Lagrangian for a particle moving in x-y plane in a constant B-field

In summary: Use the boundary conditions. z = x + iy, and both x and y are specified for two values of t. That will give you the value of z at those times. Compare that with the solution that you got above for z. That should give you Xand Y.
  • #1
spacetimedude
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Homework Statement


Not sure if the link is showing. But it's imgur.com/a/LEvd0
LEvd0

LEvd0

Homework Equations


The steps I've taken so far as written in the attempt section below is correct.

The solution provided then proceeds with letting ##z = x + iy## and setting ##\ddot z+i \omega \dot z = 0##. Then ##z(t)=X\exp(-i\omega t) + Y## where X, Y are complex.
Lastly, ##X=Aexp{(i\omega t)} * \text{something illegible}## and ##Y=D+iE## and then says this leads to the x and y equations shown in the question.

The Attempt at a Solution


We first solve the lagrangian equation by splitting x and y terms. So for x:
$$\frac{d}{dt} \frac{\partial L}{\partial \dot{x} } - \frac{\partial L}{\partial x} = 0$$
and $$\ddot x -\omega \dot y =0.$$
Similarly for y:
$$\ddot y +\omega \dot x= 0$$

I'm not quite sure what to do after this step. As explained in the previous section, the solution sets ##z=x+iy## but the steps afterwards are very unclear and hard to read.

Any help will be appreciated.
 
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  • #2
Step 1: Set z = x + iy, use the two differential equations you got for x and y, to get the differential equation for z. You can proceed from there.
 
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  • #3
Chandra Prayaga said:
Step 1: Set z = x + iy, use the two differential equations you got for x and y, to get the differential equation for z. You can proceed from there.
That's the part where I'm confused. How would I incorporate the x and y differential equations to z=x+iy?
 
  • #4
Can you multiply one of the two equations with i and then add the two equations? Will that help?
 
  • #5
Chandra Prayaga said:
Can you multiply one of the two equations with i and then add the two equations? Will that help?
Thank you.

Multiply i to the y DE and add the two equations so:
$$ \ddot x - \omega \dot y + i\ddot y +i\omega \dot x =0. $$
Rearrange and it becomes
$$\ddot z +i\omega \dot z = 0.$$
Does this then imply
$$\dot z + i\omega z = Y$$ where Y is complex?
Then solving for z leads to $$z = X\exp(i\omega t) +Y$$ where X and Y are complex (the Y here is different by a factor from Y before but still a complex).

Now, another part I am confused is, the solution says Y can always be written as Y = D+iE which I understand since it's a complex, but it also says X can be written as something like ##A\exp(iωt_a)+C\exp(iωt_b)## but I can't really read off what it says. To conform with the solutions of x and y given, what could this X be?
 
  • #6
Use the boundary conditions. z = x + iy, and both x and y are specified for two values of t. That will give you the value of z at those times. Compare that with the solution that you got above for z. That should give you X nd Y
 

1. What is the Lagrangian for a particle moving in the x-y plane in a constant B-field?

The Lagrangian for a particle moving in the x-y plane in a constant B-field is given by L = (1/2)m(vx^2 + vy^2) + q(vxBy - vyBx), where m is the mass of the particle, q is the charge of the particle, B is the magnitude of the constant magnetic field, and vx and vy are the velocities of the particle in the x and y directions, respectively.

2. How is the Lagrangian derived for a particle in a constant B-field?

The Lagrangian for a particle in a constant B-field can be derived using the Lagrangian formalism, which is a mathematical framework for analyzing the dynamics of a system. The Lagrangian is defined as the difference between the kinetic and potential energies of the system. In the case of a particle in a B-field, the kinetic energy is given by (1/2)m(vx^2 + vy^2) and the potential energy is given by q(vxBy - vyBx).

3. What is the significance of the Lagrangian for a particle in a constant B-field?

The Lagrangian for a particle in a constant B-field is significant because it allows us to analyze the motion of the particle using the principles of classical mechanics. By solving the equations of motion derived from the Lagrangian, we can determine the trajectory of the particle in the B-field.

4. How does the Lagrangian change if the B-field is not constant?

The Lagrangian for a particle in a non-constant B-field would also include a potential energy term that takes into account the changing magnetic field. This potential energy term would depend on the position of the particle and the changing magnetic field, and would be added to the kinetic energy term in the Lagrangian.

5. Can the Lagrangian be used to analyze particles in other types of fields?

Yes, the Lagrangian can be used to analyze the motion of particles in various types of fields, including electric, magnetic, and gravitational fields. It is a general mathematical tool that allows for the analysis of a wide range of physical systems.

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