# Noether's Theorem: There are conserved quantities corresponding to symmetries of position, orientation, and time, but why not velocity?

Noether's Theorem seems to be one of the most fundamental and beautiful results in all of physics. As I understand it, the fact that the laws of physics are the same independent of position, orientation, and time, leads to the conservation of momentum, angular momentum, and energy, respectively.

But the laws of physics are also independent of velocity. Why does this not lead to another conserved quantity? Or is it just Newton's third law (forces in a closed system must add to zero)?

• Mar 15 at 15:30
• You might want to look at this in a more relativistic way. The homogeneity of spacetime leads to the energy-momentum being conserved in field theory. The Isotropy leads to general relativistic version of angular momentum. That is it. The Poincaré group is the most general there is (susy being a bit under right now). That is not to say that other systems might present other invariances and then have a conserved quantity attached to it that but being limited in generality and scope. Mar 15 at 15:30
• @ Nelson Vanegas A. I interpret "homogeneity of spacetime" to mean it's the same at every place and time. I wouldn't necessarily include velocity. Is that wrong thinking? Mar 15 at 17:17
• @Qmechanic I followed your link which does seem to be exactly the same question. The answer seems to be just a statement about where the center of mass is at a particular instant in time. But in what sense is this a conserved quantity, if it doesn't apply to other instants in time? Mar 15 at 17:33

The issue here is that the statement about laws of physics being independent of velocity has been misunderstood or is not precise enough. Noether's theorem pertains explicitly, theories described through a Lagrangian. A Lagrangian in its bare bones, should contain a kinetic term of some sort. Without going to a relativistic setting, you can consider a free particle in classical mechanics, for which $$L_{\rm free} = \frac{1}{2} m \left(\frac{dx}{dt}\right)^2 = \frac{1}{2}m v^2$$ You can easily try to displace the speed $$v\rightarrow v + \Delta v$$, and you will see that the Lagrangian actually changes by an amount which should be proportional to acceleration, therefore a system described by such Lagrangian is not invariant under infinitesimal changes in velocity in general. Since this term (or similar) appears in most Lagrangians of physical systems, changes in speed do not correspond to symmetries.

This however does not mean that one cannot build a Lagrangian which is indeed symmetric under such changes. As it has been mentioned in the comments and other answers, if you interpret displacements in speed, as "boosts", then a relativistic Lagrangian does display such symmetry, although it is often more useful to think about it as an imaginary rotation, rather than a shift in velocity (if it is understood in a linear way), or in other words one is changing time and space in a very specific way, see Lorentz transformations.

I would suggest to try to follow the usual derivation of conserved currents to see the Lagrangians do not generally display a "speed" symmetry at least not in the Noether sense of symmetries. I also recommend considering the action associated to the Lagrangian above with the addition of a square root, which makes the action "speed invariant" because it will only depend on the end-points. This illustrates clearly the point that kinetic energy is frame-dependent (in classical mechanics), but the actual arc-length of the trajectory is not.

• That answer is helpful. Do you have any comment on "forces in a closed system must add to zero" as representing a conserved quantity? Mar 15 at 17:44
• If there are no external forces, the immediate consequence is that internal forces will add up to zero, but moreover, no external sources, means momentum conservation. So it is again rephrasing invariance under translations. Mar 15 at 18:39

There are conserved quantities related to Lorentz boosts. These are included in the angular momentum tensor, which is conserved if the Lagrangian is invariant under Lorentz transformations.

• @my2cents I interpret Lorentz boosts as having to do with velocity rather than angular momentum. Is there a more intuitive explanation perhaps? Mar 15 at 17:20
• @RogerWood The general Lorentz transformation includes rotations and boosts. Thee is a connection between a boost and a rotation in Minkowski space. See e.g. physics.stackexchange.com/questions/544002/… . Mar 15 at 22:17
• thanks, I understand now. Mar 16 at 2:43