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Invariant mass question

  1. Sep 1, 2016 #1
    1. The problem statement, all variables and given/known data
    At the LHC at CERN protons with an energy of 6.5 TeV (= 6.5·1012eV) each are collided with each other.To achieve the same invariant mass in a fixed target experiment, what would the energy of the proton beam have to be?

    2. Relevant equations
    E2 - p2c2 = m2c4

    E2 - p2c2= (m2c2-m2v2)/(1-v2/c2)

    3. The attempt at a solution
    Because the protons are colliding p =0. From there I used the first equation to solve for m.
    m = .00014. This is my invariant mass, right?

    Now, with the proton colliding with a wall, there is momentum to be considered, right? This is where I get stuck. I need to know the velocity of the protons to find their momentum, don't I? I thought I could us E = (mc2)/√(1-v2/c2) and solve for v but it doesn't work out.

    Any feedback on where I'm not understanding something correctly or missing something would be helpful.
     
  2. jcsd
  3. Sep 2, 2016 #2
    It will be an insane amount of energy. About 8.5*10^4 TeV.

    To figure this out, use 4-vectors
    [tex] P = (E,\vec{p}) [/tex]
    where invariant mass is just the square of the 4-vector:
    [tex]P^2 = E^2 - p^2 = m^2[/tex]

    The invariant mass of two colliding particles is:
    [tex] M^2 = (P_1 + P_2)^2 = P_1^2 + P_2^2 + 2 P_1 \cdot P_2 [/tex]
    Where [tex] P_1 = (E_1,\vec{p}_1) [/tex]
    and [tex] P_2 = (E_2,\vec{p}_2) [/tex]

    For fixed target experiment, where 1 proton is a stationary target the 4-vectors are
    [tex] P_1 = (E^*,\vec{p}^*) [/tex]
    and [tex] P_2 = (m,0) [/tex]
     
    Last edited: Sep 2, 2016
  4. Sep 2, 2016 #3

    George Jones

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    This is a nice solution, but ... from the Physics Forums Rules:

    https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/

     
  5. Sep 2, 2016 #4
  6. Sep 3, 2016 #5
    thank you! ill play with this some more tonight. I'm just seeing this so i missed the solution you posted. I'm guessing you posted a solution based on the third comment haha.
    thank you though. I'm fairly confused with this stuff.
     
  7. Sep 4, 2016 #6

    vela

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    Doesn't seem right as ##m## typically has units.

    The energy of the proton is much greater than its mass, so it's moving essentially at the speed of light. If you want to get a useful result, you need to expand using a Taylor series:
    $$\frac vc = \sqrt{1-\left(\frac{m_p}{E_p}\right)^2} \cong 1 - \frac 12 \left(\frac{m_p}{E_p}\right)^2.$$ In this case, you find the protons are moving about 3 m/s slower than light.

    Generally it's a good idea to avoid working with velocities and focus on energy and momentum instead. Note that in DuckAmuck's approach, you never need to calculate the speed of the protons.

    Another approach to this problem is to boost to the frame where one proton is at rest. You'd probably want to use rapidities if you took this approach.
     
    Last edited: Sep 5, 2016
  8. Sep 5, 2016 #7
    why is the invariant mass squared equal to (P1+P2)^2 and not (P1^2 + P2^2)?
     
  9. Sep 5, 2016 #8

    vela

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    You're looking for the invariant mass of the two-proton system, so you need to find the square of the four-momentum of the system. To get the momentum of the system, you have to sum the momenta of the parts that make up the system.

    P1^2 and P2^2 would be the invariant masses of the individual protons squared, and the sum would always be ##2m_p^2##.
     
  10. Sep 5, 2016 #9
    im a little confused on how 4-momentum works but what i have so far...

    for the colliding system
    M^2 = 2E^2

    for the system with a stationary particle
    M^2 = (E'^2 - p'^2) + (mproton)

    then set those equal to each other and solve for E'. but i still have a 'p'' hanging around.
     
  11. Sep 5, 2016 #10

    vela

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    How did you get this? It's incorrect, or you're being sloppy.

    This equation can't be right. The units don't work out.
     
  12. Sep 5, 2016 #11
    honestly, i must have no freaken idea what I'm doing then.

    m^2 = E^2 - p^2.

    E = total energy and p = total momentum. since they are colliding, their momentums cancel out. their energies are equal. so m^2 = 2E^2. this is the invariant mass of the first scenario, right? thats how i got there.

    as for the second part. I'm just trying to make sense of DuckAmuck wrote. obviously, I'm not making much sense of it.

    i actually have to solution that DuckAmuck posted initially but thats little help because i don't understand where some of the values come from. I'm just trying to work from the instructors slides now.
     
  13. Sep 5, 2016 #12
    actually, i was finally able to make sense of it.


    times like this really make me regret not changing majors two years ago haha
     
  14. Sep 5, 2016 #13

    vela

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    Is this for a particle physics class? I found the chapter on relativity in Griffith's textbook to be really clear and helpful.
     
  15. Sep 5, 2016 #14
    The class is titled "Quantum Mechanics and Relativity". The class is taught in a "non traditional" form, I guess. We were not suggested a book to purchase at the beginning of the semester.
    Which book by Griffith? Intro to QM?
     
  16. Sep 6, 2016 #15

    vela

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    Intro to Particle Physics
     
  17. Sep 6, 2016 #16
    The first one treats both particles' vectors as one vector, and so the mass you get is their *shared* invariant mass (squared). The second one is each particle's independent invariant mass (squared) added together.

    [tex] P^2 = (E,p)^2 = M^2 = E^2 - p^2 [/tex]

    which is equivalent to [tex] E^2 = M^2 + p^2 [/tex]

    Side note: in my calculations c=1 (you'll find this is a popular convention)
     
  18. Sep 6, 2016 #17
    Last edited by a moderator: May 8, 2017
  19. Sep 6, 2016 #18
    thanks guys. i have some left over from a book scholarship so ill talk to my library and see if they can order it. I'm sure they can though.

    DuckAmuck, that make total sense. thank you. i noticed all the c = 1. our instructor has been doing that since the beginning of the semester.
     
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