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Idk where, but at some point i heared the Event Horizons of a Schwarzschild black hole and Darkstar coincidentally are the same size. I have since then done a lot of stuff in GR, including Kerr black holes and wanted to revisit the Darkstar. I found the Event Horizon to actually equal $r_d = \frac{GM}{c}$

Which is of course half of a Schwarzschild Black Hole, being $r_s = \frac{2GM}{c^2}$ if memory serves me correctly. It is of course possible my "derivation" for the Darkstar is just wrong. In Cartesian coordinates my photon direction equation is;

$$\overrightarrow{d} = ||\overrightarrow{rayDir}+\left(\frac{||\overrightarrow{rayOrig}||\frac{GM}{r^2}h}{c}\right)||$$

Before yall get the pitchforks out, yes i am mixing vectors and numbers, i am not great at math. I have tried to take this equation into its component form, but it didnt work, i did something wrong.

I should note, $h$ is the step size, while $\overrightarrow{rayDir}$ is the normalized direction and $\overrightarrow{rayOrig}$ the ray position vector, both in cartesian coordinates.

My derivation, if you can call it that, is really more logic based than anything. The basic idea is to calculate the "Momentum" of our photon which is done with $\frac{||\overrightarrow{rayOrig}||\frac{GM}{r^2}h}{c}$. Momentum is direction * magnitude, so since the BH is at 0,0,0 the direction is just the current normalized position, or $||\overrightarrow{rayOrig}||$. The magnitude is the gravitational potential, $\frac{GM}{r^2}$, factored with the temporal step $h$. The $\frac{x}{c}$ bit is probably the most ad hoc aspect because my justification for it is that it checks out. If the speed of light is say double, the Horizon is half as big.

The whole party is updated as such;

$\overrightarrow{rayDir} = \overrightarrow{d}$

$\overrightarrow{rayOrig} += \overrightarrow{d}h$

which works out to be a slightly wrong version of Velocity verlet i think.

Just experimentally, i know at least for this formulation $r_d = \frac{GM}{c}$ is correct as can be seen here;

enter image description here

You can see 36 rays being shot parallel to the Horizon. The Horizon is marked with the red ring for $G=M=c=1$. This is also kinda in line with a Vox video where they said the deflection of light predicted by Newtonian mechanics is half of GR´s.

So whilst this appears to be true on a "experimental" level, i kinda struggle to see where the $\frac{GM}{c}$ comes from. Or if this is a case of me being right, for the wrong reason. My question(s) boil down to if this is a sensible approach and if my results are "true", within the bounds of what constitutes true for a Darkstar. And like maybe where $r_d = \frac{GM}{c}$ comes from.

On the side, if we actually compare the Darkstar to Schwarzschild (Kerr metric with $a = 0, M = 1, G = 1$) they are remarkably similar, if $M = 2$ for the Darkstar.

enter image description here

Alright then, thanks for reading and hopefully my math did not violate to many human rights.

EDIT;

I am working on a Visualization of what i am doing in the program, exactly. In the meantime a lot of the discussion in the comments seems to be focused around the escape velocity. Which as far as i can see is not relevant here because by normalizing everything the velocity is constant.

ErikHall
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  • Related/worth checking: Is relativistic event horizon half of Newtonian event horizon? See also this answer. – Quillo Oct 04 '23 at 12:28
  • @Quillo , thanks for the link. I should have probably included a section where i explain that i know any Darkstar is by design arbitrarily because c is not a speed limit. You have to artificially reinforce that, here using the normalization stuff. Still i think it is at least interesting to consider since last time i checked, Newton does not actually say photons can go faster than c. And didnt they know back then that the speed of light was constant ? – ErikHall Oct 04 '23 at 12:34
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    Honestly, your logic is not very clear, but I think that what you've shown is that the deflection of light works as if the black hole has half the mass - which is correct, since the Newtonian deflection is half the relativistic one. But the "event horizon" is still at $2GM/c^2$, as can be shown by just using the escape velocity formula. – Javier Oct 04 '23 at 12:43
  • In re notation: one can write $\hat{\mathbf{x}}$ to indicate the direction (i.e., "unit vector") and $\mathbf{x}$ the vector. – Kyle Kanos Oct 04 '23 at 12:48
  • @Javier - the Newtonian deflection is twice in the weak field, not close to the horizon. For a correct comparison Schwarzschild vs Newton's corpuscel theory in the strong field see here – Yukterez Oct 04 '23 at 13:05
  • @Javier sorry if it is not clear. If the deflection is half though, how can the "Horizon" be the same size ? – ErikHall Oct 04 '23 at 13:05
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    Since the radial escape velocity is the same under Schwarzschild and Newton it gives the speed of light at r=2GM/c² for both. Under Newton the escape velocity is independed of direction though, while under Schwarzschild a transverse lightray will already get caught below the photon sphere at r<3GM/c² – Yukterez Oct 04 '23 at 13:08
  • Does this answer your question? Is relativistic event horizon half of Newtonian event horizon? – Dale Oct 04 '23 at 13:34
  • Also, half the deflection angle (which is only valid in the weak file anyway) is not the same as the same deflection angle at the half radius since we don't have 1/r but 1/r². – Yukterez Oct 04 '23 at 13:35
  • According to your last edit you normalize the velocity to be constant even under Newton, which can not be done for his corspuscel theory, with that even photons accelerate. Constant photon speed is not Newton, but some mix of Newton with special relativity or ether, but that won't work or at least there's no unique way to do it to which you could compare your results. – Yukterez Oct 04 '23 at 13:37
  • With Newton's corpuscels, the light ray that starts at r=2 with v=1 should have decelerated to v=0 when he arrives at r=∞, if you normalize you get v=1 even at r=∞ for an escaping photon, which violates the Newtonian definition of the escape velocity and energy conservation. – Yukterez Oct 04 '23 at 13:42
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    IMO, this question is quite problematic. Where it is not a duplicate of the linked question it seems to be a personal theory. IMO, the question needs to be heavily edited to make it clear how it differs from the linked question, and to cite sources that show it is not based on personal theory. Especially needed is a mainstream science reference to the definition of a "darkstar" – Dale Oct 04 '23 at 13:43
  • @Yukterez i see so the weird r = GM/c thingy for the Horizon is due to me in essence doing it wrong. Since the math is layed out such that if h = 1, the distance at each step is also 1 no matter how far away the photon is. The speed is always constant. Thanks for pointing these errors out – ErikHall Oct 04 '23 at 13:46
  • @Dale , idk if i would go as far as say personal theory. The question was more related to what i did wrong to get the GM/c result. I am not proclaiming any of this to be true, rather that what i got matches what others get in some cases, but not in others. Naimly the deflection being half of what GR says, but the Horizon being GM/c instead of 2GM/c² – ErikHall Oct 04 '23 at 13:47
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    Since you speak german and the thread is already closed I might as well link my code which works with corpuscels that start with v=c relative to the source and accelerate afterwards in the Newtonian field, the Newtonian simulator is the first code on the site: https://notizblock.yukterez.net/viewtopic.php?t=15 – Yukterez Oct 04 '23 at 13:51
  • @ErikHall 1) I have never heard of a "Newtonian darkstar". So regardless of if it is a personal theory or not it needs some clear definitions from the mainstream scientific literature. As it is the "darkstar" topic sounds like some pseudo-science nonsense, not mainstream science. Certainly I don't know everything, so it could be legitimate science that I just don't know. But if I don't know it then it is also uncommon enough to need some details, like a mainstream scientific definition. 2) "check my work" questions are also off-topic here also – Dale Oct 04 '23 at 14:26
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    @Dale i should have included a link, ill concede that. But ill not get dragged into Pseudo-Science land because someone didnt know a concept, considering it is not that hard to find; https://en.wikipedia.org/wiki/Dark_star_(Newtonian_mechanics) . Furthermore, my question was not on the validity of the concept but what i did wrong to get conflicting results. – ErikHall Oct 04 '23 at 14:45
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    @ErikHall fair enough. Such a reference is sufficient. However, what you did wrong is a "check my work" question which is still off topic here. – Dale Oct 04 '23 at 14:55
  • @ErikHall The spatial part of the Schwarzschild radius is zero and thus is not comparable to the Newtonian. What you refer to as $r_s$ is not the radius, but the reduced circumference of the horizon. Not the same thing. Not even of the same type, because the Schwarzschild radial interval is timelike while the reduced circumference is spacelike. – safesphere Oct 06 '23 at 03:30

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