Can We Practically Measure the Gravitomagnetic Effect with Spinning Cylinders?

[olgerm]

$dF=\omega^2\ R\ \rho\ dR\ dA$

Yes, that is the necessary force.

How to calculate maximum $\omega$ from this?

$dF=dA\ \sigma$ and therefore
$dA\ \sigma=\omega^2\ R\ \rho\ dR\ dA$
$\sigma=\omega^2\ R\ \rho\ dR$
$\omega=\sqrt{\frac{\sigma}{R\ \rho\ dR}}$
that can't be right because $dR$ remains in. What the correct way for calculating max $\omega$?

 

[pervect]

Which was one rotation per 30 million years. Not a big effect (and Gravity Probe B had a lot of trouble measuring it).

This lines up with my ballpark β² estimate. Say orbinal speed is 17000 miles per hour and the Earth rotation is 1000. So it's about a 4 x 10^-11 effect. One turn per 30 million years is a part in 10¹¹.

For the record, I also very much doubt that a lab experiment is going to be able to replicate the GP-B results. I do think that the effort to investigate other approaches could be of educational benefit if done correctly and methodically. The primary benefit and onus of such investigation is on the poster. An important part of the investigation process is to read as much of the relevant literature as one is able, to see what others have done previously, and to be on the lookout for things that don't quite fit.

 

[Vanadium 50]

The primary benefit and onus of such investigation is on the poster. An important part of the investigation process is to read as much of the relevant literature as one is able, to see what others have done previously, and to be on the lookout for things that don't quite fit.

I see very little of that from the OP. He keeps asking us the same questions over and over and not addressing any of the issues we have brought up. In fact, I found a thread of his from five years ago where we are covering the same issues. Including the suggestion that he look at Gravity Probe B. Not much seems to have come from that.

 

[olgerm]

]Do you think $\omega=\sqrt{\frac{2\ K\ \sigma}{\rho}}$ is correct?
if it is then
$|\tau|=\frac{2\pi \rho_3 (R_2^4-R_1^4) \omega_3 B_G}{2}=\frac{\mu_G \omega_3 \rho_3 (r_2^2-r_1^2) 2\pi \rho_1 (R_2^4-R_1^4) \omega_1}{4}=\frac{2\ K\ \sigma\ \mu_G\ (r_2^2-r_1^2) 2\pi\ (R_2^4-R_1^4)}{4}$

if I assume it is correct:
$\mu_G=\frac{2\ 2\pi\ G}{c^2} \approx 9.33\ 10^{-27}\ N\ s^2/kg^2$
$K=0.5$
$\sigma=7000000000\ Pa$
$r_2=5$
$r_1=0$
$R_2=1$
$R_1=0$
$|\tau|=\frac{2\ K\ \sigma\ \mu_G\ (r_2^2-r_1^2) 2\pi\ (R_2^4-R_1^4)}{4}=2.57\ 10^{-15}\ N\ m$

Do you agree, that making this assumation my result is correct? Do you agree, that this effect is too small to be measured?
Do you think the assumation is correct?

 

[Nugatory]

that can't be right because $dR$ remains in.

A differential that you can’t get rid of is telling you that there’s an integral in your future.

 

[olgerm]

A differential that you can’t get rid of is telling you that there’s an integral in your future.

Should I integrate this side over R?

 

[Nugatory]

Should I integrate this side over R?

if you choose the right bounds of integration you will have the quantity that you will compare with the tensile strength

 

[pervect]

Here is a sketch of my argument that a large rotating object, such as the Earth, is going to be a stronger source of the gravitomagnetic field than a smaller, lab-created object.

The significance of this is that it's very hard to test for the existence of the gravitomagnetic filed of the Earth (though, as has been mentioned, it has been done) - thus, trying to do it in a pure lab envirnoment will be more difficult and probably impossible.

To start with, we need to know the basics dimensional dependency of the gravitomagnetic field. Neglecting all the various numerical factors, we find from the examples in Wiki that at the surface of a spherical body of radius R, the gravitomagnetic field strength will be proportional to

$$B_G \propto \frac{G}{c^2} \frac{m}{R} \, \omega\propto \frac{G}{c^2} \rho \, R^2 \, \omega \propto \frac{G}{c^2} \, \rho \, V^2 / \omega$$

here G is the gravitational constant and c is the speed of light - not strictly necessary to include, as they are constants, but they make the units sensible. R is the radius of the sphere, $\omega$ is it's rate of rotation, and V = $r \omega$ is the surface velocity of the sphere.

The significance of using V is that for any given material, a hoop of that material held together only by it's own strength (and not gravity) has a maximum tangential velocity before it will fail under stress. Gravity becomes important for larger objects, but for lab scale objects, this characteristic velocity is what's useful. While we are using a sphere and not a hoop, I think the failure criterion for a rotating sphere would be similar, though it would certainly be better to do a more thourough analysis.

The source of the observation about the importance of V is from the wiki page on space tethers, https://en.wikipedia.org/w/index.ph...ldid=953929239#Properties_of_useful_materials

Hypersonic skyhook equations use the material's "specific velocity" which is equal to the maximum tangential velocity a spinning hoop can attain without breaking:

From the same article, we can see that the maximum known theoretical characteristic velocity would be for single walled carbon walled nanotubes, of about 5 km/sec, with commercially available materials (single walled carbon nanotubes are not yet commerically available in usable form AFAIK) being as high as 2km/sec.

So, where does the Earth fit on this scale? The velocity at the surface of the Earth is small, so V = .5 km/sec, about 1/10 of the maximum value of V we can get in a lab. So, potentailly, the lab could be better based solely on V. However, the field stregnth is proportioanl to $V^2 / \omega$, and $\omega$ is very small for the Earth, making it the better candidate. Basically, the fact that the field strength is inversely proportional to $\omega$ tells us that larger structures, even though they rotate at a slower angular velocity $\omega$ will produce a larger gravitomagnetic field than small, faster rotating structures.

Since the Earth is already here, it's an ideal candidate for the source of the field. Perhaps another astronomical object would be even better, but the Earth is convenient because it's so nearby.

The remaining issue is how to detect the field. Measuring the torque directly would require bearings that can hold the axis of rotation stable with unreasonable accuarcy levels, and have a significant potential for introducing errors. So observing the precession directly, rather than trying to hold the direction stable and measure the force, seems far preferable. The question remains as to why an orbital approach is the best approach, rather than some surface mounted experiment. My thoughts on this are that eliminating effects due to Thomas precession , https://en.wikipedia.org/wiki/Thomas_precession, was the motivation for having the measuring instrument be in free fall, which implies an orbital experiment as opposed to a surface based experiment.

So in conclusion, larger is better as far as generating a gravitomagnetic field, and there isn't any obvious choice for a better detection scheme of said field than the GP-B satelite system - which was up to the job, but just barely.

 

[olgerm]

if you choose the right bounds of integration you will have the quantity that you will compare with the tensile strength

I still do not know how to do it. Could you just solve it for maximum angular speed?

 

[Vanadium 50]

@pervect, I like the analysis. However, there is a difference between angular momentum GR effects and gravitomagnetism, which is one particular angular momentum GR effect. Given that the OP has not commented on any of the fundamental issues that have been brought up, it's not clear what his reaction will be.

Dimensional analysis like yours shows the effect is of order β² or in more detail, β₁ x β₂. The real power of Gravity Probe B is that it's in orbit: so you have a velocity of 17000 mph there - 17x faster than the Earth's rotation. Gravity Probe B needed that factor of 17.

The perigee advance - the equivalent of Mercury's perihelion advance - was measured to be about 15x larger than Mercury's. Mercury advances more per orbit, but a LEO satellite makes many more orbits.

 

[Nugatory]

I still do not know how to do it. Could you just solve it for maximum angular speed?

To do that you need the stress as a function of the angular speed. So far you have the force needed to keep a segment of thickness $dr$ at a distance $r$ from the axis on its circular path. The radial stress at a distance $r$ from the center must be sufficient to keep every segment between $r$ and the outer edge on its circular path.

(Of course we’re trying to come up with an order of magnitude estimate here. We’re neglecting tangential stresses, and if the outer rim is moving at relativistic speeds this calculation is going to be waaay optimistic... but it will quickly tell you whether any reasonable material can come close to the sort of rotational velocity you need)

 

[olgerm]

To do that you need the stress as a function of the angular speed.

Is that correct $\sigma=\int_{r_1}^{r_2}(\omega^2\ R\ \rho\ dR)=(r_1^2-r_2^2)\ \omega^2\ \rho$?

 

[Nugatory]

Is that correct $\sigma=\int_{r_1}^{r_2}(\omega^2\ R\ \rho\ dR)=(r_1^2-r_2^2)\ \omega^2\ \rho$?

You have two arithmetic errors in the integral you did (missing a factor of 1/2 and the wrong sign), both easily correctable. But the integral still isn’t right because $dA$ is not constant, it’s a function of $r$.

 

[olgerm]

a large rotating object, such as the Earth, is going to be a stronger source of the gravitomagnetic field than a smaller, lab-created object.

how big is $B_{G\ Earth}$and how big is $B_{G\ lab}$?

 

[pervect]

how big is $B_{G\ Earth}$and how big is $B_{G\ lab}$?

It depends on the size of the rotating object in the lab. The argument basically says the bigger the object the better.

I can give you a more detailed answer if you give me figures for $\rho$, $V$, and $R$ for your version of the lab experiment. $\rho$ and V can both be estimated from what material you chose to make your lab mass out of.

The TL;DR (too long, didn't read) answer is that you'd need a lab sphere with a radius of at least 120 km to get a $B_G$ comparable to that of the Earth.

The basic formula is simple, though.

$$B_G = (some-constant) * \frac{G}{c^2} \rho \, V \, R$$

This is slightly modified (and clearer) than the expression in my original post. Here $\rho$ is the density, V = $R \omega$ is the tangential velocity of the sphere, ($\omega$ being the angular velocity of the sphere) and R is the radius of the sphere.

We don't need to know the value of the constant to compare identically shaped rotating sources. Since the Earth is roughly spherical, I am assuming that the lab source is also spherical to make the sources more comparable. You might pick up some small advantage (or potentially a disadvantage) for other shapes for your source, such as your favorite cylinder, but it probably won't be a huge advantage.

Since V for the Earth is .5 km / second, and the best plausible materials would be under 5 km/sec, you might get a factor of up to 10 with the right material. V is limited by the constraint that your rotating test mass doesn't fail due to internal stresses. See the article previously referenced on space tethers for where the requirement for limiting V, the tangential velocity, came from.

Hypersonic skyhook equations use the material's "specific velocity" which is equal to the maximum tangential velocity a spinning hoop can attain without breaking:

$$ V = \sqrt{\frac{\sigma}{\rho}}$$

$\sigma$ is the stress limit of your material, and $\rho$ is it's density. The equation is not for a rotating sphere, but for a rotating hoop, so it's only an estimate.

You probably won't get a density $\rho$ more than 5 times that of the Earth. So, optimistically, assuming you could both get high density and high V in the same material (unlikely), your lab mass would need to be 1/50 the radius of the Earth to have the same gravitomagnetic field. And that's being very optimistic, a better analysis would probably show that the required radius would be much larger.

Since the Earth's radius is about 6000 km, you'd be talking about your lab mass being a sphere 120 km in diameter to get a comparable field. And this is just silly when the Earth's field is free. Also, if you are doing your experiment on the Earth, its field would affect your result anyway, unless you took pains to elimiante it. But there's no sensible reason to try and elimiante the contribution of the Earth gravitomagnetic field rather than to take advantage of it.

 

[olgerm]

@pervect , don't forget that Earth centre is a lot further than that lab objects centre.

 

[Ibix]

He hasn't done. He's just applying the formulae in the Wikipedia article you linked to get that the equatorial field is proportional to $R^4v\rho/r^3$, where $R$ is the radius of the object and $r$ is the distance from its center to your sensor. That's obviously maximised by setting $r=R$, which recovers the formula in @pervect's post.

 

[olgerm]

Is that correct $\sigma=\int_{r_1}^{r_2}(\omega^2\ R\ \rho\ dR)=\frac{1}{2}\ (r_2^2-r_q^2)\ \omega^2\ \rho$?

You have two arithmetic errors in the integral you did (missing a factor of 1/2 and the wrong sign), both easily correctable. But the integral still isn’t right because $dA$ is not constant, it’s a function of $r$.

As I understand it is not correct. Shold I find $\sigma$ on surface of cylinder on in centre of that?

 

[olgerm]

I really think it is important experiment if it is possible. Trying to find out if it is possible. I need estimations on maximum $\omega$ and $\rho$ for that.

 

[Ibix]

Trying to find out if it is possible.

@pervect's calculation is sufficient for that. We can - just - detect $B_G$ from the Earth. $B_G\propto Rv\rho=R^2\omega\rho$ (by both pervect's calculation and yours), so for a detectable $B_G$ you require $R^2\omega\rho \approx R_E^2\omega_E\rho_E$, where the subscripted quantities are for the Earth. For something in a lab, $R\approx R_E/6\times 10^6$ and $\rho\approx\rho_E$. That tells you that you need $\omega\approx 36\times 10^{12}\omega_E$. I make that a 2m diameter solid mass of metal spinning at about 25 billion rpm.

That isn't anywhere near plausible. The rim velocity is in excess of the speed of light by a couple of orders of magnitude. (Which tells you that this semi-Newtonian calculation isn't valid, but the practical problems inherent in accelerating a macroscopic object to speeds where you need to use relativity should be obvious even without getting into the materials science.)

 

[Vanadium 50]

I really think it is important experiment if it is possible. Trying to find out if it is possible. I need estimations on maximum ω\omega and ρ\rho for that.

1. What will this tell you that Gravity Probe B will not?
2. If it's so important, why are you ignoring so many comments?

 

[olgerm]

What will this tell you that Gravity Probe B will not?

Even if it would repeat Gravity Probe B's experiment, it would be big deal.

 

[Vanadium 50]

Why not launch another one then?

 

[olgerm]

I do not understand @pervect's post.

$B_G\propto Rv\rho=R^2\omega\rho$ (by both pervect's calculation and yours), so for a detectable $B_G$ you require $R^2\omega\rho$

Where does this come from? Is it in centre of Earth or at poles?

 

[Ibix]

Where does this come from?

Your formula in your first post, stripped of some subscripts and setting the inner radius to zero, is $B_G=\mu_G\rho\omega r^2/2$. That is, $B_G\propto\rho\omega r^2$. I believe @pervect got his formula from Wikipedia, which gives a different answer to you but only by some numerical factor - so it agrees that $B_G\propto\rho\omega r^2$ and only the constant of proportionality varies with position.

Is it in centre of Earth or at poles?

It will work anywhere at the surface of the Earth, although as noted the constant of proportionality varies. As I noted in post #35 I do not believe that the gravitoelectromagnetic approximation is valid at the center of mass of the generating object, so there is no formula for the gravitomagnetic field at the center of the Earth, or anything else. Your original configuration isn't workable for this reason.

You may be able to get a factor of two or three reduction in your required $\omega$ by adjusting the configuration of your experiment. You need six or seven orders of magnitude reduction. It's not going to happen, I'm afraid.

 

[pervect]

I got my formula from Wikipedia and from another source that analyzed a charge rotating sphere, namely https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.29.pdf, and applying the techniques of dimensional analysis to that formula. Dimensional analysis means looking at how things vary, and not worrying about the numerical factors, but focusing on how scale affects the results. Rather than being abstract, we can equiavalently ask "what happens if we double everything"and/or "what happens if we increase the size of everything by some constant factor N".

Note that according to my source above, the charged rotating sphere is analyzed in Griffiths, but I did not go so far as to look up the textbook reference, but relied on the internet.

Specifically, the questions I was trying to answer is "is it true that bigger objects have a bigger magnetic field"? We could consider a bigger object with the same value of rotational speed $\omega$, but I found it convenient to analyze instead a bigger object where $\omega r$, the tangential velocity, stayed constant.

The answer to the question is "yes". And because the answer to the question is yes, we can say that since the Earth is a big object, it has a stronger gravitomagnetic field than a lab-scale object, because the Earth is a lot bigger than anything we can build in a lab.

While I used Wiki and the other source I mentioned, and while the results are the same as the OP's results, I have come to realize that the dependency on scale follows more generally from a dimensional analysis of the Biot-Savart law.

Hyperphysics, http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/Biosav.html, quotes this law as:

$$d\vec{B} =\frac{ \mu_0 I}{4 \pi r^2} d\vec{L} \times \hat{r}$$

I have taken the liberty of replacing the unit vector pointing in the direction from the current element to the point where the magentic field B is being measured with a different symbol, $\hat{r}$, rather than the symbol that the hyperphysics used.

When we hold the tangential velocity and the charge density constant, we hold the current density $j$, the current/square meter, constant for a rotating object. This is because $j = \rho v$, and we hold $\rho$ and v constant.

It's possible to analyze this in terms of holding $\omega$ constant, rather than v constant, and get equivalent results, but I've chosen to do it this way I am writing it now.

The cross sectional area of the rotating object will increase by a factor of 4 if we double it's size. This implies that the current quadruples, because $I = j r^2$. However, the distance from the center of the object to the point at which the magnetic field is being measured, r, also doubles. So for a doubling of size, $\frac{ \mu_0 I}{4 \pi r^2}$ does not vary when we hold the tangential velocity v constant, i.e. when we hold $\omega r$ constant.

This leaves us with the term $d\vec{L} \times \hat{r}$ which increases linearly with the scale factor. This happens because $\hat{r}$ is defined as a unit vector, so the only factor remaining is the length of the current element, which doubles. Thus doubling the size of the objects doubles the magnetic field when we hold $\omega r$ constant.

 

[Vanadium 50]

namely https://physicspages.com/pdf/Griffiths%20EM/Griffiths%20Problems%2005.29.pdf, and applying the techniques of dimensional analysis to that formula.

Hmmmm...

I don't think this derivation is that of a rotating charged sphere external to the sphere. At large r, the dipole term is not leading (!) because the r and θ terms are becoming infinite. Also, the energy in the field is infinite too. Following their derivation - and I had some difficulty here - it appears that at no point are they taking an integral past the surface.

Deriving it myself, I get the following (EM case, units where c=1):

$$\vec{B} = qR_0^2\omega \frac{3(\hat{\omega} \cdot \hat{r})\hat{r} - \hat{\omega}}{3r^3}$$

As I said earlier, I expect the force to go as β². The gravitational tensor "charge" is like a mass, so I want something in dimensions of mass (or energy in these units), e.g. L²/2I, which has one power of m and 2 powers of velocity. Hence my β₁ x β₂.

In this case, we get one power of velocity in the derivation of B, and one in the Lorentz force.

 

[Vanadium 50]

Let me re-order this slightly for clarity:

$$\vec{B} = \frac{qR_0^2\omega}{r^3} \left( \frac{3(\hat{\omega} \cdot \hat{r})\hat{r} - \hat{\omega}}{3} \right)$$

Now the right term in big parentheses is purely angular.

 

[pervect]

Are $\hat{r}$ and $\hat{\omega}$ both unit vectors?

 

[Vanadium 50]

Yes.

 

[olgerm]

so it agrees that $B_G\propto\rho\omega r^2$ and only the constant of proportionality varies with position.

maybe the constant of proportionality for Earth (ball) is a lot smaller than for cylinder and cylinders gravitomagnetic field is still detectable.

 

[PeterDonis]

maybe the constant of proportionality for Earth (ball) is a lot smaller than for cylinder and cylinders gravitomagnetic field is still detectable.

This is personal speculation (with no basis whatever in GR) and is off limits here.

This thread has run its course. Thread closed.