- 61

- 0

Hi!

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

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 (at the rigor of mathematics)? This would also clarify much of the above paragraph.

Thanks for any help

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

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 (at the rigor of mathematics)? This would also clarify much of the above paragraph.

Thanks for any help

Last edited: