Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I am reading a online tutorial about Lagrangian mechanics. In one section, it states that if the kinetic term in Lagrangian has no explicit time dependence, the Hamiltonian does not explicitly depends on time, so $H=T+V$. I just wonder if it is always true that $H=T+V$, why require it has no explicit time dependence?

share|improve this question
    
possible duplicate of Lagrangian and conservation of energy –  Kyle Kanos Jan 9 at 2:35
    
Thanks. I know that proof from other book. But my question is: it sounds like that H = T+V only when T and V do not have explicit time. What confusing me is asumming there is explicit time, so I cannot call T+V as H? But even it is time dependent (explicitly), H=T+V is still energy, isn't it? so why I can only call H energy when it doesn't contains any explicit time? –  user1285419 Jan 9 at 2:41
    
Which online tutorial? –  Qmechanic Jan 9 at 2:55
    
Yes $T+V$ has units of energy, as does $H$. But the formal definition of the Hamiltonian is $H=\sum\dot{q}p-L$, not $H=T+V$; this term is a special case of $L$ being independent of time. –  Kyle Kanos Jan 9 at 3:04
    
Possible duplicates: physics.stackexchange.com/q/11905/2451 and links therein. –  Qmechanic Jan 9 at 3:04

2 Answers 2

up vote 1 down vote accepted

The Hamiltonian has the very legal definition that it is the Legendre transform of the Lagrangian function. So, in any physical case to find the Hamiltonian of a system, you have to take $L = T - V$ and then perform the ritualistic Legendre transform process shifting coordinates from $q,q'$ to $q,p$. The time symmetry of the system leads to the conservation of the so called Jacobian function and NOT the Hamiltonian . the Hamiltonian will however be the total mechanical energy if thesystem is conservative and in this case the Jacobian function also equals the total mechanical energy.

The Jacobian is the mechanical one and not the mathematical one. Ref Goldstein.

share|improve this answer

I think I cannot completely agree with Torsten Hĕrculĕ Cärlemän's answer (or maybe I did not understand it). However the situation is more complicated. I try to present a more general picture below.

A Lagrangian function is a sufficiently regular function $L(t,q, \dot{q})$ (where henceforth $q$ means the $R^n$ column vector $(q^1,\ldots,q^n)^t$ and so on for other similar vaiables).

In particular I will focus on Lagrangians with the form:

$$L(t,q,\dot{q}) = \frac{1}{2}\dot{q}^t A(t,q) \dot{q} + F(t,q) \cdot \dot{q} - U(t,q) \qquad (1)$$

where $A$ is a non-singular real $n\times n$ symmetric matrix, $F$ is a vector-valued function, and $G$ is a scalar field. The dot in the RHS after $F$ $\cdot$ indicates the $R^n$ scalar product.

The form (1), essentially, is the most general case considered in mechanics, where $A$ is also positively defined. That expression even includes the case of non static constraints, or interactions with non-conservative forces (e.g., a given electromagnetic field or inertial forces).

In the simplest case: a physical system such that

(a) *the system is the statically constrained $^1$ in the inertial reference frame $K$, and

(b) it subjected to conservative interactions in $K$,

one has: $F=0$ and $A=A(q)$, $U=U(q)$ (absence of $t$), so that:

$$L(t,q,\dot{q}) = \frac{1}{2}\dot{q}^t A(t) \dot{q} - U(q) \qquad (1)'$$

In this case the Kinetic energy computed in $K$ is completely defined by $A$, while $U$ describes the potential energy. In the more general case (1), the Kinetic energy may get contributions also from $F$ and $U$, depending on the nature of the chosen coordinates and the chos

The Hamiltonian associated with a generic Lagrangian, $L$ is, by definition, the Legendre transform of $L$:

$$H(r,q, \dot{q}):= \dot{q} \cdot \nabla_{\dot{q}} L - L(t,q,\dot{q}) \quad (2)$$

(I stick to the Lagrangian formulation without introducing Hamiltonian variables since it is not relevant for this reasoning.) For a Lagrangian of the form (1), $H$ results to be (even if $F\neq 0$!):

$$H(t,q, \dot{q}) := \frac{1}{2}\dot{q}^t A(t,q) \dot{q} + U(t,q) \qquad (3)\:.$$

When the Lagrangian has the particular form as in (1)' (and is computed in the reference frame $K$), $H$ coincides to the total mechanical energy of the physical system in K, otherwise its physical meaning has to be investigated case by case.

Jacobi's theorem states that, for a generic Lagrangian function (so as in (1) but even more complicated):

$$\left.\frac{d H}{dt}\right|_{solutions\: eq. \:of \:motion} = - \frac{\partial L}{\partial t}\:.$$

You see that, computing $H$ along a solution of Eulero-Lagrange's equations, one finds a constant whenever $L$ does not depend explicitly on time. In this case $H$ is called Jacobi's constant of motion.

However it does not coincide, in general, with the total mechanical energy of the system. It happens when all forces are conservative and the coordinates hare suitably chosen so that (1)' holds true.

Here is an instructive example. Consider a reference frame $K$ rotating with uniform angular velocity directed along $z$ with respect to an inertial frame $K_0$. Let $\Omega$ be the magnitude of that angular velocity.

Suppose that a point $p$ with mass $m>0$ is constrained to stay on a smooth vertical ring, with radius $R$, at rest with $K$ in the plane $xz$ and centred on the origin of $K$. There are no forces acting on $p$, barring the irrelevant reaction due to the smooth constraint. We exploit the angle $q:= \theta$ ($\theta =0$ is the $x$ axis in $K$) to describe the position of $p$.

Moreover we use the Lagrangian evaluated with respect to $K_0$ (in order to disregard the inertial forces that instead appear in $K$). With some elementary trigonometry, we have:

$$L(t, q, \dot{q}) = (Kin.\: Energy \: in\: K) = \frac{mR^2}{2}\dot{q}^2 + \frac{mR^2\Omega^2}{2}\cos^2 q\:.$$

This Lagrangian verifies the hypotheses of Jacobi's theorem, so the Hamiltonian function:

$$H(t,q, \dot{q}) = \frac{mR^2}{2}\dot{q}^2 - \frac{mR^2\Omega^2}{2}\cos^2 q\qquad (4)$$

is conserved in time along any dynamical evolution of the point $p$.

What is the meaning of $H$?

It is not the mechanical energy in $K_0$, that is $L$ itself and $L\neq K$. Moreover that energy cannot be conserved for physical reasons: Because one has to supply energy to the system to maintain the uniform rotation in $K_0$ independently on the motion of $p$ along the ring.

The meaning of that $H$ constructed out of the Lagrangian evaluated in $K_0$, but using coordinates at rest with $K$, is the mechanical energy in $K$.

Indeed: The second term in the RHS of (4) is nothing but the potential energy of the centrifugal force. in $K$, Coriolis' force can be neglected as it is normal the the ring so its work vanishes. There are no further inertial forces in $K$ and, as already noticed, the reaction due to the ring dissipates no energy as it is normal to the ring itself like Coriolis' force. The first term in (4) is just the kinetic energy of $p$ in $K$.

Footnotes:

(1) for "statically constrained in the reference frame $K$" I mean that the Lagrangian coordinates $q^1,\ldots, q^n$ are chosen as follows. The position vector $\vec{x}_i$,in the rest space of the reference frame $K$, of the generic $i$th point of mass of the system, can be written as $\vec{x}_i= \vec{x}_i(q^1,\ldots,q^n)$ without any explicit dependence on time. In general, in fact one could have a time dependence $\vec{x}_i= \vec{x}_i(q^1,\ldots,q^n)$, if in $K$ the constraints depend on time. For instance a point $p$ requested to belong to a smooth curve with an assigned motion in $K$.

share|improve this answer

Your Answer

 
discard

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.