Retardation for a moving point charge
Derive the moving-source Jacobian and evaluate the Liénard–Wiechert potentials at the correct emission event.
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The question for today
What you’ll understand
- Solve a simple implicit retarded-time equation.
- Derive the factor one minus the line-of-sight velocity divided by light speed.
- Distinguish scalar-potential enhancement from radiated power.
A moving charge samples the past light cone at a moving location. Consequently, substituting its retarded position into a Coulomb formula is insufficient: the source-time delta function contributes a Jacobian. The Liénard–Wiechert potentials incorporate both effects for an arbitrary smooth subluminal trajectory in vacuum. They describe a prescribed source outside its worldline. They do not decide how the charge moves under its own field, and a large potential by itself does not establish radiation.
Locate the emission event
Let the trajectory be z as a function of source time, and hold the observer at x and observation time t. The retarded event is the one for which emission time plus the light travel time equals observation time. The distance in that equation is from the emission position, not the charge's present position. This distinction persists even during uniform motion. Solving the implicit equation is the first task; differentiating fields or evaluating velocities comes afterward.
Define the unit vector n from the retarded charge position toward the observer, and beta as the source velocity divided by light speed at that same event. Increasing emission time by a small amount moves the source toward the observer by its line-of-sight displacement. The reception interval is therefore shorter when the source approaches. For a timelike trajectory the reception time grows monotonically with emission time, giving a unique intersection when the trajectory and observation event admit a retarded connection.
Integrate the delta function honestly
The point charge density is a three-dimensional spatial delta supported on its worldline. Writing the retarded source integral as a space-time integral lets the spatial delta be used first. What remains is a time delta whose argument depends on the moving distance. The identity for a delta of a function divides by the absolute derivative of its argument at a root. Here that derivative has magnitude kappa, which is positive for subluminal motion.
This calculation produces an inverse kappa factor in both potentials. The vector potential is the scalar potential multiplied by the retarded velocity divided by speed squared. Keeping that SI factor prevents confusion with conventions in which light speed is set to one. All quantities in the denominator and numerator belong to the same retarded event. Combining a present velocity with an emission distance does not approximate the exact formula in a controlled way unless slow changes are independently established.
What the denominator does and does not mean
Forward directions have smaller kappa, so successive source events can produce compressed reception intervals. This geometric fact later helps explain beaming, but the potentials themselves are gauge dependent. To determine observable radiation one must differentiate with the retarded-time chain rule and isolate the transverse inverse-distance fields. Uniformly moving charges have nonzero Liénard–Wiechert potentials while carrying no acceleration radiation to infinity in an inertial frame. Potential enhancement is therefore not a radiation criterion.
The vector lab can test the dot product entering kappa. Treat its horizontal axis as the fixed sightline and its magnitude as beta, and read the horizontal projection. Reversing direction changes the sign of that projection, even though speed is unchanged. This is useful geometry, not a worldline intersection calculation. The lab does not display the moving-source distance or either potential, and it cannot transform the gauge-dependent scalar potential into a direct measure of energy flux.
Work it through
Solve an approaching-source geometry
A charge follows z(tᵣ)=0.5ctᵣ along a line. A stationary observer is at x=3 m and receives at t=0. Use c=3.0×10⁸ m/s. Find the emission time, retarded distance, and scalar potential relative to q/(4πε₀×3 m).
The retarded charge is to the left of the observer, so R=3−0.5ctᵣ. The light-cone equation becomes 0=tᵣ+3/c−0.5tᵣ.
Solve 0.5tᵣ=−3/c to obtain tᵣ=−20 ns. The source was at −3 m, making the retarded distance 6 m rather than its present separation of 3 m.
The sightline and velocity are parallel, so kappa is 0.5. Their product κR is 3 m, giving exactly the stated reference scalar potential. The enlarged emission distance cancels the Jacobian enhancement in this particular collinear geometry.
Emission occurred at −20 ns from −3 m, with R=6 m and κ=0.5. The scalar potential equals q/(4πε₀×3 m), despite its two moving-source corrections.
The living notebook
Explore the model
Interactive lab · vectors
Project velocity onto the sightline
For a fresh approaching run use Reset and set BOTH angle 0° and magnitude 0.5; predict horizontal projection +0.5 and κ=0.5. For a fresh receding run Reset and set angle 180°, magnitude 0.5; predict projection −0.5 and κ=1.5. Compute κ on paper: this vector diagram does not solve retarded time.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe line-of-sight velocity determines the reception-time Jacobian; speed alone does not.
Pause & reason
Check your understanding
Your turn
Try it yourself
At an already determined retarded event, R=8 m and the velocity has magnitude 0.6c perpendicular to n. Find κ, dtᵣ/dt, and φ as a multiple of q/(4πε₀×8 m).
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Use the projection of velocity rather than its magnitude in the Jacobian.
Compare with the solution
Perpendicular velocity has zero dot product with n, so κ=1. Both the retarded-time derivative and the potential multiplier are one. This is a statement about the specified emission geometry; it does not place the charge at its present position, and it does not imply that derivatives of the potentials have no velocity dependence.
Keep the idea
Take it with you
- Retarded distance is measured from the source emission position.
- The moving-source Jacobian supplies 1/(1−n·β).
- Uniform motion can enhance potentials without producing radiation.
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