# I Moving charges in a moving frame of reference

#### olgerm

Gold Member
Nothing is wrong with classical theory. Two beams of (like) charge repel in all frames and as such they will never be parallel.
if v=c, then $F_{electric}=-F_{magnetic}$ and the beams, do not bend in the frame where particles in beam are moving.
Even if $v \not = c$ force between the beams is different in different frames of reference.

Staff Emeritus
force between the beams is different in different frames of reference
True, but unsurprising as force is not Lorentz invariant.

Is impossible for charged particles.

There is still no problem with classical theory.

#### jartsa

Nothing is wrong with classical theory. Two beams of (like) charge repel in all frames and as such they will never be parallel.
Classical physicist:

"The coordinate acceleration of electrons is reduced in the frame where the repulsive force between electrons is reduced. And the proper acceleration of the electrons is the same as the coordinate acceleration of the electrons. Which means that moving observers can know that they are moving - from the reduced acceleration of co-moving electrons."

#### olgerm

Gold Member
True, but unsurprising as force is not Lorentz invariant.
It is a contadiction in non-relativistic physics, because of the string scenario that I described in earlier posts.

#### Blanci1

Nothing is wrong with classical theory. Two beams of (like) charge repel in all frames and as such they will never be parallel.
by classical theory I mean.. newtonian mechanics with galilean relativity plus maxwell electrodynamics which together are not very logical, so we need special relativity ideas to get over the paradox of what happens to the magnetic field and attractive force between two currents when you run along with them. Special relativity explains all keeping maxwell intact for all inertial frames but galilean ideas like force invariance must be modified.

#### olgerm

Gold Member
In the rest frame $f = (0,F_x,F_y,F_z)$, so the failure condition is $|f|=\sqrt{-0^2+F_x^2+F_y^2+F_z^2}>F_{break}$

In another frame $f'=\Lambda f$ with $\Lambda$ as defined in my earlier post. So $f'=(\gamma v F_x,\gamma F_x,F_y,F_z)$ and $|f'|=\sqrt{-(\gamma v F_x)^2+(\gamma F_x)^2 + F_y^2 + F_z^2} = |f|$. So the failure condition remains the same: $|f'|=|f|>F_{break}$.
But In the situation described in my 1. post Force in direction of speed($F_x$) is 0 in both frames of reference. And it is possible to choose such coordinatesystem, that $F_z$ is 0 in both frames of reference. But $F_y$ is not same in both frames of reference, like your post claims.

#### Dale

Mentor
The y component of the four force is the same in both frames (assuming a boost in the x direction). The y component of the three force is not, but the failure condition is covariant so it is not based on the three force, it is based on the four force.

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#### olgerm

Gold Member
I think now I understand it. Can confirm that all of the following is correct?
If I took 2 identical string and 2 identical pulling machines and brought one of the machines and one of the strings into spacecraft. Accelerated the spacecraft until it reached speed v. Then tested the strings with pulling machine on Earth and in spacecraft. then:
• whether both strings broke or both strings would not break.
• Tester on Earth would say that breaking string in spacecraft would take smaller force than to break the string on Earth.
• Tester on Earth would say pullingmachine in spacecraft applied smaller force than the machine on Earth.
• Tester in spacecraft would say that breaking string on Earth would take smaller force than to break the string in spacecraft.
• Tester in spacecraft would say pullingmachine on Earth applied smaller force than the machine in spacecraft.
• Tester on Earth would say that breaking string in spacecraft would take $1-v^2/c^2$ times smaller force than to break the string on earth.
• Tester on Earth would say pullingmachine in spacecraft applied $1-v^2/c^2$ times smaller force than the machine on earth.
• testers would not need to convert forces if they view strings and the machines as collection of pointcharges that are tied to each other with chemical bond, and are interacting with force $F=q*(E+v*B)$, because E,v and B are different in their frames of reference.

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Gold Member

#### jartsa

I am almost sure first six points are correct. @Dale ,@jartsa , jbriggs444 can you confirm that?
I can completely agree.

But some square roots seem to be missing. The change factor is the gamma. $\gamma =\frac { 1}{\sqrt {1-v^2/c^2}}$

#### Dale

Mentor
Let F be the magnitude of the four force, then:
Tester on Earth would say that breaking string in spacecraft would take smaller force than to break the string on Earth.
Tester on earth would saybthst breaking string in spacecraft would take the same F as to break the string on earth.

Tester on Earth would say pullingmachine in spacecraft applied smaller force than the machine on Earth.
Tester on earth would say pulling machine in spacecraft applied the same F as the machine in earth.

Tester in spacecraft would say that breaking string on Earth would take smaller force than to break the string in spacecraft.
Tester in spacecraft would say that breaking string on earth would take the same F as breaking the string in spacecraft.

Tester in spacecraft would say pullingmachine on Earth applied smaller force than the machine in spacecraft.
Tester in spacecraft would say pulling machine on earth applied same F as the machine in spacecraft.

Tester on Earth would say that breaking string in spacecraft would take 1-v^2/c^2 times smaller force than to break the string on earth.
Tester on earth would say that breaking string in spacecraft would take same F as to break string on earth.

Tester on Earth would say pullingmachine in spacecraft applied 1-v^2/c^2 times smaller force than the machine on earth.
Tester on earth would say pulling machine in spacecraft applied same force as the machine on earth.

testers would not need to convert forces if they view strings and the machines as collection of pointcharges that are tied to each other with chemical bond, and are interacting with force F=q∗(E+v∗B), because E,v and B are different in their frames of reference
testers would not need to convert forces if they use four-forces

#### olgerm

Gold Member
Thanks very much for explaining this to me. I had previously tried to find answer to this question from various sources outside of PhysicsForums and the asked question from many people, but never got clear answer.

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#### olgerm

Gold Member
Some source said that I should use relation
$E_2=E_1+v \times B_1$
$B_2=B_1-v/c^2\times E_1$
but that would mean that $E_2\not=\frac{q*k_q}{r^2}$ Is it correct? Does that mean that there is free electromagnetic field(electromagnetic wave(s)) in second frame of reference?
I don’t know of the top of my head. I would have to look them up.
I found it now. https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Non-relativistic_approximations. Seems that the problem can be solved with nonrelativistic physics.
But still does it mean that $E_2\not=\frac{q*k_q}{r^2}$ in some cases. for example if the bodies are moving with speed v in 1. frame of reference and with speed -v in 2. frame of reference?

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#### Dale

Mentor
I found it now. https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Non-relativistic_approximations. Seems that the problem can be solved with nonrelativistic physics.
But still does it mean that $E_2\not=\frac{q*k_q}{r^2}$ in some cases.
Most likely yes. Approximations are often wrong. That is why they are approximations.

However, I must say that I don’t understand your persistent desire to do things the hard way with inaccurate approximations that are complicated and frame variant, when you could instead use the exact quantity in the four-vector formulation which is easy and covariant. Why do you insist on doing it the hard way?

#### jartsa

But still does it mean that E2≠q∗kqr2E_2\not=\frac{q*k_q}{r^2} in some cases. for example if the bodies are moving with speed v in 1. frame of reference and with speed -v in 2. frame of reference?
Is v << c ?

Anyway, there is a (slowly) moving magnet in one frame.

If the magnetic field of the magnet is B and the velocity of the magnet is v, then

there is an Electric field E around the moving magnet: $E=v \times B$

So a still standing charge q feels a force: $F=q*v\times B$

What is this called? Induction? The v refers to the velocity of a magnet.

In another frame there is a charge moving in a magnetic field of a still standing magnet. There is a force on the charge: $F=q*v \times B$

That force is called Lorentz force, and it's a magnetic force, right? The v refers to the velocity of a charge.

The two frames agree about q and B and the magnitude of v. So the two frames seem to disagree about the direction of the force.

Okay, so it must be so that the v refers to the velocity of an observer relative to a magnet, not "the velocity of a magnet", as I said.

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#### olgerm

Gold Member
Most likely yes. Approximations are often wrong. That is why they are approximations.
However, I must say that I don’t understand your persistent desire to do things the hard way with inaccurate approximations that are complicated and frame variant
I desire to understand whether maxwells equations are compatible with classic physics.
These equations where on wikipedia do not seem just approximations, but are straightly unsound, because:
$div(\vec{E_1})=q/\epsilon_0$
$\vec{E_2}=\vec{E_1}+v\times B$
$div(\vec{E_2})=q/\epsilon_0$
$\vec{B_1}=\frac{\mu_0*q*v\times r}{4*\pi*|r|^3}$
to
$div(\vec{E_1}+v\times \frac{\mu_0*q*v \times r}{4*\pi*|r|^3})=div(\vec{E_1})$
if r is crosswise to v
$div(\vec{E_1}*(1+\frac{\mu_0*q*|v|^2}{4*\pi*|r|^2*|E_1|}))=div(\vec{E_1})$
to
$\mu_0*q*|v|^2=-4*\pi*|r|^2*|E_1|$
to
$\mu_0*q*|v|^2=-4*\pi*|r|^2*\frac{q}{4*\pi*\epsilon_0}$
to
$|v|^2=-|r|^2*c^2$ which is not True for all v and all r.

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#### Dale

Mentor
I desire to understand whether maxwells equations are compatible with classic physics.
Maxwell’s equations are fully relativistic. They are not compatible with the Galilean transform.

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