# Decayed object energy into kinetic energy

An object of mass M has half life of T, if the energy released by the decayed half(mc^2) converts into kinetic energy of the rest of object what would the speed be over time? I tried and found infinity m/s as t goes to infinity which is imposible. what is correct answer?

EDIT: supose decay is total annihilation, no other particles created. Half life is just a mesure of speed of changing mass into energy

M becomes 0.5M after T and that mass is 0.5M object's kinetic energy

• I've edited my answer responding to your edit. – Tausif Hossain Jul 6 '17 at 7:13
• "M becomes 0.5M after T and that mass is 0.5M object's kinetic energy". Not quite, you need to conserve momentum as well as energy. So some of that annihilation energy has to be released (eg as photons) in order to accelerate the remaining matter. See below for details. – PM 2Ring Jul 13 '17 at 15:44

You didn't show your equations, so it's hard to say why you got that unphysical answer. Perhaps you tried to add speeds linearly instead of using the proper relativistic formula for composition of velocities.

For collinear motion, the formula (which can easily be derived from the Lorentz transformations) is

$$w = \frac{u+v}{1+\frac{uv}{c^2}}$$

Let's say we have a body $A$ moving in the $x$ direction with constant speed $u$ according to an inertial observer $O$. A body $B$ moving in the $x$ direction that has speed $v$ in $A$'s reference frame will not have a speed of $u+v$ in $O$'s frame. Instead, its speed will be $w$ computed by the above formula.

To calculate the correct speed for your total annihilation problem, we just need to use the laws of conservation of energy and of momentum, along with the relevant formulae from Special Relativity for energy and momentum. Of course, this is a highly idealised scenario, since the photons produced in the annihilation will be emitted in random directions and it's not possible to force them to go in one direction: you'd need a reaction chamber lined with some amazing material that can reflect gamma photons perfectly, without producing waste heat (which gets emitted in random directions). But anyway...

Here are the equations we need for our calculations. Firstly, the energy–momentum relation of Special Relativity:

$$E^2 = (pc)^2 + (mc^2)^2 \tag{1}$$

where $E$ is the total energy, $p$ is the momentum, and $m$ is the (rest) mass of the object; $c$, of course, is the speed of light.

We also need the relativistic equation for momentum: $$p = mv\gamma \tag{2}$$

where $\gamma$ is the Lorentz factor: $$\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \tag{3}$$

Note that $$\gamma^2 = \frac{c^2}{c^2 - v^2} \tag{3a}$$

For a massless body (eg a photon), equation (1) simplifies to $$E = |pc| \tag{1a}$$

And by combining (1) and (2) we get $$E = mc^2\gamma \tag{1b}$$

for a body with non-zero mass. For an object at rest, that simplifies to the famous $$E = mc^2 \tag{1c}$$

Let the initial mass of the body be $m$, and its final mass (after some of it's annihilated) be $km$, where $0 < k \le 1$.

Let $E_i$ be the initial energy of the body, $E_f$ its final energy, and $E_l$ the energy of the emitted light. By conservation of energy, $$E_i = E_l + E_f \tag{4}$$

The initial momentum of the body (in the rest frame) is zero, let $v$ be its final speed and $p$ be its final momentum. By conservation of momentum, the momentum of the emitted light must be $-p$ and hence its energy is $pc$. The final mass of the object is $km$, so from (2) $p = kmv\gamma$

Thus

\begin{align}\\ E_i & = mc^2\\ E_f & = kmc^2\gamma\\ E_l & = kmvc\gamma\\ \end{align}

Note that $E_l/E_f = v/c$

Putting it all together: \begin{align}\\ E_i & = E_f + E_l\\ mc^2 & = kmc^2\gamma + kmvc\gamma\\ mc^2 & = kmc(c+v)\gamma\\ c & = k(c+v)\gamma\\ c^2 & = k^2(c+v)^2\frac{c^2}{c^2 - v^2}\\ k^2 & = \frac{c^2 - v^2}{(c+v)^2}\\ k^2 & = \frac{c-v}{c+v}\\ k^2c + k^2v & = c-v\\ v + k^2v & = c - k^2c\\ v(1 + k^2) & = c(1 - k^2)\\ v & = \frac{1 - k^2}{1 + k^2}\,c \tag{5}\\ \end{align}

Thus when $k=1/2$, $v=3c/5$

With a little more algebra, it can be shown that \begin{align}\\ \gamma & = \frac{1+k^2}{2k}\\ E_f/E_i & = \frac{1+k^2}{2}\\ E_l/E_i & = \frac{1-k^2}{2}\\ \end{align}

If the body is decaying exponentially, then it will experience constant acceleration, in the sense that if you were traveling on a spaceship powered by this process you'd experience an acceleration that feels just like a constant gravitational force. To show this, we need some calculus and the formula for composition of velocities given at the start of this answer. First, we'll rearrange equation (5) slightly.

\begin{align}\\ v & = \frac{1 - k^2}{1 + k^2}\,c\\ v & = \frac{k^{-1} - k}{k^{-1} + k}\,c\\ \end{align} Now let $k = e^{-\lambda T}$, where $\lambda$ is the decay rate and $T$ is the proper time of the object. Thus the body is undergoing exponential decay, according to clocks that are traveling with it.

\begin{align}\\ v & = \frac{e^{\lambda T} - e^{-\lambda T}}{e^{\lambda T} + e^{-\lambda T}}\,c\\ v & = c \tanh(\lambda T) \tag{5a}\\ \end{align}

Now we need to show that this is the same formula that arises for an object undergoing constant acceleration. It's sometimes said that the formulas of Special Relativity only apply to constant velocity, but that's not strictly true: they can be used for accelerating objects in flat spacetime, you just need to be careful. ;) The trick is to use a sequence of inertial reference frames that at each step match velocities with the accelerating body.

Let $a$ be the acceleration and $v$ the body's current velocity. We apply a small boost $a\Delta T$ to its velocity and use the velocity composition formula to see how much that numerically increases its velocity.

\begin{align}\\ v + \Delta v & = \frac{v + a \Delta T}{1 + va \Delta T / c^2}\\ \Delta v & = \frac{v + a \Delta T - v - v^2a \Delta T / c^2}{1 + va \Delta T / c^2}\\ \frac{\Delta v}{\Delta T} & = \frac{a (1 - v^2 / c^2)}{1 + va \Delta T / c^2}\\ \frac{\Delta T}{\Delta v} & = \left(\frac{1}{a}\right) \frac{1 + va \Delta T / c^2}{1 - v^2 / c^2}\\ \end{align}

Taking the limit as $\Delta T \to 0$, the $va \Delta T / c^2$ term vanishes, $$\frac{dT}{dv} = \left(\frac{1}{a}\right) \frac{1}{1 - v^2 / c^2}$$

Integrating, \begin{align}\\ T & = \frac{1}{a} \int \frac{dv}{1 - v^2 / c^2}\\ T & = \frac{c}{a} \tanh^{-1} \left(\frac{v}{c}\right)\\ v & = c \tanh \left(\frac{aT}{c}\right) \tag{6}\\ \end{align}

The constant of integration is zero because $v=0$ when $T=0$.

We see that equation (6) has the same form as equation (5a), with $\lambda = a/c$, so the acceleration of the object is simply $c\lambda$.

To answer your original (totally unphysical) question, where the annihilated mass is magically converted to kinetic energy of the object with nothing being emitted, we can use $mc^2 = kmc^2\gamma$, i.e., $k\gamma = 1$, and $v = c \sqrt{1-\frac{1}{\gamma^2}}$. So for $k = 1/2$ (half the original mass is annihilated), $\gamma = 2$ and $v = c \sqrt 3 / 2 \approx 0.866c$.
As the decay approaches the last atom and the energy from the original mass,$M$ all turns to kinetic energy, depending on the original mass,$M$ The equation to calculate such speed,$v$ is given below where $M$ is original mass and $m$ is mass of last remaining radioactive atom if all of $M$ is finally converted to kinetic energy.