1.3: Time Translation Invariance

1.3: Time Translation Invariance

1.3.1 Uniform Circular Motion

When \(α\), \(β\) and \(γ\) in \(α\frac{d^2}{dt^2}x(t) + β\frac{d}{dt}x(t) + γx(t) = f(t)\) do not depend on the time, t, and in the absence of an external force, that is for free motion, time enters in (\(α\frac{d^2}{dt^2}x(t) + β\frac{d}{dt}x(t) + γx(t) = f(t)\) only through derivatives. Then the equation of motion has the form.

\[α\frac{d^2}{dt^2}x(t) + β\frac{d}{dt}x(t) + γx(t) = 0\]

The equation of motion for the undamped harmonic oscillator, (1.3), has this form with α = m, β = 0 and γ = K. Solutions to (1.32) have the property that

If x(t) is a solution, x(t + a) will be a solution also.

\[\frac{d}{dt}x(t + a) = [\frac{d}{dt}(t + a)] [\frac{d}{dt'}x(t')]_{t'=t+a} = [\frac{d}{dt'}x(t')]_{t'=t+a}\]

The physical reason for (1.33) is that we can change the initial setting on our clock and the physics will look the same. The solution \(x(t + a)\) can be obtained from the solution \(x(t)\) by changing the clock setting by a. The time label has been “translated” by a. We will refer to the property, (1.33), as time translation invariance.

Most physical systems that you can think of are time translation invariant in the absence of an external force. To get an oscillator without time translation invariance, you would have to do something rather bizarre, such as somehow making the spring constant depend on time.

For the free motion of the harmonic oscillator, although the equation of motion is certainly time translation invariant, the manifestation of time translation invariance on the solution, (1.6) is not as simple as it could be. The two parts of the solution, one proportional to \(cos (ωt)\) and the other to \(sin (ωt)\), get mixed up when the clock is reset. For example,

\[cos[ω(t + a)] = (cos ωa) (cos ωt) − (sin ωa) (sin ωt).\]

It will be very useful to find another way of writing the solution that behaves more simply under resetting of the clocks. To do this, we will have to work with complex numbers.

To motivate the introduction of complex numbers, we will begin by exhibiting the relation between simple harmonic motion and uniform circular motion. Consider uniform circular motion in the x-y plane around a circle centered at the origin, \(x = y = 0\), with radius R and with clockwise velocity \(v = Rω\). The x and y coordinates of the motion are

\[x(t) = R cos(ωt − φ), y(t) = −R sin(ωt − φ),\]

where \(φ\) is the counterclockwise angle in radians of the position at \(t = 0\) from the positive x axis. The \(x(t)\) in (1.36) is identical to the \(x(t)\) in (1.6) with

\[x(0) = R cos φ , x'(0) = ωR sin φ .\]

Simple harmonic motion is equivalent to one component of uniform circular motion. This relation is illustrated in figure 1.4 and in program 1-1 on the programs disk. As the point moves around the circle at constant velocity, \(Rω\), the \(x\) coordinate executes simple harmonic motion with angular velocity \(ω\). If we wish, we can choose the two constants required to fix the solution of (1.3) to be \(R\) and \(φ\), instead of \(x(0)\) and \(x'(0)\). In this language, the action of resetting of the clock is more transparent. Resetting the clock changes the value of \(φ\) without changing anything else.

Figure 1.4: The relation between uniform circular motion and simple harmonic motion.

But we would like even more. The key idea is that linearity allows us considerable freedom. We can add solutions of the equations of motion together and multiply them by constants, and the result is still a solution. We would like to use this freedom to choose solutions that behave as simply as possible under time translations.

The simplest possible behavior for a solution \(z(t)\) under time translation is

\[z(t + a) = h(a) z(t).\]

That is, we would like find a solution that reproduces itself up to an overall constant, \(h(a)\) when we reset our clocks by \(a\). Because we are always free to multiply a solution of a homogeneous linear equation of motion by a constant, the change from \(z(t)\) to \(h(a) z(t)\) doesn’t amount to much. We will call a solution satisfying (1.38) an “irreducible³ solution” with respect to time translations, because its behavior under time translations (resettings of the clock) is as simple as it can possibly be.

It turns out that for systems whose equations of motion are linear and time translation invariant, as we will see in more detail below, we can always find irreducible solutions that

________________________

³The word “irreducible” is borrowed from the theory of group representations. In the language of group theory, the irreducible solution is an “irreducible representation of the translation group.” It just means “as simple as possible.”

have the property, (1.38). However, for simple harmonic motion, this requires complex numbers. You can see this by noting that changing the clock setting by \(π/ω\) just changes the sign of the solution with angular frequency \(ω\), because both the \(cos\) and \(sin\) terms change sign:

\[cos(ωt + π) = − cos ωt, sin(ωt + π) = − sin ωt.\]

But then from (1.38) and (1.39), we can write

\[−z(t) = z(t + π/ω) = z(t + π/2ω + π/2ω)\]

\[= h(π/2ω) z(t + π/2ω) = h(π/2ω)^2 z(t).\]

Thus we cannot find such a solution unless \(h(π/2ω)\) has the property

\[[h(π/2ω)]^2 = −1.\]

The square of \(h(π/2ω)\) is −1! Thus we are forced to consider complex numbers.⁴ When we finish introducing complex numbers, we will come back to (1.38) and show that we can always find solutions of this form for systems that are linear and time translation invariant.

________________________

⁴The connection between complex numbers and uniform circular motion has been exploited by Richard Feynman in his beautiful little book, QED.