Question

Question: Equation (2-38) show that four-momentum of a particle obeys a Lorentz transformation. If we sum momentum and energy over all particles in a system, we see that the total momentum and energy also constitute a four-vector. It follows that ${\left(E_{total}/c\right)}^{\mathbf{2}}\mathbf{-}{\mathit{P}}_{\mathbf{t}\mathbf{o}\mathbf{t}\mathbf{a}\mathbf{l}}^{\mathbf{2}}$ is the same quantity in any frame of reference. Depending on what is known, this can be a quicker route to solving problems than writing out momentum and energy conservation equations. In the laboratory frame, a particle of mass m and energy ${\mathit{E}}_{\mathbf{i}}$ collide with another particle of mass initially stationary, forming a single object of mass . (a) Determine the frame of reference where the after-collision situation is as simple as possible, then determine the invariant in that frame. (b) Calculate the invariant before the collision in the laboratory frame in terms of M and ${\mathit{E}}_{\mathbf{i}}$ . (You will need to use ${\mathit{E}}_{\mathbf{i}}^{\mathbf{2}}\mathbf{/}{\mathit{c}}^{\mathbf{2}}\mathbf{-}{\mathit{p}}^{\mathbf{2}}\mathbf{=}{\mathit{m}}^{\mathbf{2}}{\mathit{c}}^{\mathbf{2}}$ for the initially moving particle to eliminate its momentum.) Obtain an expression for M in terms of m and E_i . (c) Write out momentum and energy conservation in the laboratory frame, using ${\mathit{u}}_{\mathbf{f}}$ for the speed of the initially moving particle and for the speed of the final combined particle. Show that they give the same result for M in terms of m and ${\mathit{E}}_{\mathbf{i}}$. (Note: The identity ${\mathit{\gamma }}_{\mathbf{u}}^{\mathbf{2}}{\mathit{u}}^{\mathbf{2}}\mathbf{=}{\mathit{\gamma }}_{\mathbf{u}}^{\mathbf{2}}{\mathit{c}}^{\mathbf{2}}\mathbf{-}{\mathit{c}}^{\mathbf{2}}$ will be very handy.)

Short answer

Answer

(a) The frame of reference is the rest frame and the invariant is ${\text{M}}^{\text{2}}{\text{c}}^{\text{2}}$.

(b) The invariant is $\text{2}\left({\text{m}}^{\text{2}}{\text{c}}^{\text{2}}{\text{+ mE}}_{\text{i}}\right)$ and the expression for M is$\sqrt{\text{2}\left({\text{m}}^{\text{2}}\text{+}\frac{{\text{mE}}_{\text{l}}}{{\text{c}}^{\text{2}}}\right)}$ .

(c) The momentum and energy conservation equation are $\text{M =}\sqrt{\text{2}\left({\text{m}}^{\text{2}}\text{+}\frac{{\text{mE}}_{\text{i}}}{{\text{c}}^{\text{2}}}\right)}$ and gives the same result for M in terms of m and E_i.

Step-by-step solution

Identification of the given data

The given data can be listed below as:

• The mass of the first particle is m.
• The Energy of the particle is E_i .
• The mass of the second particle is M .

Significance of the Lorentz transformation

The Lorentz transformation is beneficial for describing a phenomenon of high speed which approaches at light’s speed. It mainly expresses the concept of light and space.

(a) Determination of the reference frame and invariant

The simplest work frame is the particle’s rest frame. However, having the velocity zero, mostly the equations are being reduced becomes handy. In the example given in the question, the after-collision mass is which works in the rest frame of the particle. Hence, there will be zero momentum and the rest energy will be the same as the invariant quantity.

$$\begin{gathered} {\left(\frac{{\text{E}}_{\text{t}}}{\text{c}}\right)}^{\text{2}}-{\text{p}}_{\text{t}}^{\text{2}}={\left(\frac{{\text{Mc}}^{\text{2}}}{\text{c}}\right)}^{\text{2}} \\ ={\text{M}}^{\text{2}}{\text{c}}^{\text{2}} \end{gathered}$$

Thus, the frame of reference is the rest frame and the invariant is ${\text{M}}^{\text{2}}{\text{c}}^{\text{2}}$.

(b) Determination of the invariant and the expression 

While working at the lab frame, the particle’s total momentum before the collision is to be identified. However, the mass of the rest particle is and the other particle has energy .

The equation of the energy is expressed as:

${\text{E}}_{\text{t}}={\text{mc}}^{\text{2}}+{\text{E}}_{\text{i}}$

The equation of the momentum is expressed as:

${\text{p}}_{\text{t}}^{\text{2}}={\text{p}}_{\text{i}}^{\text{2}}$

The equation of the energy can also be expressed as:

$$\begin{gathered} {\text{E}}_{\text{i}}^{\text{2}}={\text{m}}^{\text{2}}{\text{c}}^{\text{4}}+{\text{p}}_{\text{i}}^{\text{2}}{\text{c}}^{\text{2}} \\ {\text{p}}_{\text{i}}^{\text{2}}=\frac{{\text{E}}_{\text{i}}^{\text{2}}}{{\text{c}}^{\text{2}}}-{\text{m}}^{\text{2}}{\text{c}}^{\text{2}} \end{gathered}$$

The equation of the invariant quantity is expressed as:

$$\begin{gathered} \frac{{\text{E}}_{\text{i}}^{\text{2}}}{{\text{c}}^{\text{2}}}-{\text{p}}_{\text{t}}^{\text{2}}=\frac{{\left({\text{mc}}^{\text{2}}{\text{+E}}_{\text{i}}\right)}^{\text{2}}}{{\text{c}}^{\text{2}}}-\left(\frac{{\text{E}}_{\text{i}}^{\text{2}}}{{\text{c}}^{\text{2}}}-{\text{m}}^{\text{2}}{\text{c}}^{\text{2}}\right) \\ ={\text{m}}^{\text{2}}{\text{c}}^{\text{2}}+{\text{2mE}}_{\text{i}}+\frac{{\text{E}}_{\text{i}}^{\text{2}}}{{\text{c}}^{\text{2}}}-\frac{{\text{E}}_{\text{i}}^{\text{2}}}{{\text{c}}^{\text{2}}}+{\text{m}}^{\text{2}}{\text{c}}^{\text{2}} \\ =2\left({\text{m}}^{\text{2}}{\text{c}}^{\text{2}}+{\text{mE}}_{\text{i}}\right) \end{gathered}$$

From the result of part (a), the equation of the mass of the second particle is expressed as:

$$\begin{gathered} {\text{Mc}}^{\text{2}}=2\left({\text{m}}^{\text{2}}{\text{c}}^{\text{2}}+{\text{mE}}_{\text{l}}\right) \\ \text{M}=\sqrt{\text{2}\left({\text{m}}^{\text{2}}\text{+}\frac{{\text{mE}}_{\text{l}}}{{\text{c}}^{\text{2}}}\right)} \end{gathered}$$

Here, the assumption is that the resultant quantities are invariant and also conserved. However, in this case, both the assumptions may not be held true together.

Thus, the invariant is $2\left({\text{m}}^{\text{2}}{\text{c}}^{\text{2}}+{\text{mE}}_{\text{i}}\right)$ and the expression for M is$\sqrt{2\left({\text{m}}^{\text{2}}+\frac{{\text{mE}}_{\text{l}}}{{\text{c}}^{\text{2}}}\right)}$ .

(c) Determination of the momentum and energy conservation equations

The equation of energy is expressed as:

$$\begin{gathered} E_i=E_f \\ {\gamma }_{ui}m{c}^2+m{c}^2={\gamma }_{uf}M{c}^2 \\ {{\gamma }^2}_{ui}{m}^2{c}^4+2{{\gamma }^2}_{ui}{m}^2{c}^4+m^2{c}^4={{\gamma }^2}_{uf}{M}^2{c}^4 \end{gathered}$$ (1)

The equation of momentum is expressed as:

$$\begin{gathered} p_i=p_f \\ {\gamma }_{ui}m{u}_i={\gamma }_{uf}M{u}_f \\ {{\gamma }^2}_{ui}{m}^2{u_i}^2={{\gamma }^2}_{uf}{M}^2{u_f}^2 \end{gathered}$$

Substitute $c^2{\gamma }_u^2-c^2$ for $u^2{\gamma }_u^2$ in the above equation.

$m^2\left(c^2{\gamma }_{ui}^2-c^2\right)=M^2\left(c^2{\gamma }_{uf}^2-c^2\right)$

Multiplying the above equation by c² and subtracting the equation (1) from the above equation.

$$\begin{gathered} \left(m^2{c}^4{\gamma }_{ui}^2-m^2{c}^2\right)-\left({\gamma }_{ui}^2{m}^2{c}^4+2{\gamma }_{ui}^2{m}^2{c}^4+m^2{c}^4\right)=\left(M^2{c}^4{\gamma }_{uf}^2-M^2{c}^4\right)-\left({\gamma }_{uf}^2{M}^2{c}^4\right) \\ -2m^2{c}^4-2{\gamma }_{ui}^2{m}^2{c}^4=-M^2{c}^4 \\ M=\sqrt{2\left(m^2+\frac{m{E}_i}{c^2}\right)} \end{gathered}$$

Thus, the momentum and energy conservation equation is $M=\sqrt{2\left(m^2+\frac{m{E}_i}{c^2}\right)}$ and gives the same result for M in terms of m and E_i .