# A Lagrangian function

1. Feb 18, 2017

### Alaguraja

(L=T-V) In the Lagrangian function we saw to different type of energy conservation's. That is kinetic energy and potential energy. And I have doubt in one topic. How to define potential energy?

2. Feb 18, 2017

### jfizzix

There's at least a couple ways to answer that question.
One mathematical way is that for every force dependent on position expressible as a gradient, there is an associated potential energy.

However, the simplest is to say that potential energy refers to all forms of energy that are not kinetic (yes, really), or due to $E_{0}=mc^{2}$.
Different forms of potential energy can be discovered by using conservation of total energy, and seeing what's missing (e.g., the potential energy stored in chemical bonds giving rise to explosive chemical reactions).

3. Feb 21, 2017

### hilbert2

In general, kinetic and potential energies are not conserved, only their sum (which is usually equal to the Hamiltonian function) is. For example, if we have a planet or moon orbiting a gravitational center, the kinetic and potential energies are both conserved only if the orbit is a perfect circle (not a general ellipse). Sometimes such an orbital motion system can be describes as two point masses, but in a more advanced treatment we can include the rotational energy of the celestial bodies into the kinetic energy, and also consider the tidal forces that cause a slow loss of potential energy to viscous friction.

4. Mar 14, 2017

### zwierz

usually such questions are posed by people who study mechanics by courses of Landau Lifschitz type where the Lagrage equations are derived from the Hamilton variational principle. But if we follow more classical viewpoint then first we get know about so called the Lagrange equations of the second type
$$\frac{d}{dt}\frac{\partial T}{\partial \dot q^i}- \frac{\partial T}{\partial q^i}=Q_i,$$ with generalized forces $Q_i$. By definition these forces are potential provided there exists a function $V=V(q)$ such that $$-\frac{\partial V}{\partial q^i}= Q_i$$

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