PH450 HW2

Mark Thomson - Modern Particle Physics

PH450 HW 2

 

3.1

Given:
- $m_{\pi} = 140 \, \text{MeV}$,
- $m_{\mu} = 106 \, \text{MeV}$,
- $m_{\nu} \approx 0$.

Since momentum is conserved, the magnitudes of the momenta of the muon and the neutrino are equal, but they move in opposite directions. By applying energy conservation:

$$m_{\pi} = E_{\mu} + E_{\nu}$$

This can also be written as:

$$m_{\pi} - p = E_{\mu}$$

Where we neglect the neutrino mass, approximating $E_{\nu} = p$. Using the energy-momentum relation $E^2 = m^2 + p^2$ for both the pion and the muon, we have:

$$m_{\pi}^2 = E_{\mu}^2 + p^2 = m_{\mu}^2 + p^2$$

Now solve for $p$:

$$m_{\pi}^2 - 2m_{\pi}p + p^2 = m_{\mu}^2 + p^2$$

$$p = \frac{m_{\pi}^2 - m_{\mu}^2}{2m_{\pi}} = 30 \, \text{MeV}$$

Finally, the energy of the muon is:

$$E_{\mu}^2 = m_{\mu}^2 + p^2 = 110 \, \text{MeV}$$

 

 

 

3.2

In the rest frame of the parent particle, the momenta of the two daughter particles are equal. Energy conservation implies:

$$m_a = E_1 + E_2.$$

Writing this as $m_a - E_2 = E_1$, and squaring gives:

$$m_a^2 - 2m_a E_2 + E_2^2 = E_1^2.$$

Expanding and using the relation $E^2 = m^2 + p^2$ for the energies of the daughter particles:

$$m_a^2 - 2m_a E_2 + m_2^2 + p^2 = m_1^2 + p^2.$$

Simplifying:

$$m_a^2 + (m_2^2 - m_1^2) = 2m_a E_2.$$

Squaring again to eliminate $E_2$ leads to:

$$m_a^4 + 2m_a^2 (m_2^2 - m_1^2) + (m_2^2 - m_1^2)^2 = 4m_a^2(p^*2).$$

Expanding and simplifying further:

$$m_a^4 - 2m_a^2[(m_1 + m_2)^2 + (m_1 - m_2)^2] = 4m_a p^*2.$$

Finally, we have the desired result:

$$p^* = \frac{1}{2m_a} \sqrt{[(m_a^2 - (m_1 + m_2)^2][m_a^2 - (m_1 - m_2)^2]}.$$

 

 

3.3

The branching ratio is related to the total decay width $\Gamma$ as:

$$BR(K^+ \rightarrow \pi^+ \pi^0) = \frac{\Gamma(K^+ \rightarrow \pi^+ \pi^0)}{\Gamma},$$

where in natural units, $\Gamma = 1/\tau$, or in SI units, $\Gamma = \hbar / \tau$. Therefore, we can write:

$$BR(K^+ \rightarrow \pi^+ \pi^0) = \frac{\tau}{\hbar} \Gamma(K^+ \rightarrow \pi^+ \pi^0).$$

Now substituting the values:

$$BR(K^+ \rightarrow \pi^+ \pi^0) = \frac{1.2 \times 10^{-8}}{1.06 \times 10^{-34}} \times 1.2 \times 10^{-8} \approx 1.6 \times 10^{-19} = 21 \%.$$

 

3.4

The total number of events $N$ is determined by the integral:

$$N = \int \sigma \mathcal{L} \, dt,$$

or, equivalently, the event rate is given by:

$$\text{Rate} = \sigma \mathcal{L}.$$

Here, 1 barn is $10^{-24} \, \text{cm}^2$. Therefore, using the given values:

$$\text{Rate} = \sigma \mathcal{L} = (250 \times 10^{-39}) \times (2 \times 10^{34}) = 0.005 \, \text{s}^{-1}.$$

Thus, over five years, with 50% operational efficiency, the total number of events is:

$$N = 0.005 \times 0.5 \times 365.25 \times 86400 \times 5 \approx 394000 \, e^+ e^- \rightarrow \text{HZ} \, \text{events}.$$

 

 

3.5

In natural units:

$$\sigma = \frac{4 \pi \alpha^2}{3s} = \frac{4 \pi / 137^2}{3 \cdot 50^2} = 8.9 \times 10^{-8} \, \text{GeV}^{-2}.$$

To convert to standard units (barns), either include the factor $(\hbar c)^2$, or use the conversion factor $\hbar c = 0.197 \, \text{GeV} \cdot \text{fm}$, and note that $1 \, \text{fm}^2 = 10^{-30} \, \text{m}^2 = 10^{-2} \, \text{barn}$. Therefore:

$$\sigma = 8.9 \times 10^{-8} \, \text{GeV}^{-2} \times 0.197^2 \times 0.01 = 34 \, \text{pb}.$$

This result is approximately $10^9$ times smaller than the strong interaction proton-proton cross section.

 

 

3.6

From the definition of the cross section, the event rate is given by the flux multiplied by the cross section and the number of targets:

$$\text{Rate} = \phi \sigma N_t.$$

The total number of interactions $N_{\text{int}}$ is therefore:

$$N_{\text{int}} = \sigma N_t \int \phi \, dt.$$

Suppose the "beam" of one neutrino is confined to an area $A$, then the integrated flux is just $1 / A$. Writing the thickness of the Iron as $x$, the number of targets traversed by the "beam" is $A x n$, where $n$ is the number density of target nuclei. Hence:

$$N_{\text{int}} = \frac{1}{A} n A x = \sigma n x.$$

Taking the average mass of a nucleon to be $1.67 \times 10^{-27} \, \text{kg}$, the product of the number density and thickness is:

$$n x = \frac{7874}{56 \times 1.67 \times 10^{-27}} \times 1 \, \text{m} = 8.4 \times 10^{28} \, \text{m}^{-2} = 8.4 \times 10^{24} \, \text{cm}^{-2}.$$

Hence, the average number of interactions for a single neutrino traversing the block is:

$$N_{\text{int}} = 8.4 \times 10^{24} \cdot 8 \times 10^{-39} = 7 \times 10^{-10}.$$

Therefore, the probability of interaction is less than $10^{-9}$.

 

 

3.7

First, consider the low-energy limit of the four-vector product $p_a \cdot p_b = E_a E_b - \vec{p}_a \cdot \vec{p}_b$. In the non-relativistic limit, the particle energy and momentum can be approximated as (in natural units):

$$E \approx m \left( 1 + \frac{1}{2} \beta^2 \right) = m + \frac{1}{2} m \beta^2,$$

and to $\mathcal{O}(\beta)$:

$$\vec{p} \approx m \beta.$$

Thus, the product $p_a \cdot p_b = E_a E_b - \vec{p}_a \cdot \vec{p}_b$ can be approximated as:

$$p_a \cdot p_b \approx m_a m_b \left( 1 + \frac{1}{2} \beta_a^2 \right) \left( 1 + \frac{1}{2} \beta_b^2 \right) - m_a m_b \beta_a \cdot \beta_b.$$

Expanding this:

$$p_a \cdot p_b \approx m_a m_b \left[ 1 + \frac{1}{2} (\beta_a^2 + \beta_b^2) - \beta_a \cdot \beta_b \right].$$

To $\mathcal{O}(\beta^2)$, this becomes:

$$(p_a \cdot p_b)^2 \approx m_a^2 m_b^2 \left[ 1 + (\beta_a - \beta_b)^2 \right].$$

Thus, the Lorentz-invariant flux term becomes:

$$F = 4 \left[ (p_a \cdot p_b)^2 - m_a^2 m_b^2 \right]^{1/2}$$

which simplifies to:

$$F \approx 4 m_a m_b |\beta_a - \beta_b|,$$

or, equivalently in SI units:

$$F = 4 m_a m_b |\vec{v}_a - \vec{v}_b|.$$

As expected, in the non-relativistic limit, the flux depends on the relative velocities of the two particles.

 

3.8

Independent of the rest frame, the Lorentz-invariant flux is generally given by:

$$F = 4 \left[ (p_a \cdot p_b)^2 - m_a^2 m_b^2 \right]^{1/2}.$$

In the rest frame of $b$, we can write the four-momenta as $p_a = (E_a, 0, 0, p_a)$ and $p_b = (m_b, 0, 0, 0)$. Substituting these into the expression for $F$, we get:

$$F = 4 \left[ E_a^2 m_b^2 - m_a^2 m_b^2 \right]^{1/2}.$$

This simplifies further to:

$$F = 4 \left[ (p_a^2 + m_a^2)m_b^2 - m_a^2 m_b^2 \right]^{1/2}.$$

Simplifying the terms:

$$F = 4 p_a m_b.$$

Thus, we have shown that the flux in the frame where $b$ is at rest is:

$$F = 4 m_b p_a.$$

 

 

3.9

In the center-of-mass frame, we know that $\sqrt{s} = E_1^* + E_2^*$. Therefore:

$$(\sqrt{s} - E_1^*)^2 = E_2^2$$

Expanding this, we get:

$$s - 2\sqrt{s}E_1^* + E_1^2 = E_2^2$$

Using the relation $s - 2\sqrt{s}E_1^* + m_1^2 + p_i^2 = m_2^2 + p_i^2$, we can simplify:

$$2\sqrt{s}E_1^* = s + (m_1^2 - m_2^2).$$

Expanding both sides, we arrive at:

$$4s(p_i^2 + m_1^2) = s^2 + 2s(m_1^2 - m_2^2) + (m_1^2 - m_2^2)^2.$$

Simplifying further:

$$4sp_i^2 = s^2 - 2s(m_1^2 + m_2^2) + (m_1 + m_2)(m_1 - m_2)^2.$$

This becomes:

$$p_i^2 = \frac{1}{4s} \left[ s - (m_1 + m_2)^2 \right] \left[ s - (m_1 - m_2)^2 \right].$$

Thus, we have proven the required expression for the momentum in the center-of-mass frame.
 

 

3.10

a) With the electron masses ($m$) included, this calculation becomes more tedious. In the laboratory frame, where the proton is at rest, the four-momenta of the particles are:

$$p_1 = (E_1, 0, 0, p_1), \quad p_2 = (m_p, 0, 0, 0), \quad p_3 = (E_3, 0, p_3 \sin \theta, p_3 \cos \theta), \quad p_4 = (E_4, p_4).$$

Differentiating $E_3^2 = p_3^2 + m_e^2$ with respect to $\cos \theta$ gives:

$$2E_3 \frac{dE_3}{d(\cos \theta)} = 2p_3 \frac{dp_3}{d(\cos \theta)}.$$

Next, equating the expressions for the Mandelstam $t$ variable written in terms of the electron and proton four-momenta:

$$t = (p_1 - p_3)^2 = (p_2 - p_4)^2.$$

Expanding:

$$t = 2m_e^2 - 2p_1 \cdot p_3 = 2m_p^2 - 2p_2 \cdot p_4.$$

Simplifying:

$$2m_e^2 - 2(E_1 E_3 - p_1 p_3 \cos \theta) = 2m_p^2 - 2m_p E_4.$$

Now, differentiate with respect to $\cos \theta$:

$$-(E_1 + m_p) \frac{dE_3}{d(\cos \theta)} + p_1 p_3 \frac{dE_3}{d(\cos \theta)} \cos \theta = p_1 p_3.$$

Using this relation and rearranging gives:

$$\frac{dE_3}{d(\cos \theta)} = \frac{p_1 p_3^2}{p_3(E_1 + m_p) - p_1 E_3 \cos \theta}.$$

This is the desired result.
 

 

b) From equation (3.37) of the main text, we have:

$$\frac{d\sigma}{dt} = \frac{1}{64 \pi s p_i^2} |\mathcal{M}_{fi}|^2.$$

This can be related to the differential cross section in terms of solid angle using:

$$\frac{d\sigma}{d\Omega} = \frac{d\sigma}{dt} \frac{dt}{d\Omega}.$$

But from the relation $t = (p_4 - p_2)^2 = 2m_p E_4 - 2m_p E_1$, we get:

$$\frac{dt}{d\Omega} = 2m_p \frac{dE_3}{d(\cos \theta)}.$$

Therefore:

$$\frac{d\sigma}{d\Omega} = \frac{1}{2\pi} \cdot 2m_p \frac{dE_3}{d(\cos \theta)} \cdot \frac{1}{64\pi s p_i^2} |\mathcal{M}_{fi}|^2.$$

Substituting the expressions and simplifying:

$$\frac{d\sigma}{d\Omega} = \frac{1}{64\pi^2} \frac{p_3^2}{p_1 m_e} \cdot \frac{1}{p_3(E_1 + m_p) - E_3 p_1 \cos \theta} \cdot |\mathcal{M}_{fi}|^2.$$

This completes the proof.