Subfield: Modern Physics

$g=9.8 \mathrm{~N} / \mathrm{kg}$



$k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}$



$c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}$



$N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1}$



$\sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right)$



$1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J}$



$m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2}$



$m_{p}=1.673 \times 10^{-27} \mathrm{~kg}=938 \mathrm{MeV} / \mathrm{c}^{2}$



$\sin \theta \approx \theta-\frac{1}{6} \theta^{3}$ for $|\theta| \ll 1$



$$

\begin{aligned}

& G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\

& k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\

& k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\

& R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\

& e=1.602 \times 10^{-19} \mathrm{C} \\

& h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times 10^{-15} \mathrm{eV} \cdot \mathrm{s} \\

& (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\

& \ln (1+x) \approx x \text { for }|x| \ll 1 \\

& \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1

\end{aligned}

$$

$$

\begin{array}{lrl}

\text { velocity parameter } & \beta & =\frac{v}{c} \\

\text { Lorentz factor } & \gamma & =\frac{1}{\sqrt{1-\beta^{2}}} \\

\text { relativistic momentum } & p & =\gamma \beta m c \\

\text { relativistic energy } & E & =\gamma m c^{2} \\

\text { relativistic doppler shift } & \frac{f}{f_{0}} & =\sqrt{\frac{1-\beta}{1+\beta}}

\end{array}

$$

Extra Supplementary Reading Materials:

Relativistic particles obey the mass energy relation



$$

E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}

$$



where $E$ is the relativistic energy of the particle, $p$ is the relativistic momentum, $m$ is the mass, and $c$ is the speed of light.



A proton with mass $m_{p}$ and energy $E_{p}$ collides head on with a photon which is massless and has energy $E_{b}$. The two combine and form a new particle with mass $m_{\Delta}$ called $\Delta$, or "delta". It is a one dimensional collision that conserves both relativistic energy and relativistic momentum.

Question: a. Determine $E_{p}$ in terms of $m_{p}, m_{\Delta}$, and $E_{b}$. You may assume that $E_{b}$ is small.

Solution:
We can solve the problem exactly, approximating only in the last step. This certainly isn't necessary; we do this to illustrate a useful technique. We set $c=1$ throughout, and transform to an inertial frame where the proton is initially at rest. Before the collision,



$$

E_{p}=m_{p}, \quad E_{\gamma}=\left|p_{\gamma}\right|

$$



For the $\Delta$ particle, we have the usual relativistic relation



$$

E_{\Delta}^{2}=p_{\Delta}^{2}+m_{\Delta}^{2}

$$



By energy-momentum conservation,



$$

E_{p}+E_{\gamma}=E_{\Delta}, \quad p_{\gamma}=p_{\Delta} .

$$



Combining these results gives



$$

\left(m_{p}+E_{\gamma}\right)^{2}=E_{\gamma}^{2}+m_{\Delta}^{2} \Rightarrow E_{\gamma}=\frac{m_{\Delta}^{2}-m_{p}^{2}}{2 m_{p}}

$$



Now we need to transform back to the original frame, where the energy of the photon is $E_{b}$. We can use the Lorentz transformation for this, but it's a little easier to realize that $E=h f$ for photons, and apply the Doppler shift. Then



$$

\alpha \equiv \frac{E_{b}}{E_{\gamma}}=\sqrt{\frac{1-\beta}{1+\beta}}

$$



where $\beta$ is the velocity parameter of the proton in the inertial frame where the photon has energy $E_{b}$. Solving for $\beta$ in terms of the energy ratio $\alpha$,



$$

\beta=\frac{1-\alpha^{2}}{1+\alpha^{2}}

$$



To calculate the proton energy, we need the Lorentz factor,



$$

\gamma=\frac{1}{\sqrt{1-\beta^{2}}}=\frac{1+\alpha^{2}}{2 \alpha}

$$







Then the proton energy in the original frame is



$$

E_{p}=\gamma m_{p}=\frac{m_{p}}{2}\left(\alpha+\frac{1}{\alpha}\right)=\frac{m_{p}}{2}\left(\frac{2 m_{p} E_{b}}{m_{\Delta}^{2}-m_{p}^{2}}+\frac{m_{\Delta}^{2}-m_{p}^{2}}{2 m_{p} E_{b}}\right)

$$



which is the exact answer.



At this point, we can approximate. The second term is much larger than the first, so



$$

E_{p} \approx \frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}

$$



which was the required answer for this problem.
We now show another method that approximates throughout. We'll do everything in the lab frame, so the symbols in this solution don't mean the same things they did in solution 1. In the lab frame, energy-momentum conservation gives



$$

p_{p}-p_{b}=p_{\Delta}, \quad E_{p}+E_{b}=E_{\Delta}

$$



Squaring both expressions and dropping $E_{b}^{2}$ terms, since $E_{b}$ is small,



$$

p_{p}^{2}-2 p_{p} p_{b} \approx p_{\Delta}^{2}, \quad E_{p}^{2}+2 E_{p} E_{b} \approx E_{\Delta}^{2}

$$



Subtracting these equations gives



$$

m_{p}^{2}+2 E_{p} E_{b}+2 p_{p} E_{b}=m_{\Delta}^{2} \Rightarrow E_{p}+p_{p}=\frac{m_{\Delta}^{2}-m_{p}^{2}}{2 E_{b}}

$$



Since $E_{b}$ is small, this quantity must be large. But this means the protons are ultrarelativistic, so $E_{p} \approx p_{p}$, giving



$$

E_{p} \approx \frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}

$$



as desired.

Final answer: $\frac{m_{\Delta}^{2}-m_{p}^{2}}{4 E_{b}}$