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The three vectors in the Null Geodesic equation

  1. Sep 5, 2011 #1
    From a metric maybe the Schwarzschild, you can find g in co and contra varient forms. From that you can calculate Affinity.

    My question is from the Null Geodesic equation (ds=0) what do the three contravarient vectors represent? Do they represent the path of a planet around the sun or the path of a moon around a planet?
     
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  3. Sep 5, 2011 #2

    WannabeNewton

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    Null geodesics represent the paths of massless particles: [tex]\frac{d^{2}x^{\alpha }}{d\lambda ^{2}} + \Gamma ^{\alpha }_{\beta \gamma }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\gamma }}{\mathrm{d} \lambda } = 0[/tex] where [itex]\lambda [/itex] is an affine parameter that is not proper time. The solutions [itex]x^{\alpha }(\lambda )[/itex] will give you the path of the massless particle given the initial conditions. What three vectors are you talking about?
     
  4. Sep 5, 2011 #3

    PAllen

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    Null geodesics represent the path of light or any uncharged (really, un-forced) massless particle. Note that ds=0 only specifies that the path is lightlike, not necessarily geodesic. There is such a thing as non-geodesic null paths.
     
  5. Sep 5, 2011 #4
    The three vector I am refering to are:

    [tex]x^\beta[/tex]
    [tex]x^\gamma[/tex]
    [tex]x^\alpha[/tex]

    If I want particles with "mass" then ds<>0, "right". Like planets and moons?
     
  6. Sep 5, 2011 #5

    WannabeNewton

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    [itex]x^{\alpha }[/itex] refers to a component of x and [itex]x^{\gamma },x^{\beta }[/itex] refer to components of x that are also being summed over the christoffel symbol. Massive particles travel on time - like geodesics so for massive particles, [itex]ds^{2} >0[/itex] or [itex]ds^{2} < 0 [/itex] depending on your sign convention for time - like intervals. In the geodesic equation make the replacement [itex]\lambda \to \tau [/itex] where [itex]\tau [/itex] is proper time.
     
  7. Sep 5, 2011 #6
    I was wondering how to formulate them? If you are heliocentric and you are analyzing the orbit of Mercury. You are given L,B and R of Mercury for three separate points of observation would they correspond to:

    r, [tex]\theta[/tex] and [tex]\phi[/tex]. Also, how do you formulate the time element?
     
  8. Sep 5, 2011 #7

    pervect

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    What do you mean by L, B, and R?

    The meaning of the symbols in the geodesic equations are this:

    [itex]\alpha, \beta, \gamma[/itex] are just placeholders, that take on the values (0,1,2,3)


    So [itex]x^{\alpha}[/itex] stands for the set [itex](x^0, x^1, x^2, x^3)[/itex]

    This is a vector, it's got four elements though, not the three you might be used to. One of them is time.

    It's got four components, one of them is time, and three are space. The exact correspondence can vary, but usually it will be

    [itex]x^0[/itex] -> t [itex]x^1[/itex]->r [itex]x^2[/itex]->[itex]\theta[/itex] [itex]x^3[/itex]->[itex]\phi[/itex]

    The geodesics are parameterized in terms of an affine parameter, usually either proper time [itex]\tau[/itex], or more generally [itex]\lambda[/itex].

    For now, you can think of [itex]\lambda[/itex] as just a number. Each value of [itex]\lambda[/itex] specifies one point on the geodesic curve, sweeping it through it's allowed range generates the complete curve.

    We don't write the position as a function of time. Instead, we write the position and the time as a function of the parameter lambda.


    So, going back to the original equation

    [tex]
    \frac{d^{2}x^{\alpha }}{d\lambda ^{2}} + \Gamma ^{\alpha }_{\beta \gamma }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\gamma }}{\mathrm{d} \lambda } = 0
    [/tex]

    This is a compact way of writing four equations, one equation for each value of [itex]\alpha[/itex]

    The repeated indices, [itex]\beta[/itex] and [itex]\gamma[/itex] imply a summation.

    So it should really be written as

    [tex]
    \sum_{\beta=0}^{4} \sum_{\gamma=0}^{4}\frac{d^{2}x^{\alpha }}{d\lambda ^{2}} + \Gamma ^{\alpha }_{\beta \gamma }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\gamma }}{\mathrm{d} \lambda } = 0
    [/tex]

    but that's too much work, so people use the "Einstein summation convention" and omit the sums for repeated indices.

    So this is really a set of four differential equations.
     
    Last edited: Sep 5, 2011
  9. Sep 5, 2011 #8
    What do [tex]x^{\alpha}[/tex][tex]x^{\beta}[/tex][tex]x^{\gamma }[/tex]

    represent in real world applications? Are they three succesive points on an orbital curve with a corresponding

    r=Ecliptic Radius, R

    [tex]\theta[/tex]=Ecliptic Longitude, L

    [tex]\phi[/tex]=Ecliptic Latitude, B

    t=time?

    or are they equations of the 3 different gravitational sources affecting the orbit?

    (I am just want to know how to formulate x^4)
     
    Last edited: Sep 5, 2011
  10. Sep 5, 2011 #9

    WannabeNewton

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    If you use the [itex](t, r, \theta , \phi )[/itex] coordinate chart then the [itex]x^{\alpha }[/itex] are the components of the geodesic relative to the coordinate basis for the chart. [itex]t[/itex] is of course coordinate time, [itex]r[/itex] doesn't have the same meaning as radius in euclidean n - space rather it's just the circumferential radius, [itex]\theta [/itex] and [itex]\phi [/itex] have the same meaning they do on the 2 - sphere and that is the polar and azimuthal angle respectively.
     
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