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## Homework Statement

For a gauge function G(t,q) where

does

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- Thread starter dynamicskillingme
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For a gauge function G(t,q) where

does

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I'm working on proving how different quantities change, such as the euler lagrange equations, the generalised momenta, mechanical energy and power

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So for example for the euler lagrange I have

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correction

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If [itex]G[/itex] is a function of [itex]q[/itex] and [itex]t[/itex], then you have:

[itex]\dot{G} = \frac{\partial G}{\partial q} \dot{q} + \frac{\partial G}{\partial t}[/itex]

So in that case, [itex]\frac{\partial \dot{G}}{\partial \dot{q}} = \frac{\partial G}{\partial q}[/itex]

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Oh wow, this is painful how much I overlooked that. Thanks!

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If [itex]G[/itex] is a function of [itex]q[/itex] and [itex]t[/itex], then you have:

[itex]\dot{G} = \frac{\partial G}{\partial q} \dot{q} + \frac{\partial G}{\partial t}[/itex]

So in that case, [itex]\frac{\partial \dot{G}}{\partial \dot{q}} = \frac{\partial G}{\partial q}[/itex]

A quick question about the same function, would

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A quick question about the same function, wouldbe a true statement?

Yes, if [itex]G[/itex] is only a function of [itex]q[/itex] and [itex]t[/itex].

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Yes, if [itex]G[/itex] is only a function of [itex]q[/itex] and [itex]t[/itex].

G is a function of q and t, (G(t,q) to be exact). Could you explain why you can change the order of the derivatives in this case?

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G is a function of q and t, (G(t,q) to be exact). Could you explain why you can change the order of the derivatives in this case?

If you have a function [itex]X(t,q)[/itex]of [itex]q[/itex] and [itex]t[/itex], then [itex]\dot{X} = \frac{d}{dt} X = (\frac{\partial}{\partial t} + \frac{\partial}{\partial q} \frac{dq}{dt}) X = (\frac{\partial}{\partial t} + \frac{\partial}{\partial q} \dot{q}) X[/itex]

So [itex]\dot{G} = \frac{\partial G}{\partial t} + \dot{q} \frac{\partial G}{\partial q}[/itex]. Therefore, [itex]\frac{\partial \dot{G}}{\partial q} = \frac{\partial^2 G}{\partial t \partial q} + \dot{q} \frac{\partial^2 G}{\partial q^2}[/itex].

Similarly, [itex]\frac{d}{dt} \frac{\partial G}{\partial q} = (\frac{\partial}{\partial t} + \frac{\partial}{\partial q} \dot{q}) \frac{\partial G}{\partial q} = \frac{\partial^2 G}{\partial t \partial q} + \frac{\partial^2 G}{\partial q^2} \dot{q}[/itex]

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