# Performing a Taylor Series Expansion for Lorentz Factor

1. Sep 15, 2016

### Kunhee

1. The problem statement, all variables and given/known data
Perform a Taylor Series expansion for γ in powers of β^2, keeping only the third terms (ie. powers up to β^4). We are assuming at β < 1.

2. Relevant equations
γ = (1-β^2)^(-1/2)

3. The attempt at a solution
I have no background in math so I do not know how to do Taylor expansion. Could you help me out? Thank you very much.

2. Sep 15, 2016

### Staff: Mentor

Have you looked up Taylor series in Wikipedia?
Here is a link to the formula, which is clearer than on the English version. Just look at the formula.
Which brings me to the next question. Do you know how to differentiate a / your function (with respect to $\beta$)?

3. Sep 15, 2016

### Kunhee

Sorry, I am not sure. I have been away from all maths for a long time.
But could you help me out? What is my x and a?

4. Sep 15, 2016

### Staff: Mentor

Your $f$ is $\gamma$ and your $x$ is $\beta$. Unfortunately you didn't specify what your $a$ is, so I assume it is $a=0$.
If not mentioned otherwise it's almost always $0$. But some functions aren't defined at $0$. In these cases one cannot "automatically" take zero. However, your function is defined in $0$ and your $x = \beta$ is assumed smaller than one, so $a=0$ should be ok.

Edit: don't forget the internal differentiation: $\gamma' = \frac{d}{d \beta} (1- \beta^2)^{-\frac{1}{2}} = -\frac{1}{2}(1- \beta^2)^{-\frac{3}{2}} \, \cdot \, (-2 \beta)$

Last edited: Sep 15, 2016
5. Sep 15, 2016

### Kunhee

Okay I am solving it, one moment please!

6. Sep 15, 2016

### Kunhee

https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61 is γ = (1-β^2)^(-1/2)
x = β
and a = 0 so therefore my equation would look like this:

T [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61(x[I][I][I];a) [Broken] = [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61(a) [Broken] + [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61'(a) [Broken] (x-a) + [[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61''(a) [Broken] / 2 ] (x-a)^2 + [[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61'''(a) [Broken] / 6] (x-a)^3 ...[/I][/I][/I]

T [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61([I][I]β[I];0) [Broken] = [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61(0) [Broken] + [PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61'(0) [Broken] (x-0) + [[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61''(0) [Broken] / 2 ] (β-0)^2 + [[PLAIN]https://wikimedia.org/api/rest_v1/media/math/render/svg/132e57acb643253e7810ee9702d9581f159a1c61'''(0) [Broken] / 6] (β-0)^3 ...
[/I][/I][/I]

I am stuck here. I can find each derivative of (1-β^2)^(-1/2) but there is nowhere to plug in "a" in this
function. I apologize if I am making a very dumb mistake. Thanks for the help.

Last edited by a moderator: May 8, 2017
7. Sep 15, 2016

### Ray Vickson

Better: take $x = \beta^2$. Expand $(1-x)^{-1/2}$ in powers of $x$, then put $x = \beta^2$ later. That makes finding the needed derivatives a lot easier.

8. Sep 15, 2016

### Kunhee

I see, thanks. That's what the question meant!
But I am still confused at how a = 0 fits into this equation.
Looking at the equation, the functions do not have anywhere to plug in for a.

9. Sep 15, 2016

### Ray Vickson

If your "$a$" refers to the formula
$$f(x) = f(a) + f'(a) (x-a) + \frac{1}{2!} f''(a) (x-a)^2 + \cdots,$$
then you are dealing with the case $a = 0$

Last edited: Sep 16, 2016
10. Sep 15, 2016

### Staff: Mentor

If you take $\gamma(x) = (1-x)^{-\frac{1}{2}}$ then $\gamma'(x) = \frac{1}{2}(1-x)^{-\frac{3}{2}}$ and $\gamma'(0) = \frac{1}{2}(1-0)^{-\frac{3}{2}} = \frac{1}{2}$. Don't forget that there is a factor $(x-a)^1= (x-0)^1=x$, too. (In the second term of the series.)

11. Sep 15, 2016

### Kunhee

Just a quick question...
When do we use γ and when do we use its reciprocal α when solving for coordinates of S' frame through Lorentz transformations?
My teacher used the reciprocal but my textbook uses γ in the same equation and my length contraction is actually going reverse and
getting longer.

12. Sep 15, 2016

### Kunhee

Solving...

13. Sep 15, 2016

### Staff: Mentor

Yes. But you want to expand at $a=0$, that is in the formula $f^{(n)}(a)=f^{(n)}(0)$. So you have to substitute $0$ where $x$ was or $\beta$ is. And the exponents are still negative (although it doesn't matter in this case). You multiply with $(x-a)^n=(x-0)^n=x^n=\beta^{2n}$. The differentials of $f=\gamma$ are evaluated at $0$, not extra multiplied.

So the first term of the sum is $f(0)x^0 = 1$ the second $f'(0)x^1=\frac{1}{2}x$ the third $\frac{1}{2}f''(0)x^2=\frac{3}{8}x^2$ and so on.

Edit: Remember the formula is (with $f=\gamma$ and $x=\beta^2$) $\; Tf(x;0)= f(0)+f'(0)x+\frac{f''(0)}{2}x^2+\frac{f'''(0)}{6}x^3+ \dots$.

14. Sep 15, 2016

### Kunhee

Thanks so much...