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On the Reissner-Nordstrom Metric

  1. Nov 1, 2011 #1
    The Reissner Nordstrom metric considers charge apart from mass in its composition. Both charge and mass appear in the temporal as well as the spatial components of the metric. By considering a large amount of charge against a small amount of mass we can have an estimate the individual contribution of electric charges towards the curvature of spaceā€”in influencing clock-rates and spatial separations

    The metric suggests an interaction between charge and mass.
    We consider a curved space created by a massive charged body,having a significal amount of charge.

    Now, a test charge[mass very small] or a test mass[chargeless] will follow the same geodesic so far as the metric is concerned . In the extremization of proper-time we consider the metric properties. But we do not consider the intrinsic[fundamental] properties[like charge,mass and spin of the test particle] of the particles whose motion would be investigated by the metric.

    How does one resolve this issue?
    Last edited: Nov 2, 2011
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  3. Nov 1, 2011 #2


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    Only an uncharged particle follows a geodesic. The charged particle deviates from a geodesic due to the electric and magnetic forces on it due to the electric and magnetic fields present in the RN solution.

    That said the presenece of electric and magnetic fields in the Rieissner-Norman solution contribute to the stress energy tensor in Einstein's equation, making the resulting metric different than the vacuum metric of the Schwarzschild solution.
  4. Nov 1, 2011 #3
    You seem to say that the principle of least action is not valid for charged particles and that
    the geodesic equation is not valid here.[We are simply interested in calulating the extremized values of proper time for the travel of a charged particle between two points[spacetime points]
    Geodesic equation:
    [tex]{\frac {{d}^{2} {x^{\alpha}}}{{d }{{\tau}^{2}}}}{=}{-}{{\Gamma}^{\alpha}}_{\beta\gamma}{\frac{{d}{x^{\beta}}}{{d}{\tau}}}{\frac{{d}{x^{\gamma}}}{{d}{\tau}}}[/tex]

    The metric may produce a field:
    But so far as as the extremization of propertime is concerned there is no reason to consider the field stated above.
    The mteric[line element,ds^2] itself should be sufficient.
    Last edited: Nov 2, 2011
  5. Nov 1, 2011 #4


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    Forget exotic GR solutions. You have a charged moon (somehow). You drop a charged baseball and a neutral baseball. Do you really think they follow the same path? The latter follows a geodesic, the former doesn't because a force acts on it. Do you think forces don't exist in GR? The principle of inertia in GR is that an object not subject to force (signified by no proper acceleration) follows geodesic (if it is small enough to be considered a test body).
  6. Nov 1, 2011 #5
    What is a "FORCE" in the context of General relativity[when geodesics are being considered]? Does such a thing exist in the realm of General Relativity[simultaneous existence of force and geodecis]?
    If you consider geodesics you cannot consider forces [inertial agents] simultaneously.
    GR solutions are SOLUTIONS. Why do you call them exotic?

    [The Reissner-Nordstrom metric describes the field due to a source that contains MASS as well as CHARGE. We are simply interested in evaluating the extremized propertime path between a pair of spacetime points. Theory[in its present state] should not entertain a differential treatment towards charged and uncharged test particles in the evaluation of geodesics]
    Last edited: Nov 2, 2011
  7. Nov 2, 2011 #6


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    The action of a particle "freely falling" in a gravitational field is indeed


    The presence of gravity modifies this ds term, and the extremization of this action gives us the geodesics. (Notice that extremization of this action is the same as extremization of the proper time)

    A charged particle moving in an electric and magnetic field; however, is not a "freely falling" particle. It's action is NOT what we quoted above. Therefore, it follows a different trajectory, and not necessarily a geodesic of the metric.

    A charged particle's action is given as:

    [tex]S=\int_A^B-mcds+\frac{e}{c}\int_A^B A_\mu dx^\mu[/tex]. This second term is not present for a non-charged particle.

    Obviously forces exist in GR. Not everything is falling on a geodesic. Right now, you are not falling on a geodesic, you are sitting on your chair, or standing on the floor. The normal forces keep you from moving along the geodesic. Only freely falling bodies move along geodesics.
  8. Nov 2, 2011 #7


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    Please read any text on general relativity. Four force on a test body is simply mass times proper acceleration. A geodesic is the path of an free fall body not forced. GR eliminates gravity as a force; it doesn't eliminate other forces.

    Also, you ignore my question: do you really think the two balls follow the same path? Do you really think such a trivial thing is violation of GR?

    My comment about exotic simply refers to the fact that your issue would arise in weak field domain where GR is indistinguishable from Newtonian gravity. My wording was maybe not good.
  9. Nov 2, 2011 #8
    On Proper Time Interval:
    [Geodesics are being considered first]

    For the calculation of the proper time interval we integrate d(tau) between a pair of points [in consideration of some path]

    [tex]{Proper}{\;}{ time}{\;}{Interval}{=}\int{d}\tau[/tex] -------- (1)

    The above integral is calculated between a pair of points following some path.
    The metric properties and the path in consideration are sufficient for the evaluation of the propertime interval,with out considering properties like charge ,mass,or spin of the test particle[in the present state of theory]

    For non-geodesic paths equation (1) continues to hold-- but inertial interactions come into play.
    But you should not condsider electromagenetic interaction as an inertial interaction because Q is already present in the metric.
    [Similarly:One should not consider Schwarzschild's Geometry and Newton's law of gravitation simultaneously]
    Last edited: Nov 2, 2011
  10. Nov 2, 2011 #9


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    All this is true. So what? This is perfectly consistent with what everyone (pervect, matterwave, and I) have been saying. The uncharged particle follows a geodesic world line; the charged particle does not. Along either world line, you calculate proper time with same integral, charge being irrelevant for this calculation - but very relevant for determining the world line.
  11. Nov 2, 2011 #10
    Infinitesimal path length:


    Condition for an extremal path length is given by:

    [tex]\delta\int {ds}{=}\delta\int {(}{g}_{\mu\nu}{dx}^{\mu}{dx}^{\nu}{)}^{1/2}{=}{0}[/tex] ------------- (1)
    The limits of integration correspond to the end points of the path.
    If lambda is taken to be an arbitrary parameter along the world-line,
    we have,

    [tex]\delta\int {ds}{=}\delta\int {(}{g}_{\mu\nu}\frac{{dx}^{\mu}}{{d}\lambda}\frac{{dx}^{\nu}}{{d}\lambda}{)}^{1/2}{d}\lambda[/tex] --------- (2)

    The metric coefficients in relations (1) and (2) invlove the mass and the charge of the SOURCE[Reissner-Nordstrom metric being considered here]. They do not involve the intrinsic[fundamental]properties of the test particles[like the charge or mass of the test particle].

    So the condition for extremal path has nothing to do with the mass or charge of the test particle[in the present context of the theory]

    Both CHARGED and UNCHARGED bodies[test bodies/particles] fall within the remit of equation (1) or equation (2)
    Last edited: Nov 2, 2011
  12. Nov 2, 2011 #11


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    Have you literally read anything anyone has said? ............

    Where in Einstein's theory does it say "all particles move along geodesics"?

    It's always "freely falling particles move along geodesics".

    A charged particle, in the presence of a charged mass...is not in free fall...

  13. Nov 2, 2011 #12
    All particles are supposed to move along geodesics provided the the correct initial conditions are satisfied

    What about planetary motion in Schwarzcshild's Geometry?[it is a special case of the R-N metric]
    What is your idea/concept of free fall in relation to planetary motion,in Schwarzscild's Geometry of course?

    A Geodesic is a GEODESIC. Why do you drag in the NARROW concept of something called free fall.

    To interpret or evaluate/determine GEODESICS you have just got to consider eqations (1) or (2) of post #10 NOTHING ELSE.

    [You may try out a transformation where the Christoffel symblos reduce to zero.
    That is not supposed to go against the content/implications of equations(1) and (2) of post#10. Nor would it provide any extra information about geodesics ]

    If the amount of charge dominates over the mass in the RN metric,you may have planetary motion of a charged body round the source charge[which is of course much larger than the test charge]
    Last edited: Nov 2, 2011
  14. Nov 2, 2011 #13
    The nature Geodesics [the evaluation of the stationary path] depends on:

    1. The nature or composition of the Lagrangian.
    2. The end points.
  15. Nov 2, 2011 #14


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    Particles DO NOT always follow geodesics. Do you think that you sitting there in your chair constitutes you following a geodesic?

    Let's have a thought experiment.

    Let's assume there's a platform above the Earth's surface, say 100m tall. Assume we are in a vacuum so we don't have to worry about air or anything else like that. I am standing on this platform. At time t=0, I drop a penny off this platform, giving it no initial velocity.

    At time t=0, both mine and the penny's 4-velocities are identical.

    If it were true that all particles followed geodesics. And the penny and I have the SAME initial 4-velocity, why are they not tracing out the SAME world line?

    We know from experiment that I would remain stationary on the platform, and the penny would drop. This is not the same world line.
  16. Nov 2, 2011 #15
    The inertial interaction[the reaction in this case] is not allowing a geodesic motion.But if somebody removes the chair[hope it does not happen in my case] I would move along a geodesic--- the prankster having created sufficient initial conditions for geodetic motion.

    If you fall out of the platform trying to catch the penny,that is, if the proper initial condition is created for you ,you would definitely fall with the penny[describing a geodesic motion]
    Hope it does not happen to you!

    [Appropriate initial conditions are necessary for a geodesic motion]
    Last edited: Nov 2, 2011
  17. Nov 2, 2011 #16


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    The "thing" now allowing you to fall through your chair, and wouldn't allow me to fall off the platform is the electro-magnetic interaction between the electrons of our bodies and the electrons of the surface on which we are sitting/standing on...How is this any different than the interaction of a charged particle and a EM field that permeates the spacetime?

    If you still don't buy this argument, then unfortunately I'm out of ideas on what kind of argument would actually convince you otherwise.

    The only thing I can think is to refer you to the literature where this point keeps coming up again over and over.

    Schutz's book "A First Course in General Relativity" states, on page 172-173: "Freely falling particles move on timelike geodesics of the spacetime...By 'freely falling' we mean particles UNAFFECTED by other FORCES, such as electric fields, etc. All other forces in physics are distinguished from gravity by the fact that there are particles unaffected by them".

    Wald's General Relativity states on page 67: "The world lines of FREELY FALLING bodies in a gravitational field are simply the geodesics of the spacetime metric"

    On page 69, he explicitly goes to show that "...if the particle has (rest) mass m and charge q, and is placed in an electromagnetic field [itex]F_{ab}[/itex], it satisfies the Lorentz force equation: [tex]u^a\nabla_a u^b=\frac{q}{m}F^b_c u^c[/tex]".

    Notice that if the particle was STILL moving along a geodesic, it would imply [itex]u^a\nabla_a u^b=0[/itex]

    Well, that's all I've got...
  18. Nov 2, 2011 #17
    If somebody pulls off the chair from under me I am going to fall along a geodesic till the floor stops me.

    If somebody pushes you or if you are inadvertent you will fall out of the platform tracing out an example geodesic till the ground stops you. I dont understand how the "electrons" are going to save you.
    [Even if you are in the vicinity of a charged planet you are electrically neutral,I believe.]
    Last edited: Nov 2, 2011
  19. Nov 2, 2011 #18


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    Because geodesics only represent the worldlines of freely falling objects. Geodesics do not represent the worldlines of other objects. See p 104 the paragraph containing eq 4.8 and 4.9 http://arxiv.org/abs/gr-qc/9712019

    If you believe that a geodesic describes the motion of a non-free-falling particle (such as a charged particle in a field) then please provide a mainstream reference to that effect.

    If you cannot provide such a reference then further suggestions along those lines is overly speculative, particularly considering that you have been corrected multiple times already.
  20. Nov 2, 2011 #19
    You may go through the following:

    Incidentally an inertial frame is one where the Christoffel symbols reduce to zero value. Absence of gravity alone is not a sufficient criterion----the metric considered being the Reissner Nordstrom metric.
  21. Nov 2, 2011 #20
    It is important to keep the audience apprised [to keep them alerted] of the basic equations that are used in relation to the determination of geodesics.
    We use variation techniques/principles as shown in the above quoted portion.

    The free fall concept[a transformation to make the Christoffel Symbols zero] cannot override the basic equation governing the Geodesics.
    Last edited: Nov 2, 2011
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