# Change in intensity of electric field with constant velocity

Consider a +Q charged particle is travelling towards another test charge +Q. Now what would be the difference in electric field experienced by the test charge(avoid the gradual decrease in distance between them)? Would the field lines look compressed and effective field strength increased for the test charge?

If you are looking for an effect separate from the particle's position, at classical velocities there isn't one. The electric field is $$\mathbf{E}=\mathbf{E}(\mathbf{r},t),$$ that is, the electric field is only a function of the position $\mathbf{r}$ and the time $t$. At any given instant in time, the force a test charge 'feels' due to another charge depends only on its position $\mathbf{r}$, and not on its velocity.

This velocity independence breaks down when the charges' relative velocities approach the speed of light. If a reference frame has an electric field, a frame boosted with the respect to the reference appears to have some magnetic field. For a frame boosted by a velocity $\mathbf{v}=v_x \mathbf{\hat{x}}$ where the separation $\mathbf{r}$ between the charges is given by $\mathbf{r}=r\mathbf{\hat{x}}$ (in other words, the charges are moving directly toward each other), so that $$\mathbf{\beta}=\beta_x=v_x/c,$$ and $$\gamma=\left[1-\left(\frac{v_x}{c}\right)^2\right]^{-1/2},$$ then for an electric field in the frame of the stationary charge $\mathbf{E}=E_x\mathbf{\hat{x}}+E_y\mathbf{\hat{y}}+E_z\mathbf{\hat{z}}$ with a background magnetic field ($\mathbf{B}$), the test charge will 'see' fields $\mathbf{E}'$ and $\mathbf{B}'$ given by $$\mathbf{E}'=\gamma(\mathbf{E}+\beta_x \mathbf{\hat{x}}\times\mathbf{B}) - \frac{\gamma^2\beta_x^2}{\gamma+1}(\mathbf{\hat{x}}\cdot\mathbf{E})\mathbf{\hat{x}}\\ \mathbf{B}'=\gamma(\mathbf{B}-\beta_x \mathbf{\hat{x}}\times\mathbf{E}) - \frac{\gamma^2\beta_x^2}{\gamma+1}(\mathbf{\hat{x}}\cdot\mathbf{B})\mathbf{\hat{x}}$$

(Source, J.D. Jackson 1999, section 11.10.)

The end result is that electric fields in a rest frame look like magnetic fields from a moving frame.

Interestingly, if the test particle is moving directly toward the charge, the electric and magnetic field along its trajectory will always be the classical one and relativity will have no effect. It is only when the boost has a component perpendicular to the rest-frame fields that the boost-frame fields are different.

There is compression of the field lines at relativistic velocities, but again, only for field lines that are not parallel to the velocity. If you picture the field lines radiating out of a stationary charge, then a moving charge looks similar, but with the field lines perpendicular to the boost velocity more tightly bunched together.

A test charge "feels" only an external field. If another charge is approaching, it will experience a stronger field due to $1/R^2$ field strength dependence.

• So other than this distance dependent field strength no other possibilities are there? Dec 28 '12 at 20:28
• Well, you can consider retarded fields as more exact, if necessary (a retarded near field and radiation). Dec 28 '12 at 20:30
• Not understood, please explain it little more? Dec 28 '12 at 20:32

Yes. Look at the Lienard-Wichert formula which correctly predicts the force.

If the source was approaching the target at constant velocity when the force left, then the L-W formula simplifies to

$$F_{now} = X \frac{1+|v|}{1-|v|}\frac{\overline{n-v}}{D}$$

Where $$X$$ is a constant fudge factor involving the charges, constants, etc

$$n$$ is the unit vector in the direction from source (when the force left) to target (when the force arrives)

$$v$$ is the relative velocity of source (when the force left) and target (when the force arrives) divided by lightspeed, usually written $$\beta$$

$$D$$ is the distance from source (when the force left) to target (when the force arrives).

This is a simplified expression of the equation which works only when the velocity is directly toward the target and acceleration was zero when the force left the source. It would work when the velocity is directly away from the target with some sign changes.

Multiplying by $$1+|v|$$ increases the force, but cannot double it when $$v

Dividing by $$1-|v|$$ increases the force, and can increase it without limit as v approaches c.

You asked about compressed field lines. We can interpret the mathematical result that way.

$$\frac{1}{1-|v|}$$ is the Doppler effect we would get when the source moves toward the target at $$v$$ and the medium does not move. You can visualize that as compressed field lines.

$$1+|v|$$ is the Doppler effect we would get when the target moves toward the source at $$v$$ and the medium does not move. More field lines cross the target in a given time than would if the target was stationary.

You might wonder how the source can move at $$v$$ while the target moves at $$-v$$ both. They each move at $$v$$ relative to the hypothetical medium, but they move at $$v$$ relative to each other. This interpretation does not quite make sense in classical terms, because we can't assign any absolute velocity to the hypothetical medium. We can't get a classical explanation that will make perfect sense. However, special relativity makes as much sense as it makes.

The formula makes sure that we can't tell whether it's the source or the target that moves (though it assumes it is the source), because we have both doppler effects right there in the formula. We can interpret it either way with the other as a fudge factor.