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

[Kostik]

TL;DR Summary

Is Dirac being sloppy in writing partial derivatives as ordinary derivatives, or am I missing something?

In Dirac's "General Theory of Relativity", he shows how in the weak field, low-speed limit, Einstein's equation gives Newtonian gravity. Along the way, he takes derivatives of the spatial components of the 4-velocity $v^m = dx^m/ds$ with respect to the coordinates $x^r$ and $x^0=t$.

Of course, $v^\mu = v^\mu(x^0,x^1,x^2,x^3)$ is a vector field in spacetime, so these should be partial derivatives.

In the excerpt below, you will note (see the equation before (16.2), as well as (16.2) and (16.3)) Dirac writes $dv^m/dx^\mu$, $dv^m/dx^0$, etc. (In the equation before (16.2), he is using $\partial x^m/\partial x^r = \delta^m_r$, but this step is hidden.)

Am I correct in thinking that he really should be writing $\partial v^m/\partial x^\mu$, $\partial v^m/\partial x^0$, etc.?

Is Dirac just being sloppy in writing partial derivatives as ordinary derivatives, or am I missing something?

 

[ergospherical]

They are ordinary (total) derivatives

 

[Kostik]

They are ordinary (total) derivatives

That does not apply here: the coordinates $(x^0,x^1,x^2,x^3)$ are independent. $x^1,x^2,x^3$ are not functions of $x^0$.

 

[ergospherical]

How else do you interpret equation 16.3? e.g. analogy in 1d classical mechanics

$\frac{dv}{dt} = F(x(t))$

The solution is the trajectory (x(t), v(t)). LHS is ordinary derivative with respect to $t$. In 16.3, RHS is a 'force' derived from the function $(g_{00}^{1/2})_{,m}$ of the spatial coordinates.

 

[Kostik]

So what does $$\frac{dv^m }{dx^\mu}$$ mean? If you actually follow the calculation, it looks as if Dirac really means $\partial v^m/\partial x^0$ and $\partial v^m/\partial x^r$.

 

[PeterDonis]

The solution is the trajectory (x(t), v(t)). LHS is ordinary derivative with respect to .

But that's because $x$ and $v$ in your example are only functions of one variable, $t$.

In the case under discussion here, we have functions of multiple variables, so you can't just blithely generalize your 1-d example with functions of one variable.

 

[Kostik]

The only way I can understand the equation preceding (16.2) is:

$$\frac{dv^m}{ds} = \frac{\partial v^m}{\partial x^\mu} \frac{dx^\mu}{ds} = \frac{\partial v^m}{\partial x^0}v^0 + \frac{\partial v^m}{\partial x^r}v^r = \frac{\partial v^m}{\partial x^0}v^0 + \frac{d}{ds}\left( \frac{\partial x^m}{\partial x^r}\right)v^r = \frac{dv^m}{dx^0}v^0$$

 

[Kostik]

The only way I can understand the equation preceding (16.2) is:

$$\frac{dv^m}{ds} = \frac{\partial v^m}{\partial x^\mu} \frac{dx^\mu}{ds} = \frac{\partial v^m}{\partial x^0}v^0 + \frac{\partial v^m}{\partial x^r}v^r = \frac{\partial v^m}{\partial x^0}v^0 + \frac{d}{ds}\left( \frac{\partial x^m}{\partial x^r}\right)v^r = \frac{dv^m}{dx^0}v^0$$

On the other hand, $dv^m/dx^0$ does seem to have meaning if you write
$$\frac{dv^m}{dx^0} = \frac{d}{dx^0}\left( \frac{dx^m}{ds} \right) = \frac{d}{ds}\left( \frac{dx^m}{dx^0} \right)$$

 

[ergospherical]

You're getting hung up about notation without thinking about what the equations (e.g. 16.3) mean. Look at this example, now in 3d classical mechanics:

$\frac{d\mathbf{v}}{dt} = \mathbf{F}(\mathbf{x}(t))$

along with $\frac{d\mathbf{x}}{dt} = \mathbf{v}(t)$. Again, LHS is an ordinary derivative, and solution is $(\mathbf{x}(t), \mathbf{v}(t))$ both parameterized by this $t$. You're looking at a specific motion of the particle (i.e. derivatives taken along this specific curve).

 

[PeterDonis]

LHS is an ordinary derivative

Yes, because again, the vectors in question are functions of only one variable.

You're looking at a specific motion of the particle (i.e. derivatives taken along this specific curve).

Only if $t$ is a curve parameter instead of a single variable that the vectors are a function of. But that's not what Dirac is doing in the expressions the OP is asking about: in those expressions (as opposed to the ones where the derivative is taken with respect to $s$), he's not taking absolute derivatives along a curve. He's taking derivatives with respect to spacetime coordinates.

 

[Kostik]

You're getting hung up about notation without thinking about what the equations (e.g. 16.3) mean. Look at this example, now in 3d classical mechanics:

$\frac{d\mathbf{v}}{dt} = \mathbf{F}(\mathbf{x}(t))$

along with $\frac{d\mathbf{x}}{dt} = \mathbf{v}(t)$. Again, LHS is an ordinary derivative, and solution is $(\mathbf{x}(t), \mathbf{v}(t))$ both parameterized by this $t$. You're looking at a specific motion of the particle (i.e. derivatives taken along this specific curve).

This is simply wrong. Dirac is differentiating the 4-velocity $v^m$ w.r.t. the coordinates $x^0,x^1,x^2,x^3$. What does $$\frac{dv^m}{dx^3}$$ mean?

 

[Kostik]

On the other hand, $dv^m/dx^0$ does seem to have meaning if you write
$$\frac{dv^m}{dx^0} = \frac{d}{dx^0}\left( \frac{dx^m}{ds} \right) = \frac{d}{ds}\left( \frac{dx^m}{dx^0} \right)$$

Please ignore this, it's nonsense. Unfortunately, no longer able to edit/delete.

 

[Kostik]

Let me ask the following question. The equation preceding (16.2) reads
$$\frac{dv^m}{ds} = \frac{dv^m}{dx^\mu} \frac{dx^\mu}{ds} = \frac{dv^m}{dx^0}v^0 \qquad(*)$$ which in expanded form probably reads
$$\frac{dv^m}{ds} = \frac{\partial v^m}{\partial x^\mu} \frac{dx^\mu}{ds} = \frac{\partial v^m}{\partial x^0}v^0 + \frac{\partial v^m}{\partial x^r}v^r = \frac{\partial v^m}{\partial x^0}v^0 + \frac{d}{ds}\left( \frac{\partial x^m}{\partial x^r}\right)v^r = \frac{dv^m}{dx^0}v^0$$
Why couldn't Dirac have written $(*)$ simply:
$$\frac{dv^m}{ds} = \frac{dv^m}{dx^0} \frac{dx^0}{ds} = \frac{dv^m}{dx^0}v^0 \quad ?$$

 

[ergospherical]

Right, I see the bit you mean -- yes, in the line before 16.2 it should be $\tfrac{\partial v^m}{\partial x^{\mu}} \frac{dx^{\mu}}{ds}$ . But it's quite alright to transfer that over to $\frac{dv^m}{dx^0} v^0$, where now the $v^m = v^m(x^0)$ are viewed as functions of the parameter $x^0$ ("on shell").

 

[Kostik]

Right, I see the bit you mean -- yes, in the line before 16.2 it should be $\tfrac{\partial v^m}{\partial x^{\mu}} \frac{dx^{\mu}}{ds}$ . But it's quite alright to transfer that over to $\frac{dv^m}{dx^0} v^0$, where now the $v^m = v^m(x^0)$ are viewed as functions of the parameter $x^0$ ("on shell").

I think it is simply unwise to work with the "hybrid" quantity $$\frac{dv^m}{dt} = \frac{d}{dt}\frac{dx^m}{ds} = \frac{d}{dt} \left( \frac{dx^m}{dt} \frac{dt}{ds} \right) = \frac{d^2 x^m}{dt^2} {g_{00}}^{-1/2} \, .$$ (Note that the metric is assumed to be time independent.) Ultimately, it is $$\frac{d^2 x^m}{dt^2}$$ that we want to say something about.

 

[ergospherical]

It’s not supposed to be that confusing. You’re in the Newtonian limit anyway, so $v_m$ is your ordinary velocity components and $x^0$ is your ordinary time. (15.3) is just Newton’s second law. All of this only works because the $v_m$ are small.

 

[Kostik]

It’s not supposed to be that confusing. You’re in the Newtonian limit anyway, so $v_m$ is your ordinary velocity components and $x^0$ is your ordinary time. (15.3) is just Newton’s second law. All of this only works because the $v_m$ are small.

That's true, of course: in the weak-field, low-speed limit, $ds\approx dt$. Bu then, why doesn't Dirac just use $dv^m/ds$ to approximate $d^2 x^m/dt^2$? What's the point of using the hybrid quantity $dv^m/dt$?

Also, it's well to keep track of orders of magnitude of the "small quantity" $v^m$. If you accidentally set $ds=dt$ (i.e., $v^0=1$), (or $g_{00}=1$, then gravity disappears and you lose the Newtonian approximation!

 

[JimWhoKnew]

He's taking derivatives with respect to spacetime coordinates.

This point is often ignored in textbooks, which freely replace $\nabla_\mathbf{v}$ by $v^\mu \nabla_\mu$ .
At the present case $v^\mu$ is defined only on the curve, and is curve-dependent. If only the coordinates of an ordinary point in spacetime are given, it is not enough for determining $v^\mu$ .

For example: suppose we know the geodesic of a particle that falls radially in Schwarzschild spacetime. How can $\frac{\partial v^r}{\partial\theta}$ and $\frac{\partial v^r}{\partial\phi}$ be evaluated along the geodesic?

Edited addition:
I would expect a rigorous argument to extend $v^\mu$ to a vector field in a tube around the geodesic, and then show that the results are extension-independent.

 

[PAllen]

This point is often ignored in textbooks, which freely replace $\nabla_\mathbf{v}$ by $v^\mu \nabla_\mu$ .
At the present case $v^\mu$ is defined only on the curve, and is curve-dependent. If only the coordinates of an ordinary point in spacetime are given, it is not enough for determining $v^\mu$ .

For example: suppose we know the geodesic of a particle that falls radially in Schwarzschild spacetime. How can $\frac{\partial v^r}{\partial\theta}$ and $\frac{\partial v^r}{\partial\phi}$ be evaluated along the geodesic?

Edited addition:
I would expect a rigorous argument to extend $v^\mu$ to a vector field in a tube around the geodesic, and then show that the results are extension-independent.

I also don't like the confusion in many physics oriented presentations of the concept of absolute derivative along a curve versus covariant derivative of a tensor field . However, I assume you are well aware than many books provide just such an argument as you suggest, which allows the sloppiness. Some books (e.g. by Synge & Schild) even derive the covariant derivative from the absolute derivative, making the equivalence a matter of definition.

 

[JimWhoKnew]

However, I assume you are well aware than many books provide just such an argument as you suggest, which allows the sloppiness.

Actually, I don't recall any discussion in a textbook that attempts to justify the use of
$$u^\nu u^\mu{}_{;\nu} \, .$$
The issue is often either ignored (MTW, Wald, Carroll, Poisson, ...) or avoided (Weinberg, ...).

I'd be grateful to get some nice references that address this issue directly.

I looked at Synge & Schild, and saw that the covariant derivative is derived for differentiating tensor fields on TM. This poses no problem, of course. Do they use $u^\nu u^\mu{}_{;\nu}$ or its equivalents? (where $u^\mu$ is defined only on the curve)

Edit addition:
I looked at it again, and found that it is simple to argue that my suggestion from #18 is valid (ie. an arbitrary off-curve smooth extension can be used, and the off-curve values cancel each other "telescopically" up to $(dx^\mu)^2$ )

 

[PeterDonis]

This point is often ignored in textbooks, which freely replace $\nabla_\mathbf{v}$ by $v^\mu \nabla_\mu$ .

Please give specific examples. I suspect you are misinterpreting what is being done in the textbooks.

At the present case $v^\mu$ is defined only on the curve

And if that is true, then of course you can't write an expression for its derivative with respect to the curve parameter in terms of ordinary covariant derivatives, because you don't know what those are.

But in all of the examples I have seen in textbooks, you don't just have one curve, you have a family of curves defined by a 4-velocity field $v^\mu$ on an open region of spacetime. And in that case, yes, you most certainly can replace $\nabla_\mathbf{v}$ by $v^\mu \nabla_\mu$ along any curve in the family.

I would expect a rigorous argument to extend $v^\mu$ to a vector field in a tube around the geodesic

Yes, of course. In the case of geodesics in a known spacetime geometry, this is straightforward, since the geodesics are known everywhere.

and then show that the results are extension-independent.

They don't have to be extension independent. It's possible in principle for a given curve to be a member of more than one family of curves on an open region of spacetime. If so, then there will be more than one way to define a vector field on that region that includes the tangent vectors to the curve.

However, once you've picked a particular family of curves on an open region that includes the curve you're interested in, then the tangent vector field defined by that family of curves on that open region is obviously unique, and therefore so are the covariant derivatives defined in terms of it.

 

[martinbn]

I'd be grateful to get some nice references that address this issue directly.

Any book on differential geometry. For example John Lee's "Riemannian Manifolds: An Introduction to Curvature." The chapter on connections, the subsection on vector fields along curves.

 

[PAllen]

Actually, I don't recall any discussion in a textbook that attempts to justify the use of
$$u^\nu u^\mu{}_{;\nu} \, .$$
The issue is often either ignored (MTW, Wald, Carroll, Poisson, ...) or avoided (Weinberg, ...).

Wald, so far as I can see, only deals with cases where a field is given, and the covariant derivative is what is desired. However, they do show that parallel transport of vector depends only on values of the vector on the curve. I see no example of confusion or sloppiness in sections I just reviewed. Do you have some section in mind? [In fact, I don't see Wald bothering to use the absolute derivative at all, or handle any cases where a tensor is only defined on a curve].

I looked at Synge & Schild, and saw that the covariant derivative is derived for differentiating tensor fields on TM. This poses no problem, of course. Do they use $u^\nu u^\mu{}_{;\nu}$ or its equivalents? (where $u^\mu$ is defined only on the curve)

The very old edition of Synge & Schild I have (Tensor Calculus, 1969 printing), first defines the absolute derivative. Then they define the covariant derivative of a tensor field in such a way that it has the property that when contracted with the tangent vector along any curve, it equals the absolute derivative of the tensor along that curve - the absolute derivative depending only on values along the curve . This implies, given a tensor defined only on a curve, any extension will produce a covariant derivative with the desired property. (This is all in pp. 47-51 of my printing).

 

[JimWhoKnew]

Any book on differential geometry. For example John Lee's "Riemannian Manifolds: An Introduction to Curvature." The chapter on connections, the subsection on vector fields along curves.

Thanks. Grateful as promised!

(although extension-independence is left to the reader...)

 

[JimWhoKnew]

Wald, so far as I can see, only deals with cases where a field is given, and the covariant derivative is what is desired. However, they do show that parallel transport of vector depends only on values of the vector on the curve. I see no example of confusion or sloppiness in sections I just reviewed. Do you have some section in mind? [In fact, I don't see Wald bothering to use the absolute derivative at all, or handle any cases where a tensor is only defined on a curve].

Wald defines the geodesic equation(s) by using the off-curve directions $\nabla _a$ operating on the tangent vector (section 3.3 , first two paragraphs).

This implies, given a tensor defined only on a curve, any extension will produce a covariant derivative with the desired property.

Good point. Yet I think that textbooks who use off-curve directions, like Wald, should say something about it.

 

[PeterDonis]

Wald defines the geodesic equation(s) by using the off-curve directions $\nabla _a$ operating on the tangent vector (section 3.3 , first two paragraphs).

More precisely, he defines geodesics in terms of the derivative operator $\nabla_a$, which he defines earlier, in Section 3.1, as an operator on tensor fields.

I think that textbooks who use off-curve directions, like Wald, should say something about it.

Given that Wald explicitly defines derivative operators as operators on tensor fields, as noted above, I'm not sure what more he could do to make explicit what you're looking for.

 

[PAllen]

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.

 

[JimWhoKnew]

More precisely, he defines geodesics in terms of the derivative operator $\nabla_a$, which he defines earlier, in Section 3.1, as an operator on tensor fields.


Given that Wald explicitly defines derivative operators as operators on tensor fields, as noted above, I'm not sure what more he could do to make explicit what you're looking for.

I'm confused by your replies, so please allow me a question to hopefully help clarify (to me) what we are arguing about:

Consider a body at its rest frame in SR (Cartesian coordinates). Wald's geodesic equation 3.3.1 reads
$$T^a \nabla_a T^b=0$$
where $T^a$ is the tangent vector to the worldline ($=\delta^a_t$ on the worldline).

What is the value of
$$\nabla_x T^t$$
in this specific case?

 

[PeterDonis]

Consider a body at its rest frame in SR (Cartesian coordinates). Wald's geodesic equation 3.3.1 reads
$$T^a \nabla_a T^b=0$$
where $T^a$ is the tangent vector to the worldline ($=\delta^a_t$ on the worldline).

No, $T^a$ is the tangent vector field associated with the family of geodesics of which your chosen worldline is just one. Every one of that family of geodesics has constant spatial coordinates in your chosen frame (so each unique set of 3 spatial coordinates labels a unique geodesic--you appear to be thinking of your chosen body's worldline as the geodesic labeled by $x = 0$, $y = 0$, $z = 0$) and can be affinely parameterized by the time coordinate of your chosen frame. So every 3-surface of constant $t$ in your chosen frame is orthogonal to every geodesic in the family. Thus, your property of $T^a$ having components $\delta^a_t = (1, 0, 0, 0)$ for every geodesic.

What is the value of
$$\nabla_x T^t$$
in this specific case?

Obviously it's zero. What's your point?

 

[JimWhoKnew]

Every one of that family of geodesics has constant spatial coordinates in your chosen frame

Says who?

 

[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.