# I Moving charges in a moving frame of reference

#### olgerm

Gold Member
Hi.
If 2 bodies with charge q are in rest then both have electric force $F_1=\frac{q*q*k_q}{|\vec{r}|^2}$.
But in another frame of reference, that is moving with velocity v relative to first frame of reference, they feel both magnetic and electric force $F_2=|\vec{F_{electric}}+\vec{F_{magnetic}}|=|\vec{F_{electric}}|+|\vec{F_{magnetic}}|=|q*(\vec{E}+\vec{v}\times \vec{B})|=|q*(\frac{q*k_q}{r^2}+\vec{v}\times \frac{q*\mu_0*\vec{v}}{4*\pi*|\vec{v}|^2})|=q*(\frac{q*k_q}{|\vec{r}|^2}+\frac{q*\mu_0*|\vec{v}|^2}{4*\pi*|\vec{v}|^2})=\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2}$
So forces are not same in both frames of reference. Is it correct?
$\vec{v}$ is crosswise to $\vec{r}$.

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?

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• Parth
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#### Dale

Mentor
So forces are not same in both frames of reference. Is it correct?
Yes, that is correct.

However, there is a relativistic generalization of force called the four-force which is the same in all frames of reference.

• olgerm

#### olgerm

Gold Member
Yes, that is correct.
Thanks, where should I use the formulas
$E_2=E_1+v \times B_1$
$B_2=B_1-v/c^2\times E_1$
?
Are these correct formulas?

#### Dale

Mentor
I don’t know of the top of my head. I would have to look them up. I am assuming you can do that as well as I can

For what it’s worth, I almost never bother transforming the fields like that. If I am using multiple frames then I will use a covariant formulation of Maxwell’s equations

#### olgerm

Gold Member
If there were dynamometer between the charges: would it show reading $F_1$ or $F_2$?

#### jartsa

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?

$E_2$ can be calculated from just $E_1$ and v. The closer the v is to c, the more the electric field is distorted.

Here is a formula:
https://en.wikipedia.org/wiki/Biot–Savart_law#Point_charge_at_constant_velocity

By the way, isn't $B_1$ zero?

#### Dale

Mentor
If there were dynamometer between the charges: would it show reading $F_1$ or $F_2$?
If it’s at rest in frame 1 then it will measure $F_1$. If it is at rest in frame 2 then it will measure $F_2$.

#### Luxucs

However, there is a relativistic generalization of force called the four-force which is the same in all frames of reference.
One quick question I would appreciate clarification on about this - does this include non-inertial frames as well, or is the magnitude of the four-force only invariant in inertial frames?

#### Nugatory

Mentor
One quick question I would appreciate clarification on about this - does this include non-inertial frames as well, or is the magnitude of the four-force only invariant in inertial frames?
The four-force is frame-independent so is the same in all frames whether inertial or not.

You'll sometimes see threads here in which someone states some variant of "you need general relativity to work with non-inertial frames" and being correctly told that that is not true, that SR works just fine with non-inertial frames, you just need more mathematical tools than for inertial frames. The four-vector formalism (including four-force) is one of those tools.

• Dale

#### Luxucs

The four-force is frame-independent so is the same in all frames whether inertial or not.

You'll sometimes see threads here in which someone states some variant of "you need general relativity to work with non-inertial frames" and being correctly told that that is not true, that SR works just fine with non-inertial frames, you just need more mathematical tools than for inertial frames. The four-vector formalism (including four-force) is one of those tools.
Excellent, thank you!

#### olgerm

Gold Member
If it’s at rest in frame 1 then it will measure $F_1$. If it is at rest in frame 2 then it will measure $F_2$.
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.

#### Dale

Mentor
• davenn

#### Nugatory

Mentor
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.
You have created a surprisingly tricky problem by adding a force-measuring instrument. Before we consider that one, let's try answering the question I think you were trying to ask when you asked about connecting a "dynamometer" between the charges: What force does the particle 'feel'? To answer this, we use the four-vector formalism: we calculate the frame-independent four-force from the Faraday tensor (which combines the E and B three-vectors into a single frame-independent object that describes the electromagnetic field as a single entity). The four-force gives us the frame-independent proper acceleration through the four-vector equivalent of $\vec{F}=m\vec{a}$ and that proper acceleration is what the particle 'feels'; it's a measure of how much the worldline of the particle deviates from the free-fall inertial worldline it would follow if it weren't being pushed around by electromagnetic forces.

But if we attach a dynamometer to the particles to see what it reads? One way or another, all force-measuring devices are equivalent to attaching a spring to the object and then measuring how much the spring deflects when it is applying a force that exactly cancels the force we're looking at. That means that it changes the trajectory and speed of the object in question - so it can't tell us about the velocity-dependent Lorentz force. In effect, we now have a completely different problem: the motion of two charged particles connected by an ideal spring.

• PeroK

#### jartsa

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.

I guess those two rods must be the dynamometer. So it's actually two parallel rods that are connected by many springs.

So this dynamometer (or an observer strapped to the dynamometer) should say that the two bodies:

1: exert a small force on each other (because of their motion)
2: exert a small force on the rails (because of 1)
3: every spring experiences a large force deformation at some moment of time

So this is just another "relativity paradox" which is solved by the relativity of simultaneity.

(If there was just one spring connecting the rods, then the the dynamometer would not function as intended, because of various wave-phenomenons - except if we assumed perfectly rigid rods, in which case there would be a real paradox)

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#### olgerm

Gold Member
If it’s at rest in frame 1 then it will measure $F_1$. If it is at rest in frame 2 then it will measure $F_2$.
Which part of dynamometer must be moving fordynamomenter to show $F_2$?
Lets say we have a string that breaks if $F=\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2}$ but does not break if $F=\frac{q^2*k_q}{|\vec{r}|^2}$for dynamometer.
• As i understoof if string is tied between the charged bodies it would not break.
• if the bodies would move WITH pipes and string where attached between pipes it would not break.
• if the bodies would move IN pipes and string where attached between pipes it would break.
• if the bodies would move IN pipes and string where attached between pipes but would move as fast as the charged bodies with sines that connect it to the pipes with it would break(???).
maybe it about in which frame of reference we are not about in which frame of reference the dynamometer is in rest.

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#### olgerm

Gold Member
By the way, isn't $B_1$ zero?
it is. It is in formula, because formula is generally for EM-field not for only this setup.

#### Dale

Mentor
Which part of dynamometer must be moving fordynamomenter to show F2
The part which the dynamometer’s manufacturer indicates should be fixed and immobile. Often a base plate or a housing or frame.

Lets say we have a string that breaks if F=q2∗(kq+μ0∗|→v|2)|→r|2F=q2∗(kq+μ0∗|v→|2)|r→|2F=\frac{q^2*(k_q+\mu_0*|\vec{v}|^2)}{|\vec{r}|^2} but does not break if F=q2∗kq|→r|2F=q2∗kq|r→|2F=\frac{q^2*k_q}{|\vec{r}|^2}for dynamometer.
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. This is similar to indicating a distance or a time without identifying the frame.

#### jartsa

it is. It is in formula, because formula is generally for EM-field not for only this setup.
Okay so the equation simplifies to $E_2=E_1$, and that is not the truth. Electric field is not the same in all frames. It changes depending on how close to c the speed of the electric field is, as I said.

#### jartsa

• As i understoof if string is tied between the charged bodies it would not break.
• if the bodies would move WITH pipes and string where attached between pipes it would not break.
• if the bodies would move IN pipes and string where attached between pipes it would break.
• if the bodies would move IN pipes and string where attached between pipes but would move as fast as the charged bodies with sines that connect it to the pipes with it would break(???).

First, second and fourth are the same. String moves with bodies.

The third one is the interesting one. The moving bodies can be thought to be a moving dynamometer that is measuring how much force the string is generating. Or the string can be thought to be a dynamometer that is measuring how much force the bodies are genarating.

There is a "paradox": How can both dynamometers consider the other dynamometer to be the dynamometer that is moving?

There is a similar problem with devices that measure time. I mean in special relativity.

(If two clocks move relative to each other, how can both clocks consider the other clock to be time dilated?)

You said:
if the bodies would move IN pipes and string where attached between pipes it would break.
But the rule is that: The faster a pair of repelling charges move, the larger their magnetic attraction is.

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#### olgerm

Gold Member
The faster a pair of repelling charges move, the larger their magnetic attraction is.
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.

#### 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

Homework Helper
Gold Member
2018 Award
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

Homework Helper
Gold Member
2018 Award

"Moving charges in a moving frame of reference"

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