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In Gregory's Classical Mechanics there's a proof that when a standard system is conservative, the generalized forces $Q_j$ can be written as a potential. But I can't seem to explain some steps in the proof. It goes like this:

Let $q^A$ and $q^B$ be two points of the configuration space that can be joined by a straight line parallel to the $q_j$-axis (why is this necessary?). Then

$\int_{q_A}^{q_B} Q_jdq_j = \int_{q_A}^{q_B} (\sum_i F^S_i \cdot \frac{\partial r_i}{\partial q_j})dq_j = \sum_i \int_{C_i} F^S_i \cdot dr = V(q^A) - V(q^B) = - \int_{q_A}^{q_B} \frac{\partial V}{\partial q_j}dq_j $ and since these hold for any two points in configuration space thus described, the integrands are equal.

Some words on notation:

$F^S_i$ are the specified forces acting on the particle $P_i$.

$\int_{C_i} F^S_i \cdot dr$ means the work done by $F_i$ on a particle moving along any path connecting $q^A$ and $q^B.$

The first equality is just using the defintion, as is the last one. So I'm mainly wondering about the justification for the second and third equalities.

Edit: I suppose you can define $V := \sum_i V_i$ which explains the third equality. Is this correct?

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The condition that they can be joined by a straight line parallel to the $q_j$ axis is necessary because otherwise the first integral, $\int_{q_A}^{q_B}Q_j dq_j$ might be ill defined if there is no straight line connecting $q_A$ and $q_B$. Someone with more experience than me can probably tell you in which cases the configuration space is a simplex (in which case this is always possible) and when it might not be, thus making this requirement necessary.

In the second step, the author changes the order of integration and summation. Integration theory will tell you when that is possible mathematically, but generally speaking, for most physical situations, this can be done without worrying too much. Further he interprets the $q_j$ coordinate as a parametrisation of a special curve connecting $q_A$ and $q_B$. For conservative forces $F_i^S$, of course the integral is path independent, so we can as well write it as the integral over an arbitrary path.

The third equality is using the fact that, if an integral is path independent, there is a potential and the integral value only depends on the potential difference between beginning and end points. With your suggested definition $V=\sum_i V_i$ we arrive at the fourth expression, as all the $V_i$ are evaluated at the same points.

Edit:
Follow-Up questions by OP in the comments:

  • "do you mean to say every point on the straight line should be a possible configuration of the system (does this make any sense?)?"
    Yes, this is what I mean. I think it does not make sense to integrate over a line that is physically impossible.
  • "And am I right when I say the only reason the straight line should be parallel to the $q_j$-axis is that only the $q_j$-coordinate varies?"
    Yes, otherwise the parametrisation of the straight line from $q_A$ to $q_B$ can not be achieved using only the $q_j$ coordinate and the integral in the form above is not valid for the curve we want to integrate along.
  • "Is it then in principle enough to consider planar curves spanned by the $q_j$-axis and the points $q_A$ and $q_B$?"
    I am not quite sure I understand what you mean here. I think as long as the proof holds for any coordinate system and all of its coordinates, it should be valid in general.
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  • $\begingroup$ Thank you for your answer! Some follow-up questions: do you mean to say every point on the straight line should be a possible configuration of the system (does this make any sense?)? And am I right when I say the only reason the straight line should be parallel to the $q_j$-axis is that only the $q_j$-coordinate varies? Is it then in principle enough to consider planar curves in the plane spanned by the $q_j$-axis and the points $q^A$ and $q^B$? $\endgroup$ – Jan De Meyer Aug 2 '16 at 14:27
  • $\begingroup$ @JanDeMeyer I have edited my original answer; I hope this clarifies things a bit. $\endgroup$ – Sanya Aug 2 '16 at 14:37
  • $\begingroup$ Thank you again! What I meant with the 3rd comment was that I suppose the straight line hypotheses is in principle too strict, and that any curve parametrized by $q_j$ and connecting the points $q^A$ and $q^B$ would do. Of course this adds no physical content; I just wanted to clarify :) $\endgroup$ – Jan De Meyer Aug 2 '16 at 14:42
  • $\begingroup$ @JanDeMeyer yeah, I have to say this seems to strict to me as well. I would however claim that in any simple case where the configuration space is basically just $\mathbb{R}^n$ using a shift of origin and a rotation, a suitable coordinate system can always be found - this is however probably not always the case for any geometry. At this point, a comparison with other textbooks might be useful. $\endgroup$ – Sanya Aug 2 '16 at 14:50
  • $\begingroup$ I actually consulted Goldstein's Classical Mechanics, but his proof is completely different. Do you perhaps have any references? Possibly more abstract approaches that might answer the question? Or we can wait patiently until someone more knowledgable spots this question. $\endgroup$ – Jan De Meyer Aug 2 '16 at 15:04

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