# Why does this Lagrangian seem "asymmetric"?

1. Jun 28, 2014

### Alexandre

I got this Lagrangian on the exam and it just seems weird to me:
L = $\frac{m}{2}$(ẋ²+ẏ²) – eBẋy

What I mean by "asymmetric" is that it doesn't seem to behave in a same way on x as on y, because there is ẋ in the last term and y but there is no ẏ and x.

Deriving Euler-Lagrange equations I get:
ẍ = ωẏ
ÿ = –ωẋ

(Where ω=$\frac{eB}{m}$)

Solution to the equations as I calculated must be:
x(t) = $\frac{c_{2}}{ω}$sin(ωt) – $\frac{c_{1}}{ω}$cos(ωt) + x0
y(t) = $\frac{c_{2}}{ω}$cos(ωt) + $\frac{c_{1}}{ω}$sin(ωt) + y0

(c1 and c2 depend on initial velocities along x and y, x0 and y0 are initial positions)

This must be the Lorentz force acting on a particle with elementary charge moving with velocity v perpendicular to magnetic field B.
Here's the animation
https://www.desmos.com/calculator/r3mzstju9x

So to clarify my question and extend it a little bit, how is the Lagrantian derived (the last term obviously)? Why is there ẋy and not ẏx or yx for instance? Can eBẋy be a potential energy?

This is not a homework, I'm posting this for my curiosity. And also it doesn't seem to match the strict guidelines provided in homework subforum.

Last edited: Jun 28, 2014
2. Jun 28, 2014

### Staff: Mentor

It is some sort of potential energy - it does not depend on positions alone, so it is not a true potential. The magnetic vector potential is a related concept.
You could use -ẏx as well, or even some sum of ẋy and -ẏx. xy would be a regular potential, that does not give the right solution.

3. Jun 28, 2014

### Alexandre

Thanks for the reply! As I understand using -ẏx instead of ẋy wouldn't change the direction of the particle since there is a minus sign there on ẏx, right? And concerning the magnetic vector potential, how exactly do I derive the Lagrangian for the two dimensional case?

4. Jun 28, 2014

### stevendaryl

Staff Emeritus
A fact about Lagrangians is that adding a total derivative makes no difference.

So adding a term like $\dfrac{d}{dt} (- x y) = - x \dot{y} - y \dot{x}$ makes no difference.

5. Jun 28, 2014

### vanhees71

The point is that the magnetic part of the Lorentz force (reluctantly using SI units) is
$$\vec{F}=q \vec{v} \times \vec{B}.$$
It is velocity dependent and thus has not a potential. Nevertheless, one can derive a Lagrangian which leads to this force.

The trick is that from Maxwell's equations we know that there exists a vector potential for the magnetic field:
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Since the force is linear in $\vec{v}$, the part of the Lagrangian, $L_I$, describing the interaction with the magnetic field should be linear in $\vec{v}$ too, because setting
$$L=\frac{m}{2} \dot{\vec{x}}^2+L_I$$
via the Euler-Lagrange equations yields the EoMs
$$m \ddot{\vec{x}}=\frac{\partial L_I}{\partial \vec{x}}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L_I}{\partial \dot{\vec{x}}} \stackrel{!}{=} q \dot{\vec{x}} \times \vec{B}.$$
Thus we make the ansatz, linear in the velocity
$$L_I=\dot{\vec{x}} \cdot \vec{\alpha}(\vec{x}).$$
To evaluate the derivatives above, it's more convenient to switch to the Ricci calculus with indices, giving
$$\frac{\partial L_I}{\partial x_j}=\dot{x}_k \partial_j \alpha_k$$
and
$$\frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial L}{\partial \dot{x}_j} = \frac{\mathrm{d}}{\mathrm{d} t} \alpha_j)=\dot{x}_k \partial_k \alpha_j$$
This gives
$$\frac{\partial L}{\partial x_j}-\frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial L}{\partial \dot{x}_j} = \dot{x}_k (\partial_j \alpha_k - \partial_k \alpha_j)=\dot{x}_k \epsilon_{jkl} (\vec{\nabla} \times \vec{\alpha})_l=[\dot{\vec{x}} \times (\vec{\nabla} \times \vec{\alpha})]_j \stackrel{!}{=} q \vec{v} \times (\vec{\nabla} \times \vec{A})$$
Comparison of both sides of the equation leads finally to
$$\vec{\alpha}=\frac{q}{c} \vec{A}.$$
Thus the non-relativistically approximated Lagrangian for the motion of a particle in a magnetic field is
$$L=\frac{m}{2} \dot{\vec{x}}^2 + q \dot{\vec{x}} \cdot \vec{A}.$$
Now your Lagrangian is precisely of this form with
$$\vec{A}=-B y \vec{e}_x.$$
The curl is
$$\vec{\nabla} \times \vec{A}=B \vec{e}_z.$$
Thus, you have indeed described a particle of charge $e$ in a magnetic field in $z$ direction.

The vector potential is, however, only determined up to the gradient of a scalar field, i.e., for any scalar field $\chi$ also
$$\vec{A}'=\vec{A}+\vec{\nabla} \chi$$
leads to the same magnetic field. The interaction Lagrangian changes to
$$L_I'=q \dot{\vec{x}} \cdot \vec{A}'=q \dot{\vec{x}} \cdot (\vec{A}+\vec{\nabla} \chi)=q \dot{\vec{x}} + q \frac{\mathrm{d}}{\mathrm{d} t} \chi,$$
i.e., the new Lagrangian deviates from the old one only in a total time derivative of a function that depends only on the generalized coordinates and not the generalized velocities, which yields the same equations of motion according to the Hamilton principle, as you can derive immediately, using the Euler-Lagrange equations.

This is an example for gauge invariance in the Lagrangian formulation. The full glory of this scheme, of course, becomes apparent only in the relativistic formulation, which is the adequate description of electromagnetic phenomena.

6. Jun 29, 2014

### Alexandre

Though I find it difficult to follow at some places, I thank you all for replying!

7. Jun 29, 2014

### maajdl

Please note the symmetry of the equations of motion you have derived.

8. Jul 1, 2014

### HomogenousCow

What I think he is trying to say is that the vector potential (which appears explicitly in the lagrangian) is unphysical and thus can exhibit unphysical asymmetry. If you wish you could probably add a gradient of some function to the vector potential to make it symmetriic.

9. Jul 1, 2014

### vanhees71

The question then is, what you mean by "symmetric"? The point is that the equation of motion (variation of the action or even the action itself as in this) is gauge independent. By changing the gauge, sometimes you can make a variable cyclic which wasn't before. That's why it's clever in the present case of a constant magnetic field to use a vector potential that contains only on one variable (here $y$), because then the other two coordinates are cyclic, which makes the solution of the equations of motion easier.

10. Jul 1, 2014

### HomogenousCow

This has actually made me think, if one defines a positive charge and an upwards direction, doesn't a unique sense of left and right emerge from the magnetic interaction?
Additionally, since there is an imbalance between matter and antimatter in the universe, hasn't a natural positive charge already been picked out for us?

11. Jul 1, 2014

### dauto

No, it does not. The "right hand rule" is arbitrary you could just as well use the "left hand rule" and everything would work the same (with reversed direction for the magnetic field since its direction depends on the arbitrary choice of right hand rule. That's why the Magnetic field is a Pseudo-vector. A pseudo vector really is a anti-symmetric second rank tensor represented as a vector just for convenience.