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!

Can the equation E = pc be applied to particles?

  1. Jul 28, 2015 #1
    1. The problem statement, all variables and given/known data


    Can the equation E = pc be applied to particles? Why or why not?


    2. Relevant equations


    3. The attempt at a solution
    It can be applied to particles that DONT have a rest mass (photons, which are particles). It can not be applied to particles that have a rest mass (almost everything).
     
  2. jcsd
  3. Jul 28, 2015 #2
    OK start, but it would be more complete to consider the equations that express the relative energy of particles that do have rest mass.
     
  4. Jul 29, 2015 #3

    andrevdh

    User Avatar
    Homework Helper

    So its only applicable to photons (and gluons)?
    But you only make a stament and do not explain why it is so?
     
  5. Jul 29, 2015 #4
    Try this. Evaluate the quantity

    m2c4 - m02c4
    where of course mc2 = E
     
  6. Jul 29, 2015 #5
    Just a clarification of what gleem is saying:
    E over here is the total energy of the object. So the actual formula for this will be ##E= \gamma m_0 c^2##, where $$\gamma = \frac {1}{\sqrt{1-\frac{v^2}{c^2}}}$$. By m, gleem means the relativistic mass (an orthodox concept, really) ##m=\gamma m_0##.
    (I'm providing this clarification in case the OP accidentally uses the rest mass for the formula for E over here)
     
    Last edited: Jul 29, 2015
  7. Jul 31, 2015 #6

    andrevdh

    User Avatar
    Homework Helper

    I get up to the point where it evaluates to E2(v2/c2)?
     
  8. Jul 31, 2015 #7
    OK. Now substitute ##E=\gamma m_0 c^2## for E. What does this reduce to? Do you know the relativistic momentum expression?
     
  9. Jul 31, 2015 #8

    andrevdh

    User Avatar
    Homework Helper

    I think it comes to (pc)2?
    Which approaches the energy-momentum relation from the other side.
    How is this relevant to the original question?
    We evaluated a difference between two terms and found
    that they are related to the relativistic momentum of the entity?
     
    Last edited: Jul 31, 2015
  10. Jul 31, 2015 #9
    It is relevant because you just derived the forumula ##pc=\sqrt{{\gamma}^2m_0^2 c^4 - m_0 ^2c^4} = \sqrt{E^2-m_0^2 c^4}##, proving the fact that ##E≠pc## if ##m_0 ≠0 ##, which I believe was your original question.
     
  11. Jul 31, 2015 #10

    andrevdh

    User Avatar
    Homework Helper

    I'll think about it.
    It seems to make sense.
    Thank you.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Can the equation E = pc be applied to particles?
Loading...