Where did the 1/2 go in E=mc^2?

[olgerm]

The limit is effectively $c \to \infty$, not $c \to 0$, or, perhaps better, $(v/c) \to 0$

Yes. I corrected it now.

 

[olgerm]

If m is relativistic mass in the original post:
K and E are still not same quanities, beacuse $E=m_{relativistic}*c^2$ is rest energy plus kinetic energy and $K=\frac{m_0*v^2}{2}$ is kinetic energy.

 

[cmb]

I suggested a while back that, of course, E=mc^2 is not an equation with 'c' as a variable, it is really a conversion factor. At least in the rest frame.

I have seen the following additional explanation, which describes the connection with Newtonian mechanics. If one writes the Lorenz factor as y = 1/SQRT(1-b^2), where b = v/c (sorry no latex) then the equation in special relativity, E = ymc^2, can be expanded as a Maclaurin series.

y = 1/SQRT(1-b^2) = 1 + 1/2.b^2 + 3/8.b^4 + 5/16.b^6 ... etc

As v (i.e. b) tends to small values then the end terms of the series vanish.

The first two terms of the expansion are then

E = mc^2 + 1/2.mv^2

which is therefore the Newtonian approximation for small v.

So I suppose an alternative answer for the OP is that the "1/2" can be made to "reappear" by a Maclaurin expansion of the Lorentz factor.

 

[Ibix]

So I suppose an alternative answer for the OP is that the "1/2" can be made to "reappear" by a Maclaurin expansion of the Lorentz factor.

This is a standard way of showing that the low speed limit of relativistic kinetic energy is Newtonian kinetic energy, yes.

 

[Herman Trivilino]

The kinetic energy of a particle can be expressed as $\frac{\gamma^2}{\gamma+1}mv^2$. In the low speed limit $\gamma \approx 1$ so the fraction $\frac{\gamma^2}{\gamma+1}\approx \frac{1}{2}$. So you can see that the $\frac{1}{2}$ didn't go anywhere. It's still there!