Question

Suppose you have a collection of particles, all moving in the x direction, with energies ${\mathbf{E}}_{\mathbf{1}}\mathbf{,}{\mathbf{E}}_{\mathbf{2}}\mathbf{,}{\mathbf{E}}_{\mathbf{3}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}}$. and momenta${\mathbf{p}}_{\mathbf{1}}\mathbf{,}{\mathbf{p}}_{\mathbf{2}}\mathbf{,}{\mathbf{p}}_{\mathbf{3}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}}$ . Find the velocity of the center of momentum frame, in which the total momentum is zero.

Short answer

The velocity of the center of momentum frame is$v=\frac{c^2(p_1,p_2,p_3,...........)}{(E_1,E_2,E_3........)}$

Step-by-step solution

Expression for the center of momentum frame:

Write the expression for the center of the momentum frame.

${\overline{\mathbf{p}}}_{\mathbf{T}}\mathbf{=}\mathit{Y}\left(p_T-\frac{\beta E_T}{c}\right)$ …… (1)

Here, $\gamma$is the Lorentz contraction, $p_T$is the total momentum of the particle, $\beta$is the Lorentz factor, $E_T$is the total energy of the particle, andc is the speed of light.

Expression for the velocity of the center of momentum frame:

Write the formula for the Lorentz factor.

$\beta =\frac{v}{c}$

It is given that the total momentum is zero, i.e.,localid="1654751633192" ${\overline{p}}_T=0$

Substitutelocalid="1654752509058" $p_T=0$and $\beta \frac{v}{c}$in equation (1).

$$\begin{gathered} 0=\gamma \left(p_T-\frac{\beta E_T}{c}\right) \\ 0=\left(p_T-\frac{\beta E_T}{c}\right) \\ {p}_T=\left(\frac{v}{c}\right)\left(\frac{E_T}{c}\right) \\ v=\frac{p_T{c}^2}{E_T} \end{gathered}$$

Substitute $E_T=E_1,E_2,E_3,............$and$p_T=p_1,p_2,p_3,..........$in the above expression.

$v=\frac{c^2(p_1,p_2,p_3,.............)}{(E_1,E_2,E_3,...........)}$

Therefore, the velocity of the center of momentum frame is

$V=\frac{c^2(p_1,p_2,p_3,.............)}{(E_1,E_2,E_3,.........)}$