0
$\begingroup$

How do I show invariance under the target space Poincare transformations of the action for a relativistic string,

$$S=-\frac{1}{2 \pi \alpha'} \int{\text{d}^2 \zeta}\sqrt{-\det(\partial_{\alpha}X^{\mu}\partial_{\beta}X^{\nu}\eta_{\mu \nu})}~~?$$

This is specifically taken from West's "Introduction to strings and branes" and the world sheet reparameterisation invariance is explained; however this one isn't.

$\endgroup$
2
  • 1
    $\begingroup$ It looks to me like Lorentz symmetry is manifest while the presence of derivatives only implies translation invariance. That's all you need. $\endgroup$ – Danu May 27 '14 at 10:11
  • $\begingroup$ Related: physics.stackexchange.com/q/228111/2451 $\endgroup$ – Qmechanic Jan 24 '16 at 5:23
3
$\begingroup$

The Poincare group is really just the Lorentz group + translations. If you can show that the expression is invariant under rotations, boosts and translations, you're done.

This is fairly trivial in the case at hand since the appearance of contracted tensor indices only implies Lorentz invariance. In fact, that's one of the main reasons why we use tensor notation: To maintain manifest Lorentz invariance throughout calculations.

Lorentz invariance covers boosts and rotations, and translation invariance is immediately apparent if you realize the the expression consists of spacetime derivatives of $X^\mu$ only. Therefore, the expression is Poincare invariant.

$\endgroup$
0
0
$\begingroup$

Poincare symmetry group of d-dimensional flat spacetime (isometries of flat space) consists of ${\Lambda^{\mu}}_{\nu}$ (Lorentz transformations, i.e. satisfying $SO(d-1,1)$ algebra) and $K^{\mu}$ (translations i.e. commuting algebra). In Nambu-Goto action one interprets $X^{\mu}$ as "flat" spacetime vectors, so the action should be invariant under $X'^{\mu} = {\Lambda^{\mu}}_{\nu} X^{\nu} + K^{\mu}$. To check this note that $\partial_{\alpha} X'^{\mu} = {\Lambda^{\mu}}_{\nu} \partial_{\beta} X^{\nu}$, therefore: \begin{equation} \eta_{\mu \nu} \partial_{\alpha} X'^{\mu} \partial_{\beta} X'^{\nu} = {\Lambda^{\mu}}_{\gamma} \eta_{\mu \nu} {\Lambda^{\nu}}_{\sigma}\partial_{\alpha} X^{\gamma} \partial_{\beta} X^{\sigma} = \eta_{\gamma \sigma} \partial_{\alpha} X^{\gamma} \partial_{\beta} X^{\sigma} \end{equation} where the last equality comes from ${\Lambda^{\mu}}_{\gamma} \eta_{\mu \nu} {\Lambda^{\nu}}_{\sigma} = \eta_{\gamma \sigma}$ which is the definition of Lorentz transformations.

It is interesting to note that in Nambu-Goto action the above transformations should be considered as internal symmetries of fields $X^{\mu}(\sigma, \tau)$ on the world-sheet.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.