I cannot see how I can obtain the yellow highlighted section on the RHS from that of the LHS.

The following equation can be found in both my lecture notes(*1) (page 9, equation 2.7) and is also used as Problem 2-6 in Feynman & Hibbs, Quantum Mechanics and Path Integrals, (pg. 36),

$$ \require{color} \tag{9} \frac{d S_{cl}}{dt_b} = L(x_b,\dot{x}_b,t_b) = \colorbox{yellow}{ $\frac{\partial S_{cl}}{\partial t_b} + \frac{\partial S_{cl}}{\partial x_b}\dot{x}_b$ } $$

Below, is the relevant section of notes used to derive the expression and also my own attempt at getting the expression on the RHS.


Using (8), and integrating both sides (not entirely sure I can actually do this!),

$$ \tag{13} S_{cl} = p_b x_b $$

so using product rule and again (8) for $p_b$ in the second term,

$$ \tag{14} \frac{d S_{cl}}{dt_b} = \frac{d p_b}{d t_b}x_b + P_b\frac{d x_b}{d t_b} = \frac{d p_b}{d t_b}x_b + \frac{d S_{cl}}{d x_b}\dot{x_b} $$

However, I have now two problems:

  1. I have proper derivatives on the RHS, not partial derivatives
  2. I can only obtain $S_{cl}$ in the first term by treating $x_b$ as independent of $t_b$ and using (14).


This derivation copied directly from the notes mentioned above on page 9. The relevant equation is (8)

The action for a path $x(t)$,

$$ \tag{1} S[ x ( t ) ] = \int^{t_b}_{t_a} L\left(x,\dot{x},t\right) dt $$

where $L$ is the Lagrangian. Now, Using the principle of least action: the classical path $\bar{x}(t)$ is an extremum of the functional S,

$$ \tag{2} \left.\frac{\delta{S}}{\delta{x}} \right|_{x=\bar{x}}= 0 $$

where $\delta{S}/\delta{x}$ is the functional derivative. For a small variation of the path: $x(t) \to x(t) + \delta x(t)$:

$$ \tag{3} \delta S = S[x + \delta x] − S[x] = \int_{t_a}^{t_b} dt\ L\left(x + \delta{x},\dot{x} + \delta{\dot{x}},t\right) - L\left(x,\dot{x},t\right) $$

Using 2D Taylor expansion around $\delta{x},\delta{\dot{x}}$ we get

$$ \tag{4} \delta S = \int_{t_a}^{t_b} dt\ \left( \frac{\partial L}{\partial x}\delta x + \frac{\partial L}{\delta \dot{x}}\delta{\dot{x}} \right) +\mathcal{O}(\delta x^2) $$ Now integrating by parts,

$$ \tag{5} \delta S = \left[ \frac{\partial L}{\partial \dot{x}}\delta{\dot{x}} \right]^{t_b}_{t_a} - \int_{t_a}^{t_b} dt\ \left( \frac{d}{dt} \left( \frac{\partial L}{\partial x} \right) - \frac{\partial L}{\delta \dot{x}} \right) \delta x +\mathcal{O}(\delta x^2) $$

If the end-points of the path are fixed, i.e. $\delta x(t_a) = \delta x(t_b) = 0$, then the we obtain Lagrange’s equation for the classical path, $$ \tag{6} \frac{d}{dt} \left( \frac{\partial L}{\partial x} \right) - \frac{\partial L}{\delta \dot{x}} =0 $$

Considering the value of the action $S$ on the classical path, $S_{cl} \equiv S[\bar{x}(t)]$: $S_{cl}$ will be a function of the end points, i.e. of $x_a, t_a, x_b$, and $t_b$.

Varying the endpoint $(x_b, t_b)\to (x_b, t_b) + (\delta x_b, \delta t_b)$, but keep $(x_a, t_a)$ fixed. Lagrange’s equation holds for the classical path. We choose $\delta t_b = 0$ and remember canonical momentum $p$ conjugate to $x$ as $p = \frac{\partial L}{\partial \dot{x}}$ to get,

$$ \tag{7} \delta S_{cl} = \left[ \frac{\partial L}{\partial \dot{x}}\delta{\dot{x}} \right]^{t_b}_{t_a} = \left[ p\delta x \right]^{t_b}_{t_a} = p(t_b)\delta x_b −p(t_a)\delta x_a =p(t_b)\delta x_b $$ Hence,

$$ \tag{8} \frac{\partial S_{cl}}{\partial x_b} = p_b $$

Now considering $\frac{dS_{cl}}{dt}$. From (1) we may write:

$$ \require{color} \tag{9} \frac{d S_{cl}}{dt_b} = L(x_b,\dot{x}_b,t_b) = \colorbox{yellow}{ $\frac{\partial S_{cl}}{\partial t_b} + \frac{\partial S_{cl}}{\partial x_b}\dot{x}_b$ } $$

This gives,

$$ \tag{11} \frac{\partial S_{cl}}{\partial t_b} = L − p_b\dot{x}_b =−E_b $$ $$ \tag{12} E_b = -\frac{\partial S_{cl}}{\partial t_b} $$ where $E_b$ is the energy function or Hamiltonian.


(*1) Brian Pendleton, Quantum Theory, The University of Edinburgh, 2015


2 Answers 2


Comments to the question (v3):

  1. The boundary data $x_b$, $t_b$, $x_a$, $t_a$ are independent variables in the on-shell action $S_{\rm cl}(x_b,t_b;x_a,t_a)$, and we assume it makes sense to take partial derivatives wrt. each of them.

  2. One can in general not give a closed formula for the on-shell action $S_{\rm cl}(x_b,t_b;x_a,t_a)$ (that doesn't somehow refer to the off-shell action), only in special cases.

  3. OP's question is essentially about the proof of the Lemma in my Phys.SE answer here.

  4. It should probably be stressed that Ref. 1 implicitly assumes in the total time differentiation $$\frac{d S_{\rm cl}(x_b(t_b),t_b;x_a,t_a)}{dt_b} ~=~ L(x_b,\dot{x}_b(x_b,t_b;x_a,t_a),t_b), \qquad \tag{btwn 2.6 and 2.7 }$$ that the variation of boundary conditions is along the same classical path, see my above mentioned Phys.SE answer for details.


  1. Brian Pendleton, Quantum Theory Lecture Notes, University of Edinburgh, September 2015. The pdf file is available here.
  • $\begingroup$ This has helped me a little but still unable to see where you obtained eq. (13) in your SE answer that is linked $\endgroup$ Jan 1, 2016 at 15:58
  • $\begingroup$ Which step in eq. (13)? $\endgroup$
    – Qmechanic
    Jan 1, 2016 at 16:07
  • $\begingroup$ The bit that required the relation below that I did not know! I appreciated your detailed explanations on the linked answer though $\endgroup$ Jan 1, 2016 at 16:43

Taken directly from this link "From the Hamilton’s Variational Principle to the Hamilton Jacobi Equation", PSU-Physics Lecture Notes...

For an arbitrary function, $S = S(\mathbf{u}, \mathbf{v}, t)$,

$$ \frac{dS}{dt} = \dot{\mathbf{u}} \frac{\partial S}{\partial \mathbf{u}} + \dot{\mathbf{v}} \frac{\partial S}{\partial \mathbf{v}} + \frac{\partial S}{\partial t} $$ using, $$ \frac{\partial S}{\partial \mathbf{u}} = \left( \frac{\partial S}{\partial u_1}, \frac{\partial S}{\partial u_2}, \frac{\partial S}{\partial u_3} \right) $$

This solves the whole problem. It was simply that I did not know the above relation.

  • 1
    $\begingroup$ Another reason for the advantage of index notation, as one can write ${\rm d}S/{\rm d}t=\dot{u}_i\partial S/\partial u_i+\cdots$ for $i\in(1,3)$ (or 0,3 for 4D). $\endgroup$
    – Kyle Kanos
    Jan 1, 2016 at 16:29
  • $\begingroup$ Nice :) I'll leave it as the above to keep it straight forward to read though $\endgroup$ Jan 1, 2016 at 16:37
  • 1
    $\begingroup$ That formula is the chain rule for functions with explicit $t$-dependence, cf. e.g. this Phys.SE post. $\endgroup$
    – Qmechanic
    Jan 1, 2016 at 17:00
  • $\begingroup$ @Qmechanic thanks for naming it. I was struggling to find its proof $\endgroup$ Jan 1, 2016 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.