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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?


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use a rocket not an isotope, and you need to make the question clearer. As it is, there is no question here. – Ron Maimon Jun 3 '12 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. – Ron Maimon Jun 4 '12 at 6:07

Radioactive decay doesn't appreciably change the mass of most isotopes. So for practical purposes, you can ignore the radioactivity when calculating relativistic mass increases.

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.

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The alpha particle is 4 nucleons, and carries off a significant mass. – Ron Maimon Jun 3 '12 at 19:43
How would you replace the constant m_0 with the variable mass as it decays? Would you ... take the integral or ...? – Eiyrioü von Kauyf Jun 3 '12 at 19:59
@Ron not significant for most radioactive isotopes, at least not for these purposes - usually the mass of the nucleon is considerably more than that of an alpha particle. – David Z Jun 3 '12 at 20:57
@Eiyriou: that would be a good thing to ask as a separate question. (It has nothing to do with relativity.) – David Z Jun 3 '12 at 21:01
@DavidZaslavsky: An alpha is 2% of the mass. The other things are less than 1 part in 100,000. It is not good to lump the two, since 2% is significant mass. If there is a double alpha decay, that's 4%. If there is a spontaneous fission, it can be considered up to 50% of the mass. It is just not right to say that radioactive decay has negligible changes in mass. – Ron Maimon Jun 4 '12 at 4:14

$$m_{e0}\simeq 0.511 \frac {Mev}{c^2}$$ 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: $$p_e=m_{e0}\sqrt 3 .c\simeq 0.885 \frac {Mev}{c}$$

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