# 4-velocity transforms as a vector?

1. Jan 17, 2012

### haushofer

Hi,

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

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

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

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

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

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

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

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

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

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

2. Jan 17, 2012

### torquil

I haven't checked your calculations, but you must be careful about considering the basis of your space of 4-vectors and/or which coordinates your $\partial$s act on (new or old coords)?

3. Jan 17, 2012

### haushofer

The old coordinates of the path, I would say :) Compare it with the usual transformation of a vector V under the infinitesimal gct,

$$\delta V^{\mu}(x) \equiv V^{'\mu}(x) - V^{\mu}(x) = \xi^{\lambda}\partial_{\lambda} V^{\mu} - V^{\lambda}\partial_{\lambda} \xi^{\mu}$$

Here I just took

$$V^{\mu}(x) = \dot{x}^{\mu}(\tau)$$

4. Jan 17, 2012

### Bill_K

The issue is whether you want to calculate the change at the same physical point (3), or at points that have the same coordinate values (2).

5. Jan 17, 2012

### haushofer

Can you elaborate on that?

6. Jan 18, 2012

### haushofer

My question arises when you want to consider isometries of a certain background. If I have a particle,

$$S[x^{\rho}(\tau)] = - mc\int_{\tau_i}^{\tau_f} d\tau \sqrt{-g_{\mu\nu}(x)\dot{x}^{\mu}\dot{x}^{\nu}}$$

and I want to derive the isometries of the metric, then I consider the infinitesimal transformation (1),

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

while keeping the metric fixed. Because the metric is a function of the coordinates, this induces the transformation

$$g_{\mu\nu}(x) \rightarrow g_{\mu\nu}(x - \xi(x))$$

So Tayloring implies

$$\delta g_{\mu\nu} = -\xi^{\rho}\partial_{\rho}g_{\mu\nu}(x) \ \ \ \ \ \ \ (4)$$

which I fully understand (and which is of course NOT a general coordinate transformation!). What I don't understand completely, is: why do we use

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

to derive the isometries of the metric, and not
$$\delta \dot{x}^{\mu} = \xi^{\lambda}\partial_{\lambda} \dot{x}^{\mu} - \dot{x}^{\lambda}\partial_{\lambda} \xi^{\mu} \ \ \ \ (2)$$

?

Plugging (3) and (4) in the action implies that the Lie-derivative of the metric wrt xi should vanish, which is the expected result. I'd like to understand what goes wrong if we would use transformation (2).

7. Jan 18, 2012

### haushofer

I think I have some clearity now. It helps if one thinks in terms of sigma-models.

The point particle action defines a one-dimensional field theory. The fundamental fields are the x's (the path of the particle). We know that the x-dots transform under gct's as honest vectors, via formula (2). This is merely a field redefinition. If we do that, we must redefine the metric also. We will find that the action transforms as a scalar under such a redefinition.

In terms of sigma-models covariance is a pseudo-symmetry: it keeps the action invariant, hence it is a symmetry, but we cannot associate Noether charges to these symmetries, hence the "pseudo".

To derive isometries however conceptually we do something very different, but the confusing part is that the starting point is the same as for the field redefinitions: namely, formula (1)!

Last edited: Jan 18, 2012
8. Jan 19, 2012

### TrickyDicky

isn't this a passive versus active diffeomorphism issue? The way I understand it if you are considering a flat manifold or maybe restricting your problem to local isometries you don't need to use the Lie derivative wich is always an active diffeomorphism, because the passive-active distinction is not so relevant in flat manifolds, but if the manifold you are considering is curved and you want to physically move points in it in a not strictly infinitesimally local way then you need the Lie derivative because you are not simply changing coordinates (passive transf.), it is an active diffeomorphism. Hope this is not too confusing.

9. Jan 20, 2012

### haushofer

I can't reconcile that with my calculations I showed here; to show the isometries I explicitly do NOT act with a Lie derivative on the x-dot or the metric. I apply (3) and (4), which gives me that the Lie-derivative of the metric should vanish for these isometries.

10. Jan 20, 2012

### TrickyDicky

I guess I must have worded it confusingly, I was precisely trying to say why I think the Lie derivative must vanish for you isometries. IMO it is about the difference between a "local isometry" and an isometry, the isometries you seem to be talking about are local (you are dealing with infinitesimal transformations keeping the metric fixed).