- #1

bigguccisosa

- 24

- 3

## Homework Statement

Attached.

## Homework Equations

I am assuming the coordinate transformation is [itex] \vec{x}' = \vec{x} + \alpha\vec{\gamma} [/itex] ?

Then you have [itex] \vec{v}' = \vec{v} + \alpha\frac{d\vec{\gamma}}{dt} [/itex]

And r is the magnitude of the x vector.

## The Attempt at a Solution

Part A.

So to get the change in lagrangian, I put the primed v and x into the Lagrangian and subtracted the given lagrangian to get:

[tex] \Delta L = \frac{1}{2}m|\vec{v} + \alpha\frac{d\vec{\gamma}}{dt}|^2 + \frac{k}{|\vec{x} + \alpha\vec{\gamma}|} - \frac{1}{2}m|v|^2 - \frac{k}{r} [/tex]

So following examples from class, I expand the vector magnitude terms and neglected the second order alpha quantity, which leaves me with

[tex] \Delta L = \frac{1}{2}m(2\alpha \vec{v} \cdot \frac{d\vec{\gamma}}{dt}) + \frac{k}{\sqrt{|\vec{x}|^2 + 2 \vec{x}\cdot\alpha\vec{\gamma}}} - \frac{k}{r} [/tex]

So I thought maybe to taylor expand the first k/r term up to first order alpha, after factoring out [itex] |\vec{x}|^2 [/itex]. And since |x| is r, i can cancel out the last k/r term. That leaves me with

[tex] \Delta L = m(\alpha \vec{v} \cdot \frac{d\vec{\gamma}}{dt}) - \frac{k\alpha}{|\vec{x}|^3}(\vec{x}\cdot\vec{\gamma}) [/tex]

Now taking the derivative of gamma, and noting that dn/dt should be zero and that p cross dx/dt and dp/dt cross x should also be zero, I get for the final expression, plugging everything in

[tex] \Delta L = m\alpha[\vec{v} \cdot [\frac{d\vec{x}}{dt}\times( \vec{p}\times\hat{n}) + \vec{x} \times (\frac{d\vec{p}}{dt} \times \hat{n})]] - \frac{k\alpha}{|\vec{x}|^3} [\vec{x} \cdot [\hat{n} \times (\vec{p} \times \vec{x}) + \vec{x} \times (\vec{p} \times \hat{n})]] [/tex]

But from here, I don't really see how I could get it to equal part b, or if I even did the right process, I probably made a mistake in getting the difference, any tips?

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