# Is acceleration due to gravity a 4-vector?

pellman
We can write the geodesic condition as

$$\frac{d^2 x^\alpha}{d\tau^2}={-\Gamma^\alpha}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}$$

On the RHS we have two contravariant vectors ( or (0,1) tensors ) contracted with the connection. But the connection is NOT a tensor. So is the LHS a vector? And if so, how does that work?

##a^{\mu} = \ddot{x}^{\mu} + \Gamma^{\mu}_{\alpha\beta}\dot{x}^{\alpha}\dot{x}^{\beta}## is a 4-vector but the two individual terms on the RHS are not 4-vectors. It's very easy to show that they both fail to transform as 4-vectors but that their sum does transform as a 4-vector.

1 person
pellman
Thanks. Regarding $\ddot{x}^\mu$ ... where do I go wrong in the following?

Suppose we have a change of coordinates $x^\mu={\Lambda^\mu}_\alpha x'^\alpha$. Elements of the tangent space are vectors and likewise transform according to $V^\mu={\Lambda^\mu}_\alpha V'^\alpha$. In particular the tangent to a curve $x^\mu(\tau)$ is $dx^\mu/d\tau$ which as a tangent vector transforms as

$$\frac{dx^\mu}{d\tau} ={\Lambda^\mu}_\alpha \frac{dx'^\alpha}{d\tau}$$

but if we take the derivative of both sides of

$x^\mu={\Lambda^\mu}_\alpha x'^\alpha$

we get

$\frac{dx^\mu}{d\tau}=\frac{d}{d\tau}\left({\Lambda^\mu}_\alpha x'^\alpha\right)$

This is equivalent to the previous expression only if $\frac{d}{d\tau}{\Lambda^\mu}_\alpha =0$. In this case if we take the second derivative, we get

$$\frac{d^2 x^\mu}{d\tau^2} ={\Lambda^\mu}_\alpha \frac{d^2 x'^\alpha}{d\tau^2}$$

and so $\ddot{x}^\mu$ transforms as a vector.

?????

##x^{\mu}## are not position vectors. The transformation ##x^{\mu} \rightarrow x^{\mu'}(x^{\nu})## will not be the same as the transformation matrix ##V^{\mu} \rightarrow V^{\mu'}## which is where your error is; in fact if we write the former transformation as ##x^{\mu'} = x^{\mu'}(x^{\nu})## then ##V^{\mu'} = \frac{\partial x^{\mu'}(x^{\nu})}{\partial x^{\mu}}V^{\mu}## so clearly we are talking about different transformations for the two.

1 person
Staff Emeritus
but if we take the derivative of both sides of

$x^\mu={\Lambda^\mu}_\alpha x'^\alpha$

we get

$\frac{dx^\mu}{d\tau}=\frac{d}{d\tau}\left({\Lambda^\mu}_\alpha x'^\alpha\right)$

This is equivalent to the previous expression only if $\frac{d}{d\tau}{\Lambda^\mu}_\alpha =0$.

To simplify the notation, I'm going to drop the indices. It should be clear how to put them back.

In general, if you converting from a coordinate $x'$ to a coordinate $x$, then it's a nonlinear, continuous, invertible function:

$x = F(x')$

So,

$\frac{dx}{d \tau} = \frac{\partial F}{\partial x'} \frac{dx'}{d\tau}$

So even if $F(x')$ is nonlinear, the relationship between velocities is linear:

$\frac{dx}{d \tau} =\Lambda \frac{dx'}{d\tau}$

where $\Lambda = \frac{\partial F}{\partial x'}$

But note: the transformation between coordinates is not the same as the transformation between velocities:

$\frac{dx}{d \tau} =\Lambda \frac{dx'}{d\tau}$ versus $x = F(x')$

The only time that you can use the same matrix $\Lambda$ to transform coordinates or velocities is when:

$F(x') = \Lambda x'$

which is only possible when $\frac{\partial \Lambda}{\partial x'} = 0$

1 person
pellman
I got this one straight now, guys. Thanks!