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Let a,b be two charged particles. $$\vec{r}_a(0)=\vec{0}$$ $$\vec{r}_b(0)=r\hat{j}$$ $$\vec{v}_a(t)=v_a \hat{i}$$ $$\vec{v}_b(t)=v_b\hat{j}$$

In which both $v_a$ and $v_b$ $<<c$.

Then

$$\vec{E}_{ab}(0)=\frac{q_a}{4\pi \epsilon r^2}\hat{j}$$

$$\vec{B}_{ab}(0)=\frac{\mu q_av_a}{4\pi r^2} \hat{k}$$

$$\vec{E}_{ba}(0)=-\frac{q_b}{4\pi \epsilon r^2}\hat{j}$$

$$\vec{B}_{ba}(0)=\vec{0}$$

Note that $v_a$ and $v_b$ $<<c$ thus a and b almost obey Coulomb's law. Moreover, $\vec{j_i}(\vec{r})=q_i\delta(\vec{r}-\vec{r}_i)\vec{v}_i$ hence BS law can be applied.

Hence

$$\vec{F}_{ab}(0)=q_b(\vec{E}_{ab}+\vec{v}_b \times \vec{B}_{ab})$$ $$=\frac{q_a q_b}{4\pi \epsilon r^2}\hat{j}-\frac{\mu q_av_a v_b}{4\pi r_b^2} \hat{i}$$

But

$$\vec{F}_{ba}(0)=-\frac{q_aq_b}{4\pi \epsilon r^2}\hat{j}$$

Consequently $$\vec{F}_{ab} \ne -\vec{F}_{ba}$$

This result contradict to Newton's 3rd law!! But I cannot find any error... It troubled me.

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your first equation about r_a should say r_a=r, because you state it for all t ( time) as it is a cannot have any velocity. The difference vector between r_a ( which you wrongly call r) and r_b should be in the denominator for the force between the two. –  anna v Jan 17 '13 at 6:27
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to be clear, r_a can be 0 only at t=0 or another specific time if it is moving. And correct the above, r_a= r_i ( not r) to be correct. then r should be the vector difference between r_i and r_j. I think you are ignoring that in moving systems only an initial value at a specific time, t=0 can be given for space location. –  anna v Jan 17 '13 at 7:01
    
As @MarkEichenlaub points out, there are some issues with your analysis, but +1---a nice question and astute attempt none-the-less! –  zhermes Jan 17 '13 at 7:13
    
@annav You are right. I'm going to correct it. –  Popopo Jan 17 '13 at 13:49
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2 Answers 2

up vote 18 down vote accepted

The details of your analysis are not quite right - that's not what the electric field of a moving charge looks like, for example. This is probably because you haven't learned all the rules of electromagnetism yet. Still, the spirit of your question is hitting at an important point.

Charges do not conserve momentum and don't obey Newton's third law. You have to include the momentum of the electromagnetic field to see conservation laws hold.

There's an accessible discussion in section 8.2 of Griffiths "Introduction to Electrodynamics" if you would like a little more math.

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Oh right, moving charges do not obey Coulomb&Biot-Savart's law... So I have added a condition that they are both moving in low velocity. Besides, many thanks to your book. –  Popopo Jan 17 '13 at 15:00
    
Okay, I have read section 8.2. It seems the interaction in classical field theory is localized. Hence the forces charges feel are given by EM-field rather than other charges. So the result is not $\vec{F}_{ab} \ne -\vec{F}_{ba}$ but $\vec{F}_{Fa}\ne -\vec{F}_{Fb}$ thus does not violate the 3rd law. –  Popopo Jan 18 '13 at 3:15
    
Well, we need to include the EM-field to avoid this problem, but it seems EM-field obeys more Relativistic mechanics rather than Newton's. So does the 3rd law also holds in Relativistic mechanics? Besides, it seems quantities such as 'force' and 'acceleration' are not so clear and important in Relativistic mechanics... –  Popopo Mar 19 '13 at 2:34
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The main thing you have to start out with is to define the current of a single electron using a Delta function:

$j(\vec r')=-e\,\delta(\vec r'- \vec{r}(t)) \dot{\vec{r}}(t)$,

where $r'(t)$ is the position of the particle. Then everything else (Maxwell equations, Biot-Savart law), should work.

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Thank you very much. –  Popopo Jan 19 '13 at 13:08
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