- #1
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Hi,
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)?
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)?