Magnetic force between two parallel moving charges

[Arre]

Hi everyone,

I have been trying to find an answer to the following question on several websites on the topic, but no luck until so far. Maybe there is someone who can help me here:

question:
If two charges move parallel to each other (say in vacuum and relative to me), with equal speed, will there be a magnetic force felt by each one from the other?

All explanations I have found so far say that indeed there is one, and this is often explained by the classical experiment with two parallel currents in wires.

But, I would say that is a different situation. The two charges in my question do not move relative to each other, and hence they do not see mutual magnetic influence, right?

In the two wire setup there is relative motion of two types of charges in each wire (the ions and electrons), hence a magnetic field will develop in each wire, causing both wires to attract. Or you could say that the electron current in one wire exerts magnetic force on the ion current in the other and vice versa (I know this sounds strange but hey, it's all relative motion that counts, not?).

The explanations of the two wire experiment I found seem odd to me, since they almost always explain things by regarding the Lorentz force on the electron movement only. But since there is no relative motion between the electrons (at equal currents and wires) this should not suffice for an explanation.

Comments are very appreciated, since I am puzzled.

regards,

Arjan

 

[Meir Achuz]

The force between two moving charges must be determined relativistically.
The result is a complicate formula in advanced textbooks.
Even what is meant by "force" in this case needs clarification.
There will be a magnetic and an electric force.
It is much more complicated than for two wires.
Explaining the wire force as between two electrons is used in elementary courses because it seems easy, but it is wrong.

 

[jtbell]

In the reference frame in which the two charges are at rest, they exert electrostatic forces on each other. In a reference frame in which they are moving, they exert a combination of electric and magnetic forces on each other. The effects of those forces on the charges' motion are equivalent to each other, and are related by the relativistic Lorentz transformation between the two frames. The electric and magnetic fields produced by the charges in the two frames are also related by the Lorentz transformation, applied to a tensor whose components are the components of the electric and magnetic fields.

http://farside.ph.utexas.edu/teaching/em/lectures/node121.html

http://farside.ph.utexas.edu/teaching/em/lectures/node123.html

 

[zhanghe]

jtbell, does it mean that
different force situations will be seen by different observors, for instance
one is at rest, and the other is moving with the two charge?

 

[Phrak]

zhanghe. excuse me, but nevermind the obscure reference to force. But it is all about relativity. The inertial frame of the observer with respect to charges is involved.

Say you have this field called F generated by some source like one of your electrons. When you are stationary with respect to the electron, the parts of F that aren't zero valued all look like an electric field.

But when you are in motion with respect to the electron, some parts of F are measured as the magnetic field.

It's all really one field. What we measure as the electric and magnetic fields depend on relative motion.

 

[zhanghe]

Thank you, Phrak.
I think I get to know why the textbook says the electric and magnetic fields are united by the relativity.thanks again.

 

[Phrak]

anytime, zhanghe

 

[Philip Wood]

Start by defining what we mean by a magnetic force in a given frame of reference. It is one felt by a moving 'test' charge, and proportional to the speed of motion. Such forces are generated by other moving charged particles, 'source charges', but not by stationary ones. So there is no problem in saying that electrons moving in parallel paths at the same speed experience magnetic forces: we have a moving 'test charge' and a moving 'source charge'. They are not moving relative to one another, but they ARE moving relative to the frame of reference. That's what matters. [Of course the electrons also experience a repulsive electric-field or 'Coulomb' force. This is greater than the (attractive) magnetic force.]

If we view the electrons from a frame in which they are stationary (that is, we 'ride along with them'), then, by definition, there is no magnetic force between them. There is, of course, a (repulsive) electrostatic or 'Coulomb' force.

Now Relativity theory tells us the ('transverse') force between objects will be less in the frame in which they are both moving than when measured in a frame in which they are both stationary. So the repulsive force between the electrons will be less in the laboratory frame than in the ride-along-with frame. But that's just what we observe, only we choose to regard the repulsive force in the laboratory frame as a repulsive Coulomb force added to a weaker attractive magnetic force. In a way we've EXPLAINED the origin of magnetic forces as a frame-modification to Coulomb forces.

[A word of warning. There is another complication (which doesn't affect the essential argument above.) The ELECTRIC field force (i.e. that part of the force on a testing charge which isn't affected by the testing charge's motion) IS affected by the motion of the source charge. It's greater when the source charge is moving. So the Coulomb force between the electron moving in the laboratory is greater than when seen in the ride-along-with frame. But the increase in repulsive Coulomb force is less than, and outweighed by, the attractive magnetic force which comes into play in the lab frame!]

Sorry if this seems complicated. I'm afraid it just IS, though the complication is largely due to our insistence on classifying electromagnetic forces as electric or magnetic or both. However, it really is useful so to do - usually.

 

[gabrown]

Can i ask a related question,

If we have two charges that are moving highly relativistically in a straight line, I have been told (and looked in the textbook) that they will repel to a certain distance and stay at that distance (or the force drops to very small), now I am concerned because the book explained this as the magnetic and electric forces canceling, but in the frame of reference of the electrons there is only electric force. So what is actually happening here?

cheers

Gareth

 

[Philip Wood]

In the charges' frame (my 'ride-along-with-frame') there is, as you say, only the electric repulsion. This will get weaker the further the charges get from each other. Even when the force is so weak that the acceleration is almost zero, they'll still have their recession velocity and will go on receding for ever.

To find out what happens in the lab frame it's easiest to look at what happens in terms of frame-modified forces (first two sentences of my paragraph 3 above). Thus we can say that the resultant force in the lab frame (in which the charges are moving) will be less than in the ride-with frame. But it won't, for some finite separation of the charges, change from repulsive to attractive, or even drop to zero. Look at it this way: if the force changed direction, that would be a definite recordable event. It couldn't happen in one reference frame (the lab frame) but not in another (the ride-with frame).

I think there's confusion between this scenario and another. The resultant (repulsive) force in the lab frame is less than in the ride-with frame. But it isn't zero. However, the faster the electrons are traveling (sorry about the double 'l'; I'm English) in their parallel paths, the smaller the resultant force becomes. It approaches zero for charges' speeds approaching the speed of light. When I say 'approaches' I'm not describing a process occurring in time; I'm using it in the mathematical sense.

 

[gabrown]

I may have gotten confused. I don't have the book I looked at on me, but the one i am referring to is PROBLEMS AND SOLUTIONS ON ELECTROMAGNETISM page 581 the problem is 5019 part b. I will try and post it tomorrow when i get the book to explain where my confusion is

 

[Calluin]

Hello! I thought of this problem when I was learning Electromagnetism and read about the scenario of the two wires (the Amper definition). I just want to make sure I got it right. So the magnetic force is just a decrease in the electrostatic force caused by relativistic transformation, but we percieve the electrostatic force as constant and make up a magnetic force so that the results are equivalent? That's how I have interpreted the Philip's explanation. Sorry if I have messed up the terminology :-)

 

[Philip Wood]

Hi. What you say pretty well sums it up. But, as I said, for the two charges moving in parallel paths the analysis is rather tricky. It is extremely easy to show from Special Relativity the factor by which the repulsive force is less when the charges are viewed from the lab frame, in which the charges are moving. But, in terms of electric and magnetic forces, the reduction has to be attributed to a magnetic attractive force AND a somewhat increased repulsive electric field. [Electric field forces on a test charge are, by definition, those which are the same whether or not the particle is moving.]

Two current-carrying wires (each consisting of a stationary line of + charges and a moving line of - charges) don't experience mutual ELECTRIC field forces, only magnetic field forces. That makes life easier, but this time the relativistic analysis is more complicated (than for two point charges), but not difficult if you keep a clear head.

There is, in my opinion, a good treatment of these things in Robert Resnick's "Introduction to Special Relativity'. I'm afraid it's probably out of print and expensive secondhand.

Best wishes.

 

[pam]

The force exerted on a particle q' traveling at velocity v' a distance r from a particle q traveling at constant velocity v is given by
$$\frac{qq'[{\bf r}+{\bf v'}\times({\bf v}\times{\bf r)}]} {\gamma_v^2[{\bf r}^2-({\bf v}\times{\bf r})^2]^{\frac{3}{2}}}$$.

 

[DaTario]

If the two charges are at rest relatively to each other, then it seems natural to expect them to experience only static electric force. Magnetic force appears with relative velocity between the charge and the ids element, which, in this case, is the other moving charge. The charge, thus, is at rest respectively to this element of current.

Wire case may be explained differently foer the existence of protons at rest in the laboratory reference frame is part of the cause of the force between the wires.

best regards

DaTario

 

[Calluin]

Oh thank you very much, I'm looking forward to relativity classes I will have in 2 years. I know
just the basics of special relativity from secondary school and I'm not familiar with the necessary calculus to do it properly. But it's nice to know where the catch is, thanks!

 

[pzlded]

Magnetic force on a moving point charge is centripetal force.

Point charge movement within a curved path (windings in an inductor) increases magnetic energy per current, because the curved path increases inductance.

U$$_{M}$$ = LI$$^{2}$$

Ion migration is a function of volts, not bends versus straight wire. The increased magnetism at windings is not due to higher electron velocity.

v(drift) is a function of charge * volts / friction

 

[Philip Wood]

Being at rest relative to each other doesn't mean no magnetic forces. In the particles' 'own' frame, i.e. that in which they are both at rest, there are, I agree, no magnetic forces. In another frame (the 'lab' frame) they will both have the same velocity as each other. If the particles' separation is transverse to that velocity (as in the cases discussed above), then there IS a magnetic force between them in the lab frame. How do we know this? The key thing to remember is that the magnetic force on a charge is DEFINED as that part of the Lorentz force which is proportional to the charge's velocity. The relativistic analysis and the classical electromagnetic analysis of the case of the moving charges both predict that there IS just such a non-zero force term in the lab frame for the particles moving in parallel paths with the same velocities, even though there is no RELATIVE velocity between the particles.

 

[pzlded]

In high frequency circuit design, all bends are 90 degrees. Curves increase inductance. The inductance of the curve increases the time the wire requires to reach full current. During this time, electromagnetic energy (voltage's energy) is converting to magnetic energy in the wire, instead of being delivered to the load.

I do not know why bending a (superconductive) wire in an area outside of a voltage field instantly increases magnetic energy (velocity?) per electron (inductance) in the wire. Any curvature in electron path determines location of the center of a magnetic field.

It is hard for me to understand current flow among atoms. Any non-zero quantized mass movement requires infinite energy. For an electron to travel from one molecule to the next, a chemical bond must exist between the two atoms. For superconductivity to occur, giant molecules the length of the wire in a superconductive magnet must exist. How does an atom get a superconductive energy level (free electrons are not superconductive)?

 

[G Diddy]

So is the net force repulsive or attractive?

In the comoving frame, there's no B field but in the lab frame there is. E²-B² is conserved, so by going to the comoving frame, the
electric field is responsible for the force. In the lab frame, B exists, so E must be greater.
Is that right?

It's been a while since I took E & M.

 

[Philip Wood]

You've got it right (though I prefer to stick to SI, and include factors of c, in which case it's E^2 - c^2 B^2 which is invariant.

In detail, this is what happens... Let E be the electric field due to one charge, in the vicinity of the other charge, in the charges' own rest frame (your 'comoving frame'). This transforms to g E in the lab frame. [I write "g" for gamma = 1/root(1 - v^2/c^2).] But in the lab frame, there's also a magnetic field, of (v/c^2) g E. You'll find that these fields leave E^2 - c^2 B^2 invariant.

Now let's look at forces. In their own rest frame, the particles (of charge q) repel each other with a force of qE = F (let's say). In the lab frame, the repulsive force due to the electric field is q g E = g F. The (attractive) magnetic field force is q v (v/c^2) g E, that is
(v^2/c^2) g F. If you subtract the magnetic force from the electric force you get (1/g) F. This is less than F, but still repulsive!

This analysis in terms of fields is interesting (well, I think it is!) but quite complicated. But if you're interested only in the final result, you can reach it from a single force transformation).

 

[Per Oni]

Philip, I thought I got the hang of your explanations until you said there are different forces in the different frames. In my mind a see 2 different things happening.

Suppose I set up an experiment in which the repulsive force of the electrons in their own frame make it possible for them to just collide with the (opposite) insides of a vacuum tube in which they are travelling.
In the labs frame the repulsive force is less so that they not collide.
Where do I go wrong?

 

[Philip Wood]

Per Onl: I applaud your general line of thought: we can't have events (e.g collisions) which happen in one inertial frame, but not when viewed in another. But if the electrons collide with the tube walls in one frame, they will do so in the other. You seem to be suggesting that we can reduce the force between the electrons in their own frame until they only 'just' hit the tube walls. But as long as they are repelling AT ALL, they will hit the tube walls (even if it takes them a long time to do so). And, seen from the lab frame, they will hit the walls, though it will take them even longer to do so!

 

[Per Oni]

Suppose excluding the magnetic force and seen from the lab frame the electrons should hit the tube at point A. With inclusion they hit further along at point B.
As seen from the rest frame of the electrons and excluding relativity they also hit at A but including length contraction (of a for them moving tube) they also hit further along at B.
Is this picture correct?

 

[Philip Wood]

Haven't got time at present to think carefully about this, I'm afraid. One thing you should bear in mind - and you may be doing so - is that the ELECTRIC field is different in the two frames; the difference isn't just that there's a magnetic field in the lab frame but none in the electrons' rest frame.

 

[Per Oni]

Please take as long as needed any input is much appreciated.
In the mean time I’ll try to explain myself a bit better.

You said:

we can't have events (e.g collisions) which happen in one inertial frame, but not when viewed in another.

Therefore in both frames there need to be agreement that the collision happened say at point B (as in my earlier post).
Now as far as the electrons are concerned in their rest frame the coulomb force F is not altered. The only thing happening for them is the tube shooting past with some velocity. Therefore the only thing for them that can change a possible point of collision is the length contraction of the tube.
In the lab frame F is modified to (1/g) F with no length contraction taking place.
Therefore somehow these 2 modifications must produce the same result?

I'll try this a different way.
Say the velocity v of the electrons is so low that v^2 can be ignored, so that g can be ignored. Then there's no length contraction nor a modified E field. The only thing different in the 2 frames is the magnetic field but the resulting magnetic force is also dependent on v^2
and can also be ignored. In this case both frames agree on the same point of collision.
But can they agree when v^2 gets a lot higher?

 

[Philip Wood]

You put your point very clearly. I think that what we need to do is to sort out how forces are related to lengths and times. The force between the electrons is transverse to the relative velocity between the two frames of reference. This transverse force is defined as d(transverse momentum)/dt. But transverse momentum is invariant between frames. dt is a proper time in the electrons' frame (we can ignore any small transverse movement under the repulsive force). So the corresponding time in the lab frame is g times as long. So the force in the lab frame is 1/g times as large. [This is the force transformation argument I mentioned in hash 21 (above).] [Hope you can cope with misuse of dt as a finite time interval: this flaw can easily be removed.]

We get the same result for forces by detailed consideration of the electric and magnetic field transformations (see hash 21).

Now let's look at the mark made on the side of the tube where an electron hits it. [We ought also to imagine another mark made on an 'inner' tube, or on a parallel ruler when the electrons are released and start to move apart.] The time of flight (tau) in the electrons' own frame is near-enough a proper time. But the time of flight in the lab frame is improper and equal to g tau. So in the lab frame the start and finish marks will be g v tau apart x-wise. [v is the electrons' x-velocity in the lab frame.] But in the electrons' frame we'd say the marks were only v tau apart. This is one way of looking at the length contraction you mentioned.

I think this does answer your question. The key point is that time dilation can be viewed as the reason why transverse force must transform in the way it does, and ALSO as the reason why length contraction must occur. [I'm not saying that time dilation CAUSES either of these things, just that its linked by logic to both these things.]

 

[Per Oni]

I think this does answer your question.

Yes thanks a lot.

I realized that the traverse time was involved and wanted to calculate it.
This maths problem turned out not to be simple straight forward.

Say there’s a traverse acceleration a=K/r^2; K is a constant and r is the distance between the electrons. When t=0; r=2xD. How long does it take to collide with the tube’s wall when each electron will then have covered a traverse distance of L ?

 

[Philip Wood]

For the electrons in 'their own frame', equate gain in KE to loss in PE. After tidying and square-rooting I get dy/dt = K sqrt(1/a - 1/y) in which K = sqrt[(e^2)/(8 pi epsilon-0 m)], 2a is the separation of the electrons at time t = 0, when they're released from rest, and 2y is their separation after time t. You then separate the differential equation. The Wolfram integrator site does the y-integral for you.

For comparing the fields and forces in the two frames, I don't think much is to be gained from letting the electrons move far enough for the force between them to change significantly. But your ambitions go further than mine!

 

[G Diddy]

Hi Guys
My 'conundrum' was that I couldn't figure out how E increased in the lab frame. But E does increase perpendicular to the motion of a moving charge.
See for example 'Classical Electrodynamics' by J. D. Jackson, pg 555:
E=qr/(r³gamma²(1-beta²sin²psi))^3/2
The electric field vectors compress perpendicular to the direction of motion.

You can also use a Lorentz boost on the electromagnetic field tensor.
Cheers

 

[Philip Wood]

Exactly so. When psi = pi/2, i.e. in the transverse direction, (1 - beta^2 sin^2 psi)^3/2 becomes (1 - beta^2 )^3/2 = gamma^-3, so the transverse E in the lab frame is simply gamma times that in the rest frame of an electron. We can reach this result very simply by applying length contraction to a coin-shaped gaussian box with an electron at its centre.