Choosing which paths to draw

In[129]:=

  R

Out[129]=

                2
               b     2              2
  {{r -> -Sqrt[-- + t  - 2 t t0 + t0 ]},
                2
               c

                2
               b     2              2
    {r -> Sqrt[-- + t  - 2 t t0 + t0 ]}}
                2
               c

These solutions have r = b/c at t = t0. We'll choose positive r:

In[130]:=

  r[t_] = r /. R[[2]]

Out[130]=

        2
       b     2              2
  Sqrt[-- + t  - 2 t t0 + t0 ]
        2
       c

Since these functions are identical, it doesn't matter which one we use, so we'll use rin[r] from now on.

The constant theta0 gives the angle at which r = b/c, so we know theta = theta0 at t = t0. Just as r = b/c defines the distance of closest approach, we can call theta0 the angle of closest approach.

Now we need to choose the initial conditions. Suppose we want to draw the light cone of some spacetime event. That means we want the null geodesics to all pass through at the same time, like a flash going off at t=0.

Let's try looking at paths that pass through the space point {r = 1, theta = 0} at t = 0. First we'll solve the r=1 at t=0 condition:

In[131]:=

  r[0]

Out[131]=

        2
       b      2
  Sqrt[-- + t0 ]
        2
       c

In[132]:=

  t0rule = Solve[r[0] == 1, t0]

Out[132]=

                     2                     2
                    b                     b
  {{t0 -> -Sqrt[1 - --]}, {t0 -> Sqrt[1 - --]}}
                     2                     2
                    c                     c

Now look at the theta condition. We want theta = 0 when r = 1. This gives us a way to eliminate the parameter b in terms of theta0 and t0.

In[133]:=

  theta1[1,theta0]
  theta2[1,theta0]

Out[133]=

                  b
           ArcCos[-]
                  c
  theta0 + ---------
               c

Out[134]=

                  b
           ArcCos[-]
                  c
  theta0 - ---------
               c

Here is the result:

In[135]:=

  thrule = Flatten[Solve[theta1[1,theta0] == 0, b]]

Out[135]=

  {b -> c Cos[c theta0]}

Notice that this tells us something important about t0:

In[136]:=

  trule = PowerExpand[Expand[Simplify[t0rule /. thrule]]][[2]]

Out[136]=

  {t0 -> Sin[c theta0]}

Now here is the function for r[t] in the way we want it for plotting:

In[137]:=

  r[t_,c_,theta0_]= Simplify[r[t] /.thrule] /. trule
  

Out[137]=

                    2                      2
  Sqrt[Cos[c theta0]  + (t - Sin[c theta0]) ]

Now we need to deal with theta again:

In[138]:=

  theta1[r[t,c,theta0],theta0]

Out[138]=

  theta0 +

                                  b
     ArcCos[---------------------------------------------]
                                2                      2
            c Sqrt[Cos[c theta0]  + (t - Sin[c theta0]) ]
     -----------------------------------------------------
                               c

In[139]:=

  theta2[r[t,c,theta0],theta0]

Out[139]=

  theta0 -

                                  b
     ArcCos[---------------------------------------------]
                                2                      2
            c Sqrt[Cos[c theta0]  + (t - Sin[c theta0]) ]
     -----------------------------------------------------
                               c

One thing that Mathematica finds difficult is trigonometric manipulations. By hand it can be verified (by remembering how to do trigonometry by hand) that a better way to express the above relations is through the formula:

In[140]:=

  theta[t_,c_,theta0_] =
theta0 + ArcTan[(t - Sin[c theta0])/Cos[c theta0]]/c

Out[140]=

           ArcTan[Sec[c theta0] (t - Sin[c theta0])]
  theta0 + -----------------------------------------
                               c

We'll check this out with a plot:

In[141]:=

  Plot[theta[t,1,0],{t,-5,5}]

Out[142]=

  -Graphics-

This behaves as it should both before and after the turning point at t=0.

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