Doppler Shift & Christoffel Symbols Issues

In summary, the conversation discussed the simulation of light physics around black holes and the discovery of a fast Christoffel symbols function. However, the photon ring appeared to be flipped, causing concern about the accuracy of the calculation. The conversation also mentioned the possibility of an oversight in the code or incorrect Mathematica calculation. The included code and image showed the results of the simulation.
  • #1
Jessie24789
8
4
TL;DR Summary
I am doing simulations of black holes in GLSL and recently stopped using a general calculation in shader. Now my photon ring has a bright spot opposite the accretion disk bright spot.
About a month or two ago I started doing simulations of light physics around black holes and yesterday I got a fast Christoffel symbols function for the Schwarzschild metric in cartesian coordinates, but now the photon ring appears flipped. I feel as though it is wrong. But as I am still pretty new to general relativity, I am not sure. Would the physics allow for this to happen, or is it caused by an oversight in the code? Or is the Mathematica calculation of it incorrect?

Included below is the code I am using as well as an image showing the results of the simulation.
Screenshot 2022-12-14 184557.png

GLSL Code: 
mat4[4] SchwarzschildChristoffel(in vec4 q) {
    vec3 position = q.yzw;

    float rho = dot(position,position);
    float r2 = 0.5*(rho + sqrt(rho*rho));
    float r = sqrt(r2);
    
    float x = position.x;
    float y = position.y;
    float z = position.z;

    float f = (G / r2) * (2.0 * M * r);

    vec4 empty = vec4(0.0);
    mat4 T0 = mat4(
        empty,
        vec4(0.0, (f * r) / ((f - 1.0) * r2 - f * rho), 0.0, 0.0),
        vec4(0.0, 0.0, (f * r) / ((f - 1.0) * r2 - f * rho), 0.0),
        vec4(0.0, 0.0, 0.0, (f * r) / ((f - 1.0) * r2 - f * rho))
    );
    mat4 T1 = mat4(
        empty,
        vec4(0.0, (f * x) / (-((f - 1.0) * r2) + f * rho), 0.0, 0.0),
        vec4(0.0, 0.0, (f * x) / (-((f - 1.0) * r2) + f * rho), 0.0),
        vec4(0.0, 0.0, 0.0, (f * x) / (-((f - 1.0) * r2) + f * rho))
    );
    mat4 T2 = mat4(
        empty,
        vec4(0.0, (f * y) / (-((f - 1.0) * r2) + f * rho), 0.0, 0.0),
        vec4(0.0, 0.0, (f * y) / (-((f - 1.0) * r2) + f * rho), 0.0),
        vec4(0.0, 0.0, 0.0, (f * y) / (-((f - 1.0) * r2) + f * rho))
    );
    mat4 T3 = mat4(
        empty,
        vec4(0.0, (f * z) / (-((f - 1.0) * r2) + f * rho), 0.0, 0.0),
        vec4(0.0, 0.0, (f * z) / (-((f - 1.0) * r2) + f * rho), 0.0),
        vec4(0.0, 0.0, 0.0, (f * z) / (-((f - 1.0) * r2) + f * rho))
    );

    return mat4[4](
        T0,
        T1,
        T2,
        T3       
    );
}

vec4 ParallelTransportRateOfChange(mat4[4] christoffelSymbols, vec4 v, vec4 dxds) {
    vec4 dvds = vec4(0.0);
    dvds -= vec4(christoffelSymbols[0][0][0], christoffelSymbols[1][0][0], christoffelSymbols[2][0][0], christoffelSymbols[3][0][0]) * dxds[0] * v[0];
    dvds -= vec4(christoffelSymbols[0][0][1], christoffelSymbols[1][0][1], christoffelSymbols[2][0][1], christoffelSymbols[3][0][1]) * dxds[0] * v[1];
    dvds -= vec4(christoffelSymbols[0][0][2], christoffelSymbols[1][0][2], christoffelSymbols[2][0][2], christoffelSymbols[3][0][2]) * dxds[0] * v[2];
    dvds -= vec4(christoffelSymbols[0][0][3], christoffelSymbols[1][0][3], christoffelSymbols[2][0][3], christoffelSymbols[3][0][3]) * dxds[0] * v[3];
    dvds -= vec4(christoffelSymbols[0][1][0], christoffelSymbols[1][1][0], christoffelSymbols[2][1][0], christoffelSymbols[3][1][0]) * dxds[1] * v[0];
    dvds -= vec4(christoffelSymbols[0][1][1], christoffelSymbols[1][1][1], christoffelSymbols[2][1][1], christoffelSymbols[3][1][1]) * dxds[1] * v[1];
    dvds -= vec4(christoffelSymbols[0][1][2], christoffelSymbols[1][1][2], christoffelSymbols[2][1][2], christoffelSymbols[3][1][2]) * dxds[1] * v[2];
    dvds -= vec4(christoffelSymbols[0][1][3], christoffelSymbols[1][1][3], christoffelSymbols[2][1][3], christoffelSymbols[3][1][3]) * dxds[1] * v[3];
    dvds -= vec4(christoffelSymbols[0][2][0], christoffelSymbols[1][2][0], christoffelSymbols[2][2][0], christoffelSymbols[3][2][0]) * dxds[2] * v[0];
    dvds -= vec4(christoffelSymbols[0][2][1], christoffelSymbols[1][2][1], christoffelSymbols[2][2][1], christoffelSymbols[3][2][1]) * dxds[2] * v[1];
    dvds -= vec4(christoffelSymbols[0][2][2], christoffelSymbols[1][2][2], christoffelSymbols[2][2][2], christoffelSymbols[3][2][2]) * dxds[2] * v[2];
    dvds -= vec4(christoffelSymbols[0][2][3], christoffelSymbols[1][2][3], christoffelSymbols[2][2][3], christoffelSymbols[3][2][3]) * dxds[2] * v[3];
    dvds -= vec4(christoffelSymbols[0][3][0], christoffelSymbols[1][3][0], christoffelSymbols[2][3][0], christoffelSymbols[3][3][0]) * dxds[3] * v[0];
    dvds -= vec4(christoffelSymbols[0][3][1], christoffelSymbols[1][3][1], christoffelSymbols[2][3][1], christoffelSymbols[3][3][1]) * dxds[3] * v[1];
    dvds -= vec4(christoffelSymbols[0][3][2], christoffelSymbols[1][3][2], christoffelSymbols[2][3][2], christoffelSymbols[3][3][2]) * dxds[3] * v[2];
    dvds -= vec4(christoffelSymbols[0][3][3], christoffelSymbols[1][3][3], christoffelSymbols[2][3][3], christoffelSymbols[3][3][3]) * dxds[3] * v[3];
    return dvds;
}
float Doppler(in vec3 photonDirection, in vec3 velocity) {
    float lorentzFactor = 1.0 / sqrt(1.0 - dot(velocity, velocity));
    float doppler = lorentzFactor * (1.0 + dot(photonDirection, velocity));
}
Simulated Schwarzschild Black Hole.png
 
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  • #2
Well, that looks like an image of a Kerr black hole with an accretion disc. Using Christoffel symbols from Schwarzschild spacetime is not going to give you correct answers in a Kerr spacetime...
 
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  • #3
Ibix said:
Well, that looks like an image of a Kerr black hole with an accretion disc. Using Christoffel symbols from Schwarzschild spacetime is not going to give you correct answers in a Kerr spacetime...
I can assure you it is not a Kerr black hole. I am in the process of implementing the Christoffel symbols for a Kerr black hole, but that simulation is using the Schwarzschild metric for the spacetime and the symbols.
 
  • #4
I could not figure out how to edit the post, so I will say it here. I forgot to mention something, specifically that I tried multiplying the Christoffel symbols by 2 and that appears to fix it. But that multiply feels wrong.
 
  • #5
I would have expected an image in a Schwarzschild metric to show left-right symmetry. But perhaps you are also simulating the emissions from a rotating accretion disc and the Doppler effects of that are responsible for the asymmetry?
 
  • #6
Ibix said:
I would have expected an image in a Schwarzschild metric to show left-right symmetry. But perhaps you are also simulating the emissions from a rotating accretion disc and the Doppler effects of that are responsible for the asymmetry?
That is correct. My apologies for not making that clear in the original post.
 
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  • #7
What's the functional form of your metric? Is it:
$$\pmatrix{1-\frac{R_S}r&0&0&0\cr
0&-1-{{R_Sx^2}\over{r^2(r-R_S)}}&-{{R_Sxy}\over{r^2(r-R_S)}}&-{{R_Sxz}\over{r^2(r-R_S)}}\cr
0&-{{R_Sxy}\over{r^2(r-R_S)}}&-1-{{R_Sy^2}\over{r^2(r-R_S)}}&-{{R_Syz}\over{r^2(r-R_S)}}\cr0&-{{R_Sxz}\over{r^2(r-R_S)}}&-{{R_Syz}\over{r^2(r-R_S)}}&-1-{{R_Sz^2}\over{r^2(r-R_S)}}\cr}$$(Did this on the train - errors are possible...)
 
  • #8
Ibix said:
What's the functional form of your metric? Is it:
$$\pmatrix{1-\frac{R_S}r&0&0&0\cr
0&-1-{{R_Sx^2}\over{r^2(r-R_S)}}&-{{R_Sxy}\over{r^2(r-R_S)}}&-{{R_Sxz}\over{r^2(r-R_S)}}\cr
0&-{{R_Sxy}\over{r^2(r-R_S)}}&-1-{{R_Sy^2}\over{r^2(r-R_S)}}&-{{R_Syz}\over{r^2(r-R_S)}}\cr0&-{{R_Sxz}\over{r^2(r-R_S)}}&-{{R_Syz}\over{r^2(r-R_S)}}&-1-{{R_Sz^2}\over{r^2(r-R_S)}}\cr}$$(Did this on the train - errors are possible...)
1671127477248.png

Kind of. It is this. f is Rs / r. I even tested it to make sure the equation is correct for the Schwarzschild metric in cartesian coordinates, well technically Kerr-Schild but that is just a cartesian coordinate system. I derived the metric from Reissner-Nordström in Kerr-Schild coordinates, as that becomes Schwarzschild if the charge is zero.
 
  • #9
Ah. Mine is for "Cartesianised" Schwarzschild coordinates, which correspond to Boyer-Lindquist coordinates in the Kerr metric with zero angular momentum. I'll try starting with Kerr-Schild later.

This is why you should always specify exactly what coordinates you're using...
 
  • #10
I haven't bothered re-deriving your metric. I've just confirmed that it's spherically symmetric, vacuum, and has a time-like Killing vector field. Thus by Birkhoff's theorem it's the Schwarzschild metric, so I don't see any mistakes there.

The Christoffel symbols I get from it are rather messy and I've only made limited efforts to spot the ones that are the same. You can them a few against your code and check they're the same:
$$\begin{eqnarray*}
\Gamma^t_{tt}&=&{{R_S^2}\over{2r^3}}\\
\Gamma^t_{tx}=\Gamma^t_{xt}&=&{{\left(R_Sr+R_S^2\right)x}\over{2r^4}}\\
\Gamma^t_{ty}=\Gamma^t_{yt}&=&{{\left(R_Sr+R_S^2\right)y}\over{2r^4}}\\
\Gamma^t_{tz}=\Gamma^t_{zt}&=&{{\left(R_Sr+R_S^2\right)z}\over{2r^4}}\\
\Gamma^t_{xx}&=&{{\left(4R_Sr+R_S^2\right)x^2-2R_Sr^3 }\over{2r^5}}\\
\Gamma^t_{xy}=\Gamma^t_{yx}&=&{{\left(4R_Sr+R_S^2\right)xy}\over{2r^5}}\\
\Gamma^t_{xz}=\Gamma^t_{zx}&=&{{\left(4R_Sr+R_S^2\right)xz}\over{2r^5}}\\
\Gamma^t_{yy}&=&{{\left(4R_Sr+R_S^2\right)y^2-2R_Sr^3 }\over{2r^5}}\\
\Gamma^t_{yz}=\Gamma^t_{zy}&=&{{\left(4R_Sr+R_S^2\right)yz}\over{2r^5}}\\
\Gamma^t_{zz}&=&{{\left(4R_Sr+R_S^2\right)z^2-2R_Sr^3 }\over{2r^5}}\\
\Gamma^x_{tt}&=&{{\left(R_Sr-R_S^2\right)x}\over{2r^4}}\\
\Gamma^x_{tx}=\Gamma^x_{xt}&=&-{{R_S^2x^2}\over{2r^5}}\\
\Gamma^x_{xx}&=&-{{\left(3R_Sr+R_S^2\right)x^3-2R_Sr^3x }\over{2r^6}}\\
\Gamma^x_{xy}=\Gamma^x_{yx}&=&-{{\left(3R_Sr+R_S^2\right)x^2y}\over{2r^6}}\\
\Gamma^x_{xz}=\Gamma^x_{zx}&=&-{{\left(3R_Sr+R_S^2\right)x^2z}\over{2r^6}}\\
\Gamma^x_{yy}&=&-{{\left(3R_Sr+R_S^2\right)xy^2-2R_Sr^3 x}\over{2r^6}}\\
\Gamma^x_{zz}&=&-{{\left(3R_Sr+R_S^2\right)xz^2-2R_Sr^3 x}\over{2r^6}}\\
\Gamma^y_{tt}&=&{{\left(R_Sr-R_S^2\right)y}\over{2r^4}}\\
\Gamma^y_{tx}=\Gamma^y_{xt}=\Gamma^x_{ty}=\Gamma^x_{yt}&=&-{{R_S^2xy}\over{2r^5}}\\
\Gamma^y_{ty}=\Gamma^y_{yt}&=&-{{R_S^2y^2}\over{2r^5}}\\
\Gamma^y_{xx}&=&-{{\left(\left(3R_Sr+R_S^2\right)x^2-2R_Sr ^3\right)y}\over{2r^6}}\\
\Gamma^y_{xy}=\Gamma^y_{yx}&=&-{{\left(3R_Sr+R_S^2\right)xy^2}\over{2r^6}}\\
\Gamma^y_{yy}&=&-{{\left(3R_Sr+R_S^2\right)y^3-2R_Sr^3y }\over{2r^6}}\\
\Gamma^y_{yz}=\Gamma^y_{zy}&=&-{{\left(3R_Sr+R_S^2\right)y^2z}\over{2r^6}}\\
\Gamma^y_{zz}&=&-{{\left(3R_Sr+R_S^2\right)yz^2-2R_Sr^3 y}\over{2r^6}}\\
\Gamma^z_{tt}&=&{{\left(R_Sr-R_S^2\right)z}\over{2r^4}}\\
\Gamma^z_{tx}=\Gamma^z_{xt}=\Gamma^x_{tz}=\Gamma^x_{zt}&=&-{{R_S^2xz}\over{2r^5}}\\
\Gamma^z_{ty}=\Gamma^z_{yt}=\Gamma^y_{tz}=\Gamma^y_{zt}&=&-{{R_S^2yz}\over{2r^5}}\\
\Gamma^z_{tz}=\Gamma^z_{zt}&=&-{{R_S^2z^2}\over{2r^5}}\\
\Gamma^z_{xx}&=&-{{\left(\left(3R_Sr+R_S^2\right)x^2-2R_Sr ^3\right)z}\over{2r^6}}\\
\Gamma^z_{xy}=\Gamma^z_{yx}=\Gamma^y_{xz}=\Gamma^y_{zx}=\Gamma^x_{yz}=\Gamma^x_{zy}&=&-{{\left(3R_Sr+R_S^2\right)xyz}\over{2r^6}}\\
\Gamma^z_{xz}=\Gamma^z_{zx}&=&-{{\left(3R_Sr+R_S^2\right)xz^2}\over{2r^6}}\\
\Gamma^z_{yy}&=&-{{\left(\left(3R_Sr+R_S^2\right)y^2-2R_Sr ^3\right)z}\over{2r^6}}\\
\Gamma^z_{yz}=\Gamma^z_{zy}&=&-{{\left(3R_Sr+R_S^2\right)yz^2}\over{2r^6}}\\
\Gamma^z_{zz}&=&-{{\left(3R_Sr+R_S^2\right)z^3-2R_Sr^3z }\over{2r^6}}
\end{eqnarray*}$$Hint: quote my post to see the LaTeX source, paste it into a code editor, and search and replace will get you most of the way to valid code in your language of choice.
 
  • #11
Ibix said:
I haven't bothered re-deriving your metric. I've just confirmed that it's spherically symmetric, vacuum, and has a time-like Killing vector field. Thus by Birkhoff's theorem it's the Schwarzschild metric, so I don't see any mistakes there.

The Christoffel symbols I get from it are rather messy and I've only made limited efforts to spot the ones that are the same. You can them a few against your code and check they're the same:
$$\begin{eqnarray*}
\Gamma^t_{tt}&=&{{R_S^2}\over{2r^3}}\\
\Gamma^t_{tx}=\Gamma^t_{xt}&=&{{\left(R_Sr+R_S^2\right)x}\over{2r^4}}\\
\Gamma^t_{ty}=\Gamma^t_{yt}&=&{{\left(R_Sr+R_S^2\right)y}\over{2r^4}}\\
\Gamma^t_{tz}=\Gamma^t_{zt}&=&{{\left(R_Sr+R_S^2\right)z}\over{2r^4}}\\
\Gamma^t_{xx}&=&{{\left(4R_Sr+R_S^2\right)x^2-2R_Sr^3 }\over{2r^5}}\\
\Gamma^t_{xy}=\Gamma^t_{yx}&=&{{\left(4R_Sr+R_S^2\right)xy}\over{2r^5}}\\
\Gamma^t_{xz}=\Gamma^t_{zx}&=&{{\left(4R_Sr+R_S^2\right)xz}\over{2r^5}}\\
\Gamma^t_{yy}&=&{{\left(4R_Sr+R_S^2\right)y^2-2R_Sr^3 }\over{2r^5}}\\
\Gamma^t_{yz}=\Gamma^t_{zy}&=&{{\left(4R_Sr+R_S^2\right)yz}\over{2r^5}}\\
\Gamma^t_{zz}&=&{{\left(4R_Sr+R_S^2\right)z^2-2R_Sr^3 }\over{2r^5}}\\
\Gamma^x_{tt}&=&{{\left(R_Sr-R_S^2\right)x}\over{2r^4}}\\
\Gamma^x_{tx}=\Gamma^x_{xt}&=&-{{R_S^2x^2}\over{2r^5}}\\
\Gamma^x_{xx}&=&-{{\left(3R_Sr+R_S^2\right)x^3-2R_Sr^3x }\over{2r^6}}\\
\Gamma^x_{xy}=\Gamma^x_{yx}&=&-{{\left(3R_Sr+R_S^2\right)x^2y}\over{2r^6}}\\
\Gamma^x_{xz}=\Gamma^x_{zx}&=&-{{\left(3R_Sr+R_S^2\right)x^2z}\over{2r^6}}\\
\Gamma^x_{yy}&=&-{{\left(3R_Sr+R_S^2\right)xy^2-2R_Sr^3 x}\over{2r^6}}\\
\Gamma^x_{zz}&=&-{{\left(3R_Sr+R_S^2\right)xz^2-2R_Sr^3 x}\over{2r^6}}\\
\Gamma^y_{tt}&=&{{\left(R_Sr-R_S^2\right)y}\over{2r^4}}\\
\Gamma^y_{tx}=\Gamma^y_{xt}=\Gamma^x_{ty}=\Gamma^x_{yt}&=&-{{R_S^2xy}\over{2r^5}}\\
\Gamma^y_{ty}=\Gamma^y_{yt}&=&-{{R_S^2y^2}\over{2r^5}}\\
\Gamma^y_{xx}&=&-{{\left(\left(3R_Sr+R_S^2\right)x^2-2R_Sr ^3\right)y}\over{2r^6}}\\
\Gamma^y_{xy}=\Gamma^y_{yx}&=&-{{\left(3R_Sr+R_S^2\right)xy^2}\over{2r^6}}\\
\Gamma^y_{yy}&=&-{{\left(3R_Sr+R_S^2\right)y^3-2R_Sr^3y }\over{2r^6}}\\
\Gamma^y_{yz}=\Gamma^y_{zy}&=&-{{\left(3R_Sr+R_S^2\right)y^2z}\over{2r^6}}\\
\Gamma^y_{zz}&=&-{{\left(3R_Sr+R_S^2\right)yz^2-2R_Sr^3 y}\over{2r^6}}\\
\Gamma^z_{tt}&=&{{\left(R_Sr-R_S^2\right)z}\over{2r^4}}\\
\Gamma^z_{tx}=\Gamma^z_{xt}=\Gamma^x_{tz}=\Gamma^x_{zt}&=&-{{R_S^2xz}\over{2r^5}}\\
\Gamma^z_{ty}=\Gamma^z_{yt}=\Gamma^y_{tz}=\Gamma^y_{zt}&=&-{{R_S^2yz}\over{2r^5}}\\
\Gamma^z_{tz}=\Gamma^z_{zt}&=&-{{R_S^2z^2}\over{2r^5}}\\
\Gamma^z_{xx}&=&-{{\left(\left(3R_Sr+R_S^2\right)x^2-2R_Sr ^3\right)z}\over{2r^6}}\\
\Gamma^z_{xy}=\Gamma^z_{yx}=\Gamma^y_{xz}=\Gamma^y_{zx}=\Gamma^x_{yz}=\Gamma^x_{zy}&=&-{{\left(3R_Sr+R_S^2\right)xyz}\over{2r^6}}\\
\Gamma^z_{xz}=\Gamma^z_{zx}&=&-{{\left(3R_Sr+R_S^2\right)xz^2}\over{2r^6}}\\
\Gamma^z_{yy}&=&-{{\left(\left(3R_Sr+R_S^2\right)y^2-2R_Sr ^3\right)z}\over{2r^6}}\\
\Gamma^z_{yz}=\Gamma^z_{zy}&=&-{{\left(3R_Sr+R_S^2\right)yz^2}\over{2r^6}}\\
\Gamma^z_{zz}&=&-{{\left(3R_Sr+R_S^2\right)z^3-2R_Sr^3z }\over{2r^6}}
\end{eqnarray*}$$Hint: quote my post to see the LaTeX source, paste it into a code editor, and search and replace will get you most of the way to valid code in your language of choice.
Thank you, I appreciate the assistance. I will implement those and let you know the results. I do have a question though. What did you use to calculate them? I want to know so I can calculate them in a similar, or even the same, way for other metrics, since I do need the Christoffel symbols for the Kerr metric and a few other metrics.
 
Last edited:
  • #12
Maxima. It's a free symbolic algebra package and its ctensor library (plus a bit of ingenuity in simplifying things) will spit out what you need. It has a fairly steep learning curve, but check out my Insight article on it (tap/click the orange Insight Author badge in the header of my posts) and former PF Mentor Ben Crowell's free online GR book (www.lightandmatter.com/genrel) for some code. I can post what I used to generate the above tomorrow.
 
  • #13
Ibix said:
Maxima. It's a free symbolic algebra package and its ctensor library (plus a bit of ingenuity in simplifying things) will spit out what you need. It has a fairly steep learning curve, but check out my Insight article on it (tap/click the orange Insight Author badge in the header of my posts) and former PF Mentor Ben Crowell's free online GR book (www.lightandmatter.com/genrel) for some code. I can post what I used to generate the above tomorrow.
I actually installed Maxima before Mathematica since I do not have $300 to spend on a math program, but I could not figure out how to use Maxima, due to exhaustion from bare sleeping the nights before, so I got the free trial of Mathematica. However, once the trial runs out I shall start using Maxima so I can continue this stuff. Once again, thank you for the assistance and help. It is very much appreciated.
 
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  • #14
Here's the Maxima code. Sorry it took me longer to get to pasting this in than I thought it would - it needed a bit of cleaning, but I took the opportunity to add some comments. Hope it's helpful!
Code: 
load(ctensor);

/* Define the metric (lower indices) */
dim:4;
ct_coords:[t,x,y,z];
f:Rs/r;
lg:matrix([-1+f,f*x/r,f*y/r,f*z/r],
		[f*x/r,1+f*x^2/r^2,f*x*y/r^2,f*x*z/r^2],
		[f*y/r,f*x*y/r^2,1+f*y^2/r^2,f*y*z/r^2],
		[f*z/r,f*x*z/r^2,f*y*z/r^2,1+f*z^2/r^2]);

/* Manifestly symmetric under interchange of spatial coordinates. Check that */
/* it's symmetric under rotation about x - that is enough to confirm that    */
/* it's spherically symmetric.                                               */
R:matrix([1,0,0,0],
          [0,1,0,0],
          [0,0,cos(psi),sin(psi)],
          [0,0,-sin(psi),cos(psi)]);

lgr:R.lg.transpose(R);
trigsimp(substitute([y=yp*cos(psi)-zp*sin(psi),
			z=yp*sin(psi)+zp*cos(psi)],lgr));

/* Grind through to the Christoffel symbols */
depends(r,[x,y,z]);     /* Warn Maxima that r is a function of x, y, and z. */
cmetric(false);         /* Compute the inverse metric. */
christof(false);        /* Compute the Christoffel symbols. */

/* Simplify the Christoffel symbols. Replace dr/dx with x/r (etc) and      */
/* x^2+y^2+z^2 with r^2. Unfortunately the latter means a bit of mucking   */
/* around to make sure we don't replace an isolated x^2, y^2 or z^2. So we */
/* try replacing x^2 with r^2-y^2-z^2, then try y^2 then z^2, and keep the */
/* shortest expression.                                                    */
diffDefs:[diff(r,x)=x/r, diff(r,y)=y/r, diff(r,z)=z/r];
for i:1 thru 4 do block (
  for j:1 thru 4 do block (
    for k:1 thru 4 do block (
      [temp,shortest],
      shortest:ratsimp(subst([x^2=r^2-y^2-z^2],subst(diffDefs,mcs[i,j,k]))),
      temp:ratsimp(subst([y^2=r^2-x^2-z^2],subst(diffDefs,mcs[i,j,k]))),
      if length(args(expand(temp))) < length(args(expand(shortest))) then block(
        shortest:temp
      ),
      temp:ratsimp(subst([z^2=r^2-x^2-y^2],subst(diffDefs,mcs[i,j,k]))),
      if length(args(expand(temp))) < length(args(expand(shortest))) then block(
        shortest:temp
      ),
      mcs[i,j,k]:shortest,
      /* Print Christoffel symbols in tex form */
      print("\\Gamma^",ct_coords[k],"_{",ct_coords[i],ct_coords[j],"}="),
      tex(mcs[i,j,k])
    )
  )
);

/* Finally, generate and simplify the Ricci tensor to check this is a vacuum */
ricci(false);
for i:1 thru 4 do block (
  for j:1 thru 4 do block (
    ric[i,j]: radcan(substitute(diffDefs,ric[i,j])),
    ric[i,j]: radcan(substitute([r=sqrt(x^2+y^2+z^2)],ric[i,j]))
  )
);
cdisplay(ric);
I tend to use Maxima because I know more or less what I'm doing with it. Another thing you could look into is the python SymPy symbolic algebra package. I've been meaning to learn how to use it for a while (it's a lot more modern than Maxima) but never get round to it, but it might be a better place to start if you aren't overly invested in Maxima.
 
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  • #16
Ibix said:
Here's the Maxima code. Sorry it took me longer to get to pasting this in than I thought it would - it needed a bit of cleaning, but I took the opportunity to add some comments. Hope it's helpful!
Code: 
load(ctensor);

/* Define the metric (lower indices) */
dim:4;
ct_coords:[t,x,y,z];
f:Rs/r;
lg:matrix([-1+f,f*x/r,f*y/r,f*z/r],
        [f*x/r,1+f*x^2/r^2,f*x*y/r^2,f*x*z/r^2],
        [f*y/r,f*x*y/r^2,1+f*y^2/r^2,f*y*z/r^2],
        [f*z/r,f*x*z/r^2,f*y*z/r^2,1+f*z^2/r^2]);

/* Manifestly symmetric under interchange of spatial coordinates. Check that */
/* it's symmetric under rotation about x - that is enough to confirm that    */
/* it's spherically symmetric.                                               */
R:matrix([1,0,0,0],
          [0,1,0,0],
          [0,0,cos(psi),sin(psi)],
          [0,0,-sin(psi),cos(psi)]);

lgr:R.lg.transpose(R);
trigsimp(substitute([y=yp*cos(psi)-zp*sin(psi),
            z=yp*sin(psi)+zp*cos(psi)],lgr));

/* Grind through to the Christoffel symbols */
depends(r,[x,y,z]);     /* Warn Maxima that r is a function of x, y, and z. */
cmetric(false);         /* Compute the inverse metric. */
christof(false);        /* Compute the Christoffel symbols. */

/* Simplify the Christoffel symbols. Replace dr/dx with x/r (etc) and      */
/* x^2+y^2+z^2 with r^2. Unfortunately the latter means a bit of mucking   */
/* around to make sure we don't replace an isolated x^2, y^2 or z^2. So we */
/* try replacing x^2 with r^2-y^2-z^2, then try y^2 then z^2, and keep the */
/* shortest expression.                                                    */
diffDefs:[diff(r,x)=x/r, diff(r,y)=y/r, diff(r,z)=z/r];
for i:1 thru 4 do block (
  for j:1 thru 4 do block (
    for k:1 thru 4 do block (
      [temp,shortest],
      shortest:ratsimp(subst([x^2=r^2-y^2-z^2],subst(diffDefs,mcs[i,j,k]))),
      temp:ratsimp(subst([y^2=r^2-x^2-z^2],subst(diffDefs,mcs[i,j,k]))),
      if length(args(expand(temp))) < length(args(expand(shortest))) then block(
        shortest:temp
      ),
      temp:ratsimp(subst([z^2=r^2-x^2-y^2],subst(diffDefs,mcs[i,j,k]))),
      if length(args(expand(temp))) < length(args(expand(shortest))) then block(
        shortest:temp
      ),
      mcs[i,j,k]:shortest,
      /* Print Christoffel symbols in tex form */
      print("\\Gamma^",ct_coords[k],"_{",ct_coords[i],ct_coords[j],"}="),
      tex(mcs[i,j,k])
    )
  )
);

/* Finally, generate and simplify the Ricci tensor to check this is a vacuum */
ricci(false);
for i:1 thru 4 do block (
  for j:1 thru 4 do block (
    ric[i,j]: radcan(substitute(diffDefs,ric[i,j])),
    ric[i,j]: radcan(substitute([r=sqrt(x^2+y^2+z^2)],ric[i,j]))
  )
);
cdisplay(ric);
I tend to use Maxima because I know more or less what I'm doing with it. Another thing you could look into is the python SymPy symbolic algebra package. I've been meaning to learn how to use it for a while (it's a lot more modern than Maxima) but never get round to it, but it might be a better place to start if you aren't overly invested in Maxima.
I do intend on learning how to use SymPy, and actually started this entire journey with the Christoffel symbols by trying to learn it. I gave up and went for something simpler after seeing how much of a pain it would be to convert the output to GLSL code. I do however plan on going back to it eventually, just not now. Thank you for the Maxima code btw, it is very helpful. And the comments definitely help with that.
 
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FAQ: Doppler Shift & Christoffel Symbols Issues

What is the Doppler shift and how does it relate to Christoffel symbols?

The Doppler shift is a phenomenon in which the frequency of a wave, such as light or sound, appears to change when the source of the wave is moving relative to the observer. This effect is described by the Christoffel symbols, which are mathematical objects used to calculate the change in frequency due to the relative motion of the source and observer.

How is the Doppler shift used in scientific research?

The Doppler shift is used in a variety of scientific fields, including astronomy, meteorology, and oceanography. In astronomy, it is used to measure the velocity and distance of celestial objects. In meteorology, it is used to track the movement of weather systems. In oceanography, it is used to study ocean currents and the movement of marine life.

What is the difference between the Doppler shift and the relativistic Doppler effect?

The Doppler shift is a classical effect that describes the change in frequency of a wave due to relative motion. The relativistic Doppler effect takes into account the effects of special relativity, such as time dilation and length contraction, on the observed frequency. It is used to describe the behavior of waves traveling at high speeds, such as light.

Can the Doppler shift be used to measure the speed of objects?

Yes, the Doppler shift can be used to measure the speed of objects, as long as the relative motion between the source and observer is known. In astronomy, for example, the Doppler shift of light from a star can be used to determine its velocity towards or away from Earth.

How do Christoffel symbols affect the accuracy of Doppler shift measurements?

Christoffel symbols are important in accurately calculating the Doppler shift, as they take into account the effects of the curvature of space-time. Without considering these effects, Doppler shift measurements may be inaccurate, especially when dealing with objects moving at high speeds or in strong gravitational fields.

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