I want to visualize the energy required to continue accelerating near the speed of light. If I use some amount of energy to accelerate to 0.5c, what speed will I be going if I use the same amount of energy to accelerate again? I tried taking a stab at calculating this, but I'm stuck by the fact that I don't know enough calculus (which I think I need). I'm making some assumptions based on the Lorentz factor chart I've seen (the blue line in the image below).

lorentz factor

Lets say I've traveled from a stationary position where my Lorentz factor is 1, to a velocity where it's 1.2, which is some segment I'll call "a". If I accelerate with the same amount of energy again, I'll arrive at the end of segment "b". Does it take double the energy to reach the end of "b" as it does to reach the end of "a" from a standstill?

If this is all correct, then I think I need help coming up with the calculation for getting the velocity using lengths along the line defined by the Lorentz factor equation.

If this is not correct, then hopefully what I'm trying to accomplish makes sense at least!

Note I'm not interested in the actual energy requirements or the mass of the ship. Assume a perfect ship that accelerates without losing mass.


2 Answers 2


I'm going to do all calculations in the rest frame (the frame the rocket is in before it starts accelerating). This way, I can use the same equations that are used for particle accelerators.

If you want a particle of mass $m$ traveling at speed $v$, you need to give it a kinetic energy $K$ of $$K = \left(\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} - 1\right)mc^2 \qquad (1)$$ where $c$ is the speed of light. To find out how fast an object moves when it is given a certain amount of kinetic energy, we can solve the above equation for $v$: $$\frac{v}{c} = \sqrt{1-\left(\frac{1}{\frac{K}{mc^2}+1}\right)^2}. \qquad (2)$$ Now, we can take any kinetic energy, plug it into this equation, and get the resulting speed. I left $v/c$ on the left side since it makes more sense to talk about fractions of the speed of light in this context.

So, the amount of kinetic energy needed to get a mass up to half the speed of light is (using Equation 1): \begin{align} K &= \left(\frac{1}{\sqrt{1 - \frac{(c/2)^2}{c^2}}} - 1\right)mc^2 \\ \\ &= \left(\frac{2}{\sqrt{3}} - 1\right)mc^2 \approx 0.155mc^2 \end{align} or, about 15.5% of the rest energy of the particle.

If we add this amount of energy again, the kinetic energy will be $$K = 2\left(\frac{2}{\sqrt{3}} - 1\right)mc^2.$$ We can now plug this into Equation 2 to get the final speed: \begin{align} \frac{v}{c} &= \sqrt{1-\left(\frac{1}{\frac{2\left(\frac{2}{\sqrt{3}} - 1\right)mc^2}{mc^2}+1}\right)^2} \\ \\ &= \sqrt{1-\left(\frac{\sqrt{3}}{4-\sqrt{3}}\right)^2} \end{align} which is approximately $$\frac{v}{c} = 0.646.$$ Doubling your kinetic energy in this case gets you about 30% more speed, as opposed to 41% classically (at low speeds).

If you want to do this problem in terms of the Lorentz factor $\gamma = 1/\sqrt{1-v^2/c^2}$, it gets easier since the equation for kinetic energy in terms of $\gamma$ is $$K = \left(\gamma - 1\right)mc^2.$$ The change in kinetic energy in terms of changes in $\gamma$ is $$\Delta K = \left(\gamma_2 - 1\right)mc^2 - \left(\gamma_1 - 1\right)mc^2 = \left(\gamma_2 - \gamma_1\right)mc^2 = \left(\Delta\gamma\right)mc^2.$$ From this, we can see that it takes the same amount of additional energy to go from a Lorentz factor of 1.0 to 1.2 as to go from 1.2 to 1.4. Converting this to a velocity requires the messier math seen above.

  • $\begingroup$ Thank you so much!!! This is EXACTLY what I needed! I'm very surprised even at speeds like 0.01c you get noticeable diminishing returns on acceleration (according to the stationary observer). If I did my math right using the same amount of energy to go from 0.00c to 0.01c will only further your speed to 0.014c. Wild! $\endgroup$
    – Brian O.
    Dec 14, 2018 at 18:27
  • $\begingroup$ Oh wait, I now see that you'd expect a 41% increase at low speeds, I guess I don't understand how kinetic energy normally increases velocity even at non relativistic speeds. Haha, well now I have something else to try and understand! $\endgroup$
    – Brian O.
    Dec 14, 2018 at 18:34
  • $\begingroup$ @BrianO. At low speeds, Newtonian physics and relativity give experimentally indistinguishable answers. So, for $v \ll c$, $K = (1/2)mv^2$, so $v = \sqrt{2K/m}$. Doubling the kinetic energy gets 1.41 times the original velocity. $\endgroup$
    – Mark H
    Dec 15, 2018 at 3:04

When you talk about quantities in SR you always need to be clear about what observer you are referring to. For example do you mean the observers on the accelerating rocket experience a constant acceleration, or that the observer on Earth watching the rocket observes it to have a constant acceleration.

The reason this matters is that the accelerations observed on the rocket and on Earth are very different. This is somewhat involved to derive properly, but we can give a rough derivation as follows. Remember that acceleration is metres per second squared i.e. distance divided by time squared. But as observed from Earth distance for a fast moving observer are Lorentz contracted:

$$ L_{Earth} = \frac{L_{rocket}}{\gamma} $$

And times observed on Earth are dilated:

$$ T_{Earth} = \gamma T_{Rocket} $$

Since $a = L/T^2$ we end up with:

$$ a_{Earth} = \frac{a_{Rocket}}{\gamma^3} $$

where $\gamma$ is the Lorentz factor that you show in your graph:

$$ \gamma = \sqrt{\frac{1}{1 - v^2/c^2}} $$

So if the observers have a rocket motor that produces a steady acceleration of e.g. $1g$ the rest of us are going to observe the acceleration of the rocket to decrease to zero as the rocket approaches the speed of light.

This is why you need to be clear what power you are asking about. For a constant acceleration as measured on the rocket the power is constant since the acceleration is constant while for a constant acceleration as measured on Earth the power required would go to infinity in a finite time.

Assuming you're interested in what happens on Earth then this is actually easy to calculate. If the acceleration is constant as observed from Earth then the velocity of the rocket is just:

$$ v = at \tag{1} $$

The relativistic kinetic energy is:

$$ E = (\gamma-1)mc^2 \tag{2} $$

And the power is just $dE/dt$. So substitute equation (1) for the velocity into equation (2) then differentiate with respect to time. This is messy but straightforward.

If you're interested in finding out more about acceleration rockets in relativity then I recommend Phil Gibbs' article on the relativistic rocket.


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