# Does the mass of a particle increase with increasing velocity [duplicate]

In the chapter of magnetic effect of current I came across cyclotron limitations.

Statement:

The cyclotron cannot accelerate the particles to velocities as high as comparable to the speed of light.

Reason:

The reason is that at these velocities the mass of a particle increases with increasing velocity.

MyDoubt:

The reason for the statement is difficult for me to digest since I have never seen a particle or read about such a particle that follows this kind of reason as mentioned.

Can you provide some answers related to this?

## marked as duplicate by John Rennie, stafusa, Kyle Kanos, Mike, Whit3rdFeb 18 '18 at 10:49

• – John Rennie Feb 17 '18 at 6:59
• @johnrennie, I deleted my answer . – niels nielsen Feb 17 '18 at 7:39
• @calculus programmer, I leave you in John's capable hands. -Niels – niels nielsen Feb 17 '18 at 7:39
• @nielsnielsen I mean no disrespect to anyone.You and John are both great 👍🏻 at answering questions.Please don’t mingle in small quarrels.Students like us need teachers like you.🙂 – user184590 Feb 17 '18 at 7:43

The inertia of the electron, as measured by the change in velocity wrought by a given impulse and as seen in the cyclotron rest frame, does indeed increase with the speed of the electron.

In the early days of relativity (i.e. more than 80 to 90 years ago now) people did indeed describe this phenomenon as an increase in the electron's relativistic mass: one thought of the kinetic energy as being stored up as extra mass of the electron. This idea, although not wrong, is awkward, as I explain here. In particular, there's no way to make an increasing mass convention fit smoothly with other Newtonian concepts. Do we measure it through inertia? Well, OK, but now our definition of mass depends on the angle between the 3-force and the 3-velocity vector: a particle resists a shove more if the shove is along its direction of motion! Do we measure it through increase in total energy? That's possible, and that's what was done in the past, but, as described, it doesn't mean the same thing as inertia.

The modern description simply uses four vectors, and then Newton's second law behaves itself as it ought; it reads $\vec{F} = m_0\,\vec{A}$ where now $\vec{F}$ and $\vec{A}$ are the force and acceleration four vectors and the electron mass $m_0$ is the Lorentz invariant electron rest mass.

So now, what should our description of the cyclotron phenomenon be? Simply that the Lorentz factor $\gamma$ that relates the four acceleration to the more readily measured rates in our frame changes. Indeed:

$$\vec{A} =\frac{d\vec{U} }{d \tau} = \gamma(\vec{u}) \left(\frac{d{\gamma}(\mathbf{u})}{dt} c, \frac{d{\gamma}(\vec{u})}{dt} \vec{u} + \gamma(\vec{u})\, \vec{a} \right)$$

where $\vec{a}$ and $\vec{u}$ are the co-ordinate accelerations and velocities (as we'd measure them by our clocks and distances in the cyclotron rest frame).

The description you cite is an old-time attempt to absorb the speed dependent $\gamma$ factor into the mass and describe the phenomenon as a variation in that mass rather than leaving everything in a covariant description.

• Thank you for this. What is your take on the views expressed in math.ucr.edu/home/baez/physics/Relativity/SR/mass.html? – niels nielsen Feb 17 '18 at 8:06
• oops I forgot to flag you, will retype my comment: what is your take on math.ucr.edu/home/baez/physics/Relativity/SR/mass.html – niels nielsen Feb 17 '18 at 8:20
• @niels Nothing there is wrong, but then there was never anything that prevents you from doing physics with relativistic mass. But the authors glosses over some points that are not so convenient. For instance, the argument that by using relativitistic mass one can have the mass of the system equal to the sum of the masses of the parts depends crucially on there being zero potential energy in the system, and as soon as potential energy enters the picture you have to build a new rule. The invariant formulation uses a single rule for all systems. – dmckee Feb 17 '18 at 15:38
• thanks for taking the time to explain this to me. best regards, niels – niels nielsen Feb 17 '18 at 21:12