Massive objects moving too fast

  1. I've been puzzling over this for quite sometime and have not been able to find it anywhere here which is stunning. What if you had a structure that was moving fast enough to have it's relativistic mass be...1 kg for now. And we had this 1 kg mass zipped by something that was 0.5 kg for sake of argument. What would be the "grativistic(sp?)" forces at work here? Should I try and use Newton universal law to figure it out? or would the speed negate any other forces because it is just simply moving too fast. Here's another thought: Is gravity affected by time or is time affected by gravity?

    Thank you for reading and if it has already been said please just post a link or a few keywords I can search for. :)

  2. jcsd
  3. HallsofIvy

    HallsofIvy 41,051
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    No, that gravitational force, at a specific instant, would be calculated, in special relativity, by using Newton's gravitational formula with the distance between the objects and their masses at that instant.

    I specified "in special relativity" because, of course, general relativity has a completely different way of computing gravity.
  4. Mentz114

    Mentz114 4,510
    Gold Member

    I'm not clear on your scenario, but if it involves gravitational interaction between masses which are in relative motion, there is a newer theory than Newton's gravity, called general relativity which deals with this.

    Clocks are affected by gravity. Obviously the gravitational field at a certain place may vary if masses are moving nearby.
  5. tiny-tim

    tiny-tim 26,016
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    is this the question?

    I think what you're asking is:

    Gravitational force depends on mass and distance.

    Is that mass the ordinary mass (rest-mass) m, or, for something moving at a very fast speed v, is it the increased mass [tex]\frac{m}{\sqrt{1\,-\,v^2/c^2}}[/tex]?

    Erm … I don't actually know the answer to that! :redface:
  6. Well what I was referring to was this. In order to calculate a relitivistic mass you need to take into account its energy. Thus the fast you go the more mass you have. So what if this high speed object with its new mass passes a stationary object half its weight? What happens? Is the second object attracted to the first? I hope that helps
  7. I'm going to take a guess here. From my understanding, at relativistic speeds, an object doesn't gain mass, per se. By that I mean it doesn't gain particles or anything. Gaining mass at high speeds is a bad interpretation because it leads to weird conclusions when you talk about a mass traveling at, say, a 30 degree angle between the x and y axis. Does it gain more mass in the x direction than the y since the velocity component is higher in that direction? As you can see, different masses in different directions of the same object doesn't make much sense. But, if you interpret mass as a quantity that tells you the resistance to change in its trajectory, I think it makes more sense. Then, in my analogy about traveling at 30 degrees, its conclusion would be that it's harder to accelerate the mass in the x direction than in the y direction. So the big point is, if something is going at a relativistic speed in a particular direction, it harder to make that object go any faster in that direction than if it were going at a lower speed. In your question, I assume you're talking about some big object passing beside another big object at a high speed. I think you should break things down into vector components. Let's say object B is at the origin in the x-y plane, and object A is traveling in the y-direction and passed the x-axis at some distance "d" from the origin. Since object A is going beside object B, and the gravitational force is one of attraction, object B wouldn't cause A to go any faster in the y-direction--it's pulling it towards the origin. So the effect would be that object A would gain a velocity component in the -x direction, toward object B. Since object A is not traveling at relativistic speeds in this direction, it's normal mass would be used. If you want the situation of the gravitational pull being in the same direction as the relativistic velocity, you need General Relativity. Special Relativity does not apply to non-inertial frames. Hope this helps.
  8. tiny-tim

    tiny-tim 26,016
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    increased mass is the same for each direction

    Sorry, but this is just wrong. Momentum in each direction is rest-mass times velocity-in-that-direction times the same factor, [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], where v is the total velocity.

    The increased mass, rest-mass times [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], is the same for all directions.

    Yes, but the question is whether B's normal mass would be used. I believe the answer is no, because of the equality of inertial mass and gravitational mass, but that's general relativity, so I'd rather let someone else comment on it.
  9. You are trying to unify Special Relativity with newtonian gravitation, I guess.
    At the moment I have no idea about how doing that, but on one point I am quite sure. Newton's gravity has as framework the Euclidean space and three-vectors, while Special Relativity resides in the space-time (Minkowski space) and uses four-vectors.
    So, Newton's law cannot be used as it stands, but needs to be generalized to the four-dimensional case, as one usually does for Electromagnetism, introducing the four-potential A^{\mu}.
    However, the generalization of Newton's law to Minkowski space (if it is possible) has not to be trivial!
    If we assume it possible, then one uses four-momenta and four-forces and should be able to deal properly with newtonian gravity.

    I think I have been right!
  10. tiny-tim

    tiny-tim 26,016
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    No. All Newtonian laws can be converted into four-dimensional laws. :smile:

    You just have to be careful!

    Usually, it just involves multiplying the mass by [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex].

    In electromagnetism, for example, E and B are parts of a six-component tensor in four-dimensional space, and would be different for an observer with a different velocity, but for any particular observer the usual three-dimensional rules for E and B apply. :smile:
  11. You will be much better off applying the accepted standard of only using rest mass.
    Just ignore any impulse to apply the gamma factor to mass at all to satisfy any apparent effect you think you see from your point of view.

    Energy is defined by applying gamma to momentum no more than that is needed.
    No need to assume that the momentum value change is the result of a change in any actual mass inside the equation for momentum. Physicist today find it makes much better sense to view the momentum itself as being altered relativisticly by time and distance changes defined in the gamma factor. Those effects are what is important not the idea that it “looks as if the mass changed”. That is all that has happened, the relativistic effects are real. Not what only seems like a change in mass from our point of view.
  12. tiny-tim

    tiny-tim 26,016
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    We apply the "gamma factor", [tex]\frac{1}{\sqrt{1\,-\,v^2/c^2}}[/tex], because the Lorentz equations tell us to. That then tells us what effects will be apparent, not the other way round!

    What? :confused:

    I don't really understand what you're saying here.

    I'm quite happy to accept that there's more than one way of looking at the same mathematics, and which you choose is largely a matter of taste.

    My taste is to say that the mass increases. That's perfectly consistent with the mathematics.

    If you want to state clearly what your taste is, I'll probably be happy to agree with you that that's fine too. :smile:

    I'll just add that if you have two identical boxes containing identical flywheels, one flywheel rotating and one not, and you can't see inside the box, then you'll find that the one with the rotating flywheel is more difficult to push, and you have little alternative but to say that that box has increased mass! :smile:
  13. Electromagnetism is a very special case! In fact Einstein conceived Special Relativity starting from it (in his pioneer work "Zur Elektrodynamik bewegter Körper", "on the electrodynamics among bodies").
    He also said that Special Relativity was already contained in Maxwell equations, for which Galileo transformations give problems.
    The same problems are absent for newtonian gravity, so one has to investigate some sort of "gravitodynamics among bodies"! :-)
    Namely, which are the relativistic effects in newtonian gravity? And how do you build the gravitational four-potential?

    Anyway, in my opinion the rest mass, being a scalar, should be used as the source of the gravitational force. The gamma factor would enter through Lorentz transformations!
  14. tiny-tim

    tiny-tim 26,016
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    So far as I know, all forces are affected in the same way by the Lorentz transformation.

    No, they're there, but gravity is many many orders of magnitude weaker than electromagnetism (think how fast an electron can be accelerated form rest across a cathode ray tune, and how long gravity would take to pull it the same distance!), so nobody noticed them.
  15. Wow thanks for the response. I had no idea that this topic was so unclear. I figured that I was just missing something. I still don't know what to think but it has been a great thought provoking experience non the less!
  16. DrGreg

    DrGreg 1,986
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    There's a big problem trying to reconcile Newtonian gravity with special relativity. In Newtonian gravity, two masses exert equal and opposite forces on each other simultaneously. In relativity, there's no such thing as "simultaneous", so if the masses are moving relative to each other, you are in trouble.

    I stand to be corrected, but I believe you have no choice but to use general relativity. I'm no expert, but I recall other answers to similar questions saying that, in GR, gravity relates to energy-momentum rather than to (rest) mass. So even photons have some (tiny) gravitational influence.
  17. ...I was afraid someone was going to say that. Unfortunately I couldn't do Tensor Calculus to save my life. As a matter of fact I'm barely hanging on with regular calculus. So I guess I'll just have to take your guys' word for it. Thank you for the prompt responses! :)
  18. DrGreg

    DrGreg 1,986
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    I agree.

    (...and mine is not!)

    I would suggest to everyone that when we talk about mass, we should make it clear whether we mean "relativistic mass" (proportional to energy, including kinetic energy) or "rest mass" (="invariant mass"). The modern fashion is for "mass" to mean "invariant mass". In these forums, many readers are struggling to work out what mass is, so we should avoid ambiguity.

    Yes, it has increased invariant mass. A rotating flywheel has more invariant mass than an identical non-rotating wheel (one reason why "rest mass" isn't such a good name).

    For the benefit of anyone who doesn't know why, the invariant mass mtot of a collection of particles can be defined by

    [tex](\Sigma E)^2 = m_{tot}^2 c^4 + ||\Sigma \textbf{p}||^2 c^2 [/tex].

    which reduces to

    [tex]\Sigma E = m_{tot} c^2 [/tex]

    in the box's frame, where the total momentum is zero. For each individual particle of invariant mass m,

    [tex]E^2 = m^2 c^4 + ||\textbf{p}||^2 c^2[/tex], so [tex]E > m c^2[/tex], and [tex]m_{tot} > \Sigma m[/tex].
  19. jtbell

    Staff: Mentor

  20. tiny-tim

    tiny-tim 26,016
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    invariant mass

    As you say, invariant mass = rest mass.

    And you agree something can have "increased invariant mass".

    Which is the same as "increased rest mass"! :smile:

    Which is why my taste is to regard mass as increased in all situations!

    (And don't forget, there's nothing unusual about rotational motion contributing to invariant/rest mass: the electrons in you and me have very fast rotational motion which does contribute to increasing our "rest mass"! :smile:)

    No. If a wheel is rotating so that its rim is moving with speed U relative to the centre of the wheel, and if the whole wheel is moving parallel to its axis with speed V relative to us, then the Lorentz-increased mass (energy) of the rim is the rest-mass multiplied by:
    [tex]\Large\frac{1}{\sqrt{1 - U^2}\sqrt{1 - V^2}}[/tex]​

    Even in the box's frame, the Lorentz-increased mass (energy) of the rim is still multiplied by [tex]\Large\frac{1}{\sqrt{1 - U^2}}[/tex].
    Last edited: Feb 29, 2008
  21. thank you jtbell. I will look when my boss isn't here!
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