- #1

JD_PM

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- Homework Statement
- Find a transformation that leaves the given Lagrangian invariant.

Show explicitly that the transformation you have chosen is a symmetry of the action.

- Relevant Equations
- Please see the given Lagrangian below

The given Lagrangian is:

##L = \frac 1 2 m_1 ( \dot x_1^2 + \dot y_1^2 + \dot z_1^2) + \frac 1 2 m_2 ( \dot x_2^2 + \dot y_2^2 + \dot z_2^2) + G \frac{m_1 m_2}{\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}}##Please note: I have been inspired by the post: https://www.physicsforums.com/threa...rem-to-get-a-constant-of-motion.982721/page-2. Particularly in Orodruin's #39

OK the first is to propose a transformation 'candidate'. Let's pick rotation.

Defining rotation in 3D as follows:

$$x \to x \cos(\lambda) + y \sin(\lambda)$$

$$y \to -x\sin(\lambda) + y \cos(\lambda)$$

$$z \to z$$

By taking the partial derivative of ##x##, ##y## and ##z## with respect to ##\lambda## we get the components of the tangent vector field to such a transformation:

$$v^1 = \left.\frac{dx}{d\lambda}\right|_{\lambda = 0} = y, \qquad

v^2 = \left.\frac{dy}{d\lambda}\right|_{\lambda = 0} = -x, \qquad

v^3 = \left.\frac{dz}{d\lambda}\right|_{\lambda = 0} = 0$$

Thus the vector field is:

$$\vec v = \frac{d\vec x}{d\lambda} = y\vec e_1 - x \vec e_2$$

The point is to show if the following transformation is a symmetry of the action:

$$\vec x \to \vec x + \lambda \vec v + \mathcal O(\lambda^2)$$

The time derivative of it is

$$\dot{\vec x} \to \dot{\vec x} + \lambda \dot{\vec v} + \mathcal O(\lambda^2)$$

OK let me at this point drop masses 1 and 2 the constant G and introduce ##\gamma## (which has dimensions ##L^2/T##), so that we get a cleaner picture. We now have:

$$L = \dot x_1^2 + \dot y_1^2 + \dot z_1^2 + \dot x_2^2 + \dot y_2^2 + \dot z_2^2 + \frac{\gamma}{\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}}$$

It follows that

$$L = \dot{\vec x_1}^2 + \dot{\vec x_2}^2 + \frac{\gamma}{\sqrt{\dot{\vec x_1}^2 - \dot{\vec x_2}^2}}$$

But this does not look right; If we do the algebra we do not get ##L \to L + \mathcal O(\lambda^2)##

What am I missing?

Any help is appreciated.

Thank you.

##L = \frac 1 2 m_1 ( \dot x_1^2 + \dot y_1^2 + \dot z_1^2) + \frac 1 2 m_2 ( \dot x_2^2 + \dot y_2^2 + \dot z_2^2) + G \frac{m_1 m_2}{\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}}##Please note: I have been inspired by the post: https://www.physicsforums.com/threa...rem-to-get-a-constant-of-motion.982721/page-2. Particularly in Orodruin's #39

OK the first is to propose a transformation 'candidate'. Let's pick rotation.

Defining rotation in 3D as follows:

$$x \to x \cos(\lambda) + y \sin(\lambda)$$

$$y \to -x\sin(\lambda) + y \cos(\lambda)$$

$$z \to z$$

By taking the partial derivative of ##x##, ##y## and ##z## with respect to ##\lambda## we get the components of the tangent vector field to such a transformation:

$$v^1 = \left.\frac{dx}{d\lambda}\right|_{\lambda = 0} = y, \qquad

v^2 = \left.\frac{dy}{d\lambda}\right|_{\lambda = 0} = -x, \qquad

v^3 = \left.\frac{dz}{d\lambda}\right|_{\lambda = 0} = 0$$

Thus the vector field is:

$$\vec v = \frac{d\vec x}{d\lambda} = y\vec e_1 - x \vec e_2$$

The point is to show if the following transformation is a symmetry of the action:

$$\vec x \to \vec x + \lambda \vec v + \mathcal O(\lambda^2)$$

The time derivative of it is

$$\dot{\vec x} \to \dot{\vec x} + \lambda \dot{\vec v} + \mathcal O(\lambda^2)$$

OK let me at this point drop masses 1 and 2 the constant G and introduce ##\gamma## (which has dimensions ##L^2/T##), so that we get a cleaner picture. We now have:

$$L = \dot x_1^2 + \dot y_1^2 + \dot z_1^2 + \dot x_2^2 + \dot y_2^2 + \dot z_2^2 + \frac{\gamma}{\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2 + (z_1-z_2)^2}}$$

It follows that

$$L = \dot{\vec x_1}^2 + \dot{\vec x_2}^2 + \frac{\gamma}{\sqrt{\dot{\vec x_1}^2 - \dot{\vec x_2}^2}}$$

But this does not look right; If we do the algebra we do not get ##L \to L + \mathcal O(\lambda^2)##

What am I missing?

Any help is appreciated.

Thank you.