Is acceleration due to gravity a 4-vector?

  • Thread starter pellman
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  • #1
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We can write the geodesic condition as

[tex]\frac{d^2 x^\alpha}{d\tau^2}={-\Gamma^\alpha}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}[/tex]

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?
 

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  • #2
WannabeNewton
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##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.
 
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  • #3
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Thanks. Regarding [itex]\ddot{x}^\mu[/itex] ... where do I go wrong in the following?

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

[tex]\frac{dx^\mu}{d\tau} ={\Lambda^\mu}_\alpha \frac{dx'^\alpha}{d\tau}[/tex]

but if we take the derivative of both sides of

[itex]x^\mu={\Lambda^\mu}_\alpha x'^\alpha[/itex]

we get

[itex]\frac{dx^\mu}{d\tau}=\frac{d}{d\tau}\left({\Lambda^\mu}_\alpha x'^\alpha\right)[/itex]

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

[tex]\frac{d^2 x^\mu}{d\tau^2} ={\Lambda^\mu}_\alpha \frac{d^2 x'^\alpha}{d\tau^2}
[/tex]

and so [itex]\ddot{x}^\mu[/itex] transforms as a vector.

?????
 
  • #4
WannabeNewton
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##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.
 
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  • #5
stevendaryl
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but if we take the derivative of both sides of

[itex]x^\mu={\Lambda^\mu}_\alpha x'^\alpha[/itex]

we get

[itex]\frac{dx^\mu}{d\tau}=\frac{d}{d\tau}\left({\Lambda^\mu}_\alpha x'^\alpha\right)[/itex]

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

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 [itex]x'[/itex] to a coordinate [itex]x[/itex], then it's a nonlinear, continuous, invertible function:

[itex]x = F(x')[/itex]

So,

[itex]\frac{dx}{d \tau} = \frac{\partial F}{\partial x'} \frac{dx'}{d\tau}[/itex]

So even if [itex]F(x')[/itex] is nonlinear, the relationship between velocities is linear:

[itex]\frac{dx}{d \tau} =\Lambda \frac{dx'}{d\tau}[/itex]

where [itex]\Lambda = \frac{\partial F}{\partial x'}[/itex]

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

[itex]\frac{dx}{d \tau} =\Lambda \frac{dx'}{d\tau}[/itex] versus [itex]x = F(x')[/itex]

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

[itex]F(x') = \Lambda x'[/itex]

which is only possible when [itex]\frac{\partial \Lambda}{\partial x'} = 0[/itex]
 
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  • #6
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I got this one straight now, guys. Thanks!
 

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