# What is the mathematical function, accounting for special relativity, for a graph of velocity vs. time?

Newton believed the work done on a body is equal to a net force applied to it multiplied by its displacement. He described energy as that which is capable of getting work done. So, if the body's initial velocity is zero, kinetic energy is equal to work done. He also believed $F=ma$ so his formula for energy was as follows... $${E_k} = {m_o}ad % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaGaamyramaaBaaaleaacaWGRbaabeaaki % abg2da9iaad2gadaWgaaWcbaGaam4BaaqabaGccaWGHbGaamizaaaa % !34FA!$$For a graph with velocity on the $y$-axis and time on the $x$-axis, the area under a function would be its displacement. For Newton, and assuming the body's initial velocity is zero and under uniform force, the function was:

$$v(t) = at % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG0bGaaiykaiabg2 % da9iaadggacaWG0baaaa!343B!$$

I take it Newton thought the velocity of a body is boundless; that is, if a uniform net force is applied to a body forever, it will increase in velocity forever.

Special relativity, on the other hand, says the graph tends towards the speed of light. So, assuming an initial velocity of zero and uniform force, what would be the function for this?

Force is the rate of change of momentum, $dp/dt = F$, so with a constant force you have $p = Ft$.

Substitute in the relativistic equation for momentum,
$$p = \frac{mv}{{\sqrt{ 1 - v^2/c^2} }}$$

and then solve for $v$ to get

$$v(t) = \frac{\frac{F}{m}t}{{\sqrt{ 1+\frac{F^2t^2}{m^2c^2}}}}$$

For small times, this reduces to the linear Newtonian formula $$v = \frac{F}{m}t$$

and for large times it approaches $c$ from below, just as you expected.

• That's awesome! You wouldn't believe how many hours I've "wasted" trying to figure it out myself. On the other hand, people who know me would not be surprised. Thing is, if the function can be integrated (it looks very difficult -- if not impossible), it will tell us the work done on any interval with the formula W=F(displacement) displacement is the area under the function. Thank you so much Paul for your contribution. – Michael Lee Mar 12 '17 at 4:13

No.

Its kinetic energy is $\frac{1}{2}m v^2$ (or its relativistic equivalent, which would be $E-mc^2$). A body may not be accelerating yet still have plenty of kinetic energy: an example would be a spaceship travelling at constant speed to Mars, with $0$ acceleration but certainly quite a bit of $\frac{1}{2}m v^2$ (in the non-relativistic limit).

By the work-energy theory, the work done by a constant force over a distance $d$ is $m\times a\times d$, but this is the change in kinetic energy resulting from the work done by the constant force. Moreover, if the force is not constant, then one needs to find in general $W=\int \vec F\cdot d\vec \ell$ as the change in $E_k$.

• Okay, now I follow you. The total energy of a body has two components, that from its velocity (one half m/v^2) and another from its mass (e=mc^2). Your my Hero Zero! – Michael Lee Mar 12 '17 at 6:13