# Visual Appearance ofSuperluminal Circles

Imagine that you are hovering   h   kilometres above an immense perfectly flat plane.   A powerful searchlight out in space projects the image of a circle of radius   a down onto the plane and sweeps that image past you in a straight line, such that the true speed of the circular image is   v   and the line of the path followed by the centre of the circle passes, at its closest approach,   b km (= “the impact parameter”) from the point directly under you (your “nadir”).

Because the speed of light c is finite, the difference in delay times caused by the distance that light has to travel from different parts of the circle at various times causes a distortion in the visual appearance of the circle.   That distortion becomes more extreme the greater the speed v of the circle is.

The situation is particularly fascinating when the true speed of the circle is faster than light.   The circle then moves faster than the reflected light by which we see it.   There will therefore be a time in the circle’s journey, before which the image will not be visible to the observer at all.   When the images from that part of the journey do arrive, they will arrive in reverse order.   That image will be back-to-front and will recede towards earlier images.   At the same time, the observer will see later images travelling in the opposite direction!

In the standard Cartesian (x, y, z) coordinate system an object has a velocity   v î   parallel to the x-axis.   It travels in the plane   z = -h   (the horizontal plane h units below the observer).   The observer’s clock is calibrated in such a way that the observer sees that a reference point on that object achieves its closest approach at a distance   d   from the observer in the   x = 0   plane at observer’s time   t = 0 .

Due to the finite speed of light   c , it takes a time     for that image to reach the observer   d km   away.   The reference point actually passes through the point of closest approach at observer’s time   .   It actually takes a further time interval of     for the point to travel a further distance   x km, then another time   for that image to reach the observer.

Therefore the observer sees the reference point at location   x(t)   at time   t   where   x   and   t   are related by

Therefore the apparent position of the reference point at any time   t   is

Provided that   v < c , the expression for   x(t)   is real for all time   t , so that the reference point is visible at all times.   Note that the true position of the reference point at that instant is

In the event that the true speed of the reference point is the speed of light,   v = c , then the expression for the apparent x coordinate can be shown to be

—   there is an earliest time before which the reference point is not visible.

Letting     and taking the superluminal (faster than light) case   v > c , the expression for   x(t)   becomes

The positive square root corresponds to the forward moving image and the negative square root to the backward moving image.

These expressions for   x(t)   are real only if the expression under the square root is non-negative, so that there is again an earliest time before which the reference point is not visible at all:

Therefore the instant of first appearance for both images is     and occurs at the common location

Graphs of real and apparent x-coordinates against time are displayed here for four different true speeds   V :

 v = 0.6 c v = 0.8 c

 v = c v = 1.4 c

Now a general point P on the circle can be located relative to the centre of the circle by a single parameter, the anti-clockwise angle f made with the positive x-axis.   When this point crosses the plane x=0, the centre C of the circle will be a distance r cos f to the left of that plane.   As the circle is moving at a speed v = b c, P will pass through the plane x=0 some (r cos f) / v seconds before C does.   [See figure 3 below]

Figure 3:   The Plane   z = –h

However, the information that P has passed through the x=0 plane takes longer to arrive at the observer O than it does for C, because P is farther away.   [See figure 4.]

Figure 4:   The Plane   x = 0

From simple triangle geometry,

d 2   =   b 2 + h 2     and
dP2   =   (b + r sin f)2 + h 2   =   d 2 + 2 b r sin f + (r sin f)2

In the expression for   x(t) , when applied to the point P in place of C, “(ct + d)” is replaced by “” and the single “d 2” under the square root is replaced by d 2 + 2 b r sin f + (r sin f)2

y(t) is modified to
y(t) = b + r sin f
and z(t) doesn’t change:
z(t) = - h

This scheme has been implemented in a QBASIC simulation and its executable file.
A screenshot from this program appears below.

A negative of this screenshot is more printer-friendly!

A final refinement is to project the plane z = -h onto a hemisphere, radius h, centre O.   When flattened out onto a plane, this maps infinity onto the circle of radius   h p / 2.
With   , r   = sqrt(x2 + y2),   the projection is achieved by the transformation

This revised scheme has been implemented in a QBASIC simulation and its executable file.

Dr. G.H. George
Faculty of Engineering and Applied Science
S.J. Carew Building
Memorial University of Newfoundland