# Relativistic Mass including exponential decay

So from what I gather, relativistic mass = $m_0\gamma$ where $\gamma$ is the lorentz factor.

So if i have a mass that is .5 at rest then it is safe to say that the relativistic mass will be 1 if it goes at $\frac{\sqrt{3}}{2}c$

My question is what happens if that .5 is actually a radioactive isotope and is decaying while speeding up? Then at what speed will it approximately 'equal' 1?

Thanks

• use a rocket not an isotope, and you need to make the question clearer. As it is, there is no question here. Jun 3, 2012 at 19:44
• I think you are thinking that radioactive decay makes the mass decrease exponentially. That's not so, the decay is at some sharp time, when you detect the outgoing particle. You are really thinking of a rocket, where there is ejected mass, and you should make the question more explicit, so that you isolate the actual confusing part, because I don't know what it is right now. Jun 4, 2012 at 6:07

If you did want to take into account the very slight change in mass from radioactive decay, assuming you treat any emitted alpha or beta particles as lost mass, you'd just replace the constant $m_0$ with the variable mass of the radioactive isotope as it decays.
• How would you replace the constant m_0 with the variable mass as it decays? Would you ... take the integral or ...? Jun 3, 2012 at 19:59
$$m_{e0}\simeq 0.511 \frac {Mev}{c^2}$$ this case could be electron mass. when relativistic factor is equal two $\gamma=2$ which means the speed parameter of electron is equal to $\beta=\frac {\sqrt 3}{2}$. so relativistic electron will be: $$m_{e}=m_{e0}.\gamma\simeq 1.0 \frac {Mev}{c^2}$$. that means kinetic energy of electron is equal to rest energy of electron: $$k_e=m_{e}-m_{e0}=m_{e0} (2-1)=m_{e0}$$ and momentum of electron will be: $$p_e=m_{e0}\sqrt 3 .c\simeq 0.885 \frac {Mev}{c}$$