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Inconsistency in formulas for relativistic energy

  1. Nov 17, 2013 #1
    I'm very noob at this and am a bit confused:

    Formula 1: [itex] E_T = \gamma \cdot m c^2 [/itex]
    Formula 2: [itex] p = \gamma m v[/itex]
    Formula 3: [itex] E_T^2 = (pc)^2 + (mc^2)^2[/itex]

    Formula 3 says a particle of negligible mass can have energy, but isn't this in contradiction to
    formula 1? Unless maybe the velocity of the particle is c, such that ##\gamma## becomes infinite too?

    So if I get an assignment where I'm supposed to neglect the mass of a moving particle, I must either neglect its total energy or set its velocity to c in my calculations?

    For instance, let's say I need to calculate on myon decay: ##\pi_+ \rightarrow \mu_+ + \nu##. Do I have to set the velocity of the myon-neutrino to c? But why can't I just neglect the energy of the neutrino?
     
  2. jcsd
  3. Nov 17, 2013 #2
    Formulas 1 and 2 don't hold for massless particles.

    Also, you have to properly define all the properties.
    In Newtonian physics rest mass and velocity are basic properties and energy and momentum are derived from them.
    In relativity it's the opposite. Energy and momentum are basic properties and rest mass and velocity are derived from them. We avoid division by zero this way.
     
  4. Nov 17, 2013 #3

    Dale

    Staff: Mentor

    As haael said, formula 3 is the correct one. Formulas 1 and 2 can be derived from formula 3 in the special case that mass is non-zero.
     
  5. Nov 17, 2013 #4
    Uhmmmmm, in my book formula 3 was derived FROM formula 1 and 2, so I assumed formula 3 was built upon formula 1 and 2.

    But OK.. so formula 3 is the one which is correct for all cases, while 1 and 2 are only correct when the particle has non-neglible/non-zero mass? Is there an easy-to-understand derivation for formula 3 lying around somewhere on the web?
     
  6. Nov 17, 2013 #5

    Dale

    Staff: Mentor

    In units where c=1, the four-momentum is given by ##\mathbf{P}=(E,\mathbf{p})##. The Minkowski norm of this quantity is then ##|\mathbf{P}|^2=E^2-\mathbf{p}^2\equiv m^2##. Rearrange and you get your equation 3.
     
    Last edited: Nov 17, 2013
  7. Nov 17, 2013 #6

    PAllen

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    1 and 2 hold for anything except an exactly massless particle. For a massless particle, gamma is undefined, so any formula using it is useless for a massless particle.
     
  8. Nov 17, 2013 #7

    jtbell

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    You can solve for all the momenta and energies in any two-body decay like this one, using only energy and momentum conservation, and your "Formula 3" above (applied separately to each particle). This is most easily done in the rest frame of the initial particle (the pion in your example).

    After you have the energies and momenta, if you want the speeds of the decay products, you can get them most simply using

    $$\beta = \frac{v}{c} = \frac{pc}{E}$$
     
    Last edited: Nov 17, 2013
  9. Nov 18, 2013 #8

    stevendaryl

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    There are several different arguments leading to formulas 1-3 that all have the same conclusion, even though their starting points are very different. There is a mathematical argument that it's the only possibility, which is intuitively unsatisfying, perhaps. In addition, there is a purely classical argument (from Maxwell's equations plus the Lorentz force law) that electromagnetic waves carry energy and momentum that are related by [itex]E = pc[/itex] (which is the special case of 3 that applies when the mass is zero). Then there are several different heuristic arguments that massive particles have to satisfy equations 1&2, which imply equation 3. One derivation (I think this is original with Einstein) considers a massive box with a pulse of light bouncing back and forth between the two walls. Another derivation starts with the assumptions that (1) Energy is proportional to mass, and is conserved in collisions, (2) momentum is proportional to mass, and is in the same direction as the velocity, and is conserved in collisions, (3) in the nonrelativistic limit, kinetic energy and momentum approach their Newtonian values, [itex]\frac{1}{2} m v^2[/itex] and [itex]m v[/itex].
     
  10. Dec 1, 2013 #9
    Thanks for all the replies, but I have one more question:

    For a system of particles, is the momentum term in formula 3 the sum of the momentums, right? Well, doesn't that bring forward contradictions?

    For example: Let's say you have two electrons heading towards each-other at equal speeds (seen from inertial frame). Then the momentum for the system as a whole is zero, and thus total energy is just equal to 2 times the rest energy of an electron? But this is ridiculous, as this would mean the system contains zero kinetic energy....
     
  11. Dec 1, 2013 #10

    PAllen

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    For a system of particles you add the 4-momenta, the quantity (E,p) in units of c=1. Thus, for two oppositely moving electrons you get (2E,0) as the 4-momentum of the system. Then, m^2= E^2 - p^2 tells that the invariant mass of the system is equal to the total energy of the two electrons, with no momentum for the system. Effectively, the systems can be treated as one object with no momentum and rest mass determined by total kinetic energy and rest mass of the electrons.

    The key point is that the 'rest mass' of a system is not simply the sum of the rest mass of its components.
     
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