Okay, So i'm not really into physics or mathematics but i'm trying my best to learn some bit of how the calculations needed for the amount of antimatter needed to propel a spacecrafft to relativistic speeds.

So lets take a realistic spacecraft like the ISV Venture Star from Avatar. enter image description here

So, according to the wiki, The ISV class of ships is accelerated to 0.7c for 6 to 7 months straight from Earth to Alpha-Centauri and vice versa.

It carries approximately 250 metric tons of cargo at full capacity which included all 250 passengers, two valkyrie shuttles and the cargo racks loaded.

Let's recalculate the amount of antimatter required for a longer duration of constant acceleration, specifically 6.5 months, which is approximately 4,745 hours:

1. Spacecraft Mass: Assume a spacecraft mass (including cargo, colonists, shuttles, and structure) of 250 metric tons, which is approximately 250,000 kilograms.

2. Desired Acceleration: Let's assume the spacecraft accelerates at a constant rate for a 6.5 Months (4,745 Hours). The desired final velocity is 0.7c (70% of the speed of light).

3. Relativistic Effects: When approaching the speed of light, relativistic effects become significant. To account for this, we'll use the relativistic rocket equation:

Δv = c * tanh⁻¹(a * Δt / c)


  • Δv is the change in velocity (0.7c - 0c = 0.7c).

  • a is the constant acceleration.

  • Δt is the time duration in the spaceship's frame (4,745 hours).

4. Antimatter Propulsion Efficiency: We'll assume 100% conversion efficiency of antimatter to energy (which is an idealized assumption and may not be achievable in practice).

Now, we can solve for the required acceleration (a):

Now, we calculate the required energy (E) to achieve this acceleration:

E = (1/2) * m * Δv²

Where m is the spacecraft mass and Δv is the change in velocity.

E ≈ (1/2) * 250,000 kg * (0.7c)²

Next, we convert the energy to antimatter mass using Einstein's mass-energy equivalence (E = mc²):

m = E / c²

Plugging in the values:

m ≈ [(1/2) * 250,000 kg * (0.7c)²] / c²

m ≈ 0.000214 kilograms of antimatter

So, approximately 0.000214 kilograms (or 0.214 grams) of antimatter would be required to propel the spacecraft to 0.7 times the speed of light (0.7c) over a duration of 6.5 months with the corrected assumptions.

Is my math correct in this?

And if that's the case... THAT'S ALL THE AMOUNT OF ANTIMATTER NEEDED to propel such a big ass ship??? not even 1 kg? Is this how powerful antimatter is?

I mean of course producing it is the riskiest economic suicide ever but I never thought antimatter would be absolutely this intense.

Edit: I made a mistake. I literally calculated 0.7 the decimal as opposed to 0.7 light speed (209854720.6) and only used c as opposed to c². its indeed 61,250 kgs of antimatter for the 250,000.

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    $\begingroup$ Typing your exact equation into WolframAlpha gives 61,000 kg. I'm not sure what happened here, but this updated result is consistent with stuff I've heard about relativistic antimatter starships being, in large part, fuel mass (remember, you also need a bunch more to slow down, and that counts as the mass you're accelerating...). See Frisbee as an example. $\endgroup$
    – parasoup
    Sep 10, 2023 at 4:23
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    $\begingroup$ I got 61.25 tons of matter plus antimatter... you need to halve the amount since antimatter is annihilated with an equivalent amount of matter. $\endgroup$
    – Monty Wild
    Sep 10, 2023 at 4:30
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    $\begingroup$ Your efficiency has to be 100%. If it's just 99.99%, the 0.01% waste heat will melt the ship. $\endgroup$
    – Mike Scott
    Sep 10, 2023 at 10:39
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    $\begingroup$ You absolutely cannot use the classical kinetic energy equation when you're dealing with significant fractions of the speed of light. You must use the relativistic kinetic energy equation if you hope to be accurate. $\endgroup$
    – notovny
    Sep 10, 2023 at 13:27
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    $\begingroup$ "250 metric tons, which is approximately 250,000 kilograms" $\endgroup$
    – Fattie
    Sep 10, 2023 at 15:32

2 Answers 2


I don't think you can mark a answer to be correct if it uses classical mechanics, but whatever. $\frac{1}{2}mv^2$ definitely does not work at these speeds anymore, plus there is a factor missing.

First of all, the ISV´s from Avatar are not realistic. These antimatter engines would be screamingly radioactive, and the crew/cargo section is right in the middle of the two jets. Leaving aside the radiant heat from this, the radiation should kill everything and transmutate all structures into lead. Also, the exhaust wouldn't be blue. In fact, it would be invisible from most angles due to Doppler effects.

Anyways, both you and the Monty guy used way to much abstraction. What you are trying to do is answer a question about energy. To kind of show the thought process, we know the ship has to presumably slow down, and we can assume the deceleration phase is symetric to the acceleration phase. The total energy is just:

$E_{total} = 2E_{\Delta V}$

Ehere $E_{\Delta V}$ is the energy needed to change the velocity for the acceleration / deceleration phase.

Leaving aside abstraction, we know $E_{\Delta V}$ is a quantity of kinetic energy, so $E_{\Delta V}$ has to follow some;

$E_{\Delta V} = m_0c^2 \left( \frac{1}{\gamma}-1 \right)$ where $\gamma = \sqrt{1-\frac{v^2}{c^2}}$

This is in essence the kinetic energy, the amount of Work, the ship / propulsion system has to do. If we plug in your numbers, we get something like;

$E_{\Delta V} = 250000000kg \times 299792458\frac{m}{s}^2 \times \left( \frac{1}{\sqrt{1-\frac{(0.7 \times 299792458\frac{m}{s})^2}{299792458\frac{m}{s}^2}}} -1\right)$

$E_{\Delta V} = 9\times10^{24}$

Then of course:

$E_{total} = 2E_{\Delta V}$

$E_{total} = 2 \times 9\times10^{24}$

$E_{total} = 1.8\times10^{25}$

That is a lot of energy.

To figure out how much antimatter we need, we can once again drop abstraction and just use the equation. As $E = mc^2$ tells us, energy and mass are truly the same quantity. Since antimatter annihilation, theoretically, converts 100% of mass to energy, this is one of these times where $E = mc^2$ can be used. We know what $E$ is, we know the speed of light.

$E = mc^2$

$\frac{E}{c^2} = m$

$\frac{1.8\times10^{25}}{299792458\frac{m}{s}^2} = 200140042 kg$

$m \approx 200000 t$

Well that's a downer, but actually not that surprising. 0.7c is very fast, and at this point the kinetic energy is well into its exponential phase. Plus you need to slow down again. How long this burn lasts, by the way, is really irrelevant. To illustrate the point, if we lower the desired speed to 0.5c, we get;

$\dot{m} = 77350 t$

Antimatter works best at medium relativistic speeds. You won't get to 70% with it, well if you don't plan on 70% of your ship being antimatter.

That being said, these calculations are very wrong / idealized as a physicist might say. Annihilation is neither 100% effective, nor can be converted into a change in momentum. Indeed, annihilation produces omnidirectional radiation, most of it in the form of gamma rays and a lot of heat. You would be lucky to covert 50% of the annihilation energy into thrust. The rest will just be heat, so slap a $\frac{2}{1}$ factor on all of these. Alternatively, heating up a propellant using annihilation might yield 70% efficiency, but now you have to carry a propellant with you.

Really, to get to these speeds all you can really do is use power beaming and staging of some sort.

EDIT; I just saw you said 250 tons for the entire spacecraft. In that case:

$m \approx 200 t$

$\dot{m} = 77.3 t$

Its just linear from the perspective of the ship but no need to get into that. The math is the same, just $m_0$ changes.

EDIT #2; As quarague pointed out, this math has a flaw in that it sort of assumes the mass of the propellant is 0. Meaning the start and end mass of the ship is constant. So ill include some more stuff to give a more accurate answer;

In essence, we will be using the equation;

$\frac{M}{m} = \gamma(1+\frac{v}{c})-1 $

Note, $\frac{M}{m}$ is the mass fraction, we are interested in $M$ as $m$ is our payload / vessel mass. Solving for $M$ we get;

$M = e^{\frac{aT}{c}}-1$

Which again is a sort of proportionality constant, if $M$ is 2, that means for ever 1kg of $m$ we need 2kg of $M$ or fuel. You might notice a conspicuous $T$ in there. This is the burn time, or how long we plan to accelerate. You might also notice that the $aT$ term is just some $\frac{x}{c}$ and we actually simplify further to;

$M = e^{\frac{v}{c}}-1$

This is more illustrative since our result is independent of acceleration and Time, which conceptually makes sense. While in reality a less powerful engine will almost always be more efficient (see Ion Drives) we dont really care for that. Finally, we need to consider that we want to slow down again, which works out to;

$M = e^{\frac{2v}{c}}-1$

Using this, for $v = 0.7c$ we get $M = 3.055$. Again this is not actually true because we are basically doing the $E = mc^2$ math from above while taking acceleration into account. But this is all still 100% efficient. But this tells us that a Ship, using Antimatter, would have to have 3 times as much fuel as mass to reach 70% the speed of light and slow down again.

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    $\begingroup$ Ahh yes. thank you. The reason I marked it answered cause I didn't expect this. I got some of the maths to show and that's it for me as my world doesn't focus a lot on the deep math stuff. But you are right though that the ISV class ships are not realistic in the sense that they are not long enough. If I recall, The actual original design for the ship is actually 4KM long. Which I suspected is for the dissipation of the gamma rays the engines will fart out. James Cameron just pulled an the "unobtanium" trope to wish all the complexity away so theres that:) $\endgroup$
    – Ashimix
    Sep 10, 2023 at 15:41
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    $\begingroup$ Ashimix ... you may be (wildly) underestimating the un-realistic-ness of the ships. "original design for the ship is actually 4KM long..." It's likely that if we magically said "Oh, they are 10,000 km long" it wouldn't help matters. note that just to begin with, even with "ultimate handwaving" it would be INCREDIBLY hard to hold, contain, bottle the odd spoonful of anitmatter - far less "1000s of tons" of antimatter $\endgroup$
    – Fattie
    Sep 10, 2023 at 15:54
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    $\begingroup$ @ErikHall If I recall, The tractor-configuraiton idea liek this was actually originally from Charles Pellegrino. In which in his design, His ship is 10km long with engines on both the front and back. In terms of the "gamma ray fart death"... You have to remember that the ships engines are actually angled by 3 degrees so that the plume forms a V shape to go atleast angle it away. Do you think that that helps by a bit? $\endgroup$
    – Ashimix
    Sep 10, 2023 at 16:00
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    $\begingroup$ Your computation still misses an important aspect, namely that you need to accelerate the fuel as well. You computed 200t of antimatter to accelerate 250t of cargo but that means at the beginning you need to accelerate 450t of total mass so these 200t won't get you quite to the speed you want. This kind of computation is usually done via the 'rocket equation' $\endgroup$
    – quarague
    Sep 11, 2023 at 6:34
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    $\begingroup$ It doesn't. You compute the energy needed to accelerate 250t and then compute how many tons of antimatter that would correspond to. That is not enough. The mass of the space ship is fuel + cargo and is not constant over time. The rocket equation is a differential equation and people usually use some calculator for solving that (easy to find with google). $\endgroup$
    – quarague
    Sep 11, 2023 at 7:11

Two problems:

  1. 250 tons of cargo, not 250 tons of ship. I'd be more inclined to believe 1000 tons of ship including cargo.

  2. Somehow, your calculations were wrong... I got 61,250 kg of matter from your m ≈ [(1/2) * 250,000 kg * (0.7c)²] / c² formula, meaning that you'd need 30.625 tons of antimatter.

However, since I believe that you have the mass wrong:

m ≈ [(1/2) * 1,000,000 kg * (0.7c)²] / c² ≈ 245,000 kg

Since the 245t is both matter and antimatter, you'd need about 122.5 tons of antimatter. That's an awful lot of antimatter... and that's just to accelerate the ship.

Since we'd have to decelerate the ship at the other end of the trip, we could easily assume an equal amount of antimatteer again.

So, you would likely need the full 245 tons of antimatter.

Of course, as you used the antimatter, your ship would get lighter, and need less energy to accelerate, but I'd err on the side of having the extra fuel, just in case of emergencies or miscalculations.

I'm assuming C=300,000,000 m/s, which is a little pessimistic. This also doesn't take relativistic effects into consideration.

  • $\begingroup$ Holy shit. I made a mistake. I literally calculated 0.7 the decimal as opposed to 0.7 light speed (209854720.6) and only used c as opposed to c². its indeed 61,250 kgs of antimatte for the 250,000. Question though. when it says 61k kg of antimatter. is it literally 61,000 kgs of antihydrogen? Thats how much the antimatter propellant capsule woudl weight if it was on earth??? $\endgroup$
    – Ashimix
    Sep 10, 2023 at 5:11
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    $\begingroup$ One of the unfortunate problems with trying to stick to real physics as we understand it today to deal with future tech is that 245 tons of antimatter makes the ship pretty easy to destroy. Just puncture the tanks. Foof! $\endgroup$
    – JBH
    Sep 10, 2023 at 5:48
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    $\begingroup$ @JBH I can't imagine why they'd need FOOF (en.wikipedia.org/wiki/Dioxygen_difluoride)... surely the antimatter would be better... and worse. $\endgroup$
    – Monty Wild
    Sep 10, 2023 at 6:10
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    $\begingroup$ Since you're working with non-trivial fuel fractions, you can't just double the fuel requirements to account for deceleration, you need to use the full Tsiolkovsky rocket equation, or the relativistic variant if you're flying in a relativistic universe. $\endgroup$
    – Mark
    Sep 10, 2023 at 19:09
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    $\begingroup$ @MontyWild +1, BTW, for the chemistry joke. Heaven help us if anybody is using FOOF. $\endgroup$
    – JBH
    Sep 13, 2023 at 2:20

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