I Can We Practically Measure the Gravitomagnetic Effect with Spinning Cylinders?

[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.