Functional and total variations in einbein action

Question

I'm currently studying String theory by Becker& Becker, Schwarz textbook. The exercise 2.3 consists in verifying diffeomorphism invariance of einbein action wich is given by

$$ S_0 = \frac{1}{2} \int d\tau (e^{-1}\dot{X}^2 - m^2e), \ e= e(\tau) \tag1 $$

such that under a diffeomorphism $\tau'({\tau}), \ e(\tau) d\tau = e'(\tau')d\tau'$.

At the beginning of solution, the author states that the embedding functions $X^\mu: \mathbb{R}^2 \longrightarrow \mathbb{R}$ transform as scalar fields $(X'^\mu(\tau') = X^\mu(\tau))$under diffeomorphisms, in particular, "infinitesimal" diffeomorphisms $\tau'(\tau) = \tau - \xi(\tau)$ wich is totally reasonable to me.

The problem is that the author also states that the first-order variation on the functions should be $\delta X^\mu = X'^\mu(\tau) - X^\mu(\tau)=\xi(\tau)\dot{X}^\mu$. As I understand, this is a functional variation, such that the total variation should be $\delta_T X^\mu = \delta X^\mu - \xi(\tau)\dot{X}^\mu= 0$. Why did the author ignored the coordinate change in $X^mu$ in order to compute later the variation of the action?

There is also a problem concerning the difference between functional and total variations: the author uses the property $e(\tau) d\tau = e'(\tau')d\tau'$ to obtain

$$\delta e = \frac{d}{d\tau} (\xi e). \tag2$$ From my computations, I've obtained the above result and that the total variation of $e$ should be $\delta_T e= \dot{\xi}e$. Why are only functional variations used instead of total variations in order to show that $(1)$ is diff. invariant?

The results mentioned above can be found in Basic Concepts in String Theory by Blumenhagen, and the classical WSG Superstring Theory textbook.

Since the results are present in three separate books, I suppose I must be making a serious mistake. My knowledge of variational calculus can be summarized in chapter 4 of Classical Fields by R. Aldrovandi and J. G. Pereira.

Answer 1

Regarding the variation of $X^\mu(\tau)$ under worldline reparameterizations I'm not sure I understood what you are confused about, but the way it is derived is by noticing that under the reparameterization $\tau\mapsto \tau'(\tau)$ the fields transform as ${X'}^\mu(\tau')=X^\mu(\tau)$.

In that case considering one infinitesimal reparameterization $\tau'(\tau)=\tau-\xi(\tau)$ we have that since $\xi$ is infinitesimal $\tau(\tau')=\tau'+\xi(\tau')$ and therefore $$X'^\mu(\tau')=X^\mu(\tau'+\xi(\tau')).$$

Now denoting $\tau'$ as $\tau$ in the above equation and Taylor expanding the RHS we have $$X'^\mu(\tau)=X^\mu(\tau)+\xi(\tau)\dfrac{dX^\mu}{d\tau}\Longrightarrow \delta X^\mu(\tau)=\xi(\tau)\partial_\tau X^\mu$$

where in the last step we recognize $\delta X^\mu = X'^\mu(\tau) - X^\mu(\tau)$, the difference at same $\tau$ between transformed and unstransformed fields.

About the einbein it works in the same way. In fact $e(\tau)d\tau=e'(\tau')d\tau'$ implies the one-form transformation law $$e(\tau)\dfrac{d\tau}{d\tau'}=e'(\tau')$$ and writing $\tau =\tau'+\xi(\tau')$

$$e(\tau'+\xi(\tau'))(1+\partial_{\tau'}\xi(\tau'))=e'(\tau')$$

Now just denoting $\tau'$ as $\tau$ and Taylor expanding the einbein on the LHS gives $$(e(\tau)+\partial_\tau e(\tau)\xi(\tau))(1+\partial_{\tau}\xi(\tau))=e'(\tau)$$

fully expanding the LHS and discarding second order terms in $\xi(\tau)$ it is

$$e(\tau)+e(\tau)\partial_\tau\xi+\partial_\tau e(\tau) \xi(\tau)=e'(\tau)\Longrightarrow \delta e(\tau)=\partial_\tau (e(\tau)\xi(\tau))$$

where we again recognized the variation $\delta e(\tau)=e'(\tau)-e(\tau)$ as the difference at same $\tau$ of transformed and untransformed fields, and where we recognized a derivative of a product in the process.