Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How does GR describe acceleration from non-gravitational forces?

  1. Feb 7, 2014 #1
    I have a pretty basic question - one that should have occurred to me long ago, but I never really thought about before.

    We all know how the effects of gravity are described by the curvature of spacetime - rubber sheets and all that - as well as the equivalence of inertial and gravitational mass. Those basic concepts have always made me think of the inertia of a massive object as somehow being the result of the curvature of spacetime. First of all - is that even a correct idea? More to the point, though - how do you account for the inertia of an object that is accelerated by a non-gravitational force? Does spacetime still get curved in such a way that the object accelerates in inverse proportion to its mass, or does that come about purely in the non-relativistic way (i.e. F = ma)?

    As a concrete example, let's say we have a small mass with an electric charge in a uniform electric field, far from any other massive object, so that spacetime is relatively flat. The object will accelerate uniformly, which should mean that in some sense it's locally equivalent to a uniform gravitational field, but is there actually any curvature of spacetime?
     
  2. jcsd
  3. Feb 7, 2014 #2

    Nugatory

    User Avatar

    Staff: Mentor

    That acceleration is not equivalent to a local gravitational field, because particles with different charges will accelerate differently - a gravitational field accelerates everything equally. Thus, there's no spacetime curvature involved - instead, there's a force that is pushing the charged mass off of the geodesic path it would follow if left alone.

    Mathematically, non-gravitational forces produce non-zero four-acceleration. An object moving under the influence of gravity experiences no four-acceleration.
     
  4. Feb 7, 2014 #3

    WannabeNewton

    User Avatar
    Science Advisor

    No.

    Space-time curvature is independent of the trajectories of test particles, accelerated or not-that's why we use test particles. Freely falling test particles are described by geodesics and accelerating particles are described by other time-like space-time curves, simple as that. The equations of motion for test particcles are simply a curved space-time version of Newton's 2nd law, more precisely ##m u^{\nu}\nabla_{\nu}u^{\mu} = F^{\mu}## where ##u^{\mu}## is the 4-velocity, ##F^{\mu}## is the total 4-force, and ##m## is the rest mass.

    This is garbled word salad at best, sorry to say. You have a fundamental misunderstanding of the equivalence principle.
     
  5. Feb 7, 2014 #4

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    The question in the title of your post is based on a misconception. In GR, the *only* kind of acceleration that has any physical meaning is proper acceleration, which can only be produced by non-gravitational forces. In fact, in GR, gravity itself is not a force, so the term "non-gravitational forces" is redundant; the *only* forces there are in GR are "non-gravitational" forces.
     
  6. Feb 7, 2014 #5
    Curved spacetime doesn't create the inertial mass of a test particle in it, but the solution for the spacetime from the matter it contains does determine what constitutes a localy inertial frame Vs an accelerated one.
    The stress-energy tensor of an electric field does curve spacetime in general relativity. What you do to describe the motion of a test charge in a spacetime with curvature and an electric field is you write down the 4-vector version of Newton's 2nd law
    [itex]F^\lambda = mA^\lambda [/itex]
    The 4-force on the test charge is
    [itex]F^\lambda = qg_{\mu \nu }\frac{U^{\mu }}{c}F^{\nu \lambda }[/itex]
    Where [itex]U^{\mu } =\frac{dx^\mu }{d\tau }[/itex] is the velocity 4-vector and
    [itex]F^{\nu \lambda }[/itex] is the electromagnetic field tensor. There is both sign and units conventions in these to be aware of.
    You then write out the 4-acceleration
    [itex]A^\lambda = \frac{d^2 x^\lambda }{d\tau ^2}+\Gamma ^{\lambda} _{\mu \nu}\frac{dx^{\mu }}{d\tau }\frac{dx^{\nu }}{d\tau}[/itex]
    And so input into the law of motion you get
    [itex]qg_{\mu \nu }\frac{U^{\mu }}{c}F^{\nu \lambda }=m\left(\frac{d^2 x^\lambda }{d\tau ^2}+\Gamma ^{\lambda} _{\mu \nu}\frac{dx^{\mu }}{d\tau }\frac{dx^{\nu }}{d\tau}\right)[/itex]
    This is general relativity's law of motion for a test charge in an electromagnetic field in curved spacetime. Now if either the test mass has zero charge or the electromagnetic field is zero you get
    [itex]\frac{d^2 x^\lambda }{d\tau ^2}+\Gamma ^{\lambda} _{\mu \nu}\frac{dx^{\mu }}{d\tau }\frac{dx^{\nu }}{d\tau}=0[/itex]
    which is the geodesic equation which is saying that the acceleration 4-vector is zero
    [itex]A^\lambda = 0[/itex]
    which is general relativity's version of Newton's first law. Unacted on by 4-vector forces, a test particle will undergo geodesic motion.
     
  7. Feb 7, 2014 #6
    Thanks to all - I knew I had a pretty fundamental misunderstanding, and these responses have all helped a lot.

    I think I was suffering from the cartoon understanding of the observer in a uniformly accelerating elevator who cannot distinguish between that and a stationary elevator in a uniform gravitational field. Saying that he can't distinguish between them doesn't exactly mean that they're physically the same, does it?

    Thanks again.
     
  8. Feb 7, 2014 #7

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    It means that when you stand on the surface of the Earth you are physically accelerating upwards (just like a firing rocket in space is physically accelerating). And in free fall towards the Earth you aren't physically accelerating (just like non-firing rocket in space isn't physically accelerating). That is called proper acceleration and is what an accelerometer measures.

    So in terms of acceleration it is physically the same.
     
  9. Feb 7, 2014 #8
    I think I would have said that even before today, although I suspect I wouldn't have understood what I was saying - or slightly less than I do now, anyway.

    So, let's see - when I stand on the surface of the Earth, I am prevented from following a geodesic on my locally curved bit of spacetime by the upward force of whatever I'm standing on. Similarly, the guy in the uniformly accelerating rocket is prevented from following a goedesic on his locally flat bit of spacetime by the force imparted by the rocket.

    So they're the same in that sense, but there is no reason to think of the curvature of spacetime as being responsible for any of that, other than to say what our respective geodesics would be if we were moving inertially.

    Is that a little closer to correct?
     
  10. Feb 7, 2014 #9

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Yes, this is better. The only clarification I would make is that "what our respective geodesics would be" is not a local concept--you can only see the difference between the two geodesics (flat spacetime vs. curved spacetime) if you look beyond the small local patch of spacetime in which the comparison between "observer standing on Earth" and "observer standing inside accelerating rocket" is being made. Within that small local patch of spacetime, both geodesics look the same (because the *definition* of the "local patch of spacetime" is a region small enough that the effects of spacetime curvature are negligible).
     
  11. Feb 7, 2014 #10

    Dale

    Staff: Mentor

    I would say that is pretty close to correct. I will add a distinction that may help or may confuse, if the latter then feel free to ignore.

    As several others have mentioned the "force" of gravity is not treated as a real force in GR. Instead, it is treated as something arising from using curved coordinates.

    For example, suppose you take a piece of paper and draw a straight line on it, now imagine that you draw a set of polar coordinates and describe the line in terms of those polar coordinates. You would find that the straight line seems to "curve" when expressed in those coordinates and that its equation would involve extra terms beyond the basic linear terms.

    Similarly, in GR things like the Coriolis and centrifugal forces and other fictitious "inertial" forces arise because the time and/or space coordinates are curved. The gravitational force cannot be distinguished from those inertial forces locally. This is the essence of the equivalence principle.

    All of that dealt with curved coordinates, not curved spacetime. Curved spacetime becomes necessary when you are dealing with tidal gravity, or in other words, gravity which varies from place to place.
     
  12. Feb 7, 2014 #11
    Yes - I was speaking sloppily again. What I really meant was that rocket guy is not near a gravitating body, so his spacetime is flat (that's all I meant by "local"), and mine is very much near a massive body, so mine is curved.

    Thanks again.
     
  13. Feb 7, 2014 #12

    A.T.

    User Avatar
    Science Advisor
    Gold Member

    Locally, where tidal effects are negligible, your space-time is flat too. However, both frames are non-inertial, and can be described using curved coordinates on a flat manifold. See here:

    https://www.youtube.com/watch?v=DdC0QN6f3G4

    Note that the cone surface in the video has zero intrinsic curvature. Only over a larger area it becomes a curved saddle surface, which cannot be rolled out flat without distortions anymore.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: How does GR describe acceleration from non-gravitational forces?
Loading...