Superluminal Expanding Gas Shells

When a star explodes, it may eject a spherical shell of glowing gas, which expands very rapidly.  If we know how far away the star is, then by observing it over a period of time we can tell how fast the gas shell is expanding.

It's well known that in many cases, the shell appears to be expanding faster than the speed of light.

Of course, according to relativity theory, that can't happen.  Instead, what we're seeing must be a "trick of the light".  Due to the trigonometry of the situation, the shell appears to be expanding much faster than it really is.  In fact, if you think about what it would look like if the shell were really expanding faster than C, it's clear that it can't be what's going on:  The shell would expand all the way to Earth before we ever saw the explosion!

The cause of this effect is hard to picture.  Furthermore, it doesn't depend on the distance to the star -- "perspective" doesn't enter into it.  On this page, I've worked out the details of what's going on.

There is different but related effect, in which a star contained in a spherical nebula explodes, and a "shell of light" expands through the nebula at C.  The nebula appears out of the darkness as the shell of light passes through it, lighting it up.  The shell may appear to travel much faster than C when viewed from a distant point.  That case is actually somewhat simpler to analyze than the case of an expanding gas shell.  However, the analysis given here can't be applied to that case without some additional work.

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Description of the Problem

Please see Diagram 1.  (As usual, I've assumed c=1.)

A star is located at the origin. We are observing it from a location a large distance away along the 'y' axis (we're "above" it in the diagram).  Our exact distance doesn't matter.


Diagram 1:




The star throws off a spherical shell of glowing gas. The velocity of the shell is 'v' (0 < v < 1).

At time T=0 the shell has already expanded to diameter r₀. At that time, we can imagine a "sight plane" which contains the star itself and a "slice" of the spherical shell.  The "sight plane" contains all the photons which will reach us at the same time from the distance of the star -- in other words, the sight plane contains the image we will eventually see. As time goes by, the sight plane -- with its photons -- travels toward us at c. Additional photons which are emitted in our direction within the "sight plane" as it passes through nearer points will also reach our eyes at the same time, and may be perceived as having been emitted at the same distance as the star.  This last point is the primary origin of the illusion!

As the sight plane moves away from the star, it slices through ever nearer parts of the expanding spherical shell. However, the shell is expanding as this takes place. Since the shell's surface is perpendicular to the plane at the moment when the sight plane contains the star itself, it's clear that initially, the "image" of the shell which is contained in the plane -- and which we will eventually see -- is expanding.  And this is the rest of the origin of the illusion.

In the diagram, I have shown the sight plane at time T=t, some time after the plane passed the star. The plane is moving vertically. Two half-circles are shown, representing the expanding spherical shell at two moments.  The smaller is the sphere at time T=0, with radius r₀. The larger is the sphere at time T=t, with radius r₀ + v*t. The sight plane at time t is located at y=t in the diagram. The intersection of the sight plane with the larger sphere defines an apparent burst size. The diagonal which runs from the x axis to the sight plane shows the "illusory expansion" of the burst.

The "sight plane" consists of the photons which will reach us at one single moment in time, and it travels at c. That's faster than the expansion rate of the sphere, so at some point the intersection of the plane and the sphere will reach a maximum diameter and will then decline to zero. We need to find the maximum, because that's the size the shell will appear to us to have been at the moment when the sight plane contained the star.  The ratio of that maximum to r₀ will be the "magnification" of the image we see, relative to the actual shell size at the moment in time from which the image of the star dates.

Derivation of the Magnification

Now, some math.

At time T=t, we want to find the intersection of the expanding sphere and the sight plane that lies in the half-plane defined by z=0 and y>0 (z axis is not shown on the diagram). The sight plane is moving toward us at C which we've defined to 1, so we must have y=t.  Since the shell is expanding at v and had diameter r₀ at time t, we have



Substituting t for y and solving for x, we obtain



Multiplying out and combining terms,



We want to know the maximum value x can take as the sight plan passes through the shell, since that will tell us the radius we actually observe.  At the maximum, dx/dt = 0.  So, taking the derivative,


 

Setting the derivative to zero, we get



or finally,



Now, we need to be cautious, because (v² - 1) is negative. So, for large values of t, the value under the radical in eq (4) must also be negative; in consequence, the derivative isn't valid everywhere. It blows up at the zero of the quadratic in t under the radical in eq (4). We will take a side trip to see whether the derivative is valid in the range in which we care about it.  It's clearly positive at t=0; using the quadratic formula, we see that the roots are at:







The first term in eq (9)  contains v² - 1 which is negative. We don't care about the negative root here, and at the positive root, the other term must be negative too. We therefore choose -1, and we get



Note that this is necessarily larger than eq(6), since 1+v > v. So, the maximum we found above is between 0 and the point where the formula blows up. Good; that means our maximum is valid. Now, let's simplify eq(10) a little by dividing out (1+v), and see:

 

Since the shell is moving at v, and the "sight plane" is moving at 1, and the sight plane needs to "catch up" a distance of r₀ to get to the edge of the shell, this is indeed the time at which the sight plane reaches the edge of the shell and the intersection of the two vanishes. This makes sense!

Continuing on, let's plug t_max back into our formula for x to obtain:




or, expanding t_max,



This looks messy, and it's still not quite what we want. We want the ratio of x_max to r₀ -- the ratio of the radius we see (in the sight plane, when it arrives) to the radius of the gas shell at the moment when the sight plane contained the star. So, we divide through by r₀, and after moving (1/r₀²) under the radical, we get:



Note that this depends only on the velocity -- it does not depend on time or on the diameter of the shell. Thus, it's a fixed ratio, and we can use it to compute the apparent velocity, which will also be a simple function of the actual velocity:



Now, at this point let's revert to "real world" units in which c isn't necessarily 1. The c's in the v's in the numerators cancel, leaving us with:



We have our answer, and now we can ask a couple of interesting questions about it.

At what velocity will the shell appear to be expanding at c? Reverting to relativistic units (c=1) for a moment, from eq(16), if the apparent expansion rate is 1, then we have:



Solving for the actual velocity when the apparent velocity is 1, we find



or, putting the 'c' back in, we have



We can also ask what happens if v=c. In that case, we have

   Whoops!  It's infinite!

But actually, this makes sense! Consider that if v=c, then we see the star burst at the same moment the expanding shell reaches us. Therefore, the apparent diameter of the shell will be infinite as soon as we see the explosion: we can look 90 degrees off axis in any direction and see light from the shell shining on us, since it's passing us at that moment.