General Relativity – Computing and Interpreting Schwarzschild Black-Hole Metric Line Element in Cartesian Coordinates

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The Schwarzschild metric in Cartesian coordinates is listed on Wikipedia as:

Line element Notes
$$-{\frac {\left(1-{\frac {r_{\mathrm {s} }}{4R}}\right)^{2}}{\left(1+{\frac {r_{\mathrm {s} }}{4R}}\right)^{2}}}\,{dt}^{2}+\left(1+{\frac {r_{\mathrm {s} }}{4R}}\right)^{4}\,\left(dx^{2}+dy^{2}+dz^{2}\right)$$ $$R = \sqrt{ x^2 + y^2 + z^2 }$$ Valid only outside the event horizon: $R>r_s/4$

I am new to computing metrics, if anyone more experienced can help please. Using this Cartesian formula above to compute the path value $[t, x, y, z]$ while choosing units to be Schwarzschild radius $R_s = 1$ gives the following results:

From $[0, 4, 4, 4]$ to $[1, 5, 5, 5]$, $ds = 1.61$

From $[0, 4, 4, 4]$ to $[2, 5, 5, 5]$, $ds = 0.07$

I am unsure what this means, as both computations start and end at the exact same space coordinates, from $[4,4,4]$ to $[5,5,5]$. The particle is moving radially outward from the Schwarzschild sphere of radius $R_s = 1$, centered at the origin $[0,0,0]$.

The only difference is that the first test particle takes 1 time unit to travel the same distance, while the second particle takes a longer 2 time unit to travel the same distance.

1. Why do the ds values vary so greatly if the space endpoints are the same?

2. What does it mean to have $ds=0$ between two spacetime points?

The code used to compute the results are as follows:

let ds = straight_line_path([0, 4, 4, 4], [1, 5, 5, 5], 1000000);      

function straight_line_path (st1, st2, pixel)
{
    let dt = (st2[0]-st1[0])/pixel;
    let dx = (st2[1]-st1[1])/pixel;
    let dy = (st2[2]-st1[2])/pixel;
    let dz = (st2[3]-st1[3])/pixel;
    
    let total = 0;
    
    for (let i=0; i<pixel; i++)
    {
        total += line_schwartzchild (dt, dx, dy, dz, [st1[1]+i*dx, st1[2]+i*dy, st1[3]+i*dz]);
    }
    
    return total;
}

function line_schwartzchild (dt, dx, dy, dz, [x0,y0,z0])
{
    let r = Math.sqrt (x0*x0 + y0*y0 + z0*z0);

    let km = 1 - 1/(4*r);
    let kp = 1 + 1/(4*r);

    let a = km*km/(kp*kp);
    let b = kp*kp*kp*kp;

    let sum = Math.abs(-a*(dt*dt) + b*(dx*dx + dy*dy + dz*dz));

    return Math.sqrt(sum);
}

Am I using the wrong formula, or computing something wrong along the way?

Best Answer

In addition to @Andrea's answer about interpreting timelike, spacelike, and null paths, there is a computational issue with your method as well.

The metric, $g_{\alpha\beta}$, is not the same everywhere in spacetime. It is a function of radial distance from the center. In your calculation you evaluate the metric at the starting point ($\vec{r}_0$), and then use that value for the whole path.

$$s = \sqrt{\pm g_{\alpha\beta}(\vec{r}_0) \Delta r^\alpha\,\Delta r^\beta}$$ $$\Delta\vec{r} = \vec{r}_1 - \vec{r}_0,$$ where $\vec{r}$ is a four-vector with components $r^\alpha = (t,x,y,z)$. If the displacement is small, then this approximate calculation should be acceptable. The correct calculation would integrate $ds$ over the path.

$$s = \int_{\vec{r}_0}^{\vec{r}_1} ds = \int_{\vec{r}_0}^{\vec{r}_1} ​d\tau \sqrt{\pm g_{\alpha\beta}[\vec{r}(\tau)] \frac{d r^\alpha}{d\tau}\, \frac{d r^\beta}{d\tau}},$$ where $\tau$ is an affine parameterization of the path. For timelike paths, proper time is a common choice.

The integral accounts for the fact that $g$ takes on a different value at each point along the path.

For a spacelike path, you would use the $+$ under the square root and interpret the $s$ as the proper length separating the points. For a timelike path, you would use the $-$ and $s$ is the proper time taken.

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