the absolute derivative is acting on that same tangent, MTW is implicitly assuming a tangent field defining a family of curves in an open region, not just a single curve.

That's not what I am talking about. In section 10.4, in equations 10.20 and 10.21, MTW introduce what other authors call the absolute derivative, with a special notation that explicitly uses only values along a curve.

 

[JimWhoKnew]

Perhaps, but you chose one particular one.


Wald doesn't. You did. You defined $T^a$ to have components $(1, 0, 0, 0)$ in your chosen frame.

No, I didn't. But because of the problem described in# 39 (and your overwhelming firepower), you missed the correction "on the worldline" in #28.

For the actual case you gave, namely, flat Minkowski spacetime, no, there aren't. Once you pick one geodesic worldline in Minkowski spacetime, there is only one family of geodesics that it can possibly be a part of--the one I described. Any other geodesic outside that family will intersect your chosen geodesic and hence can't be a part of any family of geodesics that includes it.

As I said in #40, we can have a family of geodesics such that each has constant $x,z$ and a velocity $v^y=\epsilon x$. My original central geodesic is the same, $(t,0,0,0)$ , and the geodesics don't intersect. Since $\epsilon$ is assumed to be small but otherwise arbitrary, there are infinitely many families in this class, each yielding a different $\epsilon$-dependent $\nabla _x T^t$ at the central ("my") geodesic. Where does Wald provide us with the tools to single out one of these families?

Edit: corrected LaTeX expression

 

[JimWhoKnew]

Note, reviewing section 10.4 of MTW, I don't see any imprecision. Here, the absolute derivative (without using that name) is introduced with a specific notation, and the discussion appears complete to my eyes.

At the top of page 263, MTW just write
$$u^\nu u^\mu{}_{;\nu}$$
as the components of
$$\nabla _\mathbf{u} \mathbf{u}$$

 

[PAllen]

At the top of page 263, MTW just write
$$u^\nu u^\mu{}_{;\nu}$$
as the components of
$$\nabla _\mathbf{u} \mathbf{u}$$

But this is correct - the covariant derivative operator is only defined for a field, and the first of these is the component notation for the second. Then, just below that, they derive that values of u outside off the curve drop out, ending with a differential equation with just the curve parameter. The fact that this happens implies that, in retrospect, you never need to produce values of u off the curve to compute whether it is a goedesic. I'll grant that they could stand to "splash some gold paint" on this (an expression used by a professor of mine); but there is nothing imprecise or sloppy about it. Note, that the professor using this phrase was contrasting introductory textbooks to more advanced texts - noting for the latter, that you were not going to see authors providing emphasis on possibly important side points you should notice on your own.

To emphasize more: the second expression you give is meant to be defined on a field of u. It asks for the covariant change in u along the direction u.

 

[JimWhoKnew]

But this is correct - the covariant derivative operator is only defined for a field, and the first of these is the component notation for the second. Then, just below that, they derive that values of u outside off the curve drop out, ending with a differential equation with just the curve parameter. The fact that this happens implies that, in retrospect, you never need to produce values of u off the curve to compute whether it is a goedesic. I'll grant that they could stand to "splash some gold paint" on this (an expression used by a professor of mine); but there is nothing imprecise or sloppy about it. Note, that the professor using this phrase was contrasting introductory textbooks to more advanced texts - noting for the latter, that you were not going to see authors providing emphasis on possibly important side points you should notice on your own.

To emphasize more: the second expression you give is meant to be defined on a field of u. It asks for the covariant change in u along the direction u.

Suppose the connections are known, and we're given a curve $x^\mu(\tau)$ , for which we are asked to determine whether it is a geodesic (affinely parametrized). With the form
$$\frac{d^2x^\mu}{d\tau ^2} +\Gamma^\mu{}_{\nu\rho}\frac{dx^\nu}{d\tau}\frac{dx^\rho}{d\tau}=0$$
we know exactly how to proceed. But the form
$$u^\nu u^\mu{}_{;\nu}=0$$
requires the evaluation of $u^\mu{}_{,\nu}$ , and for this we need to plug in some off-curve data (since we know by now that the contraction with $u^\nu$ is extension-independent, any reasonable extension will do, but we still need to pick one for the calculation).

I think the treatment in John Lee's book, as kindly referenced by @martinbn in #22, is better.

 

[PAllen]

Suppose the connections are known, and we're given a curve $x^\mu(\tau)$ , for which we are asked to determine whether it is a geodesic (affinely parametrized). With the form
$$\frac{d^2x^\mu}{d\tau ^2} +\Gamma^\mu{}_{\nu\rho}\frac{dx^\nu}{d\tau}\frac{dx^\rho}{d\tau}=0$$
we know exactly how to proceed. But the form
$$u^\nu u^\mu{}_{;\nu}=0$$
requires the evaluation of $u^\mu{}_{,\nu}$ , and for this we need to plug in some off-curve data (since we know by now that the contraction with $u^\nu$ is extension-independent, any reasonable extension will do, but we still need to pick one for the calculation).

I think the treatment in John Lee's book, as kindly referenced by @martinbn in #22, is better.

Except that in a few lines they derive your first equation from the field form, so, no, you don’t need any extra computation, nor do you have to ever produce off curve values

Pedagogically, I agree that authors would be better off explicitly introducing the absolute derivative and using the most appropriate for each problem. But this is a matter of taste. Nothing is wrong, left out, or sloppy about how MTW and Wald do it.

 

[PeterDonis]

you missed the correction "on the worldline" in #28.

Ok. But then, according to Wald's framework, you haven't specified a vector field at all, so you can't even use Wald's equations. So your question about $\nabla_x T^t$ isn't well defined because your scenario isn't well defined.

You can't have it both ways. If you specify a well-defined scenario according to Wald's framework, then your question has a well-defined answer. If it doesn't have a well-defined answer, it's because you didn't specify a well-defined scenario according to Wald's framework in the first place, not because Wald is being sloppy.

 

[PeterDonis]

Where does Wald provide us with the tools to single out one of these families?

He doesn't. You need to specify the vector field in your specification of the scenario. If you don't, as I said in my previous post just now, that's on you, not on Wald. His framework is based on having a vector field specified in the scenario. If you don't know the vector field, you can't use his framework.

In practice, of course, you do have a vector field, because real objects aren't point particles. There will be some congruence of worldlines that describes the object, not just a single worldline. And that congruence of worldlines will defined a vector field in an open "world tube" of spacetime. Typically the congruence will be picked out by some physical constraint, such as that the object is not rotating (which means the vorticity of the congruence must be zero).

 

[PeterDonis]

the second expression you give is meant to be defined on a field of u. It asks for the covariant change in u along the direction u.

And note that this works on any curve in the congruence of curves defined by u in whatever open region it's defined on. It doesn't just work on the particular curve that one might be interested in.

 

[JimWhoKnew]

If you don't know the vector field, you can't use his framework.

I agree with that.

I prefer the extension-independence approach. In my above example, any choice for the surrounding family of geodesics will yield a finite $\nabla _x T^t$ which is multiplied by $T^x\quad (=0)$ . Had MTW and Wald bothered to dedicate a sentence or two to clarify, I would have said nothing. Especially MTW which is aimed also for beginners who don't have your decades of hindsight.

 

[PeterDonis]

I prefer the extension-independence approach.

If you're talking about an extension at all, you're talking about defining a vector field in an open region around the worldline you're interested in. Which means you're using Wald's framework or something equivalent to it, including deriving extension independence in that framework.

So I'm confused about how you think your preferred approach is different. "Different" to me would be not even using an extension at all, and working solely with quantities that only need to be defined along the worldline you're interested in, not in an open region around it. But that doesn't seem to be what you're describing.

 

[JimWhoKnew]

If you're talking about an extension at all, you're talking about defining a vector field in an open region around the worldline you're interested in. Which means you're using Wald's framework or something equivalent to it, including deriving extension independence in that framework.

Of course. I already wrote in #35 that "I don't say the books are wrong".

So I'm confused about how you think your preferred approach is different. "Different" to me would be not even using an extension at all, and working solely with quantities that only need to be defined along the worldline you're interested in, not in an open region around it. But that doesn't seem to be what you're describing.

Some implementations come with a "natural" extension, like the discussion of Raychaudhuri's equation, where the "Wald form" is very useful. Some come without, and the extension can be chosen arbitrarily. Some come with a "built in" vector field that obeys "Wald form" on certain curves, like the KVF on black holes' horizons. And some don't necessitate the "Wald form" at all, and can be worked out solely with quantities defined on the curve.

It is not different from Wald. It is different from your interpretation that there always exists a unique "natural" extension that should be used.

 

[PeterDonis]

It is different from your interpretation that there always exists a unique "natural" extension that should be used.

I made no such claim. I did say, incorrectly, that there was a unique extension in one particular case, the case of a geodesic in flat Minkowski spacetime. I was incorrectly thinking of the 1+1 case instead of the 3+1 case.

 

[JimWhoKnew]

Had MTW and Wald bothered to dedicate a sentence or two to clarify, I would have said nothing.

It turns out that Dirac does find it worthwhile to acknowledge the point. On page 46 of his GR book, he writes: "... if $v^\mu$ is defined as a continuous field function instead of having a meaning only on one world line ...".