Archive for November, 2013

Why is kinetic energy proportional to v^2?

November 23, 2013

I answered this question here, channeling this. That seems like the best answer – kinetic energy is proportional to v^2 because of Galilean relativity.

The more common answer, though, is to say that work is force times distance. Then because f=ma, we can show that the work done is proportional to the change in v^2. Ron, at least, is not satisfied with this approach, saying

The previous answers all restate the problem as “Work is force dot/times distance”. But this is not really satisfying, because you could then ask “Why is work force dot distance?” and the mystery is the same.

I’m not sure if this is the case. Suppose we assume that acceleration is some function of position, so

\mathbf{\ddot{x}} = a(\mathbf{x})

Then we have

\mathbf{\dot{x}}\mathbf{\ddot{x}} = \mathbf{\dot{x}} a(\mathbf{x})

\frac{1}{2}\frac{d}{dt}(\mathbf{\dot{x}}^2) = \frac{d}{dt} \int_{x_0}^{\mathbf{x}} a(\mathbf{x'}) ds

This shows that we can create a conserved quantity, namely \frac{1}{2}\dot{x}^2 - \int_{x_0}^{\mathbf{x}} a(\mathbf{x'}) ds, using just the knowledge f=ma, which doesn’t seem to be begging the question at all. It does, however, assume that the line integral is independent of path. So if we’re told f=ma and told that f is a gradient, we can find that kinetic energy is proportional to v^2. I suppose we ought to be able to, since from f=ma it is manifest that physics has Galilean invariance, and the thought experiment linked up top shows that energy depends on v^2.

So both explanations seem correct, but the one based on invariance certainly feels better.