- #1
mcjosep
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I was recently exploring time dilation from Gravity and from velocity and I came up with an interesting derivation that I have not seen before. I was wondering if there is a paper published showing these relationships like this before and where I could find it?
First you start with the gravitational time dilation formula where:
$$
T_1=Tsqrt(1-((2GM)/rc^2))
$$
and rather than entering r for the radius we replace r with the Schwarzschild radius formula ##((2GM)/C^2)x##
with an ##x## at the end representing how many Schwarzschild radii you are away from the center. This brings the formula to look like:
$$
T_1=Tsqrt(1-(2GM)/((((2GM)/c^2)x)c^2))
$$
Which when simplified breaks down to:
$$
T_1=Tsqrt(1-1/x)
$$
and if you make ##T=1## then you just get
$$
=sqrt(1-1/x)
$$
This is very similar to the one in many physics books ##=sqrt(1-r_0/r)## where ##r_0## is equal to the Schwarzschild radius and then r equals the radius from the center. The formula above it just makes it slightly simpler due to making ##r_0## equal to 1 and x equal to how many radii a point you are observing is from the center of the mass.
That is the gravitational time dilation side portion of this relationship. Now for the Velocity time dilation side we use a similar methodology and start with:
$$
T_0=Tsqrt(1-v^2/c^2)
$$
Now we make ##T## equal to 1, ##v## equal to 1, and ##c## to ##y## because now we are going to make ##c## a variable.
$$
T_0=sqrt(1-1/y^2)
$$
What you see now "##1/y^2##" is showing the velocity as a constant 1 and ##y## represents how much faster light is going than the velocity constant of 1. If the above were to show the fraction as ##1/5^2## then this would be the same as saying an object is going at a velocity 1/5th the velocity of light.
So now if we solve the velocity and gravitational time dilation formulas so that we can see how they dilate time to come up with the same result:
$$
sqrt(1-1/x)=sqrt(1-1/y^2)
$$
We can simplify this to
$$
x=y^2
$$
So let's say your radius from the center of the mass ##x## is equal to 4 Schwarzschild radii then the speed an object must move to get the same dilation due to velocity must be equal to half the speed of light since ##y## equals 2 and represents the speed of light going two times faster than the moving object.
I was pretty happy to see a seemingly simple yet complex set of formulas breakdown into one of the simplest algebraic formulas.
This can also be constructed geometrically as seen below:
[1]: http://i.stack.imgur.com/nSC7Y.jpg
The edge of the black circle represents the Schwarzschild radius and each point on the x-axis is another radii away from the center. The Y axis has points that are the square root of x.
Let me know what you think. Please check my math.
You can use the same methodology to break down other formulas as well.
Circular Orbit Velocity
$$
v=sqrt((GM)/r)
$$
Then add in the Schwarzschild radius formula for ##r## and get
$$
v=sqrt((GM)/(((2GM)/c^2)x)
$$
Which then simplifies to
$$
c/(sqrt(2)*sqrt(x))
$$
and again you can replace ##c## with 1 and now the formula will split out what percent the speed of light you are going.
This also works with orbital energy:
$$
E=c^2/4x
$$
Critical or escape velocity:
$$
v=c/sqrt(x)
$$
and that formula comes out with the same answer as ##y## in ##x=y^2## from the formulas above.
First you start with the gravitational time dilation formula where:
$$
T_1=Tsqrt(1-((2GM)/rc^2))
$$
and rather than entering r for the radius we replace r with the Schwarzschild radius formula ##((2GM)/C^2)x##
with an ##x## at the end representing how many Schwarzschild radii you are away from the center. This brings the formula to look like:
$$
T_1=Tsqrt(1-(2GM)/((((2GM)/c^2)x)c^2))
$$
Which when simplified breaks down to:
$$
T_1=Tsqrt(1-1/x)
$$
and if you make ##T=1## then you just get
$$
=sqrt(1-1/x)
$$
This is very similar to the one in many physics books ##=sqrt(1-r_0/r)## where ##r_0## is equal to the Schwarzschild radius and then r equals the radius from the center. The formula above it just makes it slightly simpler due to making ##r_0## equal to 1 and x equal to how many radii a point you are observing is from the center of the mass.
That is the gravitational time dilation side portion of this relationship. Now for the Velocity time dilation side we use a similar methodology and start with:
$$
T_0=Tsqrt(1-v^2/c^2)
$$
Now we make ##T## equal to 1, ##v## equal to 1, and ##c## to ##y## because now we are going to make ##c## a variable.
$$
T_0=sqrt(1-1/y^2)
$$
What you see now "##1/y^2##" is showing the velocity as a constant 1 and ##y## represents how much faster light is going than the velocity constant of 1. If the above were to show the fraction as ##1/5^2## then this would be the same as saying an object is going at a velocity 1/5th the velocity of light.
So now if we solve the velocity and gravitational time dilation formulas so that we can see how they dilate time to come up with the same result:
$$
sqrt(1-1/x)=sqrt(1-1/y^2)
$$
We can simplify this to
$$
x=y^2
$$
So let's say your radius from the center of the mass ##x## is equal to 4 Schwarzschild radii then the speed an object must move to get the same dilation due to velocity must be equal to half the speed of light since ##y## equals 2 and represents the speed of light going two times faster than the moving object.
I was pretty happy to see a seemingly simple yet complex set of formulas breakdown into one of the simplest algebraic formulas.
This can also be constructed geometrically as seen below:
[1]: http://i.stack.imgur.com/nSC7Y.jpg
The edge of the black circle represents the Schwarzschild radius and each point on the x-axis is another radii away from the center. The Y axis has points that are the square root of x.
Let me know what you think. Please check my math.
You can use the same methodology to break down other formulas as well.
Circular Orbit Velocity
$$
v=sqrt((GM)/r)
$$
Then add in the Schwarzschild radius formula for ##r## and get
$$
v=sqrt((GM)/(((2GM)/c^2)x)
$$
Which then simplifies to
$$
c/(sqrt(2)*sqrt(x))
$$
and again you can replace ##c## with 1 and now the formula will split out what percent the speed of light you are going.
This also works with orbital energy:
$$
E=c^2/4x
$$
Critical or escape velocity:
$$
v=c/sqrt(x)
$$
and that formula comes out with the same answer as ##y## in ##x=y^2## from the formulas above.