# Why does the inertial mass of an object rise when a force is exerted on it in a particle accelerator? [closed]

A particle in a cyclotron requires more and more force to maintain the same acceleration as it accelerates.

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## closed as not a real question by Manishearth♦Mar 24 '13 at 17:18

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this is a very general question. The short answer is special relativity holds because the velocities in accelerators are fractions of the velocity of light and mass grows as the velocity increases. en.wikipedia.org/wiki/Mass_in_special_relativity – anna v Mar 24 '13 at 14:45
The question itself is very broad, and will need an entire book about relativity to be given justice. Per that, I am closing it. – Manishearth Mar 24 '13 at 17:18

If you were able to apply a force indefinitely, and have the particle accelerate at a constant rate indefinitely, then the particle would at some point travel faster than light in vacuum. In fact, as its kinetic energy increases, it's effective inertial mass in the lab frame will increase so you could reasonably think that you need a larger force because there is a larger mass.

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In fact, particles can, in principle, accelerate indefinitely without exceeding c. For example, if a constant force is applied indefinitely, the particle's speed asymptotically approaches c while the momentum increases without bound. In any case, the speed will never get to or exceed c. – Alfred Centauri Mar 24 '13 at 16:57

Force is the time rate of change of momentum:

$$\vec F = \dfrac{d \vec p}{dt}$$

In Newtonian mechanics, momentum is proportional to velocity:

$$\vec p = m \vec v$$

So,

$$\vec F = m \vec a$$

But, in the context of Special Relativity, momentum is a more complicated function of velocity:

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

In fact, momentum increases to infinity as speed increases to c.

Now, when the force is parallel to the velocity, we have:

$$F = ma \dfrac{1}{(1-\frac{v^2}{c^2})^{\frac{3}{2}}}$$

When the speed is zero, we just have:

$$F_0=ma$$

So, as the speed increases, the force required to maintain the same acceleration increases:

$$F = \dfrac{F_0}{(1-\frac{v^2}{c^2})^{\frac{3}{2}}}$$

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