# How to find the continuous transformations which leave the action invariant?

Assume one has a continuous transformation of fields, and also of coordinates - in case if we consider coordinate transformations as well. Global internal symmetries, rotations, translations, dilatations, whatever, or any other. How do we generally check whether the action is invariant under such a transformation?

I feel like my question is way more basic than Noether theorem, charges, currents, etc.

To be more precise. Let's change the fields (and probably the coordinates) as: \begin{align} x &\to x + \xi(x)\\ \phi(x) &\to \phi(x) + \delta\phi(x) \end{align} where $$\xi(x)$$ and $$\delta \phi(x) = F[\phi(x),\partial\phi(x)]$$ have certain functional form. I would like to know, whether the action $$$$S = \int \operatorname{d}^D x\, \mathcal{L}[g^{\mu\nu}, \phi(x), \partial\phi(x)]$$$$ is form-invariant under such a transformation. In other words, whether it will go into something like $$$$\tilde{S} = \int \operatorname{d}^D \tilde{x}\, \mathcal{L}[\tilde{g}^{\mu\nu}, \tilde{\phi}(\tilde{x}), \partial\tilde{\phi}(\tilde{x})]$$$$ (notice no tilde above $$\mathcal{L}$$! I guess, that is form-invariance...)

So far, it seems to me that the obvious option - to calculate $$\delta S$$ does not give anything (that's what we do when deriving Noether currents). On classical trajectories, we end up with smth like $$\delta S = \partial(\ldots)$$ - which if of course correct, but does not help to answer my question at all.

I feel like my question should be related to Killing vectors... In various textbooks I found how the Lie derivatives are applied to the metric in order to determine which transformations preserve it. Seems like I'm interested in a similar procedure for the action.

As usually, any references greatly appreciated.

UPDATE

Since the original question can be answered in a brute force way - "just plug your transformations into the Lagrangian and see how it goes", let me ask it in a slightly more general way:

Given the action, what is the procedure to find all continuous symmetries which leave it form-invariant?

• – Qmechanic Nov 27 '16 at 7:42

I will answer for the 1-D case, or particle mechanics, instead of field mechanics, but the idea is the same. The approach is similar to that of getting the Killing vector field of a metric, and this approach reduces to that when applied to purely kinetic Lagrangians. The objective is to get the so called Rund-Trautman identity

## Theory and set up

Let us assume the following: the system is characterised by a configuration manifold $\mathcal Q$ of dimension $n\equiv dim \mathcal Q$ and a Lagrangian $L:T\mathcal Q \times\mathbb R\rightarrow \mathbb R$, possibly dependent on time.

Definition.- We say that a Lagrangian is (quasi-)invariant with respect to the $r-$parameter transformation $$\bar{x} = \phi(x,t) = x + \varepsilon^s\xi_s + o(\varepsilon) \simeq x^i + \varepsilon^s\xi_s^i + o(\varepsilon) ,$$ $$\bar{t} = \psi(x,t) = t + \varepsilon^s\tau_s + o(\varepsilon) \simeq t + \varepsilon^s\tau_s,$$ with $s=1,\dots,r$ ; $\simeq$ means that we drop the higher orders on $\varepsilon$ and we use an index expression; and (quasi-) refers to if there is a divergence term $dG$ or not; if and only if

$$S[x(t)]- S[\bar{x}(\bar t)] = \varepsilon^s G_s(x(t),t)|_a^b + o(\varepsilon).$$

Lemma.- A Lagrangian L is invariant under a transformation $\iff$ the following $k$ equations hold:

$$\frac{\partial L}{\partial t} \tau_s + L\frac{d \tau_s}{dt} + \frac{\partial L}{\partial x^i} \xi^i_s +\frac{\partial L}{\partial \dot{x}^i}(\frac{d \xi^i_s}{dt} - \dot x^i \frac{d\tau_s}{dt} ) = \frac{d}{dt}G_s.$$

Sketch of the Proof

We can write the invariance condition as if we change the integration over $\bar t$ to $t$ in $S[\bar x]$ as $$L\left(\bar{x}(\bar t),\frac{d}{d\bar t}\bar{x}(\bar t), \bar{t}\right)\frac{d\bar{t}}{dt} - L(x,\dot x, t) \simeq \varepsilon^s\frac{d}{dt}G_s(x,t).$$
where $G_s$ is a divergence term. After this differentiate this equation w.r.t. $\varepsilon^s$ at $\varepsilon^s=0$.

This lemma gives us a general relation between the transformation and the Lagrangian. It can be used in different ways:

1. To check if a known transformation $(\phi,\psi)$ is a symmetry of a known Lagrangian $L$, and from here derive the Noether conserved quantities.
2. If the Lagrangian is unknown, but the transformations are, you get a system of $r$ PDE's on the Lagrangian L, an can be used to impose symmetries,
3. Lastly, we can find the symmetry transformations of a given Lagrangian L, considering that the $\dot x^i$ and their powers are independent, so the coefficients in the polynomial $P(x^i)$ are a system of PDE's to obtain $\xi$ and $\tau$.

The third point of view is the one that you want to find the symmetries of a Lagrangian.

## Application

In the case of a natural Lagrangian: $$L \equiv \frac{1}{2} g_{ij}(x)\dot{x}^i\dot{x}^j - V(x)$$

we compute the derivatives $$\partial_t L =0, \;\;\;\; \partial_{k}L = \frac{1}{2}\partial_{k}g_{ij}\dot{x}^i\dot{x}^j - \partial_{k}V, \;\;\;\; \partial_{,k}L = \frac{1}{2}g_{il}\left(\dot{x}^i\delta^l_k+\dot{x}^l\delta^i_k\right)$$ Then the equations become:

$$0\cdot \tau_s + \left( \frac{1}{2} g_{ij}\dot{x}^i\dot{x}^j - V \right) \frac{d \tau_s}{dt} + \left( \frac{1}{2}\partial_{k}g_{ij}\dot{x}^i\dot{x}^j - \partial_{k}V \right) \xi^k_s + \frac{1}{2}g_{il}\left(\dot{x}^i\delta^l_k+\dot{x}^l\delta^i_k\right) \left( \frac{d \xi^k_s}{dt} - \dot x^k \frac{d\tau_s}{dt} \right) = \frac{d}{dt}G_s.$$ taking the powers of $\dot x$ as independent and demanding the equations to be satisfied always, we get the first order PDE's:

$$1 \; : \;\; V\partial_t \tau_s -\partial_k V\xi^k_s = \partial_tG_s$$

$$x^i \; : \;\; - V \partial_i\tau_s + \frac{1}{2}\left(g_{il}\partial_t\xi^l_s+g_{li}\partial_t\xi^l_s\right) = \partial_iG_s$$

$$\dot x^i \dot x^j \; : \;\; -\frac{1}{2} g_{ij} \partial_t\tau_s + \frac{1}{2}\partial_{k}g_{ij} \xi^k_s + \frac{1}{2}\left( g_{il}\partial_j\xi^l_s + g_{lj}\partial_i\xi^l_s \right) = 0$$

$$\dot{x}^i\dot{x}^j\dot x^k\; : \;\; g_{ij}\partial_k\tau_s - 2g_{ik}\partial_j\tau_s= 0$$

These are the Rund-Trautman identities, or generalised Killing equations to obtain the symmetries of the Lagrangian L. We note that for every $s=1\dots r$ has the same system of equations, and that the number of equations is highly dependant on the form of the Lagrangian.

In the next part I will illustrate one example that particularly interests me. More examples can be found in the references, particularly in [1,4].

## Example

A particle in the Hyperbolic Poincaré Disc $\mathbb D$, that has a metric $\frac{2|dz|^2}{1-|z|^2}$, has a Lagrangian $$L = \frac{\dot x^2 + \dot y^2}{(1-(x^2+y^2))^2} = \gamma^2( \dot x^2 + \dot y^2)$$ with $\gamma \equiv \frac{1}{1-(x^2+y^2)}$. Then $g_{ij}=\gamma^2\delta_{ij}$ and $V=0$. I only want time independent transformations, this means that I fix $\tau_s=0$ and $\partial_t=0$, then the R-T identities become

$$1 \; : \;\; 0 - 0 = \partial_tG_s$$

$$x^i \; : \;\; 0 = \partial_iG_s$$

$$\dot x^i \dot x^j \; : \;\; \frac{1}{2}\partial_{k}g_{ij} \xi^k_s + \frac{1}{2}\left( g_{il}\partial_j\xi^l_s + g_{lj}\partial_i\xi^l_s \right) = 0$$

$$\dot{x}^i\dot{x}^j\dot x^k\; : \;\; 0 = 0$$

we only have one set of equations, the coefficients of the square terms. Taking into account that $\partial_kg_{ij} = 2\gamma (-\gamma^2)(-2x^k)= 4\gamma^3 x^k\delta_{ij}$ we have the system $$4\gamma x_k\delta_{ij} \xi^k_s + \partial_j\xi^i_s + \partial_i\xi^j_s = 0$$

And in components $(x,y)$ we get 3 equations: $$\partial_y\xi^y_s = -2 \gamma \vec{x}\vec{\xi}_s$$ $$\partial_x\xi^x_s = -2 \gamma \vec{x}\vec{\xi}_s$$

$$\partial_x\xi^y_s + \partial_y\xi^x_s = 0$$

The second equation tells us directly that one family of solutions is given by the vector field $\vec{\xi}_s=s(-y,x)$, and it checks with the others. So rotations, that could be seen directly from the Lagrangian. If we sum the first two, we get $$\nabla\cdot \vec{\xi}_s = -4 \gamma \vec{x}\vec{\xi}_s$$

## Continuation

For the other possible symmetries you will have to wait, as I haven't computed them yet. Well, one and two can be writen as $$\frac{1}{\gamma}\partial_i(\gamma \xi^i)=0, \text{ no sum over } i$$ and unfortunatley, to fulfil both at the same time we need $$\xi = \gamma^{-1}(f_1(y),f_2(x))$$ and this doesn't fulfil whater the choice of $f_1,f_2$ se make. So it seams that we cant extract more variational simmetries this way. Because I know that there is a transformation that leaves the Lagrangian invariant: $$\Phi_{\alpha\in\mathbb R, a\in \mathbb C}(z) =\exp(i\alpha) \frac{z-a}{\bar{a}z-1}$$ leaves the Lagrangian invariant, but has not appeared entirely, only the rotational parti, not the one dependent in $a$. This is a mistery.

## Conclusion

There exists a method to find the symmetries of a Lagrangian, and it is cumbersome and involve huge PDE systems. For Lagrangian densities, check 1 and 3. Have fun and report if you find something interesting.

## Bibliography

1. Invariant Variational Problems, D.J. Logan, Elsevier

2. "Classical Noether’s theory with application to the linearly damped particle", Raphaël Leone and Thierry Gourieux (LPM) arXiv:1412.7523v2 [math-ph]

3. Emmy Noether's Wonderful Theorem Dwight E. Neuenschwander, John Hopkings University Press

4. "Variational symmetries of Lagrangians", G.F. Torres del Castillo, C. Andrade Mirón, and R.I. Bravo Rojas, Rev. Mex. Fis. E 59(2) (2013) 140.

• Whoaaaa that's cool, thanks! Will take me some time to go through it. – mavzolej Dec 4 '16 at 20:54