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## Main Question or Discussion Point

Just wondering how much validity there is to this derivation, or if it's just a convenient coincidence that this works.

We have a Lagrangian dependent on position and velocity: [tex] \mathcal{L} (x, \dot{x}) [/tex]

Let's say now that we've perturbed the system a bit so we now have:

[tex] \mathcal{L} (x + \delta x, \dot{x} + \delta \dot{x}) [/tex]

But the physics can't change from a perturbation, therefore:

[tex] \mathcal{L} (x + \delta x, \dot{x} + \delta \dot{x}) = \mathcal{L} (x, \dot{x}) [/tex]

We can expand the perturbed Lagrangian to first order for both position and velocity:

[tex] \mathcal{L} (x + \delta x, \dot{x} + \delta \dot{x}) = \mathcal{L} (x, \dot{x}) + \frac{\partial \mathcal{L}}{\partial x} \delta x + \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta \dot{x} [/tex]

This implies that:

[tex] \frac{\partial \mathcal{L}}{\partial x} \delta x + \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta \dot{x} = 0 [/tex]

which I will call "equation A".

Assuming

[tex] \delta \dot{x} = \frac{d}{dt} \delta x [/tex]

we can use the product rule:

[tex] \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right) = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x + \frac{\partial \mathcal{L} }{\partial \dot{x}} \delta \dot{x} [/tex]

which we can plug into equation A and get:

[tex] \left( \frac{\partial \mathcal{L} }{\partial x} - \frac{d}{dt} \frac{\partial \mathcal{L} }{\partial \dot{x}} \right) \delta x + \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right) = 0 [/tex]

You can see the Euler-Lagrange equation in there. All that's left to eliminate is the term:

[tex] \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right) [/tex]

You can argue that this term is 0, because complete time derivative terms can be integrated over t, and the endpoints can be chosen to have dx(endpoint) = 0.

So then you are left with:

[tex] \frac{\partial \mathcal{L} }{\partial x} - \frac{d}{dt} \frac{\partial \mathcal{L} }{\partial \dot{x}} = 0 [/tex]

Anyone with the patience to read this, what are your thoughts? Thank you.

We have a Lagrangian dependent on position and velocity: [tex] \mathcal{L} (x, \dot{x}) [/tex]

Let's say now that we've perturbed the system a bit so we now have:

[tex] \mathcal{L} (x + \delta x, \dot{x} + \delta \dot{x}) [/tex]

But the physics can't change from a perturbation, therefore:

[tex] \mathcal{L} (x + \delta x, \dot{x} + \delta \dot{x}) = \mathcal{L} (x, \dot{x}) [/tex]

We can expand the perturbed Lagrangian to first order for both position and velocity:

[tex] \mathcal{L} (x + \delta x, \dot{x} + \delta \dot{x}) = \mathcal{L} (x, \dot{x}) + \frac{\partial \mathcal{L}}{\partial x} \delta x + \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta \dot{x} [/tex]

This implies that:

[tex] \frac{\partial \mathcal{L}}{\partial x} \delta x + \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta \dot{x} = 0 [/tex]

which I will call "equation A".

Assuming

[tex] \delta \dot{x} = \frac{d}{dt} \delta x [/tex]

we can use the product rule:

[tex] \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right) = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x + \frac{\partial \mathcal{L} }{\partial \dot{x}} \delta \dot{x} [/tex]

which we can plug into equation A and get:

[tex] \left( \frac{\partial \mathcal{L} }{\partial x} - \frac{d}{dt} \frac{\partial \mathcal{L} }{\partial \dot{x}} \right) \delta x + \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right) = 0 [/tex]

You can see the Euler-Lagrange equation in there. All that's left to eliminate is the term:

[tex] \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \delta x \right) [/tex]

You can argue that this term is 0, because complete time derivative terms can be integrated over t, and the endpoints can be chosen to have dx(endpoint) = 0.

So then you are left with:

[tex] \frac{\partial \mathcal{L} }{\partial x} - \frac{d}{dt} \frac{\partial \mathcal{L} }{\partial \dot{x}} = 0 [/tex]

Anyone with the patience to read this, what are your thoughts? Thank you.