## Main Question or Discussion Point

The Absolute Derivative
In relativity we typically deal with two types of quantities: fields, which are defined everywhere, and particle properties, which are defined only along a curve or world line. The familiar covariant derivative is appropriate when we need to differentiate a field. A field is a function of all four coordinates, and the covariant derivative of ##\varphi(x)## consists of the four partial derivatives ##\partial \varphi /\partial x^\mu## plus correction terms involving the Christoffel symbols, one for each tensor index on ##\varphi##.
A particle property ##\varphi(s)##, on the other hand, is a function only of a single parameter ##s## running along the curve. In this situation, the partial derivatives of ##\varphi## with respect to the four coordinates do not exist. (Unfortunately many references miss this point!) Writing partial derivatives would require that ##\varphi## be defined everywhere in a neighborhood of the curve, which is not the case. The...

kith, atyy and Greg Bernhardt

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pervect
Staff Emeritus
Is Bill_K back, or was this edited from old posts of his?

PeterDonis
Mentor
2019 Award
It was edited from an old post of his.

stevendaryl
Staff Emeritus
I'm a little uncertain about one line from the essay. It says:

For a given timelike curve ##x^\mu(s)##, we define the vector ##v^\mu = \frac{Dx^\mu}{Ds}##...
That would seem to imply that ##v^\mu## involves connection coefficients. But it doesn't. The components of ##v^\mu## are just ##\frac{dx^\mu}{ds}##.

vanhees71
PeterDonis
Mentor
2019 Award
That would seem to imply that ##v^\mu## involves connection coefficients. But it doesn't. The components of ##v^\mu## are just ##\frac{dx^\mu}{ds}##.
From what we can tell, it was a capital ##D## in the original post by Bill_K (from years ago) that this Insight is based on. I think the resolution might be that for the particular case of ##D x^\mu / Ds##, the connection coefficient terms vanish; but I'll have to check the math.

PAllen
2019 Award
From what we can tell, it was a capital ##D## in the original post by Bill_K (from years ago) that this Insight is based on. I think the resolution might be that for the particular case of ##D x^\mu / Ds##, the connection coefficient terms vanish; but I'll have to check the math.
For general coordinates in SR, or in GR, they don’t vanish. It is correct as stated.

PeterDonis
Mentor
2019 Award
For general coordinates in SR, or in GR, they don’t vanish.
"Vanish" was the wrong word; "cancel" would have been better. But let's work a specific example: a stationary observer in Schwarzschild spacetime.

For this observer, if we assume the angular coordinates are zero for simplicity and write ##\tau## for ##s## to parameterize by proper time, we have ##x^\mu (\tau) = (t, r, \theta, \phi) = \left( \frac{\tau}{\sqrt{1 - 2M / R}}, R, 0, 0 \right)##, where ##R## is a constant independent of ##\tau##. So we have ##d x^\mu / d \tau = \left( \frac{1}{\sqrt{1 - 2M / R}}, 0, 0, 0 \right)##.

We then have

$$\frac{D x^\mu}{D \tau} = \frac{d x^\mu}{d \tau} + x^\nu \Gamma^\mu{}_{\nu \sigma} \frac{d x^\sigma}{d \tau}$$

We have two nonzero components:

$$\frac{D x^t}{D \tau} = \frac{d x^t}{d \tau} + x^r \Gamma^t{}_{r t} \frac{d x^t}{d \tau} = \frac{1}{\sqrt{1 - 2M / R}} \left( 1 + \frac{R M}{R - 2M} \right)$$

$$\frac{D x^r}{D \tau} = \frac{d x^r}{d \tau} + x^t \Gamma^r{}_{t t} \frac{d x^t}{d \tau} = \frac{M \tau}{R^2}$$

We know that the correct answer should be ##v^\mu = \left( \frac{1}{\sqrt{1 - 2M / R}}, 0, 0, 0 \right)## (i.e., that the connection coefficient terms should cancel if ##v^\mu = D x^\mu / D s## is correct), so unless I'm missing something, it does appear that @stevendaryl is right and we should define ##v^\mu = d x^\mu / d s##. But it could be that I'm missing something.

PAllen
2019 Award
Bill_k was making a general statement. It can’t be addressed with a specific case.

[edit: wait, it can’t make sense to have plain coordinates as above. So @stevendaryl is right. The tangent vector is defined by simple derivative. It is a unit vector if the parameter is proper time.

Last edited:
PeterDonis
Mentor
2019 Award
Bill_k was making a general statement. It can’t be addressed with a specific case.
If any specific case contradicts a general proposition, that general proposition is falsified.

The general proposition under consideration is that the definition of ##v^\mu## should be ##v^\mu = D x^\mu / Ds##, as it is in the Insight as it currently stands, rather than ##v^\mu = d x^\mu / ds##, as @stevendaryl suggests.

When I look at those two possibilities for a specific case, a stationary observer in Schwarzschild spacetime, it seems like the second definition, the one @stevendaryl suggests, is correct and the first one is wrong.

Unfortunately Bill_K is not around to either correct any mistake I made (if I made one) in my reasoning above and in the previous post where I did the math for the specific case and thereby show that the first definition is actually correct, or to confirm that the second definition is correct and something got garbled in the transition from his original post, many years ago, to the Insight article (which could have happened). But if you, or anyone else, can point out an error in my reasoning about which definition is correct, please do so. I want to make sure the Insight is correct, and would be happy to be shown that it is; but if it isn't, then I want to correct it.

PAllen
2019 Award
I checked a text on differential geometry to review this. @stevendaryl is right.

PeterDonis
Mentor
2019 Award
I checked a text on differential geometry to review this. @stevendaryl is right.
Thanks! Which text (and chapter/section), for reference?

PAllen
2019 Award
Thanks! Which text (and chapter/section), for reference?
Well, showing my age, it is section 1.3 (contravariant tensors) of:

Tensor Calculus by Synge and Schild, 1949

dextercioby
PeterDonis
Mentor
2019 Award
it is section 1.3 (contravariant tensors) of:

Tensor Calculus by Synge and Schild, 1949
Ok, thanks!

stevendaryl
Staff Emeritus
The confusion is that even though the coordinate ##x^\mu## is written with a raised index, it isn't actually a component of a vector (except in the special case of flat spacetime, in which case ##x^\mu## can be identified with the vector from the origin to the point).

vanhees71
Gold Member
2019 Award
The notation ##D x^{\mu}/D s## doesn't make sense for general coordinates since ##x^{\mu}## are no vector components. The only exception is if you are in a flat (i.e., affine) manifold, i.e., when used as a space-time description, Minkowski time. Then you can choose the ##x^{\mu}## as the components of the then sensible space-time vector with respect to a Galilean basis.

In the general case, I'd thus always write ##q^{\mu}## for arbitrary (local) coordinates of a general manifold (in GR it's a pseudo-Riemannian manifold with the torsion free standard connection). Then the covariant time derivative of a vector field (sic!) with vector components ##V^{\mu}## along a curve ##q^{\mu}(s)## (where ##s## is an arbitrary affine parameter) is given by
$$\mathrm{D}_s V^{\mu}=\mathrm{d}_s V^{\mu} + \Gamma^{\mu}_{\rho \sigma} \mathrm{d}_s q^{\rho} V^{\sigma},$$
where ##\Gamma^{\mu}_{\rho \sigma}## are the usual Christophel symbols. ##\mathrm{d}_s## denotes the usual derivative of the object it acts on with respect to the parameter ##s##.

Note that however the ##\mathrm{d} q^{\mu}## are vector components and thus also
$$u^{\mu}=\mathrm{d}_s q^{\mu}$$
are vector components. Thus it makes sense to use the covariant derivative to define the acceleration as
$$a^{\mu}=\mathrm{D}_{s} u^{\mu}=\mathrm{d}_s u^{\mu} + \Gamma^{\mu}_{\rho \sigma} u^{\rho} u^{\sigma}.$$
$$a^{\mu}=0.$$