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TL;DR: Why can't we write $\mathcal{L} = E - 2V$ where $E = T + V = $ Total Energy?

Let us consider the case of a particle in a gravitational field starting from rest.

Initially, Kinetic energy $T$ is $zero$ and Potential energy $V$ is $mgh$.

At any time $t$, Kinetic energy $T = \frac{m\dot x^2}{2}$ and Potential energy $V$ is $mgx$.

$$\mathcal{L} = T-V = \frac{m\dot x^2}{2}-mgx.$$

If we write $T = mgh-mgx$, the Lagrangian becomes $\mathcal{L} = T-V = mgh-2mgx$ which is independent of $\dot x$ . Here $\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot x} = 0$ while $\frac{\partial \mathcal{L}}{\partial x} = -2mg$.

Why does this simple change of form of Lagrangian not work?

I do understand that this form does not have $\dot x$ but what is the deeper reason for this to not work?

How do I know that my Lagrangian is correct (for any arbitrary problem)?

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3 Answers 3

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OP is essentially asking:

Why can't we replace the Lagrangian $L=T-V$ with $L=E-2V$ by using energy conservation $T+V=E$, where $E$ is an integration constant?

Answer: Generically an action principle gets destroyed if we apply EOMs in the action.

Specifically, OP used energy conservation $T+V=E$, which were derived from EOMs. Here it is important to understand that the stationary action principle must be defined for all (sufficiently smooth) paths. Not just the classical trajectories, which satisfy the EOMs. Note in particular, that the off-shell/virtual paths are not required to obey energy conservation.

Alternatively, it is easy to check that OP's proposed Lagrangian $L=E-2V$ would lead to wrong EOMs.

For examples, see this & this related Phys.SE posts.

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  • $\begingroup$ Understood, the Kinetic Energy and Potential Energy must be valid for any valid configuration of the system and not just the configuration under consideration. $\endgroup$
    – NiKS001
    Commented Aug 11, 2020 at 17:33
  • $\begingroup$ In this context, it is of course possible to writhe $L=E-2V$. You may argue that this definition of the Lagrangian is not very useful because it is somewhat circular. But it leads to the correct result if done properly (see my answer) $\endgroup$
    – sintetico
    Commented Aug 12, 2020 at 2:46
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic ♦
    Commented Aug 12, 2020 at 8:56
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Of course you can write $L=E-2V$, but you have to be careful with what is the total energy $E$ of the system. In the expression of the Lagrangian, the kinetic, potential, and total energy are considered to be defined instantaneously. The total energy is

$$E=\frac12m\dot x^2+mgx$$

You wrote instead that $E=mgh$. This is not the instantaneous total energy, but the initial total energy.

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  • $\begingroup$ Thanks for the response. $T = mgh - mgx$ is still instantaneous Kinetic Energy. Or am I missing something? $\endgroup$
    – NiKS001
    Commented Aug 11, 2020 at 16:59
  • $\begingroup$ no it is not. The intstaneus energy is $m\dot x/2$ $\endgroup$
    – sintetico
    Commented Aug 12, 2020 at 2:45
  • $\begingroup$ Do you mean "an arbitrary configuration" instead of "instantaneous"? Because instantaneous Kinetic Energy $T = \frac{m \dot x^2}{2} = mgh - mgx$ . $\endgroup$
    – NiKS001
    Commented Aug 12, 2020 at 17:19
  • $\begingroup$ For an arbitrary configuration $T = \frac{m \dot x^2}{2}$ is true but $T = mgh$ is not. $\endgroup$
    – NiKS001
    Commented Aug 12, 2020 at 17:22
  • $\begingroup$ Correction: "but $T = mgh - mgx$ is not." $\endgroup$
    – NiKS001
    Commented Aug 13, 2020 at 8:10
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In deriving the eqns of motion from the principle of least action, position and velocity are considered independent variables

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  • $\begingroup$ Thanks for the response. I could still rewrite the equation such that it contains $\dot x$ and $x$. How do I know which expression of Kinetic Energy and Potential Energy is correct? $\endgroup$
    – NiKS001
    Commented Aug 11, 2020 at 15:56
  • $\begingroup$ Not sure I follow. Ke is the usual expression. Also In your first post, don't forget eg the partial derivative of L with respect to position means keeping the velocitiy constant (how does L change if the position changes whilst keeping the velocity constant). As you made position and velocity dependent on each other, I guess the partial derivatives don't check out. I'm no expert though! $\endgroup$
    – Scott
    Commented Aug 11, 2020 at 18:58

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