# Is my equation exacter than Einstein's equation?

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#### simplified

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##### Is my equation exacter than Einstein's equation?
« on: 23/12/2012 06:58:03 »
I have created equation for gravitational slowing of time to avoid gravitational  length contraction.
T=T_o/(1-2GM/Rc^2)^[(1+v^2/c^2)/2]
v - speed of atomic clock in gravitation
Is this equation exacter than Einstein's equation?

#### CliffordK

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##### Re: Is my equation exacter than Einstein's equation?
« Reply #1 on: 23/12/2012 12:55:33 »
T=T_o/(1-2GM/Rc^2)^[(1+v^2/c^2)/2]

T = $$\frac{T_0}{(1-\frac{2GM}{Rc^2})^{\frac{(\frac{1+v^2}{c^2})}{2}}$$

I think.

(oops, I think I got the "1 -" in the wrong place, fixed).
« Last Edit: 23/12/2012 16:25:31 by CliffordK »

#### simplified

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##### Re: Is my equation exacter than Einstein's equation?
« Reply #2 on: 23/12/2012 18:14:33 »
T=T_o/(1-2GM/Rc^2)^[(1+v^2/c^2)/2]

T = $$\frac{T_0}{(1-\frac{2GM}{Rc^2})^{\frac{(\frac{1+v^2}{c^2}}){2}}$$

I think.

(oops, I think I got the "1 -" in the wrong place, fixed).
T=$$\frac{T_0}{(1-\frac{2GM}{Rc^2})^{\frac{c^2+v^2}{2c^2}}$$
« Last Edit: 23/12/2012 18:35:31 by simplified »

#### CliffordK

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##### Re: Is my equation exacter than Einstein's equation?
« Reply #3 on: 23/12/2012 23:28:12 »
Welcome to a bit of TEX hacking.

$$\frac{(\frac{1+v^2}{c^2})}{2}$$ = $$(\frac{1+v^2}{2c^2})$$

However, it is not the same as:
$$\frac{c^2+v^2}{2c^2}$$ = $$\frac{1}{2} + \frac{v^2}{2c^2}$$

Unless I missed something in the original equation.

I'll try to understand your equation later, but I assume you are doing something similar to the Lorentz Transformation.

Velocity is a function of time and distance, thus
Time is a function of velocity and distance.

Hmmm,
So, is T0 a constant (initial time), or is it a function (the time function independent of gravitation effects)?
M is the mass of the planet.
G is the gravitational constant.
R is distance from the center of the planet.
c is the speed of light.
v is the velocity of whatever is carrying the clock.

Of course, if you think of Earth, the planet surface where most of our clocks are is spinning and moving.

Anyway, why don't you post a little about what you're trying to accomplish, and the derivation of your equation.

#### simplified

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##### Re: Is my equation exacter than Einstein's equation?
« Reply #4 on: 24/12/2012 17:35:49 »
Cliff
1-v^2/c^2=(c^2-v^2)/c^2
1+v^2/c^2=(c^2+v^2)/c^2
What is wrong?I think you don't know the mathematical rule.Maybe the rule works only in Russia,I don't know.Even my calculator works so.

You want to know equation derivation.I don't like gravitational length contraction.If the equation is right then we don't need gravitational length contraction.Therefor I have created it.

#### CliffordK

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##### Re: Is my equation exacter than Einstein's equation?
« Reply #5 on: 25/12/2012 04:32:41 »
No...  that algebra is fine.
Thinking:
^[(1+v^2/c^2)/2] = ^[((1+v^2)/c^2)/2]

[]

See, TEX is so much easier  []

#### simplified

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##### Re: Is my equation exacter than Einstein's equation?
« Reply #6 on: 02/01/2013 19:05:03 »

Hmmm,
So, is T0 a constant (initial time), or is it a function (the time function independent of gravitation effects)?

http://en.wikipedia.org/wiki/Gravitational_time_dilation
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe

#### Raphael

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##### Re: Is my equation exacter than Einstein's equation?
« Reply #7 on: 21/01/2013 21:04:13 »
simplified what do you think of this site and its graphics in its attempt to explain gravitational length contraction?

this puzzle is found at the bottom of the page....
Quote
A bucket of water has a spring soldered to the bottom, as shown. A cork is attached the spring, and is therefore suspended under the surface of the water.

You are on top of the CN tower, holding the bucket, and step off. While falling towards the ground, do you see the cork move towards the top of the water, towards the bottom of the bucket, or stay where it is relative to the bucket and the water?

namaste
« Last Edit: 21/01/2013 21:07:09 by Raphael »

#### simplified

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##### Re: Is my equation exacter than Einstein's equation?
« Reply #8 on: 24/01/2013 17:26:13 »
Archimedes' force will increase.
delta of Archimedes' force=pv*delta of g
But force of the spring will increase too
delta of force of the spring=2delta of energy of the spring