Magnetic field at relativistic speed

[alba]

I am trying to understand magnetism and its relation to electricity:

Suppose there are two electrons traveling side by side in deep space at 1 cm distance at .99 c

In this article http://academic.mu.edu/phys/matthysd/web004/l0220.htm it is said that the ratio $F_B/F_E$ is $v^2/c^2$ for v << c. According to that formula the magnetic pull sould be .9801 times the electrostatic repulsion. Can you tell me

- 1) what is the actual value (applying relativity, since the speed is close to c) ? Is there a big difference? Intuitively it shouldn't be smaller, yet it can't possibly get greater than 1, can it?
- 2) what happens to the trajectory of the particles due to the influence of the mutual magnetic field, does if deflect from the straight line?

 

[jartsa]

Force: To get the force in the other frame, divide the center of mass force by gamma, force means here simply the force that effects the electrons. Force transforms in this simple way in this simple case.

Trajectory: The separation process is time dilated. To get the timetable of the separation process in the other frame, multiply the center of mass timetable by gamma.

 

[alba]

Force: To get the force in the other frame, divide the center of mass force by gamma, force means here simply the force that effects the electrons. Force transforms in this simple way in this simple case.

Trajectory: The separation process is time dilated. To get the timetable of the separation process in the other frame, multiply the center of mass timetable by gamma.

- 1) Is the attractive force greater than .98 or smaller? gamma is 7.09, right? center of mass is at .5 cm, what is the CoM force? The repulsive electrostatic force isn't affected at all by relativistic speed?

- 2) I am asking about the trajectories, are they bent from the straight line, besides the particles slowly separating?

 

[jartsa]

- 1) Is the attractive force greater than .98 or smaller? gamma is 7.09, right? center of mass is at .5 cm, what is the CoM force? The repulsive electrostatic force isn't affected at all by relativistic speed?

- 2) I am asking about the trajectories, are they bent from the straight line, besides the particles slowly separating?

1: I didn't check the link, until now. I am sure the total force changes by gamma, but the question wasn't about total force. But somehow I feel that the formula F=q*v*B should work even if v is relativistic.

2: A force on a projectile causes a bending of its trajectory. So trajectories are bent in this case.

 

[alba]

1: I didn't check the link, until now. I am sure the total force changes by gamma, but the question wasn't about total force. But somehow I feel that the formula F=q*v*B should work even if v is relativistic.

1) Are you saying that the actual value of magnetic force is .9801 the coulomb force and no correctionis needed? Really? So, what is all the big deal of SR?
2. does magnetic field mutually generated (Lorenz force) make the trajectories converge? does this compensate the divergence ? Can you roughly figure out what happens to the particles after a seond or two?

 

[greswd]

Use the E and B transforms for SR

 

[greswd]

https://en.wikipedia.org/wiki/Class...rmation_of_the_fields_between_inertial_frames

 

[alba]

https://en.wikipedia.org/wiki/Class...rmation_of_the_fields_between_inertial_frames

Hi greswd, I have read some wiki articles, I posted because Iam a student and aint able to do the transform,can you just tell me what happens, if the trajectories diverge and at what rate?
Thanks.

P.S. if the electrons diverge, isn't that phenomenon observable in all frames of references?

 

[greswd]

Hi greswd, I have read some wiki articles, I posted because Iam a student and aint able to do the transform,can you just tell me what happens, if the trajectories diverge and at what rate?
Thanks.

P.S. if the electrons diverge, isn't that phenomenon observable in all frames of references?

You know what the Lorentz Force Law is? It is co-variant with SR, it applies in all cases. That is how we obtain the magnitude of the force.

In relativity Force = dp/dt where p is the relativistic momentum.

1) Yes, it cannot exceed 1.

2) It will deflect.

 

[alba]

1) Yes, it cannot exceed 1.
2) It will deflect.

1) tkanks, so if without SR it is .98 it cannot increase much
2) momentum is .99*10^10 *7.098 = 7.018 *10 * 10 , right?
Lorenz force pushes the trajectories to converge, doesn't it? whereas electric exceeds magnetic force by .0199. Since you say that in the end they diverge, Lorenz force is less than .0199 , right?

 

[jartsa]

1) Are you saying that the actual value of magnetic force is .9801 the coulomb force and no correctionis needed? Really?

No, I'm just thinking about a wire loop, and the fact that the magnetic field of the loop is proportional to the current in the loop, and the current in the loop is proportional to the velocity of the moving charges in the loop, if the amount of moving charges is constant. So when we know the currents of two loops, we can calculate the magnetic force between the two loops very easily, without worrying about relativity

And we can also calculate very easily magnetic force between two straight wire segments, using Ampere's law, no relativistic corrections needed.

2. does magnetic field mutually generated (Lorenz force) make the trajectories converge? does this compensate the divergence ? Can you roughly figure out what happens to the particles after a seond or two?

What happens after one second is the same what happens after 1/gamma seconds in the other frame. In other words: What happens in the non-rest frame is a slow-motion version of what happens in the rest-frame. In other words high speed motion causes time dilation.

 

[greswd]

1) tkanks, so if without SR it is .98 it cannot increase much
2) momentum is .99*10^10 *7.098 = 7.018 *10 * 10 , right?
Lorenz force pushes the trajectories to converge, doesn't it? whereas electric exceeds magnetic force by .0199. Since you say that in the end they diverge, Lorenz force is less than .0199 , right?

Idk where you get those numbers from.

Both electric and magnetic forces are part of the Lorentz force.

The electric force will always exceed the magnetic force so they will always diverge.

 

[PeterDonis]

what happens to the trajectory of the particles due to the influence of the mutual magnetic field, does if deflect from the straight line?

This might be the best question to start with. First, look at things in the rest frame of the electrons--or at least, in their rest frame at some instant. In that frame, there is no magnetic force; there is only the electrostatic repulsion between the electrons. So in this frame, they will accelerate away from each other.

Now transform to a frame in which the electrons start out moving parallel to each other at 0.99c. Since the relative motion of the electrons must be invariant, if they diverge in one inertial frame, they must diverge in all inertial frames. So they must diverge in this new "moving" frame as well.

I am trying to understand magnetism and its relation to electricity

The above analysis should show you that one way of viewing what we call a "magnetic" field is as just what you get when you look at an electrostatic field in a frame in which the source is moving. More precisely, it's a piece of what you get.

If you look at the Wikipedia page greswd linked to, that shows the transformation of the E and B fields, you will see, if we start from a frame where the field is electrostatic--i.e., where we have nonzero E but zero B--and then transform to a different frame, where the relative velocity between the frames is perpendicular to the direction of the E field (which in this case means perpendicular to the direction of separation between the electrons), you will see that the E field increases by $\gamma$, and the B field in the new frame will be $\gamma v$ times the original E field. Since the force due to the B field has a magnitude of $v$ times the B field, the ratio $F_B / F_E$ will be $v^2$. In other words, the formula that is valid for $v << c$ is also valid for $v \rightarrow c$, as others have already pointed out.

Furthermore, if you work out the direction of the force due to the B field, you will see that it is the direction given by $v \times (v \times E)$, which, using standard identities for cross products, works out to the opposite direction from the force due to the E field. So what is happening here is that, in the moving frame, the force due to the E field increases, but a B field appears giving a force that partially opposes the force due to the E field. The net effect is that the electrons diverge in all frames, and their diverging worldlines transform just as they should by the Lorentz transformation equations.

 

[alba]

Idk where you get those numbers from.

Both electric and magnetic forces are part of the Lorentz force.

The electric force will always exceed the magnetic force so they will always diverge.

Isn't there a crucial point when the the magnetic attraction plus the Lorenz divergence will be greater than the electric repulsion? In LHC protons reach .999999991 c and therefore attraction is only 0.000000018 smaller than repulsion, can't Lorenz convergence be greater than that?

 

[Ibix]

Isn't there a crucial point when the the magnetic attraction plus the Lorenz divergence will be greater than the electric repulsion? In LHC protons reach .999999991 c and therefore attraction is only 0.000000018 smaller than repulsion, can't Lorenz convergence be greater than that?

You can always transform into the rest frame of the particles. In that frame they repel due to plain vanilla electrostatic repulsion. They must, therefore, repel in all frames, or else there is a contradiction. A frame is simply a point of view. You can't make things that are moving apart start moving together by changing point of view.

 

[alba]

http://i.stack.imgur.com/3ga0O.png

You can always transform into the rest frame of the particles. In that frame they repel due to plain vanilla electrostatic repulsion. They must, therefore, repel in all frames, or else there is a contradiction. A frame is simply a point of view. You can't make things that are moving apart start moving together by changing point of view.

That is right, so, how can repulsion (electric) become attraction (magnetic) in any other frame? Can you figure out the actual rate of divergence of electrons in all frames?

 

[Ibix]

That is right, so, how can repulsion (electric) become attraction (magnetic) in any other frame? Can you figure out the actual rate of divergence of electrons in all frames?

There is always a net repulsion. Your explanation for why this is so may vary. In one frame there is an electrostatic repulsion, while in another there is a stronger electrostatic repulsion and a weaker magnetic attraction.

Yes, you can figure out the rate of divergence in any frame. Probably the easiest way to do it is to work out the positions and/or velocities of the electrons as a function of time in their rest frame (more precisely the center of mass frame of the pair) and then transform those.

 

[alba]

Yes, you can figure out the rate of divergence in any frame. Probably the easiest way to do it is to work out the positions and/or velocities of the electrons as a function of time in their rest frame (more precisely the center of mass frame of the pair) and then transform those.

Can you do that? I can't, that's why I posted in the first place. Can you tell me if the forces are correctly representedin the picture? There are two attractive forces and only one repulsive, right? Are you saying that the convergence due to Lorenz force can never be greater than Fe-Fm?

 

[Ibix]

Not in my head. The forces in the diagram look plausible (there's a net repulsion) but I haven't checked them in detail.

Do you know how to calculate the electrostatic force between two charges? Coulomb's Law? If so, can you calculate the force on the two electrons if they are at rest (i.e. electrostatic repulsion only)?

 

[alba]

Not in my head. The forces in the diagram look plausible (there's a net repulsion) but I haven't checked them in detail.

Do you know how to calculate the electrostatic force between two charges? Coulomb's Law? If so, can you calculate the force on the two electrons if they are at rest (i.e. electrostatic repulsion only)?

The point of the whole question is: what does actually happen? do A and B diverge at a rate dictated by Coulomb or at a different rate? if so, is it .019 Coulomb or more or less? Do you have to correct that value due to SR or not?do you have do throw in Lorenz convergence or not?

Even if you can't or are not willing to worg out the actual values, can you answer any of those general questions?

 

[Ibix]

I'm trying to teach you how to do it. Can you calculate the force in the rest frame? If so, there are equations to transform forces and then derive accelerations. Edit: nothing more complex than powers required.

I would expect the rate to differ from a naive application of the Coulomb force, but I haven't actually checked that.

 

[alba]

I'm trying to teach you how to do it. Can you calculate the force in the rest frame? If so, there are equations to transform forces and then derive accelerations.

I would expect the rate to differ from a naive application odmf the Coulomb force, but I haven't actually checked that.

Please check that and let me know, only when I get a picture of what is going on, I can try my hand. I do know what is Coulomb at 1 cm, but what mass do I consider: the rest mass m =(m0 c^2) or the relativistic mass = 7 m? What gravity pull does C register from A or B 1 or 7 m?

 

[Ibix]

You don't need the mass to calculate the force - just the charge and separation.

I strongly recommend forgetting you ever heard the term "relativistic mass". It's only useful in some circumstances, and whether or not it applies with forces depends on the direction of the force compared to the velocity. Most people stopped using it decades ago, except in pop science presentations where "cool" is more important than "useful".

Gravitational pull isn't relevant here. Also, we don't know how gravity works when the source is a quantum particle.

So you've got the Coulomb force, right?

 

[greswd]

Please check that and let me know, only when I get a picture of what is going on, I can try my hand. I do know what is Coulomb at 1 cm, but what mass do I consider: the rest mass m =(m0 c^2) or the relativistic mass = 7 m? What gravity pull does C register from A or B 1 or 7 m?

everything is described here:

http://newt.phys.unsw.edu.au/einsteinlight/jw/module2_FEB.htm

 

[alba]

So you've got the Coulomb force, right?

at 1 cm itis 2.3*10^-24 N, right? don't you have to divide it by mass to get acceleration?

 

[jartsa]

at 1 cm itis 2.3*10^-24 N, right? don't you have to divide it by mass to get acceleration?

Yes, and electron's rest mass is 9.10938356*10^-31kg
And luckily our electron is at rest - but unfortunately only when it has not accelerated yet.
How about if we consider one micro seconds time, use Newton's laws, and assume an uniform electric field, when we calculate the motion of this electron?

 

[vanhees71]

You know what the Lorentz Force Law is? It is co-variant with SR, it applies in all cases. That is how we obtain the magnitude of the force.

In relativity Force = dp/dt where p is the relativistic momentum.

1) Yes, it cannot exceed 1.

2) It will deflect.

It's easier to work with manifestly covariant quantities first. In this case it's the Minkowski force
$$\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=K^{\mu}.$$
Since $p^{\mu}$ are contravariant vector components and $\tau$ is a scalar (proper time of the particle), also the right-hand side must be the contravariant components of a four vector.

Using the Hamilton principle of least action, it's clear that for the electromagnetic interaction
$$K^{\mu}=\frac{q}{cm} F^{\mu \nu} p_{\nu},$$
where $F^{\mu \nu}$ are the contravariant components of the Faraday tensor.

All contravariant vector components transform under Lorentz transformations by
$$V^{\prime \mu} = {\Lambda^{\mu}}_{\nu} V^{\nu},$$
where ${\Lambda^{\mu}}_{\nu}$ is the Lorentz-transformation matrix. E.g., for a boost in $x^1$ direction, it's
$$({\Lambda^{\mu}}_{\nu})=\begin{pmatrix} \gamma & -\beta \gamma & 0 & 0 \\
-\beta \gamma & \gamma & 0 & 0\\
0& 0 & 1 & 0 \\
0& 0 & 0 &1
\end{pmatrix}.$$
The non-covariant force $\vec{F}$ relates to the spatial components of the Minkowski Force according to
$$\vec{F}=\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=\frac{\mathrm{d} \tau}{\mathrm{d} t} \frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau} = \frac{1}{\gamma} \vec{K}.$$

 

[jartsa]

everything is described here:

http://newt.phys.unsw.edu.au/einsteinlight/jw/module2_FEB.htm

I disagree with that page. Page says force transforms as gamma squared, I say force transforms as gamma.

Also I think that as velocity of two point charges approaches c, electric and magnetic fields between the charges approach infinite strength. I came to that conclusion looking at this picture: https://en.wikipedia.org/wiki/Class.../media/File:Lorentz_boost_electric_charge.svg

 

[greswd]

It's easier to work with manifestly covariant quantities first. In this case it's the Minkowski force
$$\frac{\mathrm{d} p^{\mu}}{\mathrm{d} \tau}=K^{\mu}.$$
Since $p^{\mu}$ are contravariant vector components and $\tau$ is a scalar (proper time of the particle), also the right-hand side must be the contravariant components of a four vector.

Using the Hamilton principle of least action, it's clear that for the electromagnetic interaction
$$K^{\mu}=\frac{q}{cm} F^{\mu \nu} p_{\nu},$$
where $F^{\mu \nu}$ are the contravariant components of the Faraday tensor.

All contravariant vector components transform under Lorentz transformations by
$$V^{\prime \mu} = {\Lambda^{\mu}}_{\nu} V^{\nu},$$
where ${\Lambda^{\mu}}_{\nu}$ is the Lorentz-transformation matrix. E.g., for a boost in $x^1$ direction, it's
$$({\Lambda^{\mu}}_{\nu})=\begin{pmatrix} \gamma & -\beta \gamma & 0 & 0 \\
-\beta \gamma & \gamma & 0 & 0\\
0& 0 & 1 & 0 \\
0& 0 & 0 &1
\end{pmatrix}.$$
The non-covariant force $\vec{F}$ relates to the spatial components of the Minkowski Force according to
$$\vec{F}=\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=\frac{\mathrm{d} \tau}{\mathrm{d} t} \frac{\mathrm{d} \vec{p}}{\mathrm{d} \tau} = \frac{1}{\gamma} \vec{K}.$$

thats way too complicated for the OP

 

[alba]

Yes, and electron's rest mass is 9.10938356*10^-31kg

I am baffled, ibis said I don't need mass. Now, consider the 2 particles in LHC: mass is 7 times greater, does A feel the increased mass (through gravity) of B, even though it is at rest in A's frame.

 

[greswd]

I am baffled, ibis said I don't need mass. Now, consider the 2 particles in LHC: mass is 7 times greater, does A feel the increased mass (through gravity) of B, even though it is at rest in A's frame.

you don't need mass if you're only concerned about force and not acceleration.

 

[Nugatory]

I am baffled, ibis said I don't need mass. Now, consider the 2 particles in LHC: mass is 7 times greater, does A feel the increased mass (through gravity) of B, even though it is at rest in A's frame.

You have just found another of the many reasons why thinking in terms of mass increasing with speed just creates confusion. If you haven't already read this Insights article, do so now - the mass that you plug into Newton's $F=Gm_1m_2/r^2$ has nothing do with the speed of anything.

 

[alba]

you don't need mass if you're only concerned about force and not acceleration.

I'd just like to know what is the ratio Fm/FE after the SR corrections, and the final outcome, that is the rate at which A and B will diverge. After 32 posts , still no clue!

 

[greswd]

I'd just like to know what is the ratio Fm/FE after the SR corrections, and the final outcome, that is the rate at which A and B will diverge. After 32 posts , still no clue!

google transverse mass.

that will give you the transverse acceleration

 

[vanhees71]

thats way too complicated for the OP

But it's the most simple way!