# I Moving charges in a moving frame of reference

Mentor

#### olgerm

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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#### Dale

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.

#### olgerm

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 dont 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?

#### Dale

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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#### 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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#### olgerm

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?

#### jbriggs444

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.

#### 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. $\gamma =\frac { 1}{\sqrt {1-v^2/c^2}}$

"Moving charges in a moving frame of reference"

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