# Lagrangian of a system is independent of time

1. Sep 8, 2005

### Reshma

The Lagrangian of a system is given by:
L = T - V, T = kinetic energy of the system, V = potential energy of the system.

L is a function of the generalised coordinates for a system of N particles given by:$$L = L(q_1, q_2, ....,q_{3N},\dot{q_1}, \dot{q_2},....,\dot{q_{3N}}, t)$$
Suppose L is not an explicit function of a given coordinate qi then:
$$\frac{\partial L}{\partial q_i} = 0$$
Such coordinates are the ignorable coordinates by definition. What if L is not an explicit function of time 't'? What is the nature of such a system where the Lagrangian is independent of time?

2. Sep 8, 2005

### Berislav

Well, T-V is then a constant. What do we call such systems?

3. Sep 10, 2005

### Reshma

OK, T-V is constant with respect to time. What is conserved here....Lagrangian?

4. Sep 10, 2005

### pervect

Staff Emeritus
When the Lagrangian of a system is not a function of time, the Hamiltonian of the system is not a function of time, either.

This means that the system is conservative, i.e. it conserves energy.

Note that this is closely related to Noether's theorem. If the Lagrangian isn't a function of time, energy is conserved. If the Lagrangian isn't a function of position, momentum is conserved.

5. Sep 11, 2005

### Reshma

Hi Pervect, thank you so much for the reply!

I haven't done Hamiltonian Principle yet. Is it possible to show the energy conservation using Lagrangian equations?

6. Sep 11, 2005

### samalkhaiat

PERVECT HAS TOLD YOU WHAT YOU NEED TO KNOW! IF YOU ARE NOT FAMILIAR WITH THE HAMITONIAN, THEN IT A GOOD TIME TO KNOW SOMETHING ABOUT THIS VERY IMPORTANT OBJECT IN PHYSICS. FOR NOW THE FOLLOWING IS A GOOD START;
YOU KNOW THAT, L = T-V. WHAT YOU NEED TO KNOW IS, T+V = H = The total energy of the system. DO IT YOURSELF IT IS GOOD EXERCISE, TAKE THE TIME DERIVATIVE OF THE LAGRANGIAN, THEN USE LAGRANGE EQUATION AFTER SETTING THE PARTIAL TIME DERIVATIVE OF (L) = 0. THEN YOU GET T+V= CONSTANT.

7. Sep 11, 2005

### MalleusScientiarum

Ease off the caps lock there, buddy.

Technically speaking, the Hamiltonian is given by the set of relations:
$$p_\imath = \frac{\partial L}{\partial \dot{q}_\imath}$$
and the Legendre transform
$$H = \sum_\imath p_\imath \dot{q}_\imath - L$$
The Hamiltonian can be identified with the energy in the case that $$H$$ isn't a function of time. This further implies energy conservation. On the flip side, if the Hamiltonian is a function of time, then it can't be immediately identified as the energy.

8. Sep 16, 2005

### samalkhaiat

Hi,
Can you clarify your statement for me?
(if the Hamiltonian is .........,then it can't be "immediately" identified as the energy.)
What does "immediately" mean in your statement? And can you, in general, tell us what is the relation between energy and a time dependant Hamiltonian?
By the way, the question originally was about Lagrangian that does not explicitly depend on time. And,this condition, is necessary and sufficient for the Hamiltonian to be independent of time, as one can easily prove using your Legendre transformation.

regards

9. Sep 16, 2005

### Physics Monkey

I just want to make a note of clarification here. The fact that a Lagrangian does not depend explicitly on time does not mean the Lagrangian is constant in time. L=T-V is never conserved (unless we look at the trivial case V=0). The total time derivative of the Lagrangian is not zero. However, as has been noted, if the Lagrangian does not depend explicitly on time then the total time derivative of the Hamiltonian is zero. The Hamiltonian is conserved.

Whether the Hamiltonian is equal to the energy (T+V) of the system depends on the structure of the kinetic energy and treatment of the constraints. I can elaborate if anyone is interested.

10. Sep 16, 2005

### samalkhaiat

I want to know the relation between the total energy and the total "time-dependant" Hamiltonian in a general physical system.
I thought that for any physical system;

Energy = H = H(free) + H(interaction),

or did I think wrong?

11. Sep 16, 2005

### Physics Monkey

In a mechanical system, the total energy is given by E=T+V but the Hamiltonian H does not necessarily have to equal this quantity. An example:

The system is a bead confined to move on a rod rotating in the xy plane at angular frequency $$\omega$$ with the bead attached to the origin by a spring. Using the constraint that $$\dot{\theta} = \omega$$, I can write the Lagrangian as $$L = T - V = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \omega^2 - \frac{1}{2} k r^2$$ The key thing to note is that part of the kinetic energy only depends on r. The conjugate momentum is $$p = m\dot{r}$$ and so the Hamiltonian is $$H = p \dot{r} - L = \frac{1}{2} m \dot{r}^2 - \frac{1}{2} m r^2 \omega^2 + \frac{1}{2} k r^2$$ The Hamiltonian does not equal the energy $$E = \frac{1}{2} m \dot{r}^2 + \frac{1}{2} m r^2 \omega^2 + \frac{1}{2} k r^2$$. However, the Hamiltonian is still conserved (since L did not depend explicitly on time) while the energy is not conserved as you can easily check.

What is the general relation? Suppose I can decompose the kinetic energy into three terms, $$T_0$$ $$, T_1$$, and $$T_2$$ which are zeroth, first, and second order in the generalized velocities. In the above example, $$T_0 = \frac{1}{2} m r^2 \omega^2$$, $$T_1 = 0$$, and $$T_2 = \frac{1}{2} m \dot{r}^2$$. The Lagrangian is $$L = T_0 + T_1 + T_2 - V$$. One can show with a few lines of algebra that the Hamiltonian is given by $$H = T_2 - T_0 + V$$. This means the Hamiltonian only equals the energy if $$T = T_2$$, that is as long as the kinetic energy is quadratic in the velocities.

12. Sep 17, 2005

### samalkhaiat

Hi,
Where did the energy go in you example? don't we need the total energy of the system to be conserved?
Notice that in your example, We have;

dE/dt = d(mw^2.r^2)/dt. So,

E - mw^2.r^2 = H = conserved quantity(E')

E - 2T(o) = H = E'.

Is it wrong if we take E' to be the total energy?

I know that the problem with your example arose because the equtions which define the generalized coordinates;
x =r cos(wt), y =r sin(wt)
depend explicitily on the time and the rotation angle was (ignored).
Any way, thank you, you made me do some work on classical mechanics which I have not done for long long time. The point is this ;
If the equations which define the generalized coordinates do not depend explicitly on the time, and if the potential does not depend on the velocities, then

H = T + V = the total energy.

Notice that if you include the rotation's angle as the other generalized coordinate, then the above equation does hold.

regards

13. Sep 17, 2005

### Physics Monkey

In my example, energy is not conserved. The constraint that the particle rotates with a given frequency does work on the system. The constraint can do work because it is time dependent. The energy of the system is actually not conserved because something (the thing rotating the rod) is doing work on it.

However, the Hamiltonian is conserved as it must be since the Lagrangian does not depend explicitly on time (and all my coordinates in the Lagrangian are unconstrained).

Suppose, however, that we do as you suggest and include the angle as a (constrained) generalized coordinate. You are correct that the Hamiltonian is now the total energy, but the total energy is still not conserved and now neither is the Hamiltonian. What happened? Well the Lagrangian still does not explicitly depend on time, but this only implies that Hamiltonian is conserved if the the generalized coordinates appearing in the Lagrangian are unconstrained. If you include the constrained angular coordinate in the Lagrangian as a generalized coordinate, then the theorem is no longer true. You have to use Lagrange multipliers or some other technique to deal with the constrained system i.e. to find equations of motion.

Both approaches are valid, and they each have their uses. However, the non-conservation of energy has nothing to with how I define my generalized coordinates, etc. My point is that the Hamiltonian does not necessarily have to be equal to the total energy. The Hamiltonian depends on the detailed structure of the kinetic energy, the way in which you treat the constraints, etc.

14. Sep 17, 2005

### samalkhaiat

You are right the drive mechanism does work = mw^2.r^2 on the system and the quantity

E' = E - mw^2.r^2 = H, is like a "free energy" in thermodynamics. Can it be regarded as the total energy of the system = drive mechanism + bead?

I caused some confusion,when I asked to include the rotation angle! I meant changing your example to unconstrained one; say a bead slides on the inner surface of a downward vertex cone with the obvious change in the potential.
I agree with every thing you said and I can add another easy to prove thing,that is;
If L = T-V, and if T is a homogeneous quadratic function of the generalized velocities, then H = 2T - (T-V) = T + V.

What I needed to know was the following; In a "general" physical system, with time dependant Hamiltonian (ex. H(total) = H(free)+H(interaction),as in perturbation theory)
what is the relation between the total energy and the total hamiltonian.