I'm referring here to invariance of the Lagrangian under Lorentz transformations.

There are two possibilities:

  • Physics does not depend on the way we describe it (passive symmetry). We can choose whatever inertial frame of reference we like to describe a physical system. For example, we can choose the starting time to be $t_0=0$ or $t_0=4$ (connected by a translation in time $t \rightarrow t' = t + a_0$). Equivalently it does not matter where we put the origin of our coordinate system (connected by a translation in space $x_i \rightarrow x_i' = x_i + a_i$)) or if we use a left-handed or a right-handed coordinate system (connected by a parity transformation). Physics must be independent of such choices and therefore we demand the Lagrangian to be invariant under the corresponding transformations.
  • Physics is the same everywhere, at any time (active symmetry). Another perspective would be that translation invariance in time and space means that physics is the same in the whole universe at any time. If our equations are invariant under time translations, the laws of physics were the same $50$ years ago and will be tomorrow. Equations invariant under spatial translations hold at any location. Furthermore, if a given Lagrangian is invariant under parity transformations, any experiment whose outcome depends on this Lagrangian finds the same results as an equivalent, mirrored experiment. A basic assumption of special relativity is that our universe is homogeneous and isotropic and I think this might be where the justification for these active symmetries comes from.

The first possibility is really easy to accept and for quite some time I thought this is why we demand physics to be translation invariant etc.. Nevertheless, we have violatíon of parity. This must be a real thing, i.e. can not mean that physics is different if we observe it in a mirror. Therefore, when we check if a given Lagrangian is invariant under parity, we must transform it by an active transformation and do not only change our way of describing things.

What do we really mean by symmetries of the Lagrangian? Which possibility is correct and why? Any reference to a good discussion of these matters in a book or likewise would be aweseome!

  • $\begingroup$ This obviously doesn't answer your question (which is too philosophical for me to comprehend), but here is an interesting point. Symmetries of equations in general are NOT applicable to its individual solutions (only to a set of all solutions). It is what physicists usually call symmetry breaking, and it causes confusion. For example, take a classical particle propagating in Minkowski space-time. The solution (particle trajectory) is obviously not Lorentz-invariant (there is a special frame of reference) despite all equations being Lorentz-invariant. $\endgroup$ Commented Jan 21, 2015 at 10:46
  • $\begingroup$ see this answer physics.stackexchange.com/q/158398 $\endgroup$
    – Phoenix87
    Commented Jan 21, 2015 at 11:26

4 Answers 4


I might be wrong in this, but despite the similiarities, these two things you describe are sternly different I think.

Your first point is related to the so-called "general covariance". It is something that is in effect all the time. You don't see any kind of coordinate grid anywhere when you look outside the window, and you don't see any spatial or temporal point designated as a "starting point" (ignoring stuff like the big bang now etc.), therefore, it is only logical that such constructs exist only to help describing stuff mathematically, so physics should be independent of coordinates.

The second thing you say doesn't always happen. Example, if you have an explicitly time-dependent Lagrangian, then time displacements will NOT leave the Lagrangian invariant, and energy won't be conserved (with that said, in reality, energy IS conserved, but for example, if you have friction then you generally say it is not conserved, since it gets removed from the sum of kinetic and potential energies).

Likewise, if you have a spherically symmetric potential field, then rotations of the physical system will leave the Lagrangian invariant, since the potential field is equal for all rotations around the fixed origin. BUT if you have a cylindrically symmetric potential field, whose axis is the $z$ axis, then rotations around $z$ will leave the Lagrangian invariant, but rotations around the $x$ or $y$ axes will NOT leave the Lagrangian invariant.

  • $\begingroup$ Thanks for your answer, I agree with everything you say. Nevertheless, I'm trying to understand what physicists commonly mean when they say a given Lagrangian must be symemtric under a given transformation. For example the various Lagrangians of the standard model are required to be invariant under all Lorentz transformations. Do we require this, because of the first or the second reason I outlined above? $\endgroup$
    – jak
    Commented Jan 23, 2015 at 9:39

Symmetry is an invariance of an object - most often a vector space or manifold - with respect to a transformation. For example, we say that an object has reflexional symmetry about a plane iff it is left unchanged by a reflexion in that plane.

More abstractly, we often think of any bijective map as a reversible "preservation" of information, so that an object is left essentially unchanged ("invariant") by the action of the map. So sometimes you'll see the word symmetry as almost synonymous with bijection.

A Lagrangian has symmetry if it is left unchanged by a bijective map imparted to the manifold / configuration space whereon the Lagrangian is defined. The symmetry is continuous if the transformation is a member of a family (Lie group) of such bijections which is both parameterized by Euclidian co-ordinates and such that the parameterization is a continuous function of the map-composition operation.


We mean both, although sometimes at different times.

The meaning depends on how we intend to apply Lagrangian mechanics. We may mean to say "We are measuring this system in this way, but we can show that it doesn't matter which way we measured it because the systems' behavior is independent of our measurement system." And that's all fine and good. It's useful to be able to say that people's measurements of our system can only disagree with our measurements in a consistent way.

We also say "we predict that the behavior of system X will be thus, because of these symmetries with system Y." For example, we often leverage the symmetries of dimensionless constants to do scaled down wind tunnel modeling and state the predicted behavior of another object which has yet to be measured.

In the former way, science permits us to measure and share our measurements, with a certainty that it will be useful to others. In the latter way, we use science to make predictions about what is to come.


Don't be confused. Symmetry means that equations in physics are invariant under certain transformations. You can then simplify or do other things with them. As an example take the Dirac lagrangian with which you can derive the Dirac equation for a free field of spin 1/2 :

$ \boldsymbol{\mathfrak{L}}= - m \widetilde{\Psi}\Psi+ i\widetilde{\Psi} \gamma _{\mu} \partial ^{\mathbb{\mu}}\Psi= \widetilde{\Psi}(i\gamma _{\mu} \partial ^{\mathbb{\mu}}-m) \Psi$

In this lagrangian there is an internal symmetry, because the lagrangian is invariant under $U(1)$ transformations:

$ \Psi \mapsto {\Psi}'= e^{ia}\Psi $

(just a rotation by the complex number $e^{ia}$)

$\Rightarrow \widetilde{\Psi} \mapsto {\widetilde{\Psi}}^{'} = \Psi^{'\dagger} \gamma_{0}=(e^{ia} \Psi)^{\dagger} \gamma_{0} = \widetilde{\Psi}e^{-ia} $

$ \boldsymbol{\mathfrak{L}}= - m {\widetilde{\Psi}}^{'} \Psi^{'}+ i {\widetilde{\Psi}}^{'} \gamma _{\mu} \partial ^{\mathbb{\mu}} \Psi^{'}$

$ =- m (\widetilde{\Psi}e^{-ia}) (e^{ia} \Psi) + i(\widetilde{\Psi}e^{-ia}) \gamma _{\mu} \partial ^{\mathbb{\mu}} (e^{ia} \Psi) $

$ = - m \widetilde{\Psi} \Psi \underbrace{(e^{-ia} e^{ia})}_{1} + i \widetilde{\Psi} \gamma _{\mu} \partial ^{\mathbb{\mu}}\Psi \underbrace{(e^{-ia} e^{ia})}_{1} $

$ =- m \widetilde{\Psi}\Psi+ i\widetilde{\Psi} \gamma _{\mu} \partial ^{\mathbb{\mu}}\Psi= \widetilde{\Psi}(i\gamma _{\mu} \partial ^{\mathbb{\mu}}-m) = \boldsymbol{\mathfrak{L}} $

There can be other symmetries with which you can manipulated equations in physics and simplify them or change them to some other form out of which you can get new insights.

  • $\begingroup$ Don't be confused is a great advice to someone who is confused. $\endgroup$ Commented Jan 12, 2023 at 17:44

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