1
$\begingroup$

Today in my physics class, my lecturer said something which confused me. He said:

"Newtonian tidal forces are reinterpreted as a manifestation of curvature in General Relativity".

Now I know what tidal forces are (an effect of the force of gravity), a good example is the cause of the waves on the ocean because of the tidal forces with the moon. However I do not see how this shows curvature in the GR sense.

$\endgroup$
4
  • 1
    $\begingroup$ Do you know what curvature in the GR sense even means? $\endgroup$
    – ACuriousMind
    Commented Nov 6, 2014 at 12:48
  • 2
    $\begingroup$ Yes I know of the Riemann curvature tensor. In GR particles follow geodesics and there is some work with second derivatives though I'm sort of hoping for a less mathematical explanation... $\endgroup$
    – doomdada
    Commented Nov 6, 2014 at 12:51
  • $\begingroup$ i think the meaning is that the relative motion of earth wrt to the moon , changes earth's (gravitational metric) curvature, thus tidal forces are manifested. The whole point about GR is this equivalemce: "gravity=acceleration=curvature" $\endgroup$
    – Nikos M.
    Commented Nov 6, 2014 at 13:00
  • 2
    $\begingroup$ Did you think of asking your teacher what he meant? $\endgroup$
    – Floris
    Commented Nov 6, 2014 at 14:27

2 Answers 2

4
$\begingroup$

Have a look at my answer to How to explain centripetal force in terms or relativity because much of the discussion there is relevant.

Consider what we mean by a tidal force. Suppose you're floating around in space and you arrange a number of marbles around you so they lie on the surface of a perfect sphere. Now monitor the shape of the surface marked out by those marbles. If the shape changes with time from a sphere to an ellipsoid you would conclude that there must be a force acting on the marbles to pull them apart. In the Newtonian interpretation this is the tidal force.

Now consider the GR interpretation, and this is where the discussion of geodesics in the answer I linked above comes in. Each marble follows a geodesic. In flat spacetime geodesics that are originally parallel remain parallel, so if the marbles are initially stationary with respect to each other they remain stationary with respect to each other, and the sphere does not change shape. However if spacetime is curved then initially parallel geodesics may not remain parallel, but can diverge or converge. Because each marble is following a different geodesic, and the geodesics might not remain parallel, the marbles may move apart and the sphere change shape. No force is acting - it's just that the individual marbles are following different geodesics.

And this is (I would guess) what your lecturer means. Newton would see the sphere change shape and conclude there must be a force acting. Einstein would see the sphere change shape and conclude that spacetime was curved.

$\endgroup$
2
  • $\begingroup$ So spacetime is curved. The mass reaches out (somehow or other) and curves It. Is It curved by ... "a force"? :) $\endgroup$
    – Fattie
    Commented Nov 6, 2014 at 16:10
  • $\begingroup$ @JoeBlow: general relativity gives us a set of differential equations that relate the spacetime curvature to the mass/energy density. And these equations work in the sense that their predictions match with experiment. However if you ask how does mass curve spacetime? or what's actually going on? then that's a question you need to address to a philosopher of science not a physicist. GR tells that it does, and how to calculate how much it does, but it doesn't tell us why it does. $\endgroup$ Commented Nov 6, 2014 at 16:21
0
$\begingroup$

curvature produces relative acceleration of geodesics because of equation of geodesic deviation (relation between riemann tensor and relative accleleration of geodesics) and also newtonian theory of gravity predicts tide-producing acceleration which explain two particle with separation x parallel to r ,cause evaluating relative acceleration and x prependecular to r ,have a stated amount of relative acceleration .so relative acceleration caused by curvature .. But why we use these thing instead of force , you can find the answer in weightlessness , actually there isnt any real force in a free fall particle See page 29 and 30 gravitation mtw
for tide-producing acceleration

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.