Rigorous treatment for deforming functions in string theory

[arestes]

Hi!
Reading some string theory books I always find that the introductory chapters discuss the relativistic free particle.
Then they go on about showing that the action
[itex]S=-m\int^{t_1}_{t_2} dx = -m \int^{t_1}_{t_2}d\tau\sqrt{-\frac{dx^\mu}{d \tau} \frac{dx^\nu}{d \tau} \eta_{\mu\nu}}[/itex]
is invariant under "infinitesimal reparametrizations"
[itex] \tau \rightarrow \tau' = \tau + \xi(\tau) [/itex],
for this they just Taylor expand
[itex] x'^{\mu}(\tau')=x^{\mu}(\tau) [/itex]
around [itex]\tau[/itex] and drop terms of order [itex]O(\xi(\tau)^2)[/itex] to find the function "shift"
[itex]\delta x^\mu (\tau)= x'^\mu (\tau)- x^\mu(\tau) = -\xi(\tau)\partial_\tau x^\mu(\tau)[/itex]
Two things I find annoying (even though that's how I learned it as a physicist):

1) Expanding [itex] x'^{\mu}(\tau')=x^{\mu}(\tau) [/itex] we get [itex]x'^\mu(\tau)+ \xi(\tau)\partial_\tau x'^\mu(\tau)=x^\mu(\tau)[/itex] therefore
[itex]\delta x^\mu (\tau)= x'^\mu (\tau)- x^\mu(\tau) = -\xi(\tau)\partial_\tau x'^\mu(\tau)[/itex] which is not the above result. The justification given in some lecture notes (this trick is also widely used in General Relativity) recall that
[itex] x'^\mu(\tau)= x'^\mu(\tau'-\xi(\tau))=x'^\mu(\tau')-\xi(\tau)\partial_{\tau'} x'^\mu(\tau')=x^\mu(\tau)-\xi(\tau)\partial_{\tau'} x'^\mu(\tau')[/itex].

However, when taking the tau derivative to this:
[itex] \partial_\tau x'^\mu(\tau)=\partial_\tau x^\mu(\tau)-\partial_\tau\xi(\tau)\cdot\partial_{\tau'} x'^\mu(\tau') - \xi(\tau)\cdot\partial_\tau\partial_{\tau'} x'^\mu(\tau') [/itex].

Now, multiplying by [itex]-\xi(\tau)[/itex] to get the shift of the function:
[itex] \delta x^\mu (\tau)= -\xi(\tau)\partial_\tau x^\mu(\tau)+\xi(\tau)\cdot \partial_\tau\xi(\tau)\cdot\partial_{\tau'}x'^\mu(\tau') [/itex] where I ommited the third term as it is quadratic in xi. However, the term that has the derivative of xi cannot be ommited since the derivative of an "infinitesimal" quantity doesn't necessarily have to be infinitesimal. Something being small does not imply that its derivative is small.

As you see, this all boils down to the heuristic treatment that Physics books give to the "infinitesimal" variation.

2) How can I reformulate rigorously the idea of a "shift" of the function [itex] x^\mu [/itex], maybe in terms of pushforwards and such? This would also clarify much of the above paragraph.

Also, is this the right subforum to post this topic?
Thanks for any help

 

[pat_connell]

.Yes, this is the right forum to ask your question!Regarding your first point, the heuristic treatment of "infinitesimal" variations is common in Physics, but you are correct that it can be problematic. One way to make it more rigorous is to consider the limit of infinitesimal variations, keeping track of higher order terms and taking the limit appropriately. Regarding your second point, the idea of a "shift" of a function can be formulated rigorously in terms of a pushforward. A pushforward is a mathematical operation which takes a function (or vector field) defined on a space (e.g. a manifold) and pushes it forward along a map to another space. In this context, the map would be the infinitesimal variation of the parameter (e.g. the coordinate). You can find more information about pushforwards in any good book on differential geometry or topology.