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The two point charges are as below:

enter image description here

Then the Coulomb force between them is:

$$F_1=\frac{q^2}{4\pi\epsilon_0 r^2}$$

When the inertial reference frame moves left at $v$(or the point charges move right at $v$), then there will be magnetic field according to Biot–Savart law.

The Lorenz force between the two point charge in the vertical direction will be nonzero. The magnitude of the Lorentz force is:

$$f_L=\frac{\mu_0 q^2v^2}{4\pi r^2}$$

In wiki on Lorentz force, it says that:

The electric and magnetic fields are dependent on the velocity of an observer, so the relativistic form of the Lorentz force law can best be exhibited starting from a coordinate-independent expression for the electromagnetic and magnetic fields,[27] \mathcal{F}, and an arbitrary time-direction

Considering $F_1$ and $f_L$ have opposite directions and optical velocity $1/c^2=\mu_0\epsilon_0$, the magnitude of the force $F_2$ between the two charges becomes:

$$F_2=F_1-f_L= \left(1-\frac{v^2}{c^2}\right)F_1$$

Am I right?

enter image description here

Per the calculation here, it is easy to obtain the time dilation factor $\displaystyle\sqrt{1-\frac{v^2}{c^2}}$ of an event that the two point charges separate to a distance of $r$.

The nonlinear ODE:

$$\dfrac{d^2r(t)}{dt^2}=\dfrac{\left(1-\dfrac{v^2}{c^2}\right)q^2}{2\pi m\epsilon_0\left(R+r(t)\right)^2}$$

with initial/boundary conditions $$r(t)=0,\quad r'(t)=0$$

Solution:

$$ \because\qquad\qquad\qquad\qquad\qquad\qquad\qquad $$ $$\int {\rm{Left }}\cdot dr=\int\frac{d^2r}{dt^2}\cdot {dr}=\int\frac{d\left(\dfrac{dr}{dt}\right)\cdot dr}{dt}=\dfrac{1}{2}\int d\left(\left(\dfrac{dr}{dt}\right)^2\right)=\dfrac{1}{2}\left(\dfrac{dr}{dt}\right)^2$$

$$\int{\rm Right}\cdot dr=\int\dfrac{\left(1-\dfrac{v^2}{c^2}\right)q^2 \cdot {dr}}{2\pi m\epsilon_0\left(R+r\right)^2}=\dfrac{\left(1-\dfrac{v^2}{c^2}\right)q^2 {r}}{2\pi m\epsilon_0\left(R+r\right)R}$$

$$\therefore \dfrac{dr}{dt}=\sqrt{\dfrac{\left(1-\dfrac{v^2}{c^2}\right)q^2 {r}}{\pi m\epsilon_0\left(R+r\right)R}}$$

$$\therefore{dt}=\sqrt{\dfrac{\pi m\epsilon_0\left(R+r\right)R}{\left(1-\dfrac{v^2}{c^2}\right)q^2 {r}}}dr$$

$$\therefore\int{dt}=\sqrt{\dfrac{\pi m\epsilon_0}{\left(1-\dfrac{v^2}{c^2}\right)q^2 }}\int\sqrt{\frac{\left(R+r\right)R}{r}}dr$$

$$\Delta t(r)=\frac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}\sqrt{\dfrac{\pi m\epsilon_0}{q^2 }}\left(\sqrt{r R (r+R)}-\frac{1}{2} R^{3/2} \left(\log R-2 \log \left(\sqrt{r+R}+\sqrt{r}\right)\right)\right)$$

which indicates:

The time interval that the two point charges separate a further distance $r$ in the moving reference frame is $\displaystyle\frac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}$ that in the static reference frame where $v=0$.

Does this means Lorentz transformation for time is natural for an event involving both electric and magnetic laws without the need of considering special relativity?

Can a full version of Lorentz transformation be derived in such an approach?

P.S.

I found this question with no accepted answer is similar, but I made some further calculation and extension here.

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    $\begingroup$ LCFactorization: If $r$ (the separation between $A$ and $B$) is changing while $v$ (the speed at which both $A$ and $B$ are moving wrt. a suitable reference system) remains constant then that's ... difficult for me to comprehend. On the larger, implied question whether geometric relations can be derived from "laws" of (electro-)dynamics: Formally, perhaps, yes; AFAIU. However, approaching the problem experimentally, we always need to determine geometric relations first, and separately, in order to find out in the first place e.g. whether $A$'s and $B$'s charges were equal (as you presumed). $\endgroup$ – user12262 Oct 18 '15 at 8:34
  • $\begingroup$ The Coulomb force is not given by $F_1$ in the moving frame. The electric field is different. $\endgroup$ – Rob Jeffries Jan 21 '17 at 11:16
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The answer is no, the Coulomb force between two point charges does not change, unless you change the distance between them. If the charges move in parallel, then the Coulomb equation no longer applies. You must use the Lorentz equation, instead. In other words, if the reference frame is moving and the charges are stationary, then the Coulomb equation applies. If the reference frame is stationary and the charges are moving, then the Lorentz equation applies. You can not apply both equations at the same time!

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  • $\begingroup$ The question was whether the force changed, not whether the Coulomb force changed. The net force is given by the Lorentz force always and it is not frame invariant. $\endgroup$ – Rob Jeffries Jan 21 '17 at 13:06
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Does this means Lorentz transformation for time is natural for an event involving both electric and magnetic laws without the need of considering special relativity?

It means that Lorentz transformations follow from electric and magnetic laws. It is how special relativity was discovered in the first place.

Your derivation looks correct. Maxwell's equations are not invariant under Newtonian transformations, only under Lorentz transformations.

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I would answer "yes". The Couloumb force $qE$ changes, and here's why: In frame at rest B=0, while in the moving frame (') B' is not zero. Since $E^2-B^2$ is Lorentz invariant: E' must be larger than E. (Which is the familiar relativistic squishing of fields--like in a transition-radiation-detector).

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