Physics and all that

Physics for the mathematical non-physicist

Conservation Laws Part I (Classical Particles)

   Zachary Raines
 Sep 29, 2017
   21 min read
symmetry lagrangian
Conservation Laws

This will hopefully be Part I of a multi-part series on the relationship between symmetry and conservations laws in Physics. More specifically it’s a bit of a high speed survey of aspects of that topic that I think are neat and also understandable. I won’t assume much physics knowledge, but I will assume a certain level of mathematical sophistication. The aim is for this to be understandable by a mathematically educated non-physicist. Unfortunately, every physicist thinks they’re being understandable and rarely is that true. I’ll do my best.

What are conserved quantities?

A discussion of conservation laws best starts with some definition of what is meant by that. Consider some classical particle As a matter of terminology, when I say an equation is for a classical particle what I really mean is that this equation governs the motion of the center of mass of some body. subject to net force F(x,˙x,t). The evolution of such a particle is governed classically by Newton’s equations, i.e. m¨x=F(x,˙x,t).

Here F is the force, x is a vector describing the position of the particle, m is the particle mass, and ˙f denotes the derivative of f with respect to time. I will use the notations dfdt, ddtf, and ˙f interchangeably. It is also worth noting multiple dots indicate multiple derivatives, i.e. ¨f=d2fdt2 This is just Newton’s second law, ‘Force equals mass times acceleration.’

Suppose we can define some functional I[x(t)] such that for any trajectory xi(t) which solves the Newton equation we necessarily have that ddtI[x(t)]=0. Any quantity that satisfies this condition is said to be a conserved quantity of the problem or a constant of the motion.

Why might we care about such things? One answer is to appeal by example to things that one already believes to be useful: energy, momentum, angular momentum are all often conserved quantities when solving certain physical problems. More generally, the conserved quantities allow us to reduce the work of solving Newton’s equation and to classify the types of solutions that can exist.

Let’s consider a simple example: a particle moving vertically under the influence of gravity. In this case, Newton’s equation is m¨y=−mg,

y is the particle’s height from the ground, m is the particle mass, and g is the acceleration due to gravity (or in other terms the strength of the gravitational field). Let us rewrite this with all non-zero terms on the left, and multiply by ˙y: m˙y¨y+mg˙y=ddt(12m˙y2+mgy)=0.
Defining the quantity in brackets to be E(t)=E[y(t)] we see that ˙E=0 for any solution y(t). If you’ve taken physics before, E should be familiar as the kinetic energy of a particle plus the gravitational potential energy. Knowing that E is conserved allows us to calculate things like the maximum hieght the particle can reach without actually needing to solve any differential equations. Since E is a constant, we can only maximize y2 by minimizing ˙y2. This is when the kinetic term 12m˙y2 is 0. For initial conditions y(0)=y0 and ˙y(0)=v0 we can equate the initial energy to the energy when the kinetic term is zero and we find ymax=y0+v202g
where of course this can only be realized if v0 is positive.

Conservative Forces

We can in fact further generalize the above quantity E by introducing the idea of a conservative force. We say that a force is conservative iff (if and only if) it can be written as the gradient of a scalar function i.e. F=−∇V.

If this is true we call V the potential or potential energy of the force. In that case we have for the Newton equation m¨x=−∇V[x]
where we have generalized to an N dimensional position x. We now dot ˙x into both sides of the equation and make use of the chain rule (in reverse) to obtain. m˙x⋅¨x+˙x⋅∇V=ddt(12m|˙x|2+V[x])⏟E=0.
This is the energy of a particle feeling a conservative force and as we can see, it is conserved (thus the name). Conservative forces encompass a large percentage of what is covered in basic Physics courses. This makes conservation of energy a powerful tool in many problems. Notable non-conservative forces do, however, exist such as friction and air resistance. Yet at a fundamental level energy is always conserved within fundamental theories of Physics (as far as we know) for reasons that we will see later

A common way of picturing a particle feeling a conservative force is a ball rolling along some landscape of hills

The potential profile of some conservative force. The force will point 'downhill' with a magnitude equal to the slope of the curve.

The behavior of the particle more or less follows our intuition of what the ball should do (neglecting friction) because, of course, gravity is a conservative force. For an object near the surface of a massive body (e.g. the Earth) we can write the force due to gravity as −mgˆy=−∇(mgy) where ˆy is a unit vector in the vertical direction. If we now suppose that the particle smoothly rolls along some surface with a profile given by the function y=f(x) we see that it will have a potential energy V(x)=−mgf(x)

which for suitable f(x) corresponds to the potential of whatever force we wish to discuss.

A toy example

As an example to test things out on as we introduce them, we will consider the venerable Simple Harmonic Oscillator (SHO). If you’re already familiar with simple harmonic motion feel free to skip this part.

The system

The classic SHO problem consists of a point particle of mass m subject to a conservative potential V=12mωx2.

The prototypical example is block attached to a spring in 1 dimension. For a spring near it’s equilibrium length the restoring force supplied by the spring is F=−kx where k is some constant describing the springiness and x is the distance of the block from its equilibrium position. This is known as Hooke’s Law.

The prototypical block and spring system: a couple dots and spiral.

We may immediately see that such a force is conservative. To wit We’ll see in a second why I write k as mω2 above. F=−∇(12kx2)=−∇V⟹V(x)=12kx2

Why is this potential considered such a staple in physics? Well let’s suppose we have any system with a stable equilibrium. Well what does it mean to be in stable equilibrium? Imagine that we place a particle at that equilibrium and perturb it slightly. If we push the system slightly away from this equilibrium position the force of the system will push it back. For a conservative force this is equivalent to saying that a point of stable equilibrium is one where the potential energy V is at a local minimum (convince yourself of this).

Now for our generic system near a stable equilibrium, we can expand the potential in a Taylor series Or the suitable multidimensional Taylor series. Let’s not muck about with multiindex notation. as V=V0+V1x+12V2x2+⋯

where x is the deviation from the equilibrium point. V0 doesn’t matter since it doesn’t affect the force (why?), V1=0 since we are at a minimum, and V2>0 because, again, we are at a minimum. Thus, in the vicinity of a stable equilibrium, potentials tend to look like a SHO problem. This includes the case of the mass and spring, the pendulum, and the LC circuit amongst others.

Solution

Looking at the case of the spring system we can see why I named the constants as I did. In this case Newton’s equation is m¨x=−ddx(12mω2x2)=−mω2x.

The solutions to this differential equation are of the form x(t)=Acos(ωt)+Bsin(ωt),
with A and B being determined by the initial conditions of the problem. Thus, we see that ω can be identified with the natural frequency of the system’s response.

Now for our toy example, we’re going to consider something slightly more complicated: consider a ball tied to spring which can move radially and axially with respect to the spring. The spring will provide a potential in the radial direction but the system is totally free in the angular direction. This setup can be described in terms of the potential Vtoy=12mω20(|r|−r0)2=12mω20(√x2+y2−r0)2.

Potential profile of the SHO (left) and a schematic depiction of a physical realization (right).

We’ll return to this model later as test ground to illustrate some of our conclusions, but for now lets move on to more theory.

Lagrangian Mechanics

For the purpose of investigating conserved quantities it turns out that there is a more useful framework of looking at classic mechanics. In fact, one can reproduce all classical physics from the following formalism

There is a functional S=∫dtL(t) of the trajectory x(t) called the action (L is called the Lagrangian). The classical trajectory of the particle is the one at which S is stationary.

This goes by the name of Hamilton’s Principle or the principle of least action. It really should be called the principle of stationary action or perhaps the principle of extremal action, but what can we do? This leads to a theory, known as Lagrangian mechanics, in which all properties of the system can be obtained from a single master object, the action.

Alternatively, one may consider Lagrangian mechanics in the following way. The Lagrangian of a system corresponds to a (sufficiently nice) rule for assigning a number to every possible history (past, present, and future) of the system. The trajectory that the particle will actually follow is the one such its assigned number is extremized Again, really we just need the action to be stationary. We’ll see shortly how to define that. (consistent with our boundary conditions). We can then say that different physical systems just correspond to different rules for assigning numbers to histories. In this way Lagrangians can be seen as complete descriptions of physical systems.

The general form of the Lagrangian varies from system to system but for a point particle feeling a conservative force (F=−∇V) the Lagrangian is L=T−V,

where T is the kinetic energy of the particle (usually 12m˙x2) and V is the potential energy of the particle.

To see how this relates to Newtonian mechanics requires a detour through the calculus of variations. Given my definition (and notes) above you might ask: how does one find a stationary ‘point’ of a functional? For a single variable function f at an extremum or inflection point (excluding boundary points) we have dfdx|x=x0=0.

We can apply the same idea to the notion of a stationary trajectory. We say that ˉq(t) is a stationary function if for any smooth and bounded function η(t) which leaves our boundary conditions invariant. ddϵS[ˉq(t)+ϵη(t)]=0.

The addition of η can be thought of as performing some infinitesimal deformation of the trajectory.

An infinitesimal deformation of the trajectory q(t). Obviously, I can't draw an actually infinitesimal deformation, but use your imagination. Also note that I can't easily depict changes in the rate at which the trajectory is traversed, but that's also important. An infinitesimal deformation of the contour

This provides a simple, intuitive way of bootstraping the notion of a derivatives and extrema into the space of trajectories.

Note that we have changed symbols to q(t) for our trajectory. This is to indicate that q doesn’t need to be a Euclidean position vector and instead can be an set of generalized coordinates. q for example could be the angle a pendulum makes with the vertical.

The connection to Newtonian mechanics

Let’s see what such considerations generally lead to for a Lagrangian which depends on the position and velocity of a particle (as well as potentially time). What follows is a somewhat tedious derivation of a general result. In actual work we only use the final product, but the derivation is worthwhile to follow for a couple of reasons. The most pertinent is that it illustrates the same sort of techniques that will be used a little later on to prove (a weaker version) of Noether’s theorem.

S[ˉq(t)+ϵη(t)]=∫badtL[ˉq(t),˙ˉq(t),t]=∫badtL[ˉq(t)+ϵη(t),˙ˉq(t)+ϵ˙η(t),t].

Taking the derivative with respect to ϵ and using the chain rule we find ddϵS[ˉq(t)+ϵη(t)]|ϵ=0=∫badt(∂L∂qη(t)+∂L∂˙q˙η(t))

where it’s understood that the partial derivative are evaluated at ˉq(t). We now integrate the second term by parts. The boundary term vanishes because, by definition, η(t)=0 at the boundary points. We are then left with 0=∫badtη(t)(∂L∂q−ddt[∂L∂˙q]).
Now η(t) is an arbitrary function so the expression in parenthesis must independently vanish (except for maybe a set of measure 0 etc.). This leads us to the famous Euler-Lagrange equation: ∂L∂q−ddt[∂L∂˙q]=0.
So the stationary trajectory of the action is the trajectory that satisfies the Euler-Lagrange equation (for the boundary conditions imposed at the boundaries of the action integral). Let us then see what the Euler-Lagrange equation looks like for a clasical particle. For this case we use

T=12m˙x2V=U[x]L=12m˙x2−U[x]

and upon taking the derivatives

−dUdx=ddt(m˙x).

Recall that for a conservative potential, the associated force is F=−∇U. So finally we arrive at Netwon’s equation: m¨x=F(x).

We have shown that, at least for the case of a particle in a conservative potential, Lagrangian mechanics subsumes the usual Newton’s laws that we learn about in intro to Physics. But I said that Lagrangian mechanics was a somewhat more useful framework from the perspective of symmetry. How is that?

Cyclic Coordinates

As a quick observation, let us suppose that the Lagrangian depends on ˙q but not q. What does the Euler-Lagrange equation for this system tell us? ∂L∂q0−ddt[∂L∂˙q]=0.

In this case we immediately see that ∂L∂˙q is a conserved quantity. Just this result covers a lot of conservations laws we run into in everyday life such as conservation of linear and angular momentum (what choice of coordinates do think gives these results). This suggests a mode of thought were we search for a set of coordinates where our conservation laws are made manifest. In light of this we might wonder under what conditions we can extract conserved quantities from the Lagrangian. It turns out that this can be done quite generally, so let’s see how it works.

Noether’s Theorem

Now that we’ve established the efficacy of the Lagrangian formalism (at least to some degree), we can look at what is one of my favorite results in all of Physics. It goes by the name of Noether’s theorem after it’s discoverer, mathemetician Emmy Noether. Stated loosely Noether’s theorem is

Every continuous symmetry of the Lagrangian has an associated conserved quantity.

Let us suppose we can deform the trajectories from q(t) to q′(t′) but the action remains invariant. Specifically we assume that we shift the time of the trajectories t→t′=t+δt and we shift the trajectory itself by a small amount to q′(t′)=q(t)+η(t)=q(t′)+˜η(t′), where by defining ˜η we provide ourselves some way to decouple the effect of shifting time from that of deforming the trajectory. By assumption, the action of both trajectories is the same, i.e. S=∫badtL[q(t),˙q(t),t]=S′=∫b+δta+δtdt′L[q′(t′),˙q‘(t′),t′].

Let’s consider the case of infinitesimal change of the trajectories (although the invariance of the action may well hold for arbitrary values). We will thus work to only linear order in δt and η. First we expand the limits of S′

S′≈∫badt′L[q′(t′),˙q′(t′),t′]+L[q′(b),˙q′(b),b]δt−L[q′(a),˙q′(a),a]δt=∫badt′(L[q′(t′),˙q′(t′),t′]+ddt[L[q′(t),˙q′(t),t]δt]).

Renaming t′→t we can write the invariance of the action as ∫badt′(L[q′(t),˙q‘(t),t]−L[q(t),˙q(t),t]+ddt[L[q′(t),˙q‘(t),t]δt])=0.

Expanding to linear order in δt and ˜η we have ∫badt′(∂L∂q˜η(t)+∂L∂˙q˙˜η(t)+ddt[L[q′(t),˙q‘(t),t]δt])=0.
We may now integrate the second term by parts to obtain ∫badt([∂L∂q−ddt∂L∂˙q]˜η(t)+ddt[∂L∂˙q˜η(t)L[q′(t),˙q‘(t),t]δt])=0.
Let’s now consider what this implies if we take q(t)=ˉq(t) to be a trajectory which satifies Hamilton’s principle. From above we then know that ˉq satisfies the Euler-Lagrange equations so the first term in the above integral must vanish. Additionally, going back to our definition we can see that to linear order in δt we have ˜η(t)=η(t)−(˙η(t)+˙q(t))δt≈η(t)−˙q(t)δt
We are then left with ∫badtddt[∂L∂˙q(η−˙ˉqδt)+Lδt]=0
In general the infinitesimal transformation we have described can be written as some infinitesimal parameters ϵ times finite transformations. So δt=ϵT, η=ϵQ. We then have −∫badtddt[(∂L∂˙q˙ˉq−L)T+∂L∂˙qQ]ϵ=0
This holds in general for any boundary points and so the integrand must identically vanish. We then wind up with the conservation law ddt[(∂L∂˙q˙ˉq−L)T+∂L∂˙qQ]=0.
Note that the choice of overall sign here is somewhat arbitrary. It has been chosen to match the established convention in the literature.

Let us quickly recapitulate: if the two trajectories q(t) and q′(t)=q(t−ϵT)+ϵQ(t−ϵT) always lead to the same value of the action i.e. S[q]=S[q′], then we have that the quantity (∂L∂˙q˙ˉq−L)T+∂L∂˙qQ

is conserved along any trajectory of this system.

That’s all well and good, but it’s fairly abstract. Let’s consider some specific examples. Specifically let’s look at a free particle feeling a conservative force. So the Lagrangian is L=12m˙q2−V[q]. One can readily verify that the action does not change if we shift time by an value T. This system is said to have time translation invariance. As a consequence, due to Noether’s theorem, we find

0=ddt[∂L∂˙q˙q−L]T=ddt[m˙q2−(12m˙q2−V)]T=ddt[12m˙q2+V]T=dEdtT.

where we have defined E=12m˙q2+V. Since T is totally arbitrary we have that E is conserved. Furthermore, E is nothing but the energy of the particle. Therefore, conservation of energy is nothing but the manifestation of the time translation invariance of the problem. More generally the function H=∂L∂˙q˙q−L

(and its generalization to many components) is known as the Hamiltonian. Technically, the Hamiltonian is the Legendre transform of the Lagrangian. In short in defining the Hamiltonian we change variables from (q,˙q) to (q,p=∂L∂˙q). Then we define H(q,p)=pq−L(q,p). And for any system where the Lagrangian is not explicitly time dependent, the Hamiltonian is conserved. This is why, as I explained, energy is conserved in general in Physics. The fundamental Lagrangian of Physics is as far as we know time-independent (c.f. the Standard Model Lagrangian). Therefore, as long as you’re counting all parts of the universe (easy I know) energy should be conserved. This provides one a whole new means for describing the dynamics of systems called Hamiltonian Mechanics, which I won’t get into here.

Let’s take a look at another example. Suppose we consider a particle in free space, subject to no external potential. The Lagrangian L=12m˙q2 depends only on the time derivative of the trajectory. We can therefore shift q(t) by any amount and the action is invariant. What do find then from Noether’s theorem? Well all terms drop out other than 0=ddt∂L∂˙qQ=ddt[m˙q]Q=dpdtQ=0

Since this is true for any Q we have that the quantity p=m˙q, the momentum, is conserved. This is the same result we obtained from examing cyclic coordinates. There is a small benefit here in that this result doesn’t depend on any clever choice of coordinates to obtain the result. So conservation of momentum is a statement that the problem is translationally invariant. In fact all of the common conservation laws that we run into in physics are related to symmetries (even conservation of electrical charge). Unfortunately, the more interesting and exotic conservation laws require more general formulations of Noether’s theorem. In particular, we’ll need to introduce Lagrangian mechanics for fields. Hopefully, this will be covered in Part II. As another example conservation of angular momentum mr×˙r is due to rotational invariance.

Let’s see what these concepts look like in our test system. Let us recall that we are imagining something like a ball attached to a springlike piston, which experiences a force in the radial direction, but is totally free to move tangentially to its pivot point. As we discussed before the this system is described by a conservative force with potential V=12mω2(√x2+y2−r0)2.

Here we’ll make the common Intro to Physics assumption of a massless spring While there’s obviously no such thing as a massless spring, including the spring mass complicates the problem and doesn’t really give any insight into the issues we’re interested in here. Then the kinetic energy of the problem is given simply by T=12m(˙x2+˙y2)
where m is the mass of the ball on the end of the spring. Since this is a conservative force problem, the Lagrangian is just L=T−V=12m(˙x2+˙y2)−12mω2(√x2+y2−r0)2.

Obviously this Lagrangian is invariant under time translation. By a straightforward generalization to multiple coordinates Noether’s theorem then tells us that E=∂L∂˙xx+∂L∂˙yy−L

is a conserved quantity. In fact working this out we find that this is just 12m(˙x2+˙y2)+12mω2(√x2+y2−r0)2=T+V.
So, as expected, the energy is conserved. What else is conserved? Consider the following. Writing the position of the mass as

x=[xy]

we can rewrite the Lagrangian as L=12m|˙x|2−12mω2(|x|−r0)2.

Suppose then we rotate the entire system by an angle θ through the poin (0,0). We can accomplish by transforming x→R(θ)x where R(θ) is the rotation matrix

(cosθ−sin(θ)sin(θ)cos(θ)).

This has the effect of also rotation ˙x→R(θ)˙x. Now, since R(θ) is an orthogonal matrix it preserves the norms of vectors and therefore the Lagrangian is unchanged under the is transformation. From Noether’s theorem we must then have a conserved charge. In order to find it, we consider the infinitesimal version of R(θ). For a an infinitesimal angle δθ we have

R(δθ)=1+(0−δθδθ0)+O(δθ)2.

Thus the rotation corresponds to the deformation

x→x−δθyy→y+δθx.

From Noether’s theorem, again generalized to multiple coordinates, we have a conserved quantity

Lz=∂L∂˙x(−y)+∂L∂˙yx=mx×˙x⋅ˆz.

This quantity is known as the angular momentum. As an exercise you may with to show that for a spherically symmetric 3-dimensional system the (pseudo)-vector The name pseudovector denotes the fact that the angular momentum behaves like a regular vector under rotations but not spatial reflections. L=mx×˙x is conserved.

Next time

Now that we’ve seen a bit about how symmetries and conservation laws behave in the classical mechanics of a single body we can move on to some more general cases. In the next post we’ll try to make our way toward considering how this relationship generalizes to quantum phenomena and what the consequences of that are. See you next time.