Conservation law for Newton mechanics

Question

Suppose potential is given by $V(x) = -int^x F(x')dx'$,where $F$ is force. To show energy $E(x,v)= frac{1}{2}mv^2 +V(x)$ is conserved when Newton's laws holds.
We need to use the fact that $V(x)$ is differentiable,and we know $V(x)$ given in closed interval is differentiable if and only $V(x)$ is absolutely continuous, and if $F(x)in L^1[a,b]$, $V(x)$ is absolute continuous.
So does this imply conservation law holds generally, even if $F(x)$ is a delta function? (since it is integrable.)
By the way, I feel a bit uncomfortable with the derivation, like this, since it always ignores the condition that is needed to let something hold. But sometimes in physics, it indeed considers the special case, that some condition may not hold.
For example, delta potential in QM does not seem to be a good "smooth" function, but it's taken into consideration.

Han-Kwang Nienhuys · Answer

In 1D (potential and force is a function of $x$ only), I'd say it holds. A Dirac delta function is an abstraction that generally doesn't exist in reality (because infinite forces don't exist), but is an approximation of "a peak that is so narrow that I don't really care about the width, but only about the area under the peak".
For example, around an n-p junction in a diode, or near an electrode in a conductive solution, there will be a very sharp jump in the electrical potential. Depending on what you're interested in, you could approximate the step in potential as a discontinuity (derivative is a delta function) or not.
I recommend against defining the potential as the antiderivative of the force. Although it works for a 1D case, it breaks for 2D or higher dimensions. For example,
$$ mathbf{F}(x, y) = (x^2 + y^2)(y, -x) $$
is continuous and differentiable, but it is not a conservative force field for e.g. a loop around the origin:
$$oint mathbf{F}cdot dmathbf{r} neq 0.$$
If $V(mathbf{r})$ is a potential, then $mathbf F=-nabla V$ is always correct.

Conservation law for Newton mechanics

One Answer

Add your own answers!

Ask a Question