The equation E = pc applies exclusively to massless particles, such as photons and gluons, and cannot be applied to particles with rest mass. The discussion clarifies that for particles with rest mass, the correct energy equation is E = γm₀c², where γ is the Lorentz factor. The derivation shows that E ≠ pc when rest mass (m₀) is non-zero, confirming the limitations of the equation in relation to relativistic momentum. This conclusion is essential for understanding the energy-momentum relationship in relativistic physics.
PREREQUISITESStudents of physics, particularly those studying relativity and particle physics, as well as educators looking to clarify the energy-momentum relationship for their students.
Just a clarification of what gleem is saying:gleem said:where of course mc2 = E
OK. Now substitute ##E=\gamma m_0 c^2## for E. What does this reduce to? Do you know the relativistic momentum expression?andrevdh said:I get up to the point where it evaluates to E2(v2/c2)?
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.andrevdh said: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?