# Gravitomagnetic experiment

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1. Jul 20, 2015

### olgerm

I have thought about an experiment which to demostrate gravitomagntism: spinning massive cylinder creates gravitomagnetic field above it. Oscillating pendulum above the cylinder departs from its trajectory because of "gravito-Lorents" force.
To calculate magnetic force above the cylinder I use formula:
$\vec B=\frac{G*4*π}{c^2}*\frac{\vec{v}×\vec{l}*m}{[\vec{l}]^3}$
where:
B is magnetic field in some point A.
G is gravitational constant.
c is speed of light.
v is speed of moving pointmass at point C.
l is vector $l=\vec{AC}$
m is mass of pointmass .

To get gravitomagnetic field by whole cylinder integrate B over volume, because every point of cylinder may be seen as pointmass.

for circle:
$\vec B(x, y, h, r)=\int_0^{2*π} (\frac{G*4*π*ρ}{c^2}*\frac{\vec{v}×\vec{l}*r}{[\vec{l}]^3} *dα)$

for disc:
$\vec B(x, y, h, R)=\int_0^R (\int_0^{2*π} (\frac{G*4*π*ρ}{c^2}*\frac{\vec{v}×\vec{l}}{[\vec{l}]^3}*r* dα)dr)$

And finally for cylinder:
$\vec B(x, y, R, H, H_0)=\int_{H_0}^{H+H_0} (\int_0^R (\int_0^{2*π} (\frac{G*4*π*ρ}{c^2}*\frac{\vec{v}×\vec{l}}{[\vec{l}]^3}*r* dα)dr)dh)$
where:
R is cylinder radius.
H is cylinder height.
H0 is (vertical) distance between pendulum and cylinder.

$\vec{v}=ω*(Sin(a)*r, Cos(a)*r, 0)$
$\vec{l}=(x - Cos(a)*r, y - Sin(a)*r, h)$
where ω is angular velocity of cylinder.
so $\vec v×\vec l = (h*r*Cos(a),h*r*Sin(a), -r*x*Cos(a) - r*y*h*Sin(a) + r^2)$ and
$|\vec{l}|^3=(h^2+ (x - r*Cos(a))^2+ (y - r*Sin(a))^2)^{3/2}$

$\vec B(x, y, R, H, H_0)=\int_{H_0}^{H+H_0} (\int_0^R (\int_0^{2*π} (\frac{G*4*π*ρ}{c^2}*\frac{(h*r*Cos(a),h*r*Sin(a), -r*x*Cos(a) - r*y*Sin(a) + r^2)*r*ω}{(h^2 + (x - r*Cos(a))^2 + (y - r*Sin(a))^2)^{3/2}}* dα)dr)dh)$

$\vec B(x, y, R, H, H_0)=\frac{G*4*π*ρ*ω}{c^2}*\int_{H_0}^{H+H_0} (\int_0^R (\int_0^{2*π} (\frac{(h*r*Cos(a),h*r*Sin(a), r-x*Cos(a) - y*Sin(a))*r^2}{(x^2+y^2+r^2+h^2-2*r(x*Cos(a)+y*Sin(a)))^{3/2}}* dα)dr)dh)$

$\vec B(x, y, R, H, H_0)=\frac{G*4*π*ρ*ω}{c^2}*\int_{H_0}^{H+H_0}(\int_0^R((\frac{h*π*r*x*(\sqrt{h^4*(x^2+y^2-r^2)^2+2*h^2*(r^2+x^2+y^2)}-(h^2+r^2+x^2+y^2))}{(x^2+y^2)*\sqrt{h^4+(x^2+y^2-r^2)^2+2*h^2*(r^2+x^2+y^2)}},\frac{h*π*r*y*(\sqrt{h^4+(x^2+y^2-r^2)^2+2*h^2*(r^2+x^2+y^2)}-(h^2+r^2+x^2+y^2))}{(x^2+y^2)*\sqrt{h^4 + (x^2+y^2-r^2)^2+2*h^2*(r^2+x^2+y^2)}},???)*dr)dh)$

Is my equation correct?
Can anyone simplify that equation for me?
How would be best to observe/measure gravitomagnetic field ,with pendulum changing it´s trajectory (because of gravitomagnetic field´s vertical component), pendulum changing it´s oscillation period(because of gravitomagnetic field´s horizontal component), or with two cylinders pulling/pushing each other like normal magnets?
May this experiment work?
Has such experiment ever been done before?

2. Jul 20, 2015

### Dr. Courtney

When thinking of an experiment, one needs to estimate the magnitude of an effect.

How large a cylinder of what mass and dimensions will it take to cause a measurable effect?

3. Jul 20, 2015

### olgerm

I need to simplify that equation to know how big cylinder and how big(fast) angular velocity I need.

density ρ is probably not more than about 14000*kg/m3
$\frac{4*π*G}{c^2}≈9.396706053583*10^{-27}$

How big(strong) gravitomagnetic field should be to be observable?

Last edited: Jul 20, 2015
4. Jul 20, 2015

### Staff: Mentor

You might want to read up on Gravity Probe B, a satellite experiment that detected the gravitomagnetism due to the Earth's rotation:

https://einstein.stanford.edu/

The "frame-dragging effect" is gravitomagnetism. (The "geodetic effect" is another GR effect that is not present in Newtonian gravity, but it is present for non-rotating gravitating bodies.) As you will see if you dig into the numbers, even with an object the size of the Earth, the effects being detected are very small. I'm not sure you could detect them using objects small enough to fit in a laboratory.

5. Jul 29, 2015

### olgerm

At time t=0 pendulum is in it´s amplitude position φ=θ and z=0. If I measure gravitomagnetic field ,with pendulum changing it´s trajectory (because of gravitomagnetic field´s vertical component), then angel between pendulum original horizontal moving direction and pendulum horizontal moving direction after time t is $β=\iint dt^2*M(t)/b(t)$

M is torque. $M_{x-z}(t)=F_{gravito-Lorentz}(t)*(x(t)^2+z(t)^2)^{1/2}$
b is moment of inertia. $b_{x-z}(t)=m*(x(t)^2+z(t))^2)$
Fgravito-Lorentz is gravito-Lorentz force. $F_{gravito-Lorentz}(t)=m*4*v(t)*B(t)$

So:
$β(t)=\iint dt^2*4*v(t)*B(t)/(x(t)^2+z(t)^2)^{1/2}$

Pendulum equation of motion is $φ(t)=θ*sin(\sqrt{\frac{g}{l}*t})$
so movingpiontmass equation of motion is
$\begin{cases} x(t)=sin(φ)*l=sin(θ*sin(\sqrt{\frac{g}{l}*t}))*l\\ y(t)=l*(cos(φ)*-1)=cos(θ*sin(\sqrt{\frac{g}{l}*t}))*l-l\\ \end{cases}$

and B(t)=B(x(t), y(t), R, H, H_0)
$v_{x-y}(t)=\frac{dx(t)}{dt}= \frac{g*θ*Cos(\sqrt{(g*t)/l})*Cos(θ*Sin(\sqrt{(g*t)/l}))}{2*\sqrt{(g*t)/l}}$

So:
$β(t)=\iint dt^2*4*\frac{G*4*π*ρ*ω}{c^2}*\int_{H_0}^{H+H_0} (\int_0^R (\int_0^{2*π} (\frac{(r-x*Cos(a) - y*Sin(a))*r^2}{(x^2+y^2+r^2+h^2-2*r(x*Cos(a)+y*Sin(a)))^{3/2}}* dα)dr)dh)*\frac{g*θ*Cos(\sqrt{(g*t)/l})*Cos(θ*Sin(\sqrt{(g*t)/l}))}{2*\sqrt{(g*t)/l}}$

that is my estimation of magnitude of effect.
May this experiment work?

6. Jul 29, 2015

### gianeshwar

I find your mind boggling calculations impressive!
Despite that why are you praying for it to work.It will work if it is bold enough.
As I recall an event in Einstein,s life when his General theory was proved right.A generalist asked Einstein about what if you were proved wrong,then he replied that I would have felt sorry for the poor lord.

7. Jul 31, 2015

### Khashishi

Try plugging in actual numbers. Throw it into Mathematica or similar and see what you get.