Magnetic Forces from Special Relativity

It is possible to transform the speeds of both the fixed positive charges in the wire as well as the moving negative charges in the wire to a frame where the test particle is stationary. Then the magnetic forces on the test particle disappear and we find that we are left with a net electric charge on the wire segments that are parallel to the direction of the test particle motion and the direction of the current on that wire segment.

*Figure 1: Current loop and test charge as viewed in
laboratory frame. The velocity of the
test charge is v _{t} so its beta is v_{t}/c. The velocity of the negative charge current
is v_{I}*.

Looking at Figure 1 we see that prior to transformation:

$$\begin{array}{l}{v}_{t}=-\left|{v}_{t}\right|\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}top}=-\left|{v}_{I}\right|\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}bottom}=\left|{v}_{I}\right|\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}left}=-\left|{v}_{I}\right|\widehat{y}\\ {v}_{I\text{\hspace{0.17em}}right}=\left|{v}_{I}\right|\widehat{y}\end{array}$$ |
(1.1) |

After the transformation to set the test particle speed, v_{t},
to zero we have:

$$\begin{array}{l}{v}_{t}=\left|{v}_{t}\right|\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}top}=\frac{\left(\left|{v}_{t}\right|-\left|{v}_{I}\right|\right)}{1-\frac{\left|{v}_{I}{v}_{t}\right|}{{c}^{2}}}\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}bottom}=\frac{\left(\left|{v}_{t}\right|+\left|{v}_{I}\right|\right)}{1+\frac{\left|{v}_{I}{v}_{t}\right|}{{c}^{2}}}\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}left}=-\left|{v}_{I}\right|\widehat{y}+\left|{v}_{t}\right|\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}left}=\left|{v}_{I}\right|\widehat{y}+\left|{v}_{t}\right|\widehat{x}\end{array}$$ |
(1.2) |

Let us define some value of
the various Lorentz factors, γ.

$$\begin{array}{l}{\gamma}_{t}=\frac{1}{\sqrt{1-\frac{{v}_{t}^{2}}{{c}^{2}}}}\equiv \frac{1}{\sqrt{1-{\beta}_{t}^{2}}}\\ {\gamma}_{I}=\frac{1}{\sqrt{1-\frac{{v}_{I}^{2}}{{c}^{2}}}}\equiv \frac{1}{\sqrt{1-{\beta}_{I}^{2}}}\\ {\gamma}_{I\text{\hspace{0.17em}}left}=\frac{1}{\sqrt{1-\frac{{v}_{I\text{\hspace{0.17em}}left}^{2}}{{c}^{2}}}}\equiv \frac{1}{\sqrt{1-{\beta}_{I\text{\hspace{0.17em}}left}^{2}}}\\ {\gamma}_{I\text{\hspace{0.17em}}right}=\frac{1}{\sqrt{1-\frac{{v}_{I\text{\hspace{0.17em}}right}^{2}}{{c}^{2}}}}\equiv \frac{1}{\sqrt{1-{\beta}_{I\text{\hspace{0.17em}}right}^{2}}}\end{array}$$ |
(1.3) |

$$\begin{array}{l}{\beta}_{I\text{\hspace{0.17em}}top}=\frac{{\beta}_{t}-{\beta}_{I}}{1+{\beta}_{I}{\beta}_{t}}\\ {\beta}_{I\text{\hspace{0.17em}}bottom}=\frac{{\beta}_{t}-{\beta}_{I}}{1+{\beta}_{I}{\beta}_{t}}\end{array}$$ |
(1.4) |

To compute the charge density for these betas we must obtain their Lorentz boost $\gamma $.

$$\begin{array}{l}\frac{1}{{\left({\gamma}_{I\text{\hspace{0.17em}}top}^{}\text{'}\right)}^{2}}=1-{\left(\frac{{\beta}_{t}-{\beta}_{I}}{1-{\beta}_{I}{\beta}_{t}}\right)}^{2}\\ \frac{1}{{\left({\gamma}_{I\text{\hspace{0.17em}}bottom}\text{'}\right)}^{2}}=1-{\left(\frac{{\beta}_{t}+{\beta}_{I}}{1+{\beta}_{I}{\beta}_{t}}\right)}^{2}\end{array}$$ |
(1.5) |

After a bit of algebra we obtain the results:

$$\begin{array}{l}\frac{1}{{\left({\gamma}_{I\text{\hspace{0.17em}}top}^{\text{'}}\right)}^{2}}=\frac{(1-{\beta}_{I}^{2})(1-{\beta}_{t}^{2})}{{\left(1-{\beta}_{I}{\beta}_{t}\right)}^{2}}\\ \frac{1}{{\left({\gamma}_{I\text{\hspace{0.17em}}bottom}^{\text{'}}\right)}^{2}}=\frac{(1-{\beta}_{I}^{2})(1-{\beta}_{t}^{2})}{{\left(1+{\beta}_{I}{\beta}_{t}\right)}^{2}}\end{array}$$ |

Using definitions of the gammas of t and I equation (1.6) becomes:

$$\begin{array}{l}\frac{1}{{\left({\gamma}_{I\text{\hspace{0.17em}}top}^{}\text{'}\right)}^{2}}=\frac{1}{{\gamma}_{I}^{2}{\gamma}_{t}^{2}{\left(1-{\beta}_{I}{\beta}_{t}\right)}^{2}}\\ \frac{1}{{\left({\gamma}_{I\text{\hspace{0.17em}}bottom}^{}\text{'}\right)}^{2}}=\frac{1}{{\gamma}_{I}^{2}{\gamma}_{t}^{2}{\left(1+{\beta}_{I}{\beta}_{t}\right)}^{2}}\end{array}$$ |
(1.7) |

and then the final expression for the gamma of the charge density becomes:

$$\begin{array}{l}{\gamma}_{I\text{\hspace{0.17em}}top}\text{'}={\gamma}_{I}{\gamma}_{t}(1-{\beta}_{I}{\beta}_{t})\\ {\gamma}_{I\text{\hspace{0.17em}}bottom}\text{'}={\gamma}_{I}{\gamma}_{t}(1+{\beta}_{I}{\beta}_{t})\end{array}$$ |

The positive charge spacing is contracted by the factor $\frac{1}{{\gamma}_{t}}$ and the negative charge spacing is contracted by the factor $\frac{1}{{\gamma}_{I}\text{'}}$. Prior to transformation let the positive charge per unit length be $+{\lambda}_{0}^{}$ and the negative charge per unit length be $-{\lambda}_{0}^{}$ .

You should recall that the negative charge was moving **prior** to transformation, and therefore
charge spacings were contracted by $\frac{1}{{\gamma}_{I}}$ prior to the transformation so that the factor
$\frac{1}{{\gamma}_{I}}$ must be multiplied by ${\lambda}_{-}$ to get the effective negative charge density.

$$\begin{array}{l}{\gamma}_{I\text{\hspace{0.17em}}top}\text{'}=\frac{{\gamma}_{I}{\gamma}_{t}(1-{\beta}_{I}{\beta}_{t})}{{\gamma}_{I}}\\ {\gamma}_{I\text{\hspace{0.17em}}bottom}\text{'}=\frac{{\gamma}_{I}{\gamma}_{t}(1+{\beta}_{I}{\beta}_{t})}{{\gamma}_{I}}\end{array}$$ |
(1.9) |

Then the new **net** charges per unit length
will be

$$\begin{array}{l}{\lambda}_{top}={\gamma}_{t}{\lambda}_{+}-{\gamma}_{I\text{\hspace{0.17em}}top}\text{'}{\lambda}_{-}={\gamma}_{t}{\lambda}_{0}\left(1-(1-{\beta}_{I}{\beta}_{t})\right)\\ ={\gamma}_{t}{\beta}_{I}{\beta}_{t}{\lambda}_{0}\\ {\lambda}_{bottom}={\gamma}_{t}{\lambda}_{+}-{\gamma}_{I\text{\hspace{0.17em}}bottom}\text{'}{\lambda}_{-}={\gamma}_{t}{\lambda}_{0}\left(1-(1+{\beta}_{I}{\beta}_{t})\right)\\ =-{\gamma}_{t}{\beta}_{I}{\beta}_{t}{\lambda}_{0}\end{array}$$ |

Please note in equation (1.10) that the sign of ${\beta}_{I}$ in the top side of the loop was opposite from that in the bottom side of the loop. Looking at equation (1.8), that fact also results in quite different charge densities in the top and bottom sides of the loop.

When the law for transformation of the force from the charge's moving frame to the laboratory frame

$${F}_{test\text{\hspace{0.17em}}ch\mathrm{arg}estationary}={\gamma}_{t}{F}_{moving}$$ |
(1.11) |

is taken into account, the ${\gamma}_{t\text{\hspace{0.17em}}}$ factor in equation (1.10) is factored out and the new electric force is the same as the magnetic force before transformation.

Assume that we have linear charge distributions in all 4 legs of a current carrying loop and that we've transformed the charge velocities to make the test charge near the top leg stationary.

To be definite, let's assume that the total number of
negative charges in the loop is N_{c} and, since the loop wire has no net
charge, the number of positive charges is also N_{c}. Since total charge number is a scalar, the
transformation discussed above cannot change this. The height of the loop remains the same
during the test particle transformation but its width is contracted by the
factor 1/γ_{t}. We will first compute the number of positive
charges in the loop sides and the top and bottom of the loop. We have the equation:

$$\begin{array}{l}{\lambda}_{0}^{+}=\frac{{N}_{c}}{2(h+w)}\\ {\lambda}_{sides}={\lambda}_{0}^{+}\\ {\lambda}_{T+B}={\lambda}_{0}^{+}{\gamma}_{t}\\ \delta {s}_{sides}=\frac{1}{{\lambda}_{sides}}=\frac{2(h+w)}{{N}_{c}}\\ \delta {s}_{T+B}=\frac{1}{{\lambda}_{T+B}}=\frac{2(h+w)}{{N}_{c}{\gamma}_{t}}\end{array}$$ |

It is obvious that the rate or charge flow has to be constant around the loop including at the corners where there are discontinuities in the charge per unit length. That would dictate that the following equation be valid

$${v}_{n+1}=\frac{{\lambda}_{n+1}}{{\lambda}_{n}}{v}_{n}$$ |

where n corresponds to the loop side number counted in the direction of charge flow.

Recall from Figure 1 that both v_{t} and v_{I}
in the top of the loop are both shown going to the left which is the -x
direction. Therefore, before
transformation, we write:

$$\begin{array}{l}{v}_{t}=-\left|{v}_{t}\right|\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}top}=-\left|{v}_{I}\right|\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}bottom}=\left|{v}_{I}\right|\widehat{x}\end{array}$$ |
(1.14) |

As observed in the frame of the stationary test particle these speeds become:

$$\begin{array}{l}{v}_{t}=\left|{v}_{t}\right|\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}top}=\frac{\left(\left|{v}_{t}\right|-\left|{v}_{I}\right|\right)}{1-\frac{\left|{v}_{I}{v}_{t}\right|}{{c}^{2}}}\widehat{x}\\ {v}_{I\text{\hspace{0.17em}}bottom}=\frac{\left(\left|{v}_{t}\right|+\left|{v}_{I}\right|\right)}{1+\frac{\left|{v}_{I}{v}_{t}\right|}{{c}^{2}}}\widehat{x}\end{array}$$ |

The transformed speed of all positive charges is **v _{t}** as given in equation (1.15)
. Relative to the stationary test
particle, the speed of the negative charges in the left side of the loop is $-\left|{v}_{I}\right|\widehat{y}+\left|{v}_{t}\right|\widehat{x}$ and that in the right side of the loop is $\left|{v}_{I}\right|\widehat{y}+\left|{v}_{t}\right|\widehat{x}$. The transformed speeds of the negative
charges in the top side and bottom side of the loop are given by equation (1.15).