Classic and Relativity Kinetic Energy

[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

 

[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.

 

[jtbell]

I tried this, and it doesn't work. I get different answers.
I am just wondering what is amiss here.

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

 

[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.

 

[robphy]

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}

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.

 

[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"

 

[rbj]

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}

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}}<br /> <br /> \quad = e^{-\frac{1}{2} \log\left( 1-\frac{v^2}{c^2} \right)}}<br /> <br /> \quad \approx 1 - \frac{1}{2} \log \left( 1-\frac{v^2}{c^2} \right)<br /> <br /> \quad \approx 1 - \frac{1}{2} \left( -\frac{v^2}{c^2} \right)

for |v| &lt;&lt; |c|.

get's the same thing.

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

 

[dextercioby]

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

Daniel.

 

[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}

 

[dextercioby]

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

Daniel.

 

[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

 

[rbj]

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.

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.

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}

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|.

 

[Jheriko]

but...

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

 

[pjn2000]

Is this equation for Relativistic Kinetic Energy really proven?

 

[Meir Achuz]

Is this equation for Relativistic Kinetic Energy really proven?

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

 

[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.

 

[DrStupid]

Is this equation for Relativistic Kinetic Energy really proven?

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

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