According to this paper:

around equation (1)-(2), the DBI action

$$S = \int d^4 x\Big(-\lambda\sqrt{1 + (\partial \pi)^2} + \lambda\Big)\tag{1}$$

is invariant under the non-linearly realized symmetry whose infinitesimal form is

$$\delta_v\pi(x) = v_{\mu}x^{\mu} + \pi(x)v^{\alpha}\partial_{\alpha}\pi(x),\tag{2}$$

in the sense that the Lagrangian changes by a total derivative. I am having trouble confirming that this is true. I find that

$$\delta_v \mathcal{L} = \frac{\partial^{\mu}\pi\partial_{\mu}\delta_v\pi}{\sqrt{1 + (\partial \pi)^2}} = \frac{1}{\sqrt{1 + (\partial \pi)^2}}\partial^{\mu}\pi\Big(v_{\mu} + v^{\alpha}\partial_{\mu}\pi \partial_{\alpha}\pi + v^{\alpha}\pi\partial_{\mu}\partial_{\alpha}\pi\Big) = \frac{v^{\mu}}{\sqrt{1 + (\partial \pi)^2}}\Big(\partial_{\mu}\pi + \partial_{\mu}\pi\partial_{\beta}\pi\partial^{\beta}\pi + \pi\partial^{\beta}\pi\partial_{\beta}\partial_{\mu}\pi\Big),$$

which as far as I can tell, is not a total derivative. Moreover, in equation (5) the paper claims that any scalar $P$ constructed from $$g_{\mu \nu} = \eta_{\mu\nu} + \partial_{\mu}\pi\partial_{\nu}\pi\tag{2b}$$ should transform like

$$\delta_v P = v^{\alpha}\pi(x)\partial_{\alpha}P.\tag{5}$$

Since it only depends on the determinant of $g_{\mu \nu}$, the Lagrangian is a such a scalar, and I am finding that it does not transform in this fashion. Moreover even if it did, $\delta_v \mathcal{L}$ would only be a total derivative for constant $\pi$, which cannot be correct.


1 Answer 1

  1. The DBI action (1) is $$ \begin{align} S~=~&\int\!d^4x~{\cal L}, \cr {\cal L}~=~&\lambda(1-\sqrt{|g|}), \cr |g|~=&~-g~=~-\det g_{\mu\nu}~=~\det (\eta^{-1}g)^{\mu}{}_{\nu} ~=~\prod_n\lambda_n~=~ 1+(\partial \pi)^2,\end{align}\tag{1} $$ with metric $$ \begin{align} g_{\mu\nu}~=~&\eta_{\mu\nu}+\partial_{\mu}\pi ~\partial_{\nu}\pi, \cr (g^{-1})^{\mu\nu}~=~&(\eta^{-1})^{\mu\nu}-\frac{\partial^{\mu}\pi~ \partial^{\nu}\pi}{1+(\partial \pi)^2}.\end{align}\tag{2b}$$ To deduce the determinant (1) for $(\eta^{-1}g)^{\mu}{}_{\nu}$ note that $\partial_{\nu}\pi$ is a eigenvector with eigenvalue $1+(\partial \pi)^2$, and all the orthogonal eigenvectors carry eigenvalue $1$.

  2. Ref. 1 shows that the infinitesimal transformation $$\delta_v g_{\mu\nu}~\stackrel{(2)+(2b)}{=}~\ldots ~\stackrel{(4)}{=}~({\cal L}_{\xi}g)_{\mu\nu} \tag{3}$$ of the metric tensor is a Lie derivative wrt. a vector field $$ \xi^{\mu}~=~\pi v^{\mu}, \qquad v^{\mu} \text{ is independent of }x.\tag{4} $$

  3. We calculate that the change in the Lagrangian density $$ \delta_v{\cal L}~\stackrel{(1)}{=}~-\lambda \delta_v \sqrt{|g|}, \tag{A}$$ where $$\begin{align} \delta_v \sqrt{|g|}~=~&-\frac{\delta_v g}{2\sqrt{|g|}}\cr ~=~&\ldots~=~\frac{1}{2} \sqrt{|g|}(g^{-1})^{\mu\nu}\delta_v g_{\mu\nu}\cr ~\stackrel{(3)}{=}~&\ldots~=~\frac{1}{2} \sqrt{|g|}\left((g^{-1})^{\mu\nu}\xi[ g_{\mu\nu}] + 2\partial_{\mu}\xi^{\mu}\right) \cr ~=~&\ldots~=~\partial_{\mu}\left(\xi^{\mu}\sqrt{|g|}\right) \end{align}\tag{B}$$ is a total spacetime derivative, as Ref. 1 claims. Therefore, the infinitesimal transformation $\delta_v$ is a quasi-symmetry.

  4. The result (B) is completely dictated by the fact that $\sqrt{|g|}$ is a scalar density. Similar any scalar function $P$ must transform as $$ \delta_v P~=~ {\cal L}_{\xi}P~=~\xi[P],\tag{5} $$
    as Ref. 1 claims.


  1. C. de Rham & A.J. Tolley, arXiv:1003.5917.
  • $\begingroup$ Thanks! 1. To summarize, it looks like your equation (B) reduces to the computation in the question. 2. However, it's clearly much easier to guess that it would be a total derivative from the geometric perspective. 3. Also, I guess your equation (5) does not apply to the Lagrangian density exactly because it is a density and not a scalar. $\endgroup$ Aug 7, 2020 at 15:37
  • $\begingroup$ Thanks for the correction of eq. (2b). 1. There seems to be missing a determinant in the question formulation. 2. Yes. 3. Yes. $\endgroup$
    – Qmechanic
    Aug 8, 2020 at 8:48
  • $\begingroup$ Regarding the missing determinant, I believe the expansion of $\text{det}g_{\mu \nu}$ terminates for a single scalar field, that is $\text{det} (1 + \partial_{\mu}\pi\partial_{\nu}\pi) = 1 + (\partial \pi)^2$. So I think the action as written is correct, without any implicit determinant. For multiple scalars the determinant is necessary. $\endgroup$ Aug 9, 2020 at 19:21
  • $\begingroup$ My bad. You're right. I updated the answer. $\endgroup$
    – Qmechanic
    Aug 10, 2020 at 9:11

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