Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

A thought experiment:

Suppose we are moving in a spacecraft (A), carrying an excess of electrons on spaceship surface, but we can't go outside to measure it, and it's moving at constant speed $v$, with respect to other distant ship (B).

When (A) pass in front of the ship (B), suddenly the two metal ships start to approach at increasing speeds, it's because of the force from magnetic field generated by the relative speed of moving charges .

We have to curb and redirect the ship, and of course solve the original problem, but we don't know about the excess of electrons on our surface, a scientist on board make hypothesis some minutes before the impact and thinks that the problem is a lack of electrons! (and uses some command to increment them, worsening the crash)

First question:

How a moving (and electrically charged) observer at constant speed could explain metals coming to him?

Meanwhile on the (B) ship, technicians thinks their ship surface is not charged at all, and the force is because their metal structure itself, they thinks (A) ship is the charged one, and they don't know what to do because they are neutral, and action should be taken at the (A) ship to neutralize it because attraction would occurs regardless of the sign of electric charge.

Second issue:

As it has not sense to ask which ship is moving at speed v, because it's relative, does it have sense to ask about charge sign? Is there some way to tell which sign is the own charge? It's possible to know inside a system the sign of the outside visible charge?

Help those ships captains... before they crash!. =)

share|improve this question
    
Ok, that's right, but shipbuilding is not the target of the question, you could think it's plastic, nanomaterial or whatever, they are moving charged bodies and their internal reference system is the topic, how from a charged body at constant speed can metals comming to him be explained, how from a charged body can be known the absolute sign of its own external charge, or others etc.. Those are the central themes, sorry if some details about the thought experiment were misleading, I could reword if were necesary, thanks –  HDE Oct 4 '11 at 13:22
    
I suspect that the answer you are seeking is in this short document (by Hans de Vries) The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics –  Helder Velez Oct 6 '11 at 20:27
add comment

1 Answer

up vote 0 down vote accepted
+50

You are working under the impression that the magnetic field from a moving charge is just like the magnetic field from a bar-magnet, so that it pulls pieces o metal in. Then you are getting confused, because in the rest-frame of the moving charge, there is no magnetic field. So how can the motion of the metal be explained?

A moving charge's magentic field, does not go into and out of the charge like a bar-magnet's. It must go around the line of motion in circles, like the field from a current carrying wire, and this is required by symmetry. The magnitude of the field falls off as $1/r^2$ where r is the distance from the moving point.

But in addition, there is a $1/r^2$ electrostatic field, which has dominant effects. The electric and magnetic effects only become comparable as the spaceship gets close to the speed of light.

Whatever effect you ascribe to the magnetic field in the moving frame, you ascribe to the electric field when the ship is stationary. So the metal pulled into you by a combination of magnetic/electric field gradients is now pulled in by just an electric field gradient. The combined electric-magnetic field combination is called the F tensor, and the transformations required for these quantities to avoid any paradoxes of relative motion were worked out by Einstein in 1905, in the special relativity paper.

As for determining the sign of the charge, you can't do it just be looking at the motion of macroscopic objects, because you don't know their charge. But if you see some stray particles wandering by your ship, they will deflect outward or inward according to the charge of your ship, and the light particles are negatively charged electrons, the heavy ones are positively charged nuclei.

share|improve this answer
    
""magnetic field from a bar-magnet, so that it pulls pieces o metal in"" Including metals like Bismuth? –  Georg Oct 6 '11 at 11:14
    
@Georg: I assume Bismuth is not magnetic. Who cares? This question is about E and B transforming into each other under motion, so whatever Bismuth feels in E, it feels in E and B moving transformed. –  Ron Maimon Oct 6 '11 at 15:54
    
@RonMaimon It's a good answer, a central point I would like to bold (and that I think you already noticed) is the difference between a single moving charged reference (ship) and a current, like in a wire, the problem to my mind is that if you think about many many ships one behind the other all as diferent reference frames, then ships would work as a current in a wire, will attract metals, and everyone will detect those getting closer to the "ships queue".But they would explain this "because of others ships", not self. -The "need" of many- is what drives me crazy.How much are needed?..*why?*.. –  HDE Oct 6 '11 at 16:38
    
@HDE: the ship will attract metals, but it's electric and magnetic field will both attract them to the same degree, in that they together make an invariant attractive force. –  Ron Maimon Oct 6 '11 at 16:44
2  
@Georg I think that Bismuth is not related with the question (If it's related please clarify), I think that's the meaning of "who cares?" Bismuth is diamagnetic ok, I see no conection with the question, "metal" is mention to refer ferromagnetic, and even that's not the point, but the interaction is, and the curious thing (at least for me) that in the frame of reference where the field is generated, it can't be detected. –  HDE Oct 6 '11 at 19:58
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.