I've recently come across a strange result when comparing the Hamiltonian and Lagrangian formulations of classical mechanics.
Suppose we are working in the regime where we can say the Hamiltonian $H$ is equal to the total energy $$H=T+V.\tag{1}$$ That is, the constraints are holonomic and time-independent, and the potential is $V=V(q)$ where $q$ a the generalized position vector $q=(q_1,q_2,\ldots,q_n)$. Let $$L=T-V\tag{2}$$ be the Lagrangian.
Now, the Euler-Lagrange equations tell us
$$\frac{d}{dt}\frac{\partial L}{\partial \dot{q_\sigma}} – \frac{\partial L}{\partial q_\sigma} = 0,\tag{3}$$
for the generalized coordinate $q_\sigma,$ with $\sigma\in\{1,\ldots,n\}$.
We also know that the conjugate momenta are defined by $p_\sigma = \frac{\partial L}{\partial \dot{q_\sigma}}$. So this equation tells us
$$\dot{p_\sigma} – \frac{\partial L}{\partial q_\sigma} = 0.\tag{4}$$
In the Hamiltonian formalism, we know that
$$\dot{p_\sigma} = -\frac{\partial H}{\partial q_\sigma}.\tag{5}$$
Combining these gives
$$\frac{\partial H}{\partial q_\sigma}=-\frac{\partial L}{\partial q_\sigma}.\tag{6}$$
Now, this seems very strange because in the regime we are considering, this implies that
$$\frac{\partial (T+V)}{\partial q_\sigma}=-\frac{\partial (T-V)}{\partial q_\sigma}\Rightarrow \frac{\partial T}{\partial q_\sigma}=0. \tag{7}$$
Of course, there are many examples where this is not true. I.e., simply consider the free particle analyzed using polar coordinates. Then we have
$$H = L = T = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2),\tag{8}$$
and so
$$\frac{\partial T}{\partial r } \neq 0.\tag{9}$$
What is the explanation for this strange discrepancy? Am I making a silly mistake somewhere?
Best Answer
The problem is that the Lagrangian and the Hamiltonian are functions of different variables, so you must be exceedingly careful when comparing their partial derivatives.
Consider the differential changes in $L$ and $H$ as you shift their arguments:
$$dL = \left(\frac{\partial L}{\partial q}\right) dq + \left(\frac{\partial L}{\partial \dot q}\right) d\dot q$$
$$dH = \left(\frac{\partial H}{\partial q}\right) dq + \left( \frac{\partial H}{\partial p}\right) dp$$
Finding $\frac{\partial L}{\partial q}$ corresponds to wiggling $q$ while holding $\dot q$ fixed. On the other hand, finding $\frac{\partial H}{\partial q}$ corresponds to wiggling $q$ while holding $p$ fixed. If $p$ can be expressed a function of $\dot q$ only, then these two situations coincide - however, if it also depends on $q$, then they do not, and the two partial derivatives are referring to two different things.
Explicitly, write $p = p(q,\dot q)$. Then using the chain rule, we find that
$$dH = \left(\frac{\partial H}{\partial q}\right) dq + \left(\frac{\partial H}{\partial p}\right)\left[\frac{\partial p}{\partial q} dq + \frac{\partial p}{\partial \dot q} d\dot q\right]$$
So, if we shift $q$ but hold $\dot q$ fixed, we find that
$$ dL = \left(\frac{\partial L}{\partial q} \right)dq$$ while $$ dH = \left[\left(\frac{\partial H}{\partial q} \right) + \left(\frac{\partial H}{\partial p}\right)\left(\frac{\partial p}{\partial q} \right)\right]dq$$
If $L(q,\dot q) = H(q,p(q,\dot q))$ as in the case of a free particle, then we would find that
$$dL = dH$$ so $$\left(\frac{\partial L}{\partial q}\right)= \left(\frac{\partial H}{\partial q} \right) + \left(\frac{\partial H}{\partial p}\right)\left(\frac{\partial p}{\partial q} \right)$$
We can check this for the free particle in polar coordinates, where $$L = \frac{1}{2}m(\dot r^2 + r^2 \dot \theta^2)$$ $$ H = \frac{p_r^2}{2m} + \frac{p_\theta^2}{2mr^2}$$ $$ p_r = m\dot r \hspace{1 cm} p_\theta = mr^2 \dot \theta$$
for the left hand side,
$$ \frac{\partial L}{\partial r} = mr \dot \theta^2$$
For the right hand side, $$ \frac{\partial H}{\partial r} = -\frac{p_\theta^2}{mr^3} = -mr\dot\theta^2$$ $$ \frac{\partial H}{\partial p_\theta} = \frac{p_\theta}{mr^2} = \dot \theta$$ $$ \frac{\partial p_\theta}{\partial r} = 2mr\dot \theta$$ so $$ \frac{\partial H}{\partial r} + \frac{\partial H}{\partial p_\theta} \frac{\partial p_\theta}{\partial r} = -mr\dot \theta^2 + (\dot \theta)(2mr\dot \theta) = mr\dot \theta^2$$
as expected.
Your mistake was subtle but common. In thermodynamics, you will often find quantities written like this:
$$ p = -\left(\frac{\partial U}{\partial V}\right)_{S,N}$$
which means
This reminds us precisely what variables are being held constant when we perform our differentiation, so we don't make mistakes.