# Energy-Momentum Lorentz Transformation

Suppose a micrometeorite of mass $10^9$ kg moves past Earth at a speed of $0.01c$. What values will be measured for the momentum of the particle by an observer in a system $S'$ moving relative to Earth at $0.5c$ in the same direction as the micrometeorite?

# Book Solution

According to the book the momentum of the micrometeorite as measured by the Earth Observer is $$p_x = mu_x = 10^9 * 0.01c~ \rm kg \cdot ms^{-2}$$

# My solution

I argue the following. Define the momentum and Energy of the micrometeorite in the micrometeorite frame as $p_m$ and $E_m$ and are $0$ and $mc^2$ respectively (since in its frame its velocity is $0$).

Applying Lorentz Transformation from $A_m$ frame to $A_e$ (Earth frame and $0.01c$ is the frame velocity) gives us $$p_e = \gamma_{0.01} \left(p_m +\frac{v}{c^2}E_m\right)$$ Plugging in the respective values gives us

$$p_e = \gamma_{0.01}\frac{v}{c^2}E_m = \gamma_{0.01}mv \neq mu_x$$

Why is my approach to calculate the momentum in Earth frame is wrong?

• Are $v$ and $u_x$ the same thing?
– Mike
May 6, 2018 at 13:57
• Yes $v$ and $u_x$ are same thing. $u_x$ as given in the book is $0.01c$. And I have defined $v$ as frame velocity which is $0.01c$. May 6, 2018 at 13:58
• 10^9 kg is a "micrometeorite"?
– DWin
May 6, 2018 at 15:26

My guess is just that when they quote the mass, they are giving the "relativistic mass" measured relative to Earth (which is just $\gamma$ times the rest mass) — rather than giving the rest mass, which is what you're assuming. I think your assumption is quite reasonable, but I also think the book's interpretation is reasonable as well; they're just different ways of talking about things.
Moreover, the difference really is quite small. If you look at the significant digits, your answer and the book's are the same. In this case, $\gamma = 1.00005$, so multiplying the book's answer by that doesn't change it significantly.