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Kinetic energy of an object 
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#1
Jul1614, 01:40 PM

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The kinetic energy of an object can be expressed as
[tex]KE=mc^2(\frac{1}{\sqrt{1\frac{v^2}{c^2}}}1)[/tex] (1) For speeds much lower than the speed of light however, we know that [tex]KE=\frac{1}{2}mv^2[/tex] (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 nonrelativistic 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. 


#2
Jul1614, 01:43 PM

C. Spirit
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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.



#3
Jul1614, 02:04 PM

P: 21

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.



#4
Jul1614, 02:26 PM

C. Spirit
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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 nonrelativistic i.e. Newtonian, the second and higher order terms will therefore be negligible because they parametrize the higher order relativistic nature of the system. 


#5
Jul1614, 03:01 PM

P: 21

Your second reply is much better and more explanatory but I'm still wondering something.
Hope my question(s) are better understood now. 


#6
Jul1614, 03:20 PM

C. Spirit
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#7
Jul1614, 04:16 PM

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When Einstein and others in the early twentieth century put forth an argument that ##mc^2(1\gamma)## was more accurate (I'm using the common convention ##\gamma=(1v^2/c^2)^{1/2}## which greatly cleans up the formulas) one of the obvious first questions was how to reconcile this result with the centuries of accumulated experimental evidence that support the ##mv^2/2## result. The answer, of course, was that the relativistic formula reduced to the classical formula when v was small compared with c; and doing the binomial and Taylor expansions was the way to show that. So if I'm understanding your question properly, the answer is that the motivation for the Taylor/binomial expansion was to demonstrate that the two equations agree when v is small compared with c. 


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