This seems a very simple question - and I guess it will turn out to be so. It's a given that accelerating particles with mass towards the speed of light takes more and more energy (ultimately an an infinite amount to reach the speed of light). But as the particle's speed is relative to a given reference frame, why couldn't a particle at rest say on Earth, be considered to be moving at say 50% c compared to some remote galaxy. So it seems like the energy requirements for acceleration should also be relative - but clearly this is not the case. Would appreciate an answer as to what must be an obvious flaw in the foregoing "logic".

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    $\begingroup$ Kinetic energy is relative to the observer's reference frame. $\endgroup$ Jan 28, 2015 at 8:14
  • $\begingroup$ Don't bring remote galaxies into the picture, as there are remote galaxies moving faster than $c$ w.r.t. to us. The cosmic speed limit only applies to local (flat spacetime) observers. $\endgroup$
    – JEB
    Mar 4, 2018 at 18:11

2 Answers 2


why couldn't a particle at rest say on Earth, be considered to be moving at say 50% c compared to some remote galaxy.

Yes,the particle on earth can be moving at speed 50%c with respect to some remote galaxy.

Now If I set up a laboratory in that galaxy and try(somehow) to accelerate the particle further ofcourse I need less energy to accelerate the particle to near the speed of light(for example something like 0.9c or more than that ) in compared to a laboratory on earth(where the particle is at rest) because the particle is already moving at half of the speed of light with respect to the remote galaxy.

yes,energy needed to accelerate a particle near to the speed of light is relative in that sense because of the fact that speed is relative.

EDIT: If the speed of a particle in your frame is for example 2m/s and you want to accelerate it to 4m/s then you need less energy than mine if in my frame of reference the speed of the particle is 6m/s and I want to accelerate it to 8m/s.

This is an example from this earlier post Accelerating particles to speeds infinitesimally close to the speed of light?

Suppose you have got an electron (m = 9.1 × 10-31kg) to 99.99% of speed of light. This is equivalent to providing 36 MeV of kinetic energy. Now suppose you accelerate "a little more" by providing yet another 36 MeV of energy. You will find this this only boosts the electron to 99.9975%c. Say you accelerate "a lot more" by providing 36,000,000 MeV instead of 36 MeV. That will still make you reach 99.99999999999999%c instead of 100%.

  • $\begingroup$ Thanks for the response but I had a different angle on this. My thinking was that the closer a particle gets to the speed of light, the more energy it takes to increase its speed. So the point I am trying to understand is not what energy it would take to get it to the speed of light (which is infinite in both cases - albeit a "larger" infinity in the "at rest" perspective) but a given incremental increase seems to require a different energy level from the two reference points. How is this possible ? $\endgroup$ Jan 29, 2015 at 21:37
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    $\begingroup$ @Mark Kelleher :yes,If speed increases the mass(I mean relativistic mass) increases,so that you need more energy to accelerate it further. For example, to increase the speed from 2m/s to 4 m/s you need less energy than to increase the speed from 4m/s to 6m/s. see this post.physics.stackexchange.com/questions/1557/… $\endgroup$
    – Paul
    Jan 30, 2015 at 5:14
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    $\begingroup$ @Mark Kelleher: Question is seem to a good question and the ans is ''relativistic'' mass increases with speed.In a frame where the speed is less ,relativistic mass is also less and its relatively easy to accelerate the particle further. Is this the ans you are looking for? $\endgroup$
    – Paul
    Jan 30, 2015 at 12:38

If you have an object with mass $m$, and it's moving with speed $v$, then you're going to measure its kinetic energy as:

$$ T = E - mc^2 = (\gamma - 1)mc^2 $$

$$ T = [\frac 1 {\sqrt{1-\frac{v^2}{c^2}}}-1] mc^2$$

If you add some energy $\Delta E$, then $v$ increases, and as you point out other observers with see different velocities changes and different energy changes even: how is this possible?

Energy is just not a relativistic scalar--which means, it is different in different reference frames.

It is still useful because it is part of a 4 vector, the four-momentum:

$$ p_{\mu} = (E/c, {\bf \vec p}). $$

The four vector is the same in all reference frames--by which I mean: it is the four momentum for all inertial observers. Of course, the components change for different observers, per the Lorentz transformation:

$$ p'^{\mu} = \Lambda^{\mu}_{\nu}p^{\nu},$$

but the object itself remains the four-momentum, and all observers agree on its length:

$$ ||p|| = \sqrt{p_{\mu}p^{\mu}}=\sqrt{E^2/c^2-||{\bf p}||^2}=mc.$$

Note that the four momentum can be written in terms of the four velocity:

$$ p^{\mu} = mu^{\mu},$$

with the four velocity given by:

$$u^{\mu} = \gamma(c, {\bf \vec v})$$


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