Question #9be0d

1 Answer
Jul 15, 2015

This equation is an approximation of the relativistic energy of a particle for low velocities.

Explanation:

I'm assuming some knowledge about special relativity, namely that the energy of a moving particle observed from an inertial frame is given by #E=gammamc^2#, where #gamma=1/sqrt(1-(v/c)^2)# the Lorentz factor. Here #v# is the velocity of the particle observed by an observer in an inertial frame.

An important approximation tool for physicists is the Taylor series approximation. This means we can approximate a function #f(x)# by #f(x)approxsum_(n=0)^N(f^((n))(0))/(n!)x^n#, the higher #N#, the better the approximation. In fact, for a large class of smooth functions this approximation becomes exact as #N# goes to #oo#. Note that #f^((n))# stands for the nth derivative of #f#.

We approximate the function #f(x)=1/sqrt(1-x)# for small #x#, we note that if #x# is small, #x^2# will be even smaller, so we assume we can ignore factors of this order. So we have #f(x)approxf(0)+f'(0)x# (this particular approximation is also known as the Newton approximation). #f(0)=0# and #f'(x)=1/(2(1-x)^(3/2))#, so #f'(0)=1/2#. Therefore #f(x)approx1+1/2x#.

Now we note that #gamma=f((v/c)^2)#. Indeed if #v# is small relative to #c#, which it will be in day to day situations, the approximation holds, so #gammaapprox1+1/2(v/c)^2#. Substituting this in the equation for the total energy of a particle gives #Eapproxmc^2+1/2mv^2#. This gives us the kinetic energy #E_("kin")=E-E_"rest"approxmc^2+1/2mv^2-mc^2=1/2mv^2# for low velocities, which is consistent with classical theories. For higher velocities it is wise to use more terms from the Taylor series, ending up with so called relativistic corrections on the kinetic energy.