# I Moving charges in a moving frame of reference

#### jartsa

They would move with same speed in in the setup with pipes as in setup without pipes. Pipes are just for transferring radial force from moving bodies to the string.
You said the string breaks in the third scenario. What is the reasoning behind that? J.C.Maxwell would say that the net force between the bodies approaches zero as the speed of the bodies approaches c.

#### PeroK

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I do not understand that. if dynamometer is tied between 2 moving bodies then it shows one result, but if it is in rest(for example 2 rods towards which the bodies are sliding) it gives another result. It does not seem right.
What if you replace the dynamometer with a clock? The clock might ultimately keep time using EM forces. The moving clock must feel exactly the same forces as the stationary clock and must tick at the same rate. Perhaps the idea of time dilation does not seem right either?

#### olgerm

Gold Member
The part which the dynamometer’s manufacturer indicates should be fixed and immobile. Often a base plate or a housing or frame.
There is no paradox here; this is just an incomplete specification of the breaking condition. Forces are frame variant so you have to specify which frame is the breaking condition defined in.
I think strings and other bodies break under some ultimate mechanical pullingforce irrespective of whether they are moving or not. The force needed to break can be measured measured anywhere.

#### olgerm

Gold Member
Can this question be answered with non-relativistic physics?

#### PeroK

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Can this question be answered with non-relativistic physics?
No.

#### olgerm

Gold Member
You said the string breaks in the third scenario. What is the reasoning behind that? J.C.Maxwell would say that the net force between the bodies approaches zero as the speed of the bodies approaches c.
I thought that because string is moving in frame of chraged bodies, then in feels force $F_2$, that is sufficient to break it.

#### Dale

Mentor
I think strings and other bodies break under some ultimate mechanical pullingforce irrespective of whether they are moving or not. The force needed to break can be measured measured anywhere.
This is incorrect. Force is frame variant. That is simply a fact of physics. The same force measured in different frames gives different results. You cannot simply wish that fact away.

#### jartsa

I thought that because string is moving in frame of chraged bodies, then in feels force $F_2$, that is sufficient to break it.
Yes that is quite reasonable. As the speed of the string approaches c the forces between charges in the string approach zero.

Well, I guess that when these two things that wrestle with each other have a zero relative speed, then they exert equal forces on each other, and when their relative speed increases, then they of course feel their own strength to remain the same, while the force from the other thing is felt to decrease or increases or stay constant. But because of the symmetry of this scenario the change must be the same for both things, I mean the transformation of the force from the other thing must be the same for both things.

So if the string does break, then the two bodies are pulled together by the string.

If that thing above is hard to follow, I mean that if the string says "I won this wrestling match thanks to the weakening of the opponent caused by motion", then the two bodies will say "I won this wrestling match thanks to the weakening of the opponent caused by motion". Because of symmetry.

If that sounds impossible, then maybe it never happens. So in that case it must be so that the thing made of two charged bodies will say: "the string is getting weaker as it accelerates, but for some reason I can not cause the string to break by pushing on these rods."

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#### olgerm

Gold Member
This is incorrect. Force is frame variant. That is simply a fact of physics. The same force measured in different frames gives different results. You cannot simply wish that fact away.
So with typical ultimate stress test we can calculate the stress in needed to break body if it is in rest(not moving)?
How to calculate how strong foce is needed to break the same body if it is moving with speed v?

#### jartsa

So with typical ultimate stress test we can calculate the stress in needed to break body if it is in rest(not moving)?
How to calculate how strong foce is needed to break the same body if it is moving with speed v?
When F is perpendicular to v:

$F' = \frac { F } { \gamma }$

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

When F and v are aligned:

$F' = F$

http://www.sciencebits.com/Transformation-Forces-Relativity

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#### jartsa

Thanks, but should it not be $F´=F*\gamma$, because in frame where the object moves the force is greater

Well to me it seems that in frame where the object moves the force is smaller.

What force exactly is larger in the frame where the object moves?

#### olgerm

Gold Member
What force exactly is larger in the frame where the object moves?
$\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2}>\frac{q*q*k_q}{|\vec{r}|^2}$

#### jartsa

$\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2}>\frac{q*q*k_q}{|\vec{r}|^2}$

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.

The electric field strength increases by gamma, so electric force increases by gamma. So
$F_{electric} = \gamma * \frac {k* q_1*q_2}{r^2}$

And now the magnetic field. $B=\frac {1}{v^2}v \times E$

That E there is the increased field: $\gamma * E_{rest}$

So the magnetic force is $q*v \times B$

Total force = $|F_{electric}| - |F_{magnetic}|$

Now I can see that this calculation will result in decrease of force and I think that this the right way to calculate.

I got my formulas from here:
https://en.wikipedia.org/wiki/Biot–Savart_law#Point_charge_at_constant_velocity

That complicated equation for E simplifies to multiplying Erest by gamma in this case.

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#### Dale

Mentor
OK, let's work this through explicitly with covariant notation using the (+---) sign convention and units where c=1.

The electromagnetic field tensor is: $$F_{\mu\nu}=\left( \begin{array}{cccc} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \\ \end{array} \right)$$ and the Lorentz four-force is $$f_{\mu}=q F_{\mu\nu} u^{\nu}$$For a charge, q, at rest in a pure E field we have$$f_{\mu}=q F_{\mu\nu} u^{\nu} = q \left( \begin{array}{cccc} 0 & -E_x & -E_y & -E_z \\ E_x & 0 & 0 & 0 \\ E_y & 0 & 0 & 0 \\ E_z & 0 & 0 & 0 \\ \end{array} \right) (1,0,0,0) = (0, q E_x, q E_y, q E_z)$$
If we boost to a frame where the charge is moving with velocity v in the x direction then we can write the Lorentz transform matrix as $$\Lambda = \left( \begin{array}{cccc} \gamma & v \gamma & 0 & 0 \\ v \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$ so we can immediately write that the four-force in the other frame is $$f'_{\mu}=\Lambda f_{\mu} = (\gamma v q E_x, \gamma q E_x, q E_y, q E_z)$$
If we are masochists and don't want to do things the easy way then we can boost the field tensor and the four-velocity and get $$f'_{\mu} = q F'_{\mu\nu} u'^{\nu}$$ $$= q \left( \begin{array}{cccc} 0 & -E_x & -\gamma E_y & -\gamma E_z \\ E_x & 0 & \gamma v E_y & \gamma v E_z \\ \gamma E_y & -\gamma v E_y & 0 & 0 \\ \gamma E_z & -\gamma v E_z & 0 & 0 \\ \end{array} \right) (\gamma,\gamma v,0,0)$$ $$= (\gamma v q E_x, \gamma q E_x, q E_y, q E_z)$$

Now from this tedious exercise we notice a few things. First, the field tensor which has no B field in the unprimed frame has a mixture of E and B fields in the primed frame. Second, the Lorentz force law holds in both frames with the modified field and modified velocity. Third, since for small forces the spacelike part of the four-force is $\gamma^2$ times the three-force, we have that the component of the three force in the direction of the boost is smaller by a factor of $1/\gamma$ and the transverse components are smaller by a factor of $1/\gamma^2$. Finally, if you have a breakage force in a rest frame, then you can simply boost that four-force to a different frame to get the breakage condition in that other frame.

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#### 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.

#### olgerm

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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#### olgerm

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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#### Dale

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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"Moving charges in a moving frame of reference"

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