Gravitational experiment

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  • #1
olgerm
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I was thinking about an experiment to demonstrate gravitomagnetic effect. I did my calculations using gravitomagnetic model. It is not as accurate as general relativity, but GR should give similar predictions. I do not know if it would be possible to to this experiment in real life(are there enougth accurate sensors and tought materials).
installations consists of three spinning cylinders. first to cylinders are for creating a magnetic field. last one is for detecting gravitomagnetic field. last cylinder under axis 90 degrees angle compared to first two cylinders.

gravitimagnetic field created by first two cylinders right between the cylinders: ##B_G=\frac{\mu_G*\omega_1*\rho_1*(r_2^2-r_1^2)}{2}##
  • ##\omega_1## is angular speed of 1. and 2. cylinder.
  • ##\mu_G##is gravitomagnetic constant ##\mu_G=\frac{2*2\pi*G}{c^2} \approx 9.33*10^{-27}*N/kg^2*s^2=9.33*10^{-27}*s/kg##
  • ##\rho_1## is density of 1. and 2. cylinder.
  • ##r_2## is 1. and 2. cylinders outer radius.
  • ##r_1## is 1. and 2. cylinders inner radius.
torque on third cylinder because of gravitomagnetic effect is crosswise to its angular speed and angular speed of first two cylinders.
##\tau=\frac{2\pi*\rho_3*(R_2^4-R_1^4)*\omega_3*B_G}{2}=\frac{\mu_G*\omega_3*\rho_3*(r_2^2-r_1^2)*2\pi*\rho_1*(R_2^4-R_1^4)*\omega_1}{4}##
  • ##R_1## is 3. cylinders inner radius.
  • ##R_2## is 3. cylinders outer radius.
  • ##\rho_3## is density of 3. cylinder.
  • ##\omega_3## is angular speed of 3. cylinder.

Wat you think what is the highest value for ##\tau## we could practically get?
20200509_201340.jpg
 
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  • #2
Vanadium 50
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9.331345047580891 is nonsense. If you want people to take you seriously, put a little more effort in your posts.

Also, did you put any numbers in your equation at all? You have around twenty-seven orders of magnitude you need to pick up. How do you attempt to do that? Again, if you want people to take you seriously, put a little more effort in your posts.
 
  • #3
olgerm
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Also, did you put any numbers in your equation at all?
Yes. But it is hard to guess biggest practically achievable angular speed and smallest detectable torque. Maybe you can estimate these numbers.
 
  • #4
Vanadium 50
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The first thing I would do is plug in reasonable numbers and get an idea of what I needed to see the effect. If you aren't willing to put in at least that much effort, why should I?
 
  • #5
Ibix
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Vanadium50 has a point - you seriously need to proofread. There's an error in your last equation (you expand ##B_G##but don't remove it), and you never define ##\rho##, ##\rho_2## or ##\omega_2## and you do define ##\rho_1## and ##\rho_3## which you don't use. You are also putting in asterisks for some reason. All they do is make your equations wide enough that they are difficult to read on a narrow screen like a phone.

Plug in sensible numbers for the sizes and densities. That should get you to ##\tau=A\omega\omega_2##, with a number in place of the ##A##. Then look up lab centrifuges and torsion balances and see if you are anywhere near.
 
  • #7
Vanadium 50
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More nonsense.

628318.5307179586
Even more.

So you need a 5m centrifuge made of lead? Your torque has no units? Your cylinders have no lengths?
 
  • #9
Ibix
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Please learn to format numbers sensibly. Don't quote numbers to more than 2 or 3 significant figures. Giving more precision just makes your readers' lives difficult and is completely unjustifiable in a back-of-the-envelope calculation like this. Get rid of the asterisks. Do you put them in writing? Do you buy flour in 1*kg bags, or 1 kg bags?

How long do the cylinders have to be? If they're 1m long they weigh approximately ##1.6\times 10^6\mathrm{kg}##. Each one's kinetic energy once spun up (##E=I\omega^2/2##, ##I=m(r_1^2+r_2^2)/2##) is about ##8\times 10^{18}\mathrm J## so, taking Wikipedia's 22.5GW power output for the Three Gorges dam, you need its entire output for about 11 years per cylinder, neglecting frictional losses. The metal alone (accepting PeterDonis' inference of osmium) would cost about $20bn per cylinder (fifty million troy ounces at $400 per troy ounce), assuming there's that much osmium available to buy without distorting the market.

I was kind of assuming that if you were looking at lab centrifuges you'd think about the typical masses such a thing could handle - a few grams at most.
 
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  • #10
olgerm
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I was kind of assuming that if you were looking at lab centrifuges you'd think about the typical masses such a thing could handle - a few grams at most.
I put into the highest values to see if the torque with this method is close to detectable and it is. Even with smaller cylinders and lighter materials it might still be detectable. What you estimate to be smallest torque measurable?
 
  • #11
pervect
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Have you evaluated your proposal to see if the cylinders are physically capable of rotating that fast without exploding failing under stress from centrifugal force?

That would be important in a practical test. If I'm understanding your figures right - .25 meters radius (no sense in quoting more than 2 significant figures, as others have mentioned), with w=100000 radians, second, the velocity at the outer edge of the cylinder is 25 kilometers/second, over twice escape velocity from Earth. So if it did explode fail under stress, some of the fragments would escape the Earth.

I do strongly suspect that your cylinders will not be strong enough to rotate that fast, but it would be best if you do the calculation for yourself.
 
  • #12
olgerm
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I do not have the equipment to do this experiment.

Have you evaluated your proposal to see if the cylinders are physically capable of rotating that fast without exploding failing under stress from centrifugal force?
Not saying I could make this experiment with given values. What you think were the biggest ##B_G## and ##\tau## I could get in real life?
 
  • #13
Vanadium 50
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Your equation is hard to read.
Your equation is almost certainly wrong.
Your terms are undefined.
Your units are missing.
Your numbers have nonsensical precision.
Your chosen numbers come out of nowhere.

If you're unwilling to fix all that, why should we spend our time figuring out what you mean?
 
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  • #14
Ibix
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Your equation is almost certainly wrong.
Per the link in the OP, the equations for gravitoelectromagnetism are Maxwell's equations with a bit of relabelling. A spinning ring is analogous to a current loop, from which the field on axis (equation 6 here) is perpendicular to the plane of the ring and has magnitude $$B_z=\frac{a^2 I}{2c^2\epsilon_0\left(z^2+a^2\right)^{3/2}}$$where ##z## is the distance out of the plane of the ring and ##a## is the radius of the ring and I have substituted ##\mu_0=1/c^2\epsilon_0##.

Looking back at the gravitoelectromagnetism page, we just need to replace ##\epsilon_0## with ##1/4\pi G## and ##I## with ##J_mdA##, where ##J_m=\rho v=\rho\omega a## is the mass current density and ##dA=dzda## is an elementary area. So that gets us a gravitomagnetic field from an elementary ring of$$B_g=\frac{2\pi G a^3 \rho\omega}{c^2\left(z^2+a^2\right)^{3/2}}$$Given that there's no length in the formula in the OP, it's probably either a thin annulus or an infinitely long cylinder producing this field. The equation appears to be divergent in the latter case (edit: nope - see my next post), but the former gives us$$\begin{eqnarray*}
B_g&=&\int_{r_1}^{r_2}\frac{2\pi G a^3 \rho\omega}{c^2\left(z^2+a^2\right)^{3/2}}da\\
&=&\frac{\omega\rho \mu_G}{2}\left(\frac{r_2^2+2z^2}{\sqrt{r_2^2+z^2}}-\frac{r_1^2+2z^2}{\sqrt{r_1^2+z^2}}\right)
\end{eqnarray*}$$where ##r_1## and ##r_2## are the inner and outer radii of the annulus and I have substituted ##\mu_G=4\pi G/c^2##. That, indeed, does not look much like the OP's formula. In the case that ##z=0##, however, it simplifies to $$B_g=\frac{\mu_G\rho\omega}{2}(r_2-r_1)$$which is similar to, but not identical to, the OP's formula (which has squares on the ##r_1## and ##r_2##). But this would imply that the OP's experimental setup is completely wrong - we aren't talking about spinning cylinders, but a single thin spinning annulus where ##\rho## is an areal mass density, not a volumetric one.

@olgerm - I really think you need to show us your derivation before we proceed any further.
 
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  • #15
PeterDonis
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the equations for gravitoelectromagnetism are Maxwell's equations with a bit of relabelling
And a few differences in the constants in the equations, yes. Also, these equations are approximations, not exact, whereas Maxwell's Equations are exact for electromagnetism.

I really think you need to show us your derivation before we proceed any further.
Indeed.
 
  • #16
Ibix
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The equation appears to be divergent in the [case of an infinitely long cylinder]
Hm. Tried again with Maxima on my phone instead of my laptop and it says the integral doesn't diverge, and leads to the OP's original equation. Don't know what I did wrong earlier.

But that implies we're supposed to be considering an infinitely long hollow cylinder, rather than two cylinders. Perhaps a small gap between two half-infinite cylinders is the plan? Presumably they don't actually have to be infinitely long, but some sensitivity analysis around the length and gap size is probably in order. I'm going to go out on a limb and say it needs to be longer than its radius for the formula in the OP to apply - so multiply all my energy and cash values by at least five.
 
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  • #17
olgerm
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Given that there's no length in the formula in the OP, it's probably either a thin annulus or an infinitely long cylinder producing this field.
True.
$$\begin{eqnarray*}
B_g&=&\int_{r_1}^{r_2}\frac{2\pi G a^3 \rho\omega}{c^2\left(z^2+a^2\right)^{3/2}}da\\
&=&\frac{\omega\rho \mu_G}{2}\left(\frac{r_2^2+2z^2}{\sqrt{r_2^2+z^2}}-\frac{r_1^2+2z^2}{\sqrt{r_1^2+z^2}}\right)
\end{eqnarray*}$$where ##r_1## and ##r_2## are the inner and outer radii of the annulus and I have substituted ##\mu_G=4\pi G/c^2##. That, indeed, does not look much like the OP's formula. In the case that ##z=0##, however, it simplifies to $$B_g=\frac{\mu_G\rho\omega}{2}(r_2-r_1)$$which is similar to, but not identical to, the OP's formula (which has squares on the ##r_1## and ##r_2##)
if you subsistude z to 0 the formula has squares!
 
  • #18
Ibix
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if you subsistude z to 0 the formula has squares!
The formula you quote, describing the field from a thin annulus, does not. A different formula, that for an infinitely long hollow cylinder, does have squares. Which setup are you talking about?
 
  • #19
olgerm
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Derivation:

gravitimagneticfield:
I used formula ##B=\mu*n*I## from http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html .
using similarities between gravitomagnetism and electromagnetism: ##B_G=\mu_G*n*I_G=\frac{d^2m}{dl*dt}*\mu_G=\mu_G*\frac{d^2m}{dl*dt}=\mu_G*\int_{r_1}^{r_2}(dr*\rho*v(r))=\mu_G*\int_{r_1}^{r_2}(dr*\rho*\omega*r)=\mu_G*\rho*\omega*\int_{r_1}^{r_2}(dr*r)=\frac{\mu_G*\rho*\omega*(r_2^2-r_1^2)}{2}##

magnetic moment:
I used formula ##m ={\frac {1}{2}}\iiint _{V}\mathbf {r} \times \mathbf {j} \,{\rm {d}}V## from https://en.wikipedia.org/wiki/Magnetic_moment .
using similarities between gravitomagnetism and electromagnetism: ##M_G={\frac {1}{2}}\iiint _{V}\mathbf {r} \times \mathbf {j} \,{\rm {d}}V##
##|\tau|={\frac {1}{2}}\iiint _{V}r *\frac{d^2m}{dl*dt}* dV=\frac {1}{2}\iiint _{V}r * \rho*v* dV=\frac {1}{2}\int_V r^2*\rho*\omega* dV=\frac {1}{2}\int_{R_1}^{R_2}h*r^2*\rho*\omega*2\pi*r*dr=\frac {h*\rho*\omega*2\pi}{2}\int_{R_1}^{R_2}(r^3*dr)=\frac{2\pi*\rho*(R_2^4-R_1^4)*h*\omega_2}{8}=\frac{2\pi\rho(R_2^4-R_1^4)h\omega_2}{8}##

torque:
I used formula ##\tau=M\times B## from https://en.wikipedia.org/wiki/Magnetic_moment .
using similarities between gravitomagnetism and electromagnetism: ##\tau=4*M_G\times B_G##
 
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  • #20
Ibix
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So you are considering an infinitely long rotating cylinder, with a short rotating cylinder perpendicular to it. Where does the formula for torque come from? And lose the asterisks! It's taken twenty posts to get to a derivation you should have put in the OP, and you continue to make life difficult for us by putting in unnecessary symbols that make your formulae hard to read. This is standard notation - please e polite enough to use it.
 
  • #21
olgerm
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So you are considering an infinitely long rotating cylinder, with a short rotating cylinder perpendicular to it. Where does the formula for torque come from? And lose the asterisks! It's taken twenty posts to get to a derivation you should have put in the OP, and you continue to make life difficult for us by putting in unnecessary symbols that make your formulae hard to read. This is standard notation - please e polite enough to use it.
I edited last post. Now it answers your questions.
 
  • #22
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Thread closed for Moderation...

After further Mentor discussion, this version of this thread will remain closed.
 
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