I have several questions:
who came up with this and is there any analysis of this math and physics?
Does this relationship have any realizations or conclusions that can be drawn?
here are some links for those interested
Here is the math:
$$(1)\qquad E = m{c}^2$$ $$(2)\qquad \Delta t_1 = \frac{\Delta t_2}{\sqrt{1-\left({\left(\frac{v}{c}\right)}^2\right)}}$$
for easily seeing the binomial expansion of $(2)$
$$let \quad x = {-\left({\left(\frac{v}{c}\right)}^2\right)}$$
rearranging of $(2)$ $$\Delta t_1 = \frac{\Delta t_2}{\sqrt{1+x}}$$
$$\frac{\Delta t_1}{\Delta t_2} = \frac{1}{\sqrt{1+x}}$$
$$\frac{\Delta t_1}{\Delta t_2} = \left({1+x}\right)^{-1 \over 2}$$
$$(3) \qquad \left({1+x}\right)^{-1 \over 2}$$
Applying the binomial expansion on equation $(3)$ to get the following: Fractional power binomial expansion link
$$\left({1+x}\right)^{-1 \over 2} = 1{x}^0 + -\frac{1}{2}{x}^1 + \frac{3}{8}{x}^2 + -\frac{5}{16}{x}^3 + \frac{35}{128}{x}^4+ \cdots$$
essentially the coefficients are $${n \choose q}\quad or \quad{-\frac{1}{2} \choose q}$$ where n is $-\frac{1}{2}$ and $q$ incriments up by 1 starting at 0, I think.
also
$$ -1 \lt x \lt 1$$
now substituting $$x = {-\left({\left(\frac{v}{c}\right)}^2\right)}$$
$$\left({1-\left({\left(\frac{v}{c}\right)}^2\right)}\right)^{-1 \over 2} = 1\bullet{\left({-\left(\frac{v}{c}\right)}^2\right)}^0 + -\frac{1}{2}\bullet{\left({-\left(\frac{v}{c}\right)}^2\right)}^1 + \frac{3}{8}\bullet{\left({-\left(\frac{v}{c}\right)}^2\right)}^2 + -\frac{5}{16}\bullet{\left({-\left(\frac{v}{c}\right)}^2\right)}^3 + \frac{35}{128}\bullet{\left({-\left(\frac{v}{c}\right)}^2\right)}^4+ \cdots$$
"$\bullet$" symbol here is just a multiply,because the parenthesis are getting crowded here. So removing the "$\bullet$" and the parenthesis and simplafying we get this:
$$\left({1-{\left(\frac{v}{c}\right)}^2}\right)^{-1 \over 2} = 1 + \frac{1}{2}{\left(\frac{v}{c}\right)}^2 + \frac{3}{8}{{\left(\frac{v}{c}\right)}^4} + \frac{5}{16}{{\left(\frac{v}{c}\right)}^6} + \frac{35}{128}{{\left(\frac{v}{c}\right)}^8}+ \cdots$$
then from here if you take equation $(1)$ and multply it with this approximation of equation $(3)$ we get:
$$(E)(\frac{\Delta t_1}{\Delta t_2}) = (1)(m{c}^2) +(\frac{1}{2}{\left(\frac{v}{c}\right)}^2)(m{c}^2) +(\frac{3}{8}{\left(\frac{v}{c}\right)}^4)(m{c}^2) +(\frac{5}{16}{\left(\frac{v}{c}\right)}^6)(m{c}^2) +(\frac{35}{128}{\left(\frac{v}{c}\right)}^8)(m{c}^2) + \cdots$$
simplifying a bit
$$(E)(\frac{\Delta t_1}{\Delta t_2}) = m{c}^2 +\frac{1}{2}m{v}^2 +\frac{3}{8}m\frac{v^4}{c^2} +\frac{5}{16}m\frac{v^6}{c^4} +\frac{35}{128}m\frac{v^8}{c^6} + \cdots$$
now $$ -1 \lt -{\left(\frac{v}{c}\right)}^2 \lt 1$$ $$ 1 \gt {\left(\frac{v}{c}\right)}^2 \gt -1$$ $$ c \gt v \gt ic$$
My Analysis:
My initial analysis of this sees the Kinetic energy portion of the 1/2 fraction and I do not know why that is there. Is their some other meaning for the other parts of the equation?
The units for this are going to be energy (Joules) and the farther elements only get small and less significant to the major contribution of the first 5 elements.
Rewording my questions:
Did I make a mistake, or am I allowed to mathematically multiply these two equations?
Are their any conclusions to be drawn from this math?
I know someone else came up with this, where can I learn more?
Also what does the boundary condition of v mean?
Thanks for any input into this! (let me know if there are other tags for this question)
