Transformation question; first order shift of a scalar field

[llorgos]

Hi to all! I have the following transformation

[itex] \tau \to \tau' = f(\tau) = t - \xi(\tau). [/itex]

Also I have the action

[itex] S = \frac{1}{2} \int d\tau ( e^{-1} \dot{X}^2 - m^2e) [/itex]

where [itex] e = e(\tau) [/itex]. Then in the BBS String book it says that

$$ {X^{\mu}}' ({\tau}') = X^{\mu}(\tau)$$

and that the first order shift is

$$ \delta X^{\mu} = {X^{\mu}}'(\tau) - X^{\mu}(\tau) = \xi(\tau)\dot{X}^{\mu}$$

Can someone explain why this is true? How can I realize it?

Then it say similarly for the ##e(\tau)## that

$$ \delta e = e'(\tau) - e(\tau) = \frac{d}{d\tau} (\xi e) $$

again, I cannot see how this comes!

Thank you in advance.

 

[pat_connell]

The first equation $$ {X^{\mu}}' ({\tau}') = X^{\mu}(\tau)$$ follows from the fact that the transformation is a reparametrization. The definition of a reparametrization is that it preserves the form of a function, thus in this case it preserves the form of the string coordinates. To prove this, just take a look at the definitions: \begin{align*}{X^{\mu}}' ({\tau}') &= X^{\mu}\left(f^{-1}(\tau')\right)\\&= X^{\mu}\left(t - \xi(\tau)\right)\\&= X^{\mu}(\tau).\end{align*}As for the second equation, note that by definition $\delta X^{\mu}$ is the change in $X^{\mu}$ due to the transformation, so you can just calculate the difference between the two sides and set it equal to $\delta X^{\mu}$. Again, using the definitions of the transformation and $\delta X^{\mu}$, you get\begin{align*}\delta X^{\mu} &= {X^{\mu}}'(\tau) - X^{\mu}(\tau)\\&= X^{\mu}\left(f^{-1}(\tau')\right) - X^{\mu}(\tau)\\&= X^{\mu}\left(t - \xi(\tau)\right) - X^{\mu}(\tau)\\&= \xi(\tau) \dot{X}^{\mu}.\end{align*}For the third equation, note that again, $\delta e$ is the change in $e$ due to the transformation. Thus, you can calculate the difference between the two sides and set it equal to $\delta e$. Again, using the definitions of the transformation and $\delta e$, you get\begin{align*}\delta e &= e'(\tau) - e(\tau)\\&= e\left(