# Moving charges in a moving frame of reference

jartsa
the spacelike part of the four-force is ##\gamma^2## times the three-force

Maybe it's ##\gamma## times the three-force?

Because I suspected that there must an error, googled "spacelike part of four-force", and first thing I checked says four-force and three-force differ by gamma.

Gold Member
The calculation is wrong then, because it's a wrong result that the force between the bodies increases as speed increases.
Let's assume those charge bodies move side by side, that simplifies calculations.
Yes, you are right, I found the mistake in my 1. post.
If 2 bodies with charge q are in rest then both have electric force ##F_1=\frac{q^2*k_q}{|\vec{r}|^2}##.
But in another frame of reference, where rhe bodies are moving with velocity v, they feel both magnetic and electric force ##F_2=
|\vec{F_{electric}}+\vec{F_{magnetic}}|=
|q*(\vec{E}+\vec{v}\times \vec{B})|=
q*(|\vec{E}|-|\vec{v}\times \vec{B}|)=
q*(\frac{q*k_q}{r^2}-|\vec{v}\times \frac{q*\mu_0*(\vec{v}\times \vec{r})}{4*\pi*|\vec{r}|^3}|)=
q*(\frac{q*k_q}{r^2}-\frac{q*\mu_0*(|\vec{r}|*|\vec{v}|^2)}{4*\pi*|\vec{r}|^3}|)=
\frac{q^2}{|\vec{r}|^2}*(k_q-\frac{|\vec{v}|^2*\mu_0}{4*\pi}|)=
\frac{q^2*(k_q-\mu_0*|\vec{v}|^2/4/\pi)}{|\vec{r}|^2}##

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Gold Member
##F' = \frac { F } { \gamma } ##
## \gamma = \frac {1} { \sqrt {1-v^2/c^2} } ##
##F_1=\frac{q^2*k_q}{|\vec{r}|^2}##
##F_2=\frac{q^2}{|\vec{r}|^2}*(k_q-\frac{|\vec{v}|^2*\mu_0}{4*\pi}|)##
That gives almost the same ratio between forces in 2 frames of reference as considering additional magnetic force in moving frame, but not exactly same because##\frac{|F_1|}{|F_2|}=
\frac{q^2*k_q}{|\vec{r}|^2}/(\frac{q^2}{|\vec{r}|^2}*(k_q-\frac{|\vec{v}|^2*\mu_0}{4*\pi}))=
k_q/(k_q-\frac{|\vec{v}|^2*\mu_0}{4*\pi})=
k_q/(k_q-\frac{|\vec{v}|^2}{4*\pi*\epsilon_0*c^2})=
k_q/(k_q-\frac{|\vec{v}|^2*k_q}{c^2})=
1/(1-\frac{|\vec{v}|^2}{c^2})=
\frac{1}{1-\frac{|\vec{v}|^2}{c^2}}=
\gamma^2##
but ##\frac{F}{F´}=\gamma##

Should it not be that ##\frac{F}{F´}=\frac{F_1}{F_2}##

By the way it is very cool and unexpected way to find lorentz factor in ratio of forces in different frames of reference.

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Mentor
Maybe it's ##\gamma## times the three-force?

Because I suspected that there must an error, googled "spacelike part of four-force", and first thing I checked says four-force and three-force differ by gamma.
Could you link to that source? I certainly could have made an error, but sometimes there is a difference in what relativistic factors are considered.

jartsa
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Gold Member
##F' = \frac { F } { \gamma } ##
## \gamma = \frac {1} { \sqrt {1-v^2/c^2} } ##
It does not explain why force in moving frame is ##\frac{q^2*(k_q-\mu_0*|\vec{v}|^2/4/\pi)}{|\vec{r}|^2}##.
##F:\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \not =\frac{q^2*(k_q-\mu_0*|\vec{v}|^2/4/\pi)}{|\vec{r}|^2}##

Lets say we measure force that is needed to break the string when it is not moving and get result ##F_{break}##. then tie this string between the moving bodies.
Observers in whom frame of reference bodies are moving and in whom frame of reference bodies are not moving must agree on whether the string broke or not.

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Mentor
Lets say we measure force that is needed to break the string when it is not moving and get result F_{break}. then tie this string between the moving bodies.
Observers in whom frame of reference bodies are moving and in whom frame of reference bodies are not moving must agree on whether the string broke or not.
Didn't I already clear this question for you? The string breaking is an invariant fact, so it can only depend on invariant quantities. The Minkowski norm of the four-force is such an invariant quantity, which is equal to the Euclidean norm of the three force in the rest frame.

I can write down the math if needed, but hopefully the statement is clear.

Gold Member
Didn't I already clear this question for you? The string breaking is an invariant fact, so it can only depend on invariant quantities. The Minkowski norm of the four-force is such an invariant quantity, which is equal to the Euclidean norm of the three force in the rest frame.

I can write down the math if needed, but hopefully the statement is clear.
Sorry if I do not understand it quick enougth. I do not want waste your time, but I don't have anywhere else to ask it. Jartsa said that it is possible to calculate force needed to break the string if it is moving if we know the force needed to break it if it is in rest, with formula ##F´_{break}(v)=F_{break}*\sqrt{1-v^2/c^2}##.
Since the fact whether the string broke or not must be same in both frames of reference: ##\forall_v(F_{break}>F(v) \iff F´_{break}(v)>F´(v))##
subsistuting jartsas formula and my equation for force in frame where the bodies are moving(##\frac{q^2*(k_q-\mu_0*|\vec{v}|^2/4/\pi)}{|\vec{r}|^2}##) and in frame where they are are in rest(##\frac{q^2*k_q}{|\vec{r}|^2}##) into this equation:
##\forall_v(F_{break}>\frac{q^2*k_q}{|\vec{r}|^2} \iff F_{break}*\sqrt{1-v^2/c^2}>\frac{q^2*(k_q-\mu_0*|\vec{v}|^2/4/\pi)}{|\vec{r}|^2})##
knowing that ##\mu_0=\frac{4*\pi*k_q}{c^2}## it can be simplified to
##\forall_v(F_{break}>\frac{q^2*k_q}{|\vec{r}|^2} \iff F_{break}*\sqrt{1-v^2/c^2}>\frac{q^2*k_q*(1-|\vec{v}|^2/c^2)}{|\vec{r}|^2})##
by simplifying it more:
##\forall_v(F_{break}>\frac{q^2*k_q}{|\vec{r}|^2} \iff F_{break}>\frac{q^2*k_q*\sqrt{1-v^2/c^2}}{|\vec{r}|^2})##
and more:
##\forall_v(\frac{q^2*k_q}{|\vec{r}|^2}=\frac{q^2*k_q*\sqrt{1-v^2/c^2}}{|\vec{r}|^2})##
and more:
##\forall_v(1=\sqrt{1-v^2/c^2})##
and more:
##\forall_v(v=0)##
and more:
False

So it seems that whether my equation for force in frame where the bodies are moving or jartsas transformation must be wrong. Which one is wrong and what equation should be instead of that?

Mentor
Jartsa's transformation doesn't mean what you think it means. This is because in Newtonian physics ##F=ma=dp/dt## but in relativistic physics ##ma\ne dp/dt##. However, beyond that Jartsa's transformation is not complete because the various components of the force do not transform the same way, so you cannot simply multiply a force by a scalar to get a transformed force. This is why he explicitly put the important caveat for transverse forces only.

The correct way to do this is to use the four-force. I showed in my previous post how the electromagnetic four-force transforms correctly so that it doesn't matter if you directly calculate the four-force using electromagnetism in the primed frame or if you simply boost the four-force from the unprimed frame into the primed frame. Thus, for clarity I will focus on the transformation of the four-force in general rather than on the specific details with Coulomb's law, but everything below applies to the electromagnetic four-force in all respects also.

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}##.

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olgerm
jartsa
Lets say we measure force that is needed to break the string when it is not moving and get result FbreakF_{break}. then tie this string between the moving bodies.
Observers in whom frame of reference bodies are moving and in whom frame of reference bodies are not moving must agree on whether the string broke or not.

The breaking force that is weakened by the motion, breaks the string that is weakened by the motion.

I say it again: The force that is trying to break the string is weakened, it breaks the string that is weakened.

If we zoom inside the moving string we can see lots of electric charges that are moving to the same direction. That means that there are lots of magnetic forces ##F_{magnetic}## between the charged particles. I leave it as an exercise for the reader to calculate how those magnetic forces change the strength of the string.

Oh yes, the electric forces ##F_{electric}## are changed too. Perhaps the reader would be better off using four-forces, or special relativity's transformation of three-forces instead of these "magnetic forces".

Or we could use the relativity principle and say that the strength of the moving string must change as much as the strength of the breaking force, so that no change is observed in the frame of the string.

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Gold Member
Since it was said that this question can not be answered with nonrelativistic physics, is the distance ##|\vec{r}|## between the moving bodies even same in both frames of reference?

Homework Helper
The velocity of the movement was stipulated to be at right angles to the separation vector. So yes, the separation is both constant within a frame and is the same in both frames in this setup.

From the original post:

##\vec{v}## is crosswise to ##\vec{r}##.

However, if we are talking about charges that are free to move in the direction of their relative separation and a time interval sufficient for them to do so significantly then the opposite holds true. Time dilation means that accelerations are not directly comparable.

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Blanci1
A somewhat simpler analogous paradox" is when you have 2 parallel beams of charge like cathode rays … in the frame comoving there is no magnetic attraction between parallel currents as in definition of amp. unit. Something is evidently wrong with classical theory if you try to apply it to both frames.. that is why they they thought there was a preferred ether frame for electromagnetism. With relativity of spacetime you get length contraction and extra electrostatic charge to and relativistic stuff so it all works out fine. Amazing.o

Staff Emeritus
. Something is evidently wrong with classical theory if you try to apply it to both frames

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

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.

if v=c

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."

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.

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.

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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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}}##

olgerm
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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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/Class...al_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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Mentor
I found it now. https://en.wikipedia.org/wiki/Class...al_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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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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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.