How Close to Light Speed Do Protons Move in the LHC?

[spaghetti3451]

Homework Statement

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?

Homework Equations

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?

 

[Orodruin]

Yes.

 

[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?

 

[Orodruin]

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 $\epsilon^2$ ...)

 

[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.

 

[Orodruin]

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.

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.

 

[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?

 

[Orodruin]

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 $2(1-\epsilon) +\epsilon^2$ ...)

 

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

 

[Orodruin]

No, you made a sign error.

 

[PeroK]

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

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

 

[spaghetti3451]

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

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

 

[spaghetti3451]

No, you made a sign error.

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

 

[PeroK]

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

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.

 

[Orodruin]

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

Have a closer look at what comes before this expression:

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

 

[spaghetti3451]

Have a closer look at what comes before this expression:

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.

 

[spaghetti3451]

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.

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?

 

[PeroK]

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?

It does look right.

 

[vela]

Yes, that's right.

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

 

[spaghetti3451]

Yes, that's right.

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

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

 

[vela]

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

 

[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.

 

[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.

 

[spaghetti3451]

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

 

[PeroK]

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

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

 

[spaghetti3451]

Got it!

But isn't it better use $\frac{v}{c}=(1-\epsilon)$ so that we can be more precise in our final answer?

 

[PeroK]

Got it!

But isn't it better use $\frac{v}{c}=(1-\epsilon)$ so that we can be more precise in our final answer?

There's nothing wrong with the $1-\epsilon$ idea but it seemed neat to exploit the fact that the two velocities are the same.

For part b) you get directly: $v' = c(1 - \frac{1}{\gamma^2})^\frac{1}{2}(1 - \frac{1}{2\gamma^2})^{-1}$

I was hoping that might simplify, but I think you have to expand it out.

 

[spaghetti3451]

In this case,

$v = c\Big(1 - \frac{1}{\gamma^2}\Big)^\frac{1}{2}\Big(1 - \frac{1}{2\gamma^2}\Big)^{-1}$

$= c\bigg(1+\frac{(1/2)}{1!}\Big(-\frac{1}{\gamma^{2}}\Big)+\frac{(1/2)(-1/2)}{2!}\Big(-\frac{1}{\gamma^{2}}\Big)^{2}\bigg)\bigg(1+\frac{(-1)}{1!}\Big(-\frac{1}{2\gamma^{2}}\Big)+\frac{(-1)(-2)}{2!}\Big(-\frac{1}{2\gamma^{2}}\Big)^{2}\bigg)$

$=c\Big(1-\frac{1}{2\gamma^{2}}-\frac{1}{8\gamma^{4}}\Big)\Big(1+\frac{1}{2\gamma^{2}}+\frac{1}{4\gamma^{4}}\Big)$

$=c\Big(1-\frac{1}{2\gamma^{2}}-\frac{1}{8\gamma^{4}}+\frac{1}{2\gamma^{2}}-\frac{1}{4\gamma^{4}}+\frac{1}{4\gamma^{4}}\Big)$

$=c\Big(1-\frac{1}{8\gamma^{4}}\Big)$

$=c\bigg(1-\frac{1}{8}\Big(\frac{mc^{2}}{E}\Big)^{4}\bigg)$

$=c\bigg(1-\frac{1}{8}\Big(\frac{938\ \text{MeV}}{7\ \text{TeV}}\Big)^{4}\bigg)$

$=c\bigg(1-\frac{1}{8}\Big(\frac{938 \times 10^{-6}}{7}\Big)^{4}\bigg)$

$=c(1-4.0 \times 10^{-17})$.

However, using the $v=c(1-\epsilon)$ approach, we have $v = \frac{2c}{2+\epsilon^{2}}$ with $\epsilon = 6.7 \times 10^{-5}$,

so $v = \frac{c}{1+\epsilon^{2}/2}=c(1+\frac{\epsilon^{2}}{2})^{-1} = c\Big(1-\frac{\epsilon^{2}}{2}) = c(1-\frac{(6.7 \times 10^{-5})^{2}}{2}\Big) = c(1-2.2 \times 10^{-9})$.

The two answers are different by many orders of magnitude.

 

[PeroK]

That second method, using the gamma factor, looks rock-solid to me. It shows that you end up with an answer to the order of the 4th power of gamma. Which is the order of $\epsilon^2$.

In the first method you used the approximation $1-\epsilon$ in the calculations for part b) and dropped the term in $\epsilon^2$ (which wasn't needed for part a). If you put that term back in, you should get the same answer as in the second method.

You may also want to check that value for $\epsilon$. Shouldn't it be approx $10^{-8}$?

 

[spaghetti3451]

I see. I show the equivalence below:

I have
$v'=\frac{2v}{1+\frac{uv}{c^{2}}} = \frac{2c(1-\epsilon)}{1+\frac{c^{2}(1-\epsilon)^{2}}{c^{2}}} = \frac{2c(1-\epsilon)}{1+(1-\epsilon)^{2}}=\frac{2c}{(1-\epsilon)^{-1}+(1-\epsilon)}=\frac{2c}{(1+\epsilon+\epsilon^{2})+(1-\epsilon)}=\frac{2c}{2+\epsilon^{2}}= \frac{c}{1+\epsilon^{2}/2}=c(1+\frac{\epsilon^{2}}{2})^{-1} = c\Big(1-\frac{\epsilon^{2}}{2})$.

Now, using the value of $v=c\Big(1-\frac{1}{2}\Big(\frac{938\times 10^{-6}}{7}\Big)^{2}\Big)$ from part (a), we have $\epsilon =\frac{1}{2}\Big(\frac{938\times 10^{-6}}{7}\Big)^{2} = 9.0 \times 10^{-9}$.

Therefore, plugging in the value of $\epsilon$ into $v= c\Big(1-\frac{\epsilon^{2}}{2})$, we get $v=c(1-4.0 \times 10^{-17})$.

Thanks!