Why is the Lorentz Force always perpendicular to velocity?

[JiuBeixin]

TL;DR

Exploring the relativistic origin of Lorentz force. I'm deriving how a transverse force (from length contraction) and a longitudinal force (from relativistic momentum compensation) combine to stay perpendicular to velocity.

I represent the conductor as two parallel lines: stationary protons and electrons moving at velocity $v$. A test object with a positive charge also moves at velocity $v$ parallel to the electrons.
In frame $S'$ (the rest frame of the object), the object experiences a repulsive force (the origin of the Lorentz force) due to the length contraction of the moving proton line.

If I introduce an additional vertical velocity $u$ to the object in the frame $S$ the rest frame of the conductor), the entire conductor appears to move upward in frame $S'$. Since length contraction only occurs in the direction of relative motion, this vertical shift seemingly shouldn't create any additional charge density imbalance, so there isn't any extra force.

Yet, someone told me that because of a kind of Transformation of Forces, the force in frame $S$ would be vertical, but WHT and HOW it transform?

PS: I am a high school student and not a native English speaker, so please excuse any grammar issuee and awkward phrase :)​

 

[A.T.]

If I introduce an additional vertical velocity u to the object in the frame S the rest frame of the conductor), the entire conductor appears to move upward in frame S′.

Not just upward, but to the right (like it was moving before) and upward. So you have to apply the Lorentz transformation along that diagonal direction.

Since length contraction only occurs in the direction of relative motion, this vertical shift seemingly shouldn't create any additional charge density imbalance, so there isn't any extra force.

The Lorentz transformation is non-linear, so you cannot decompose it into independent dimensions like with Galilean transformations.

 

[TSny]

If I introduce an additional vertical velocity $u$ to the object in the frame $S$ the rest frame of the conductor), the entire conductor appears to move upward in frame $S'$. Since length contraction only occurs in the direction of relative motion, this vertical shift seemingly shouldn't create any additional charge density imbalance, so there isn't any extra force.

Consdider the case where the test charge moves perpendicularly away from the conductor ("wire") in the lab frame $S$ and suppose the charge has no motion parallel to the wire. Let $S'$ be the frame moving with the test charge. In both $S$ and $S'$, there is no imbalance of charge density in the wire. In frame $S$ there is no electric field produced by the conductor, only a magnetic field.

In frame $S'$ there are both an electric field and a magnetic field . See this thread for a discussion of how the electric field arises in $S'$ even though there is no charge imbalance. The electric field in $S'$ is parallel to the wire. Thus, in this frame there is an electric force on the test charge parallel to the conductor and no magnetic force (because the test charge is not moving in this frame).

Frame $S$ also finds a force on the test charge parallel to the wire. But, in this frame the force is interpreted as a magnetic force.

 

[PeterDonis]

perpendicular to velocity.

In general this is not the case. The Lorentz force is $F = q \left( \vec{E} + \vec{v} \times \vec{B} \right)$ (modulo some constant factors that depend on your choice of units). The $\vec{v} \times \vec{B}$ component will always be perpendicular to $\vec{v}$, but the $\vec{E}$ component does not have to be. For a simple example, consider an electron that starts out at rest between two oppositely charged plates, as in a capacitor: in the rest frame of the plates, the EM field is pure $\vec{E}$ and the force on the electron will be parallel to its velocity once it starts moving towards the positively charged plate.

 

[A.T.]

The Lorentz force is $F = q \left( \vec{E} + \vec{v} \times \vec{B} \right)$

The OP might be using a different naming convention:

Some sources refer to the Lorentz force as the sum of both components, while others use the term to refer to the magnetic part alone.

 

[SiennaTheGr8]

The OP might be using a different naming convention:

Notably, Zangwill uses the "Lorentz force = magnetic force" convention.

 

[A.T.]

Notably, Zangwill uses the "Lorentz force = magnetic force" convention.

This is also how I learned it in school, while the other component was called "Coulomb force"

 

[pervect]

It seems to me that if you define the Lorentz force as being only due to the magnetic field, the Lorentz force is perpendicular to the magnetic field because you defined it that way. Perhaps there's some way to rephrase the question in a more covariant way that doesn't treat the electric and magnetic fields differently - that would likely be more illuminating.

 

[PeterDonis]

Perhaps there's some way to rephrase the question in a more covariant way

There's no way to define "Lorentz force" as being solely due to the magnetic field and have the definition be covariant. And if we define it the way I defined it, which is the only covariant way to define a force on a charged object due to the EM field, then the answer to the question is what I said in post #4--in general, the force is not always perpendicular to the velocity.

 

[Sagittarius A-Star]

Perhaps there's some way to rephrase the question in a more covariant way that doesn't treat the electric and magnetic fields differently

One could start from the equation for the Lorentz 4-force and from that it is rest mass-preserving.

Source

 

[PeterDonis]

the equation for the Lorentz 4-force

Which is equivalent to what I wrote in post #4, i.e., it is not just the magnetic force. So what I said in post #4 would apply if one takes this approach.

 

[pervect]

There's no way to define "Lorentz force" as being solely due to the magnetic field and have the definition be covariant. And if we define it the way I defined it, which is the only covariant way to define a force on a charged object due to the EM field, then the answer to the question is what I said in post #4--in general, the force is not always perpendicular to the velocity.

Indeed, yes. At this point I'm not sure how familiar the OP is with the concept of covariance and what we call the covariant fomulation of electromagnetism. In case they either were familiar with it or were potentially interested in learning about it, I thought I'd give them a nudge in that general direction.

 

[pervect]

One could start from the equation for the Lorentz 4-force and from that it is rest mass-preserving.
Source

Your source got trimmed - but I do find the argument that the idea that the electromagnetic force should preserve the rest mass of a particle does imply that if we formulate the 4-force as a tensor, said tensor should be anitsymmetric. In terms of 4-forces and 4-vectors, it implies that the idea of preservation of rest mass requires that the 4-force should be orthogonal to the 4-velocity. This formulation of limitations on the form of the force law that make it preserve rest mass may be slightly more approachable as it requires 4-vectors rather than tensors. However, the original post was not written in a covariant framework, and I'm not sure how to translate this observation into a non-covariant form or how helpful it will be to the OP. We can hope it will be useful to them though.

 

[JiuBeixin]

Not just upward, but to the right (like it was moving before) and upward. So you have to apply the Lorentz transformation along that diagonal direction.


The Lorentz transformation is non-linear, so you cannot decompose it into independent dimensions like with Galilean transformations.

Thanks yoru reply. But I still have question. As I know, Lorentz Transformation is a frame tranformation. How could it affect the direction of force? And, I only know a little about Lorentz Transformation, so do you know where can I learn more about Lorentz Transformation?

 

[JiuBeixin]

It seems to me that if you define the Lorentz force as being only due to the magnetic field, the Lorentz force is perpendicular to the magnetic field because you defined it that way. Perhaps there's some way to rephrase the question in a more covariant way that doesn't treat the electric and magnetic fields differently - that would likely be more illuminating.

You are right. Now I do not treat the electric and magnetic fields differently. I veiw lorentz force as a special electric force but caused only when related moving occurs. And I didn't say that Lorentz force is perpendicular to the magnetic field.

 

[JiuBeixin]

One could start from the equation for the Lorentz 4-force and from that it is rest mass-preserving.

View attachment 370250​

Source

LOL, this is to difficult for me XD

 

[JiuBeixin]

In general this is not the case. The Lorentz force is $F = q \left( \vec{E} + \vec{v} \times \vec{B} \right)$ (modulo some constant factors that depend on your choice of units). The $\vec{v} \times \vec{B}$ component will always be perpendicular to $\vec{v}$, but the $\vec{E}$ component does not have to be. For a simple example, consider an electron that starts out at rest between two oppositely charged plates, as in a capacitor: in the rest frame of the plates, the EM field is pure $\vec{E}$ and the force on the electron will be parallel to its velocity once it starts moving towards the positively charged plate.

In my model, the conductor is electrically neutral in the lab frame ($S$), so there is no net electric field ($\vec{E} = 0$). Therefore, the $q\vec{E}$ term vanishes, leaving only the magnetic component

 

[PeterDonis]

In my model, the conductor is electrically neutral in the lab frame ($S$), so there is no net electric field ($\vec{E} = 0$). Therefore, the $q\vec{E}$ term vanishes, leaving only the magnetic component

In the lab frame, yes, this is the case. But it is not the case in the rest frame $S'$ of the object. In that frame there is a nonzero electric field.

 

[SiennaTheGr8]

There's no way to define "Lorentz force" as being solely due to the magnetic field and have the definition be covariant.

As an interesting(?) aside, we can "pick out" the magnetic force for a particular observer in a manifestly covariant way (I know this isn't what you meant).

With the field tensor $F^{\mu \nu}$ and the "projection tensor" $P^{\mu}_{\nu} = \delta^{\mu}_{\nu} - u^{\mu} u_{\nu}$ for observer four-velocity $u^{\mu}$ (where $u^{\mu} u_{\mu} = 1$), the tensor $B^{\rho \omega} = F^{\mu \nu} P_{\mu}^{\rho} P_{\nu}^{\omega}$ for the $u^{\mu}$ observer is $F^{\rho \omega}$ but with $F^{0 \omega} = F^{\rho 0} = 0$ (i.e., with the electric-field components "zeroed out"). Then for a particle with charge $q$ and four-velocity $v^{\nu}$, the four-vector $q B^{\rho \omega} v_{\omega}/(v^{\nu} u_{\nu})$ for the $u^{\mu}$ observer has components $(0, q \vec v \times \vec B)$ (where $\vec v$ is the particle's 3-velocity).

 

[PeterDonis]

we can "pick out" the magnetic force for a particular observer in a manifestly covariant way

Yes, because you can express both the observer's 4-velocity and the overall EM field as covariant objects (a 4-vector and an antisymmetric 2nd rank tensor).

But of course if you switch to a different 4-velocity by changing which "observer" you use (or changing the state of motion of an "observer"), what looked like a pure "magnetic force" for the original observer will no longer look like one for the new observer.

 

[Herman Trivilino]

It seems to me that if you define the Lorentz force as being only due to the magnetic field, the Lorentz force is perpendicular to the magnetic field because you defined it that way.

Whether you call it the Lorentz force or the magnetic force is just a naming convention. But the fact that it's perpendicular to the magnetic field is an observed behavior of Nature, it doesn't seem like a convention to me.

 

[PeterDonis]

the fact that it's perpendicular to the magnetic field

But "perpendicular to the magnetic field" is not a covariant concept, because "the magnetic field" is not a covariant concept. Only the full EM field tensor is covariant.

You can covariantly define a notion of "perpendicular to an object's velocity", as @SiennaTheGr8 posted, but that's not the same as "perpendicular to the magnetic field".