Energy equal to time component of 4-momentum?

[marschmellow]

I'm sure this question gets brought up a lot, but I can't figure out why this is true. Everywhere I look, people simply equate the two as though it's some axiom, but never an explanation for why. It seems to me like E≠m/√(1-v^2) in general.

 

[Dale]

It is correct. Can you think of an example where E≠m/√(1-v^2).

 

[Meir Achuz]

Energy is the time component of the momentum 4-vector. Using that fact, and the fact that v=p/E, you can derive your E equation.

 

[marschmellow]

Sure, if one particle's frame measures another particle's velocity to be c/2 with rest mass of 2 kilograms, then its kinetic energy is 1/2 * 2 kg * (1.5*10^8 m/s)^2, which is not equal to its time component of four-momentum, which is 2 kg / √(3/4). The units don't even match up. Now my understanding of special relativity is completely self taught, so I might be missing something huge here.

 

[jtbell]

if one particle's frame measures another particle's velocity to be c/2 with rest mass of 2 kilograms, then its kinetic energy is 1/2 * 2 kg * (1.5*10^8 m/s)^2

No. In SR, kinetic energy is not $\frac{1}{2}mv^2$, but rather

$$K = \left[\frac{1}{\sqrt{1 - v^2/c^2}} - 1 \right] m c^2$$

in which by "m" I mean the "invariant mass" which is often called "rest mass."

When v << c, $K = \frac{1}{2}mv^2$ is a very close approximation to the equation above.

 

[jtbell]

Also, in the definition of the four-momentum, (E/c, p_x, p_y, p_z), E is the "total" energy, i.e. E = K + mc^2, not the kinetic energy K alone.

 

[The_Duck]

The reason to define the momentum 4-vector this way is that this way if you have the momentum 4-vector in one frame and you want to know it in another frame in relative motion to the first frame, you just multiply the momentum 4-vector by the Lorentz transformation matrix. This is why we like 4-vectors: their values in different frames are simply related to each other by the Lorentz transformation matrices.

 

[atyy]

There's no reason to this. It is a new definition of energy which explains why the old definition of energy is applicable at low speeds. Why this new quantity is also called energy is simply a matter of convention, mostly having to do with physicists believing that "energy" should be conserved.

 

[Parlyne]

There's no reason to this. It is a new definition of energy which explains why the old definition of energy is applicable at low speeds. Why this new quantity is also called energy is simply a matter of convention, mostly having to do with physicists believing that "energy" should be conserved.

It's more than just a belief. Noether's theorem shows that for any symmetry of a physical system, there is a conserved quantity. For classical situations with time translation symmetry, the conserved quantity is energy in the usual non-relativistic sense. By direct analogy, in relativity we call the quantity conserved as a result of time translation symmetry energy, as well.

 

[atyy]

It's more than just a belief. Noether's theorem shows that for any symmetry of a physical system, there is a conserved quantity. For classical situations with time translation symmetry, the conserved quantity is energy in the usual non-relativistic sense. By direct analogy, in relativity we call the quantity conserved as a result of time translation symmetry energy, as well.

Yes, that's a good way to explain the choice of generalization.