# Classic and Relativity Kinetic Energy

1. Nov 15, 2006

### Kakorot

It is said that at low speeds the relativistic kinetic energy formula will give you the same answers as the classical KE formula. I tried this, and it doesn't work. I get different answers.
I am just wondering what is amiss here.

How would one derive the Classical KE from the Relativistic KE?

Thanks for the help,
Matt

2. Nov 15, 2006

### dextercioby

Well

$$KE=m_{0}c^{2}\left(\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} -1\right)\simeq m_{0}c^{2}\left(1+\frac{v^{2}}{2c^{2}}-1\right) =m_{0}\frac{v^{2}}{2}$$

Daniel.

3. Nov 15, 2006

### Staff: Mentor

Show us an example where you get different results (show your work!) and someone can probably tell you where you went wrong.

4. Nov 15, 2006

### Jheriko

Just to clarify what dextercioby said, the approximation comes from the binomial expansion
$$(x+y)^r=\sum_{k=0}^{\infty} {r \choose k}x^ky^{r-k}$$
$$(x+y)^r=\sum_{k=0}^{\infty} \frac{r!}{k!\,(r-k)!}x^ky^{r-k}$$
which we can use to expand
$$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=({1-\frac{v^2}{c^2}})^{-0.5}$$

The first few terms of the expansion are
$$1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}+\frac{5}{16}\frac{v^6}{c^6}+\ldots$$

So we have
$$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \simeq 1+\frac{v^2}{2c^2}$$
which can be used, as shown above, to show that the classical expression for kinetic energy approximates the relativistic expression.

Last edited: Nov 15, 2006
5. Nov 15, 2006

### robphy

As written, you'd probably have to explain how to use r= -1/2 in the general formulas you've written.

It may be conceptually simpler to emphasize the more fundamental Taylor series expansion.

6. Nov 15, 2006

### Jheriko

ah, i forgot about the non-integer r in the 'r choose k'..

$${r \choose k}={1 \over k!}\prod_{n=0}^{k-1}(r-n)=\frac{r(r-1)(r-2)\cdots(r-(k-1))}{k!}$$

according to http://en.wikipedia.org/wiki/Binomial_theorem" [Broken]

Last edited by a moderator: May 2, 2017
7. Nov 15, 2006

### rbj

you have a small problem assuming an analytic continuation that a formula that is shown valid for integer r is also valid for a non-integer. it is true, you just didn't make the connection. one that doesn't start with the binomial expansion but uses Taylor series is

$$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \left( {1-\frac{v^2}{c^2}} \right)^{-\frac{1}{2}} \quad = e^{-\frac{1}{2} \log\left( 1-\frac{v^2}{c^2} \right)}} \quad \approx 1 - \frac{1}{2} \log \left( 1-\frac{v^2}{c^2} \right) \quad \approx 1 - \frac{1}{2} \left( -\frac{v^2}{c^2} \right)$$

for $|v| << |c|$.

get's the same thing.

i was also gratified that dextercioby used the symbol $m_0$ which seems to be out of vogue.

Last edited: Nov 15, 2006
8. Nov 16, 2006

### dextercioby

There's an analytic continuation of the binomial formula using the Eulerian gamma function, but yeah, i still prefer Taylor's polynomial.

Daniel.

9. Nov 16, 2006

### Jheriko

Sorry, I didn't mean to add confusion. I just thought it seemed quite mysterious in the first post that $$\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\simeq1+\frac{v^2}{2c^2}$$ and wanted to explain where that came from.

btw in the Taylor series example given above, why use $$e^{-\frac{1}{2}}log(1-\frac{v^2}{c^2})\simeq1-\frac{v^2}{2c^2}$$? This seems quite mysterious to me since we can write $$e^{-\frac{1}{2}}e^{log(1-\frac{v^2}{c^2})} \equiv e^{-\frac{1}{2}}(1-\frac{v^2}{c^2})$$, which is exact.

There is nothing wrong with doing it this way of course... I just don't like the idea of throwing away precision just to make the answer fit. $$e^{-\frac{1}{2}}$$ is more like 0.6 than $$\frac{1}{2}$$

Last edited: Nov 16, 2006
10. Nov 16, 2006

### dextercioby

You've made a mistake on the e^(ab) expansion. It's not e^a e^b

Daniel.

11. Nov 16, 2006

### Jheriko

Ah! Thanks! Pretty fundamental mistake I made there, a^b+c = a^b a^c.

I thought it was odd that someone would consider $$0.6\simeq\frac{1}{2}$$. Still, it leaves the question (for me at least) of where that approximation comes from?

EDIT: Never mind I just worked it out... I think it comes from the power series for $$e^x$$

12. Nov 16, 2006

### rbj

but
$$e^{-\frac{1}{2}}e^{log(1-\frac{v^2}{c^2})}$$
is not the same as
$$e^{-\frac{1}{2}log(1-\frac{v^2}{c^2})}$$.

allowing analytical extension from derivatives of integer power functions to non-integer, we know that

$$(1+x)^\alpha \approx 1 + \alpha x$$
for small |x|.

so there is a simple expression for what we need. i did the log - exp thing to avoid making that analytical extension since they were defined for noninteger arguments.

yes, there were two applications of approximation instead of just one. but i didn't need to make an assumption of analytical extension and both approximations are good for small |x|.

Last edited: Nov 16, 2006
13. Nov 17, 2006

### Jheriko

Sorry about that, I got confused about exponentiation rules, as stated above.

14. Sep 9, 2011

### pjn2000

Is this equation for Relativistic Kinetic Energy really proven?

15. Sep 9, 2011

### clem

Yes, millions of times each day at the LHC and other laboratories.

16. Sep 9, 2011

### PAllen

Hey, I always just use as every day fact that for small x:

sqrt(1+x) := 1 + x/2

1/(1+x) := 1-x

The former can be derived as trivially as:

(1+x)^2 = 1 + 2x + x^2

the latter from sum of geometric series, or even just subtracting and verifying the error is order x^2.

17. Sep 10, 2011

### DrStupid

And it results from Lorentz transformation (which is also proven experimentally).