I am trying to derive Hamilton's equations of motion without using Lagrange's method but am left with an additional factor of $1/2$. Where am I going wrong? Please note this in not a homework excercise, so full answers are appreciated.

My derivation is as follows:


For an Hamiltonian $ H $, given by

\begin{equation} H(q,p) = T(q,p) + U(q), \end{equation}

where $ T $ and $ U $ are the total kinetic energy and total potential energy of the system, respectively; $ q $ is a generalised position and; $ p $ is a generalised momentum. Using this notation, Hamilton's equations of motion are

\begin{align} \dot{q} &= \frac{\partial{H}}{\partial{p}},\\ \dot{p} &= -\frac{\partial{H}}{\partial{q}}. \end{align}

We know that

\begin{equation*} T = \frac{1}{2} m v^2 \end{equation*}

recalling that $ p = mv $

\begin{equation} T = \frac{p v}{2}, \end{equation}

and given that

\begin{equation} v = {\mathrm{d}{q}\over\mathrm{d}{t}} \equiv \dot{q}, \end{equation}

we find that

\begin{equation} T = \frac{p \dot{q}}{2}. \end{equation}


\begin{equation*} U = - W = - F q, \end{equation*}

where $ W $ is the work done. Given that

\begin{equation} F = {\mathrm{d}{p}\over\mathrm{d}{t}} \equiv \dot{p}, \end{equation}

we find that

\begin{equation} U = -\dot{p} q. \end{equation}

This, our Hamiltonian is

\begin{equation} H(q,p) = {p\dot{q}\over 2} - q\dot{p} \end{equation}

which has an infinitesimal change

\begin{equation} \mathrm{d}{H} = {\dot{q}\over 2}\,\mathrm{d}{p} - \dot{p}\,\mathrm{d}{q}, \end{equation}

In the following steps we divide through by the elements of $ \mathrm{d}{q} $ and $ \mathrm{d}{p} $, keeping in mind that the two are linearly independent and therefore do not depend on one-another:

\begin{align} {\partial{H}\over\partial{p}} &= {\dot{q}\over 2}\,\underbrace{{\mathrm{d}{p}\over\mathrm{d}{p}}}_{=1} - \dot{p}\,\underbrace{{\mathrm{d}{q}\over\mathrm{d}{p}}}_{=0},\\ {\partial{H}\over\partial{p}} &= {\dot{q}\over 2}.\ \ \ \ \ (*) \end{align}

\begin{equation} {\partial{H}\over\partial{q}} = {\dot{q}\over 2}\,\underbrace{{\mathrm{d}{p}\over\mathrm{d}{q}}}_{=0} - \dot{p}\,\underbrace{{\mathrm{d}{q}\over\mathrm{d}{q}}}_{=1}, \end{equation}

\begin{equation} {\partial{H}\over\partial{q}} = - \dot{p}. \end{equation}

Clearly equation (*) is not the correct equation of motion for Hamiltonian mechanics. Where did I go wrong?

  • $\begingroup$ You need to eliminate all references to $\dot q$ before taking derivatives. $\endgroup$ Sep 6, 2016 at 21:21

2 Answers 2


For an Hamiltonian $ H $, given by

\begin{equation} H(q,p) = T(q,p) + U(q), \end{equation} the Hamilton eqs. are

\begin{align} \dot{q} &= \frac{\partial{H}}{\partial{p}}=\frac{\partial{T}}{\partial{p}},\\ \dot{p} &= -\frac{\partial{H}}{\partial{q}}=-\frac{\partial{U}}{\partial{p}}, \end{align}

while that the kinetic term written with the momenta $p$ is given by $$ T=\frac{1}{2}m\dot{q}^2=\frac{p^2}{2m} $$ Recall $p=m\dot{q}$. Then you get that $\frac{\partial{T}}{\partial{p}}=p/m$ and Hamilton equations became \begin{align} \dot{q} &= \frac{p}{m},\\ \dot{p} &= -\frac{\partial{H}}{\partial{q}}=-\frac{\partial{U}}{\partial{q}}. \end{align}


As for your difficulty, take the kinetic energy term to be $$ K = \dfrac{p^2}{2m}$$ and you will get the correct answer.

However there are a few more issues with your derivation like you wrote $$F= p/t $$ etc. Please resolve them.


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