#### Bill_K

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
Related Special and General Relativity News on Phys.org

#### Greg Bernhardt

• vanhees71

#### pervect

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

#### PeterDonis

Mentor
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
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

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
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

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
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

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

#### PeterDonis

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

#### PAllen

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
it is section 1.3 (contravariant tensors) of:

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

#### PAllen

Ok, thanks!
I discovered it is available online:

#### 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
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
$$a^{\mu}=\mathrm{D}_{s} u^{\mu}=\mathrm{d}_s u^{\mu} + \Gamma^{\mu}_{\rho \sigma} u^{\rho} u^{\sigma}.$$
$$a^{\mu}=0.$$

"Precession in Special and General Relativity - Comments"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving