# Thread: How come Light can not escape a Black hole

1. Originally Posted by Jilan
OK, I think I'm getting there. So am I following this right: If I am standing at the event horizon and shine my torch up the light will initially go straight up and then start veering off following a geodesic.
Nope, it will go straight up.

If I am slightly inside the event horizon this geodesic will eventually return to the BH
Yes.

and if I am slightly outside it will still veer off and curve but the geodesic never brings it back?
Nope, it is much more complicated than that. The trajectory may be radial or it may be parabolic or it may be a circle.
You need to study GR and grok the math. It is all described in the "Photon trajectories in a gravitational field" chapter(s). Any good book on GR has it. I linked an excellent free chapter on the subject a few posts before. I suggest that you read it.

2. Originally Posted by Markus Hanke
MODERATOR NOTE : Right Farsight, I think this has gone far enough. I don't know if you really thought that you were going to fool anyone here into believing that any of what you present has any relation to Einstein's relativity; I allowed you to post on this forum because I was hoping you were prepared to discuss this in earnest, and perhaps be man enough to see sense and admit you are wrong on at least some of these very basic points. As it turns out I was misguided - it started off well enough, but we are now just going around in circles, which is meaningless and a waste of time for everyone concerned, including yourself. My bad for giving you the benefit of the doubt. Let me use the example of our latest exchange on this thread to demonstrate just how far off the mark you really are. From Einstein's very own text "The meaning of relativity" :

(1) The condition ds=0, which you said has no connection to reality, is formula (22) on page 30, where Einstein discusses the propagation of light. You told me I used that condition in error - well then, so did Einstein.
(2) The geodesic equations are formula (90) on page 84, together with the discussion as to how and why these describe the trajectory of freely falling particles. You called this smoke and mirrors, and a rabbit out of a hat - well, Einstein was quite a magician, then.
(3) You keep going on about variable speeds of light being the meaning of GR - there is no mention of this whatsoever anywhere in Einstein's above referenced text. Funny, given the title of Einstein's book.

When you first started posting here again a few months back, I explicitly reiterated to you that this is a mainstream forum, and to please respect that. You didn't. Not only are you rejecting basic mathematics, you are now rejecting Einstein's very own words, distorting it to fit your personal ideas about what GR should be, not what it is. Heck, you are even rejecting basic calculus, such as the covariant derivative ! There can be no discussion about mainstream physics on that basis; there simply is no common ground here, if you just decide anything and everything we calculate for you to be meaningless. That's not physics.

So here is how it is going to be - we will be going back to the previous arrangement, and you will henceforth restrict yourself to posting exclusively in the "Personal Theories" section, because that is what your ideas are : personal theories. They have nothing to do with Einstein's relativity, or mainstream physics. In particular, you will not respond to questions posted in the main forums with your personal ideas - such posts will be moved immediately. To be perfectly honest, I see no future here for you - this forum is just not for you, it is for mainstream science. You are free to disagree and reject the mainstream and have your own opinions, but please don't expect anyone to take those seriously, and this forum is most certainly not the right platform to try and proliferate such ideas.

Don't bother sending me PMs - I can open them, but I cannot reply or compose new messages, due to the forum software issues we are experiencing, so don't waste your time. In any case, the options are clear - keep posting in personal theories, or leave. It's up to you now.
Markus: you are a moderator, you permit the most appalling abuse, and now you're censoring me claiming I reject Einstein, because I've challenged your absolutely inadequate "explanation" that is no explanation at all. Shame on you. You bring physics into disrepute.

3. Originally Posted by Farsight
Markus: you are a moderator, you permit the most appalling abuse, and now you're censoring me claiming I reject Einstein, because I've challenged your absolutely inadequate "explanation" that is no explanation at all.
The fact that you are grossly ignorant doesn't mean that the explanation is wrong, it simply means that you are an ignoramus pretending to be a "scientist"

Shame on you. You bring physics into disrepute.

"Mans reach should always exceed his grasp". I believe in this philosophy.

I reject the concept of "anyone who understands physics" should not pursue any alternative to accepted doctrine. I also reject the mind-set of "only we know about how the Universe works, and

no "outsider" should dare to question us". I also reject the concept of "calculus rules all things"...such a mind-set is invalid on it's face, a "closed-loop" thinking that enumerating an

intangible force such as gravity "defines all that can be known of it".

(Thanks for reading!) p.s. I think you should put me on "ignore". I'm far too contentious w/ regard to theory.

5. Originally Posted by Farsight
Markus: you are a moderator, you permit the most appalling abuse, and now you're censoring me claiming I reject Einstein, because I've challenged your absolutely inadequate "explanation" that is no explanation at all. Shame on you. You bring physics into disrepute.
You abuse this forum.

6. Originally Posted by Beer w/Straw
You abuse this forum.

Correct.

I wonder how many times I have read this exact same excuse of "We must protect students!" used on every site in existence, by the same trolls...so they can "justify" a nasty comment

or some ego trip of their own!!! So...someone who calls someone else a "liar" and an "a-hole" now rises up as a "legitimate source of criticism!!!"

Now that's funny!

Two "posters of pithy comments" who accuse someone else of being "unworthy!!!!!" (At least "Farsight" has something to say that just might have some value, as opposed to

"quoting the party line" as if it were "Holy Writ!!!")

.....

IF "all things are known" then why is this site or any other even in existence? There is "NOTHING TO DEBATE".

8. I am a student!

9. Originally Posted by x0x
Nope, it will go straight up.

Yes.

Nope, it is much more complicated than that. The trajectory may be radial or it may be parabolic or it may be a circle.
You need to study GR and grok the math. It is all described in the "Photon trajectories in a gravitational field" chapter(s). Any good book on GR has it. I linked an excellent free chapter on the subject a few posts before. I suggest that you read it.
Thanks xox I will grok it proper, sorry what number post was it in ? All this relativity business is quite new to me and very amazing. (I didn't have you down as a Heinlein fan! )

10. Originally Posted by Jilan
If I am standing at the event horizon and shine my torch up the light will initially go straight up and then start veering off following a geodesic.
Well, almost, but not quite. What happens is that the photon travels always "straight", but because space-time is curved, this concept of "straight" is not the same as the concept of "straight" utilised by a distant observer. You must remember here that we are talking about geodesics in space-time, and in the vicinity of a black hole, both measurements of time and distance are different than the ones performed somewhere for away. For example, for a one solar mass black hole, to get from a point at r=4km to a point r=5km ( a distance of 1km as determined by a far-away observer ) from the black hole, a photon has to actually travel 1723km of proper distance ! This discrepancy is just the failure of geodesics to be "straight" in the far-away observer sense, and that is the reason why that far away observer determines the coordinate speed of light to be much lower than c. His concept of "radially straight" just isn't the same concept of "straight" as in the immediate vicinity of the black hole. This does not mean that time stops at the event horizon, or that the radially emitted photon "veers off" the left/right/up/down, it means only that to get to the event horizon and back one has to undertake a journey that is a little more complicated than just a straight line in the Euclidean sense. Of course then if there is angular momentum involved, it gets even more complicated, and you get spiral-like patterns about the black hole ( see Wolfram Demonstrations project I referenced earlier ).

If I am slightly inside the event horizon this geodesic will eventually return to the BH
If you are slightly inside the event horizon, your radial coordinate becomes time-like, so travelling into the future means you travel away from the event horizon and towards the centre of the black hole - that is because there are no stationary observers there. Thus nothing can ever escape a black hole - because nothing can ever travel into the past, only into the future, and in the future lies the centre of the black hole.

If I am at the event horizon and shine my torch sideways the light achieves orbit at that radius.
Yes, this is called the photon sphere, and is found at r=3M ( geometrised units ).

Afterthought: if I stand on the earth and fire my torch upwards it also veers off but only slightly?
The same principle applies to the Earth or any other body, just to a much lesser degree. Null geodesics not being straight radial lines in the Euclidean sense ( i.e. in the sense of a very far away flat space-time observer ) leads to a slight delay in total travel time, as compared to the same scenario without the central mass. This is called the Shapiro delay, and can be directly measured. For the earth the effect would be very small.

11. 2 out of 4 is not bad. So given that the light from my torch is emitted upwards at the event horizon would it appear to me to escape or not?

Ps I am especially pleased that I got #3 right.

Very good...I got it.

I think your comments to Markus were uncalled for, like "shame on you, bringing physics into disrepute"...this is a cheap shot, and you know it.

I think has called your posits into question because "they just don't fit or work" with accepted mainstream (you know this yourself!) So why call him on it? I agree w/ put it into "personal

theories" just as would w/ my own stuff<(definitely NOT "mainstream!")

....

How do you blame a "Mod" or "Admin." for doing what they are supposed to do? (50% of the "blame" here is on you, you must know that)

....

I sincerely hope you keep POSTING!!! I like reading your stuff...whether I agree or not! (I don't like censorship, just because something is different doesn't mean it has no value)

As was said when I was young...Keep on Truckin', baby!

14. 2 out of 4 is not bad. So given that the light from my torch is emitted upwards at the event horizon would it appear to me to escape or not?
It can definitely escape - whether it does or not depends on exactly where you emit it, and at what angle.

15. It is certainly a sight to see, Farsight denying Einstein. But that aside, I will turn to my main business. I think that the OP's question can be expressed as follows:

Can an ingoing particle emit a particle that escapes?

I will use the Schwarzschild solution with timelike metric signature: +----. That makes 4-velocity u satisfy u.u = +1, and 4-momentum p satisfy p.p = + m2 for mass m. The affine parameter I will take to be (proper time) / m, making

(4-momentum) = d (coordinates) / d (affine parameter)

The outgoing particle's energy as observed from the ingoing particle must always be positive: Eobs = (uin.pout) > 0

I will use that constraint to address the question.

The 4-momentum of a particle following a geodesic is

for energy E, angular momentum L, and bearing angle χ.

The observed energy is

Let's consider inside the Schwarzschild radius, r < 2M. Will it be that Pr > Px there? If so, then the outgoing particle will have negative observed energy. One can select the orbits to maximize Px: both orbits' bearings equal, meaning that the orbits are coplanar and in the same direction. It's easier to do the calculation with the squares of Pr and Px; since Pr > 0, that is OK. We calculate Pr2 - Px2 and see how it compares to 0.

The two masses, min and mout, increase Pr, so I will ignore then. The difference of squares becomes

where

That difference becomes

and is thus never less than zero.

Thus, inside the Schwarzschild radius, an ingoing particle can never emit an outgoing one. So once you fall past it, nothing you broadcast can escape it.

16. Originally Posted by Gerry Nightingale

I wonder how many times I have read this exact same excuse of "We must protect students!" used on every site in existence, by the same trolls...so they can "justify" a nasty comment
Where is the nastiness in those comments? They point out the truth. Farsight is literally lying in many of his posts. He is usually deceiving people by selectively quoting parts of documents without addressing the clear statements in those same documents that address and deny his position. He also alternately claims that his position is the real physics and that his position is rejected by physicists; he is presenting an alternative to today's physics but he is angry that he present these ideas in an alternate theory venue.

Arguing dishonestly is not debate.

17. That's an interesting approach to the question, Ipetrich. I like it ! It's always fascinating to see how different people approach the same question from other angles

18. Originally Posted by Markus Hanke
That's an interesting approach to the question, Ipetrich. I like it ! It's always fascinating to see how different people approach the same question from other angles
Thanx. I will concede that it is something of a brute-force method.

I've thought of doing that for the Kerr solution, but that would require even more complicated calculations.

As to the OP's specific question, one must note that in our Universe, black holes form from gravitational collapse, making all particles in them ingoing. Inside the Schwarzschild radius, they cannot emit any outgoing particles, so nothing can escape. That makes our Universe's black holes fit only the upper half of a Kruskal diagram.

Reminds me of Kruskal–Szekeres coordinates - Wikipedia, something that suggests a simpler proof of my result. In the upper quadrant, all the forward timelike geodesics are ingoing, thus no ingoing particle can emit an outgoing one.

19. Originally Posted by Ipetrich
Thanx. I will concede that it is something of a brute-force method.
I think the brute force method was mine - calculating the geodesics themselves

I've thought of doing that for the Kerr solution, but that would require even more complicated calculations.
That would be an algebraic nightmare, no doubt.

Reminds me of Kruskal–Szekeres coordinates - Wikipedia, something that suggests a simpler proof of my result. In the upper quadrant, all the forward timelike geodesics are ingoing, thus no ingoing particle can emit an outgoing one.
Yup, this is the elegant and preferred solution - no maths required

Soooo, someone can use ANY adjective they like to describe someone else as a person...as long as they qualify it with "you ignore the real truth as it is known to be!" and comments of

"you are a liar!" is okay also, in response to Farsight's observations! (am I missing the rationales here? The ones that justify calling someone an "a-hole" on an open forum?)

.....

"Calculus Rules All". Okay. Then please tell me why Mercury does not fall into the Sun, or fly off into the distance? Use all the calculus at your command to explain this simple, real feature

of the Solar System! (no...don't respond w/ "the eccentricities are calculated and known"...this is NOT what I want a mathematical explanation for)

My "lesser mortal self" wants an answer (calculus or otherwise) of the above question...go ahead, I will understand ANY equation you can write that explains the "why?"

21. Gerry, why do you regard orbits as being so unstable? The planets are not on an unstable equilibrium. A little push one way or another will change the orbit a bit but it won't destabilise it in the way you are describing.

22. Originally Posted by Gerry Nightingale

Soooo, someone can use ANY adjective they like to describe someone else as a person...as long as they qualify it with "you ignore the real truth as it is known to be!" and comments of

"you are a liar!" is okay also, in response to Farsight's observations! (am I missing the rationales here? The ones that justify calling someone an "a-hole" on an open forum?)

.....

"Calculus Rules All". Okay. Then please tell me why Mercury does not fall into the Sun, or fly off into the distance? Use all the calculus at your command to explain this simple, real feature

of the Solar System! (no...don't respond w/ "the eccentricities are calculated and known"...this is NOT what I want a mathematical explanation for)

My "lesser mortal self" wants an answer (calculus or otherwise) of the above question...go ahead, I will understand ANY equation you can write that explains the "why?"

Gerry, why don't you just go away. You are unpleasant.

Farsight, has not "observed" anything. Yet, you demand from people to explain something that is observed; like the orbit of Mercury.

Go away.

And yes, Farsight, is very dishonest. It hurts to read his crap. It makes me not want to participate at all. Same as you.

23. In reply to #122 post.

Gai cocknif en yom!

Cheers! No...I don't regard any "orbit" as "unstable". Mercury exhibits extremely well-known "perturbations" w/regard to it's orbit round the Sun, that's why I used it as an example

of "calculus" is unable to rationalize the "why"...a physically demonstrable feature of empirical reality that cannot be explained by "equation".

(by the way, yes there ARE small variations involved in any orbit, but they are "self-correcting" much like a gyroscope)

(Thanks for reading!) Cheerio! (yikes...you actually read the crud that I write!)

25. Originally Posted by Gerry Nightingale
Then please tell me why Mercury does not fall into the Sun, or fly off into the distance?
It's because all orbiting bodies are in free fall, and hence follow time-like geodesics in space-time. These can be explicitly calculated, but the calculation is long and tedious, so I don't think anyone here will bother typing up all the LaTeX code for you. However, if you want to look up the explicit calculation, I can reference you to the appropriate sources - the Mercury perihelion precession is one of the standard tests of GR, so the necessary calculations are listed in most textbooks on the matter. The calculation proceeds along the same lines as the geodesic calculation I posted for Farsight, just without the photon condition ( ds=0 ) of course. Technically speaking the Mercury-Sun system is a fully relativistic 2-body problem, but since the mass of Mercury is very much less than that of the sun, we can for most purposes neglect this and just consider a mass in free fall towards a stationary central body.

26. Originally Posted by Markus Hanke
It's because all orbiting bodies are in free fall, and hence follow time-like geodesics in space-time. ...
I don't think you understood why Gerry asked that question. Someone said Calculus Rules All and Gerry challenged that person to explain why Mercury doesn't fall into the sun of fly off into the distance using only calculus and thus no general relativity or physics at all.

However the fact that orbiting bodies are in free fall, and hence follow time-like geodesics in space-time doesn't explain why why Mercury does not fall into the Sun, or fly off into the distance. In fact all bodies that are in free-fall around the sun follow time-like curves, some fall into the sun, some fly off into the distance and some are in orbit. So clearly your response doesn't answer the question. The question is answered by stating that the stable orbits have a total energy which is less than or equal to zero. They "why" is seen by doing the math out and following the logic when doing the math.

27. In fact all bodies that are in free-fall around the sun follow time-like curves, some fall into the sun, some fly off into the distance and some are in orbit.
Ok, you have a point here...I should have added the appropriate boundary conditions to my remark about the geodesics. That seems obvious to me, but it mightn't be obvious to someone else. Point taken

The question is answered by stating that the stable orbits have a total energy which is less than or equal to zero.
I would have formulated this in terms of the effective potential, as in Taylor/Wheeler - but that's just because that's the way I learned it.

28. Originally Posted by Markus Hanke
Ok, you have a point here...I should have added the appropriate boundary conditions to my remark about the geodesics. That seems obvious to me, but it mightn't be obvious to someone else. Point taken

I would have formulated this in terms of the effective potential, as in Taylor/Wheeler - but that's just because that's the way I learned it.
I notice that you have a strong tendency to respond to questions in GR in the language of differential geometry and tensor analysis even when the answer would be better described and more easily understood without the math. Please note that I'm not criticizing you. Each to their own, right? I'm just curious that's all. So may I ask why? Thanks.

Thanks for the reply...but as Physicist pointed out "the issue is not orbit" but by what mechanism is "angular-momentum' as well as "orbital decay" being denied by two separate

gravimetric components? I think the answer begins and ends w/gravity as a "canceling factor" w/regard to Mercury's orbital fixation.

......

I think there is strong evidence that there may be more than is supposed w/ regard to gravity...and I believe "axial rotation" and "orbit" are the "keys that open a door we didn't know

was there", but that belongs in a "personal theories" thread.

30. Originally Posted by Physicist
I notice that you have a strong tendency to respond to questions in GR in the language of differential geometry and tensor analysis even when the answer would be better described and more easily understood without the math
Yes, you are absolutely right, I have that tendency. Personally I would disagree that scenarios where GR is involved are more easily understood without differential geometry - I prefer to keep the more global geometric principles of GR strictly separate from the local Newtonian physics of forces and accelerations. I think mixing these up just leads to misconceptions, unless you know exactly what you are doing.

There is no deeper reason I think, it is largely a personal preference. You see, I am a very visual person - I was born with a rare genetic abnormality that leaves me with an eyesight which exceeds that of the average person, especially in darkness, and allows me to focus on objects further away than most others can. Also, I am a synesthetic, so even spoken words and abstract concepts are connected with specific colours, shapes and sensations for me, further enhancing my being a visual person. It is very difficult for me to even imagine what the world is like for someone without this "synesthetic sense" ( that would be flat and featureless, I think ), but I dare say my perception is sufficiently different from that of other people to be able to say I see things from a different angle of sorts. For example, in your pen and paper scenario, I don't consider just a 1-dimensional trajectory of the falling pen - instead I see the pen and table as 3-dimensional objects, and the process of falling as an extension of these objects in time, so I intuitively see a 4-dimensional hypervolume of space-time with the world-sheet of the pen in it. And while the table may be small enough to be considered "local" and Minkowskian, the 4-volume spanned by it and the time it takes for the pen to fall most certainly is not ( in fact it is of the order ~10^8 m^4 ! ).

Anyway, to make a long story short, in order to understand things I always find ways to visualise them, so I have a special affinity to geometry in general. I have over the years developed geometric interpretations and visualisation for most of the important objects in GR ( i.e. the various tensors etc ), so differential geometry suits me just fine, and I think it is fully in the spirit of GR

31. but as Physicist pointed out "the issue is not orbit" but by what mechanism is "angular-momentum' as well as "orbital decay" being denied by two separate
gravimetric components?
To me, Mercury's orbit is a static twisted helix in space-time; the "twist" is just the perihelion precession, and the exact shape of the helix is determined by the geodesic equation and the appropriate boundary conditions. There are no "separate gravimetric components" - I am not even sure what it is you are referring to by this term.

Cheers!

W/regard to "separate gravimetric components", I meant the gravimetric-fields of the Sun and Mercury...I assumed you would inherently know this, since the question involves two

separate physical components. I was not referring to anything "imagined" from my mind. Are you stating flatly that "there is no distortion of the spacetime-continuum caused by the

physical presence of the Sun? Or any other body? (in this event I'm undone, and there is nothing to explain of "what I mean")

Saying there are no "separate components" denies "orbit" as a true function, as well as any "boundary conditions", i.e., "you cannot advocate the one while denying the other".

I don't understand the substance of your reply...are you stating "I don't see it" or "I will not acknowledge this "separate" condition of gravity as "real".

(take a look at the equation you include w/all your responses...is it a "true function" or no?)

33. I meant the gravimetric-fields of the Sun and Mercury
Ok, sorry...I wasn't sure what you meant. Usually what happens in cases this like, because the mass of Mercury is very much smaller than the mass of the sun, Mercury's own gravitational influence as well as that of the other planets is neglected since this greatly simplifies the calculations. If we disregard Mercury's mass and the rest of the solar system, we can approximate the scenario as being the Schwarzschild metric, and explicitly calculate everything we need in analytically closed form. Of course it is also possible to consider the gravitation of both Sun and all the rest of the solar system - but then we have a fully relativistic many-body problem at hand, which can be treated only numerically with the aid of computers. It can be done, but is not really very instructive when it comes to explaining the basic principles of GR, and as it turns out the gravitational contributions of Mercury itself and the other problems are very very small, and don't really justify the extra effort. The standard Schwarzschild approach yields an excellent approximation to the empirical data.

Does this make sense ?

34. One can write down equations of motion for objects with arbitrary masses using a post-Newtonian expansion. Its small parameters are (GM)/(rc2) and (v/c)2, but they have similar sizes and are effectively one parameter. Most GR effects observed to date have been to first order in these parameters, with gravitational radiation having order 2.5.

For Solar-System and compact-binary tests, one treats the post-Newtonian terms as perturbations of Newtonian-gravity motion, and there are well-established celestial-mechanics techniques for doing so. One can integrate the equations numerically, or else one can handle post-Newtonian terms alongside Newtonian ones as perturbations of two-body orbits and analytically integrate over each orbit.

Cheers and Hello!

Yes, your reply makes perfect numerical sense! (I bet you thought I wouldn't!)

My real problem w/Mercury as well as our own Moon is that from the standpoint of "mutual attraction" (disregarding angular momentum from the smaller orbital body) the orbit should be

be subject to a constant decay to the point of orbit function "collapse". (remember, I am discounting momentum as a factor in this problem...I am considering attraction only, however

negligible it may be) If a given gravimetric-field of smaller body is constant, as well as the larger "parent" body...then orbit of the smaller body should be subject to decay, as per the mandate

mathematical progression...yet the lessor body never yields to the greater bodies' gravitational influence.

Why?

For instance, one could say "the velocity of the smaller bodies' orbit "cancels out" the predictable decay rate" and I would accept this as the state-of-condition re: orbit, but then I must

include the rationale of "what is providing the impetus for the predictable velocity of the lesser orbital?"

It could reasonably then be stated that our Moon is, in effect "falling" at oblique angles toward Earth, and the Moon's own momentum is "holding it in check due to it's inherent velocity", a

sort of "whirlpool outer-ring" condition...the Moon cannot "fall in" toward Earth, nor can it's orbital velocity allow escapement from the "ring".

The question then presents itself once again of "Why?"

If the "whirlpool ring" scenario is at least partially correct...then what is the "substance" of the rest of the "whirlpool?" There seems no evidence of it, at least not in any real sense.

Perhaps it could be postulated mathematically...but my skills are not strong enough in this area>

36. Hello Hagara,
*Light can only travel in straight lines through space.( Rectilinear propagation)
*Mass can bent and twist space.
*For a blackhole, the mass is concentrated at a very small region in space.
*This causes every straight line in space to bent and lead to the center of the blackhole.

This is how I think it is.
HubbleSite: Black Holes: Gravity's Relentless Pull interactive: Encyclopedia

37. Originally Posted by Gerry
yet the lessor body never yields to the greater bodies' gravitational influence.
Actually, this is true only so long as we approximate the scenario via Schwarzschild metrics, i.e. we take the central body to be perfectly stationary, and only the smaller body to orbit. In reality though both bodies orbit around a common barycenter - this gives the system a non-vanishing quadrupole moment, so it will emit gravitational radiation and the orbits will slowly decay as a result. The issue here is that for "light" bodies such as the Earth-Moon or Sun-Mercury this effect is too tiny by many orders of magnitude to be relevant within any reasonable timeframe, so it is usually discounted.

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