# Kinetic energy of an object

1. Jul 16, 2014

### Tranceform

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.

2. Jul 16, 2014

### WannabeNewton

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. Jul 16, 2014

### Tranceform

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. Jul 16, 2014

### WannabeNewton

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.

5. Jul 16, 2014

### Tranceform

Your second reply is much better and more explanatory but I'm still wondering something.
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?

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.

6. Jul 16, 2014

### WannabeNewton

The binomial expansion is just a special case of the Taylor 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.

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.

7. Jul 16, 2014

### Staff: Mentor

A bullet or an armor-piercing artillery round moves at about .00001 of the speed of light, and the fastest missiles and rockets move only about ten times faster than that. Try plugging these values into your two formulas and see how different the results are; the differences are smaller than even our most sensitive instruments can detect. Thus, for several centuries, between Newton's discovery of classical mechanics and Einstein's discovery of relativity, we believed that the $mv^2/2$ expression was exactly correct, and there was no shortage of experiments that agreed with this conclusion, to the limits of experimental accuracy.

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 $\gamma=(1-v^2/c^2)^{-1/2}$ 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.