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

[Kostik]

TL;DR

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?