kislotd

2022-07-23

What is the correct frame of reference for determining the magnetic force on a charge?

If two charges are both stationary in a given inertial frame, F1, then neither charge should experience a magnetic force due to the presence of the other charge (qv = 0). If we accelerate one charge, but not the other, then again, neither charge should experience a magnetic force, since only one charge has a non-zero velocity as measured in that inertial frame, meaning the other, stationary charge will experience no force in the magnetic field of the moving charge.

Now imagine that we are riding along as part of another inertial frame, F2, and that the first inertial frame discussed above that contains the two electrons F1, is traveling at a relative velocity of v from our perspective (i.e., it’s moving faster than us). Now imagine that we fire two electrons from our inertial frame F2, towards F1, one that travels at a velocity of v from our perspective, thereby traveling at the same velocity as the electron that appeared stationary in F1, and another that travels at the same velocity as the second, "faster" electron.

From our perspective in F2, both electrons have a non-zero velocity: the “slow one” traveling at a velocity of v, and the “fast one” traveling a bit faster than that. From our perspective, in F2, both electrons should experience a magnetic force of attraction due to their non-zero velocities in non-zero magnetic fields, which would change the path of those electrons from the perspective of both inertial frames.

However, from the perspective of F1, the “slow” electron is stationary, and should not experience any magnetic force in any magnetic field.

This seems to not make sense - what would happen as an experimental matter?

If two charges are both stationary in a given inertial frame, F1, then neither charge should experience a magnetic force due to the presence of the other charge (qv = 0). If we accelerate one charge, but not the other, then again, neither charge should experience a magnetic force, since only one charge has a non-zero velocity as measured in that inertial frame, meaning the other, stationary charge will experience no force in the magnetic field of the moving charge.

Now imagine that we are riding along as part of another inertial frame, F2, and that the first inertial frame discussed above that contains the two electrons F1, is traveling at a relative velocity of v from our perspective (i.e., it’s moving faster than us). Now imagine that we fire two electrons from our inertial frame F2, towards F1, one that travels at a velocity of v from our perspective, thereby traveling at the same velocity as the electron that appeared stationary in F1, and another that travels at the same velocity as the second, "faster" electron.

From our perspective in F2, both electrons have a non-zero velocity: the “slow one” traveling at a velocity of v, and the “fast one” traveling a bit faster than that. From our perspective, in F2, both electrons should experience a magnetic force of attraction due to their non-zero velocities in non-zero magnetic fields, which would change the path of those electrons from the perspective of both inertial frames.

However, from the perspective of F1, the “slow” electron is stationary, and should not experience any magnetic force in any magnetic field.

This seems to not make sense - what would happen as an experimental matter?

phinny5608tt

Beginner2022-07-24Added 17 answers

I understand that it looks like a paradox, but it actually isn't. The transformation laws of forces and the electromagnetic field make it work. What in one frame is just a magnetic field will in another frame include an electric field.

Suppose that an electron is in rest in a pure magnetic field. It will then experience no force. According to another frame the electron is moving and should experience a force from the magnetic field. And so it does. But in this frame there will also be an electric field that will also give a force, which is equal in magnitude but opposite to the one from the magnetic field. Thus the total electromagnetic force on the electron will be zero.

Suppose that an electron is in rest in a pure magnetic field. It will then experience no force. According to another frame the electron is moving and should experience a force from the magnetic field. And so it does. But in this frame there will also be an electric field that will also give a force, which is equal in magnitude but opposite to the one from the magnetic field. Thus the total electromagnetic force on the electron will be zero.

Ishaan Booker

Beginner2022-07-25Added 2 answers

That is a very insightful question, which leads to the roots of special relativity!

Let’s consider a simpler problem first. What are the EM fields around a single point charge?

Well, if the charge is at rest, there will be the $1/{r}^{2}$ electric field and no magnetic field. If the particle is moving on the other hand, then there are both a electric and magnetic field. But a charge that looks stationary to me might not look stationary to you! We have learned that different observers will not agree on the field configuration, in general.

So now we add a second particle in the mix. As you pointed out, if we don’t agree on the velocities, and we don’t agree on the fields, we don’t agree on the forces. That seems a conundrum, since we must be able to agree on something, otherwise what’s the point of physics?

Say that I do all the calculations from my perspective (using all the velocities, positions, fields in my frame), and I obtain the trajectories ${x}_{1}(t),{x}_{2}(t)$ for the two particles. You do the same from your perspective and obtain trajectories ${x}_{1}^{\prime}(t),{x}_{2}^{\prime}(t)$

If we now want to compare results, we need to translate your findings (trajectories) from my frame of reference to your frame of reference. If we do this correctly, then we see that even though we don’t agree on the nature of the forces in action (magnetic or electrical), we will agree on the trajectories the particles take.

This is analogous to the way that a rotating observer sees a centrifugal force whereas an inertial observer does not see that force. They don’t agree on the forces, but if they transform properly between the frames, they will agree on the trajectories.

Why do I say that we have to transform correctly? Naively, if at t=0 our axis are the same, but you are moving at velocity v with respect of me, I would want to write:

$x\prime (x)=x+vt$

This is called a Galilean transformation, and turns out that it doesn’t work. If we try to compare trajectories this way, we will disagree. We also say that maxwell’s equations are not invariant under Galilean transformations. That is, if you replace x by x+vt in Maxwells equations, you get very different looking equations. But Maxwell’s equations are invariant under a different kind of transformation, called the Lorentz transformations. These are a little more complicated, and they involve changing both the x and t. People knew that Maxwells equations were invariant under Lorentz transformations, and they puzzled over their meaning for a while, until somebody realised that they meant that time flows differently for different observers! This is special relativity.

So the correct way to do it, is that I calculate trajectories ${x}_{1}(t),{x}_{2}(t)$ and you calculate ${x}_{1}^{\prime}({t}^{\prime}),{x}_{2}^{\prime}({t}^{\prime})$, notice that your time variable is different from mine, and then use the Lorentz transformations to compare. Then we will agree.

Let’s consider a simpler problem first. What are the EM fields around a single point charge?

Well, if the charge is at rest, there will be the $1/{r}^{2}$ electric field and no magnetic field. If the particle is moving on the other hand, then there are both a electric and magnetic field. But a charge that looks stationary to me might not look stationary to you! We have learned that different observers will not agree on the field configuration, in general.

So now we add a second particle in the mix. As you pointed out, if we don’t agree on the velocities, and we don’t agree on the fields, we don’t agree on the forces. That seems a conundrum, since we must be able to agree on something, otherwise what’s the point of physics?

Say that I do all the calculations from my perspective (using all the velocities, positions, fields in my frame), and I obtain the trajectories ${x}_{1}(t),{x}_{2}(t)$ for the two particles. You do the same from your perspective and obtain trajectories ${x}_{1}^{\prime}(t),{x}_{2}^{\prime}(t)$

If we now want to compare results, we need to translate your findings (trajectories) from my frame of reference to your frame of reference. If we do this correctly, then we see that even though we don’t agree on the nature of the forces in action (magnetic or electrical), we will agree on the trajectories the particles take.

This is analogous to the way that a rotating observer sees a centrifugal force whereas an inertial observer does not see that force. They don’t agree on the forces, but if they transform properly between the frames, they will agree on the trajectories.

Why do I say that we have to transform correctly? Naively, if at t=0 our axis are the same, but you are moving at velocity v with respect of me, I would want to write:

$x\prime (x)=x+vt$

This is called a Galilean transformation, and turns out that it doesn’t work. If we try to compare trajectories this way, we will disagree. We also say that maxwell’s equations are not invariant under Galilean transformations. That is, if you replace x by x+vt in Maxwells equations, you get very different looking equations. But Maxwell’s equations are invariant under a different kind of transformation, called the Lorentz transformations. These are a little more complicated, and they involve changing both the x and t. People knew that Maxwells equations were invariant under Lorentz transformations, and they puzzled over their meaning for a while, until somebody realised that they meant that time flows differently for different observers! This is special relativity.

So the correct way to do it, is that I calculate trajectories ${x}_{1}(t),{x}_{2}(t)$ and you calculate ${x}_{1}^{\prime}({t}^{\prime}),{x}_{2}^{\prime}({t}^{\prime})$, notice that your time variable is different from mine, and then use the Lorentz transformations to compare. Then we will agree.

The magnetic field inside a long straight solenoid carrying current

A)is zero

B)increases along its radius

C)increases as we move towards its ends

D)is the same at all pointsWhich of the following units is used to express frequency?

Hertz

Watt

Newton

PascalWhat does tangential force produce?

How Many Different $f$ Orbitals Are There?

What is $\frac{e}{m}$ value for electrons ?

Which of the following correctly describes the magnetic field near a long straight wire?

A. The field consists of straight lines perpendicular to the wire

B. The field consists of straight lines parallel to the wire

C. The field consists of radial lines originating from the wire

D. The field consists of concentric circles centered on the wireIs energy directly proportional to frequency?

What is a compass? How is a compass used to find directions?

Magnetic field lines never intersect each other because

There will be two directions of the field at the same point.

Feild lines repel each other

Field lines follow discrete paths only

If field lines intersect they create a new magnetic field within the existing fieldTwo parallel wires carry currents of 20 A and 40 A in opposite directions. Another wire carrying a current antiparallel to 20 A is placed midway between the two wires. The magnetic force on it will be

Towards 20 A

Towards 40 A

Zero

Perpendicular to the plane of the currentWhat is the wavelength of the light emitted when an electron in a hydrogen atom undergoes transition from the energy level n = 4 to energy level n = 2? What is the colour corresponding to their wavelengths? (Given ${R}_{H}=109677c{m}^{-1}$)

A) 486 nm, Blue

B) 576 nm, Blue

C) 650, Blue

D) 450 nm, BlueTwo identical conducting spheres A and B are separated by a distance greater than their diameters the spheres carry equal charges and electrostatic force between them is F a third identical uncharged sphere C is first brought in contact with A, then with B and finally removed as a result, the electrostatic force between A and B becomes

Why is alternating current used in homes?

Which of the following radiation has the shortest wavelength.

X-ray

Infra red

microwave

ultravioletHow to find the local extrema for $f\left(x\right)=5x-{x}^{2}$?