1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Special Relativity !

  1. Nov 15, 2009 #1

    bon

    User Avatar

    Special Relativity URGENT!

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

    The galaxy is about 10^5 light years across and the most energetic cosmic rays known have energies of the order of 10^19 eV. How long would it take a proton with this energy to cross the Galaxy as measured in the rest frame of (i) the galaxy (ii) the proton

    2. Relevant equations



    3. The attempt at a solution

    So I've done (i) and got the right answer (10^5 years) since to the best accuracy of my calculator, the particle is travelling at the speed of light....

    For part (ii) I want to use lorentz transform, but i can't seem to get my calculator to take the accuracy..

    basically: ct' = gamma(ct-Bx) where B is v rel..

    now i found gamma to be (1.06699... x 10^10) which means B is root(1 - 8.798...x10^-21)
    now on my calculator this gives a value of B of 1, which obviously means t' = 0...The actual answer gives t' as just over 4 minutes...

    I'm wondering if there is another way to work out t'..how can i compute it to get the answer required!?

    THANKS :) !
     
  2. jcsd
  3. Nov 15, 2009 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Special Relativity URGENT!

    Hint: From the proton's frame, what's the distance across the galaxy? (All you need is gamma.)


    (Don't give all your threads the same title! :smile:)
     
  4. Nov 15, 2009 #3
    Re: Special Relativity URGENT!

    Are you making sure to convert the distance in light years to meters & standardizing your units?
     
  5. Nov 15, 2009 #4

    bon

    User Avatar

    Re: Special Relativity URGENT!

    You shouldn't need to convert it all to meteres..
     
  6. Nov 15, 2009 #5

    bon

    User Avatar

    Re: Special Relativity URGENT!

    Ok so x' = x/gamma, but then how do i work out t', because v' = 0? since it is in the rest frame of the proton...

    also, was i right in using e = Mc^2/gamma for the first part, rather than E^2 = m^2 c^4 + p^2 m^2??

    When do i know when to use the former and when to use the latter?
     
  7. Nov 15, 2009 #6

    bon

    User Avatar

    Re: Special Relativity URGENT!

    anyone?
     
  8. Nov 15, 2009 #7

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Special Relativity URGENT!

    In the proton's frame it's the galaxy that is moving.

    I'd use KE = mc^2(gamma - 1).

    It's just a matter of convenience.
     
  9. Nov 15, 2009 #8

    bon

    User Avatar

    Re: Special Relativity URGENT!

    Shouldn't it come to the same though?

    And how can you use this? You aren't told its KE, you are told its total E..

    Also, surely by doing it your way with x', you encounter the same problem since v(rel) is the same value i couldn't work out accurately enough previously...

    for this reason, my answer turns out to be 4.93mins, when it should be 4.96 mins..

    how can I get 4.96 mins?

    thanks again for your help..
     
  10. Nov 15, 2009 #9
    Re: Special Relativity URGENT!

    Did you use the exact mass of the proton?

    Anyway, the problem says that the energy is of the order of 10^19 eV. So, I would not even bother to get the last decimal correct. I would solve this problem as follows. I would not bother looking up the precise mass of the proton, as I already know that it is of the order of one GeV.

    Whether the 10^19 eV is supposed to be the total energy or just the inetic energy is irrelevant. I can already tell that the gamma factor is of the order of 10^10. And that then means that the proper time for the proton is 10^(-5) years = 0.00365 days (including extra decimals for clarity). Converting to hours is easy if you use that 24 is of the order of 25 and 25*4 = 100. So we find that the proper time is about 0.36/4 hours = 0.09 hours = of the order of 5 minutes.
     
  11. Nov 15, 2009 #10

    bon

    User Avatar

    Re: Special Relativity URGENT!

    Okay thanks - useful to see, but unfortunately not what I've been asked to do.
     
  12. Nov 15, 2009 #11
    Re: Special Relativity URGENT!

    I know you don't need to but if you convert everything into metres/seconds it's alot more structured and quicker overall.

    your answer is extremeley close given the huge inaccuracies, check the mass of the proton to another significant figure maybe, also did you use 365 or 365.25 for number of days in a year? on second thought that wouldn't make much difference at all really, perhaps your value for gamma is on the short side...
     
    Last edited: Nov 15, 2009
  13. Nov 15, 2009 #12
    Re: Special Relativity URGENT!

    I also find 4.93 minutes using the exact proton mass and 365.26 days/year. To answer your question "how can I get 4.96 mins", you have to go beyond special relativity and apply some simple logic.

    Consider the ratio: 4.96/4.93 = 1.006

    If you take the proton mass you used and multiply that with 1.006, you get the desired answer.

    The proton mass is 1.67262158 × 10^(-27) kg

    Multiplying by 1.006 gives:

    1.68 × 10^(-27) kg

    The neutron mass is:

    1.67492729 × 10^(-27) kg

    We see that the mass equals the neutron mass to almost 3 significant figures. The fact that the proper time of 5 minutes is less than the neutron's half life suggests to me that the person who formulated the problem originally formulated it in terms of neutrons, perhaps asking if the neutron could cross the galaxy. He could have changed his mind (perhaps because asking that already gives away the clue that the proper time would be very small) but forgot to replace the neutron mass by the proton mass.
     
  14. Nov 15, 2009 #13

    Doc Al

    User Avatar

    Staff: Mentor

    Re: Special Relativity URGENT!

    When they tell you the energy of some particle, usually they are talking about its KE. But to the accuracy you need, it doesn't matter.

    The answer is, of course, close enough to c that it doesn't matter for calculating the time of travel. If, for some reason, you wanted to find a more accurate value for the speed, rather than calculate it directly, calculate its difference from c. Replace beta with β = 1 - x, where x << 1. Then you can find x given gamma. You'll find x to be so small that using c is good enough.

    Beats me. Perhaps they used a different value for the rest mass of the proton. Regardless, your answer is well within the accuracy of your data.

    What textbook are you using?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Special Relativity !
  1. Special Relativity (Replies: 9)

  2. Special Relativity (Replies: 3)

  3. Special Relativity (Replies: 1)

Loading...