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Hello, I am just starting Lagrangian mechanics and on a conservation problem but I am stuck. I have part of the solution (i think) but I am not sure how to complete it.

In an infinite homogeneous plane. Find all the components of momentum and angular momentum that are conserved.

Obviously there are:

[tex] \frac{\partial L}{\partial q_i} = f_i [/tex]

[tex] p_i = \frac{\partial L}{\partial \dot{q}_i}[/tex]

[tex] f_i = {\dot{p}_i}[/tex]

I said that in the infinite homogeneous plane (in the xy plane) since the potential energy doesn't depend on x or y displacement alone these coordinates did not change the Lagrangian. Therefore:

[tex] L = L(x, y, z, \dot{x}, \dot{y}, \dot{z}) = L(x+e, y+e, z, \dot{x}, \dot{y}, \dot{z})[/tex]

Where e is a change in the coordinates (I'm assuming you can displace them equally and still get back the Lagrangian. If I'm wrong please correct me) .

So since the Lagrangian is translationally invariant I said:

[tex] f_x = \frac{\partial L}{\partial x} = 0 =\dot{p}_x [/tex]

If [tex] \dot{p}_x = 0[/tex], the rate of change of the momentum in the x-direction is constant ( [tex] p_x = constant [/tex] ).

Likewise for the y-direction of momentum:

[tex] f_y = \frac{\partial L}{\partial y} = 0 =\dot{p}_y [/tex]

[tex] p_y = constant [/tex]

Does this show, sufficently, mathematically that the x and y components of momentum are conserved?

My central difficulty is showing that, since the infinite homogeneous plane is centrally symmetric about the z-axis, show the z-component of angular momentum, M_z is conserved. I'm unclear on how to do that.

-Thanks

Edit: P.S. the reason I have trouble with the M_z comonent is because I think of it as M_z = x*p_y - y*p_x and since p_x and p_y are constant and the coordinates are invariant under translation that means M_z is constant. But I don't know if that right because it doesn't seem to make sense for the field of 2 points (whos like connecting them is on the z-axis). p_y and p_x are not conserved there. I don't know hwo to show this, or show it with a general Lagrangian like L = T - U or something.

## Homework Statement

In an infinite homogeneous plane. Find all the components of momentum and angular momentum that are conserved.

## Homework Equations

Obviously there are:

[tex] \frac{\partial L}{\partial q_i} = f_i [/tex]

[tex] p_i = \frac{\partial L}{\partial \dot{q}_i}[/tex]

[tex] f_i = {\dot{p}_i}[/tex]

## The Attempt at a Solution

I said that in the infinite homogeneous plane (in the xy plane) since the potential energy doesn't depend on x or y displacement alone these coordinates did not change the Lagrangian. Therefore:

[tex] L = L(x, y, z, \dot{x}, \dot{y}, \dot{z}) = L(x+e, y+e, z, \dot{x}, \dot{y}, \dot{z})[/tex]

Where e is a change in the coordinates (I'm assuming you can displace them equally and still get back the Lagrangian. If I'm wrong please correct me) .

So since the Lagrangian is translationally invariant I said:

[tex] f_x = \frac{\partial L}{\partial x} = 0 =\dot{p}_x [/tex]

If [tex] \dot{p}_x = 0[/tex], the rate of change of the momentum in the x-direction is constant ( [tex] p_x = constant [/tex] ).

Likewise for the y-direction of momentum:

[tex] f_y = \frac{\partial L}{\partial y} = 0 =\dot{p}_y [/tex]

[tex] p_y = constant [/tex]

Does this show, sufficently, mathematically that the x and y components of momentum are conserved?

My central difficulty is showing that, since the infinite homogeneous plane is centrally symmetric about the z-axis, show the z-component of angular momentum, M_z is conserved. I'm unclear on how to do that.

-Thanks

Edit: P.S. the reason I have trouble with the M_z comonent is because I think of it as M_z = x*p_y - y*p_x and since p_x and p_y are constant and the coordinates are invariant under translation that means M_z is constant. But I don't know if that right because it doesn't seem to make sense for the field of 2 points (whos like connecting them is on the z-axis). p_y and p_x are not conserved there. I don't know hwo to show this, or show it with a general Lagrangian like L = T - U or something.

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