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B Gravity and a spinning object

  1. Jul 20, 2017 #1
    This is a just for fun question. I saw the movie "Contact" and they built a machine that generated gravity by spinning rapidly. I assume this comes form general relativity. Is this correct? My question then is this. Is there a simple formula that can be used in which something moving in an orbit / accelerating can have its mass increase as a result of increase in acceleration. I would love to have that equation and put it into my graphing calculator. If you could show example that would be great.
     
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  3. Jul 20, 2017 #2

    PeterDonis

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    I think you are referring to generating a wormhole by spinning rapidly, correct?

    As far as the physics involved in that scene in "Contact", Kip Thorne in Black Holes and Time Warps describes how Sagan consulted him about that, and Thorne suggested using a wormhole, which would have to be held open by some form of exotic matter (i.e., matter that violates technical conditions called "energy conditions" that all ordinary matter obeys--it is not known whether such exotic matter can actually exist). As I understand it, the machine portrayed in the movie was supposed to be a creating the wormhole, but I don't know that the spinning part was based on any actual physics; certainly the wormhole solution that Thorne describes is not spinning and is not produced by spinning anything rapidly.
     
  4. Jul 20, 2017 #3

    Ibix

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    The book suggests that The Machine creates a target for an alien wormhole generator to lock on to, which is why it never produces a wormhole again. We never see an actual wormhole generator.

    As Peter says, it's just science fiction.
     
  5. Jul 20, 2017 #4
    This spinning thing seems to have been around in SciFi for a number of years, the engine rooms of many of the ships contain spinning things. I suspect that the idea goes back to old discussions about the contraction of the rim of a spinning disc in SR and some people concluded from this that the spinning and the resultant contraction meant that spacetime had to be curved by this apparatus.

    Cheers
     
  6. Jul 27, 2017 #5
    @cosmik debris

    Yes, you hit on it, except for its not Special Relativity its General Relativity. That is what gave Einstein the idea for General Relativity. I read that in a book God's Equation. A spinning disc causes the curvature of space time. Apparently this is not a well known or accepted reason for the cause of gravity. Has anyone ever weighed a spinning sphere or disk and see if it gets heavier?
     
  7. Jul 27, 2017 #6

    Nugatory

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    The spinning disk problem that @cosmik debris mentions most certainly does not involve general relativity. Google for "Ehrenfest disk" for more information.
    As with all popularizations, you have to be careful about taking it too seriously. In this case, it seems to have rather seriously misled you.
    I doubt that anyone has ever actually performed this measurement because the effect would be small and the practical difficulties in performing the measurements to the required level of accuracy would be enormous. However, there is no doubt whatsoever that a rotating disk will weight very slightly more than the same disk when not rotating (it would be good exercise to calculate approximately how much - the classical expression for kinetic energy and ##E=mc^2## will suffice), as many related experiments have been done and yield results that match the predictions of special relativity. This effect has nothing to do with the rotation causing curvature of spacetime - you could get the same effect by adding energy to the disk by heating it.
     
  8. Jul 27, 2017 #7
    @Nugatory You made my day. Thanks for the info. I was hoping for a good strong statement about the subject and an idea about how physicists nowadays view this concept of rotation as it relates to mass. So are you saying that the rotating disk weighs more simply because we have added energy to the system via energy of increased motion? More energy, equals more mass because mass and energy are basically equivalent. I want to make sure I understand what you are saying Thanks.
     
  9. Jul 27, 2017 #8

    Nugatory

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    That's pretty much it.
     
  10. Jul 27, 2017 #9
    Excellent Thanks...
     
  11. Jul 27, 2017 #10
    It has at least more mass due to the additional energy. The resulting weight is not as easy to preditc.
     
  12. Jul 27, 2017 #11
    So what is the formula for predicting the extra weight or mass of a spinning disk?
     
  13. Jul 27, 2017 #12

    Nugatory

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    I gave you a very big hint in #6, so big that you're not going to get another one until you've at least thought about it on your own for a day or so.
     
  14. Jul 27, 2017 #13
    In #6 you told him how to estimate the mass but you excluded the additional effect on the weight.
     
  15. Jul 27, 2017 #14
    @Nugatory E=mc^2 is not the right formula in my humble opinion. I assume that is your hint. E=mc^2 does not account for acceleration factor. No gamma factor either. A spinning disk is a motion of acceleration this fact should not be ignored. A better hint would be F=ma. or Lorentz formula for the increase in mass. Although Lorentz formula does not account for acceleration. I was just hoping there existed another formula besides Newton's formula and Einstein's formula.
     
  16. Jul 27, 2017 #15
    For the mass it is. You just need to know the energy to use it. With the classical estimation (as suggested by Nugatory) it is quite easy but it might become tricky in relativity.
     
  17. Jul 27, 2017 #16

    PAllen

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    Why, unless you get beyond the domain of weak field GR? As long as the disk is in equilibrium (thus producing a stationary metric) and its self gravity is not significant, the theory of Komar mass would mandate that the invariant mass is the weight for the disk as a test body in some overall filed.
     
  18. Jul 28, 2017 #17
    Because

    [itex]E_{rot} = {\textstyle{1 \over 2}}J \cdot \omega ^2[/itex]

    doesn't work in relativity. You need to assume the disc to be sufficiently (but not infinite) rigid and than to integrate the relativistic kinetic energy.
     
  19. Jul 28, 2017 #18
    @DrStupid Hey that formula looks real interesting. Lets say we have a disk that has a 1 meter diameter and it weighs a 1 kilogram and it is spinning at 10000 rpm.
    How much heavier would it be?
    Can you show me how to use your formula above so I can put it in my TI Calculator. I would really love to see this worked out don't skip any steps. what is the w^2 represent? What is the E unit and the J is that Joules? I am just off on summer vacation and am learning how to put equations into my TI calculator.
     
  20. Jul 28, 2017 #19
    J is the moment of inertia. In case of a disc it is

    [itex]J_{disc} = {\textstyle{1 \over 2}}m \cdot r^2[/itex]

    And ##\omega## is the angular velocity. It is proportional to the frequency f according to

    [itex]\omega = 2 \cdot \pi \cdot f[/itex]

    With

    [itex]m = 1\;kg[/itex]
    [itex]r = 1\;m[/itex]
    [itex]\omega = 2 \cdot \pi \cdot 10000\;\min ^{ - 1} = 2 \cdot \pi \cdot \frac{{10000}}{{60}}\;s^{ - 1}[/itex]

    you get

    [itex]E_{kin} = {\rm 274156}\;{\rm J}[/itex]

    and therefore

    [itex]\Delta m = {\rm 3}{\rm .05} \cdot {\rm 10}^{{\rm - 12}} \;kg[/itex]

    That's quite hard to measure.
     
  21. Jul 28, 2017 #20

    jbriggs444

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    That "w" is actually the greek letter omega ("##\omega##"). It is the angular rotation rate, typically in radians per second. There are 2pi radians in a circle.

    The E is energy and would typically be measured in Joules. (*)

    The J is the "moment of inertia" of the spinning disk. (I am more used to seeing it as "I" rather than "J"). The moment of inertia is computed based on the mass of the spinning disk and its radius. If you are familiar with integral calculus, it is the integral over the disk of an incremental mass element multiplied by the square of the distance of that mass element from the center of rotation. If you are not familiar with integral calculus, you can look up a formula.

    For a disk of uniform density, mass m and radius r spinning around its center, the moment of inertia is ##\frac{1}{2}mr^2## The mass would typically be measured in kilograms and the radius in meters. The moment of inertia would then be in units of kilogram meter2.

    That should be enough information so that you can first calculate the moment of inertia of a 1 kg disk 1 meter in diameter.
    You should also be able to convert 10000 rpm to radians per second.
    Finally, you should be able to use that information to compute the energy of the spinning disk.

    Here on these forums, we do not believe in solving problems for you. We believe in helping you solve them for yourself.

    (*) The basic formula works with any coherent system of units. Centimeters, grams, seconds and ergs works just as well as meters, kilograms, seconds and Joules. Feet, slugs, seconds and foot-pounds works too.

    With a system of units where the unit of energy is not given by unit mass times unit distance squared per unit time, the formula is almost correct, but needs a constant added to account for the unit choice.
     
    Last edited: Jul 28, 2017
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