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Can a particle moving below the speed of light be accelerated more and more until it is travelling at say c/2? IF so does it behave like electro-magnetic radiation?

user128932
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  • No. particles that have mass cannot. More can be found here http://physics.stackexchange.com/questions/1686/why-does-the-relativistic-mass-of-an-object-increase-when-its-speed-approaches – o0BlueBeast0o Apr 16 '14 at 03:12

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No, it cannot. The momentum of the particle must be finite in any reference frame where the momentum (for a massive particle) is given by

$$\vec p = \frac{m\vec v}{\sqrt{1 - \frac{v^2}{c^2}}} $$

Note that the momentum is undefined for $v = c$ or, to put it another way, the momentum goes to infinity as $v \rightarrow c$.

So, just as it would be impossible in Newtonian mechanics to accelerate a particle to 'infinite' momentum, it is impossible in relativistic mechanics.

The difference being that, in Newtonian mechanics, momentum and velocity are proportional whereas in special relativity, they aren't.

Alfred Centauri
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  • Can a particle traveling less than c be accelerated to arbitrarily near c without reaching c? – user128932 Apr 18 '14 at 02:27
  • @user128932, a particle at rest in one frame of reference is travelling arbitrarily close to $c$ in another. Motion is relative, not absolute. – Alfred Centauri Apr 18 '14 at 02:43
  • Well, can a particle traveling at about (1/2) c be accelerated arbitrarily close to c all within the same frame of reference? – user128932 Apr 18 '14 at 05:21
  • @user128932, to accelerate a particle arbitrarily close to c requires arbitrarily large energy. Assuming the final speed is ultra-relativistic, the required energy is approximately $\Delta E \approx mc^2 \frac{\beta}{\sqrt{1 - \beta^2}}$ which is unbounded as $\beta \rightarrow 1$ where $\beta = \frac{v}{c}$. – Alfred Centauri Apr 18 '14 at 13:38
  • Would this mean for all particles that are not leptons (if that's the right term) the speed of light is not the limiting speed of most physical events but that limit is slightly less than c? (as it wounld require arbitrarily large energy to get near c) – user128932 Apr 19 '14 at 04:58
  • Regarding the Lorentz contraction formula ( if that's what it's called); if v = c/2 then p = (m v)/(square root of 3/4) ; forgive notation. – user128932 Oct 18 '14 at 03:02