Momentum $p = \nabla S$

Question

My book mentions the following equation:

$$p = \nabla S\tag{1.2}$$ where $S$ is the action integral, nabla operator is gradient, $p$ is momentum.

After discussing it with @hft, on here, it turns out book must have meant impulse is only such at initial and final points.

$$ \frac{\partial S_{cl}}{\partial x_2} = \left.\frac{\partial L}{\partial \dot x}\right|_{x_{cl}}(t_2) = p_2 $$ and also $$ \frac{\partial S_{cl}}{\partial x_1} = -\left.\frac{\partial L}{\partial \dot x}\right|_{x_{cl}}(t_1) = p_1. $$

While this is definitely clear, I still feel doubtful as I asked my professor very abruptly(couldn't have chance to have conversation) and he said that $p = \nabla S$ not only applies to initial and final points, but everywhere.

I think, this must be correct. If we imagine that potential energy of the system is not dependent on time, $\frac{\partial L}{\partial \dot x}$ would give the non-time dependent function which means impulse or speed is always the same. Hence we can conclude $\frac{\partial S_{cl}}{\partial x} = p_x$ (note, i wrote with respect to x, not x2).

Q1: Is the assumption correct?

Q2: If the potential energy is still non-time dependent and kinetic energy is given by the $\frac{1}{2}mv^2$, i don't think $\frac{\partial S_{cl}}{\partial x} = p_x$ wouldn't hold true. I'm trying to see the case when to be careful about this. Let's assume potential energies are never time dependent. Would I still need to be careful? Would be good if you could summarize the thoughts.

Answer 1

While this is definitely clear, I still feel doubtful as I asked my professor very abruptly(couldn't have chance to have conversation) and he said that $p = \nabla S$ not only applies to initial and final points, but everywhere.

When we apply the variational principle to determine the classical path, we usually think of the endpoints as fixed. That is, we fix $x_1 = x(t_1)$ and $x_2 = x(t_2)$ where $x(t)$ is any path.

However, once we have determined the classical path (the path that extremizes the action), we can then consider what happens when the endpoint $x_2$ varies.

I will denote the classical path by $$ x_{cl}(t)\;, $$ where $x_{cl}(t_1) = x_1$ and $x_{cl}(t_2) = x_2$.

And I will denote the classical action by: $$ S_{cl} = S[x_{cl}(t)] = \int_{t_1}^{t_2} L(x_{cl}(t), \dot x_{cl}(t))dt $$

By varying the endpoint (currently called "$x_2$") we can show (as we did in the previous answer) that: $$ p(t_2) = \frac{\partial S_{cl}}{\partial x_2}\;. $$

Now, here is the interesting part. As long as we stay on the classical path, we can consider any point after the initial point $x_1$ to be the endpoint. That is, since we are now considing $x_2$ to be variable, we can, if we like, refer to it as "$x$" instead of "$x_2$" as long as we do not get confused with our previous notation $x(t)$.

Therefore, what your professor means is that $$ p = \frac{\partial S_{cl}}{\partial x} $$ can be applied to any point $x$ on the classical path. Where we have renamed the "endpoint" to be "$x$" rather than "$x_2$," since we are now able to let that "endpoint" vary to be anywhere we want on the classical path (not literally anywhere, but anywhere on the classical path, which is the path the particle actually takes).

Basically, what we are doing is letting $x_2$ vary (along the classical path) and calling it $x$ instead of $x_2$.

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Hence we can conclude $\frac{\partial S_{cl}}{\partial x} = p_x$ >(note, i wrote with respect to x, not x2).

Q1: Is the assumption correct ?

Yes, this is right, given that your $p_x$ means the momentum at the point $x$ on the classical path.

Answer 2

Eq. (1.2) depends on context:

1. If $S(q_f,t_f;q_i,t_t)$ is the on-shell action, then eq. (1.2) is eq. (11) in my Phys.SE answer here where a proof is given.

2. If $S(q,P,t)$ is Hamilton's principal function, then eq. (1.2) is one of the properties of $S(q,P,t)$ being a canonical transformation of type 2.