A mathematically illogical argument in the derivation of Hamilton's equation in Goldstein

Question

In the book of Goldstein, at page 337, while deriving the Hamilton's equations (canonical equations), he argues that

The canonical momentum was defined in Eq. (2.44) as $p_i = \partial L / \partial \dot q_i$; substituting this into the Lagrange equation (8.1), we obtain

$$ \dot p_i= \frac{\partial L}{\partial q_i} \tag{8.14}$$

so Eq. (8.13) can be written as

$$ dL = \dot p_i dq_i + p_id \dot q_i + \frac{\partial L}{\partial t}dt \tag{8.13′}$$

The Hamiltonian $H(q,p,t)$ is generated by the Legendre transformation

$$ H(q,p,t) = \dot q_i p_i - L(q, \dot q, t), \tag{8.15}$$

which has the differential

$$ dH = \dot q_i d p_i - \dot p_i d q_i - \frac {\partial L}{\partial t}, \tag{8.16} $$

where the term $p_i d \dot q_i$ is removed by the Legendre transformation. Since $dH$ can also be written as

$$ dH = \frac{\partial H}{\partial q_i}d q_i + \frac{\partial H}{\partial p_i}d p_i + \frac{\partial H}{\partial t}d t, \tag{8.17} $$

However, if $H$ is defined to be a function of $q,p,t$, then how can we define $H(q,p,t) = \dot q *p - L(q,\dot q,t)$, i.e $\dot q$ is not an argument of $H$ whereas it is in its definition.

Moreover, when he is taking the differential of $H$, he argues that $pd\dot q$ is removed, but he does not say why.

I mean mathematically speaking this whole argument is plan wrong, as far as I can see, so assuming that it is not the case, what am I missing in here ?

Answer 1

However, if $H$ is defined to be a function of $q,p,t$, then how can we define $H(q,p,t) = \dot q *p - L(q,\dot q,t)$, i.e $\dot q$ is not an argument of $H$ whereas it is in its definition.

As usual in a Legendre transform, the above expression for $H$ should be understood as a shortened notation for $$ H(q,p,t) = \dot q(q,p,t) \cdot p-L(q,\dot q(q,p,t),t) $$ where $\dot q(q,p,t)$ is obtained by inverting the definition of $p$ $$ p = \frac{\partial L}{\partial \dot q}(q, \dot q, t) $$ to obtain the function $\dot q(q,p,t)$.

Answer 2

$\boldsymbol{\S\:}\textbf{A. In General}$

Consider a real function $\:f\left(x\right)\:$ of a real variable $x \in \left[\alpha,\beta\right]$ with continuous 1st and 2nd derivatives. Suppose that its 2nd derivative is everywhere negative so that its graph in the $\:xy-$plane is as in Figure-01. From every point of the graph, we have a tangent line.

Now, the graph of the function could be sketched by the family of the tangent lines, see Figure-02. We say that this curve (graph) is the envelope of the family of the tangent lines. From this fact we note that we could define the function $\:f\left(x\right)\:$ by the family of its tangent lines. Indeed, as shown in Figure-03, if from the angle $\:\theta\:$ of any tangent line we know the point where this line intersects the $\:y-$axis, let $\:\boldsymbol{-}\omega\:$ (the minus sign used for future purposes), then we would have an equivalent definition of the function $\:f\left(x\right)$. So, we must have the function $\:\omega\left(\theta\right)$. For the domain of angle $\:\theta\:$ we have from Figure-03 as example

\begin{equation} \theta \in \left[\theta_1,\theta_2\right] \quad \text{where} \quad \theta_1\boldsymbol{=}\min{(\theta_\alpha,\theta_\beta)}\quad \text{and} \quad \theta_2\boldsymbol{=}\max{(\theta_\alpha,\theta_\beta)} \tag{A-01}\label{A-01} \end{equation}

Instead of using the angle $\:\theta\:$ we equally well use the variable $\:u\boldsymbol{=}\tan\theta\boldsymbol{=}\dfrac{\mathrm df}{\mathrm dx}$. For the domain of $\:u\:$ we have
\begin{equation} u \in \left[u_1,u_2\right] \quad \text{where} \quad u_1\boldsymbol{=}\min{(\tan\theta_\alpha,\tan\theta_\beta)}\quad \text{and} \quad u_2\boldsymbol{=}\max{(\tan\theta_\alpha,\tan\theta_\beta)} \tag{A-02}\label{A-02} \end{equation}

From Figure-03 we have \begin{equation} y\boldsymbol{+}\omega\boldsymbol{=}\tan\theta \cdot x\boldsymbol{=}u \cdot x \tag{A-03}\label{A-03} \end{equation} so \begin{equation} \boxed{\:\:\omega\left(u\right)\boldsymbol{=}u \cdot x\boldsymbol{-}f\left(x\right)\vphantom{\dfrac{a}{b}}\:\:} \tag{A-04}\label{A-04} \end{equation} Now looking in above equation it seems mathematically illogical the argument that the function $\:\omega\:$ doesn't depend on the variable $\:x\:$ and must we write \begin{equation} \omega\left(u,x\right)\stackrel{???}{\boldsymbol{=}}u \cdot x\boldsymbol{-}f\left(x\right) \tag{A-05}\label{A-05} \end{equation} But this is not this case here because from \eqref{A-04} \begin{equation} \dfrac{\partial\omega}{\partial x}\boldsymbol{=}u \boldsymbol{-}\dfrac{\partial f}{\partial x}\boldsymbol{=}\dfrac{\mathrm df}{\mathrm dx} \boldsymbol{-}\dfrac{\mathrm df}{\mathrm dx}\boldsymbol{=}0 \tag{A-06}\label{A-06} \end{equation} that is $\:\omega\:$ is independent of $\:x$. It depends only on $\:u\:$ that's why we write $\:\omega\left(u\right)$.

In Figure-04 this fact is explained graphically : Suppose that a value $\:u\in \left[u_1,u_2\right]\:$ is given. This is like to give a direction, that is a line $\:\varepsilon\:$ at an angle $\:\phi\boldsymbol{=}\arctan(u)$. We find a unique line $\:\varepsilon_t\:$ tangent to the curve-graph of $\:f\left(x\right)\:$ and parallel to $\:\varepsilon\:$ which intersects the $\:y-$axis at $\:\boldsymbol{-}\omega(u)$. Beyond the value of the independent variable $\:u\:$ there is no need of any value of $\:x$. To the contrary, this value of $\:x\:$ is determined underground automatically from the contact point of the tangent line $\:\varepsilon_t\:$ with the graph.

We call the function $\:\omega\left(u\right)\:$ the Legendre transform of the function $\:f\left(x\right)\:$ with respect to the variable $\:x$.

Note that differentiating \eqref{A-04} with respect to $\:u\:$ we have \begin{equation} x\boldsymbol{=}\dfrac{\mathrm d\omega\left(u\right)}{\mathrm du} \tag{A-07}\label{A-07} \end{equation} So, the function $\:f\left(x\right)\:$ and its Legendre transform with respect to $\:x\:$, that is the function $\:\omega\left(u\right)$, fulfill the following set of equations \begin{align} f\left(x\right) \boldsymbol{+}\omega\left(u\right) & \boldsymbol{=}u \cdot x \tag{A-08a}\label{A-08a}\\ u & \boldsymbol{=}\dfrac{\mathrm df\left(x\right)}{\mathrm dx} \tag{A-08b}\label{A-08b}\\ x & \boldsymbol{=}\dfrac{\mathrm d\omega\left(u\right)}{\mathrm du} \tag{A-08c}\label{A-08c} \end{align}

If in above equations we interchange the roles as follows \begin{align} f & \boldsymbol{\rightleftarrows} \omega \tag{A-09a}\label{A-09a}\\ x & \boldsymbol{\rightleftarrows} u \tag{A-09b}\label{A-09b} \end{align} then equations \eqref{A-08a},\eqref{A-08b} and \eqref{A-08c} give respectively \begin{align} \omega\left(u\right)\boldsymbol{+} f\left(x\right)& \boldsymbol{=}x \cdot u \tag{A-10a}\label{A-10a}\\ x & \boldsymbol{=}\dfrac{\mathrm d\omega\left(u\right)}{\mathrm du} \tag{A-10b}\label{A-10b}\\ u & \boldsymbol{=}\dfrac{\mathrm df\left(x\right)}{\mathrm dx} \tag{A-10c}\label{A-10c} \end{align}

But this set of equations is identical to that of (A-08) : The function $\:f\left(x\right)\:$ is the Legendre transform of $\:\omega\left(u\right)$ with respect to $\:u$. That is application of two successive Legendre transformations returns the initial function.

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$\boldsymbol{\S\:}\textbf{B. Classical Mechanics - Lagrange and Hamilton functions}$

In Classical Mechanics the Euler-Lagrange equation of motion for one degree of freedom is \begin{equation} \dfrac{\mathrm d}{\mathrm d t}\left(\dfrac{\partial L}{\partial\dot q}\right)\boldsymbol{-}\dfrac{\partial L}{\partial q}\boldsymbol{=}0 \tag{B-01}\label{B-01} \end{equation} where \begin{align} L\left(q,\dot q,t\right) & \boldsymbol{\equiv}\text{the Lagrange function} \tag{B-02a}\label{B-02a}\\ q & \boldsymbol{\equiv}\text{the generalized coordinate} \tag{B-02b}\label{B-02b}\\ \dot q & \boldsymbol{\equiv}\dfrac{\mathrm d q}{\mathrm d t} \tag{B-02c}\label{B-02c} \end{align} For the Legendre transform of the Lagrange function $\:L\left(q,\dot q,t\right)\:$ with respect to the independent variable $\:\dot q\:$ we replace all Variables, Functions and Differential Operators in $\:\boldsymbol{\S\:}\textbf{A}\:$ as follows \begin{align} \text{Variables}\:\:\: : \:\:\:& \left. \begin{cases} x\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} \dot q\\ u\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} p \end{cases}\right\} \tag{B-03a}\label{B-03a}\\ \text{Functions}\:\:\: : \:\:\:& \left. \begin{cases} f\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} L\\ \omega\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} H \end{cases}\right\} \tag{B-03b}\label{B-03b}\\ \text{Operators}\:\:\: : \:\:\:& \left. \begin{cases} \dfrac{\mathrm d \hphantom{x}}{\mathrm d x}\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} \dfrac{\partial \hphantom{x}}{\partial \dot q}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \dfrac{\mathrm d \hphantom{u}}{\mathrm d u}\!\!\! &\!\!\! \boldsymbol{-\!\!\!-\!\!\!-\!\!\!\rightarrow} \dfrac{\partial \hphantom{p}}{\partial p} \end{cases}\right\} \tag{B-03c}\label{B-03c} \end{align} Equations \eqref{A-08a},\eqref{A-08b} and \eqref{A-08c} give respectively \begin{align} H\left(q,p,t\right)\boldsymbol{+} L\left(q,\dot q,t\right) & \boldsymbol{=}p\,\dot q \tag{B-04a}\label{B-04a}\\ p & \boldsymbol{=}\dfrac{\partial L\left(q,\dot q,t\right)}{\partial \dot q} \tag{B-04b}\label{B-04b}\\ \dot q & \boldsymbol{=}\dfrac{\partial H\left(q,p,t\right)}{\partial p} \tag{B-04c}\label{B-04c} \end{align} So the Legendre transform of the Lagrange function $\:L\left(q,\dot q,t\right)\:$ with respect to the independent variable $\:\dot q\:$ is the Hamilton function $\:H\left(q,p,t\right)\:$, where from \eqref{B-04a} \begin{equation} H\left(q,p,t\right) \boldsymbol{=}p\,\dot q\boldsymbol{-} L\left(q,\dot q,t\right) \tag{B-05}\label{B-05} \end{equation} In the spirit of the discussion in $\:\boldsymbol{\S\:}\textbf{A}\:$ the Hamilton function $\:H\left(q,p,t\right)\:$ is independent of the variable $\:\dot q$, it depends on the independent variable $\:p\boldsymbol{\equiv}\text{the generalized momentum}$.

Equation \eqref{B-05} yields \begin{equation} \dfrac{\partial H\left(q,p,t\right)}{\partial q}\boldsymbol{=}\boldsymbol{-}\dfrac{\partial L\left(q,\dot q,t\right)}{\partial q} \tag{B-06}\label{B-06} \end{equation} From this equation and the definition of $\:p$, see equation \eqref{B-04b}, the Euler-Lagrange equation of motion \eqref{B-01} gives \begin{equation} \dot p \boldsymbol{=}\boldsymbol{-}\dfrac{\partial H\left(q,p,t\right)}{\partial q} \tag{B-07}\label{B-07} \end{equation} Equations \eqref{B-04c} and \eqref{B-07} together constitute the Hamilton equations of motion \begin{equation} \text{Hamilton equations of motion}\:\:\: : \:\:\: \left. \begin{cases} \dot q & \!\!\boldsymbol{=}\boldsymbol{+}\dfrac{\partial H\left(q,p,t\right)}{\partial p}\vphantom{\dfrac{a}{\dfrac{a}{b}}}\\ \dot p & \!\!\boldsymbol{=}\boldsymbol{-}\dfrac{\partial H\left(q,p,t\right)}{\partial q} \end{cases}\right\} \tag{B-08}\label{B-08} \end{equation}

Answer 3

The Formula Goldstein has given (8.15) is not a definition of the Hamiltonian (because you are right in that, the Formula depends on $\dot{q}$, which is not an argument of the Hamiltonian. However, the formla can be understood as an equation we want $H$ to satisfy if the variable $p$ satisfies \begin{align} p = \frac{\partial L}{\partial \dot{q}}(q, \dot{q}, t) \end{align}

Unlike suggested in the previous version of this answer, $p$, $q$ and $\dot{q}$ can be independent variables in these equations.

By that now is also clear why the $p \dot{q}$ vanishes here:The differential of the term $\dot{q} p$ in formula 8.15 cancels with the one from the differential of $L$ in 8.13.

Written down: \begin{align} dH = d \dot{q} p - \dot{q} dp - dL \end{align}
With $dL$ from 8.13, you arrive at the same formula Goldstein arrives at.

Important Note from my side: Goldstein argues with the Legendre Transform here when talking about why the differential vanishes. In fact, the way he "defined" $H$ is a Legendre Transformation. However, since he started to define $H$ without making use of the term "Legendre-Transform", he could have argued without it later on as well when talking about the differentials. As I did, you can perfectly understand why $d \dot{q} p$ vanishes without making use of the term "Legendre-Transformation". Conversely, when Goldstein writes that $d \dot{q} p$ vanishes because of the "Legendre-Transformation", he implicitly means exactly what I wrote down.

Answer 4

Right, $\dot{q}$ is not an agrument of $H$ and you will only see it afterwards, but it is a function of time so you must make $dH$ from the definition (8.15) keeping it in mind and you must use $dL$ from (8.13'). Then you arrive at the right Hamiltonian differential. Note, eq. (8.16) misses a factor $dt$ at the last term (a typo).

Answer 5

First lets try out the legendre transformation on a particular example.

$$ L = \frac12 m \dot{q}^2 - V(q), $$

according to Goldstein the hamiltonian for this system is,

$$ H = \dot{q} p - L,$$

initially we think of $p$ and $\dot{q}$ as being independent variables. If we take $\partial H / \partial \dot{q}$ we will get,

$$ \frac{\partial H}{\partial \dot{q}} = p - \frac{\partial L}{\partial \dot{q}},$$

if we now constrain ourselves to the surface $p=\frac{\partial L}{\partial \dot{q}} $ we find that the partial derivative of $H$ with respect to $\dot{q}$ vanishes.

For the purposes of calculating the dynamics then we would constrain our results,

$$ H\Big|_{p=m\dot{q}} = \Big( p\dot{q} - L(\dot{q},q)\Big) \Big|_{p=m\dot{q}}$$

$$ H\Big|_{p=m\dot{q}} = \Big( \frac{p^2}{m} - L(p/m,q)\Big) \Big|_{p=m\dot{q}}$$

$$ H\Big|_{p=m\dot{q}} = \Big( \frac{p^2}{m} - \frac{p^2}{2m} + V(q) \Big) \Big|_{p=m\dot{q}}$$

$$ H\Big|_{p=m\dot{q}} = \Big( \frac{p^2}{2m} + V(q) \Big) \Big|_{p=m\dot{q}}$$

This kind of "constraining our variables after the fact" practice is very common in Classical Mechanics.