Build a photograph from emission and arrival times

The preceding chapter followed two flashes in words. Here we need trigonometry, uniform motion and a careful choice of clocks. We work in flat spacetime with a distant observer, and neglect cosmic expansion across the example. The two compact knots leave the origin simultaneously and move at the same constant speed in opposite directions.

Choose the clocks and axes

Put the jet along the source-frame vertical axis $z$ . Positive $z$ points along the approaching jet. The observer lies in the $x$ – $z$ plane. The unit vector $\mathbf n$ points from the source toward that observer, and the angle $\theta$ runs from the approaching axis to $\mathbf n$ . The knot's speed is $\beta c$ , where $c$ is light speed. Let $s=+1$ label the approaching knot and $s=-1$ the receding knot.

The source-frame clock reads $t$ . A clock travelling with either knot measures proper time $\tau$ . Special relativity relates them by

\begin{aligned} \Gamma&=\frac{1}{\sqrt{1-\beta^2}},\\ \tau&=\frac{t}{\Gamma}. \end{aligned}

The telescope clock needs a convenient zero. Subtract the fixed light-travel time from the origin to the observer, and call the remaining arrival time $T$ . This subtraction removes a common offset; it does not remove the changing distance as the knots move. All three times have units of time. They are different physical quantities.

Subtract the distance saved by the photon

After source time $t$ , the knot is at $\mathbf r_s=s\beta ct\,\hat{\mathbf z}$ . Its displacement toward the observer is the dot product $\mathbf n\cdot\mathbf r_s$ . A dot product takes the component of one vector along another. The light path is shorter by that displacement, so

T=t-\frac{\mathbf n\cdot\mathbf r_s}{c} =t(1-s\beta\cos\theta).

A nearby observer would require the exact distance to each event. The expression above uses the distant-observer limit, where the directions from the whole source are effectively parallel. It is the usual far-field geometry of a compact jet image.

To build an image at a chosen telescope time, invert the relation first:

t_s(T)=\frac{T}{1-s\beta\cos\theta}.

Each side now has its own emission time. Feeding a common source time into both sides would instead build a simultaneous source-frame diagram.

At one arrival time, source emission times differ, with the approaching knot older. — Analytic positions at the same arrival time for the default example. The approaching knot's light was emitted much later than the receding knot's light. The figure distinguishes the common telescope time from the two source times and from the sky-projected distances.

Project onto the sky

The sky plane is perpendicular to $\mathbf n$ . Choose its first unit vector $\mathbf e_1$ along the projected approaching axis. With the coordinates above, one convenient oriented basis is

\begin{aligned} \mathbf n&=(\sin\theta,0,\cos\theta),\\ \mathbf e_1&=(-\cos\theta,0,\sin\theta),\\ \mathbf e_2&=(0,-1,0). \end{aligned}

These vectors have unit length, are mutually perpendicular, and satisfy $\mathbf e_1\times\mathbf e_2=\mathbf n$ . A positive first sky coordinate points toward the approaching side. Both knots have zero second sky coordinate in this straight-line example. Their first coordinates at arrival time $T$ are

x_{\perp,s}(T)=\frac{s\beta cT\sin\theta}{1-s\beta\cos\theta}.

The slope of a position-versus-arrival-time graph is the signed apparent velocity. Dividing by light speed gives

\begin{aligned} b_+&=\frac{\beta\sin\theta}{1-\beta\cos\theta},\\ b_-&=-\frac{\beta\sin\theta}{1+\beta\cos\theta}. \end{aligned}

The negative sign tells us which way the receding feature moves across the image. When comparing positive speed magnitudes, use $a=b_+$ and $r=-b_-$ .

Work one equal-arrival image

Choose $\beta=0.95$ , $\theta=20^\circ$ , and an arrival time $T=2T_0$ , where $T_0=L_0/c$ for an arbitrary reference length $L_0$ . The arrival denominators are about $0.1073$ and $1.893$ . The source emission times are about $18.6T_0$ and $1.06T_0$ . Their projected coordinates are about $6.06L_0$ and $-0.343L_0$ .

Both knots have the same Lorentz factor, about $3.20$ . Their proper ages at these emission events are therefore about $5.82T_0$ and $0.330T_0$ . There is no inconsistency in those unequal ages: their photons have different distances left to travel.

This example also checks units. A projected position is a speed times a time. The apparent speeds, about $3.03c$ and $0.172c$ in magnitude, multiply the common arrival time of two units to give the image coordinates. The source speeds multiply the separate emission times to give the physical positions before projection.

Find the fastest apparent direction

For a fixed actual speed, the approaching apparent speed rises and then falls as the viewing angle increases from zero. Differentiating its numerator and denominator gives

\frac{\mathrm d b_+}{\mathrm d\theta} =\frac{\beta(\cos\theta-\beta)}{(1-\beta\cos\theta)^2}.

The derivative is the small change in apparent speed per small angle change. It vanishes when $\cos\theta=\beta$ . At that angle the maximum is $\Gamma\beta$ . A measured apparent speed $a$ therefore requires at least $\Gamma\geq\sqrt{1+a^2}$ under this ballistic interpretation. A lower bound is useful even when one image sequence cannot determine the viewing angle uniquely.

Exercises with explained answers

Set the viewing angle to a right angle. What happens to the source ages and the two signed apparent speeds?
At the default viewing angle, double the arrival time. Which quantities double, and which stay fixed?
A calculation gives equal source ages on the two sides at nonzero arrival time, despite a strongly approaching jet. Which clock has probably been used to assemble the image?

Answers

At a right angle, the displacement toward the observer vanishes. Both emission times equal the arrival time, and the signed apparent speeds are plus and minus the actual speed in light-speed units.

Doubling arrival time doubles both emission times, proper ages and projected positions. Constant speed and angle leave the Lorentz factor, Doppler factors and apparent-speed slopes unchanged. A non-evolving emitting patch also keeps its intensity unchanged in this model.

Equal ages usually mean the code selected a common source time before drawing both positions. To construct a telescope image, solve the arrival-time equation separately for each side and project those events.

Sources

Urry and Padovani (1995), Appendix A, A1 and A4, apparent-speed maximum and the pattern/bulk assumption. Author derivation.
Urry and Padovani (1995), Appendix B, B1–B2, for intervals measured in the emitter rest frame and by the observer. The other pair of clocks.

Try the prediction

Compare the ages on one image

At the default parameters, predict which knot is older when their light arrives together.
Choose Pause, then Save this view. Read the two emissionTime values at the same arrivalTime.
Compare each along-jet sky position with its emission time. Explain why equal arrival times do not give equal source ages.