# Co-ordinate vs. true acceleration

• masudr

#### masudr

In G.R. we do physics in any kind of frame, inertial or not. To compensate for that, we get nice little Christoffel symbols in all our derivatives. And straight lines become geodesics.

But is there truly no way to distinguish between co-ordinate acceleration and "real" acceleration? Or is there no such thing as real acceleration, since we can always transform it away? Or is is that in a reference frame where we did transform the acceleration of a particle away, geodesics would no longer be straight lines? But if the curvature tensor's components all vanished, then we know that the space is Minkowski, and hence we can distinguish between these two types of acceleration.

Hmm, I'm not sure if I've made sense here. Is every question [posed above] answerable by a resounding "yes" or have I made a mistake somewhere?

But if the curvature tensor's components all vanished, then we know that the space is Minkowski, and hence we can distinguish between these two types of acceleration.
It's all in the Riemannian.

Garth

But is there truly no way to distinguish between co-ordinate acceleration and "real" acceleration? Or is there no such thing as real acceleration, since we can always transform it away?

No coordinate transformation will tranform away the real 4-acceleration of an observer. Let $\left\{ x^\mu \left( \tau \right) \right\}$ be the coordinate description of an observer's worldline parametrized by proper time. Then the observer's 4-velocity is

$$u = \frac{dx^\mu}{d\tau} \frac{\partial}{\partial x^\mu},$$

and 4-acceleration is

$$\begin{equation*} \begin{split} a &= \nabla_u u \\ &= u^\mu \nabla_\mu \left( u^\nu \frac{\partial}{\partial x^\nu} \right) \\ &= u^\mu \left[ \left( \nabla_\mu u^\nu \right) \frac{\partial}{\partial x^\nu} + u^\nu \nabla_\mu \frac{\partial}{\partial x^\nu} \right] \\ &= u^\mu \left[ \frac{\partial u^\nu}{\partial x^\mu} \frac{\partial}{\partial x^\nu} + u^\nu \Gamma^{\alpha}_{\nu \mu} \frac{\partial}{\partial x^\alpha} \right] \\ &= u^\mu \left[ \frac{\partial u^\alpha}{\partial x^\mu} + u^\nu \Gamma^{\alpha}_{\nu \mu} \right] \frac{\partial}{\partial x^\alpha}. \end{split} \end{equation*}$$

If an observer's 4-acceleration is zero in one coordinate system (freely falling obsrever), it is zero in all coordinate systems; if an observer's 4-acceleration is non-zero in one coordinate system, it is non-zero in all coordinate systems.

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The operative word here of course is "4-acceleration".

In the OP it sounded as if 3-acceleration was being talked about.

Hmm, I'm not sure if I've made sense here.
Was that the cause of your confusion masdur?

Garth

The operative word here of course is "4-acceleration".

In the OP it sounded as if 3-acceleration was being talked about.

But, $1 = g \left( u , u \right)$ gives that

$$0 = \nabla_u g \left( u , u \right) = g \left( \nabla_u u , u \right) + g \left( u , \nabla_u u \right),$$

so

$$0 = g \left( a , u \right).$$

Thus, since the 4-velocity $u$ is timelike, the 4-acceleration $a$ is spacelike. Therefore, if the 4-acceleration is non-zero, the three "spatial" (assuming that three of the coordinates are spacelike and one is timelike) coordinates of the 4-acceleration cannot be simultaneously transformed away.

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$$\begin{equation*} \begin{split} a &= \nabla_u u \\ &= u^\mu \nabla_\mu \left( u^\nu \frac{\partial}{\partial x^\nu} \right) \\ &= u^\mu \left[ \left( \nabla_\mu u^\nu \right) \frac{\partial}{\partial x^\nu} + u^\nu \nabla_\mu \frac{\partial}{\partial x^\nu} \right] \\ &= u^\mu \left[ \frac{\partial u^\nu}{\partial x^\mu} \frac{\partial}{\partial x^\nu} + u^\nu \Gamma^{\alpha}_{\nu \mu} \frac{\partial}{\partial x^\alpha} \right] \\ &= u^\mu \left[ \frac{\partial u^\alpha}{\partial x^\mu} + u^\nu \Gamma^{\alpha}_{\nu \mu} \right] \frac{\partial}{\partial x^\alpha}. \end{split} \end{equation*}$$