- #26

- 19

- 0

Work = $\int \frac{d\; \vec{p}}{dt} \bullet d \vec{x} = m \int \frac{d^2 \vec{x}}{dt^2} \bullet d \vec{x}$.

But the kinetic energy is $\int \vec{p} \bullet d \vec{v} = m \int \frac{d \vec{x}}{dt} \bullet dfrac{d \vec{x}}{dt}$

It seems as though this is saying that the first derivative squared is equal to the second derivative of x with respect to t.

The same thing happens in the Schrodinger equation, when p is squared to set up the kinetic energy. The momentum operator in coordinate space is squared, and this is taken to be equivalent to the second derivative with respect to x, which is true when the function is the plane wave, but not in general. But, since all wave functions can be expanded in a fourier series or transform, we can get away with this seeming slip of the pen.

What is it about energy that, regardless of the functional dependence of the position with respect to time, we can unequivocally say that the first derivative squared is equal to the second derivative?