Thought experiment: charged particles

[davidge]

Suppose we have two charged particles on the laboratory and two observers A and B.

$A$ is inside one of the charges (never mind how)
$B$ is sitting at the laboratory

In the lab reference frame we accelerate the particles. According to $A$ there will be only electric attraction or repulsion, depending on the sign of the charges. But according to $B$ there also will be a magnetic force due to the movement of the particles.

Now suppose we stop the particles from moving (according to lab frame) and we measure their relative distance. Both observers $A$ and $B$ have to find out the same distance, but since $F = ma$, in the frame that there were two forces (electric and magnetic forces) there was a greater acceleration and consequently the charges would be more distant from each other.

How do we solve this problem?

 

[David Lewis]

Even if acceleration is the same and starts at the same time for both particles, A and B will not agree on what at the same time means.

 

[Dale]

A's frame is non inertial. None of the usual reasoning holds. I am not even sure that there is a way even in principle to separate electric and magnetic fields in such a frame.

 

[davidge]

Even if acceleration is the same and starts at the same time for both particles, A and B will not agree on what at the same time means.

You are correct, after starting this thread I considered the situation more carefully. See below

I am not even sure that there is a way even in principle to separate electric and magnetic fields in such a frame.

So are you saying that even that the particles don't move at all in $A$'s frame, we still would have a magnetic field in that frame?

Consider the following:

In classical theory the time and (consequently) acceleration of the particles will be the same in both frames. So if $F = ma$ is still valid in this thought experiment, then the force of interaction between the particles is the same in both frames. But in $B$'s frame there is a additional force, namely the magnetic force. The solution is that the electric field measured in $B$'s frame differs from the one measured in $A$'s frame. It's not surprising, because I guess since before Relativity coming up, physicists knew that a electrostatic field -the one that we can measure using Coulomb' theory- is different from the electric field of a moving charge.

Now, let $t_o$, $E_o$, $m_o$, $a_o$ be the time, electric field, mass and acceleration of a particle in $A$'s frame and quantities without a subscript be the quantities in $B$'s frame. Additionaly, let the particle charge be $q$ -it's the same in both frames-.

We know that $dt = \gamma dt_o$ and so, $a = a_o / \gamma$. If $F = ma$ is true, then $E = E_o / \gamma$.
Therefore for $B$ we have $q( \gamma E_o + v \times B) = m a_o / \gamma$. Divinding by the force in $A$'s frame and arranging terms, we get
$$-v^2 \ \gamma \ \vec{E_o} = \vec{v} \times \vec{B}$$ where I've taken $\vec{v}$ to be the velocity of both particles relative to the lab.

Is this correct?

EDIT: Just realized that the electric field in $B$ would be $E_o / \gamma$, and so the conclusion is that the magnetic field should be zero in the lab frame. So is relativity showing us that the magnetic field doesn't exist?

 

[Dale]

So are you saying that even that the particles don't move at all in A's frame, we still would have a magnetic field in that frame?

No, I am saying that in non-inertial frames there isn't always something that you can even identify as a magnetic field.

In classical theory the time and (consequently) acceleration of the particles will be the same in both frames.

This is not true in non-inertial frames.

So if F=ma is still valid

It is not valid in non-inertial frames, that is what makes it non-inertial.

We know that dt=γ dt_o and so, a=ao/γ. If F=ma is true, then E=Eo/γ.

None of this is necessarily correct in non-inertial frames. You actually have to work out each one of these expressions to see if they are valid in the specific non-inertial frame under consideration. You simply cannot use arguments based on inertial frames when working with non-inertial frames.

 

[davidge]

You simply cannot use arguments based on inertial frames when working with non-inertial frames

Why the particle frame is non inertial? As its velocity is constant, I don't see why it's non inertial.

 

[Dale]

In the lab reference frame we accelerate the particles.

As its velocity is constant

Please clarify

 

[Nugatory]

Why the particle frame is non inertial? As its velocity is constant, I don't see why it's non inertial.

Because earlier you said "In the lab reference frame we accelerate the particles.", which is not consistent with the lab frame being inertial and the frame in which the particles are at rest also being inertial and the particle frame having a constant velocity. Only one of them can be inertial, and it will be the one in which objects at rest do not experience proper accelerations.

 

[davidge]

Oh, ok. What I mean is in the lab frame we accelerate the particles up to a velocity $v$, possibly with their velocities going from zero to $v$ in a very short time.

 

[Dale]

So then the reference frame of the particles is non inertial. That is A's frame. The velocity is not constant.

 

[Nugatory]

Oh, ok. What I mean is in the lab frame we accelerate the particles up to a velocity $v$, possibly with their velocities going from zero to $v$ in a very short time.

If you're looking at the problem that way, you can limit your attention to the period of time between the two accelerations. Solve the problem once using an inertial frame in which the lab is at rest, and again using an inertial frame in which the particles are at rest after they have accelerated (which is to say that before the first and after the second acceleration they are moving along with the lab). The amount of time that the particles are in motion and subject to magnetic forces will be different in the two frames, the magnitude of the electrical field will be different in the two frames, the magnitude of the magnetic field will be different in the two frames... And all of these differences will come together in such a way that the invariants such as proper acceleration and proper time under acceleration are the same.

 

[davidge]

If you're looking at the problem that way, you can limit your attention to the period of time between the two accelerations. Solve the problem once using an inertial frame in which the lab is at rest, and again using an inertial frame in which the particles are at rest after they have accelerated (which is to say that before and after the acceleration they are moving along with the lab). The amount of time that the particles are in motion and subject to magnetic forces will be different in the two frames, the magnitude of the electrical field will be different in the two frames, and the magnitude of the magnetic field will be different in the two frames... And all of these differences will come together in such a way that the invariants such as proper acceleration and proper time under acceleration are the same.

This is what I did in post #4. Or that analysis of the problem isn't correct?

So then the reference frame of the particles is non inertial. That is A's frame. The velocity is not constant.

The velocity is constant after some time

 

[Nugatory]

This is what I did in post #4.

Is it? There you assumed that the times and the coordinate accelerations would be the same in both frames.

 

[davidge]

Is it? There you assumed that the times and the coordinate accelerations would be the same in both frames.

No, please read it again.

 

[David Lewis]

In classical theory the time and (consequently) acceleration of the particles will be the same in both frames.

Do you mean classical Newtonian theory of motion, or classical (non-quantum) Relativity?

 

[davidge]

Do you mean classical Newtonian theory of motion, or classical (non-quantum) Relativity?

non - relativistic theory, ie. Newtonian theory

 

[Dale]

The velocity is constant after some time

Constant after some time isn't the same as constant. If you want to specify an inertial frame then you should say "moving uniformly" from the beginning.

Now suppose we stop the particles

Plus you add this in, to further make the scenario non inertial.

Both observers A and B have to find out the same distance,

Why? Inertial frames don't have to agree on distance. So are you going back to a non inertial frame whose distances are defined the same as the lab frame?

since F=ma, in the frame that there were two forces (electric and magnetic forces) there was a greater acceleration and consequently the charges would be more distant from each other.

So now you are going back to inertial frames where distances are in fact different.

Your scenario is hopelessly muddled now. I would recommend starting over in a new thread with a much more carefully arranged and described scenario.

 

[SlowThinker]

Perhaps it would be easier to describe and analyze if you used the principle of equivalence backwards:
Two charges are resting in an upper floor of a tall building.
Then the support is removed and the charges fall to the ground, where they stop.
It makes it clear that the charges don't feel a magnetic field at any time.

@davidge , you may want to write out a few intermediate steps of your derivation, I don't see how the "arranged" equation follows from what is said before.
Also if you start another thread, don't name the lab frame B, or use the phrase "frame B" instead of just "B".

 

[Dale]

@davidge it is perfectly acceptable to not attach a reference frame to the charge. You can simply state that there is a lab frame and a moving frame, both inertial, with relative velocity v. Then you can fully describe the acceleration and deceleration in one inertial frame, and transform into the other inertial frame.

 

[davidge]

Thanks Dale and SlowThinker.
I was looking the Feynman Lectures available on Caltech Website, where this topic is discussed. I will read it and start another thread if some doubt remain.