Mass in special relativity

I have just got a query about how this equation works if its right.

We have Newtonian Physics saying $F=ma$,

According to the 'Mass in special relativity' the mass changes according to $$m= \dfrac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}.$$

So, our

$$F=\dfrac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} \cdot \dfrac{dv}{dt}.$$

$v^2$ isn't constant. It is moving with an uniform acceleration. So, I find out the average velocity,

Since the $v=at$, Cause: $u=0$

$$\dfrac{\int_{t_1}^{t_2}f(t)dt}{t_2-t_1}=v_{avg}$$

$$\dfrac{a(t_2^2-t_1^2)}{2(t_2-t_1)}=v_{avg}$$

My Force equation becomes $$F=\dfrac{m_0}{\sqrt{1-\frac{\dfrac{a^2(t_2^2-t_1^2)^2}{4(t_2-t_1)^2}}{c^2}}} \cdot \dfrac{dv}{dt},$$

or simply

$$\dfrac{m_0}{\sqrt{1-\frac{v_{avg}^2}{c^2}}} \cdot \dfrac{dv}{dt}.$$

What happens when I travel faster than light? Or equal the speed of light. My mass doesn't become '$\infty$' for sure. What will be the correction factor if my acceleration is not uniform.

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By the way, if something isn't a question, it doesn't belong on this site. Now, it sounds like this is a question, but it's not clear exactly what it is you're asking. Try this: can you rephrase your question in a way that doesn't reference traveling faster than light, or at the speed of light? – David Z May 11 '13 at 9:19
– Dimensio1n0 Jul 27 '13 at 11:51

You can't travel faster than light. And anything with mass can't travel at the speed of light, so if you set $v=c$ in any of those equations you need $m_0=0$, and so you would be talking about a photon or some other massless particle. As $v\to c$ in those equations, the force goes to infinity, so you can never reach $c$ for $m_0\ne 0$.

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the force goes to infinity - and more than that, some important integrals of force go to infinity as well, that is, the work and the impulse of force, which would go to the energy and to the momentum of the accelerated body. There would be mathematically possible situation when function goes to infinity but its integral does not. – firtree May 11 '13 at 7:52
What happens when acceleration is not uniform? What will be the correction factor? – Inceptio May 11 '13 at 8:02
What happens when acceleration is not uniform? - The force is calculated from the acceleration value taken at the same instant, so it does not care if it is uniform or not. And integrals of force are tied to the energy and the momentum of the body, which are calculated from the speed and do not depend on the mode this speed has been reached. No way to get round the limitation - that becomes obvious if you consider the 4-vector geometrical picture. This is like acceleration is similar to walking on the surface of the planet, and speed $v>c$ is like just another planet, hanging over your head. – firtree May 11 '13 at 9:04

Your second equation is incorrect. In both Newtonian Mechanics and Relativistic Mechanics, force is the time rate of change of momentum:

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

Where, in special relativity, momentum is:

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

and $m$ is the invariant mass.

Thus, when the time derivative is taken, by the product rule:

$\vec F = m(\gamma \vec a + \dot \gamma \vec v)$

Note that, in general, the force vector is not parallel with the acceleration vector!

If the force vector is always parallel with the velocity vector, the force equation simplifies to:

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

Now, when you write:

It is moving with an uniform acceleration

you must actually be more specific. It appears that you're thinking about coordinate acceleration however, there is also the acceleration as measured by accelerometers (the proper acceleration) and this distinction is often not appreciated by SR "newbies".

While it is possible for the proper acceleration to be uniform, it isn't possible for the coordinate acceleration to be uniform as that would require unlimited force.

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The power must be $\dfrac{1}{2}$ ? Your equation seems dimension-ally incorrect – Inceptio May 11 '13 at 13:50
@Inceptio, if you asking about the exponent in the final equation, note that the Lorentz factor is dimensionless. – Alfred Centauri May 11 '13 at 14:03
Sorry. Didn't check it properly. – Inceptio May 11 '13 at 14:05

Maybe the thing you are seeking for is a dynamics of a tachyon. I will briefly introduce it. Below I use $m=\mathrm{const}$ as a rest mass or simply the mass (mass changing with speed is a pedagogical mistake), and $c=1$ - a usual convention to simplify formulas without information loss. $\gamma$ is an abbreviation for $1/\sqrt{1-v^2}$ and is called Lorentz factor (for the chosen $v$).

Mass $m$ of a particle plays that role that it fixes the relation between particle's energy and momentum: $E^2-|\vec{p}|^2=m^2$. Energy and momentum, taken together, make an energy-momentum four-vector, with an energy value counted as a 4th component (usually called 0th, and written at the beginning, in a $(E,\vec{p})$ or $(E,p_x,p_y,p_z)$ manner; graphically it is usually drawn vertical). This four-vector is important because it is tangent to the world line - a trajectory in space-time - and hence its direction gives a velocity and the speed: $v=p/E$.

A force applied to the particle deals with its energy-momentum four-vector. It tryes to change it. But the energy-momentum vector has a constraint shown above, so force cannot give it any value - only allowed ones. In the meantime, within that constraint, any value is reachable, and in many ways - with uniform acceleration, non-uniform, non-unceasing, and so on.

So we have to look closer at the energy-momenum values. For a non-zero mass, $E^2-|\vec{p}|^2=m^2$ describes the two-sheet hyperboloid (only an upper sheet is taken to get the correct energy sign). The energy-momentum vector cannot tilt lower than to slope ratio $E/p=1$ (actually it is always even higher, $E/p>1$), hence the limitation on the speed: it cannot be more than 1 (in our convention, that is the speed of light $c$). But we can consider a theoretical possibility of a particle that is different from the very beginning, $E^2-|\vec{p}|^2=\mathrm{const}<0$, that would formally give $m$ an imaginary value. Such a particle is called tachyon (though no such particles are found yet, and most physicists believe there would not be found any in the future - because of advanced reasons). It would "live" on the one-sheet hyperboloid, instead of the two-sheet one, and would never become "normal".

All the formulas for such particle stay the same, if we just assign $m$ and $\gamma$ imaginary values - and in fact imaginarities cancel in many formulas. For example, for the acceleration collinear with the velocity, $$F=m\gamma^3\frac{dv}{dt}$$ and for the acceleration perpendicular to the velocity, $$F=m\gamma\frac{dv}{dt}$$ (a total formula just decomposes the total force into collinear and perpendicular parts). Since $m$ and $\gamma$ are both imaginary, both $F$ and $dv/dt$ end up being real.

Besides that, tachyon shows some peculiar properties, following from its definition and formulas. It cannot move slower than light. It can move with infinite speed. It can change its direction of motion with respect to time - it can be accelerated to "faster than infinity", that would appear as a large velocity in the opposite direction, and in the preceding instants of time. In fact, an observer would see two tachions approaching each other and then disappearing. Or else, two tachions could appear from nowhere and start to move away from each other. (Remember that this picture ignores some other aspects and issues.)

Now some formulas so you can work with them yourself. SR particle mechanics is written with respect to the proper time, which given by $$d\tau^2=dt^2-(dx^2+dy^2+dz^2)\qquad\qquad\gamma\,d\tau=dt$$ Four-force is (by definition) $$f^\mu=(\gamma P,\gamma\vec{F})$$ where $P$ is a mechanical power of that force. Four-velocity is (by definition) $$u^\mu=(\gamma,\gamma\vec{v})$$ We would also need a derivative $$\dfrac{d}{d\tau}=\dfrac{dt}{d\tau}\dfrac{d}{dt}=\gamma\dfrac{d}{dt}$$ And now, the four-acceleration is (by definition) $$a^\mu=\dfrac{du^\mu}{d\tau}$$ (you can find it in explicit form), and the Newton's second law is $$f^\mu=\dfrac{dp^\mu}{d\tau}=m\dfrac{du^\mu}{d\tau}$$ (with $p^\mu$ meaning an energy-momentum four-vector). You can find the explicit equations for spatial values $\vec{F}$ and $\vec{a}$.

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