# I Is the Christoffel symbol orthogonal to the four-velocity?

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1. May 27, 2017

### Sonderval

Consider a force-free particle moving on a geodesic with four-velocity $$v^\nu$$.
The formula for the four-acceleration in any coordinate system is
$$\frac{dx^\mu}{d\tau} = - \Gamma^\mu_{\nu\lambda} v^\nu v^\lambda$$
Since the four-acceleration on the left side is orthogonal to the four-velocity, this implies
$$\Gamma^\mu_{\nu\lambda} v^\nu v^\lambda v_\mu=0$$
Is this correct? I've never seen this equation anywhere. (Perhaps because it is trivial?)

It seems to imply that for any coordinate system with 0 as a time-like coordinate
$$\Gamma^0_{00}=0$$
because I can always find a particle which is at rest in this system (i.e. with a four-velocity of (1,0,0,0) ). Is that a valid conclusion?

2. May 27, 2017

### DrGreg

No it isn't. The 4-acceleration $a^\mu$ is
$$a^\mu = \frac{\text{D}x^\mu}{\text{d}\tau} = \frac{\text{d}x^\mu}{\text{d}\tau} + \Gamma^\mu{}_{\nu\lambda} v^\nu v^\lambda$$

3. May 27, 2017

### Orodruin

Staff Emeritus
Not even that, it is
$$a^\mu = \frac{\text{D}v^\mu}{\text{d}\tau},$$
the OP should have a $v^mu$ instead of the $x^\mu$ in the derivative. Of course you are right in that you need to use the covariant derivative to define the 4-acceleration.

4. May 27, 2017

### DrGreg

Whoops! Of course, you are right, I should have said
$$a^\mu = \frac{\text{D}v^\mu}{\text{d}\tau} = \frac{\text{d}v^\mu}{\text{d}\tau} + \Gamma^\mu{}_{\nu\lambda} v^\nu v^\lambda = \frac{\text{d}^2x^\mu}{\text{d}\tau^2} + \Gamma^\mu{}_{\nu\lambda} v^\nu v^\lambda$$

5. May 27, 2017

### Sonderval

Sorry, yes, I garbled up on the left side, it should have been
$$\frac{dv^\mu}{d\tau}$$
For a force-free particle that moves on a geodesic (as stated in the OP), the covariant derivative is zero, isn't it?
But still, in any coordinate system, the equation
$$\frac{dv^\mu}{d\tau} v_\mu$$
should hold, or is that wrong as well?

6. May 27, 2017

### DrGreg

Yes, and the covariant derivative is the 4-acceleration.
No, it is $a^\mu v_\mu$ that is zero.

7. May 27, 2017

### Orodruin

Staff Emeritus
You are missing the fact that the metric is not constant. It does not hold that
$$\frac{dv^\mu}{dt} v_\mu = \frac{dv_\mu}{dt} v^\mu.$$
In general
$$0 = \frac{dv^2}{dt} = \frac{d(g_{\mu\nu} v^\mu v^\nu)}{dt} = 2v_\mu \frac{dv^\mu}{dt} + v^\mu v^\nu \frac{dg_{\mu\nu}}{dt} = 2v_\mu \frac{dv^\mu}{dt} + v^\mu v^\nu v^\rho \partial_\rho g_{\mu\nu}.$$
Of course, this becomes much easier with the covariant derivative
$$0 = \frac{dv^2}{dt} = \frac{dg(v,v)}{dt} = \nabla_v g(v,v) = (\nabla_v g)(v,v) + 2g(\nabla_v v, v) = 2g(a,v),$$
where $a = \nabla_v v$ is the 4-acceleration, i.e., $a\cdot v = 0$. Note that in local coordinates, you would have
$$a^\mu = v^\nu \nabla_\nu v^\mu = v^\nu (\partial_\nu v^\mu + \Gamma^\mu_{\nu\rho} v^\rho) = \frac{dv^\mu}{dt} + v^\nu v^\rho \frac{1}{2}g^{\mu\sigma}(\partial_\nu g_{\sigma\rho} + \partial_\rho g_{\sigma\nu} - \partial_\sigma g_{\nu\rho}).$$
Multiplying by $v_\mu$ would then give
$$v_\mu a^\mu = v_\mu \frac{dv^\mu}{dt} + v^\mu v^\rho v^\nu \frac{1}{2} \partial_\rho g_{\mu\nu} = 0.$$

Edit: Bloody indices ...

8. May 27, 2017

### stevendaryl

Staff Emeritus
I think you mean: $\frac{d^2 x^\mu}{d\tau^2} = - \Gamma^\mu_{\nu\lambda} v^\nu v^\lambda$

The more substantial mistake here is that the left-hand side is not the acceleration, except in the special case where $\Gamma^\mu_{\nu \lambda} = 0$.

With a lot of the concepts in General Relativity (or Special Relativity, for that matter), it's worth understanding what the analogous concepts are for the more familiar cases of Euclidean geometry and nonrelativistic kinematics.

For a particle traveling at constant velocity in the 2-dimensional Euclidean plane, the description in terms of Cartesian coordinates is quite simple:

$\frac{d^2 x}{dt^2} = 0$
$\frac{d^2 y}{dt^2} = 0$

But now, describe the same motion in polar coordinates $(r, \theta)$ using the transformations $x = r cos(\theta), y = r sin(\theta)$. In these coordinates, the motion doesn't look as simple:

$\frac{d^2 r}{dt^2} = r (\frac{d\theta}{dt})^2$
$\frac{d^2 \theta}{dt^2} = -\frac{2}{r} \frac{dr}{dt} \frac{d\theta}{dt}$

So in curvilinear coordinates $x^j$, the condition that the particle is traveling at constant velocity is not simply

$\frac{d^2 x^j}{dt^2} = 0$

$\frac{d^2 x^j}{dt^2} = - \sum_{k, l} \Gamma^j_{kl} \frac{dx^k}{dt} \frac{dx^l}{dt}$

For polar coordinates, $\Gamma^r_{\theta \theta} = -r$, $\Gamma^\theta_{r \theta} = \Gamma^\theta_{\theta r} = - \frac{1}{r}$

9. May 27, 2017

### vanhees71

However, if you use $\tau$ as the proper time you have by definition an affine parameter, namely
$$g_{\mu \nu} v^{\mu} v^{\nu}=g_{\mu \nu} \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}=c^2.$$
This implies
$$g_{\mu \nu} v^{\nu} \mathrm{D}_{\tau} v^{\mu}=0,$$
where the covariant derivative is defined as given above
$$\mathrm{D}_{\tau} v^{\mu} = \frac{\mathrm{d} v^{\mu}}{\mathrm{d} \tau} + {\Gamma^{\mu}}_{\rho \sigma} v^{\rho} v^{\sigma}=a^{\mu}$$
defines the proper acceleration.

The reason is that by definition in a pseudo-Riemannian space (as is spacetime in GR) the metric components fulfill
$$\nabla_{\rho} g_{\mu \nu}=0,$$
i.e., the affine connection used to define the covariant derivative (and parallel transport) is compatible with the metric. If the space is assumed to be torsion free (which is usually also implies by using a pseudo-Riemannian manifold) this determines the connection and thus the Christoffel symbols uniquely.

That's of course the same as what's written in #7, only translated to physicists' language.

10. May 27, 2017

### Orodruin

Staff Emeritus
Just to point out that $g(a,v) = 0$ is not dependent on the requirement of the connection being torsion free, all that we needed for this conclusion was that the connection is metric compatible, i.e., $\nabla_v g = 0$ (actually, the metric being parallel along the geodesic is sufficient ... but if we want it as a general statement for any geodesic, we need metric compatibility).

11. May 27, 2017

### vanhees71

True, compatbility of the metric is sufficient. In your prelast formula you however used the connection of the torsion free pseudo-Riemannian manifold.

12. May 27, 2017

### Orodruin

Staff Emeritus
Indeed I did, the deviation $\tilde \Gamma_{ab}^c = \Gamma_{ab}^c - \bar\Gamma_{ab}^c$ from the Levi-Civita connection (with connection coefficients $\bar \Gamma_{ab}^c$) satisfies
$$\tilde \Gamma_{cab} = \tilde \Gamma_{ab}^d g_{dc} = - \tilde \Gamma_{ac}^d g_{bd} = -\tilde \Gamma_{bac}.$$
It therefore follows that
$$v_\mu v^\nu v^\rho \tilde \Gamma_{\nu\rho}^\mu = v^\mu v^\nu v^\rho \tilde\Gamma_{\mu\nu\rho} = 0$$
due to the anti-symmetry of $\Gamma_{\mu\nu\rho}$. The deviation from the Levi-Civita connection therefore gives no contribution to $a\cdot v$.

13. May 27, 2017

### Sonderval

@All
Thanks a lot. So I confused a coordinate-based quantity (the coordinate acceleration on a geodesic) with a physical quantity (four-acceleration). Seems I keep making this mistake...

14. May 27, 2017

### Sonderval

@Orodruin
Sorry for not having a clue, but could you explain (or link to some explanation) what the differences are between
the different Gammas (with tilde and overbars)?

15. May 27, 2017

### Orodruin

Staff Emeritus
I was just showing that the difference ($\tilde \Gamma_{ab}^c$) between the connection coefficients of a general metric compatible connection ($\Gamma_{ab}^c$) and those of the Levi-Civita connection ($\bar\Gamma_{ab}^c$) does not contribute to what I was computing and that it is therefore perfectly fine to use the Levi-Civita expression even if the connection is not torsion-free. I introduced the notation in the post.

16. May 27, 2017