# Gravitomagnetic experiment

• I
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## Main Question or Discussion Point

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.

What you think what is the highest value for $B_G$ and $\tau$ we could practically get?

Derivation of equations
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: $M_G={\frac {1}{2}}\iiint _{V}\mathbf {r} \times \mathbf {j} \,{\rm {d}}V$
$|M_G|={\frac {1}{2}}\iiint _{V}r \frac{d^2m}{dl dt} dV=\frac {1}{2}\iiint _{V}r \rho v dV=\frac {1}{2}\int_V r^2 \rho \omega dV=\frac {1}{2}\int_{R_1}^{R_2}h r^2 \rho \omega 2\pi r dr=\frac {h \rho \omega 2\pi}{2}\int_{R_1}^{R_2}(r^3 dr)=\frac{2\pi \rho (R_2^4-R_1^4) h \omega_2}{8}=\frac{2\pi\rho(R_2^4-R_1^4)h\omega_2}{8}$.

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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Gold Member
What youyhink what would the highest $\rho$ and $\omega$ values practically possible?

PeterDonis
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What youyhink what would the highest ρ\rho and ω\omega values practically possible?
What do you think? Have you looked up properties of any materials you might use, like density and tensile strength? Have you calculated what the stress in the material would be at various rotational frequencies?

Staff Emeritus
2019 Award
highest value for τ
How long is a piece of string?

Clearly the smallest torque you could measure on a watchspring is smaller than the smallest torque you could measure on a planet.

Ibix
Much more readable than last time, thank you. However, note that this is incorrect
$\mu_G$is gravitomagnetic constant $\mu_G=\frac{2 2\pi G}{c^2} \approx 9.33\ 10^{-27}N/kg^2s^2 It should be$\mu_G=4\pi G/c^2\approx 9.33\times 10^{-27}\mathrm{N kg^{-2}s^2}$. And some use of the paragraph maths mode ( instead of$) and the eqnarray environment would also help. So instead of this
$|M_G|={\frac {1}{2}}\iiint _{V}r \frac{d^2m}{dl dt} dV=\frac {1}{2}\iiint _{V}r \rho v dV=\frac {1}{2}\int_V r^2 \rho \omega dV=\frac {1}{2}\int_{R_1}^{R_2}h r^2 \rho \omega 2\pi r dr=\frac {h \rho \omega 2\pi}{2}\int_{R_1}^{R_2}(r^3 dr)=\frac{2\pi \rho (R_2^4-R_1^4) h \omega_2}{8}=\frac{2\pi\rho(R_2^4-R_1^4)h\omega_2}{8}$
you get$$\begin{eqnarray} |M_G|&=&{\frac {1}{2}}\iiint _{V}r \frac{d^2m}{dl dt} dV\\ &=&\frac {1}{2}\iiint _{V}r \rho v dV\\ &=&\frac {1}{2}\int_V r^2 \rho \omega dV\\ &=&\frac {1}{2}\int_{R_1}^{R_2}h r^2 \rho \omega 2\pi r dr\\ &=&\frac {h \rho \omega 2\pi}{2}\int_{R_1}^{R_2}(r^3 dr)\\ &=&\frac{2\pi \rho (R_2^4-R_1^4) h \omega_2}{8}\\ &=&\frac{2\pi\rho(R_2^4-R_1^4)h\omega_2}{8} \end{eqnarray}$$

I haven't had a chance to think about the details of what you wrote - I will have a look. Obvious comments given what we worked out last time are that you are assuming that your cylinders are infinitely long and that your small cylinder is infinitely small in an infinitely narrow gap between the two cylinders. How sensitive is your maths to the lack of infinities in reality?

etotheipi and berkeman
Ibix
As I noted, you need to think about the field from a pair of finite cylinders with a finite separation. That'll get you a revised $B_G$ and hence a revised expression for $\tau$. Incidentally, where did the 4 come from in the last expression in your OP?

But this new $\tau$ will depend on a lot of variables - $\rho_1$, $r_1$, $r_2$, $L$ (the length of cylinders 1 and 2), $\delta$ (half the gap between cylinders 1 and 2), $\omega_1$, $\rho_3$, $R_1$, $R_2$, $h$, and $\omega_2$. You need to constrain these somehow. A few constraints are easy:
• At least for a first pass I'd set $r_1=R_1=0$.
• You can set $R_2=\delta$, so the perpendicular cylinder fits exactly between the co-axial ones - don't worry about clearances at this stage.
• Unless you want to go into details of the off-axis field, I'd just require $h\ll 2r_2$ - maybe set $h=r_2/2$.
A more complex constraint is that you don't want your cylinders to disintegrate. You need to look at the stresses in a spinning cylinder (Google is your friend) and require that they be below the yield stress of the material. That'll give you an expression for you $\omega$s in terms of the sizes, densities, and yield stresses of your cylinders.

Then you just play around. Look up some material densities and yield stresses and plug them in. That should mean that you now have an expression for $\tau$ in terms of just three variables $r_2$, $L$, $R_2$. You can add another constraint by insisting that the volumes of the cylinders are some constant determined by your budget for buying materials - so now you have a function of only two variables (times a few different materials). Plug a range of values in and see if you can find the combination that gives you the best $\tau$ - optimisation algorithms will help, but you could just brute force it.

Optionally, you could plug in a maximum energy for spinning the cylinders, which would limit the range of cylinder sizes possible given your rotation rate. If you don't do that, definitely check that you don't have stupid energy requirements!

That'd be my approach, anyway. Then you could see what kind of $\tau$s you can actually generate on cylinders of known mass, and we can think about whether detecting them is plausible at all.

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Gold Member
As I noted, you need to think about the field from a pair of finite cylinders with a finite separation.
For simplicity I assuma that 1. and 2. cylinders are much larger than 3. cylinder.

How sensitive is your maths to the lack of infinities in reality=
Not very sensitive.

Gold Member
What do you think? Have you looked up properties of any materials you might use, like density and tensile strength? Have you calculated what the stress in the material would be at various rotational frequencies?
I do not know for sure how to calculate maximum rotational speed from tensile strength. Maybe someone who has some experience could give me order of magnitude.

berkeman
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Ibix
Nugatory
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I do not know for sure how to calculate maximum rotational speed from tensile strength. Maybe someone who has some experience could give me order of magnitude.
I won’t do that, but I will tell you how to arrive at such an estimate for yourself.

consider a small volume at the edge of the cylinder, surface area $dA$ and thickness $dR$. What is its volume and mass? What is the radial acceleration needed to keep it on its circular path? What force is needed to produce that acceleration? That force is more or less evenly distributed across the surface area... you can take it from here

Ibix
Ibix
For simplicity I assuma that 1. and 2. cylinders are much larger than 3. cylinder.
Ok. But your torque depends on $R_2^4$. Insisting that $R_2$ be small seems like a mistake to me.
Not very sensitive.
Perhaps you could show your maths for this? My quick calculation suggests that for a fixed mass of metal I can get an order of magnitude variation in $B_G$ by varying the aspect ratio of the cylinders.

Gold Member
${\displaystyle {\frac {E}{I}}=K\left({\frac {\sigma }{\rho }}\right),}$ is not helpful because it is about kinetic energy not generated GM-field.

$B_G=\frac{\mu_G \rho \omega (r_2^2-r_1^2)}{2}$, but
$E_{kinetic}=\frac{\rho\ \omega^2\ 2\pi (r_2^4-r_1^4)\ h}{8}$

B_G is proptional to $r^2$, but kinetic energy is propotional to $r^4$.

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Ibix
And rotational kinetic energy depends on...

Staff Emeritus
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On top of everything else, I am not sure that conceptually this is correct. Which way does the third cylinder want to rotate? And why?

(If you say "because of the right-hand rule" as part of your answer, my next question is "and where does that come out of GR?")

Gold Member
And rotational kinetic energy depends on...
You mean $\frac{E}{I}=\frac{\omega^2}{2}$
so $\omega=\sqrt{\frac{2\ K\ \sigma}{\rho}}$?
It seems wrongbeacause omega does no depend of r.
Could you just write correct expression for maximum $\omega\ r^2$?

Gold Member
Which way does the third cylinder want to rotate? And why?
because of the right-hand rule. I use Gravitoelectromagnetic model, which is approximation for GR where gravitational fields are small.

PeterDonis
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because of the right-hand rule.
As @Vanadium 50 was hinting in post #15, a human choice of convention for how to represent things, which is what the right hand rule is, cannot have a physical effect. So this answer cannot be correct. The right-hand rule can't make things rotate a certain way.

Ibix
It seems wrongbeacause omega does no depend of r.
Good point - that does seem odd. You could check another source, or you could work through Nugatory's method (which was more or less what I had in mind when I proposed this above).
Could you just write correct expression for maximum $\omega\ r^2$?
No, because I haven't worked through the maths myself. Happy to check yours when you've done it.

Staff Emeritus
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which is approximation for GR where gravitational fields are small.
Articles are important. It is an approximation, not the approximation.

I am not convinced this is a valid approximation. I'd want to see an explanation for the direction that Cylinder 3 turns.

Staff Emeritus
2019 Award
Which way does the third cylinder want to rotate? And why?
If you say "because of the right-hand rule" as part of your answer, my next question is "and where does that come out of GR?"
because of the right-hand rule
I didn't think I'd need to say this, but OK, "where does that come out of GR?"

Ibix
I am not convinced this is a valid approximation. I'd want to see an explanation for the direction that Cyliner 3 turns.
Depends on the sense of rotation of cylinders 1/2 and 3, I think. You could look at pairs of ingoing and outgoing geodesics near the pole of a Kerr black hole, I guess?

I'm also slightly concerned that this is applying a non-coaxial torque to a spinning cylinder - i.e. a gyroscope. In other words you'd need to look for precession, which is likely to look very like vibration from the spinning...

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I'd like to know where it comes from.

What I suspect is happening is that the rotation causes terms to cancel at first order (for every mass element moving at v, there is one moving at -v) so the terms we neglected in getting to geomanetism become important. But let's start with the simpler question: clockwise or counter-clockwise.

Staff Emeritus
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And for extra fun, if I reverse all three spins, which way does it rotate?

Gold Member
And for extra fun, if I reverse all three spins, which way does it rotate?
Torque on 3. cylinder is crosswise to angular speeds of 1. , 2. and 3. cylinder.

$\tau=4 M_G\times B_G$
$|\tau|=\frac{2\pi \rho_3 (R_2^4-R_1^4) \omega_3 B_G}{2}=\frac{\mu_G \omega_3 \rho_3 (r_2^2-r_1^2) 2\pi \rho_1 (R_2^4-R_1^4) \omega_1}{4}$

What I suspect is happening is that the rotation causes terms to cancel at first order (for every mass element moving at v, there is one moving at -v)
No, because opposite moving sides have opposite displacement from centre. Same way that solenoid magnetic field is not canceling out altougth there are opposite current on different sides from centre.

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