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I am wondering if I could build an ion thruster, that is capable of exit velocity bigger that the speed of light. If I start with the equation for calculating exit velocity $v_e$ of a particle of charge and mass $q$ and $m_p$ respectively accelerated by a voltage difference $V$ :

$$v_e = \sqrt{\frac{2Vq}{m_p}}$$

and then use a voltage of 1x1024 volts and a proton with charge and mass of 1.6x10-19 Coulomb and 1.7x10-27 kg I get a velocity of 1.4x10+16 meters per second.

That's turns out to be much faster than the speed of light.

How does my calculation stack up against known physics? Would this work? If so, why not?

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  • $\begingroup$ (Someone with a better understanding might be able to explain this, but AFAIK you cannot go faster than the exhaust velocity anyways and that exhaust velocity for an ion thruster is slower than the speed of light, ergo can't go faster.) $\endgroup$
    – DarkDust
    Dec 21, 2018 at 13:02
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    $\begingroup$ @DarkDust "you cannot go faster than the exhaust velocity" maybe that's not correct. $v/v_e=\ln(m_f/f_0)$ is a pain, tyrannical even, but it has no hard limit. $\endgroup$
    – uhoh
    Dec 21, 2018 at 13:14
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    $\begingroup$ Okay, so now show us the design for your yottavolt power supply. $\endgroup$
    – Russell Borogove
    Dec 21, 2018 at 17:40
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    $\begingroup$ If you take a breakdown field strength of 20 MV per meter, you need a distance of 50 Pm (yes petameter) for isolation of 1x10^24 volts. But a non relativistic calculation of an exhaust velocity of 1.4x10^16 m/s is wrong anyway. $\endgroup$
    – Uwe
    Dec 22, 2018 at 15:00
  • $\begingroup$ @DarkDust "you cannot go faster than the exhaust velocity" with an engine that uses the atmosphere as the majority of it's reaction mass $\endgroup$
    – user20636
    Apr 11, 2019 at 15:41

3 Answers 3

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The expression $v_e = \sqrt{\frac{2Vq}{m}}$ is a non-relativistic approximation. This is quite valid when the exhaust velocity is small compared to the speed of light, which is the case for ion thrusters made to date (exhaust velocity is on the order of $10^{-4}c$). A more precise expression is $${v_e}^2\left(1+\frac{2Vq}{mc^2}\right) = \frac{2Vq}m$$ No matter how much one raises the voltage, the relative exhaust velocity will not exceed the speed of light.

Using the values in the question, $V=10^{24}$ volts, $q$ is the electron charge, and $m$ is the proton mass, the non-relativistic expression results in an exhaust velocity of about 44 million times the speed of light. The non-relativistic approximation is completely invalid in this regime because these values make $\frac{Vq}m \cong 10^{15}c^2$. Instead of 44 million times the speed of light, these values result in an exhaust velocity that is a shave less than the speed of light (about $0.99999999999999975\,c$ -- a very close shave indeed).

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  • $\begingroup$ Hmmm. Sorry, It seems that it doesn´t make a sense $\endgroup$
    – David
    Jan 11, 2019 at 20:26
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    $\begingroup$ @David, as DarkDust put it, matter cannot move faster than the speed of light in vacuum. Whenever you run across an expression that implies otherwise, it's wrong. Oftentimes the problem lies in applying a non-relativistic expression to a domain where relativistic effects are important. The expression in your question is a non-relativistic simplification and is extremely close to correct for small voltages. But it's rather incorrect for high voltages. $\endgroup$
    – David Hammen
    Jan 12, 2019 at 7:19
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Matter cannot move faster than the speed of light in vacuum. Nothing you try to come up with is changing that. (If you do come up with a peer-reviewable proof that it's possible you're in for a Nobel prize.)

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    $\begingroup$ Agreed, I think it would be good to identify the flaw in the OP's equation $\endgroup$
    – Jack
    Dec 21, 2018 at 13:13
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    $\begingroup$ I'm afraid my math/physics-fu isn't strong enough for that. I just happened to know that the limit is absolute and cannot be overcome and that's it. ^_^ $\endgroup$
    – DarkDust
    Dec 21, 2018 at 13:17
  • $\begingroup$ To add to this, we have a vital question. What if we keep trying to accelerate? According to doctor Neil Degrasse Tyson, you will gain mass when you put in additional energy to accelerate past light speed since the energy has to go somewhere. Quite interesting. $\endgroup$
    – mberna
    Feb 9, 2022 at 15:13
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Your equation is non-relativistic, and hence only works for small numbers.

What happens at relativistic speeds is that, from our point of view, an object will shrink in length, increase its mass, and time will run slower on it. As it approaches light speed, its length will approach zero, its mass will approach infinity, and time will approach stasis. You can find explanations in a lot of popular books on relativity.

In that equation, you're adding up a lot of little adjustments. A charge potential of 10^24 volts is effectively two potentials of 5x10^23 volts put together. You can view it as going through 10^24 one-volt potentials. Each of these is adding a little momentum to the ion. At non-relativistic speeds, a change in speed is proportional to change in momentum, but that doesn't hold at relativistic speeds.

What happens is that, when the ion is going well below light-speed, adding a certain amount of momentum gives a fixed amount of speed, and the math is simpler if we solve for speed rather than momentum. You could take your equation and add a multiplier of q on both sides, and you'd have an equation for momentum, but anyone using the equation would cancel out the qs because they don't help mathematically.

However, the effective mass of the ion is always increasing with relative speed, and so a given shot of momentum will produce less and less additional speed as the ion gets faster. As the ion gets really close to light-speed, it will increase its momentum almost exclusively by increasing its mass and only a very little increase in speed.

This doesn't mean the equation you've used is useless, because you can use it to figure out the momentum of the ion as it leaves the ship, and by conservation of momentum you can tell how much it pushes the ship.

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