# Gyroscopes curving space by their energy

1. Dec 22, 2015

### Doug Brown

many properties of gyroscopic motion are non-intuitive because I think the standard physical features are not correctly identified by observers. Some who know little quickly claim that this or that law of physics is violated and some new discovery is afoot. I tend to think that all laws of physics are being obeyed but the features are not properly identified by most observers who are too quick to claim knowledge of the underlying dynamics

While I know from certain posts I have seen on this GR forum that I have lot to learn about GR and curvature of space and that mass/energy causes curvature of space from the non-vanishing of the energy-stress tensor alone, wouldn't it be the case that a spinning disc or gyroscope or flywheel is concentrating energy in a small space in such a way as to curve space? Wouldn't this explain precession motion as resulting from the center of mass of objects following curved space lines made curved by the increase in energy density local to the gyroscope? So the spinning top falls due to gravity not directly toward the center of the earth but rather in a helical path around the axis pointing to the center of the earth, and this helical path is the result of the energy concentration which is in the energy-stress tensor curving the space in the first place?

Therefore, can or has a full mathematical GR analysis of gyroscopes been done to explain behaviours and derive behaviours from gyroscopic motion and combinations of gyroscopes in relative motion?

2. Dec 22, 2015

### Staff: Mentor

It's possible in principle, sure. And a spinning disc or gyroscope that concentrates enough energy in a small enough space will have a stress-energy tensor that causes the appropriate spacetime curvature. (It's spacetime curvature, btw, not space curvature.)

In practice, however, no ordinary spinning disc or gyroscope has anywhere near enough stress-energy to cause measurable spacetime curvature. So this can't be the explanation of precession.

You don't need GR to analyze an ordinary gyroscope; precession of a gyroscope can be analyzed just fine using Newtonian gravity. (Of course "Newtonian gravity" here is really the Newtonian approximation to GR, but the point is that there are no uniquely GR effects involved.)

3. Dec 23, 2015

### Staff: Mentor

I doubt that we have any materials which could handle the stresses involved in rotating so fast that relativistic effects became important.

4. Dec 23, 2015

### Doug Brown

thank you for clarifications

Now, given the role normalized velocity plays in the Lorentz factor $\gamma = {1\over \sqrt{ 1 - \beta^2 }}$ where $\beta = v/c$ is the light-speed normalized velocity making clear when relativistic effects are important in Special Relativity, what is this equivalent factor for General Relativity?

Now, given this equivalent factor for GR, which I assume is some value of Energy density, what value must the energy density of our toy gyroscope (just a spinning disc) be greater than in order for curvature of space to be achieved at some macroscopic level noticable to any observer in our laboratory?

In gyroscopic physics discussions it is often stated that "there is a lot of energy stored in this particular gyro right now". Well can we put together this calculation here in this thread and see how big must the disc be, how fast must its rotation be in order for an observably macroscopic curvature of spacetime to be detected in the laboratory even if that laboratory has to be much bigger than CERN's LHC.

5. Dec 23, 2015

### Staff: Mentor

That means "a lot of energy for a mechanical device". It's not enough for relativistic effects to even be a rounding error.
For the rotational energy of the disk to produce detectable curvature effects, it must be order-of-magnitude comparable to the rest energy of the disk. That rest energy is given by the famous $E=mc^2$ where $m$ is the rest mass of the disk; that mass must be large enough to be detectable by an Eotvos-style experiment. We're talking about something like a disk with a radius of fifteen centimeters turning at a few billion revolutions per minute and storing as much energy as several hundred multi-megaton nuclear weapons.

Last edited: Dec 24, 2015
6. Dec 24, 2015

### Doug Brown

ok great, thank you very much for nailing down those magnitudes. Now I'd like to try to use the provided orders of magnitude in some mathematics and then plot some relationships so your 15 cm is just one point on a curve in a plot, and your few billion rpm is one point on a curve. Does PhysicsForums (PF) allow insertion of figures & plots like it allows insertions of latex equations? if someone knows how to insert a graph of some sort, I'd like to write down some equations now and then add some plots based on some of the equations. Any PF graphics insertion help would be super. Thanks

Formalism to deduce laboratory values for parameters

Given a disc of uniform mass m, radius r (any thickness) spinning an angle $\alpha$ in $s$ seconds at angular velocity $\omega_{\alpha} = {\alpha \over s}$ its moment of inertia around its center axis is $I = {1\over 2} m r^2$ giving it an angular momentum of $L=I \omega_{\alpha}={1\over 2} { \omega_{\alpha} m r^2}$.

If this spinning disc is now placed at the end of a rod of length $x$ (along an $x$ axis) fixed at the origin of a coordinate system, a torque $\tau = x \times F$ is created by the downward force of gravity $F=mg$. In this coordinate system $\theta=0$ defines the $z$ axis, while $\theta=\pi/2$ defines the azimuthal $x-y$ plane where $\phi=0$ defines the $x$ axis.

The disc's angular momentum $L$, gravity $g$ and the constraint that the disc is fixed a length $x$ from an origin then causes a precession about the polar axis with an azimuthal angular velocity $\omega_{\phi} = {\tau \over L}= 2 {x g \over { \omega_{\alpha} r^2 }}$ assuming $\omega_{\alpha} \gg \omega_{\phi}$.

The rotational energy of this system is then $E_R= E_{\alpha} + E_{\phi} = {1\over 2} I ({\omega_{\alpha}}^2 + {\omega_{\phi}}^2)$. Since $\omega=2\pi f$ where $f$ is the frequency in rpm the rotational energy becomes $E_R=\pi^2 m r^2 (f_{\alpha}^2 + f_{\phi}^2)$.

So now if the rotational energy $E_R$ is chosen to rival its rest mass energy $E_0=mc^2$ then $E_R=E_0$. This equality is $\pi^2 m r^2 (f_{\alpha}^2 + f_{\phi}^2) = mc^2$.

But now it looks like $m$ drops out and we are left with some relation from which we can solve for $r$ as some function $h(f_{\alpha},f_{\phi})$ or either of the $f$'s as some function $j(r)$ but that seems like barking up the wrong tree.

Rather a total energy $E_T=E_R+E_0$ seems to be more relevant and this is what must be set to your value of "as several hundred multi-megaton nuclear weapons".

In any event, if the $f$s are in hertz = cycles per $s$, $r$ in $cm$, $c$ in $cm/s$ do these units come out right? what can be done with this equality? what's the next step in the thought experiment of deriving actual laboratory values for $m$, $r$, $x$, $f_\alpha$ and $f_\phi$ to force this gyroscope system to curve spacetime explicitly?

Last edited: Dec 24, 2015
7. Dec 24, 2015

### Staff: Mentor

Where does the gravity come from? Are you trying to figure out the precession of the gyroscope, or how much it curves spacetime? An external source of gravity is irrelevant for the latter question, and how much the gyroscope curves spacetime is irrelevant for the former question. Even if the gyroscope does contain enough energy to produce non-negligible spacetime curvature, that curvature won't make the gyroscope precess. Precession is caused by some externally applied torque; there's no way a gyroscope can apply a torque to itself.

The $f_{\phi}$ term is irrelevant here, for the reasons given above. Also, it looks like you've made a slight mistake in deriving the $f$ version. If $I = \frac{1}{2} m r^2$ and $E_R = \frac{1}{2} I \omega_{\alpha}^2 = \frac{1}{4} m r^2 \omega_\alpha^2$ and $\omega_\alpha = 2 \pi f_\alpha$, then I get $E_R = \frac{1}{4} m r^2 4 \pi^2 f_\alpha^2 = \pi^2 m r^2 f_\alpha^2$.

$x$ is irrelevant, for the reasons given above. So is any "rotational energy" from precession (what you are calling $\omega_{\phi}$); "energy" due to the motion of the gyroscope (or any other object) relative to something else doesn't contribute to the spacetime curvature produced by the object.

The spacetime curvature produced by the gyroscope is due to its stress-energy tensor, as I said in post #2 above. Nothing else contributes. So you need to figure out the stress-energy tensor of the gyroscope. I don't have an explicit expression handy for this, but the rest energy $m c^2$ and the rotational energy $\frac{1}{2} I \omega_{\alpha}^2$ will definitely both contribute to it. Other contributions will be the angular momentum and the pressure and stresses within the gyroscope, caused by its rotation.

Also, note that the rule of thumb Nugatory gave (rotational energy same order of magnitude as rest energy) was the condition for the rotational energy of the gyroscope to cause non-negligible spacetime curvature compared to the rest energy. But this assumes that the rest energy itself is large enough to cause non-negligible spacetime curvature. For an ordinary-sized gyroscope, you would need extremely sensitive measurements to detect the spacetime curvature produced by its rest energy $m c^2$ (for example, you could do a Cavendish-style experiment with torsion balances to measure the gravitational attraction of the gyroscope). But even for an ordinary-sized gyroscope, say 10 kg, the rest energy $m c^2$ is enormous: about $10^{18}$ Joules for a 10 kg gyroscope, or about 250 megatons of TNT. In other words, that enormous amount of rest energy produces a very tiny amount of spacetime curvature, that requires extremely sensitive measurements to detect.

Further, the rotational energy can't actually be equal to the rest energy, because that would make the outer edge of the gyroscope move faster than light. The linear velocity of the gyroscope's outer edge is $\omega_\alpha r$, and that must be less than $c$, so we must have $\omega_\alpha < c / r$. That sets a limit on the rotational energy of $E_R = \frac{1}{4} m r^2 \omega_\alpha^2 < \frac{1}{4} m c^2$, or 1/4 of the rest energy. So for the above 10 kg gyroscope, the maximum rotational energy would be about $2.5 \times 10^{17}$ Joules, or about 62 megatons of TNT. (Note that this fraction is independent of the radius of the gyroscope--if we increase the radius, we decrease the maximum allowed angular velocity in the same proportion.)

Last edited: Dec 24, 2015
8. Dec 24, 2015

### Doug Brown

Thank you for additional comments. Still digesting your comments but I can see at outset some of my context is not clear to you. I am thinking of a real disc in my hands in my lab and the Formalism I wrote is for a table top gyroscope set up where I have a) a spinning disc, and b) it is has a rod through its center one end of which is suspended by a string from the ceiling or it could be in a frame with acting gimbals.

Now, if I want to consider $only$ the disc alone I can take $x=0$ (which implies $f_\phi=0$) but if I take $x\ne0$ then there will be gravity because we are in a real lab, there will be torque by gravity and $that$ will cause precession (again here on earth with my hands nearby). This is the system I want to consider and make calculations about which is why I set up that formalism to make concrete all the values we are now going to be getting explicit about.

So while it might not serve other aspects of the calculation you are thinking about, I need those to be included in the system I am calculating about because this is related to a device I can actually build which helps me think about it. Now, I will go back to your comments and continue digesting them. Thank you for your detailed specific input!

9. Dec 24, 2015

### Staff: Mentor

I just assumed that $E_R$ and $E_0$ would both be on the order of a kilogram or so to get that result. That's getting down towards the lower limit of detectable gravitational effects.

There are enough practical problems here that any calculation you do is only going to be accurate within an order of magnitude or so. For example, no real energy-storing gyro is a uniform disk; the classical formulas for moment of inertia and rotational kinetic energy cannot be trusted at relativistic speeds (although for values of $\gamma=1/\sqrt{1-v^2/c^2}$ not too far from one the error will be not be too far from the value of $\gamma$); and probably some other stuff that I haven't though about.

Because it is completely impossible to build such a thing (just for grins, you might calculate the centrifugal forces at the rim of a 15cm radiius object rotating at a few billions of RPMs; and do remember that eventually all that energy is going to be released explosively unless you can find a way of safely dissipating the equivalent of a medium-large nuclear war) order-of-magnitude calculations around some reasonably chosen parameters will deliver most of the physical insight here.

10. Dec 24, 2015

### Doug Brown

The gravity comes from Earth gravity since my discs are in my lab and I want to hang the disc from a string from the ceiling of the lab. Therefore, this will be considered to be in the already curved spacetime of my lab space, but still I think we can still talk about the curvature coming from the spinning disc and I want to calculate the actual values for all the parameters to make this curvature non-negligible

11. Dec 24, 2015

### Doug Brown

This comment seems to be more to my initial query. I have been assuming that if an enormous amount of energy were stored in the spinning disc then that energy would indeed be capable of curving spacetime. If that is not the case at all, then I need to learn more here on this very point.

12. Dec 24, 2015

### Doug Brown

oh yes you're right, I see my error of not including the 2π in the square, thank you for the correction!

13. Dec 24, 2015

### Doug Brown

Ok I see your point. I will now build $T^{\mu\nu}$ for the gryoscope element by element (this will take some time, and perhaps not soon, but I do want to come back to this and fill out this matrix in full -- and will do so in this thread in LaTeX). I still want to add plots of the various quantities still also so if anyone knows how to insert plots into threads on PF please inform.

14. Dec 24, 2015

### Doug Brown

while $E_0$ is a constant, $E_R$ can take on any values we tell it to take on, so it does not have to be assumed to be any particular value like a kilogram or so. In fact I want to set it to whatever the minimum value it must be in order to curve spacetime, and I hope the rest of this analysis assigning specific functional forms to the tensor elements of $T^{\mu\nu}$ will take us there. We will have all of $m$, $r$, $f_\alpha$ as knobs to set getting us there and if $x\ne0$ then $f_\phi$ as well.

Your guidance is very helpful and crucial, thank you very much for contributing it.

Last edited: Dec 24, 2015
15. Dec 24, 2015

### Doug Brown

I expect after we get further in this analysis we will certainly be calculating what any Lorentz factor value we end up with. If we want to keep it near 1 then by all means we will try. As stated in post #13 I will be building out all the elements of $T^{\mu\nu}$ inserting anywhere I can the various functional quantities I have described in the formalism above and hopefully some of those will be in the tensor elements of $T^{\mu\nu}$

Last edited: Dec 24, 2015
16. Dec 24, 2015

### Doug Brown

Thank you for these very crucial contributions. This nails exactly the point Nugatory made about the spinning disc not satisfying all the non-relativistic relations assumed and provides conversions to some of the relevant units I need. This is the heart of the matter. Need time for further digesting :)

17. Dec 24, 2015

### Staff: Mentor

Yes, and in this calculation, the fact that the gyroscope is hanging from a string is irrelevant. All you need is the gyroscope's rest mass, radius, and rate of rotation. (Actually, in general, you would need its energy density as a function of position, and you would not be able to assume that it was a perfect sphere, but we can assume that its density is constant and that it is perfectly spherical as a reasonable idealized case.)

As the numbers I gave show, an "enormous amount of energy" by everyday standards can still be negligible (or almost so) as far as producing spacetime curvature is concerned.

18. Dec 24, 2015

### Staff: Mentor

Any non-zero values will curve spacetime to some extent so there is no minimum - but it sounds as if you're looking for an effect that is large enough to detect. For that, you need kilograms (as opposed to, for example, grams or micrograms), so you might as well assume a kilogram-sized mass as you're considering the general scale of the mechanical system. If you decide later that you want the relativistic effects to be five times as large, you can dial the mass up by a factor of five.

19. Dec 24, 2015

### Staff: Mentor

As the discussion progresses, it sounds more as if you are looking for a general understanding of the stress energy tensor of a rotating body and how it affects the curvature of spacetime around it. For that, you can look at the relativistic gyroscopes that nature has already helpfully scattered throughout the universe - rotating collapsed stars. Google for "Kerr metric" to find the details of the solution and see how the various physical properties of the rotating body contribute.