First of all, I have read other questions, and seen how an object gains mass just by gaining a lot of speed. But here, I am talking about losing some of that mass to convert the value into kinetic energy.

Let's use the starship Enterprise, firstly, because it is physically massive and well known, also because because I was just watching a documentary about physics on netflix, and it made me want to watch Star Trek. The person on the documentary said that mass and energy are equal (which is well known), but he also said that if someone were to be turned into pure energy, they would produce a massive amount (I never knew just how tremendous this value was), but that it was probably impossible to do.

I tried to look up special relativity with acceleration, but I could not understand it very well, and was unable to come to a conclusion [some places even said it was foolish to use acceleration in special relativity, and I did not know what to think].

The ship is supposed to combine matter and antimatter in order to have a net energy gain of 100% (which I am not asking anyone to validate), and can achieve relativistic speeds (obviously, because it travels between star systems with ease). If we say that the ship itself weighs something like one million kg [just something huge], how much energy would it have to use to accelerate to a any given relativistic speed? Also, would the ship itself be converted into kinetic energy just by traveling this fast, since it is so large?

If a larger value for acceleration proves to be impractical for humans, please address it nonetheless. Does it make a difference how fast you accelerate when considering this problem?

  • $\begingroup$ Related (especially part c): physics.stackexchange.com/a/109062/44126. Note that objects accelerated by external fields, like solar sails or electrons in accelerators, don't have to give up mass. $\endgroup$ – rob Aug 21 '14 at 3:35

If the ship is accelerating in a vacuum under its own power, it has to be a rocket (because momentum is conserved) and has to lose mass (because mass-energy is conserved). The theoretically most efficient type of rocket is a photon rocket, basically because light (or anything that moves at $c$) gives you the most bang for your buck in terms of momentum per unit energy. As mentioned in the article, the maximum speed is a simple function of the ratio of the rocket's initial and final rest mass. Solving for that ratio gives $$\frac{m_i}{m_f} = \sqrt{\frac{1+v/c}{1-v/c}}$$

So if you wanted to accelerate to, say, 0.5c, the ratio would be $\sqrt{3}$, and about $1 - 1/\sqrt{3} \approx 42\%$ of your ship's total mass would have to be fuel. If you want to decelerate again at the other end of your trip, the same ratio applies again, which means the total mass ratio is now $3$, and $1 - 1/3 = 2/3$ of your ship's total mass has to be fuel. For an ideal photon rocket, the answer is independent of the rate at which you accelerate.

If the ship's initial mass is $10^6\,\mathrm{kg}$, it will weigh only $3\cdot10^5\,\mathrm{kg}$ after the trip. The rest will be light with a total energy, in the initial and final rest frame, of $mc^2 \approx 6\cdot10^{22}\,\mathrm{J} \approx 2\cdot10^{16}\,\mathrm{kWh}$. Compare the world's current yearly energy consumption of about $2\cdot10^{13}\,\mathrm{kWh}$.

Real rocket ships are vastly less efficient than this.

  • $\begingroup$ Ahhh I see, I never thought of it like that! Yeah, the ship would be heavier from it's fuel wouldn't it. So it's energy wouldn't come from the mass of the actual working part. But, if the ship grows lighter as the fuel burns, do we have to use a different equation? Thanks for the answer, I feel smarter. $\endgroup$ – user56829 Aug 22 '14 at 0:45

The full form of Einstein's Mass-Energy equivalence relation looks like this:

$$E^{2} = m^{2}c^{4} + m^{2}p^{2}$$

where $m$ is the object's rest mass (the mass of the object when it is at rest relative to the frame of reference).

But you can get this from the normal $$E = mc^{2}$$ if you realize that mass is not constant - i.e. the equation should really be $E = m(v)c^{2}$ where mass now changes depending on what velocity you are moving at in your frame of reference. $m(v)$ is corrected for my the Lorentz transformation: $$m(v) = \dfrac{m_{rest}}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \to E = \dfrac{m_{rest}c^{2}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

A little bit of algebra gets you to the form of the equation above, with $p = mv$.

What this equation really tells you is that an object has intrinsic energy based on two things: one, simply having mass gives an object a huge amount of energy (with the value $mc^{2}$ - $c$ is very large so $mc^{2}$ is a lot of energy) and two, based on how fast it is moving relative to the frame of reference. Moving faster in the frame of reference gives the object more energy, but it does not take away mass to do so.

So, no. An moving/accelerating object does not give up any of it's "physical mass" (by this I assume you meant rest mass).


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