Energy for acceleration towards speed of light is relative?

Question

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".

Answer 1

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.

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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%.

Answer 2

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:

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