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I'm trying to understand to get the following formula (first formula on pg 33) in Altland Simons second edition:

$$\Delta S \simeq \int d^m x (1 + \partial_{x_\mu} \, (\omega_a \, \partial_{\omega_a} \, x_{\mu} \, )) \, L \, (\phi^i + F_a^i \omega_a \, , \, (\delta_{\mu \nu} - \partial_{x_{\mu}}\, (\omega_a \, \partial_{\omega_a} \, x_{ \nu} \, )) \, \partial_{x_{\nu}} \, (\phi^i + F_a^i \, \omega_a)) \\ - \int d^m x \, L \, (\phi^i(x), \, \partial_{x_{\mu}} \, \phi^i (x))$$

Specifically how do we get the expression $\delta_{\mu \nu} - \partial_{x_{\mu}}\, (\omega_a \, \partial_{\omega_a} \, x_{\nu} \, )$?

This is related to the PSE question here, but I'm confused about the step before the question asked in that link. The summary given there is great and I have included it below:

Suppose we have a transformation: $$x^\mu \to (x^{\prime})^{\mu} = x^\mu + f^\mu_a \omega^a(x)$$ and $$\phi^i(x)\to (\phi^{\prime})^i =\phi^i(x) + F^i_a \omega^a(x)$$ then we can compute the action difference $$\Delta S = \int_V d^m x^\prime \mathcal{L}(\phi^\prime(x^\prime),\partial_{x^\prime} \phi^\prime(x^\prime))-\int_V d^m x \mathcal{L}(\phi (x),\partial_x \phi (x))$$ where we can express everything in terms of $x$ by using the transformation formulas and the Jacobi determinant.

where $f^\mu_a = \frac{\partial \, x_{\, \mu}}{\partial \, \omega_{\, a}} \, \Big\rvert_{\omega = 0}\, \,$.

I am confused about the case $\mu \neq \nu$. Then we have:

$$\frac{\partial \, x'_{\mu}}{\partial \, x_{\nu}} = \left(\frac{\partial}{\partial \, x_{\nu}} \right) \left(\omega_a \frac{\partial \, x_{\mu}}{\partial \, \omega_{a}}\right) = \left(\frac{\partial \, \omega_a}{\partial \, x_{\nu}} \frac{\partial \, x_{\mu}}{\partial \, \omega_{a}}\right),$$

where we get the last equality because $\partial x_{\mu} \, / \, \partial x_{\nu} = 0$ as $\mu \neq \nu$.

$$\frac{\partial \, x_{\nu}}{\partial \, x'_{\mu}} = \frac{1}{\frac{\partial \, x'_{\mu}}{\partial \, x_{\nu}}} = \frac{1}{\frac{\partial \, \omega_a}{\partial \, x_{\nu}} \frac{\partial \, x_{\mu}}{\partial \, \omega_{a}}} = \frac{\partial \, x_{\nu}}{\partial \, \omega_a} \frac{\partial \, \omega_{a}}{\partial \, x_{\mu}} = \partial_{x_{\mu}}\, (\omega_a \, \partial_{\omega_a} \, x_{\nu} \, ),$$

again where we get the last equality because $\partial x_{\mu} \, / \, \partial x_{\nu} = 0$ as $\mu \neq \nu$.

But the expression $\delta_{\mu \nu} - \partial_{x_{\mu}}\, (\omega_a \, \partial_{\omega_a} \, x_{\nu} \, ) = - \partial_{x_{\mu}}\, (\omega_a \, \partial_{\omega_a} \, x_{\nu} \, )$ for $\mu \neq \nu$ has a minus sign!

i) What am I doing wrong?

ii) How can I get the correct answer?

Physics Enthusiast
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1 Answers1

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I think your problem is using $\frac{\partial x_\nu}{\partial x'_\mu} = \frac{1}{\frac{\partial x'_\mu}{\partial x_\nu}}$. This is equivalent to saying $$ \frac{\partial x'_\mu}{\partial x_\nu}\frac{\partial x_\nu}{\partial x'_\mu} = 1 $$ where no sum is implied. However, the correct relation is $$ \sum_\nu \frac{\partial x'_\mu}{\partial x_\nu}\frac{\partial x_\nu}{\partial x'_\lambda} = \delta_{\mu\lambda} $$ that is the Jacobian matrices are related by a matrix inverse not a scalar inverse.

As you wrote above, $$ \frac{\partial x'_\mu}{\partial x_\nu} = \delta_{\mu\nu} + \sum_a \frac{\partial x^\mu}{\partial \omega_a} \frac{\partial \omega_a}{\partial x^\nu} = \delta_{\mu\nu} + \frac{\partial}{\partial x^\nu}\sum_a \omega_a\frac{\partial x^\mu}{\partial \omega_a} . $$

To find the inverse we may then write to lowest order in $\omega_a$, $$ \frac{\partial x_\nu}{\partial x'_\mu} \approx A_{\nu\mu} + B_{\nu\mu}. $$ And imposing $$ \delta_{\mu\lambda} =\sum_\nu \frac{\partial x'_\mu}{\partial x_\nu}\frac{\partial x_\nu}{\partial x'_\lambda} = \sum_\nu\left( \delta_{\mu\nu} + \frac{\partial}{\partial x^\nu}\sum_a \omega_a\frac{\partial x^\mu}{\partial \omega_a} \right)\left(A_{\nu\lambda} + B_{\nu\lambda}\right) $$ order by order $$ \delta_{\mu\lambda} = A_{\mu\lambda} \\ 0 = B_{\mu\lambda} + \sum_\nu A_{\nu\lambda} \frac{\partial}{\partial x^\nu}\sum_a \omega_a\frac{\partial x^\mu}{\partial \omega_a} = B_{\mu\lambda} + \frac{\partial}{\partial x^\lambda}\sum_a \omega_a\frac{\partial x^\mu}{\partial \omega_a} \implies B_{\mu\lambda} = -\frac{\partial}{\partial x^\lambda}\sum_a \omega_a\frac{\partial x^\mu}{\partial \omega_a}. $$

Thus, $$ \frac{\partial x_\nu}{\partial x'_\mu} \approx \delta_{\mu\nu} - \frac{\partial}{\partial x^\mu}\sum_a \omega_a\frac{\partial x^\nu}{\partial \omega_a} $$ as in Altland and Simons.

Zach R.
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