Conservation of energy

Status: Done
Confidence: Certain

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Last night it occurred to me that, although I knew Landau’s derivation of conservation of energy in Lagrangian mechanics, I had never seen it done from Newton’s third law. A few minutes of scribbling produced a proof; a few more produced the following exposition that might be of use to some poor physics student somewhere:

The overarching theme is Noether’s theorem. Consider some problem described by a differential equation (or an action principle, but for now let’s stay with differential equations), which has some continuous symmetry. The rotations that leave a square unchanged are a discrete symmetry. All displacements in time of a system which does not depend on any particular position in time are a continuous symmetry. Essentially, if we can construct a continuous function from the transformations to the real numbers, then it is a continuous symmetry.

Here’s the plan: change coordinates in the differential equation until the symmetry transformations we’re interested in are represented by a displacement in a single coordinate. For time displacement, this is simple. For spatial displacement, we use Cartesian coordinates with one axis along the direction of displacement. For rotation, we use polar coordinates.

Shift the coordinate infinitesimally, and Taylor expand to first order. For any solution of the differential equation, the zeroth order expansion cancels. Taylor’s remainder theorem makes the second and higher orders smaller than the first order, so to have invariance, the first order must vanish. The invariance must hold for different infinitesimal displacements, so the first order coefficient must vanish, and we may multiply this by anything that is not zero to get a cleaner form at the end. The integral of the first order coefficient with respect to the coordinate is a constant, then we integrate everything in sight by parts and occasionally use the original differential equation to substitute for things until we find a convenient form. The last part sounds vague, and it is. I don’t know if there is some algebraic structure that will yield an algorithm for getting clean, useable forms for conservation laws. How did I find this particular set of integrations and multipliers? I played around until it worked.

Conservation of energy comes from invariance under a displacement in time. Begin with Newton’s third law, F(x(t),t)=mD2x(t)F(x(t), t) = m D^2 x(t). DD here is the total time derivative. This is not the same as the partials, as DF(x(t),t)=0F(x(t),t)Dx(t)+1F(x(t),t)DF(x(t),t) = \partial_0 F(x(t),t) \cdot Dx(t) + \partial_1 F(x(t), t) (where n\partial_n is the partial derivative with respect to the nnth argument).

Let tt+δtt \rightarrow t + \delta t. Then

mD2x(t+δt)=F(x(t+δt),t+δt)m \cdot D^2 x(t+\delta t) = F(x(t+\delta t), t+\delta t)
= { Taylor expand xx and FF to first order }
mD2x(t)+mδtD3x(t)=F(x(t),t)+δtDF(x(t),t)m \cdot D^2 x(t) + m\cdot \delta t \cdot D^3 x(t) = F(x(t), t) + \delta t \cdot DF(x(t), t)

The first terms on each side are equal by the original differential equation, and they cancel. Now we multiply by sides by x(t)x(t) and rearrange to get

δt[mx(t)D3x(t)x(t)DF(x(t),t)]=0\delta t [ m \cdot x(t) \cdot D^3 x(t) - x(t) \cdot DF(x(t), t) ] = 0

This must hold for different, nonzero values of δt\delta t, so the quantity in brackets must be zero.

Aside. There is a subtlety to note: it was also zero before we threw in the x(t)x(t), so it remains zero after we multiply by x(t)x(t), so long as the particle isn’t at infinity. If we were not already guaranteed this, then we would have to set up the coordinates where x(t)x(t) was never zero. This requires symmetry under spatial displacement (which yields conservation of momentum), and the fact that in any finite time interval there are points which are not visited by the path. The whole of the path over infinite time may visit all the points in the space (this is called ergodicity), but we can break the whole of time into connected subsets, each of which have an unvisited point, prove it on each subset, and then stitch them together. Spaces where we can do this slicing and stitching are called manifolds, and we’ll use this trick later. End Aside.

Now we integrate with respect to time. Integration by parts on the first term yields (everything is multiplied by mm, but we’ll just put it back in later):

x(t)D3x(t)dt\int x(t) \cdot D^3 x(t) dt
= { integration by parts, D(x(t)D2x(t))=D3x(t)+Dx(t)D2x(t)D(x(t) \cdot D^2 x(t)) = D^3 x(t) + Dx(t) \cdot D^2 x(t) }
D(x(t)D2x(t))dtDx(t)D2x(t)dt\int D(x(t) \cdot D^2 x(t)) dt - \int Dx(t) \cdot D^2 x(t) dt
= { first integral: fundamental theorem of calculus; second: integrate by parts }
x(t)D2x(t)D(12(Dx(t)Dx(t)))x(t) \cdot D^2 x(t) - \int D( \frac{1}{2}(Dx(t) \cdot Dx(t)) )
= { fundamental theorem of calculus }
x(t)D2x(t)12Dx(t)Dx(t)x(t) \cdot D^2 x(t) - \frac{1}{2} Dx(t) \cdot Dx(t)

The first term is going to be kinetic energy. The other term will cancel after we analyze the second integral.

x(t)DF(x(t),t)dt\int x(t) \cdot DF(x(t), t) dt
= { integration by parts }
D(x(t)F(x(t),t))dtDx(t)F(x(t),t)dt\int D(x(t) \cdot F(x(t),t)) dt - \int Dx(t) \cdot F(x(t), t) dt
= { Newton’s 3rd law: F(x(t),t)=mD2x(t)F(x(t), t) = m D^2 x(t) }
D(x(t)mD2x(t))dtDx(t)F(x(t),t)dt\int D(x(t) \cdot m D^2 x(t)) dt - \int Dx(t) \cdot F(x(t), t) dt
= { Fundamental theorem of calculus }
mx(t)D2x(t)Dx(t)F(x(t),t)dtm x(t) \cdot D^2 x(t) - \int Dx(t) \cdot F(x(t), t) dt

Now we combine these, putting that mm we neglected in the first part back in. mx(t)D2x(t)m x(t) \cdot D^2 x(t) terms do indeed cancel and we get

12Dx(t)Dx(t)Dx(t)F(x(t),t)dt=\frac{1}{2} Dx(t) \cdot Dx(t) - \int Dx(t) \cdot F(x(t), t) dt =constant

There’s conservation of energy. Now a few remarks to tie things up:

The second term looks unfamiliar. Try changing variables so that position is independent and time depends on it. This must be done carefully! If the path x(t)x(t) intersects itself, we break the integration into a sum of integrals over interval of time where the path does not self-intersect, and change variables separately in each one so that t(x)t(x) is well defined. On any such interval II, we get a term that looks like

IxF(x,t(x))dx\int_I x \cdot F(x, t(x)) dx

which is more familiar.

Next, if our force has the form F(x,t)=0U(x,t)F(x, t) = - \partial_0 U(x, t), then the second integral takes the form

U(x(t),t)+1U(x(t),t)dt-U(x(t), t) + \int \partial_1 U(x(t), t) dt

For potentials UU which are independent of time, this is particularly useful as the second integral vanishes. For many oscillating potentials, the integral takes on an average value over long times.