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Chapter 5: Problem 6

Two particles with rest masses \(m_{1}\) and \(m_{2}\) move collinearly in some inertial frame, with uniform velocities \(u_{2}\) and \(u_{2}\), respectivcly. They collide and form a single particle with rest mass \(m\) moving at velocity \(u\). Prove that $$ m^{2}=m_{1}^{2}+m_{2}^{2}+2 m_{1} m_{2} \gamma\left(u_{1}\right) \gamma\left(u_{2}\right)\left(1-u_{1} u_{2} / c^{2}\right) $$ and also find \(u\). [Hint: for the first part, use a four-vector argument, or a result of Section \(30 .]\)

Short Answer

Expert verified
Question: Prove the equation for the rest mass, \(m^2\), of the new particle formed by the collision of two particles with rest masses \(m_{1}\) and \(m_{2}\) and velocities \(u_{1}\) and \(u_{2}\). Also, find the expression for the velocity \(u\) of the new particle. Answer: The equation for the rest mass of the new particle is given by: \[ m^2 = m_{1}^{2} + m_{2}^{2} + 2 m_{1} m_{2} \gamma(u_{1}) \gamma(u_{2}) \left( 1 - u_{1} u_{2} / c^{2} \right) \] The expression for the velocity \(u\) of the new particle is: \[ u = \frac{m_1 u_1 + m_2 u_2}{m \sqrt{1 + \frac{(m_1 u_1 + m_2 u_2)^2}{c^2 m^2}}} \]

Step by step solution

01

Understanding Four-Momentum Conservation

In special relativity, four-momentum is a four-vector that combines a particle's energy and momentum into a single quantity. The conservation of four-momentum states that the total initial four-momentum of a system is equal to the total final four-momentum. Since the particles are moving collinearly, we only need to consider the x-component of the four-momentum. On the left side, we have the four-momentum of particles 1 and 2, and on the right side, the four-momentum of the new particle.
02

Writing the Energy-Momentum Equation

Using the conservation of four-momentum, we can write the energy-momentum equation. For the initial state, particles 1 and 2 have four-momentum \((E_1, p_{1x})\) and \((E_2, p_{2x})\), respectively. In the final state, the new particle has four-momentum \((E, p_x)\). The equation becomes: \[ (E_1 + E_2, p_{1x} + p_{2x}) = (E, p_x) \]
03

Calculating Energy and Momentum Terms

To find the energy and momentum terms, we can use the following formulas: \[ E_i = \frac{m_i c^2}{\sqrt{1 - (u_i / c)^2}} \] \[ p_{i} = \frac{m_{i} u_{i}}{\sqrt{1 - (u_i / c)^2}} \] where \(E_i\) and \(p_i\) are the energy and momentum of the ith particle, respectively. We can substitute these expressions into the energy-momentum equation.
04

Substituting Expressions and Combining Terms

By substituting the energy and momentum expressions into the energy-momentum equation, we have: \[ \left( \frac{m_1 c^2}{\sqrt{1 - (u_1 / c)^2}} + \frac{m_2 c^2}{\sqrt{1 - (u_2 / c)^2}}, \frac{m_{1} u_{1}}{\sqrt{1 - (u_1 / c)^2}} + \frac{m_{2} u_{2}}{\sqrt{1 - (u_2 / c)^2}} \right) = \left( \frac{m c^2}{\sqrt{1 - (u / c)^2}}, \frac{m u}{\sqrt{1 - (u / c)^2}} \right) \] Then, we can square both sides of the equation and combine terms to obtain the desired equation for \(m^2\).
05

Squaring and Simplifying

After squaring both sides of the energy-momentum equation and simplifying, we obtain the following equation: \[ m^2 = m_{1}^{2} + m_{2}^{2} + 2 m_{1} m_{2} \gamma(u_{1}) \gamma(u_{2}) \left( 1 - u_{1} u_{2} / c^{2} \right) \] This proves the first part of the problem. Now we need to find the velocity \(u\) of the new particle.
06

Finding u Using Momentum Conservation

From the conservation of four-momentum equation, we have: \[ \frac{m_{1} u_{1}}{\sqrt{1 - (u_1 / c)^2}} + \frac{m_{2} u_{2}}{\sqrt{1 - (u_2 / c)^2}} = \frac{m u}{\sqrt{1 - (u / c)^2}} \] We can solve for \(u\) by dividing both sides by \(mc\) and cancelling out common factors. This results in: \[ u = \frac{m_1 u_1 + m_2 u_2}{m \sqrt{1 + \frac{(m_1 u_1 + m_2 u_2)^2}{c^2 m^2}}} \] We have now found the velocity \(u\) of the new particle.

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Most popular questions from this chapter

A rocket propels itself rectilinearly by giving portions of its mass a constant fbackward) velocity \(U\) relative to its instantaneous rest frame. It continues to do so until it attains a velocity \(V\) relative to its initial rest frame. Prove that the ratio of the initial to the final rest mass of the rocket is given by $$ \frac{M_{\mathrm{i}}}{M_{\mathrm{f}}}=\left(\frac{c+V}{c-V}\right)^{\frac{c}{2 U}} $$

A particle of rest mass \(m\) decays from rest into a particle of rest mass \(m\) ' and a photon. Find the separate energies of these end products. [Answer: \(c^{2}\left(m^{2} \pm m^{\prime 2}\right) / 2 m\). Hint: use a four-vector argument.]

Planck's constant \(h\) has the dimensions of action (energy \(x\) time or momentum \(\times\) distance) which suggests that the action of any periodic phenomenon may have to be a multiple of h. Accordingly Bohr constructed a model of the hydrogen atom in which the action of the single orbiting electron was quantized, requiring \(2 \pi r m v=n h\), \(n=1,2, \ldots\), where \(m\) is the mass of the electron, \(v\) its speed, and \(r\) the radius of the orbit. This led to a hydrogen spectrum which fitted the then known facts. Show that Bohr's hypothesis (1913) is equivalent to the assumption that a permissible orbit must contain an integral number of de Broglie electron waves.

A particle with four-momentum \(P\) is observed by an observer who moves with four-velocity U \(_{0}\). Prove that the energy of the particle relative to that observer is \(\mathbf{U}_{0} \cdot \mathbf{P}\).

Consider a head-on elastic collision of a "bullet' of rest mass \(M\) with a stationary 'target' of rest mass \(m\). Prove that the post-collision \(\gamma\)-factor of the bullet cannot exceed \(\left(m^{2}+M^{2}\right) / 2 m M\). This means that for large bullet energies (with \(\gamma\)-factors much larger than this critical value), almost the entire energy of the bullet is transferred to the target. [Hint: if \(\mathbf{P}, \mathbf{P}^{\prime}\) are the pre-and post-collision four-momenta of the bullet, and \(\mathbf{Q}, \mathbf{Q}^{\prime}\) those of the target, show, by going to the \(\mathrm{CM}\) frame, that \(\left(\mathbf{P}^{\prime}-\mathbf{Q}\right)^{2} \geqslant 0\); in fact, in the CM frame \(\mathbf{P}^{\prime}-\mathbf{Q}\) has no spatial components.] The situation is radically different in Newtonian mechanics, where the pre- and post-collision velocities of the bullet are related by \(u / u^{\prime}=(M+m) /(M-m)\). Prove this.
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