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?

 

[WannabeNewton]

$a^{\mu} = \ddot{x}^{\mu} + \Gamma^{\mu}_{\alpha\beta}\dot{x}^{\alpha}\dot{x}^{\beta}$ 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.

 

[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&#039;^\alpha}{d\tau^2}<br />

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

?

 

[WannabeNewton]

$x^{\mu}$ 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.

 

[stevendaryl]

but if we take the derivative of both sides of

x^\mu={\Lambda^\mu}_\alpha x&#039;^\alpha

we get

\frac{dx^\mu}{d\tau}=\frac{d}{d\tau}\left({\Lambda^\mu}_\alpha x&#039;^\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&#039; to a coordinate x, then it's a nonlinear, continuous, invertible function:

x = F(x&#039;)

So,

\frac{dx}{d \tau} = \frac{\partial F}{\partial x&#039;} \frac{dx&#039;}{d\tau}

So even if F(x&#039;) is nonlinear, the relationship between velocities is linear:

\frac{dx}{d \tau} =\Lambda \frac{dx&#039;}{d\tau}

where \Lambda = \frac{\partial F}{\partial x&#039;}

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

\frac{dx}{d \tau} =\Lambda \frac{dx&#039;}{d\tau} versus x = F(x&#039;)

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

F(x&#039;) = \Lambda x&#039;

which is only possible when \frac{\partial \Lambda}{\partial x&#039;} = 0

 

[pellman]

I got this one straight now, guys. Thanks!