# Is there an intuitive explanation for why Lorentz force is perpendicular to a particle's velocity and the magnetic field?

The Lorentz force on a charged particle is perpendicular to the particle's velocity and the magnetic field it's moving through. This is obvious from the equation:

$$\mathbf{F} = q\mathbf{v} \times \mathbf{B}$$

Is there an intuitive explanation for this behavior? Every explanation I've seen simply points at the equation and leaves it at that.

I can accept mathematically why $\mathbf{F}$ will be perpendicular to $\mathbf{v}$ and $\mathbf{B}$ (assuming the equation is correct, which it is of course). But that doesn't help me picture what's fundamentally going on.

Trying to create an analogy with common experiences seems useless; if I were running north through a west-flowing "field" of some sort, I wouldn't expect to suddenly go flying into the sky.

I'm hoping there's a way to visualize the reason for this behavior without a deep understanding of advanced theory. Unfortunately, my searching for an explanation makes it seem like something one just has to accept as bizarre until several more years of study.

-
As usual, there's people giving over-your-head non-intuitive explanations designed to show off their knowledge, rather than answering the question. – Larry Harson May 29 '12 at 12:06

How about this? Assume that the Lorentz force is not perpendicular to $v$ (that is $v$ has a nonzero component parallel to $F$). The force acts on the charge causing it to accelerate...which in turn increases the force ($qvB$ where $v$ is the component parallel to $F$) this in turn increases the acceleration which increases the force and so on ad infinitum. This would obviously violate conservation of energy and therefore $v$ must be perpendicular to $F$. The same argument explains why $B$ must be perpendicular to $v$.

-
No, this doesn't rule out that F could have a component anti-parallel to v. – Retarded Potential Mar 2 at 5:45
@RetardedPotential then the velocity would slow down, without the energy being dissipated into heat. Still violates the first law. – Nathaniel Mar 2 at 9:11
Also if F had an antiparallel component I could just change the sign of the charge could I not? – Ryan Mar 2 at 17:47
How does this argument explain why $B$ has to be perpendicular to $F$? (You meant $F$ at the end? $B$ does not have to be perpendicular to $v$.) – Retarded Potential Mar 2 at 22:25
@Ryan you're right about flipping the charge, I withdraw my initial objection. – Retarded Potential Mar 2 at 22:25
show 5 more comments

Here's an example from Schwartz's Principles of Electrodynamics, based on relativity. (Have I just disqualified this answer as "not intuitive"? Carrying on regardless...)

Imagine an infinite, straight wire with a constant current that is composed of equal numbers of positive and negative charges flowing in opposite directions (so the wire's net charge density is 0).

Now add a particle with charge q moving parallel to the wire with constant velocity v (lab frame K). What's the force on the particle?

To answer, Schwartz transforms the problem to the rest frame K' of the particle. Applying the corresponding Lorentz boost to the wire's charge-current four-vector, one finds that in K' it has a non-zero charge density. The charge is therefore attracted towards the wire by an electrostatic field.

(There are two assumption/empirical facts here: 1) in the particle's rest frame, the force is given by the electric field as the particle sees it, and 2) because the charge and current distributions are time independent, one can calculate the electric field in K' by the usual integration-over-charge-density approach.)

Boosting back to the lab frame, one finds the answer to the question, which is that the moving charge feels a force perpendicular to its velocity.

If you now separately calculate the B-field for the current by the usual formula, you also find that the force calculated above satisfies F=q vxB.

Of course, the above is just one example (and would not be applicable if the particle were moving towards the wire instead of parallel to it, because assumption 2 would be violated.) There's a more elaborate induction process to get to the full relativistic apparatus. However, the above example does establish the existence of a force perpendicular to a charged particle's velocity and the magnetic field.

-

I'll give you here a very short argument based on quantum mechanics. This argument actually has a profound physical and historical origin which I'll try to give in the sequel. In quantum mechanics, the velocity operator of a particle in an external magnetic field is (under minimal coupling):

$\mathbf{v} = \frac{p - q \mathbf{A}}{m}$

Where $\mathbf{A}$ is the vector potential. This implies that the magnetic field is given using the canonical commutation relations:

$\mathbf{B} = - i m^2 \mathbf{v} \times \mathbf{v}$

Now, it is not difficult to check that

$\mathbf{v}.\mathbf{F} \propto \epsilon_{ijk} [v_i, [v_j, v_k]]$

which vanishes by the Jacobi identity ($\mathbf {F}$ is the Lorentz force), which is the required orthogonality relation.

This "derivation" is actually a part of a very interesting argument by Feynman, behind which there is a very interesting story. Actually, the argument doesn't require quantum mechanics but only the notion of Poisson brackets. Feynman didn't take the argument seriously and didn't publish it. It was not until 1990 after his death when this argument was published (in his name) by Dyson: F.Dyson,Am.J.Phys.58,209(1990).

Feynman's argument is very profound because it allows the derivation of the whole Maxwell theory (together with the Lorentz force equation) starting from very simple and basic assumptions:

1. The canonical Posisson brackets of the position and velocity.

2. Minimal coupling: The electromagnetic force on a charged point particle depends on the position and velocity only.

Please, see the introductory section in the following article by Carinena, Ibort, Marmo and Stern.

One of the generalization of this procedure is the derivation of the Yang-Mills equations according to the same principles

-

### Pseudovector argument

There is an intuitive argument, but the first thing to do is to take the Poincare dual of B. In 3 dimensions, there is an epsilon tensor $\epsilon_{ijk}$ which is invariant--- it doesn't change under rotations. It has $\epsilon_{123}=1$ and all interchanges give a minus sign, so that the value of $\epsilon$ is zero of two of the indices are equal, and the sign of the permutation to get to 123 if they are all different. The epsilon tensor contracted with three vectors $v_1,v_2,v_3$ gives the signed area spanned by the parallelepiped they form. Because the signed area is the determinant of the matrix of v's put together as 3 columns, it changes sign under reflection of all three coordinate axes.

The fundamental quantity in electromagnetism is $B_{\mu\nu}=\epsilon_{\mu\nu\sigma}B^{\sigma}$, the epsilon tensor contracted with B. This thing is a rank 2 antisymmetric tensor. Because the $\epsilon$ tensor is invariant, an antisymmetric tensor is equivalent to a vector under rotations, but it isn't equivalent under reflections. The reason is that although a vector changes sign under reflections, a tensor doesn't. This is also true of B--- it's a pseudovector, if you reflect a space with a current carrying wire, the direction of B does not reverse.

The fact that B is fundamentally a tensor, not a vector, means that when it is interacting with a particle with velocity v, it can only form a force when one of the indices are contracted with something. The only thing one can contract with is the velocity, so you get $B_{\mu\nu}v^\mu$ as the force, and this is $v\times B$

In relativity, this is seen to be the only natural thing, since the E and B fields together make an antisymmetric 2-tensor, and the four-Lorentz force is this tensor contracted with the 4-velocity. This form is so natural and intuitive, that it does not require a detailed justification.

### More physical restatement of the argument

The above is sort-of formal sounding, but it is just saying this: the magnetic field doesn't change sign under reversing the coordinates of space. To see this physically, consider a solenoid of current stretching along the z axis from -a to a, with the current mostly in the x-y plane along each winding, and reflect this solenoid in the x-y-z axes. Reflecting x reverses the current, reflecting y gets it back to where it started, and reflecting z doesn't change the solenoid.

Since the current is the same, B is the same! So the B from a solenoid doesn't change under reflection. So the force on a particle can't be along the direction of B, because force reverses direction under reflection and B doesn't. The force can only be in the direction of a quantity which does reverse direction, and the simplest such quantity is $v\times B$. Under a reflection, v reverses direction and B doesn't, so the Lorentz force properly reverses.

This argument assumes reflection symmetry, which is a symmetry of electromagnetism, but is actually not a fundamental symmetry in our universe. The same reflection argument shows that magnetic charge is not properly symmetric with electric charge, since magnetic charge changes sign under reflection (reflect all the coordinates with the charge at the origin--- the field moves to a new location, but points in the same direction, so the sense of the magnetic charge is reversed). This property means that magnetic monopoles were an early sign that nature is not parity invariant, and may explain why Dirac was not surprised when the weak interactions were shown to violate parity.

The other assumption is that the force is the simplest reflection invariant combination of B and v. If you abandon the idea that the force is linearly proportional to B, there are more complicated combinations that also work to give a reflection invariant force-law. These combinations generally fail to obey conservation of energy.

In order to have automatic energy conservation (and an automatic phase space with the symplectic properties), you should derive your equations of motion from the action.

### Hamiltonian argument and gauge invariance

The best argument is from the concept of momentum potential (or vector potential). Like the energy has a potential energy added to it, which is $e\phi$, the momentum has a potential added which is $eA$, where A is the vector potential.

The Lagrangian for the interaction is

$$mA\cdot \dot{x}$$

Which makes the conjugate momentum $mv + eA$, so that the kinetic energy is $(p-eA)^2\over 2m$ and the potential energy is $e\phi$. The Hamilton equations for this energy give the Lorentz force law. The Hamilton equations are:

$$\partial_t p = \partial_x {(p-eA)^2\over 2m + \phi}$$ $$\partial_t x = {p - eA \over m}$$

And combining the equations to a second order equation for the acceleration of x gives the Lorentz force law. The same replacement in the Hamiltonian, $p$ to $p-eA$, works in relativity to give the correct four dimensional Lorentz force law.

The identification of B with $\nabla \times A$ can be justified from the invariance of the equations under adding a gradiant to A. The classically physical part of A is its curl, and this is sensible to identify with the B in Maxwell's equations.

This argument is fundamentally sound, because it doesn't depend on reflection invariance (any argument relying on reflection symmetry is really bogus, since we know this is not a symmetry of nature in any fundamental sense), and is correct quantum mechanically when you interpret the gauge invariance as a freedom in the local phase redefinition of a charged particle wavefunction. It's only drawback is that it requires some familiarity with Hamilton's principle.

-
how is this intuitive? there's maths left, right and center – Larry Harson May 29 '12 at 11:33
@LarryHarson: This is intuitive math. Intuitive doesn't mean not precise. – Ron Maimon May 29 '12 at 16:18
It's not intuitive math if it's more complicated than the mathematical expression it's trying to justify!! – Larry Harson May 29 '12 at 18:18
@LarryHarson: It's more complicated, but less arbitrary. – Ron Maimon May 29 '12 at 19:25
@LarryHarson: What do you mean? The force is perpendicular to "v" and to "B", and one must argue this without relativity (since it predates relativity, and inspires it). It only becomes non-arbitrary when you have relativity. Perhaps the right thing is to say "relativity" right at the start, but it won't justify why 19th century folks were sure they understood it before Einstein. If you think this is a bad answer, perhaps I'll agree--- the reflection issue is no longer considered fundamental, and the Hamilton formulation might be arbitrary too (but I don't think so). – Ron Maimon May 29 '12 at 20:51
show 3 more comments

The Lorentz force is orthogonal to the velocity which is equivalent to the proposition that the force does no work on the charged particle; it only changes the direction of the velocity, not its magnitude.

The force is also orthogonal to the magnetic field. It follows from the formula and this fact – and the whole formula – may be derived in various methods, e.g. from the special theory of relativity.

This feature – the force's being perpendicular to the field – makes the magnetic field different from the electrostatic and gravitational fields. It is different in this respect: unlike the electrostatic and gravitational field, the field strength isn't a gradient of any "scalar potential". But there is no paradox. Different effects in Nature may follow different mathematical formulae and they often do.

If you have a problem with that, just appreciate that $(B_x,B_y,B_z)$ which looks like a vector is just a shorthand for $(F_{yz},F_{zx},F_{xy})$, three components of an antisymmetric tensor with three indices (the components are subsets of the relativistic tensor $F_{\mu\nu}$ which also contains the electric field).

For example, if $\vec B=(0,0,B_z)$, then the nonzero third component may be written as $B_{z}=F_{xy}$ and instead of an arrow in the $z$-direction, you may visualize the field by an oriented loop (with an arrow) in the $xy$-plane (to which the $z$-directed vector is normal). It's the same information.

This loop in the $xy$ plane really tells you what the magnetic field does to the charged particles: it rotates their velocities clockwise (or counter-clockwise, depending on the sign of $B_z$ and the charge $Q$) in the $xy$-plane.

-
It would help the op enormously if you could show how the electric force of a charge in a static electric field becomes both electric and magnetic in another frame which should only take a few lines. – Larry Harson May 29 '12 at 12:20
How about if I was to say... pretty please - with cherries on the top? – Larry Harson May 29 '12 at 19:57
Dear Larry, I am not sure how your proposed addition would answer the original question. For the Lorentz transformation of the electromagnetic field, see en.wikipedia.org/wiki/… - The cross products in the transformation are indeed related to the question but I personally believe that the explanation of the origin and consequences of these transformation laws is more complicated than what the OP was originally asking, so as an attempt to demystify an issue, it's like beating a flu virus with the HIV virus. – Luboš Motl May 30 '12 at 10:28
I do appreciate the effort you've clearly put into your answer. Unfortunately this is a bit beyond my knowledge, but I will be trying my best to understand it. Thank you. – Stephen Jennings May 31 '12 at 4:06