# Conservation Laws and Symmetry

The toughest of topics in physics, like Quantum Mechanics, Relativity, String theory, can be explained in layman words and many have done so. Though there is no substitute to the understanding a theory in all it's mathematical detail, the idea can more or less be driven home. However, I seem not to be able to find such an explanation for Noether's theorem. I've tried my hand at Group Theory and found quite perplexing. I've got stuck with generators, and this and that ....

So here is my question. Is it possible to explain without using heavy mathematics, why a conservation law arises from a symmetry under some transformation?

I'm in school, so my "heavy" might not be the same as yours. Thanks.

• Noether's Theorem is a mathematical theorem, so without heavy mathematics, we cannot explain why it says what it says. But we can say that when there is a symmetry wherein a given transformation leaves a system unchanged, Noether's Theorem says that this means we can do some fancy math and always find a quantity whose time derivative vanishes. That is, a quantity that doesn't change with time. Being a mathematical theorem, asking us to explain why it says that without using heavy math is like asking us to explain the Pythagorean theorem without using any math.
– Jim
Aug 24 '15 at 17:57
• I can't think of an example that can be easily explained with simpler math. Every example of Noether's theorem being used has pretty much the same complexity. It's like going to Subway and getting a sandwich. The ingredients may be different, but it's always about the same level of complexity for them to make your sandwich. A good starting point for learning the math would be calculus, differential equations, then tensor calculus.... Actually, there's a lot of math that goes into it. Just learn all the math you come across and don't know. Can't go wrong with that
– Jim
Aug 24 '15 at 18:22
• If you go the pure math way and tackle the mathematician's formalism of Noether's theorem, you need to learn a lot more math. If you approach it through QFT and use physics to learn how it works, it's much simpler. How much tensor calculus do you know? Because for me, that was about all that was necessary. Mind you, everything makes more sense in a formal learning environment.
– Jim
Aug 24 '15 at 18:33
• The smallest and clearest explanation I know of is here. Meanwhile, a great question (+1): it would be wonderful to come up with a clearer explanation! Baez's explanation shows that there is not as much to it as one would think at first. Do you know about "minimum action" ideas (dynamics that minimize a Lagrangian)? Aug 25 '15 at 7:09
• Related: physics.stackexchange.com/q/4959/2451 , physics.stackexchange.com/q/21572/2451 and links therein. Aug 31 '15 at 15:20

A helpful yet elementary answer may do the trick, If you are familiar with the Euler-Lagrange equation then it will be straight forward and you can skip ahead a little. If not then you have to accept that there is an equation in physics that generalises classical mechanics called the Euler-Lagrange equation. For a particle moving in one dimension under a conservative force it is written, $$\frac{d}{dt}\bigg(\frac{\partial T}{\partial \dot x}\bigg)+\frac{\partial V}{\partial x}=0$$ Where $T$ is the kinetic energy of the system and $V$ is the potential energy, $x$ is the particles position and $\dot x=\frac {d}{dt}x$ is the velocity of the particle. We define the momentum of the particle to be, $$p:=\frac{\partial T}{\partial \dot x}$$ And you will note that we can now write the Euler-Lagrange equation as, $$\frac{d}{dt}(p)+\frac{\partial V}{\partial x}=0$$ This is Newton's second law of motion. The momentum is changed by the action of a force on the particle, if there are no forces then the time derivative of the momentum is zero. If the time derivative is zero then the momentum does NOT change as time evolves and will have the same value at the end of the experiment as it did at the beginning.
In this way the Euler-Lagrange equation has given us a conservation law for $p$ only when $\partial V/\partial x=0$. The invariance of the potential with respect to $x$ leads to a conservation law.
In general we do not write the Euler-Lagrange equation for a one dimensional particle. The general form is written, $$\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot x}\bigg)-\frac{\partial L}{\partial x}=0$$ Where $L(x,\dot x)=T(\dot x)-V(x)$ is the Lagrangian of the system. Check that this will give the above stated equation. In general if the Lagrangian for a particular system is not a function of $x$ then you can clearly see that, $$\frac{\partial L}{\partial \dot x}=constant$$ Since the time derivative vanishes. When the Lagrangian is not a function of $x$ we say that the Lagrangian has a symmetry. When the Lagrangian has a symmetry, there is a conservation law.