Hooke's Law, Phase Space and Classical Geometry

Question

Hooke's Law tells us that $m\ddot{x} = -kx$. We can apply the chain rule to rewrite $\ddot{x}$ as follows:

$$\frac{\operatorname{d}\!^2x}{\operatorname{d}\!t^2} = \frac{\operatorname{d}\!v}{\operatorname{d}\!t}=\frac{\operatorname{d}\!x}{\operatorname{d}\!t}\frac{\operatorname{d}\!v}{\operatorname{d}\!x} = v\frac{\operatorname{d}\!v}{\operatorname{d}\!x}$$

Substituting this into $m\ddot{x} = -kx$, separating variables, and then integrating gives:

$$mv^2+kx^2 = c$$

for some constant $c$ to be determined by initial conditions. If we assume that the spring is stretched to a length $\lambda$ to the right before it is released then we have $v=0$ when $x=\lambda$ and so $c=k\lambda^2$. Hence:

$$mv^2+kx^2 = k\lambda^2$$

For all values of $m>0$, $k>0$ and $\lambda>0$, this gives the equation of an ellipse in the $xv$-plane. Ellipses have two special points, named foci. The Earth follows a more-or-less elliptical path around the Sun, which sits at a focus. The Moon follows a more-or-less elliptical path around the Earth, which sits at a focus.My questions is: What is the physical significance of the foci of $mv^2+kx^2=k\lambda^2$

One characterisation of the foci, say $\phi_1$ and $\phi_2$, is that the sum of the distances $\operatorname{d}(\phi_1,p) + \operatorname{d}(p,\phi_2)$ is constant for all points $p$ of the ellipse.

Answer 1

To formalize Michael Brown's comment: Suppose we want to change the units that we are working in to ones that are naturally set by the problem. We can write the dimensional variables $x$ and $t$ as $x = S_x \bar{x}$ and $t = S_t \bar{t}$ where $S_x$ and $S_t$ are the length and time scales respectively. These scales carry the dimensions of the quantity and the variables with bars are dimensionless variables, ie just pure numbers. Obviously, in the SI system $S_x$ is 1 meter and $S_t$ is 1 second.

Let us substitute our expressions for $x$ and $t$ into the equation that you found: $$mv^2+kx^2 = m \big(\frac{dx}{dt}\big)^2 +k x^2 = m \frac{S_x^2}{S_t^2} \big(\frac{d \bar{x}}{d \bar{t}}\big)^2 + k S_x^2 \bar{x}^2 = m \frac{S_x^2}{S_t^2} \bar{v}^2 + k S_x^2 \bar{x}^2 .$$

We are free to use whatever unit system that we like, so let's choose $S_x = \lambda$ and $S_t = \sqrt{m/k}$. Substituting this above gives $ k \lambda^2 \bar{v}^2 + k \lambda^2 \bar{x}^2 = k \lambda^2 $ or $$ \bar{v}^2 + \bar{x}^2 = 1,$$ which is just the equation of a circle of radius 1.

So, just by changing the units we have moved the distance between the foci to 0. This implies that the distance between the foci itself can't be a physical quantity because we set it to 0 just by rescaling. We cannot ever recover the original answer in the SI unit system from this one.

However, the distance between the foci can be related to a physical quantity. If the semimajor axis of the ellipse is labeled $a$ and the semiminor axis labeled $b$ then the area enclosed by the ellipse is $\pi a b$. In your equation in the original units (assuming without loss of generality that $k/m < 1$) $a = \lambda$ and $b = \lambda \sqrt{k/m}$ so the area is $ \pi \lambda^2 \sqrt{\frac{k}{m}}$. This area is known as the symplectic area, and it is a conserved quantity for Hamiltonian systems. Interestingly, even for Hamiltonians that are time dependent, the symplectic area is still conserved.

We can write the area of an ellipse in terms of the distance between the foci. If the distance between the foci is $c$, then the semiminor axis is defined to be $b^2 \equiv a^2 - c^2$, so the area can be written as $\pi a b = \pi a \sqrt{a^2 - c^2}$. You can check that by using the scales we considered earlier, you can convert the symplectic area for the original case to the symplectic area for the rescaled case (the circle) meaning that the symplectic area is a perfectly physical quantity.

If you are looking for something you can measure, then the period of motion, $T$, of the object along the curve can also be found from the symplectic area. $ T = \frac{dA}{dE}$ where $A$ is the symplectic area and $E$ is the energy.

In short, the locations of the foci and the distance between them are themselves not physical quantities. However, they can be related to some physical quantities, namely the symplectic area and the period of motion.

Answer 2

As you said, one way (arguably the way) in which the foci of an ellipse are special is that the sum of the distances from the two foci to any point on the ellipse is a constant. But how do you define distance in phase space, anyway? The formula you might naively think to use, namely

$$r^2 = \Delta x^2 + \Delta v^2$$

doesn't work because $\Delta x^2$ and $\Delta v^2$ have different units, so you can't add them together. In the lingo, phase space is not a metric space: there's no meaningful concept of the distance between two arbitrary points in phase space. So even though the equation $mv^2 + kx^2 + k\lambda^2$ happens to look a lot like the equation of an ellipse, it doesn't actually define a geometrical ellipse, and there are no foci to speak of.

Of course, you can make a coordinate transformation on phase space that turns it into a metric space. All you have to do is change things so that the two coordinates use the same units. A general way to do this is to make the transformations

$$\begin{align}x&\to d_x = \frac{x}{x_0} & v&\to d_v = \frac{v}{v_0}\end{align}$$

In other words, instead of having the actual length and velocity be your coordinates, you express them as multiples of a standard length $x_0$ and a standard velocity $v_0$, and use those multiplying factors, which are dimensionless numbers, as your coordinates. When you do this, the equation becomes

$$\begin{align}m(d_v v_0)^2 + k(d_x x_0)^2 &= k\lambda^2 \\ \frac{d_v^2}{\frac{k}{m}\bigl(\frac{\lambda^2}{v_0^2}\bigr)} + \frac{d_x^2}{\bigl(\frac{\lambda^2}{x_0^2}\bigr)} &= 1 \end{align}$$

For convenience, I'll introduce the variable $\omega = \sqrt{\frac{k}{m}}$, in terms of which this becomes

$$\frac{d_v^2}{(\omega\lambda/v_0)^2} + \frac{d_x^2}{(\lambda/x_0)^2} = 1$$

The foci of this ellipse are located at

$$d_v = \pm\sqrt{\biggl(\frac{\omega\lambda}{v_0}\biggr)^2 - \biggl(\frac{\lambda}{x_0}\biggr)^2}$$

(let's just assume the square root is real). But you can make this number come out to be any value you want, because you have complete freedom to choose the units $v_0$ and $x_0$. So the locations of the foci are physically meaningless. This is what Michael was saying in the comments.