# Kinetic energy of an object

by Tranceform
Tags: energy, kinetic, object
 P: 22 The kinetic energy of an object can be expressed as $$KE=mc^2(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1)$$ (1) For speeds much lower than the speed of light however, we know that $$KE=\frac{1}{2}mv^2$$ (2) The second expression can be derived from the first one using binomial expansion of the term with the square root in the denominator. I can see how the math behind this works, so that is not my question. What I'm wondering is rather WHY (again, I don't mean the mathematical reason but rather a physical reason) doing this operation on the equation for relativistic kinetic energy leads to the equation non-relativistic kinetic energy? I mean if we look at the first expression, the term with the square root has the exponent -(1/2). This exponent is apparently constant, not variable. So from where/what would one get the idea of binomally expanding it like it was variable? Where does the intuition behind this come from? Hope my question is understandable.
 C. Spirit Sci Advisor Thanks P: 5,620 It's just a Taylor expansion to first order in ##\frac{v}{c}##. If ##v \ll c## then this first order approximation will be very accurate, as is the case in Newtonian mechanics.
 P: 22 This unfortunately doesn't answer my question. Whether the derivation is made using binomal expansion or Taylor expansion doesn't answer my question. The question is rather: why would you do (again I'm not looking for a mathematical but physical reason) any expansion - of a seemingly constant term - to start with? Hope this clarifies the question.
 C. Spirit Sci Advisor Thanks P: 5,620 Kinetic energy of an object I've already answered your question. You're looking for an approximation of the exact kinetic energy in the case where ##v\ll c## i.e. you are looking for the Newtonian limit of the exact kinetic energy. Obviously the most natural way to do this is to Taylor expand the kinetic energy in powers of ##v/c## and throw away all second order and higher terms. There isn't anything deep to this. The parameter ##v/c## characterizes the extent to which your system is relativistic; it is the dimensionless characteristic velocity scale of the system. If you systematically expand the kinetic energy in powers of ##v/c## you get more and more relativistic contributions to the kinetic energy for each higher order term in ##v/c##. For a system that is non-relativistic i.e. Newtonian, the second and higher order terms will therefore be negligible because they parametrize the higher order relativistic nature of the system.
P: 22
Your second reply is much better and more explanatory but I'm still wondering something.
 Quote by WannabeNewton Obviously the most natural way to do this is to Taylor expand the kinetic energy in powers of ##v/c## and throw away all second order and higher terms.
I wonder why you use words like "obviously" unless you clarify what is so obvious about it? I have looked in two different physics books where this derivation is done and in both cases they use binomal expansion and not Taylor expansion, so apparently the authors of these books didn't think the Taylor expansion was so "obvious" to use. Also you are using words like "natural" which could do with some explanation? What is that you think makes Taylor expansion "the most natural" way to do the expansion?

 Quote by WannabeNewton The parameter ##v/c## characterizes the extent to which your system is relativistic; it is the dimensionless characteristic velocity scale of the system.
So why is only this parameter charaterizing and not mc^2? To me these parameters(?) seem like constant terms in both cases, so I still wonder what is the reasoning behind doing an expansion of a seemingly constant term?

Hope my question(s) are better understood now.
C. Spirit
Thanks
P: 5,620
 Quote by Tranceform I have looked in two different physics books where this derivation is done and in both cases they use binomal expansion and not Taylor expansion, so apparently the authors of these books didn't think the Taylor expansion was so "obvious" to use.
The binomial expansion is just a special case of the Taylor expansion.

 Quote by Tranceform Also you are using words like "natural" which could do with some explanation? What is that you think makes Taylor expansion "the most natural" way to do the expansion?
I'm not really sure how to properly explain this. It's just a matter of experience. As a physics student, resorting to the Taylor expansion is as second-nature as breathing. It's the absolute easiest way to get a power series expansion of a function in terms of a perturbation parameter. You can't do physics without it.

 Quote by Tranceform So why is only this parameter charaterizing and not mc^2? To me these parameters(?) seem like constant terms in both cases, so I still wonder what is the reasoning behind doing an expansion of a seemingly constant term?
How will that characterize the velocity scale of the system? The velocity of the system is ##v## and the speed of light is ##c## so we form a dimensionless expansion parameter ##v/c## to characterize the extent to which the system is relativistic. The rest energy won't do that.