# Lie Brackets and Bulletproof Vests

A bullet punctures a wall made of two thin layers \(A\) and \(B\) of different materials (see Figure 1). Does the exit velocity change if we turn the two-layered “vest” inside out, as in the figure? It turns out that it does, and the difference \(v_2-v_1\) of exit velocities is given by a certain Lie bracket. Let us assume, perhaps unrealistically, that the bullet’s deceleration is a given function of its velocity, \(dv/dt = a(v)\); for example, one could take \(a(v)= -k v ^2\). The velocity considered as a function of the position \(x\) (rather than the time) satisfies the differential equation \(dv/dx = f(v)\) with \(f(v)=a(v)/v\). Indeed,

\[\frac{dv}{dx} = \frac{dv}{dt} \frac{dt}{dx} = \frac{a(v)}{v} = f(v). \tag1\]

Similarly, for the second material we have \(dv/dx = g(v)\) for some other function \(g\). We thus have two vector fields, \(f\) and \(g\), in the one-dimensional *velocity space*; the position \(x\) plays the role of time, since we eliminated \(t\) in \((1)\).

**Figure 1.**Does the exit velocity change if the two layers are permuted? Figure credit: Mark Levi.

*I claim that the effect of swapping \(A\) and \(B\) is given by the Lie bracket:*

\[v_2-v_1= (g ^\prime f - f^\prime g)hk+ \ldots= [g, f] hk + \ldots, \tag2 \]

for small \(h\) and \(k\). Here, \(\ldots = o (h ^2 + k ^2 )\).

The Lie bracket \([g, f]\) can be the difference between life and death (although this difference is small).

### Proof

Let \(F^x\) denote the flow of the vector field \(f\). In other words, \(v_0\) becomes \(F^x v_0\) after the bullet travels distance \(x\) in material \(A\); the flow \(G^x\) is defined similarly for material \(B\). Then

\[v_1 = G^kF^hv_0, \: \: \: v_2=F^hG^k v_0,\]

and

\[v_2-v_1 = F^hG^k v_0-G^kF^hv_0.\]

This difference is given by the right side of \((2)\), as verified by a Taylor expansion of \(G^x v_0\) and \(F^x v_0\) to second order; for example,

\[F^x v_0 = v_0 + f x +\frac{1}{2} f ^\prime f x^2 + \ldots.\]

I omit further details.

### An example

If \(f(v)=- v ^ \alpha\), \(g (v)= -v^ \beta\), then we get \([f,g]= ( \beta - \alpha )v^{ \alpha + \beta -1}\), and so only for \(\alpha = \beta\) is the "vest" reversible.

I conclude with a trivial but curious observation, worth mentioning in an ordinary differential equation or mechanics course or a faculty lounge: if \(dv/dt = f(v)\), then \(K = \frac{v ^2 }{2}\) satisfies \(dK/dx = f(v)\). Thus, \(dv/dt=dK/dx\). *Velocity changes with respect to time exactly like kinetic energy changes with respect to distance*.

As a special case of this remark, *the kinetic energy decays exponentially with the distance* for the commonly-assumed quadratic drag \(f = -c v ^2\). Indeed, we have \(dK/dx = f(v) = -2cK\).