MassEnergy
Discussion
momentum and energy separately
These ideas are completely disorganized. Keep that in mind when reading this.
Relativity has a different equation for (almost) everything. It's like classical physics just isn't good enough. There's a different one for time (time dilation) and a different one for space (length contraction) and now there's a different one for momentum (relativistic momentum) and another different one for energy (relativistic energy).
The equation for relativistic momentum looks like this…
p =  mv 
√(1 − v^{2}/c^{2}) 
When v is small (as it is for the kinds of speeds we deal with in everyday life) the denominator is approximately equal to one and the equation reduces to its classical version…
v ≪ c  ⇒  p ≈ mv 
which is as it should be. Relativity doesn't replace classical physics, it supplements it. All equations in special relativity should reduce to classical equations at low velocities. This is known as the correspondence principle.
The equation for relativistic energy looks like this…
E =  mc^{2} 
√(1 − v^{2}/c^{2}) 
Applying the correspondence principal to give us the classical equations is not so easy here. Once again, at low speeds the denominator is one, but the numerator we're left with is something new. Something with no classical counterpart. Something famous.
v ≪ c  ⇒  E ≈ mc^{2} 
This equation says that an object at rest has energy, which is why it is sometimes called the rest energy equation. It also says that the reason an object at rest has any energy at all is because it has mass, which is why this equation is also known as the massenergy equivalence.
Let's try a more sophisticated approach using the binomial series (sometimes called the Taylor series).
(a + b)^{n} = 

⎛ ⎝ 
n  ⎞ ⎠ 
a^{n − k}b^{k}  
k 
If we apply this to the denominator of the energy formula…
√(1 − v^{2}/c^{2})
where…
a = 1  b = −v^{2}/c^{2}  n = −½ 
we get…
E = mc^{2}  ⎛ ⎝ 
1 +  1  v^{2}  +  3  v^{4}  +  5  v^{6}  +  35  v^{8}  +  63  v^{10 }  +…  ⎞ ⎠ 

2  c^{2}  8  c^{4}  16  c^{6}  128  c^{8}  256  c^{10} 
Multiply all the terms in the expansion by mc^{2}. The zeroeth term is the rest energy.
E_{0} = mc^{2}
The first term is the classical equation for kinetic energy.
E_{1} =  1  mv^{2} 
2 
The higher order terms are corrections to the classical kinetic energy equation that become more and more noticeable as the speed approaches the speed of light.
The energy added to an object to take it from an initial speed of zero to a final speed of something is its kinetic energy.
K = E − E_{0}
Subtitute.
K =  mc^{2}  −  mc^{2} 
√(1 − v^{2}/c^{2})  √(1 − v_{0}^{2}/c^{2}) 
Let initial velocity equal zero.
K =  mc^{2}  −  mc^{2} 
√(1 − v^{2}/c^{2})  1 
Factor out like terms and in the end we get an equation for relativistic kinetic energy that looks like this in expanded notation…
K =  ⎛ ⎝  1  − 1  ⎞ ⎠  mc^{2} 
√(1 − v^{2}/c^{2}) 
…and like this is compact notation.
K = (γ − 1)mc^{2}
momentum and energy together
The momentum equation…
p =  mv 
√(1 − v^{2}/c^{2}) 
and the energy equation…
E =  mc^{2} 
√(1 − v^{2}/c^{2}) 
have a common thing — the Lorentz factor, a.k.a. the relativistic gamma…
γ =  1 
√(1 − v^{2}/c^{2}) 
which means they can be written in a more compact form like this…
p = γmv  E = γmc^{2} 
For no immediately apparent reason, start with this expression…
E^{2} − p^{2}c^{2}
Replace energy and momentum with their gamma versions like this…
γ^{2}m^{2}c^{4} − γ^{2}m^{2}v^{2}c^{2}
The identity rule allows us to multiply the second term by 1 in the form of c^{2}/c^{2}.
γ^{2}m^{2}c^{4} − γ^{2}m^{2}v^{2}c^{2}(c^{2}/c^{2})
Using the associative and commutative properties of multiplication, move things around in the second term.
γ^{2}m^{2}c^{4} − γ^{2}m^{2}(v^{2}/c^{2})(c^{2}c^{2})
Simplify a little.
γ^{2}m^{2}c^{4} − γ^{2}m^{2}(v^{2}/c^{2})c^{4}
And pull out like terms.
γ^{2}m^{2}c^{4}(1 − v^{2}/c^{2})
Notice that the stuff in parentheses is the reciprocal of γ^{2}, which means everything cancels out except the stuff in the middle…
m^{2}c^{4}
Thus…
E^{2} − p^{2}c^{2} = m^{2}c^{4}
or…
E^{2} = p^{2}c^{2} + m^{2}c^{4}
This is the relativistic energymomentum relation. For massed particles at rest we get the famous massenergy relationship or the rest energy equation…
v = 0  ⇒  E = mc^{2} 
For massless particles like photons…
m = 0  ⇒  E = pc 
 Does E^{2} = p^{2}c^{2} + m^{2}c^{4} also show that…
 massless particles must travel at the speed of light.
 particles traveling at the speed of light must be massless.
 Can E^{2} = p^{2}c^{2} + m^{2}c^{4} also be thought of as a version of Pythagoras' theorem?
 The squared terms correspond to the sides of a right triangle.
 Are the angles in this triangle anything?
 Is energymomentum a four dimensional vector with
 pc as the spatial component?
 mc^{2} as the temporal component?