Physics Puzzle Solution: Missing Energy of Traveling Waves

April 13th, 2005 | Categories: Physics Puzzles - Solutions

This is the solution to this puzzle.

The short answer is that energy is stored in the rope in two forms: potential (by stretching it transversely) and kinetic (by changing its shape). Even when the two waves cancel out the energy is still there, stored in the kinetic term (the rope’s tension), waiting to spring back into the potential term. No energy is lost, really.

How we see this mathematically? Let’s begin by deriving a conservation equation from the wave equation in one dimension. The wave equation is given by:

c^2 \frac{\partial^2 p}{\partial x^2} = \frac{\partial^2 p}{\partial t^2}

We can derive a conservation equation from this by multiplying both sides by the time-derivative of p (and dividing by c^2),

\frac{\partial p}{\partial t} \frac{\partial^2 p}{\partial x^2} = \frac{1}{c^2} \frac{\partial p}{\partial t} \frac{\partial^2 p}{\partial t^2} = \frac{1}{2c^2} \frac{\partial}{\partial t} \left( \frac{\partial p}{\partial t} \right)^2

Next, note that (you can verify this by carrying out the differentiation on the RHS and comparing it to the LHS):

\frac{\partial p}{\partial t} \frac{\partial^2 p}{\partial x^2} = \frac{\partial}{\partial x} \left[ \frac{\partial p}{\partial t} \frac{\partial p}{\partial x} \right] - \frac{1}{2} \frac{\partial }{ \partial t} \left( \frac{\partial p}{\partial x} \right)^2

Substituting this back into the previous equation, we get an expression of the following form:

\frac{\partial E}{\partial t} + \frac{\partial J}{\partial x} = 0

This has the form of a continuity equation, with the energy density, E, given by

E = \frac{1}{2c^2} \left( \frac{\partial p}{\partial t} \right)^2 + \frac{1}{2} \left( \frac{\partial p}{\partial x} \right)^2

and the energy flux, J, by

J = -\frac{\partial p}{\partial x} \frac{\partial p}{\partial t}

In the expression for E, the first term corresponds to the kinetic energy, and the second to the potential energy. It is quite possible to have p=0 for all x at some instant in time and still have nonzero energy via the time-derivative of p (the kinetic energy).

We note in passing an interesting property. When you have two waves traveling on the same string - call them u(x,t)=u(x+ct) and w(x,t)=w(x-ct) - the kinetic and potential terms become (using p(x,t) = u(x+ct) + w(x-ct)):

\frac{1}{2c^2} \left( \frac{\partial p}{\partial t} \right)^2 = \frac{\left( -c w'(x-ct) + c u'(x+ct) \right)^2 }{2c^2} \frac{1}{2} \left( \frac{\partial p}{\partial x} \right)^2 = \frac{\left(w'(x-ct) + u'(x+ct) \right)^2 }{2}

Expanding and adding the two terms up causes the cross terms u’w’ to cancel out, leaving us with

E(x,t) = E_w(x,t) + E_u(x,t)

with (for example):

E_w(x,t) = \frac{1}{2c^2} \left( \frac{\partial w}{\partial t} \right)^2 + \frac{1}{2} \left( \frac{\partial w}{\partial x} \right)^2

Similarly, E_u depends solely on u(x,t). In other words, the energy density at each point is the sum of the individuals waves’ energies - no interference terms at all. So, it appears the energy density obeys a similar superposition rule as well!

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