Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

According to equation (6) on the first page of some lecture notes online, the above equation is used to prove the virial theorem. For rectangular coordinates, the relation

$$ 2T~=~\sum_i p_{i}\dot{q}^{i} $$

is obvious. How would I show it holds for generalized coordinates $q^{i}$?

share|cite|improve this question
I've just found an excelled reference in p.259 of the 5th edition of Thornton and Marion's Classical Dynamics. I'll read it tonight and hopefully be able to answer my own question by then. If anyone wants to chime in before then, I'd be more than thankful. – G.P. Burdell Mar 22 '13 at 7:27
up vote 2 down vote accepted

Your answer looks complicated and I feel it misses the point. The fact that $2T=\sum_l p_l \dot{q}_l$ is fundamentally due to Euler's homogeneous function theorem. This states that $$ \text{if }f(\alpha \mathbf{x})=\alpha^k f( \mathbf{x})\text{, then } \mathbf{x}\cdot\nabla f(\mathbf{x})=kf(\mathbf{x}). $$ When stated like that, I would even hesitate to add a proof - simply differentiate with respect to $\alpha$ and set it to $1$.

The theorem holds for Lagrangians that depend quadratically and homogeneously in the velocities. This means specifically that $L(q_l,\dot q_l)=T(\dot q_l;q_l)-V(q_l)$, and $T(\alpha \dot q_l;q_l)=\alpha^2 T(\dot q_l;q_l)$. (I use the notation $T(\dot q_l;q_l)$ to emphasize that the $q_l$ dependence is as parameters, with $T$ being fundamentally a homogeneous quadratic function of the $\dot q_l$.) The homogeneous function theorem then states that $$2T(\dot q_l;q_l)=\sum_l\dot q_l \frac{\partial T}{\partial \dot q_l}.$$ Since $V$ is independent of the $\dot q_l$, each partial derivative equals the corresponding momentum $p_l=\frac{\partial L}{\partial \dot q_l}$.

Note in particular that your first assumption (that the functional dependence of $T$ on the $\dot q_l$ be time-independent) is not necessary.

share|cite|improve this answer

Ok so the answer is more straightforward and tedious than I expected.

In rectangular coordinates, the kinetic energy $T$ of $n$ particles is:

$$ T=\frac{1}{2}\sum_{\alpha=1}^{n}\sum_{i=1}^{3}m_{\alpha}\dot{x}_{\alpha,i}^{2} $$

where $x_{\alpha,i}$ is a function of generalized coordinates:

$$ \begin{eqnarray} x_{\alpha,i} &=& x_{\alpha,i}(q_{j},t),\hspace{12pt}j=1,2,...,s\\ \dot{x}_{\alpha,i} &=& \sum_{j=1}^{s}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\dot{q}_{j}+\frac{\partial x_{\alpha,i}}{\partial t} \end{eqnarray} $$

So $\dot{x}_{\alpha,i}^{2}$ becomes

$$ \dot{x}_{\alpha,i}^{2}=\sum_{j,k}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial q_{k}}\dot{q}_{j}\dot{q}_{k} + 2\sum_{j}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial t}\dot{q}_{j} + \left(\frac{\partial x_{\alpha,i}}{\partial t}\right)^{2} $$

The total kinetic energy is:

$$ T=\sum_{\alpha}\sum_{i,j,k}\frac{1}{2}m_{\alpha}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial q_{k}}\dot{q}_{j}\dot{q}_{k} + \sum_{\alpha}\sum_{i,j}m_{\alpha}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial t}\dot{q}_{j} + \sum_{\alpha}\sum_{i}\frac{1}{2}m_{\alpha}\left(\frac{\partial x_{\alpha,i}}{\partial t}\right)^{2} $$

If $x_{\alpha, i}$ has no explicit time dependence (as is the case if the system is scleronomic), then $\frac{\partial x_{\alpha ,i}}{\partial t}=0$, so the total kinetic energy becomes:

$$ \begin{eqnarray} T &=& \sum_{\alpha}\sum_{i,j,k}\frac{1}{2}m_{\alpha}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial q_{k}}\dot{q}_{j}\dot{q}_{k}\\ &=& \sum_{j,k}\left(\sum_{\alpha}^{n}\sum_{i}^{3}\frac{1}{2}m_{\alpha}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial q_{k}}\right)\dot{q}_{j}\dot{q}_{k}\\ T &=& \sum_{j,k}a_{jk}\dot{q}_{j}\dot{q}_{k} \hspace{12pt} \square \end{eqnarray} $$

Taking the partial derivative of $T$ with respect to $\dot{q}_{l}$ gives us:

$$ \begin{eqnarray} \frac{\partial T}{\partial \dot{q}_{l}} &=& \sum_{k}a_{lk}\dot{q}_{k} + \sum_{j}a_{jl}\dot{q}_{j}\\ \dot{q}_l\frac{\partial T}{\partial \dot{q}_{l}} &=& \sum_{k}a_{lk}\dot{q}_{k}\dot{q}_{l} + \sum_{j}a_{jl}\dot{q}_{j}\dot{q}_{l}\\ \end{eqnarray} $$

Summing over all $l$ gives us:

$$ \sum_{l}\dot{q}_l\frac{\partial T}{\partial \dot{q}_{l}} = \sum_{k,l}a_{lk}\dot{q}_{k}\dot{q}_{l} + \sum_{j,l}a_{jl}\dot{q}_{j}\dot{q}_{l} $$

All indices are dummy indices so:

$$ \sum_{l}\dot{q}_l\frac{\partial T}{\partial \dot{q}_{l}} = 2\sum_{j,k}a_{jk}\dot{q}_{j}\dot{q}_{k} = 2T \hspace{12pt} \square $$

If the potential doesn't depend on the generalized velocities, then:

$$ \begin{eqnarray} \sum_{l}\dot{q}_l p_{l} = \sum_{l}\dot{q}_l\frac{\partial L}{\partial \dot{q}_{l}} &=& \sum_{l}\dot{q}_l\frac{\partial (T-U)}{\partial \dot{q}_{l}}\\ &=& \sum_{l}\dot{q}_l\frac{\partial T}{\partial \dot{q}_{l}}\\ \sum_{l}\dot{q}_l p_{l} &=& 2T \hspace{12pt} \blacksquare \end{eqnarray} $$

So I guess the equation I originally asked does hold in general, given the two assumptions:

  • No explicit time dependence in the coordinate transformation
  • No explicit velocity dependence in the potential term of the Lagrangian
share|cite|improve this answer

The equation $2T=\sum_{i=1}^n p_i\dot{q}^i$ holds both in Lagrangian and Hamiltonian formalism for large classes of systems.

  1. In Lagrangian formalism, it holds for Lagrangians of the form $L(q,\dot{q},t)=T(q,\dot{q},t)-V(q,t)$, where the kinetic energy is of the form $T=\frac{1}{2}\sum_{i,j=1}^n \dot{q}^i m_{ij}(q,t)\dot{q}^j$. Now use the Lagrangian definition of momentum $p_i = \frac{\partial L}{\partial \dot{q}^i}$.

  2. In Hamiltonian formalism, it holds for Hamiltonians of the form $H(q,p,t)=T(q,p,t)+V(q,t)$, where the kinetic energy is of the form $T=\frac{1}{2}\sum_{i,j=1}^n p_i m^{ij}(q,t)p_j$. Now use Hamilton's equations $\dot{q}^i = \frac{\partial H}{\partial p_i}$.

In both cases, we are secretly using the Euler homogeneous vector field $\sum_{i=1}^n\dot{q}^i\frac{\partial}{\partial \dot{q}^i}$, which counts the number of $\dot{q}$s, as Emilio Pisanty also points out in his answer.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.