1

I want to know why Mass-Energy equivalence comes from special relativity:

I see Einstein's Special Relativity as providing corrections to Classical Mechanics, when the speed of particles becomes close to the speed of light.

Why is the Mass-Energy equivalence explicit only when particles get close to the speed of light?

PhyEnthusiast
  • 2,908
  • 1
    You are actually asking "why momentum and energy are a four vector under lorentz transformations".http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html . see also http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/releng.html – anna v Jan 01 '18 at 10:51
  • 1
    Possible duplicates: https://physics.stackexchange.com/q/43813/2451 , https://physics.stackexchange.com/q/178960/2451 and links therein. – Qmechanic Jan 01 '18 at 11:37

2 Answers2

0

A simple way to state the equivalence between energy and mass of a particle is to extend the concept of the Newtonian three-momentum to a Lorentzian four-vector.
$p^\mu = m dx^\mu / d\tau$
where:
$\mu = 0, 1, 2, 3$
$p^\mu$ = four-momentum
$x^\mu$ = (ct, x, y, z)
$m$ = rest mass
$\tau$ = proper time
As $dt / d\tau = \gamma$, and showing correct units you define $\gamma m c = E / c$,
where:
$\gamma = 1 / \sqrt{1 - v^2 / c^2}$ Lorentz factor
$v$ = Newtonian velocity
Hence:
$p^\mu = (E/c, p^i)$
where:
$i$ = 1, 2, 3
By contracting the four-momentum with itself you get an invariant:
$p_\mu p^\mu = -E^2 / c^2 + p^2$ (1)
where:
$\eta_{\mu\nu} = diag(-1, 1, 1, 1)$ Lorentzian metric
$p^2 = p_i p^i$
In the rest frame of the particle you have:
$(p_\mu p^\mu)_0 = -E^2_0 / c^2 = -m^2 c^2$ (2)
From which you have:
$$E_0 = m c^2$$
Being an invariant, by equating (1) and (2):
$E^2 = p^2 c^2 + m^2 c^4$ (3)
Which relates the energy in a generic inertial reference frame to the rest mass of the particle and its momentum. What is particularly valuable is that it works for a massless particle (photon) as well.
Note: if you Taylor expand (3) vs. the velocity for a non-relativistic motion and neglect higher-order terms you get the Newtonian kinetic energy of the particle plus its rest energy. This is a confirmation of the procedure.

Michele Grosso
  • 3,781
0

"Why is the Mass-Energy equivalence explicit only when particles get close to the speed of light?"

I don't know quite what you mean by 'expicit', but the mass of a helium nucleus is getting on for 1% less than the sum of the masses of 2 protons and 2 neutrons when separated from each other. The mass discrepancy is $\frac{1}{c^2}$ times the energy needed to separate the nucleons from each other. Here is one of many example of $E=mc^2$ in action at low speeds.

Philip Wood
  • 35,641
  • So why does it require relativity to explain $E = mc^2$ – PhyEnthusiast Jan 30 '19 at 07:06
  • By studying what the Lorentz transforms imply for dynamics at all speeds (in particular collision dynamics), we discover that there are implications for the nature of mass and energy that must apply even at low speeds. The actual argument won't fit into this space, but is essentially that in any interaction, even inelastic, 4-momentum is conserved, including the time component, $\Sigma \gamma m c,$ and therefore $\Sigma \gamma m c^2.$ This must be the particles' total energy, so if KE ($\Sigma (\gamma m -m)c^2$) is lost, $\Sigma mc^2$ is gained and so on. – Philip Wood Jan 30 '19 at 10:46