# Homework Help: Special relativity and colliders

1. Dec 31, 2015

### spaghetti3451

1. The problem statement, all variables and given/known data

a) The LHC was designed to collide protons together at 14 TeV centre-of-mass energy. How many kilometres per hour less than the speed of light are the protons moving?

b) How fast is one proton moving relative to the other?

2. Relevant equations

3. The attempt at a solution

a) We assume that the two beams are coming towards their common centre-of-mass with equal and opposite momenta, so that the centre-of-mass is at rest. In that case, if the centre-of-mass energy is 14 TeV, then the total energy of the system is 14 TeV. The centre-of-mass frame is the same as the laboratory frame, and so that the energy of each beam in the laboratory frame is 7 TeV.

Am I correct so far?

2. Dec 31, 2015

### Orodruin

Staff Emeritus
Yes.

3. Dec 31, 2015

### spaghetti3451

Then, $E = \gamma mc^{2}$
$\implies v = c \sqrt{1-\Big(\frac{mc^{2}}{E}\Big)^{2}} = c \sqrt{1-\Big(\frac{938\ \text{MeV}}{7\ \text{TeV}}\Big)^{2}} = c \sqrt{1-\Big(\frac{938\times 10^{-6}}{7}\Big)^{2}} = c \Big(1-\frac{938\times 10^{-6}}{14}\Big)$

So, the difference required is $=\frac{938\times 10^{-6}}{14}c = 20,000\ \text{ms}^{-1} = 72,000\ \text{kms}^{-1}$.

b) In the laboratory, let proton A be coming from the left with velocity $u$, and proton B be coming from the right with velocity $v_{x}$. Let's called this the unprimed frame.

Now, consider the frame where proton A is at rest and proton B is coming from the right with velocity $v'_{x}$. Let's call the primed frame.

So, using the formula for velocity addition, $v'_{x}=\frac{v_{x}-u}{1-\frac{uv_{x}}{c^{2}}}=$

Now, $u =$ velocity of proton A in the laboratory frame $= c(1-\epsilon)$, where $\epsilon = (6.7 \times 10^{-5})$

$v_{x} =$ velocity of proton B in the laboratory frame $= - c(1-\epsilon)$, where $\epsilon = (6.7 \times 10^{-5})$.

Therefore, $v'_{x}=\frac{c(1-\epsilon)+c(1-\epsilon)}{1+\frac{c^{2}(1-\epsilon)^{2}}{c^{2}}} = \frac{2c(1-\epsilon)}{2-2\epsilon+\epsilon^{2}}$.

Do I just plug in numbers now?

4. Dec 31, 2015

### Orodruin

Staff Emeritus
Yes. Note that you can make life easier for yourself by only keeping the leading contribution in $\epsilon$ without losing any relevant accuracy.

(Also note that the leading contribution goes as $\epsilon^2$ ...)

5. Dec 31, 2015

### spaghetti3451

But, plugging in numbers, I find that I need to keep all the terms in in the denominator, otherwise the answer is either negative, or above the speed of light, etc.

6. Dec 31, 2015

### Orodruin

Staff Emeritus
This is why I specifically mentioned that the leading contribution goes as $\epsilon^2$. I suggest dividing both numerator and denominator by $1-\epsilon$ and then starting to expand the resulting expression in small quantities.

7. Dec 31, 2015

### spaghetti3451

So, I have $\frac{2c(1-\epsilon)}{2-2\epsilon+\epsilon^{2}}=\frac{2c}{(2-2\epsilon+\epsilon^{2})(1-\epsilon)^{-1}}=\frac{2c}{(2-2\epsilon+\epsilon^{2})(1+\epsilon)}=\frac{2c}{(2-2\epsilon+\epsilon^{2}+2\epsilon-2\epsilon^{2}+\epsilon^{3})}=\frac{2c}{(2-\epsilon^{2}+\epsilon^{3})}$.

Am I on the right track?

8. Dec 31, 2015

### Orodruin

Staff Emeritus
Yes, although I suggest getting rid of that $\epsilon^3$ as well as it will not do anything for you.

(I also found it more straight-forward to write the denominator as $2(1-\epsilon) +\epsilon^2$ ...)

9. Dec 31, 2015

### spaghetti3451

Alright, so I have $\frac{2c(1-\epsilon)}{2-2\epsilon+\epsilon^{2}} = \frac{2c}{(2(1-\epsilon)+\epsilon^{2})(1-\epsilon)^{-1}}= \frac{2c}{(2(1-\epsilon)+\epsilon^{2})(1+\epsilon)}= \frac{2c}{2(1-\epsilon)(1+\epsilon)+\epsilon^{2}(1+\epsilon)}$.

Using $(1+\epsilon)(1-\epsilon)=(1-\epsilon^{2})$ and ignoring the term in $\epsilon^{3}$, we have

$\frac{2c}{2(1-\epsilon^{2})+\epsilon^{2}} = \frac{2c}{2-\epsilon^{2}} = c(1-\frac{\epsilon^{2}}{2})^{-1}=c(1+\frac{\epsilon^{2}}{2})$.

I presume that the reason for ignoring higher order terms in the expansions of $(1-\epsilon)^{-1}$ and $2(1-\epsilon)(1+\epsilon)+\epsilon^{2}(1+\epsilon)$ in the denominator is that, the higher order terms would have been taken to the numerator during the expansion of $({2-\epsilon^{2}+\cdots})^{-1}$ and so would have been ignored there as well.

But now $c(1+\frac{\epsilon^{2}}{2}) > c$!

10. Dec 31, 2015

### Orodruin

Staff Emeritus
No, you made a sign error.

11. Dec 31, 2015

### PeroK

I'm struggling a little to see that $20,000\ \text{ms}^{-1} = 72,000\ \text{kms}^{-1}$

12. Dec 31, 2015

### spaghetti3451

Oh right! That should be $\text{km h}^{-1}$. Silly typo!!!!

13. Dec 31, 2015

### spaghetti3451

I've reviewed my calculations a couple of times, but still don't find a sign error.

14. Dec 31, 2015

### PeroK

In any case, the figure of $20,000\ \text{ms}^{-1}$ looks a bit high. You lost a square root in the very first line.

15. Dec 31, 2015

### Orodruin

Staff Emeritus
Have a closer look at what comes before this expression:

16. Dec 31, 2015

### spaghetti3451

I understand exactly what you've said. I also understand where I went wrong. I illustrate both approaches - mine and yours - below.

Yours:

$\frac{2c(1-\epsilon)}{2-2\epsilon+\epsilon^{2}}=\frac{2c(1-\epsilon)}{2(1-\epsilon)+\epsilon^{2}}=\frac{2c}{2+\epsilon^{2}(1-\epsilon)^{-1}}=\frac{2c}{2+\epsilon^{2}(1+\epsilon)}=\frac{2c}{2+\epsilon^{2}}$.

Mine:

$\frac{2c(1-\epsilon)}{2-2\epsilon+\epsilon^{2}}=\frac{2c}{(2-2\epsilon+\epsilon^{2})(1-\epsilon)^{-1}}=\frac{2c}{(2-2\epsilon+\epsilon^{2})(1+\epsilon+\epsilon^{2})}=\frac{2\epsilon}{2+2\epsilon+2\epsilon^{2}-2\epsilon-2\epsilon^{2}+\epsilon^{2}}=\frac{2c}{2+\epsilon^{2}}$.

My mistake was in not taking the second order term in $\epsilon$ in the expansion of $(1-\epsilon)^{-1}$. You should always take as many terms in the binomial expansion as is needed to give the leading contribution.

Your approach, however, is more elegant.

17. Dec 31, 2015

### spaghetti3451

I did, in fact , lose a square in the very first line.

Here's a corrected attempt:

$E= \gamma mc^{2}\implies v = c \sqrt{1-\Big(\frac{mc^{2}}{E}\Big)^{2}} = c \sqrt{1-\Big(\frac{938\ \text{MeV}}{7\ \text{TeV}}\Big)^{2}} = c \sqrt{1-\Big(\frac{938\times 10^{-6}}{7}\Big)^{2}} = c \bigg[1-\frac{1}{2}\Big(\frac{938\times 10^{-6}}{7}\Big)^{2}\bigg]$

So, the difference is $\frac{c}{2}\Big(\frac{938\times 10^{-6}}{7}\Big)^{2} = 3\ \text{m s}^{-1} = 10\ \text{km h}^{-1}$.

Does it look alright?

18. Dec 31, 2015

### PeroK

It does look right.

19. Dec 31, 2015

### vela

Staff Emeritus
Yes, that's right.

To avoid some unnecessary algebra in the future, you might want to remember the relation $\frac vc = \frac{pc}{E}$.

20. Dec 31, 2015

### spaghetti3451

Thanks!

How might I be able to find the momentum $p$ so that I am able to use $\frac{v}{c} = \frac{pc}{E}$ to find $v$?

Should I use $p = \sqrt{E^{2}-mc^{2}}$?

21. Dec 31, 2015

### vela

Staff Emeritus
Yes. If you do that, you'll see you end up with the same expression for $v$ that you used.

22. Dec 31, 2015

### spaghetti3451

Alright, let me do the calculation again.

$\frac{v}{c}=\frac{pc}{E} \implies v = c\frac{pc}{E} = c\frac{\sqrt{E^{2}-(mc^{2})^{2}}}{E}=c\sqrt{1-\Big(\frac{mc^{2}}{E}\Big)^{2}}=c\Big(1-\frac{1}{2}\big(\frac{mc^{2}}{E}\big)^{2}\Big)$, in line with we had before.

23. Dec 31, 2015

### PeroK

If you use expressions involving the $\gamma$ factor, part b) should come out more easily:

For part a) you have $\frac{v^2}{c^2} = 1- \frac{1}{\gamma^2}$

And for part b)

$v' = \frac{2v}{1+ v^2/c^2} = \frac{2v}{2 - 1/\gamma^2}$ etc.

24. Dec 31, 2015

### spaghetti3451

So, do you mean finding the $\gamma$-factor from part (a) and using it in part (b)?

25. Dec 31, 2015

### PeroK

You're more or less given the $\gamma$ factor by being given the energy.