Hamiltonian Formulation Questions and Answers

Q: Hamilton's principal function S and Hamilton's characteristic function W for a conservative system, are related as :

The relation is \(S(q_{k}, t) = W(q_{k}) - Et\).For a conservative system the Hamiltonian does not depend on time explicitly, so in the Hamilton-Jacobi equation \(H(q, \partial S/\partial q) + \partial S/\partial t = 0\) the time dependence separates. With \(H = E\), a constant:\[ \begin{aligned} \frac{\partial S}{\partial t} &= -E \\ S &= W(q) - Et \end{aligned} \]where W, the characteristic function, satisfies the time-independent equation \(H(q, \partial W/\partial q) = E\).The options with +Et have the wrong sign, and those adding or subtracting E alone are dimensionally inconsistent, since S has the dimensions of action (energy × time).

Q: A particle of mass m is falling freely vertically in a uniform gravitational field. The Hamilton-Jacobi equation describing its motion is :

The equation is the one in \(z\) alone:\[ \begin{aligned} \frac{1}{2m}\left(\frac{\partial s}{\partial z}\right)^{2} + mgz \\ + \frac{\partial s}{\partial t} = 0 \end{aligned} \]The Hamilton-Jacobi equation is \(H(q, \partial S/\partial q) + \partial S/\partial t = 0\). A particle falling vertically has one coordinate, \(z\), and Hamiltonian \(H = \frac{p_{z}^{2}}{2m} + mgz\). Replacing \(p_{z}\) by \(\frac{\partial S}{\partial z}\) gives the third option.The second option is the general three-dimensional form; for a purely vertical fall the \(x\) and \(y\) terms do not enter the motion described. The first uses the horizontal momenta only, and the last omits \(\frac{\partial S}{\partial t}\), which belongs in the time-dependent equation.

Q: The Hamiltonian of a free particle is \(H = \frac{p^{2}}{2m}\). The phase space will be :(Where, m, p and E are mass, momentum and energy of the particle, respectively)

The phase-space trajectory is a straight line parallel to the q-axis, intersecting the p-axis.For a free particle the Hamiltonian does not depend on \(q\), so Hamilton's equations give \(\dot{p} = -\partial H/\partial q = 0\): the momentum is constant. With energy \(E\),\[ \begin{aligned} p &= \pm\sqrt{2mE} = \text{constant} \\ \dot{q} &= p/m \end{aligned} \]so the point moves along a horizontal line \(p = \sqrt{2mE}\) (or \(-\sqrt{2mE}\)) in the \((q, p)\) plane, cutting the \(p\)-axis at that value while \(q\) increases steadily.A line parallel to the \(p\)-axis would mean fixed position with changing momentum, the opposite of free motion. Sloped lines would mean \(p\) changes with \(q\); \(\pm\sqrt{2mE}\) is the intercept, not a slope.

11.

Hamilton's principal function S and Hamilton's characteristic function W for a conservative system, are related as :

A.\(S(q_{k}, t) = W(q_{k}) - E\)
B.\(S(q_{k}, t) = W(q_{k}) + E\)
C.\(S(q_{k}, t) = W(q_{k}) + Et\)
D.\(S(q_{k}, t) = W(q_{k}) - Et\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The relation is \(S(q_{k}, t) = W(q_{k}) - Et\).

For a conservative system the Hamiltonian does not depend on time explicitly, so in the Hamilton-Jacobi equation \(H(q, \partial S/\partial q) + \partial S/\partial t = 0\) the time dependence separates. With \(H = E\), a constant:

\[ \begin{aligned} \frac{\partial S}{\partial t} &= -E \\ S &= W(q) - Et \end{aligned} \]

where W, the characteristic function, satisfies the time-independent equation \(H(q, \partial W/\partial q) = E\).

The options with +Et have the wrong sign, and those adding or subtracting E alone are dimensionally inconsistent, since S has the dimensions of action (energy × time).

Discuss in Forum

12.

A particle of mass m is falling freely vertically in a uniform gravitational field. The Hamilton-Jacobi equation describing its motion is :

A.\(\frac{1}{2m}\Big[\left(\frac{\partial s}{\partial x}\right)^{2}\) \(+ \left(\frac{\partial s}{\partial y}\right)^{2}\Big]\) \(+ mgz + \frac{\partial s}{\partial t} = 0\)
B.\(\frac{1}{2m}\Big[\left(\frac{\partial s}{\partial x}\right)^{2}\) \(+ \left(\frac{\partial s}{\partial y}\right)^{2}\) \(+ \left(\frac{\partial s}{\partial z}\right)^{2}\Big]\) \(+ mgz + \frac{\partial s}{\partial t} = 0\)
C.\(\frac{1}{2m}\Big[\left(\frac{\partial s}{\partial z}\right)^{2}\Big]\) \(+ mgz + \frac{\partial s}{\partial t} = 0\)
D.\(\frac{1}{2m}\Big[\left(\frac{\partial s}{\partial z}\right)^{2}\Big] + mgz = 0\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The equation is the one in \(z\) alone:

\[ \begin{aligned} \frac{1}{2m}\left(\frac{\partial s}{\partial z}\right)^{2} + mgz \\ + \frac{\partial s}{\partial t} = 0 \end{aligned} \]

The Hamilton-Jacobi equation is \(H(q, \partial S/\partial q) + \partial S/\partial t = 0\). A particle falling vertically has one coordinate, \(z\), and Hamiltonian \(H = \frac{p_{z}^{2}}{2m} + mgz\). Replacing \(p_{z}\) by \(\frac{\partial S}{\partial z}\) gives the third option.

The second option is the general three-dimensional form; for a purely vertical fall the \(x\) and \(y\) terms do not enter the motion described. The first uses the horizontal momenta only, and the last omits \(\frac{\partial S}{\partial t}\), which belongs in the time-dependent equation.

Discuss in Forum

13.

The Hamiltonian of a free particle is \(H = \frac{p^{2}}{2m}\). The phase space will be :

(Where, m, p and E are mass, momentum and energy of the particle, respectively)

A.A straight line parallel to q-axis, intersecting the p-axis
B.A straight line parallel to p-axis, intersecting the q-axis
C.A straight line with slope \(\sqrt{2mE}\)
D.A straight line with slope \(-\sqrt{2mE}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The phase-space trajectory is a straight line parallel to the q-axis, intersecting the p-axis.

For a free particle the Hamiltonian does not depend on \(q\), so Hamilton's equations give \(\dot{p} = -\partial H/\partial q = 0\): the momentum is constant. With energy \(E\),

\[ \begin{aligned} p &= \pm\sqrt{2mE} = \text{constant} \\ \dot{q} &= p/m \end{aligned} \]

so the point moves along a horizontal line \(p = \sqrt{2mE}\) (or \(-\sqrt{2mE}\)) in the \((q, p)\) plane, cutting the \(p\)-axis at that value while \(q\) increases steadily.

A line parallel to the \(p\)-axis would mean fixed position with changing momentum, the opposite of free motion. Sloped lines would mean \(p\) changes with \(q\); \(\pm\sqrt{2mE}\) is the intercept, not a slope.

Hamiltonian Formulation Questions and Answers

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