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Equivalence principle so important?

  1. Jun 18, 2012 #1
    Hello Forum,

    According to general relativity, objects in a gravitational field behave similarly to objects within an accelerating enclosure. For example, an observer will see a ball fall the same way in a rocket as it does on Earth, provided that the acceleration of the rocket provides the same relative force.

    But even without mentioning relativity, Newton described gravitation as a force field that accelerates object downward...so the connection between acceleration and gravitation was already made...
    What is the difference? Is it simply that Einstein does not mention the word "force"?

    Also, what is so special about the equivalence between inertial and gravitational mass? What is the big deal? What should I appreciate about that result?

    Einstein was thinking that in free fall it does not feel his own weight...what is so special about that thought? He would not feel his own weight simply because he would be in free fall. Weight is always there, but its perception exists only if there is a support surface....(contact force)...

    thanks,
    fisico30
     
  2. jcsd
  3. Jun 18, 2012 #2

    haushofer

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    In Newtonian gravity you indeed have an equivalence principle; under an arbitrary (possibly time dependent) linear acceleration

    [tex]
    \delta x^i = \xi^i (t) \,,
    [/tex]

    the gravitational potential phi transforms as

    [tex]
    \delta \phi = \ddot{\xi}(t) x_i
    [/tex]

    and an observer can use an acceleration to erase gravity.

    However, in GR these transformations are extended. Now, ALL observers, not just the ones which are linearly accelerating, can (locally in spacetime!) erase every appearance of gravity. This motivates the use of differential geometry, and makes the identification of gravity as spacetime geometry more direct.

    It's a necessity for the equivalence priniple to hold.
     
  4. Jun 18, 2012 #3

    A.T.

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    In Newtonian mechanics gravity is a interaction force (aka "real force"). In GR gravity is a inertial force (aka "fictitious force" or "pseudo force"):
    http://en.wikipedia.org/wiki/Fictitious_force
    The less different quantities you have the better. It makes physics simpler that they are the same.
    How do you know this, if you can't measure it for free falling objects?
     
  5. Jun 18, 2012 #4

    stevendaryl

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    It's actually possible to rewrite Newtonian physics so that gravity is a fictitious force. This is the Newton-Cartan theory. What makes this possible is exactly the equivalence of inertial and gravitational mass.
     
  6. Jun 18, 2012 #5
    What is special about it as haushofer said is that is key for identifying our gravitational universe with a certain geometric object, a pseudo-Riemannian curved manifold and that is pretty important for a theory.
    So simply put a Lorentzian (a type of pseudo-Riemannian manifold) curved 4-manifold looks locally like flat minkowski spacetime (in the same way a Riemannian manifold looks localy like Euclidean space) and this mathematical condition is captured physically by the Equivalence principle in its simplest formulation.
     
  7. Jun 18, 2012 #6

    Nugatory

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    That statement would work a lot better if you were to say "mass" instead of "weight" here.
     
  8. Jun 19, 2012 #7

    haushofer

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    Can you elaborate on that? I don't see directly why one wouldn't be able to rewrite Newtonian gravity in Newton-Cartan form if inertial and gravitational mass were different.

    I would say that in the usual Newtonian formulation one can also regard gravity as an inertial (fictitious) force; one can always apply a linear acceleration to go to a non-inertial frame and erase the appearance of gravity. You don't need Newton-Cartan theory for that, right?

    Newton-Cartan is just a geometric reformulation in which the equations of motion are invariant under general coordinate transformations (while the solutions are not, of course); I just regard such a rewriting as some sort of Stückelberg trick.
     
    Last edited: Jun 19, 2012
  9. Jun 19, 2012 #8

    stevendaryl

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    The point about "fictitious forces" is that a test mass moving under the influence of fictitious forces (and no non-fictitious forces) obeys the geodesic equation

    dU[itex]\alpha[/itex]/ds = - [itex]\Gamma^{\alpha}_{\beta\gamma}[/itex] U[itex]\beta[/itex] U[itex]\gamma[/itex]

    where U[itex]\beta[/itex] = dx[itex]\beta[/itex]/ds, and where s is an affine parametrization of the path: x[itex]\beta[/itex](s). Note that there is no mass anywhere, and that's because being a geodesic is a property of the path, not a property of the particle following the path.

    On the other hand, the Newtonian equations of motion for a test particle of mass m are:

    d[itex]^{2}[/itex]x[itex]^{i}[/itex]/dt[itex]^{2}[/itex] = 1/m F[itex]^{i}[/itex]

    These two equations seem worlds apart, but they can be seen to be equivalent by the following identifications:

    1. x[itex]^{0}[/itex] = t (so U[itex]^{0}[/itex] = 1)
    2. s = t
    3. [itex]\Gamma^{i}_{00}[/itex] = 1/m F[itex]^{i}[/itex] (for i=1,2,3)
    4. [itex]\Gamma^{\alpha}_{\beta\gamma}[/itex] = 0 (otherwise)

    This would seem like a pointlessly complicated way to write things, except for the fact that in this form, the equations of motion have the same form in every coordinate system. Although in inertial cartesian coordinates, most of the components of [itex]\Gamma^{\alpha}_{\beta\gamma}[/itex] are zero, they become nonzero if you switch to accelerated coordinates (in which case, it includes the "g" forces), or rotating coordinates (in which case, it includes the centrifugal and Coriolis forces).

    Because connection coefficients depend only on the coordinate system, not specific particles, it only works if 1/m F[itex]^{i}[/itex] is independent of the mass.

    There are three components of the Newton-Cartan theory: first, your observation that you can get rid of gravity by switching to an accelerated coordinate system, second, rewriting the equations of motion in terms of general coordinates, and third, coming up with a differential equation for computing the curvature (the curvature tensor can be computed from the connection coefficients [itex]\Gamma^{\alpha}_{\beta\gamma}[/itex] just as they are in GR).

    But the first step requires the equivalence of inertial and gravitational mass. If they were not equal, then that means that different objects would accelerate at different rates, which would mean that you could not get rid of gravity by changing to an accelerated coordinate system. You could make the acceleration of one object equal to zero, but then the acceleration of other objects would be nonzero.
     
  10. Jun 19, 2012 #9

    haushofer

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    Yes, but already here I can use the equivalence of both masses, and use this to write

    [tex]
    F^i = m \partial^i \phi
    [/tex]

    and eliminate the mass from the EOM. Also, I wouldn't say you need the geodesic equation to define inertial forces; for me, an inertial force is a force which you can set to zero by going to an appropriate frame of reference. But I see your point, thanks!
     
  11. Jun 19, 2012 #10
    One of the reasons the EPE is so important is that it allows you to transform complicated problems that would require the mathematical machinery of GR into simpler to solve problems that require only knowledge of accelerated motion in SR.
     
  12. Jun 20, 2012 #11

    haushofer

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    So, to give a concrete answer to your question: the difference is that Einstein used the equivalence principe to its full extent (ALL observers are equivalent, not just inertial observers without gravity, or linearly accelerating observers in the presence of gravity) in order to write gravity via differential geometry as spacetime curvature, thereby extending Newton's theory :)
     
  13. Jun 20, 2012 #12

    stevendaryl

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    The geodesic equation describes what the motion looks like in an arbitrary coordinate system. In flat spacetime, you can always find a coordinate system (I would say "coordinate system" rather than "frame of reference") such that [itex]\Gamma[/itex]αβγ is zero locally. You can't make it zero everywhere if there is gravity, except in the special case of uniform gravity.
     
  14. Jun 30, 2012 #13
    So, the equivalence principle states that we cannot distinguish between a situation where we are in an accelerating frame of reference in a region of space with zero gravity or in a stationary frame of reference inside a gravitational field.
    That reasoning leads also to the equivalence between inertial and gravitational masses since the same mass m is used in the arguments....

    So gravity can be interpreted as being inside a noninertial frame of reference, correct?

    Everything that has energy falls downward with the same acceleration in a gravitational field.
     
  15. Jun 30, 2012 #14
    I think comparing electric and gravitational forces is useful here. imagine an infinite plane of charge and a gravitating mass one. an electron is falling under the influence of each of them.
    you can move in an accelerated frame that cancels the electric force and another frame that cancels the gravitational force (in Newtonian mechanics).

    as a consequence of the equality between inertial and gravitational mass the frame which canceled the gravitational force on the electron will simultaneously cancel it for a billiard ball. but the one that cancelled the electric force only works for the charge of the electron and no other charge.
     
  16. Jun 30, 2012 #15

    jcsd

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    In Newtonian physics all test particles are affectedby pseudo-forces, but it's not necessarily the case that all test particles are affected by gravity. The equivalence prinicple says that they are.
     
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