# Gravitomagnetic experiment

• I
Gold Member
@pervect , dont 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.

Gold Member
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?

Gold Member
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.)

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etotheipi
Staff Emeritus
2019 Award
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?

Gold Member
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.

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Staff Emeritus
2019 Award
Why not launch another one then?

Gold Member
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.

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pervect
Staff Emeritus
I got my formula from Wikipedia and from another source that analyzed a charge rotating sphere, namely https://physicspages.com/pdf/Griffiths EM/Griffiths Problems 05.29.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.

Staff Emeritus
2019 Award
namely https://physicspages.com/pdf/Griffiths EM/Griffiths Problems 05.29.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.

Staff Emeritus
2019 Award
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.

vanhees71
pervect
Staff Emeritus
Are ##\hat{r}## and ##\hat{\omega}## both unit vectors?

Staff Emeritus
2019 Award
Yes.

Gold Member
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
Mentor
2019 Award
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.