# Question

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.

# Attempt

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).

# Derivation

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.

# References

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

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.

References:

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

• 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). Jan 1, 2016 at 16:29
• Nice :) I'll leave it as the above to keep it straight forward to read though Jan 1, 2016 at 16:37
• That formula is the chain rule for functions with explicit $t$-dependence, cf. e.g. this Phys.SE post. Jan 1, 2016 at 17:00
• @Qmechanic thanks for naming it. I was struggling to find its proof Jan 1, 2016 at 17:02