Classic and Relativity Kinetic Energy

In summary, the relativistic kinetic energy formula gives the same results as the classical kinetic energy formula at low speeds. However, when tested, the results were different which raises the question of how to derive the classical kinetic energy formula from the relativistic one. After some discussion and clarification, it was determined that the classical formula can be approximated from the relativistic formula using a binomial expansion or a Taylor series expansion. However, this approximation may not be exact and could result in a loss of precision.
  • #1
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
 
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  • #2
Well

[tex] 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} [/tex]

Daniel.
 
  • #3
Kakorot said:
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.
 
  • #4
Just to clarify what dextercioby said, the approximation comes from the binomial expansion
[tex](x+y)^r=\sum_{k=0}^{\infty} {r \choose k}x^ky^{r-k}[/tex]
[tex](x+y)^r=\sum_{k=0}^{\infty} \frac{r!}{k!\,(r-k)!}x^ky^{r-k}[/tex]
which we can use to expand
[tex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=({1-\frac{v^2}{c^2}})^{-0.5}[/tex]

The first few terms of the expansion are
[tex]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[/tex]

So we have
[tex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \simeq 1+\frac{v^2}{2c^2}[/tex]
which can be used, as shown above, to show that the classical expression for kinetic energy approximates the relativistic expression.
 
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  • #5
Jheriko said:
Just to clarify what dextercioby said, the approximation comes from the binomial expansion
[tex](x+y)^r=\sum_{k=0}^{\infty} {r \choose k}x^ky^{r-k}[/tex]
[tex](x+y)^r=\sum_{k=0}^{\infty} \frac{r!}{k!\,(r-k)!}x^ky^{r-k}[/tex]
which we can use to expand
[tex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=({1-\frac{v^2}{c^2}})^{-0.5}[/tex]

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

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

according to http://en.wikipedia.org/wiki/Binomial_theorem"
 
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  • #7
Jheriko said:
Just to clarify what dextercioby said, the approximation comes from the binomial expansion
[tex](x+y)^r=\sum_{k=0}^{\infty} {r \choose k}x^ky^{r-k}[/tex]
[tex](x+y)^r=\sum_{k=0}^{\infty} \frac{r!}{k!\,(r-k)!}x^ky^{r-k}[/tex]
which we can use to expand
[tex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=({1-\frac{v^2}{c^2}})^{-0.5}[/tex]

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

[tex]\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) [/tex]

for [itex] |v| << |c| [/itex].

get's the same thing.

i was also gratified that dextercioby used the symbol [itex]m_0[/itex] which seems to be out of vogue.
 
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  • #8
There's an analytic continuation of the binomial formula using the Eulerian gamma function, but yeah, i still prefer Taylor's polynomial.

Daniel.
 
  • #9
Sorry, I didn't mean to add confusion. I just thought it seemed quite mysterious in the first post that [tex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\simeq1+\frac{v^2}{2c^2}[/tex] and wanted to explain where that came from.

btw in the Taylor series example given above, why use [tex]e^{-\frac{1}{2}}log(1-\frac{v^2}{c^2})\simeq1-\frac{v^2}{2c^2}[/tex]? This seems quite mysterious to me since we can write [tex]e^{-\frac{1}{2}}e^{log(1-\frac{v^2}{c^2})} \equiv e^{-\frac{1}{2}}(1-\frac{v^2}{c^2})[/tex], 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. [tex]e^{-\frac{1}{2}}[/tex] is more like 0.6 than [tex]\frac{1}{2}[/tex]
 
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  • #10
You've made a mistake on the e^(ab) expansion. It's not e^a e^b

Daniel.
 
  • #11
Ah! Thanks! Pretty fundamental mistake I made there, a^b+c = a^b a^c.

I thought it was odd that someone would consider [tex]0.6\simeq\frac{1}{2}[/tex]. 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 [tex]e^x[/tex]
 
  • #12
Jheriko said:
Sorry, I didn't mean to add confusion. I just thought it seemed quite mysterious in the first post that [tex]\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\simeq1+\frac{v^2}{2c^2}[/tex] and wanted to explain where that came from.

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

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

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

[tex] (1+x)^\alpha \approx 1 + \alpha x [/tex]
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. [tex]e^{-\frac{1}{2}}[/tex] is more like 0.6 than [tex]\frac{1}{2}[/tex]

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|.
 
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  • #13
rbj said:
but...

Sorry about that, I got confused about exponentiation rules, as stated above.
 
  • #14
Is this equation for Relativistic Kinetic Energy really proven?
 
  • #15
pjn2000 said:
Is this equation for Relativistic Kinetic Energy really proven?
Yes, millions of times each day at the LHC and other laboratories.
 
  • #16
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
clem said:
pjn2000 said:
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).
 

Related to Classic and Relativity Kinetic Energy

1. What is the difference between Classic and Relativity Kinetic Energy?

Classic and Relativity Kinetic Energy are two different ways of understanding the concept of energy in physics. Classic Kinetic Energy is based on Newton's laws of motion and describes the energy of an object in terms of its mass and velocity. Relativity Kinetic Energy, on the other hand, is based on Einstein's theory of relativity and takes into account the effects of an object's velocity on its mass. This means that as an object approaches the speed of light, its mass increases and therefore its kinetic energy also increases.

2. How is Classic Kinetic Energy calculated?

Classic Kinetic Energy is calculated using the formula E = 1/2 * m * v^2, where E is the kinetic energy, m is the mass of the object, and v is the velocity of the object. This formula assumes that the object is moving at a relatively slow speed compared to the speed of light.

3. How is Relativity Kinetic Energy calculated?

Relativity Kinetic Energy is calculated using the formula E = m * c^2 * (1/sqrt(1-v^2/c^2)-1), where E is the kinetic energy, m is the mass of the object, c is the speed of light, and v is the velocity of the object. This formula takes into account the increase in mass as the object approaches the speed of light.

4. Are there any real-life applications of Relativity Kinetic Energy?

Yes, Relativity Kinetic Energy is important for understanding the behavior of particles at high speeds, such as those found in particle accelerators. It is also used in the field of astrophysics to study the behavior of objects in space, such as stars and galaxies, which can reach speeds close to the speed of light.

5. Can Classic Kinetic Energy and Relativity Kinetic Energy be used interchangeably?

No, Classic and Relativity Kinetic Energy are based on different theories and therefore cannot be used interchangeably. Classic Kinetic Energy is accurate for everyday speeds, while Relativity Kinetic Energy is necessary for high-speed objects. Using the wrong formula can lead to incorrect calculations and predictions.

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