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Question about value of the metric tensor and field strength

  1. May 13, 2013 #1
    Is it the value of the metric tensor that determines the strength of a gravitational field at a specific point in spacetime?
     
  2. jcsd
  3. May 13, 2013 #2

    WannabeNewton

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    This is somewhat of a subtle topic in GR. In Newtonian mechanics, if we are given a gravitational potential ##\varphi##, then we say the acceleration due to gravity, of a test particle in the potential, is given by ##a = -\nabla\varphi##. So, interpreting "strength of the gravitational field" as the acceleration of a test particle placed in the field, the strength would be determined by the potential ##\varphi##.

    The problem in GR is that for general space-times, the concept of a gravitational potential is not well-defined. Moreover, in GR, any particle that is in free fall will always have vanishing acceleration because a free falling particle travels on geodesics which satisfy ##u^{b}\nabla_{b}u^{a} = 0## where ##u^{a}## is the 4-velocity of the test particle. What we can do, however, is "associate" certain geometrical quantities in GR with quantities in Newtonian mechanics that involve the gravitational potential and field when looking at limiting cases of GR. This is best seen in the linearized approximation to GR.

    In the linearized approximation, we write the metric tensor as ##g_{ab} = \eta_{ab} + h_{ab}## where ##h_{ab}## is a perturbed metric that represents "small deviations from flatness". In writing things like the geodesic equation, we throw out all terms that are second order or higher in ##h_{ab}##; the quantity ##\bar{h}_{ab} = h_{ab} - \frac{1}{2}\eta_{ab}h## is usually defined in order to simplify calculations. Thus if we were to adopt a global inertial coordinate system for the Minkowski metric, and consider the Newtonian limit where we use the slow motion approximation ##|u^{0}|>>|u^{i}|##, we find that in these coordinates, the geodesic equation ##u^{\nu}\nabla_{\nu}u^{\mu} = 0## becomes ##\ddot{x}^{\mu} = -\Gamma ^{\mu}_{00} = -\partial^{\mu}\varphi## where ##\varphi \equiv -\frac{1}{4}\bar{h}_{00}##. Notice how the above has essentially the exact same form as the acceleration due to gravity in Newtonian mechanics. It would then be natural, in the Newtonian limit of GR, to associate the usual gravitational potential with ##\bar{h}_{00}## and the acceleration with ##\Gamma^{\mu}_{00}##.

    However it should be noted that this similarity in the form of the equations of motion does not shadow the fact that the viewpoints regarding gravitation offered by GR and Newtonian mechanics are still quite different. The above correspondences may be clear in the Newtonian limit but in the general theory, one must stick to the notion of gravitation as being a consequence of the curvature of space-time, a curvature that is induced by mass-energy distributions. Things like the trajectories of test particles are then described in terms of geometrical quantities as opposed to gravitational potentials and gravitational fields.

    In order to further stress the above, let's consider another important quantity that involves gravitation. In Newtonian mechanics the tidal forces of a gravitational field can be measured by attaching a separation vector ##\xi## between two nearby particles and seeing how ##\xi## evolves; the tidal acceleration will then be given by ##-(\xi\cdot \nabla)\nabla\varphi##. In GR however, the tidal acceleration due to the effects of gravitation as described by space-time curvature is given by ##-R_{abc}{}{}^{d}u^{a}u^{c}\xi^{b}## where ##u^{a}## is the 4-velocity of the nearby particles. Notice how in the latter we are describing the tidal effects using the geometry of space-time (in particular the Riemann curvature tensor ##R_{abcd}##) whereas in the former we are describing the tidal effects using the gravitational potential.
     
  4. May 13, 2013 #3
    [edit: I see wannabe posted while I was composing....mine is a more basic introduction....I'm rather simple minded according to my wife!!]

    sort of, but not really: the term 'metric' or 'metric tensor' is generic....different kinds are listed here:

    http://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)#Other_metrics

    the 'metric' is a distance measure...not the source of gravity [see below]

    You are perhaps thinking of the stress energy momentum tensor [SET}specific to general relativity and the Einstein Field Equations:

    http://en.wikipedia.org/wiki/Stress-energy_tensor


    which is the source of gravity.....So not just mass but also things like pressure and energy also contribute to gravity.

    The 'gravitational field' is also a term with mixed meanings....Einstein says 'gravity' is curvature but just how one measures or determines curvature is not so simple. There are a number of different measures and no one captures all the components....

    For a general introduction, not too complicated, try starting here from Figure 4 ....see if you like the approach....

    http://www.jimhaldenwang.com/black_hole.htm

    Note where he writes:
    which links the 'metric tensor' to the SET.

    edit: You can think of the metric as the dependent expression, the SET as the independent one, analogous to y = f(x) ....in other words, the SET [mass,energy,pressure,etc] determines the metric [distance measure, curvature dependent]...
     
    Last edited: May 13, 2013
  5. May 13, 2013 #4
    here is the source I was looking for: ... a quote I kept from a discussion in these forums...from a highly regard classic textbook [Misner, Thorne, Wheeler]:

     
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