- #1
mcjosep
- 35
- 0
I've been doing some recreational physics and wanted to see if I could come up with a formula for energy/time equivalence. I decided to start with the formula for gravitational time dilation.
$$
y=Z\sqrt{1-(2GM/rc^2)}
$$
This formula shows what time would look like to an outside observer (infinitely far) looking in at a watch within a gravitational field. $Z$ is the time of the observer and $y$ is the slow moving time in a gravitational field at radius from the mass $r$.
I modified the formula a little to look for energy change, this is probably where a mistake was made.
$$
(1-(y*5.39106*10^{-44})))=1*\sqrt(1-(2*6.67384*10^{-11}*(x/299792458^2))/(1.616199*10^{-35}*299792458^2))
$$
what you see in the above formula to the left of the equal sign (assuming we are comparing a single second for the top formula $Z$ to the dilated time) is 1 second - x*Planck time.
to the right of the equal sign everything is the same except rather than the $M$ for mass in the top formula we used $E/c^2$ from the mass energy equivalence formula $E=mc^2$ so the $x$ in the above formula represents Energy.
For radius, $r$, I fixed it at the Planck length.
So when I entered the formula above into my favorite tool WolframAlpha to solve for $x$ it returned.
$$
x=1.05457*10^-34*y-2.84263*10^-78*y^2
$$
what this is showing is
$$
E=\hbar*y-\hbar*P_t*2*y^2
$$
(Reduced Planck constant * the amount of plank time units you want to take away from one second) - (Reduced Planck constant * 2 * Planck time * the amount of plank time units you want to take away from one second ^2) equals the multiple at which energy would change.
I believe the "2" is in the formula above because at the same rate time would be effected by gravitational time dilation so would length, so you multiply the Planck time by 2 and you get the effect of both being "dilated" equaling the rate at which energy changes. This is just a guess, I know that that gravitational length contraction is a disputed subject.
I put the formula above so that you could drop it in wolfram yourself and see the answer, I do not have all the steps it took to derive. I was mainly surprised how my modified formula turned into a formula with two constants and a variable.
I was also surprised how it charted out. Let me know what you think.
[Modified formula][1]
[Derived Formula][2] [1]: http://www.wolframalpha.com/input/?...^2))/(1.616199*10^-35*299792458^2)) for x&f=1
[2]: http://www.wolframalpha.com/input/?i=1.05457x10^-34 y-2.84263x10^-78 y^2&lk=1&a=ClashPrefs_*Math-
$$
y=Z\sqrt{1-(2GM/rc^2)}
$$
This formula shows what time would look like to an outside observer (infinitely far) looking in at a watch within a gravitational field. $Z$ is the time of the observer and $y$ is the slow moving time in a gravitational field at radius from the mass $r$.
I modified the formula a little to look for energy change, this is probably where a mistake was made.
$$
(1-(y*5.39106*10^{-44})))=1*\sqrt(1-(2*6.67384*10^{-11}*(x/299792458^2))/(1.616199*10^{-35}*299792458^2))
$$
what you see in the above formula to the left of the equal sign (assuming we are comparing a single second for the top formula $Z$ to the dilated time) is 1 second - x*Planck time.
to the right of the equal sign everything is the same except rather than the $M$ for mass in the top formula we used $E/c^2$ from the mass energy equivalence formula $E=mc^2$ so the $x$ in the above formula represents Energy.
For radius, $r$, I fixed it at the Planck length.
So when I entered the formula above into my favorite tool WolframAlpha to solve for $x$ it returned.
$$
x=1.05457*10^-34*y-2.84263*10^-78*y^2
$$
what this is showing is
$$
E=\hbar*y-\hbar*P_t*2*y^2
$$
(Reduced Planck constant * the amount of plank time units you want to take away from one second) - (Reduced Planck constant * 2 * Planck time * the amount of plank time units you want to take away from one second ^2) equals the multiple at which energy would change.
I believe the "2" is in the formula above because at the same rate time would be effected by gravitational time dilation so would length, so you multiply the Planck time by 2 and you get the effect of both being "dilated" equaling the rate at which energy changes. This is just a guess, I know that that gravitational length contraction is a disputed subject.
I put the formula above so that you could drop it in wolfram yourself and see the answer, I do not have all the steps it took to derive. I was mainly surprised how my modified formula turned into a formula with two constants and a variable.
I was also surprised how it charted out. Let me know what you think.
[Modified formula][1]
[Derived Formula][2] [1]: http://www.wolframalpha.com/input/?...^2))/(1.616199*10^-35*299792458^2)) for x&f=1
[2]: http://www.wolframalpha.com/input/?i=1.05457x10^-34 y-2.84263x10^-78 y^2&lk=1&a=ClashPrefs_*Math-