# Dyson Spheres and the Speed of Light

Many years ago when I was living in Switzerland, one of my co-workers and friends there was almost as much of a physics nerd as I would like to think I am, and we were wondering about different ways to shoot things off at the speed of light. As it turns out, you can end up in some pretty impressive energy realms when you start moving things really fast (for instance: Oh-My-God particles that are cosmic rays (subatomic particles) with kinetic energies similar to macroscopic objects like baseballs!)

The Relativistic Energy Problem
The first piece of interest occurs due to the amazingly high accelerations. As everyone learns in high school physics, objects have kinetic energy as a result of their movement. This is approximated by the well known formula:

$E_{kn} = \frac{1}{2}mv^2$

The key word here is of course approximation: this formula is only close to the true kinetic energy at the incredibly low speeds that we are used to on a day to day basis. Einstein’s relativity (specifically the special one) provides a much better calculation that holds as speeds approach the speed of light. It adds the Lorentz Factor to the equation (represented by the greek letter gamma: γ).

Figure 1: Newtonian and Relativistic Kinetic Energy from 0% to 90% of the speed of light.

$\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$
$E_{kr} = mc^2\cdot(\gamma-1)$

Interesting! As v approaches c(the speed of light in a vacuum), the denominator of gamma approaches zero, so gamma (and hence the kinetic energy) approaches infinity! In plot form, this looks like Figure 1. As you can see, the Newtonian equation is a pretty good approximation until 40-50% of c at which point the true kientic energy starts to diverge rapidly. This is also the reason that the speed of light is the ‘universe’s speed limit’: when v = c, the v2/c2 term becomes 1, γ becomes infinite, and the kinetic energy becomes infinite as well. We don’t have enough energy to move things that fast!

So what kind of energy does this really come out to? Lets use the space shuttle as our basic interstellar rocket ship – it may as well be used for something! According to The Only Information Source on the Internet, it has a weight of 2,030 tonnes, or 2,030,000 kg. Accelerating it to 50% of c would require 2.82×1022 joules of energy (about 28.2 zettajoules). By comparison the total planetwide energy consumption in 2008 was about 4.74×1020 joules (474 exajoules). So to accelerate one space shuttle to 50% of c we would have to devote every joule of energy used on the planet to the task for about 60 years! To get to 90% of c requires 2.36×1023 J, or 498 years of annual world energy use! There must be a better way…

Energy Sources

Figure 2: Kinetic energy of a 2030 tonne spaceship from 0% to 60% of c with some reference points

While 1022 J or 1023 J is certainly a lot of energy, it is still within “earthy” orders of magnitude. For instance, the total estimated energy in the world’s coal reserves is 21 zettajoules, and the total energy striking the earth from the sun every year is 5.5 yottajoules! Figure 2 shows the “low-speed” (up to 60% of c) energy profile and some of the equivalent energy sources/sinks along the way.

But what if we want to accelerate things to exceptionally high speeds? Say 99% of the speed of light, or 99.999% of c?! For the sake of clarity, lets re-design our speed-kinetic energy plot a little bit and use an inverse-logarithm plot. In this view, the first 90% of speed is crunched into one logarithmic division, the next 9% of speed in another division, the next 0.9% of speed into another and so on. This is useful as it shows the really high speeds clearly. The y-axis (energy) has also been changed to log-scale so that the massive changes in kinetic energy can be clearly visualized – a straight line on this log-log-1 plot is actually an exponential increase in energy. Figure 3 shows this plot with some reference points. The classical newtonian energy is almost flat in this view which is expected: the equation is of order two and thus a small change (percentage-wise) in the velocity results in twice that change in energy consumption. This isn’t a big change when the change in velocity is only 0.01% or 0.0001%. Conversely the relativistic energy shows massive exponential changes in energy for these small changes in velocity.

Figure 3: Inverse log plot of very high fractions of the speed of light for a 2,030 t vessel.

So how fast could we go? A Dyson-sphere is a hypothetic megastructure that completely encircles a star, say the sun. Since it intercepts everything radiated from the sun, it theoretically captures all of the energy radiated from the star. Of course in reality the efficiency of the sphere would be less than 100%, and a significant of the energy would be radiated as heat from the outside of the sphere. For the purposes of this discussion though, lets pretend we have a 100% efficient Dyson Sphere. We would capture about 3.86×1026 J every second: enough to accelerate us to 99.999989% of the speed of light every second (still assuming that our ship weighs 2,030 t)! Of course the ship itself may not survive the acceleration – but that’s really beside the point!

What if we were happy going 99% of the speed of light? That still gets us to Alpha Centauri (~4.4 ly) in 4 years and 5.3 months… For this feat we need about 1.1×1024 J, or about the amount of energy from the sun that reaches the earth in 2.4 months.

One last idea! Anyone whose anybody knows that the USS Enterprise from Star Trek runs on anti-matter. The ideal energy source since it can convert all of its mass into energy when it comes into contact with its regular-matter equivalent. Einstein’s most famous equation shows how much energy we can produce with a 100% efficient anti-matter reactor:

$E=mc^2$

This means that is has an energy density of c2 J/g, or about 9×1016 J/g. That’s amazing! So if our 2,030,000 kg ship was outfitted with an anti-matter reactor and we wanted to accelerate it to, say 99% the speed of light so that it could transition into warp drive, how much mass is needed?

$\gamma=\frac{1}{\sqrt{1-(v/c)^2}}$
$\gamma=\frac{1}{\sqrt{1-0.99^2}}=7.09$

$E_k=mc^2\cdot(\gamma-1)$
$E_k=(2,030,000)c^2\cdot(6.09)=1.1\times10^{24}\ \mathrm{J}$

$m=\frac{E}{c^2}$
$m=\frac{(1.1\times10^{24}\ \mathrm{J})}{c^2}=12,360,000\ \mathrm{kg}$

Uh-oh! So the amount of mass needed to produce our accelerating energy is more than the weight of the ship itself! And the situation is made even worse by the fact that an equal amount of regular matter is needed to react with the anti-matter. So 24,720,000 kg of mass are required to accelerate a 2,030,000 kg ship to 99% of the speed of light… doesn’t seem promising!

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