[markisus]

Solutions to linear differential equations are exponentials, and this may be what the author meant. In a state space model, we can write Newton's law under gravity if we let y(t) = [x(t), x'(t)]^T, let A = [ 0, 1; 0 0], let b = [0, -g]^T, and set y'(t) = A y(t) + b.

Indeed, the solution is y(t) = exp(A t) (y0 + int_0^t exp(-As) b ds).

In Einstein's theory, we have a generalization which says a stone's path through space-time follows a geodesic of the metric. In differential geometry, the exponential function, by definition, sends tangent vectors to geodesic curves.

So the author is right, under this interpretation.

[amluto]

> So the author is right, under this interpretation.

Huh?

You can say “state space,” invoke the general solution to inhomogeneous ODEs, and write it out with integrals and matrix exponentials, and the answer is still quadratic. Unless you consider t^n e^(0t) to be exponential. (Hint: what are the eigenvalues of A? One could start by calculating A^2.)

Which leads to one of the most important lessons from all of physics: you can take a problem, use a different technique to solve it, and you get the same answer! It’s magic.

[markisus]

I consider exp(At) to be an exponential expression in A. Depending on A, you may get a matrix whose entries are quadratic in t, as in this case. For another case, you may have trigonometric entries in t, as in the case of A = [0, 1; -1, 0].

Maybe you would personally say that neither of these cases is truly or essentially exponential, since we have more recognizable closed forms. But then you should also commit to saying exp(it) = cos(t) + i sin(t) is not truly exponential. I would find that a little strange, but to each his own.

Anyway, I agree that you get the same answer in the end, as you must.

[amluto]

I would say that a mass on a spring or a pendulum moves sinusoidally or harmonically (approximately, anyway), not exponentially. And I wouldn’t say that a ball falling through the air or sitting on the ground is moving exponentially, either, even if one could get at it via a matrix exponential.