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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 $$dK/dx = f(v) = -2cK$$.

Mark Levi ([email protected]) is a professor of mathematics at the Pennsylvania State University.