How To Go Faster Than Light
Bottom Line
For those with limited attention spans: yes, in this universe, with a powerful enough rocket you really can go anywhere in the universe as quickly as you like, in your own lifetime, without resorting to any medical tricks like suspended animation. Einstein’s theory of relativity won’t stop you from getting there, the same day even. The hard part is just getting enough energy — and working out the math.
Speed Limits
One big downer — if you can call it that — most people take away from Einstein’s theory of special relativity is that nothing can go faster than light. We are a species that likes to explore, after all, and the idea that there is a depressingly slow speed limit imposed on us by nature makes it very difficult to journey through the galaxy.
Science fiction often addresses this either by inventing a device to “warp” space-time (Star Trek), or by adding a few extra dimensions to the universe and bypassing normal space by jumping into hyperspace (Star War).
A lot of this comes, I think, from a confusion about how the universe actually works, as described by Einstein’s theory of relativity. (Note: given the recent creationist attempt to color the word “theory” as meaning something tentative, I prefer to use Richard Dawkin’s coined word “theorum” — similar to theorem — as indicating a theory that is so well established by overwhelming evidence that it might as well be an undebatable mathematical theorem).
The Confusion
Here is the deal: while it is true that to people watching from earth a spacecraft can never be observed to go faster than light, that doesn’t mean that the passengers on the spacecraft have the same experience. In fact, what Einstein’s theory would say is that as far as the passengers can tell, it seems like they can go as fast as they like. Due to the relativistic “warp” of space-time as you approach the speed of light \(c\), the passengers experience time much more slowly and their “effective” speed as they travel through space appears to be much greater than light.
Let’s do the numbers.
Some Terminology
In the “Star Trek” series they used the “warp \(N\)” terminology to refer to “effective” speeds that were \(N\) times the speed of light \(c\), so that “Warp Two” for example was twice the speed of light or \(2c\). In that series they had a special “warp drive” that bent space-time around so that they would go faster, but the actual fact is that in our everyday world just the mere act of going faster by any means actually warps space-time.
The Warp Equation
I plan on using the trekkie terminology (and standard relativity) to state and prove the following interesting fact:
| The Warp Equation |
| If you have a payload with mass \(m_{payload}\), and a means of converting matter into kinetic energy with 100% efficiency, then the mass \(m_{fuel}\) of fuel needed for you to travel at an effective speed of Warp \(\omega\) where \(\omega > 0\) is given by$$ m_{fuel} = {\omega}^2 m_{payload} $$ |
So for example, in order to travel at Warp 2, a person of mass 80 kilograms would require 320 kilograms of (say) a proton-antiproton fuel in order to travel at that effective speed. That is roughly equivalent to 6,400 Megatons of TNT. Coincidentally, that is almost exactly the combined explosive power of all nuclear weapons now on our planet. That is a hell of a lot of energy, but the point to be made is that is within the bounds of our current technology.
The fact that you have to square the warp factor to get the amount of energy to go that speed makes perfect sense. Even in classical Newtonian physics, the energy related to going at velocity \(v\) is given by
$$E = \frac{1}{2}mv^2$$
so doubling the velocity \(v\) on the right hand side multiplies the energy by four. The fact that the energy happens to be equivalent to four times your payload’s mass comes from Einstein.
The way in which we’ll prove this is to first calculate how much matter is needed to attain an observed velocity v, and then figure out what the relationship is between the observed velocity, and what effective velocity the passenger actually experiences. Note: I have no doubt that there is probably an easier way to derive this formula. But this is the one I came up with and it isn’t all that complicated.
Conversion of Matter to Kinetic Energy
Let’s start with Einstein’s equation:
$$E=mc^2$$
What we are going to do is to use this equation, together with the law of conservation of energy, to compute how much matter it takes to accelerate a payload \(m\) to (observed) velocity \(v_{o}\). Now as the observed velocity \(v_o\) approaches the speed of light, the relativistic mass of the payload becomes:
$$m_{relative} = \frac{m_{payload}}{\sqrt{1-(\frac{v_o}{c})^2}}$$
Now Einstein’s equation for energy represents both the energy of the mass at rest, together with the (kinetic) energy of the mass in motion. And so, if this mass was put into motion by the conversion (at rest) of a certain mass \(m_{fuel}\), where
$$m_{fuel} = \alpha m_{payload}, where \alpha > 0$$
Then since energy is conserved we can relate the conversion of the mass \(m_{fuel}\) into motion \(v_o\) by:
$$ (m_{payload}+m_{fuel})c^2 = E_{rest} = E_{moving} =\frac{m_{payload}}{\sqrt{1-(\frac{v_o}{c})^2}} c^2$$
so dividing both sides by \(m_{payload}c^2\)
$$ 1 + \alpha = \frac{1}{\sqrt{1-(\frac{v_o}{c})^2}}$$
squaring both sides and solving for \(v_o\) we get the following rule:
| Matter to Velocity Conversion |
| For a payload of mass \(m\) and a ratio \(\alpha > 0\), if fuel \(m_{fuel}=\alpha m\) is converted to kinetic energy, the observed velocity \(v_o\) of the body will be$$v_o = (\sqrt{\frac{\alpha}{1+\alpha}})c$$ |
This jibes with what Einstein said about observed velocities, as the right hand side will never be greater than the speed of light \(c\). As the ratio \(\alpha \rightarrow \infty\), the velocity goes to \(c\), so we can get as close to \(c\) as we like — but no further.
Velocity – Observed and Effective
So now we come to the idea of “effective” velocity. The weirdness of relativity comes from the fact that as the observed velocity \(v\) of ship approaches the speed of light, the passenger’s own time-scale is compressed by what’s called the Lorentz-FitzGerald contraction, according to the formula
$$t_{effective} = t_{observed}\sqrt{1-(\frac{v_o}{c})^2}$$
(From this point on we will just write \(t_e\) and \(t_o\) for \(t_{effective}\) and \( t_{observed}\) respectively) Then given a fixed distance \(\Delta x_o\) as measured by the observers on earth, the effective velocity as experienced by the passengers when traversing that segment of space over their time \(\Delta t_e\) is:
$$v_e = \frac{\Delta x_o}{\Delta t_e} = \frac{\Delta x_o}{\Delta t_o\sqrt{1-(\frac{v_o}{c})^2}}$$
which in turn simplifies to this formula for converting observed to effective velocity:
| Observed to Effective Velocity |
| $$ v_e = \frac{v_{o}}{\sqrt{1-(\frac{v_o}{c})^2}}$$ |
The important thing to note about this concept of “effective” velocity is that we are computing the ratio of observed change in our distance with our own experienced change in time. Suppose for example that over centuries the earth residents went out and placed “mile markers” along your path to your destination. Then once you got up to warp speed, not only would your sense of time be compressed, but distances that you measure out with your tape measure would also be compressed by the same Lorentz-FitzGerald contraction. Consequently, if you were going at Warp 2, for example, your sense of time elapsed going from one mile marker to the next would be cut in half, but also (this is the point) the distance between those two fixed mile markers as the earth-people laid them out would also appear to you to be half. So if you were instead to calculate your velocity by dividing your measured distance by your time, you would never get a value at or greater than c.
All Together Now
So if we start with fuel \(\alpha m\) which we use to accelerate our mass \(m\) to the observed velocity \(v_o\), we can use the two formulas we just derived to express the effective velocity \(v_e\) as a function of \(\alpha\). We can rewrite the “Matter to Velocity” formula as
$$ (\frac{v_o}{c})^2 = \frac{\alpha}{1+\alpha}$$
So our effective velocity formula simplifies the bottom of the fraction to
$$v_e = v_{o}\sqrt{1+\alpha}$$
and then substituting the formula again for \(v_o\) we see that our fuel mass \(\alpha m\) gives us an effective velocity of:
$$v_e = c\sqrt{\alpha}$$
Thus if we have defined velocity “Warp \(\omega\)” to be \(\omega c\), then we can write
$$\omega = v_e / c = \sqrt{\alpha}$$
So that to attain an effective velocity of Warp \(\omega\) we must use a fuel-payload ratio of \(\alpha = \omega^2\), ie
$$m_{fuel} = \omega^2 m_{payload} $$
which is exactly the “Warp Equation” we were to prove. QED
Let’s Do the Time-Warp Again
It should be pointed out that of course to the observers on earth, even though you are going at an effective speed of Warp \(\omega\) you will never appear to be going faster than \(c\) and so it will take you a long time to get where you are going. You, however, will not experience that, and so you will effectively be travelling through time much faster than your friends at home. How much faster? According to our formula above relating \(t_e\) to \(t_o\), and expressing that in terms of the warp factor \(\omega\), we can show that the time-warp you experience will be:
$$t_e = \frac{t_o}{\sqrt{1+\omega^2}}$$
And so, in our example, the 80kg person travelling at Warp 2 will feel like they’ve reached their destination in \( 1/ \sqrt{5} \) of the earth time, ie getting them there in about 0.44 of the time observed on earth, and exactly twice the time it would take light to appear to get there.
So, not only can you go as fast as you like, you can also travel as far in the future as you like. For example, to travel 1000 years into the future, just get in a spaceship armed with 1000 times your own mass in matter-antimatter fuel, and then travel at Warp 1000 for one year. When you reach your destination, one year will have passed for you, and 1000 years (plus a little bit) will have passed on earth.
Of course, then you’ll have to get back to earth, so good luck with that.
The Buckaroo Banzai Principle
| The Buckaroo Banzai Principle |
|
No matter where you go — there you are. |
The point of this exercise is that if you really understand what Einstein said, the idea should be that there is no absolute frame of reference. What this means is that even if you are travelling at 99.999 % the speed of light relative to the earth, as far as you know everything still looks and feels like Newton’s physics, where F = ma and you can always accelerate faster and faster. And not only that, but if you are heading for a specific location, the faster you go, the faster you will get there.
No Free Lunches
Now having said that, there are some consequences that the universe may unleash should you decide to try to go Warp 100. This is because even though the physics of your spaceship will be the same even at this insane speed, you are also surrounded by the gases in your local galaxy, as well as all of the light from stars that are visible to you. And even though from the earth much of this light is nice, low-energy visible spectrum, and even though that light will still be reaching you at the speed of light, it’s relative energy is radically different when you are plowing through that light at Warp 100. In fact, what you will be observing is a massive Doppler-shifting into the deep blue/ultraviolet of all light coming at you in the direction you are headed (and conversely, red-shifted looking back towards earth). Some of this light may be equivalent to the powerful cosmic rays that hit the earth, and which were generated by massive explosions or quasars just after the Big Bang. The energy in these photons may be enough to kill you all by themselves, especially at Warp 100. You may need a very large and thick radiation shield, along with all the extra energy to carry that shield along with you and your ship.
And so as we already should have known, there are no free lunches. At least it is nice to know that a faster-than-light lunch is available, should one choose to pay the price.