
#1
Aug1606, 06:44 PM

P: 2,268

Magnetic fields arise because of moving charges which is the existence of a current (given by Ampere's law).
Alternatively, a magnetic dipole is set up as a result of a charge moving around in a circular loop. So it follows that magnetism occurs from moving charges moving in different ways. But people say that magnetism arise because of special relativity. How does SR come into it since it dosen't matter how fast the charges are moving? 



#2
Aug1606, 07:49 PM

P: 2,265





#3
Aug1706, 02:10 AM

P: 2,268

your premise in this last question is wrong. where did you get
that?[/QUOTE] I just assumed this. I know light must travel at speed of light but charges which are electrons can't reach that speed. Plus most of the time they don't travel at those speeds unless when they are in a particle accelerator. I certainly wouldn't expect electrons in my computers to be at speed of light or would I? Mass movement of electrons at 0.001speed of light would also constitute a current hence generate a magnetic field somewhere wouldn't it? 



#4
Aug1706, 05:42 AM

Sci Advisor
HW Helper
P: 11,863

How does magnetism occur?
Special relativity comes in connection to the em tensor, not with the relevance of the speed of charged particles...
Daniel. 



#5
Aug1706, 06:38 AM

P: 2,050

Think of any situation, say a testcharge moving beside a currentcarrying wire, and use special relativity to consider the situation from the testcharge's frame of reference. In this frame (where the testcharge is stationary for the moment) the electroncurrent in the wire is moving at a different velocity than the wire itself (which constitutes a current of positive charged atoms). Special relativity says that the Lorentzcontraction of the spacing between the current electrons is therefore different to the contracted spacing between positive atoms in the wire. Since the densities of positive and negative charges are hence different in the testcharge's frame of reference, there is an electrostatic force on the testcharge. When you look at such situations from a different frame of reference (eg. from the point of view where both the electroncurrentvelocity is equal and opposite to the velocity of the atoms composing the wire, so that positive and negative charge densities are in balance), the force you just calculated for the testcharge can no longer be explained electrostatically: In this frame of reference, it is called the "magnetic" force (although really it just arose from SR). 



#6
Aug1706, 07:16 AM

Admin
P: 21,628

Electrons do not move at the speed of light, but the electric field generated by the charge does in a sense. Maxwell's equations can be derived from SR, to which dextercioby alluded. 



#7
Aug1806, 01:14 AM

P: 2,265

The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of [itex] \lambda [/itex] ) and some nonzero mass per unit length of [itex] \rho [/itex] separated by some distance [itex] R [/itex]. If the lineal mass density is small enough that gravitational forces can be neglected in comparison to the electrostatic forces, the static nonrelativistic repulsive (outward) acceleration (at the instance of time that the lines of charge are separated by distance [itex] R [/itex]) for each infinite parallel line of charge would be: [tex] a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} [/tex] If the lines of charge are moving together past the observer at some velocity, [itex] v [/itex], the nonrelativistic electrostatic force would appear to be unchanged and this would be the acceleration that an observer traveling along with the lines of charge would observe. Now, if special relativity is considered, the clock of the observer moving along with the lines of charge would be ticking at a relative rate (ticks per unit time or 1/time) of [itex] \sqrt{1  v^2/c^2} [/itex] from the pointofview of the stationary observer because of time dilation. Since acceleration is proportional to (1/time)^{2}, the atrest observer would observe an acceleration scaled by the square of that rate, or by [itex] {1  v^2/c^2} [/itex], compared to what the moving observer sees. Then the observed outward acceleration of the two infinite lines as viewed by the stationary observer would be: [tex] a = \left(1  v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} [/tex] or [tex] a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R}  \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e  F_m }{\rho} [/tex] The first term in the numerator, [itex] F_e [/itex], is the electrostatic force (per unit length) outward and is reduced by the second term, [itex] F_m [/itex], which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors). The electric current, [itex] i_0 [/itex], in each conductor is [tex] i_0 = v \lambda [/tex] and [tex] \frac{1}{\epsilon_0 c^2} [/tex] is the magnetic permeability [tex] \mu_0 = \frac{1}{\epsilon_0 c^2} [/tex] because [tex] c^2 = \frac{1}{ \mu_0 \epsilon_0 } [/tex] so you get for the 2^{nd} force term: [tex] F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R} [/tex] which is precisely what the classical E&M textbooks say is the magnetic force (per unit length) between two parallel conductors, separated by [itex] R [/itex], with identical current [itex] i_0 [/itex]. so you can look at it two ways: one is purely classical where the magnetic force exists and has a separate origin from the electrostatic force and effects of relativity are not considered. the other is where there is only the electrostatic force but the effects if special relativity are considered. both results appear the same to the "stationary" observer. 



#8
Aug1806, 01:14 AM

P: 2,265

no matter what i do, i can't get LaTeX equations to render. what is wrong?




#9
Aug1806, 01:15 AM

Mentor
P: 11,230

I posted some very simple LaTeX a little while ago, which failed with the same mesages. I'm 99.9% sure it's just a temporary glitch in the forum software. I won't be surprised if our stuff actually appears correctly sometime tomorrow (oops, later today).




#10
Aug2006, 11:13 AM

P: 2,265

site down for 2+ days and this problem. i hope they figure it out. 



#11
Aug2006, 08:23 PM

P: 2,265

The simplest hypothetical experiment would be two identical parallel infinite lines of charge (with charge per unit length of [itex] \lambda [/itex] ) and some nonzero mass per unit length of [itex] \rho [/itex] separated by some distance [itex] R [/itex]. If the lineal mass density is small enough that gravitational forces can be neglected in comparison to the electrostatic forces, the static nonrelativistic repulsive (outward) acceleration (at the instance of time that the lines of charge are separated by distance [itex] R [/itex]) for each infinite parallel line of charge would be: [tex] a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} [/tex] If the lines of charge are moving together past the observer at some velocity, [itex] v [/itex], the nonrelativistic electrostatic force would appear to be unchanged and this would be the acceleration that an observer traveling along with the lines of charge would observe. Now, if special relativity is considered, the clock of the observer moving along with the lines of charge would be ticking at a relative rate (ticks per unit time or 1/time) of [itex] \sqrt{1  v^2/c^2} [/itex] from the pointofview of the stationary observer because of time dilation. Since acceleration is proportional to (1/time)^{2}, the atrest observer would observe an acceleration scaled by the square of that rate, or by [itex] {1  v^2/c^2} [/itex], compared to what the moving observer sees. Then the observed outward acceleration of the two infinite lines as viewed by the stationary observer would be: [tex] a = \left(1  v^2 / c^2 \right) \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} [/tex] or [tex] a = \frac{F}{m} = \frac{ \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R}  \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} }{\rho} = \frac{ F_e  F_m }{\rho} [/tex] The first term in the numerator, [itex] F_e [/itex], is the electrostatic force (per unit length) outward and is reduced by the second term, [itex] F_m [/itex], which with a little manipulation, can be shown to be the classical magnetic force between two lines of charge (or conductors). The electric current, [itex] i_0 [/itex], in each conductor is [tex] i_0 = v \lambda [/tex] and [tex] \frac{1}{\epsilon_0 c^2} [/tex] is the magnetic permeability [tex] \mu_0 = \frac{1}{\epsilon_0 c^2} [/tex] because [tex] c^2 = \frac{1}{ \mu_0 \epsilon_0 } [/tex] so you get for the 2^{nd} force term: [tex] F_m = \frac{v^2}{c^2} \frac{1}{4 \pi \epsilon_0} \frac{2 \lambda^2}{R} = \frac{\mu_0}{4 \pi} \frac{2 i_0^2}{R} [/tex] which is precisely what the classical E&M textbooks say is the magnetic force (per unit length) between two parallel conductors, separated by [itex] R [/itex], with identical current [itex] i_0 [/itex]. so you can look at it two ways: one is purely classical where the magnetic force exists and has a separate origin from the electrostatic force and effects of relativity are not considered. the other is where there is only the electrostatic force but the effects of special relativity are considered. both results appear the same to the "stationary" observer. 



#12
Aug2106, 12:33 AM

P: 2,268

I haven't read the above derivation in detail. But I think I have a better understanding of why magnetism just is electricity or vice versa, depending on one's reference frame.
But I have a qualitative question. With the example of two infinite paralle wires carrying the same (positive charge) current in the same direction, there is an inward attraction between the two wires. Now pretend we sit on a charge and move with the current. There must still be an inward force attracting me to the opposite wire but now I must use my electrostatic equations to calculate this force since I see the positive charge opposite me as stationary and no magnetic field exists according to me. The problem is in order to be attracted to this stationary charge, that charge must turn into a negative charge or the charge I am sitting on has turned negative. But how can this happen! Why is it that when I change my frame of reference, the charge on one wire must change? That seems absurd. What is wrong? 



#13
Aug2106, 11:40 AM

P: 185





#14
Aug2106, 12:20 PM

P: 2,265

now, for the stationary observer, if you do not consider the effect of SR, then what explains the reduced rate of repulsive acceleration between the two lines of charge? in classical physics, we would say that there are two forces acting on the lines of charge. one is the electrostatic repulsion which would be the same whether the lines are moving or not and the other would be an electromagnetic attraction that is quantitatively less than the repulsion until the speed of the lines whizzing past the stationary observer reaches the speed of light. in that case, the magnetic attraction is just equal to the electrostatic repulsion according to classical physics. according to SR, the clock of the moving observer (you), has come to a stop from the POV of the stationary observer. of course, those lines can't move that fast from anyone's POV. 



#15
Aug2206, 02:28 AM

P: 2,268

I have seen experiments where when a current is turned on in two conductors, they move toward each other (i.e. aluminium foil). This would be the case where the two conductors have the same charges and so instead of repelling less, they attract? 



#16
Aug2206, 10:12 AM

P: 2,265

in the SR POV, there is no magnetic force, there is only electrostatic, but because of time dilation, the lines appear to the "stationary" observer to be moving apart at a reduced acceleration compared to if they weren't moving past the observer. but if the electrons are made to move for some reason, then there are magnetic forces remaining that aren't cancelled. now how to look at this from the SR POV (leaving out the magnetic force but keeping the electrostatic force), i am not sure how this would be done because the "moving" observer (moving along with the electrons) would see the electrons as stationary but see the protons moving in the opposite direction. somehow both observers would have to see the lines or wires moving toward each other at some rate and not repelled. but i am not sure how to set it up. that's why i left the thought experiment as two lines of charge and not neutral wires with both positive and negative charges. it's easier to get a handle on that and it does offer some explaination of the phenomenon but there is both electrostatic and electomagnetic forces from the classical POV (and electrostatic from the SR POV). 



#17
Aug2306, 02:15 AM

P: 2,268





#18
Aug2306, 02:26 AM

P: 2,265




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