# OPERA experiment - Special Relativity

1. Dec 4, 2015

### Isaac Pepper

1. The problem statement, all variables and given/known data
In 2011, researchers at the OPERA experiment thought they had seen neutrinos with mass m and energy E = 28 GeV moving faster than light. The baseline between the source and the detector was 731 km, and the neutrinos seemed to arrive 60.7 ns early, compared to the maximum physical velocity c. (a) Using $m^2c^4 = E^2 - p^2c^2$with p the momentum of the neutrinos, show that $v = \frac{\delta E}{\delta p}= c\sqrt{1-\frac{m^2c^4}{E^2}}$. (b) Calculate the mass of such faster-than-light neutrinos in $GeV/c^2$

2. Relevant equations
$p=\gamma mv$
$m^2c^4=E^2-p^2c^2$

3. The attempt at a solution
(a) $m^2c^4=E^2-p^2c^2$
I've tried many calculations and never managed to find the correct result. I don't really understand the $\frac{\delta E}{\delta p}$.
Am I right in trying to convert $p$ to $mv$ or is it $p=\gamma mc$?
By using $p=mv$, I manage to get :
$$m^2c^4=E^2-m^2v^2c^2$$
$$\frac{E^2-m^2c^4}{m^2c^2}=v^2$$
$$\frac{E^2}{m^2c^2}-c^2=v^2$$
$$v^2=c^2(\frac{E^2}{m^2c^2}-1)$$
$$v=c\sqrt{\frac{E^2}{m^2c^4}-1}$$
But this isn't using $\frac{\delta E}{\delta p}$ and the result isn't quite right. What am I missing?

(b) I'm kind of fuzzy on this - to me, photons travelling at the speed of light are massless, and if I apply the given formula to find the mass of a neutrino travelling faster than light...Well I hit a dead-end because it would be less than massless? Atleast that's the way I see it.
I've managed to work out the speed of the neutrino if it arrives 60.7ns earlier than light, which would be $v=300007473 m.s^{-1}$ assuming that $c=3.0*10^8 m.s^{-1}$
All calculations including a gamma factor would not work to me, as $1-\frac{v^2}{c^2}$ would be less than 0
so the square root in the gamma factor would be less than 0... How would I then calculate the mass of the faster-than-light neutrino?

Thanks to everyone for the amazing help!

2. Dec 4, 2015

### Isaac Pepper

Any help would be greatly appreciated, I'm still stuck on the question. Sorry for the double post.

3. Dec 4, 2015

### davenn

Hi there Issac

you did understand that the experiment gave the incorrect answer because of equipment failure ?

and NOTHING with mass can travel at c, let alone faster

Dave

4. Dec 4, 2015

### Isaac Pepper

I do understand that, which is partly why I don't understand what the question is asking me...
Any ideas on how to solve this? I don't even understand how to do part (a) at this point

5. Dec 4, 2015

### PeroK

For part a), why not start by differentiating (wrt $p$):

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

PS For b) Part a) gives you an equation for mass without the $\gamma$ factor. So, you could calculate an imaginary mass.

Last edited: Dec 4, 2015
6. Dec 4, 2015

### Isaac Pepper

Ok, even then I'm not sure how to differentiate this expression with respect to p
I can re-arrange to get $E=\sqrt{m^2c^2+p^2c^2}$ but I don't know how to differentiate that wrt p

7. Dec 4, 2015

### PeroK

Perhaps time to revise differentiation? It's easier, in any case, to leave it as $E^2$ and differentiate implicitly (if that's something you know how to do).

8. Dec 4, 2015

### Staff: Mentor

Have you calculated it? I'm curious whether my 9.756 i MeV are correct?

9. Dec 4, 2015

### Isaac Pepper

Do you mean for part (b) ?

I have done $$v=c\sqrt{1-\frac{m^2c^4}{E^2}}$$
$$v^2=c^2-\frac{m^2c^6}{E^2}$$
$$m^2=\frac{E^2(c^2-v^2)}{c^6}$$
$$m=\frac{1}{c^2}\sqrt{E^2-\frac{E^2v^2}{c^2}}$$
Plugged in numbers and got : $$m=\frac{28}{(3*10^8)^2}i\sqrt{4.98*10^{-5}}=\frac{28}{(3*10^8)^2}i*7.06*10^{-3} GeV/c^2$$
It's probably wrong at this point, I'm totally lost and have given up on (a) :/

Last edited: Dec 4, 2015
10. Dec 4, 2015

### Staff: Mentor

I'm not sure whether my calculation is correct. I kept $c$ as long as possible, set $c + ε = \frac{d}{1-t}$ with $d = 1$ lightsecond and $t=60.7 ns$. Then I broke all down to $m = 28 GeV \cdot c^{-1} \cdot \sqrt{-2cε - ε^2}$ from the solution in (a) and substituted $ε$ by $t$ until I got a factor $\sqrt{ 1- \frac{1}{(1-t)^2}}$. I didn't start with actual numbers not until this term. But there have been so many rearrangements that I'm not sure anymore.

I solved part (a) with PeroK 's hint. $\frac{∂}{∂p} E^2 = 2 E\cdot \frac{∂}{∂p}E = \frac{∂}{∂p}(m^2 c^4) + \frac{∂}{∂p} (p^2c^2) = 0 + 2pc^2$ and $pc^2 E^{-1} = c \sqrt{1- \frac{m^2 c^4}{E^2} }$ is just another expression for $E^2 = m^2 c^4 + p^2 c^2$.

EDIT: My $GeV$ are actually $GeV / c^2$. How did you get $v$?

Last edited: Dec 4, 2015
11. Dec 4, 2015

### Isaac Pepper

If the neutrino arrived $60.7*10^{-9}s$ before light would, then $t_2-t_1=\frac{d}{v}-\frac{d}{c}=-60.7*10^{-9}s$
In which case $t_2-t_1+\frac{d}{c}=\frac{d}{v}$
$$v=\frac{d}{t_2-t_1+\frac{d}{c}}=\frac{731*10^3}{-60.7*10^{-9}+\frac{731*10^3}{3*10^8}} = 300 007 473.5 m.s^{-1}$$
Atleast that's what I thought!

EDIT : I follow your implicit differentiation, but how do you end up with $pc^2E^{-1} = c\sqrt{1-\frac{m^2c^4}{E^2}}$ ?

12. Dec 4, 2015

### Staff: Mentor

Yep, that's my calculation of $v = \frac{d}{1-t}$, too (since $d/c$ is numerically just $1$).

$\frac{p^2c^2}{E^2} = \frac{E^2 - m^2 c^4}{E^2} = 1 - \frac{m^2c^4}{E^2} ⇒ \frac{pc}{E} = \sqrt{1 - \frac{m^2 c^4}{E^2}} ⇒ \frac{pc^2}{E} = c \cdot \sqrt{1 - \frac{m^2 c^4}{E^2}}$

13. Dec 4, 2015

### Isaac Pepper

Fair enough, I know the calculations I did after it were wrong though, so ignore them, I'm sure you have the right answer ! I will look through what I have done in the morning to try and work it out. Thank you for all your help, both of you !

14. Dec 4, 2015

### Staff: Mentor

Until here we are identically. Pulling $E^2$ out of the root gives us $m = 28GeV \cdot \sqrt{1 - \frac{v^2}{c^2}}$

Sorry for you, you are not wrong. I am. $\frac{c}{v} = 1-\frac{c}{d}Δt$ where $d=731,000 m$ and $Δt = 60.7 ns$.
(I calculated $\frac{c}{v} = 1-Δt$ which is the wrong velocity.)

Your correct result is $\frac{v}{c} = 1+ 2.4912 * 10^{-5}$
Therefore $m = 28 GeV \cdot \sqrt{1 - (1 + 2.4912 * 10^{-5} )^2} = 28 GeV \cdot 7.06 \cdot 10^{-3}i = 197.68 i MeV$

EDIT: If $28 GeV$ is already divided by $c^2$ as usual. Forgot some i.

Last edited: Dec 4, 2015
15. Dec 4, 2015

### Isaac Pepper

Yeah, that seems to my solution too, though I divided it by $c^2$ so my answer was $$\frac{28GeV}{(3*10^8)^2}*i*7.06*10^{-3}=i*2.20*10^{-18} GeV/c^2$$

16. Dec 4, 2015

### Staff: Mentor

A like this kind of poorness: two footsore guys made it together over the goal line ...