Math Help - Einstein's Theory

1. Einstein's Theory

Einstein's theory of relativity predicts that a body of mass m0 at rest has mass: m = m0/√(1-(v^2/c^2)) , where c is the velocity of light and v is the velocity of the body.
Use a series to obtain the approximation: mc^2
m0c^2 + 1/2m0v^2

Other Info that may be needed:

1. Mass energy formula: E = mc^2
2. Kinetic energy of a body at rest: v = 1/2m0v^2 (where mass is m0 and velocity is v)

2. Originally Posted by jules027
Einstein's theory of relativity predicts that a body of mass m0 at rest has mass: m = m0/√(1-(v^2/c^2)) , where c is the velocity of light and v is the velocity of the body.
Use a series to obtain the approximation: mc^2 m0c^2 + 1/2m0v^2

Other Info that may be needed:

1. Mass energy formula: E = mc^2
2. Kinetic energy of a body at rest: v = 1/2m0v^2 (where mass is m0 and velocity is v)

All this requires is that you expand:

$
\frac{m_0}{\sqrt{1-(v^2/c^2)}}
$

using the binomial theroem or McLaurin series multiply by $c^2$ and truncate the series after the first two terms.

RonL

3. Originally Posted by CaptainBlack
All this requires is that you expand:

$
\frac{m_0}{\sqrt{1-(v^2/c^2)}}
$

using the binomial theroem of McLaurin series multiply by $c^2$ and truncate the series after the first two terms.

RonL
How would I go about setting this up? Thanks so much!!!

4. Originally Posted by jules027
How would I go about setting this up? Thanks so much!!!
The Maclaurin series for f(x) is given by

$f(x)=f(0)+f'(0)x+\frac{f''(0)x^2}{2!}+\frac{f''(0) x^3}{3!}+...+\frac{f^{(n)}(0)x^n}{n!}=\sum_{n=0}^{ \infty}\frac{f^{(n)}(0)x^n}{n!}$

so letting $f(?)=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{c\cdot{m_0}}{\sqrt{c^2-v^2}}$

assuming $c>0$ so that $|c|=c$

Also I dont know what this is in respect to? What is the constant and what is the variable?

5. Originally Posted by Mathstud28
The Maclaurin series for f(x) is given by

$f(x)=f(0)+f'(0)x+\frac{f''(0)x^2}{2!}+\frac{f''(0) x^3}{3!}+...+\frac{f^{(n)}(0)x^n}{n!}=\sum_{n=0}^{ \infty}\frac{f^{(n)}(0)x^n}{n!}$

so letting $f(?)=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{c\cdot{m_0}}{\sqrt{c^2-v^2}}$

assuming $c>0$ so that $|c|=c$

Also I dont know what this is in respect to? What is the constant and what is the variable?
It doesn't say anything about the variables or constants. The question is exactly what I had written. I'm confused. Now how will that Maclaurin series get me the approximation that I am asked to find?

6. Originally Posted by jules027
It doesn't say anything about the variables or constants. The question is exactly what I had written. I'm confused. Now how will that Maclaurin series get me the approximation that I am asked to find?
Give a show of good faith that $f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)x^n}{n!}$ then

Now they are equal due to the upper index of the summation being infinity

but I could say $f(x)\approx\sum_{n=0}^{m}\frac{f^{(n)}(0)x^n}{n!}, where\text{ }m\ne{\infty}$

So if You wanted two terms as CB suggested you can say that

$f(x)\approx\sum_{n=0}^{1}\frac{f^{(n)}(0)x^{n}}{n! }$

7. I am so sorry about this. your probably getting frustrated at me. To show work in my problem, I am not sure how to go from the McLaurin series to being able to show that the series is the approximation. How did you arrive at the fact that the McLaurin series is the approximation? Sorry again but I need to show tons of work for these problems.

8. Originally Posted by jules027
I am so sorry about this. your probably getting frustrated at me. To show work in my problem, I am not sure how to go from the McLaurin series to being able to show that the series is the approximation. How did you arrive at the fact that the McLaurin series is the approximation? Sorry again but I need to show tons of work for these problems.
You see. I dont know physics well to see if their is another way of doing this problem, but if CaptainBlack is right about this then you need to do what I said. The only problem is, if you dont know what a power series(do you?) is then you can hardly put it down as your work. It will be suspicious. The whole premise of the Maclaurin series is that if a function is equal to a polynomial at a point and increases at the same rate as the polynomial then the function is equal to that polynomial. That is the basic premise of Maclaurin series. It is a lot more math than I can show in this thread. It requires a lot of prerequisite math knowledge. What math class are you currently enrolled in?

9. Im in Calc II. I know about power series but I am just not getting the jist of this problem. I understand about the series that I have to use. I am just confused on what to plug into it. Do I do a ratio or root test to it? I am just confused on how I can use that series to approximate Einstein's theory.

10. Originally Posted by jules027
Im in Calc II. I know about power series but I am just not getting the jist of this problem. I understand about the series that I have to use. I am just confused on what to plug into it. Do I do a ratio or root test to it? I am just confused on how I can use that series to approximate Einstein's theory.
You are confusing stuff. The ratio or root test would be used to find the convergence or interval of convergence of the series instead, this approximates it. Have you learned about Taylor series yet? If so a Maclaurin series is a Taylor series centered at 0

11. Originally Posted by CaptainBlack
All this requires is that you expand:

$
\frac{m_0}{\sqrt{1-(v^2/c^2)}}
$

using the binomial theroem of McLaurin series multiply by $c^2$ and truncate the series after the first two terms.

RonL
Hey jules027,

Captain has correctly specified what you have to do, and Mathstud28 continued with the definition of McLaurin series to help you.

$f(x) = \frac1{\sqrt{1-x^2}}$

Clearly f(0) = 1,
Then do a little computation to obtain f''(0) = 1.

Now from Mathstud28's formula
$f(x)=f(0)+f'(0)x+\frac{f''(0)x^2}{2!}+.....$

So $\frac1{\sqrt{1-x^2}} \approx 1 + \frac{x^2}{2!}$

Now $mc^2 = \dfrac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}}\\\\\\\\\\ (*)$

But from the previous approximation if we let $x = \frac{v}{c}$, we get:
$\dfrac1{\sqrt{1- \frac{v^2}{c^2}}} \approx 1 + \frac{v^2}{2c^2}$

Substituting this in $(*)$ we get,
$mc^2 = m_0 c^2 \left(1 + \frac{v^2}{2c^2} \right)$

$mc^2 = m_0 c^2 + \frac{mv^2}{2}$

Done

12. Originally Posted by Isomorphism
Hey jules027,

Captain has correctly specified what you have to do, and Mathstud28 continued with the definition of McLaurin series to help you.

$f(x) = \frac1{\sqrt{1-x^2}}$

Clearly f(0) = 1,
Then do a little computation to obtain f''(0) = 1.

Now from Mathstud28's formula
$f(x)=f(0)+f'(0)x+\frac{f''(0)x^2}{2!}+.....$

So $\frac1{\sqrt{1-x^2}} \approx 1 + \frac{x^2}{2!}$

Now $mc^2 = \dfrac{m_0 c^2}{\sqrt{1 - \frac{v^2}{c^2}}}\\\\\\\\\\ (*)$

But from the previous approximation if we let $x = \frac{v}{c}$, we get:
$\dfrac1{\sqrt{1- \frac{v^2}{c^2}}} \approx 1 + \frac{v^2}{2c^2}$

Substituting this in $(*)$ we get,
$mc^2 = m_0 c^2 \left(1 + \frac{v^2}{2c^2} \right)$

$mc^2 = m_0 c^2 + \frac{mv^2}{2}$

Done
Thank you very much! I had no idea what was to be taken as a variable...although I was going to do it yoru way considering CaptainBlack said about the binomial theorem...but I didnt want to lead the poster astray