B Misuse of Velocity Addition: Troubleshooting

[Kairos]

Something seems wrong with my use of velocity addition:

A fly of rest mass $m_{0}$ in your reference frame (say a platform) is posed in a train passing with a velocity $v$ relative to the platform. The fly mass is now for you $m_{1} = m_{0} \gamma(v)$. Now in the train the fly is flying towards the front of the train with a velocity $v$ relative to the reference frame of the train, so $m_{2} = m_{1} \gamma (v) = m_{0} \gamma^2 (v) =\frac{m_{0}}{1-(v/c)^2}$. This result is different from $m_{2} = m_{0} \gamma(V)$ where $V$ is the composition of the train and fly velocities $V=\frac{2 v}{1+(v/c)^2}$, which gives if I am not mistaken $m_{2} = m_{0} \sqrt{\frac{1+(v/c)^2}{1-(v/c)^2}}$. What is wrong?

 

[Ibix]

First, a couple of points:

A fly of rest mass $m_{0}$ in your reference frame

Rest mass is an invariant. You do not need to specify a frame - everyone will agree it.

The fly mass is now for you $m_{1} = m_{0} \gamma(v)$.

"Mass" without any qualifier is usually taken to mean the invariant mass. I can infer here that you mean the relativistic mass, but strictly this statement is wrong as you've written it. The fly's mass is $m_0$, full stop. You should write "the fly's relativistic mass" if you want to talk about relativistic mass. It's also worth noting that relativistic mass has been a deprecated concept in mainstream physics for many decades now, and it's better not to use it. However, you could observe that the total energy of the fly is $E_1=E_0\gamma(v)=m_0c^2\gamma(v)$.

This result is different from $m_{2} = m_{0} \gamma(V)$ where $V$ is the composition of the train and fly velocities $V=\frac{2 v}{1+(v/c)^2}$, which gives if I am not mistaken $m_{2} = m_{0} \sqrt{\frac{1+(v/c)^2}{1-(v/c)^2}}$. What is wrong?

Getting to your question, what's wrong is your expectation that gamma factors should multiply. The energy of an object is the $t$ component of its four momentum (give or take a factor of $c$). In its rest frame, the four momentum is a four vector $(E_0/c,0,0,0)$ - that is, it has rest energy and its three momentum is zero. If I then boost the fly to some speed $v$ I can apply the Lorentz transforms to the four momentum and deduce that, in the frame of the train where the fly is doing $v$, the fly's four momentum is $(\gamma E_0/c,-\gamma v E_0/c^2,0,0)$. To get the fly's four momentum in my (platform) frame I need to boost again - but this time the $x$ component of the four momentum is already non-zero, so the $t$ component of the resulting vector is not simply $\gamma$ times what it was in the train frame.

The underlying reason for this is that rapidities add, not velocity or gamma. This turns out to be the Minkowski geometry equivalent of the Euclidean statement that angles add but gradients of lines don't.

 

[Kairos]

Thank you for your recommendations on the appropriate terms to use and for the explanation. Concerning my question, is the solution obtained with the velocity composition correct? that is $E_{2} = m_{0} c^2 \gamma (V)= m_{0} c^2 \sqrt{\frac{1+(v/c)^2}{1-(v/c)^2}}$

 

[Ibix]

I agree your expression for $V$ but I don't get the square root in the expression for $\gamma(V)$ (i.e., I think you've written $\sqrt{\gamma(V)}$).

 

[Kairos]

OK I'll redo the calculation
thanks a lot!