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rigorous treatment of infinitesimal reparametrizations

+ 3 like - 0 dislike
139 views

my first post :) I am asking this directed to mathematicians or mathematical physicists since I don't like the usual physics approach.

Reading some string theory books I always find that the introductory chapters discuss the relativistic free particle (see Lüst-Theisen's Lectures on String Theory, or Becker-Becker-Schwarz, page 21, exercise 2.3).

Then they go on about showing that the action $S=-m\int^{t_1}_{t_2} ds = -m \int^{t_1}_{t_2}d\tau\sqrt{-\frac{dx^\mu}{d \tau} \frac{dx^\nu}{d \tau} \eta_{\mu\nu}}$ is invariant under "infinitesimal reparametrizations" $ \tau \rightarrow \tau' = \tau + \xi(\tau) $, for this they just Taylor expand $ x'^{\mu}(\tau')=x^{\mu}(\tau) $ around $\tau$ and drop terms of order $O(\xi(\tau)^2)$ to find the function "shift" $\delta x^\mu (\tau)= x'^\mu (\tau)- x^\mu(\tau) = -\xi(\tau)\partial_\tau x^\mu(\tau)$ Two things I find annoying (even though that's how I learned it as a physicist):

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

However, when taking the tau derivative to this: $ \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') $.

Now, multiplying by $-\xi(\tau)$ to get the shift of the function: $ \delta x^\mu (\tau)= -\xi(\tau)\partial_\tau x^\mu(\tau)+\xi(\tau)\cdot \partial_\tau\xi(\tau)\cdot\partial_{\tau'}x'^\mu(\tau') $ 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 $ x^\mu $, maybe in terms of pushforwards and such (at the rigor of mathematics)? This would also clarify much of the above paragraph.

Thanks for any help


This post imported from StackExchange Mathematics at 2015-05-10 11:08 (UTC), posted by SE-user arestes

asked Jun 10, 2013 in Mathematics by arestes (15 points) [ revision history ]
edited May 10, 2015 by Dilaton

2 Answers

+ 4 like - 0 dislike

$\xi(\tau)$ infinitesimal means $\xi(\tau) = \epsilon \xi_0(\tau)$ for some finite $\xi_0(\tau)$ and $\epsilon$ infinitesimal. Then all computations have to be expanded in powers of $\epsilon$. In particular we have $\partial_\tau \xi_\tau = \epsilon \partial_\tau \xi_0(\tau)$ which is of first order in $\epsilon$. The key point is that when you compute a differential, you have to introduce such parameter $\epsilon$ and if you are differentiating a function defined on a space of paths parametrized by some $\tau$ then $\epsilon$ has nothing to do with $\tau$. In particular, any derivative with respect of $\tau$ of an expression linear in $\epsilon$ will be still linear in $\epsilon$.

All that is purely rigorous mathematics (it is essentially the definition of the derivative). The tangent space to the space of maps $\tau \mapsto x(\tau)$ is the space of maps $\tau \mapsto \xi_0(\tau)$ where $\xi_0(\tau)$ is a vector field at the point $x(\tau)$ for every $\tau$.

answered May 10, 2015 by 40227 (4,660 points) [ revision history ]
+ 1 like - 0 dislike

To get the correct result, repeat your argument with $\epsilon\xi(\tau)$ in place of $\xi(\tau)$, with fixed $\xi(\tau)$ and only $\epsilon$ as infinitesimal.

Alternatively, show invariance under noninfinitesimal diffeomorphisms $z(\tau)$, with finite reparameterizations $x^\mu(\zeta(\tau))$, which doesn't involve any limiting process. Then deduce the infinitesimal invariance from that.

answered May 10, 2015 by Arnold Neumaier (12,385 points) [ revision history ]

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