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Apparent contradiction: Two charges moving parallel to each other

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jose 892
#1
Mar22-12, 05:41 PM
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Hello I need help with a problem: If I see two identical charges moving in the same direction parallel to each other with the same constant velocity, my intuition tells me that the magnetic field generated by their movemente will cause them to attract much like what happensa with two wires with current flowing in the same direction. However doesn't this mean that if I put myself in the frame of reference of one of the charges y will see two seamingly at rest equal charges attracting? This cannot be because laws of physics must be the same in all inertial frames of reference so... I am clueless.
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jtbell
#2
Mar22-12, 05:55 PM
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Have you tried calculating the forces in the two situations? Don't forget that in the first situation there will also be an electric force between the two charges (in addition to the magnetic force).
jose 892
#3
Mar22-12, 06:02 PM
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jtbell: In the first situation there is a magnetic force and the electric force, If you make them go fast enough you can make the magnetic force be bigger than the electric force, in this case they will come closer until the electric repulsion is equal to the magnetic attraction. In the frame of reference of the moving charges they both seem at rest so it would look like two equally charged particles that are at rest and do not repel, which still doesn't make sense. :s

DaleSpam
#4
Mar22-12, 06:26 PM
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Apparent contradiction: Two charges moving parallel to each other

Quote Quote by jose 892 View Post
If you make them go fast enough you can make the magnetic force be bigger than the electric force
It might be instructive to calculate exactly how fast is "fast enough".

Can you give a guess as to what speed that might be? Assuming that physics doesn't have contradictions and that the situation you describe would be such a contradiction then a reasonable guess would be that the speed required is impossible to attain. Can you think of any such speeds that are impossible for massive particles to attain and might play an important role in electromagnetic phenomena?
jose 892
#5
Mar22-12, 06:55 PM
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Ok I didn't think of this before but it really doesn't matter f the magnetic force is weaker than the electric force, what is important is that it isn't zero, and it is acting against the electrostatic repulsion. This would mean that if you out yourself in the rame of reference of one of the moving charges, where they are not moving relative to each other, you would see them repel, yes, but with a weaker force than the one expected due to the electrostatic repulsion acting alone. This would mean that coulomb's law has two different constants in two inertial frames of reference which cannot be. There must be something wrong in my reasoning but I can't figure out what it is.
DaleSpam
#6
Mar22-12, 07:03 PM
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You are honing in on a very key realization. That is that forces are not frame invariant.

This is not the case in Newtonian mechanics, but once scientists started working with EM and Maxwell's equations this kind of problem began to appear. In the end, it took Einstein's special relativity to resolve the issue.

Because of how lengths and times transform in relativity, forces also transform such that coulomb's law has the same constant in every inertial frame.
jose 892
#7
Mar22-12, 07:12 PM
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Could you be more specific in your answer? Do they experience magnetic attraction or not? We seem to agree that coulomb's law constant should stay the same in al inertial frames of reference. Now the situation I described in my last post where I put myself in the inertial frame of the charges it seems they experience a smaller force than expected, which cannot be if the constant remained the same. I guess the only answer would be that the do not experience magnetic attraction in the frame of reference where they appear to be moving, however I don't know why wouldn't they.
jtbell
#8
Mar22-12, 11:56 PM
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Quote Quote by jose 892 View Post
If you make them go fast enough you can make the magnetic force be bigger than the electric force
Have you actually calculated the speed needed to make the two forces equal?

[added] I haven't done it for two point charges, but I've done it for two (very long) parallel line charges, which obviously look like line currents when in motion. Use the formulas for the electric field from a long straight line-charge, and the magnetic field from a long straight line-current.

[Aside to DaleSpam: You're probably thinking, "aha, apply the Lorentz transformation to the four-current, to get the charge density and current in the "moving" frame." Surprisingly, you get the same result if you simply treat the charge and current as being related non-relativistically between the two frames! I've done it both ways.]
Rap
#9
Mar23-12, 09:59 AM
P: 789
Quote Quote by jose 892 View Post
Could you be more specific in your answer? Do they experience magnetic attraction or not? We seem to agree that coulomb's law constant should stay the same in al inertial frames of reference. Now the situation I described in my last post where I put myself in the inertial frame of the charges it seems they experience a smaller force than expected, which cannot be if the constant remained the same. I guess the only answer would be that the do not experience magnetic attraction in the frame of reference where they appear to be moving, however I don't know why wouldn't they.
Coulomb's law only holds for charges at rest in your frame. The minute they start moving, you get relativistic effects. As long as their velocity is much less than the speed of light, the relativistic corrections can be ignored.

The relativistic thing that is going on here is time dilation. Anyone moving with respect to you seems to have their clocks running slow.

If you are looking at two charges coming together in a reference frame that is moving with respect to those charges, then compared to a person at rest with those charges, if all you use is Coulombs law, then you will disagree with each other. The person in the rest frame will say the charges meet after a certain time interval t, according to Coulombs law, and he will be right. You, in the moving frame, will say, no, his clocks are running slow, that time he measured is too long, its shorter than that, they came together faster than that. In other words, you will say that Coulomb's law is wrong, the force between the particles is larger than what Coulomb's law says. Then, if you include the magnetic force, you will see that the magnetic force accounts for that extra force, and everything comes out right. The person in the rest frame won't include a magnetic force, and everything comes out right for him too.
DaleSpam
#10
Mar24-12, 07:21 AM
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Quote Quote by jose 892 View Post
Could you be more specific in your answer? Do they experience magnetic attraction or not? We seem to agree that coulomb's law constant should stay the same in al inertial frames of reference. Now the situation I described in my last post where I put myself in the inertial frame of the charges it seems they experience a smaller force than expected, which cannot be if the constant remained the same. I guess the only answer would be that the do not experience magnetic attraction in the frame of reference where they appear to be moving, however I don't know why wouldn't they.
Quote Quote by jose 892 View Post
Could you be more specific in your answer? Do they experience magnetic attraction or not?
They do in the frame where they are moving. They do not in the frame where they are not moving.

Quote Quote by jose 892 View Post
We seem to agree that coulomb's law constant should stay the same in al inertial frames of reference.
Yes.

Quote Quote by jose 892 View Post
Now the situation I described in my last post where I put myself in the inertial frame of the charges it seems they experience a smaller force than expected, which cannot be if the constant remained the same.
No, this is not correct. They experience the force as expected by Coulomb's law in their frame.

Quote Quote by jose 892 View Post
I guess the only answer would be that the do not experience magnetic attraction in the frame of reference where they appear to be moving, however I don't know why wouldn't they.
No, Maxwell's equations hold in all frames, so in the frame where they are moving they do experience magnetic attraction. It sounds like I am contradicting myself, doesn't it?

We have two apparently contradictory facts:
1) Maxwell's equations are valid in all reference frames
2) Maxwell's equations predict a different force on the charges in different frames

The contradiction was not resolved until Einstein developed the theory of special relativity. In special relativity he realized that in order for these two facts to not contradict each other it had to follow that moving clocks tick slowly (time dilation) and moving rods are shorter (length contraction).

So suppose we have a pair of charges which are interacting as described above and they are initially at rest wrt each other but are then allowed to accelerate away from each other. Two observers will measure the acceleration of the charges and use f=ma to determine the force on the charges. One observer, Stan, is stationary wrt the charges. The other observer, Molly, is moving wrt the charges.

Stan and Molly both measure the acceleration and Stan gets a result consistent with Coulomb's law, and Molly gets a result consistent with Coulomb's law - the magnetic force as predicted by Maxwell's equations. Two observers measuring the acceleration of the same objects get different results. Stan looks at Molly's clock and ruler and notices that the clock ticks slower and the ruler is shorter than his. The amount by which her clocks and rulers are altered is just enough to account for her measurement of the magnetic force.

Hope this helps.
Redbelly98
#11
Mar25-12, 08:42 AM
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Quote Quote by jose 892 View Post
Could you be more specific in your answer? Do they experience magnetic attraction or not?
The short answer is that the two charges experience repulsion -- in all reference frames. Note my lack of an adjective like "magnetic" or "electrostatic".

The value of that repulsive force depends on the frame of reference.
harrylin
#12
Mar25-12, 09:06 AM
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Quote Quote by jtbell View Post
Have you actually calculated the speed needed to make the two forces equal?
[..] Surprisingly, you get the same result if you simply treat the charge and current as being related non-relativistically between the two frames! I've done it both ways.]
Can you please elaborate? I fully agree with "surprisingly", for I don't think that you can get the same result in classical physicals... Don't you get zero net force at a certain speed (should be around the speed of light) and the normal Coulomb attraction in the co-moving frame?
Rap
#13
Mar25-12, 03:46 PM
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Quote Quote by harrylin View Post
Can you please elaborate? I fully agree with "surprisingly", for I don't think that you can get the same result in classical physicals... Don't you get zero net force at a certain speed (should be around the speed of light) and the normal Coulomb attraction in the co-moving frame?
If you treat the Coulomb force as being the same at any speed, then you won't get the right answer. Coulomb's law only holds for charges at rest.
vanhees71
#14
Mar26-12, 04:44 AM
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Indeed the most simple calculation is to first solve the problem in the reference frame, where the charges are at rest. This gives you simply the Coulomb force between the two point charges (supposed for simplicity you have particles without a magnetic moment :-)). The you use the appropriate Lorentz transformation to evaluate the force in the frame where both charges are moving with constant velocity in parallel.

It is very important to calculate the right quantity, namely the covariant Minkowski force defined by

[tex]m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=\frac{q_1}{c} F_{\mu \nu}(x) \frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}.[/tex]

Here, [itex]x^{\mu}(\tau)[/itex] is the world line parametrized in terms of the proper time of charge 1, [itex]m[/itex] is its rest mass (a Lorentz scalar!), [itex]q_1[/itex] its charge (a scalar!), and [itex]F_{\mu \nu}[/itex] is the antisymmetric electromagnetic field-strength tensor of the field caused by the other charge.

Of course, you can also calculate this field and the force in any frame of reference you like, but it's much easier to do the calculation in the rest frame of charge 2 first and then boost to the frame, where it's moving. It's a good exercise to check that these both calculations give the same result.

Of course, this only works right when you do the calculation relativistically, i.e., with Lorentz boosts!
jtbell
#15
Mar26-12, 08:56 AM
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Quote Quote by harrylin View Post
Can you please elaborate?
Consider two long straight "line charges" with linear charge density λ C/m, separated by distance d. They repel each other with the following force per unit length:

$$f_E = \frac {\lambda^2}{2 \pi \epsilon_0 d}$$

Set those two line charges in motion along their length with speed v (or you move yourself with speed v).

Without taking relativity into account, the moving linear charge density becomes a current: I = λv. The (attractive) magnetic force per unit length between two long straight currents separated by distance d is:

$$f_B = \frac {\mu_0 I^2}{2 \pi d} = \frac {\mu_0 \lambda^2 v^2}{2 \pi d}$$

To find the speed at which the net force is zero, set the two forces equal to each other and solve for v. I'll let you do it.

Taking relativity into account, we work with the charge-current four-vector which has only two components in this situation (letting the lines lie along the x-axis): ##J^\mu = (c \rho, J, 0, 0)## where J is the current density in A/m2 and ρ is the volume charge density in C/m3. Between different inertial reference frames, J and ρ are related by a Lorentz transformation exactly the same way as x and t:

$$J^\prime = \gamma (J - v \rho)$$
$$\rho^\prime = \gamma \left( \rho - \frac{vJ}{c^2} \right)$$

In the frame where the line charges are stationary, J = I/A = 0 and ρ = λ/A, where A is the cross-section area of the charges. I'll leave it to you to apply the Lorentz transformation, along with I' = J'A and λ' = ρ'A, to verify that

$$I^\prime = - \gamma I = - \gamma v \lambda$$
$$\lambda^\prime = \gamma \lambda$$

(The "-" sign simply reflects the fact that if you start to move in the +x direction, the line charges move in the -x direction in your reference frame.)

Using these to find v which makes the net force zero, the γ's cancel and we have the same result as before.
pervect
#16
Mar26-12, 06:27 PM
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I'm surprised nobody has mentioned that relativistically speaking, the component of the electric field transverse to the direction of motion is boosted by a factor of beta.

So in the rest frame, a charge sees only an electric field , of magnitude E_0, repelling it.

In the moving frame, a charge sees a higher electric field repelling it, of magnitude [itex]\beta[/itex] E_0, plus a magnetic field attracting it.

The first thought one might have is that these two effects cancel out, but this isn't quite the case.

Rather than get into a lot of detail, I'd rather just say that various posters have given various pieces of the solution to the problem in this thread, but the issue is complex enough that, without tensors, it would be best to go to an E&M book like Griffith's to get all the details.

It's also a great opportunity to learn at least a little bit about tensors, because they greatly aid the process of ensuring that the laws of physics are framed in such a manner that you get equivalent answers in all different reference frames. Following this line of approach, one can simply explain that the E&M field transforms as a tensor (the Faraday tensor), and hence the force also transforms as a tensor (another, different tensor - the so-called four-force).

Most E&M books at this level will at least mention tensors, along with presenting the alternative non-tensor approach. And it's a good place to start to get familiar with tensors.

Three forces may be more familiar, but they aren't tensors, and so they make even a simple problem like the above a lot more complicated than it is once you know about tensors. Learning about tensors is an investment of time, but it's a wise investment for anyone who is serious about physics.
julian
#17
Mar29-12, 08:31 PM
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Feynman does this sort of calculation in The Feynman Lectures on Physics, Vol II, 13-6. There he points out that you have to take into account that transverse forces change under a Lorentz transformation. Feynman writes quite a lot on the fields of moving charges.
Bob S
#18
Mar29-12, 09:11 PM
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Quote Quote by Rap View Post
If you treat the Coulomb force as being the same at any speed, then you won't get the right answer. Coulomb's law only holds for charges at rest.
Coulomb's Law holds for both moving and stationary charges. See the Lorentz transformations of transverse E and B fields at (see bottom) http://pdg.lbl.gov/2011/reviews/rpp2...-relations.pdf

Mutual repulsion of parallel charged particles in high intensity beams is very important in high current particle accelerators. The particles in a tightly bunched beam repel one another and defocus the beam. The repulsive "space charge" Coulomb forces are always stronger than the attractive magnetic forces, and cancel only for highly relativisitic beams.
The mutual magnetic and electric forces were derived in post # 5 of
http://www.physicsforums.com/showthr...lectric+forces


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