Why Does Relativistic Kinetic Energy Differ at Low Velocities?

[JamesClarke]

Hi, I`ve been dabbling in some basic special relativity and when your deriving the famous equation

E = mc^2 you get to this equation just before it.

KE = mc^2/sqrt1-v^2/c^2 - mc^2

ie KE = Etotal - Erest


but when I try assign a mass and velocity of one i get a kinetic energy of 0 on my calculator, (since 1/c^2 is very small) instead of 1/2 (from 1/2mv^2) I thought the classical and relativistic equations would be nearly equal since the masses and velocities are very small. Why is it that this kinetic energy equation only gives relatively(pardon the pun) reasonable answers when the velocity starts to approach the speed of light?

 

[jtbell]

Your calculator probably doesn't have enough significant digits to do this calculation properly. In order to get good results from the relativistic kinetic energy formula with v << c, you have to use many significant digits in your calculation, because the two terms are very very very very nearly equal and almost cancel each other out when subtracting.

With v = 1 m/s, I estimate that you would need at least 18 significant figures to see a nonzero result, and at least 20 significant figures to get a somewhat accurate result.

 

[collinsmark]

Alternately, you could also use velocities much larger than 1 m/s. For example, try 100,000 m/s. At least increase the velocity to the point where the truncation errors don't occur. (On my calculator, it starts to be sort of, almost, accurate at v > 10,000 m/s. Other calculators may vary.)

--------------

If you're mathematically inclined (and you are familiar with calculus), try taking the Taylor series expansion of

K.E. = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} mc^2 - mc^2
for v near zero. You'll find that it does reduce to (1/2)mv², for v near zero.

(You can also use Wolfram Alpha to help you with this.)

 

[JamesClarke]

Thanks for the help. Never even thought of taylor expansion and you can see why it approximates if you do.