Energy loss at the LHC

If a proton is being kept at a constant speed, then it must have a constant energy, if it's losing energy at a rate $P$ , then in order to maintain a constant energy it must receive the same power $P$ (from the LHC).

Let $P_T$ denote the total power needed to keep the bunch at a constant speed, then as there are $1.15 \times 10^{11}$ protons, $P_T= 1.15 \times 10^{11} \times P$ . $\because P=\frac{2Ke^2p^4}{3m^4c^3r^2}$ , it follows that $P_T=\frac{1.15 \times 10^{11} \times 2Ke^2p^4}{3m^4c^3r^2}$ .

We have already been told the values of $K$ , $e$ and $m$ , we will take $c$ to be $299792458~ms^{-1}~$ so we need to work out $p$ and $r$ , since the formula only contains $p^4$ , we will work that out instead.

Calculating $r$ :

The LHC is a circle of circumference $27~km$ or $27000~m$ . The circumference of a circle is $2 \pi r$ , so $r=\frac{13500}{\pi}$ .

Calculating $p^4$ :

From the energy-momentum relation we have $E^2=(pc)^2+(mc^2)^2$ . Rearrangement yields $p^2=\frac{E^2-m^2c^4}{c^2}$ or $p^4=(\frac{E^2-m^2c^4}{c^2})^2$ . We are told from the question that each proton has an energy of $1.12$ microjoules, so $E=1.12 \times 10^{-6}$ .

Now we can just put these numbers back into our formula, this gives: $\frac{1.15 \times 10^{11}\times 2 \times 9 \times 10^9 \times (1.6 \times 10^{-19})^2 \times \left( \frac{(1.12 \times 10^{-6})^2-(1.67 \times 10^{-27})^2 \times 299792458^4}{299792458^2}\right)^2}{3 \times (1.67 \times 10^{-27})^4 \times 299792458^3 \times (\frac{13500}{\pi})^2}$ .

Evaluating this yields $0.889165359$ or $0.889$ to three significant figures.

Note: An answer of $0.885$ for this question can be obtained by letting $c=3 \times 10^8~ms^{-1}$ , doing so yields $0.884868371 \approx 0.885$ .

This is relatively straight forward, the only quantities and equations not given are relativistic momentum and rest mass of proton:

$E^2=p^2c^2+m_0^2c^4$

$m_0=93.8*10^6 eV$

The momentum p of one proton with energy $1.15*10^{11}$ is $3.73*10^{-15}$

$P*1.15*10^{11}=0.885$

Consider a single proton in the LHC. It radiates power at the rate of $P= \frac{2Ke^2p^4}{3m^4c^3r^2}$ For the proton to move in constant speed, the amount of power each proton receives from L.H.C must be equal to $P$ . Let the number of protons revolving be $n$ , so the total power L.H.C has to provide the bunch of protons will be: $P_{tot}= n \times P= n \times \frac{2Ke^2p^4}{3m^4c^3r^2}$
We know the values of $K$ . $e$ , $c$ , and $m$ . The only unknown quantities are $r$ and $p$ . We need to determine them.

The radius of the circle is simply its circumference divided by $2 \pi$ , so $r= \frac{L}{2 \pi}$ where $L$ is the length of the circle ( $L= 27 \text{ km}$ in the question).

Let the energy of each proton be $E$ ( $E= 1.12 \text{ mJ}$ in the question). From Einstein's formula, we have: $E= \sqrt{(mc^2)^2 + (pc)}$ $\implies p= \sqrt{\frac{E^2-m^2c^4}{c^4}}$ Plugging these values, we obtain: $P_{tot} = n \times \frac{2 K e^2 \left ( \frac{E^2-m^2c^4}{c^4} \right )^2}{3m^4c^3 \left ( \frac{L}{2\pi } \right )^2}$ Plugging the numerical values gives $P_{tot} \approx \boxed{0.885 \text{ Watts}}$ .