Consider a single scalar field $\phi$ on a manifold $\mathcal{M}$. Suppose in $\{x^\mu\}$ co-ordinates, the Lagrangian density is $\mathcal{L}(\phi, \frac{\partial \phi}{\partial x^\mu})$. This means that in $\{x'^\mu\}$ co-ordinates, the Lagrangian density is $$J\mathcal{L}(\phi, \frac{\partial \phi}{\partial x'^\nu}\frac{\partial x'^\nu}{\partial x^\mu}),\tag{1}$$ where $J$ is the determinant of the Jacobian for the co-ordinate transformation from $\{x^\mu\}$ to $\{x'^\mu\}$ i.e. $$J = \det|\frac{\partial x^\mu}{\partial x'^\nu}|.\tag{2}$$

The Euler-Lagrange equations in $\{x'^\mu\}$:

$$\frac{d }{d x'^\mu}\left(\frac{\partial (J\mathcal{L})}{\partial \left(\frac{\partial \phi}{\partial x'^\mu}\right)}\right) - J\frac{\partial \mathcal{L}}{\partial \phi} = 0\tag{3}$$

is normally obtained by varying $J\mathcal{L}$ with respect to $\phi$. Can it be obtained instead by from the Euler-Lagrange equations in $\{x^\mu\}$:

$$\frac{d }{d x^\mu}\left(\frac{\partial \mathcal{L}}{\partial \left(\frac{\partial \phi}{\partial x^\mu}\right)}\right) - \frac{\partial \mathcal{L}}{\partial \phi} = 0\tag{4}$$

simply by a change of co-ordinates $\{x^\mu\} \rightarrow \{x'^\mu\}$?


2 Answers 2


Under general coordinate transformations $$ x^{\mu}\quad\longrightarrow\quad x^{\prime \nu}~=~f^{\nu}(x) \tag{A},$$ the Lagrangian density $$ {\cal L}\quad\longrightarrow\quad{\cal L}^{\prime}~=~\frac{{\cal L}}{J}, \qquad J~:=~\det M, \qquad M^{\nu}{}_{\mu}~:=~\frac{dx^{\prime\nu}}{dx^{\mu}}, \tag{B}$$ transforms as a density; the field
$$ \phi~=~ \phi^{\prime} \tag{C}$$ is a scalar; the derivative $$ \phi_{,\mu}~:=~\frac{d\phi}{dx^{\mu}}~=~\frac{d\phi}{dx^{\prime\nu}} M^{\nu}{}_{\mu}\tag{D}$$ is a one-form/covector; the de Donder polymomentum density $$ \pi^{\mu}~:=~\frac{\partial {\cal L}}{\partial \phi_{,\mu}} \quad\longrightarrow\quad \pi^{\prime\mu} ~=~ J^{-1}M^{\nu}{}_{\mu} \pi^{\mu} \tag{E}$$ transforms as a vector density; and the functional derivative/Euler-Lagrange (EL) expression $$ \frac{\delta S}{\delta \phi}~=~\frac{\partial {\cal L}}{\partial \phi} - \frac{d\pi^{\mu}}{dx^{\mu}} \quad\longrightarrow\quad J^{-1}\frac{\delta S}{\delta \phi} \tag{F} $$ transforms as a density, cf. OP's question.

Check of eq. (F): That the first term $\frac{\partial {\cal L}}{\partial \phi}$ in eq. (F) transforms as a density is obvious. The transformation property of the second term $\frac{d\pi^{\mu}}{dx^{\mu}}$ in eq. (F) can be checked via the following calculation: $$\frac{d\pi^{\prime \nu}}{dx^{\prime \nu}} ~\stackrel{(E)}{=}~\frac{dx^{\mu}}{dx^{\prime \nu}}\frac{d}{dx^{\mu}}\left(J^{-1} \frac{dx^{\prime \nu}}{dx^{\lambda}} \pi^{\lambda} \right) ~=~J^{-1}\frac{d\pi^{\mu}}{dx^{\mu}} + \frac{d(J^{-1})}{dx^{\lambda}}\pi^{\lambda} + J^{-1}(M^{-1})^{\mu}{}_{\nu} \frac{d M^{\nu}{}_{\mu}}{dx^{\lambda}} \pi^{\lambda} $$ $$\stackrel{(H)}{=}J^{-1}\frac{d\pi^{\mu}}{dx^{\mu}} - J^{-1} \frac{d\ln\det M}{dx^{\lambda}}\pi^{\lambda} + J^{-1} \frac{d{\rm tr}\ln M }{dx^{\lambda}} \pi^{\lambda}~\stackrel{(I)}{=}~J^{-1}\frac{d\pi^{\mu}}{dx^{\mu}} ,\tag{G} $$ where we used $$ \mathrm{d} {\rm tr} \ln M ~\stackrel{M={\bf 1}-A}{=}~ - \mathrm{d} {\rm tr} \sum_{n=1}^{\infty} \frac{A^n}{n} ~=~ - {\rm tr} \sum_{n=1}^{\infty} A^{n-1} \mathrm{d}A ~\stackrel{M={\bf 1}-A}{=}~ {\rm tr} (M^{-1} \mathrm{d}M), \tag{H}$$ and the well-known formula $$\ln\det M ~=~{\rm tr}\ln M\qquad\stackrel{M=e^A}{\Leftrightarrow}\qquad \det e^A ~=~e^{{\rm tr}A}.\tag{I}$$ $\Box$

  • $\begingroup$ Thank you very much for this very insightful answer. 1. Can you tell me why $(M^{-1})^\mu_\nu \frac{dM^\nu_\mu}{dx^\lambda} = \frac{d \text{tr} \ln M}{dx^\lambda}$, and why $\frac{d\ln \det M}{dx^\lambda} = \frac{d \text{tr} \ln M}{dx^\lambda}$? 2. Furthermore, I don't understand why you have edited all the partial derivatives in my original formulation of the question into direct derivatives. I think the partial derivative version is correct? $\endgroup$
    – Si Chen
    Dec 2, 2018 at 19:57
  • $\begingroup$ 1. I updated the answer. 2. See my Phys.SE answer here. $\endgroup$
    – Qmechanic
    Dec 2, 2018 at 20:13

To answer this simply write your action in a coordinate invariant way. All fields $\partial_\mu\phi$ will need to be contracted by the metric or some other tensor, so that the total Lagrangian density is a scalar multiplied by $\sqrt{-g}$. Also the partial derivatives $\partial_\mu \phi$ can also be replaced by covariant derivatives $\nabla_\mu \phi$ since $\phi$ is a scalar.

For instance, for a Klein Gordon field we have $$S=\int d^4x \sqrt{-g} \frac{1}{2}\left(g^{\mu\nu}\nabla_\mu\phi\nabla_\nu\phi -m^2\phi^2\right)=\int d^4 x \sqrt{-g} L(\phi,\nabla_\mu\phi)$$

This $L$ is more convenient because it is a scalar and doesn't change by Jacobians (the $\sqrt{-g}$ takes care of that).

Note the metric $g$ can be flat, but the point of doing this is it's easy to see how things change under coordinate transformations.

Now find the equations of motion by doing integration by parts with the covariant derivative, which is more convenient because it is metric compatible and we have the identity $$\sqrt{-g}\nabla_\mu V^\mu = \partial_\mu(\sqrt{-g}V^\mu)$$ for any vector $V$.

So the equations of motion are $$\nabla_\mu\left(\frac{\partial L}{\partial \nabla_\mu \phi}\right)-\frac{\partial L}{\partial \phi}=0$$

For the Klein-Gordon example this is, $$g^{\mu\nu}\nabla_\mu\nabla_\nu\phi + m^2\phi=0$$ In flat space this is the usual $$\eta^{\mu\nu}\partial_\mu\partial_\nu \phi +m^2\phi=0,$$ but you see that you can't just change the coordinates in the partial derivatives in this expression because that forgets about the connection coefficients inside the covariant derivatives.

In your first way of writing the equations of motion with $J$ the connection coefficients will appear when you take partial derivatives of $J$ (due to that identity involving derivatives of $\sqrt{-g}$ I wrote above).


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