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I have a very basic question, which confuses me, probably due to sloppiness.

A particle follows a path [itex]x^{\mu}(\tau)[/itex] through spacetime. We know that the 4-velocity at a point,

[tex]

\dot{x}^{\mu}(\tau) \equiv \frac{d x^{\mu}}{d\tau}(\tau)

[/tex]

is the tangent vector to the path at that point. Now we perform the infinitesimal coordinate transformation

[tex]

\delta x^{\mu} = - \xi^{\mu}(x) \ \ \ \ \ (1)

[/tex]

On the one hand I would say that, being a vector, the 4-velocity transforms as

[tex]

\delta \dot{x}^{\mu} = \xi^{\lambda}\partial_{\lambda} \dot{x}^{\mu} - \dot{x}^{\lambda}\partial_{\lambda} \xi^{\mu} \ \ \ \ (2)

[/tex]

(It's just the Lie derivative). On the other hand, I can differentiate (1) and use the chain rule to get

[tex]

\delta \dot{x}^{\mu} = -\dot{\xi}^{\mu}(x) = -\dot{x}^{\lambda}\partial_{\lambda}\xi^{\mu}(x) \ \ \ \ \ (3)

[/tex]

How to reconcile (2) and (3)? What happened to the first term of (2)?

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# 4-velocity transforms as a vector?

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