# Conservation of energy when the Lagrangian includes a potential function

When proving that the homogeneity of time leads to the conservation of energy,

(This is the proof from Landau for the case when there is no field present.)

(Uses the Einstein's summation convention)

$$\frac{dL}{dt} = \frac{\partial L}{\partial q_i} \dot{q_i} + \frac{\partial L}{\partial \dot{q_i}} \ddot{q_i} + \frac{\partial L}{\partial t}$$

But, $\partial L / \partial t = 0$ assuming the homogeneity of time and so,

$$\frac{dL}{dt} = \frac{\partial L}{\partial q_i} \dot{q_i}$$

Using the Lagrange-Euler equation, (summing over all possible $i$'s) $$\frac{\partial L}{ \partial q_i} = \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q_i}} \Big)$$ and so, $$\frac{dL}{dt} = \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q_i}} \Big) \dot{q_i}$$

$$0 = \frac{d}{dt} \Big( \frac{\partial L}{\partial \dot{q_i}} \dot{q_i} - L\Big)$$

which gives,

$$\frac{d(E)}{dt} = 0$$

One assumption not mentioned above, that I believe is important, is that $\ddot{q_i} = 0$ which makes sense because in the lack of a field, a particle won't accelerate which again ties back to the statement that $L$ does not depend on time as then $\dot{q_i}$ would be constant and so $L = T - U = T-0 = T$ where $T$ is a quadratic function of $\dot{q_i}$ which is a constant would lead to the proof as shown above.

However, when there is a field $U = U(q_i)$ present, even when it is constant and does not depend on time, I'm not sure how $L$ should also be independent of time. Because now that a field is present, $\dot{q}$ would depend on time and so $\ddot{q} \not = 0$ and so I'm confused as to how to proceed with the proof. The situation is further complicated by the fact that because $L$ depends on $\dot{q}$ and because $\dot{q}$ depends on time, $L$ should depend on time and so the assumption that $\partial L / \partial t = 0$ wouldn't hold true.

Any pointers would be really appreciated.

## 2 Answers

Well I don't think you necessarily need that assumption.

Begin as you did with

$$\frac{dL}{dt}=\frac{\partial L}{\partial q}\dot q +\frac{\partial L}{\partial \dot q}\ddot q +\frac{\partial L}{\partial t}$$

And with $\frac{\partial L}{\partial t}=0$ by assumption.

Now using the Euler Lagrange replace $\frac{\partial L}{\partial q}$ by $\frac{d}{dt}\frac{\partial L}{\partial \dot q}$. The we have:

$$\frac{dL}{dt}=\frac{d}{dt}\left (\frac{\partial L}{\partial \dot q}\right)\dot q +\frac{\partial L}{\partial \dot q}\ddot q$$

And we recognise the Rhs by the product rule to be $\frac{d}{dt}\left (\frac{\partial L}{\partial \dot q}\dot q\right)$

Which implies that

$$\frac{d}{dt}\left (\frac{\partial L}{\partial \dot q}\dot q -L\right)=0$$

• Thanks a lot! But why can we still assume that $\partial L / \partial t = 0$? Now that there is a non-zero field, won't the particles accelerate thereby making the velocity time dependent in-turn making the kinetic energy and then the Lagrangian time-dependent? – tripatheea Jan 9 '15 at 2:44
• @AashishTripathee You seem to be confused about partial differentiation. If viewed as a mathematical object $L:\mathbb{R}^{2n+1}\to\mathbb{R}$ where $n$ is the number of coordinates $q$, the meaning is simple. If not viewed in that way, partial differentiation is more an "arbitrary" splitting up of coordinates, which will be valid so long as your choice is consistent. – user12029 Jan 9 '15 at 4:32
• @AashishTripathee In most cases with a potential, $\frac{dL}{dt}\neq 0$ but $\frac{\partial L}{\partial t}=0$. – user12029 Jan 9 '15 at 4:34
• Ah, it makes sense now. Thank you so much @NeuroFuzzy and snulty. – tripatheea Jan 9 '15 at 4:51

Energy is conserved in the system when the Lagrangian doesn't depend explicitly on time. The potential field that you have introduced cannot be explicitly dependent on time if you want conserved energy. The important distinction is between explicit and implicit dependence on time. All systems implicitly depend on time because as it passes the system evolves through different configurations.

When the particles have some velocity and there is a potential that varies in space then there will be accelerations as the particles moves between different potentials. This means that the velocity is position dependent not time dependent. Accelerations can occur without explicit dependence on time in the potential field and therefore without explicit dependence on time in the Lagrangian.

As long as there is no explicit dependence on time $\frac{\partial L}{\partial t}=0$