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After the tremendous help from @hft on my previous question, after thinking, new question popped up.

I want to compare how things behave when we do: $\frac{\partial S}{\partial t_2}$ and $\frac{\partial S}{\partial x_2}$. One is with respect to position and one with time.

It's easy to see that changing $\delta t_2$ results in new classical path, from which I understand the derivation. The more interesting question is, what happens when changing $x_2$? It should still cause a path change - even though time endpoints is the same, if $x_2$ changed, then in the same time, object should cover different distance - definitely path changing.

So by following the logic I said, I got:

$$S_2 - S_1 = \delta S = \delta x_b \frac{\partial L}{\partial x}(t_b) = p_2.$$

This was derived by $S1$ (original action) - $S2$ (new resulting path action). ofc, in this way, $t2$ was fixed.

Now, I wonder, how do I derive $\frac{\partial S}{\partial x} = p$? This is different since this is the formula for any point of $x$. The difficulty I have is if we change any $x$ on our path, ofc, this results in a new classical path, but I can't do anymore:

$$S_2 - S_1 = \delta S = \int_{t_1}^{t_2} [L(r, r') - L(r - \eta, r' - \eta')] dt$$

and I seem to be struggling now, I could solve it when for change of $x2$, but for change of any $x$, I'm stuck.

Also, my second question would be that feynman says that changing time endpoint results in a new classical path, but doesn't say the same when changing position endpoint. Why? changing position endpoint still results in a new classical path even though time endpoint is fixed.

Qmechanic
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Giorgi
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1 Answers1

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  1. On one hand, the variation of the on-shell action $S(q_f,t_f;q_i,t_i)$ wrt. the endpoint positions $q_i$ and $q_f$ is actually the simplest as the proof only involves vertical variation $\delta q$ with no horizontal variation $\delta t=0$, cf. eq. (11) in my Phys.SE answer here.

  2. On the other hand, the variation of the on-shell action $S(q_f,t_f;q_i,t_i)$ wrt. the endpoint times $t_i$ and $t_f$ is more challenging as the proof involves both vertical and horizontal variations, cf. eq. (12) in my Phys.SE answer here.

The classical path is typically changed in both cases.

Qmechanic
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  • so I am right that even position endpoint causes classical path to change. While I get that, I need to see why dS/dx= p for all points(not just for position endpoint variantions). https://physics.stackexchange.com/a/773015/366606 This kind of answers it, but @hft means i think that it does not result in new classical path, but it does. I understand endpoint differentiation result but I again repeat - i dont get how dS/dx = p for all points – Giorgi Jul 25 '23 at 11:14
  • I agree: Variations other than endpoint variations of the on-shell action do not make sense. – Qmechanic Jul 25 '23 at 12:15
  • The book actually mentions it without derivation and I'm having a hard time deriving it. you also agree that even if we change position endpoint, we still get left with a new classical path, right ? if so, do you have any idea what @hft means here - physics.stackexchange.com/a/773015/366606 ? I thought I had understood it, but clearly, don't get it – Giorgi Jul 25 '23 at 12:20