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Asymptotic Freedom in QCD and...

Discussion in 'Pseudoscience' started by RajeshTrivedi, Sep 14, 2017.

1. RajeshTrivediValued Senior Member

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It is produced at r = 0, so your answer that it will travel to r = 0 does not hold, and "slightly in the future" also makes no sense.
There is something serious here, and some one with deeper understanding of GR must jut in.

Let us do with some figures already thrown in one of the posts.
Consider a one million solar mass object, just below its event horizon. That means the object at this stage is a sphere of radius 3 million Kms. Assume that it is uniform density. The innermost 3.24 solar Mass part of this object is a sphere of 44000 Kms (much larger than its Schwzrschild Radius of 10 Kms).

Now we take a different scenario. Say a photon is produced at r = 44000 kms point facing radially away. The mass inside r = 44000 is just 3.24 Solar Mass, even as per GR (and Newtonian too) the outer mass (beyond r > 44000) is not going to have any gravitational impact on this photon. So I see no reason why this photon even though directed outward, will travel to r = 0? It may not cross r = 3 million kms, but there seems no restriction on its motion inside that. The key takeaway is 3.24solarmass / 44000 Kms object is not an object inside its event horizon, so why a photon should behave like it is inside a BH?

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I'm certainly not one with a deeper understanding of GR, but NotEinstein seems to be doing OK....
What I do know is that the escape velocity at the EH equals "c" and inside the BH, spacetime can be conceptualized to be falling faster then "c". Which of course violates nothing, as spacetime is not inhibited by the universal speed limit which only applies to mass/energy. Therefor obviously any photon produced inside the EH has only one direction to go...in fact the only path inside a BH is one directional towards the singularity.
This may explain it better then I......
https://arxiv.org/pdf/gr-qc/0411060.pdf

The river model of black holes:
Andrew J. S. Hamilton ∗ and Jason P. Lisle
JILA and Dept. Astrophysical & Planetary Sciences,
Box 440, U. Colorado, Boulder CO 80309, USA

This paper presents an under-appreciated way to conceptualize stationary black holes, which we call the river model. The river model is mathematically sound, yet simple enough that the basic picture can be understood by non-experts. In the river model, space itself flows like a river through a flat background, while objects move through the river according to the rules of special relativity. In a spherical black hole, the river of space falls into the black hole at the Newtonian escape velocity, hitting the speed of light at the horizon. Inside the horizon, the river flows inward faster than light, carrying everything with it. We show that the river model works also for rotating (Kerr-Newman) black holes, though with a surprising twist. As in the spherical case, the river of space can be regarded as moving through a flat background. However, the river does not spiral inward, as one might have anticipated, but rather falls inward with no azimuthal swirl at all. Instead, the river has at each point not only a velocity but also a rotation, or twist. That is, the river has a Lorentz structure, characterized by six numbers (velocity and rotation), not just three (velocity). As an object moves through the river, it changes its velocity and rotation in response to tidal changes in the velocity and twist of the river along its path. An explicit expression is given for the river field, a six-component bivector field that encodes the velocity and twist of the river at each point, and that encapsulates all the properties of a stationary rotating black hole.
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5. RajeshTrivediValued Senior Member

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What you say is right, but only when the object is fully collapsed to r = 0, in this case all the points inside the main event horizon will be critically curved or having spacetime falling faster than c as you refer.

My reference is when the object is just at its event horizon. In this case only the outer surface of the object is at c, none of the internal points will be having escape velocity c or at none of the internal points the spacetime (if any) would be falling faster than c. Under such circumstances if the photon is emitted say at r = 0 (or in ex at post #61 at r = 44000), then it is not bound by BH restrictions, it can travel outward (and of course core is collapsing inward) till it encounters a point where escape velocity becomes c. It does not contradict your reference as such.

7. NotEinsteinRegistered Senior Member

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So you don't understand basic GR results, or English. I called it "falling back": if I shoot a cannon from the ground up, the cannon ball will travel from h = 0 (h: height) to h = some number, and then "fall back" to h = 0. It'll reach h = 0 again "slightly in the future".

Look at the picture I provided, and start drawing light cones. Those are basic skills in GR (heck, special relativity!). If you cannot do that, you cannot meaningfully talk about how photons behave in a GR environment.

(Please stop using "Kms", which means "kilometer seconds". It's just "km", or if you insist: "Km's".)

It's going to redshift, so technically you're wrong.

You've significantly changed the scenario: this photon starts outside the event horizon. I (obviously) never claimed that all photons starting outside the event horizon will travel to r = 0.

If you know what GR said, you'd know it will cross r = 3 million km.

I was talking about photons produced inside the event horizon, so this question is a misinterpretation of what I said.

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I did come in on this late, but really, what you say makes no sense. When the mass collapses to its Schwarzchild radius/EH, then all of that mass is inside and all of it will continue to collapse at least to where GR fails us......
Any photon emitted anywhere inside the EH, has only one geodesic path to travel as already detailed. Imagine a fish swimming at "c" against a current ">c": The fish aint gonna get anywhere fast, is he?
And by the way, any photon emitted just on the EH directly radially away, will appear to hover there forever from the point of view of any local frame....all other photons emitted less then directly radially away, will arc back and fall into the BH, again from a local frame of reference.

9. RajeshTrivediValued Senior Member

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These two statements show that you have not understood my point.

Lets rehash it;

1. One Solar Mass object has a Schwarzschild radius of around 3 km.
2. For one million solar mass it will be 3 million km.
3. For 3.24 Solar Mass it will be around 10 km.

Now when a one million solar mass object is just beneath its event horizon (that means a sphere of radius 3 million km in size), at that time a photon emitted at its surface will not be visible or will not be able to move beyond 3 million km point. But at this moment, the inner most 3.24 solar mass part is of the size 44000 km much larger than the innermost part's schwarzschold radius of 10 km. Therefore the photon emitted at r = 0 or at r = 44000 kms, can travel towards radially outward even though both these points are well within original EH of 3 million.

This point is essential because in my proposal the energy gets released at the center (around r = 0), so it has to travel away from that till it encounters a point where escape velocity is c.

Last edited: Nov 15, 2017
10. RajeshTrivediValued Senior Member

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Does not matter if you came early or late, as long as you understood the point.
I agree that as per GR once the mass falls below its EH, it will continue to collapse.
But I do not agree with the bold color part in the quote.
This is correct only if the object has collapsed fully, in that case each and every point beyond r = 0 will have escape velocity c, so nothing can even move away even from r = 0 and all infalling material will have only one path that is towards r = 0.

But I am talking about the transient stage, the collapse of an object from r = EH to r = 0. You see the difference? During this process (for cores > 3.24 Soalr Mass) the quarks compression will happen, this will release the energy at r = 0 as proposed and this photon may travel away from r = 0 towards the infalling surface of the core till this photon encounters a point where escape velocity is c. If you agree with this, then we can move further.....If you disagree tell me why.

11. NotEinsteinRegistered Senior Member

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Point me to a worldline in the figure I provided that allows a neutron of quark to remain significantly above r = 0 for an extended period of time.
And I've asked you multiple times now to calculate the geodesic. Calculate using the metric and your density distribution what the photon does. Some hand-wavy "it travels until it reaches the escape velocity becomes c" is not how one does GR. At best, it's Newtonian.

I currently have no strong argument against this specific scenario, because we need to actually evaluate what GR predicts. And please note that without that GR prediction, you also have no strong argument in favor of this specific scenario; that's what I've been pointing out to you. Newtonian notions are not necessarily valid when strong gravity is present, or when the speeds approach the speed of light. But let's go through the calculations together then, because apparently you are unwilling or unable to provide them on your own.

First, a note: it becomes immediately clear the photons cannot leave the surface of the object: the "escape velocity" has to be c or larger there. So we are talking about photons in a medium, which complicates matters (pun intended). The speed of light in a medium is lower than that in a vacuum. So the escape velocity actually doesn't have to reach c for this to happen. Yes, yes, a minor nitpick.

In order to calculate the geodesics, we're going to need the exact metric ($g_{\mu\nu}$) and density distribution ($\rho$) you are using (and their time dependence, if any). Can you provide that, please?

Another note: We'll also have to prove the given density distribution is actually stable; if the entire thing collapses into a singularity anyway, the photons will be dragged into it as well, as my posted picture illustrates.

(Note that this discussion doesn't actually matter that much: even if some shell of "hovering photons" is created somewhere in the event horizon, when viewed from outside the event horizon, it looks just the same as a black hole with a singularity, due to the no-hair theorem.)

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Again, all you are doing is guessing. We have absolutely no observational evidence about what happens inside the EH, other then what GR tells us, and you cannot deny that so far GR has a damn great track record. And going on what you are saying, that energy would probably go into creating another quark anti quark pair as per the answer given here.....
https://www.quora.com/How-do-the-st...electromagnetic-or-gravitational-interactions

And of course you are failing to take into consideration that we have strong evidence for the existence of BHs.

13. RajeshTrivediValued Senior Member

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So there is a point here which needs to be resolved. Thanks.
We do not really know if GR notion is necessarily valid in strong Gravity.
You start the calculations because I could get no reference of any GR treatment of a photon emitted at r= 0 when the core is just at its EH and collapsing. And I have no clue how to initiate such scenario in GR.

The next step is to breach back the original EH and come out.

14. RajeshTrivediValued Senior Member

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Which part you find guessing? Its well within established Physics. There is no magic that the matter would vanish once inside EH.

GR does not tell us anything about state of matter, inside or outside EH.
GR talks about geometry of spacetime only, not about the physiology of matter/energy.

I am not denying that GR has a damn great track record. I am also aware of the Nobel given for GW discovery under GR. That only proves that the GR is the most acceptable prevalent theory of gravity.

15. NotEinsteinRegistered Senior Member

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But we do know that Newtonian physics is certainly not valid in strong gravity. If your argument now is that GR might not be valid, and that's why you're not using it, then you have nothing to work with.
However, your argument here is disingenuous: you use the speed of light as a maximum speed limit, and you are using the Schwarzschild radius. Both are GR constructs, so clearly you accept that we have to use GR.

At the very least provide the density distribution you are using; without that, there's literally nothing to work with.

Yes, and until these calculations have been done, you cannot claim that your conclusions are valid.

I don't know what you mean by this?

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Your guess that the increases in energy when quarks are separated, stops a BH forming....your guess that anything ever gets out of an EH. [see the links and answers given ]
Did I say that? I said that GR tells us that once the Schwarzchild radius is reached, further collapse is compulsory, at least up to the quantum/Planck level where GR is no longer applied. Then I suggested as per the answers in the links I gave that the energy you claim stopping BHs forming, actually probably go into creating another quark/anti quark pair.
Yet you are claiming on a public science forum, open to all, that BHs do not form.

So tell me, what is formed? and why not take it through the proper channels if you believe the majority of cosmologists out there are fools.

https://physics.stackexchange.com/questions/143145/quarks-falling-in-a-black-hole
"In the case of a black hole singularity and quark triplets ( neutrons and protons, pairs are quark antiquark i.e. mesons) falling into a black hole , the neutrons and protons falling in acquire energy from the gravitational energy of the black hole and at some point in energy will start interacting with each other. The strong force will produce quark-antiquark pairs i.e. mesons, and other particle antiparticle pairs. As more energy is gained by the fall, the strongly interacting mass will reach the quark gluon plasma stage" :

"I would imagine that, the energy that pulls the quark pair into the black hole is intrinsic energy (ie it comes from the black hole), and the quarks created through 'spaghettification' would only be proportional to that energy pulling on them. Therefore, seeing as energy is equal to mass and vice versa this new mass would not be created, rather transformed from the energy of the black hole. The black hole never gains more mass/energy than the original quark pair because it already contained the mass/energy in its system that would be then created into more quark pairs. I hope that makes sense"

Last edited: Nov 16, 2017
17. NotEinsteinRegistered Senior Member

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877
(Forgive the undirected manner in which the below post is structured; I was thinking while typing, and didn't invest the time to clean it up.)

And this is easy to prove if we assume that the inside object is mainly composed of matter. Let's assume no outside input, spherical symmetry (thus also no rotation), and no (net) charge. This means the total mass of everything inside the event horizon will be constant, and thus the Schwarzschild radius will remain constant. But we can imagine a "Schwarzschild region": a spherical shell in which the gravitational pull of the inside object is "so strong, even light can't escape". Its outer boundary is the Schwarzschild radius. Its inner radius would have to be calculated, but let's use a minimum (the calculated radius will be equal or smaller to this value): we shrink the radius from the outer bound inwards as long as there is no matter in the spherical shell at that radius. In other words, we construct the largest shell possible, where the outer bound is given by the Schwarzschild radius, and such that there is no matter inside this Schwarzschild region.

This means the particle with the largest radius (from the center of the object) will be on this inner bound. (Actually, since we've assumed spherical symmetry, it will be a shell of particles.) Now, the gravitational pull on this outer particle(shell) is at least so strong that light cannot escape ("escape velocity is c or higher"), which means matter stands no chance at all: these outer particles will start falling inwards. There is no possible other worldlines for them, no pressure or force can prevent this. Likewise, there are no worldliness for massive particles that can reach this radius, so the shell cannot be replenished.

When this outer shell collapse inwards (some irrelevant distance), and nothing to replace it, we can extend the Schwarzschild region: the outer shell of matter is now gone, and so we can make the inner bound a bit smaller.

As you might have guessed already: this process is repeatable. It only breaks down when one reaches r = 0, i.e. the singularity, or until GR breaks down (Planck scales).

Conclusion: any spherical symmetric object made purely out of matter cannot be stable inside an event horizon it creates by itself.
(Note that this is not a GR-calculation, but a hand-wavy argument based on the properties of the Schwarzschild metric.)

18. NotEinsteinRegistered Senior Member

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But let's take this one step further: what about photons (or other massless particles)? Let's make one adjustment: we define the inner bound of the Schwarzschild region as also taking into account photons, and steering clear of those. This will make its radius a tad higher (the Schwarzschild region will start out a bit smaller), but nothing spectacular (we are talking about an infinitesimal distance here: instead of the inner bound being an inclusive border, it will now be exclusive).

But there's a problem: the photons are "on the edge" of this inner bound, so the "escape velocity is c". Can they just hover there?

Time to switch to Kruskal–Szekeres coordinates. Let's use this picture:
https://en.wikipedia.org/wiki/Krusk...le:Kruskal_diagram_of_Schwarzschild_chart.svg
(Remember especially that light-cones in this depiction works as in special relativity.)

Any photons emitted from r < 1 will be emitted towards the center: it is impossible to find a light-cone in region II where the light-like lines of the light-cone allow a rise (or even constancy) in the value of r. In other words, only photons emitted at r = 1 can stay at r = 1. In fact, that's their only option!

Now throw a single neutron (or any tiny bit of energy) into the event horizon. This mass/energy will traverse the Schwarzschild region, and fall towards the inside object, increasing its mass. The Schwarzschild radius grows the tiniest bit, and what previously was r = 1, now becomes r = 0.99999… All photons that were at r = 1 now start falling towards the inside object. Since there is no radiation emitting matter or anything at r = 1 (only stuff falling into the object that passes by r = 1), there will no longer be any photons at r = 1.

Conclusion: Even if one has many photons trapped at r = 1, the tiniest addition of mass will cause all these photons to fall towards the inside object. Such a situation is meta-stable at best.

Note that photons trapped at r = 1 can in general not be orbiting, because they are below the lower bound for any stable orbit:
https://en.wikipedia.org/wiki/Photon_sphere#Derivation_for_a_Schwarzschild_black_hole

Is it possible for the inside object to produce photons that reach the inner bound of the Schwarzschild radius? Well, let's think about this.
- If there is not already some mass (energy) there, the Kruskal–Szekeres diagram clearly says that you can't reach r = 1.
- If there is matter there, that matter will collapse, as per the first part of this post.

So the only possibility is for photons (energy) to already be there. This can happen due to new in-falling material emitting radiation at exactly the right time (and angle), or because the collapse catches up to these photons. The first scenario isn't really relevant in this thread, and would only be a tiny contribution (plus the in-falling material will add to the central mass, so it's a self-defeating process), so let's focus on the second. The scenario thus is a collapsing inside object, made partially out of matter, but it is emitting photons. Immediately we hit a problem: the Kruskal–Szekeres doesn't allow any photons to be emitted outward from the surface, and the photons emitted inside the object will be traveling inside a medium, not a vacuum. My GR isn't good enough to handle that, though, so I'm going to assume these effects to be negligible. (Tips and pointers in this regard are welcome!)

In the described scenario, our Schwarzschild region extends up to the outermost photons. But remember that the lower bound of the region was chosen such that the region is a vacuum (devoid of mass). This is a too strong restriction in general, as already pointed out when I introduced the region. Due to spherical symmetry and Birkhoff's theorem, we have something similar as the shell theorem in Newtonian physics: mass at radii larger than the particle don't affect the spacetime curvature (technically, this is only valid in a static situation, but let's hand-wavingly call our situation adiabatic). As long as the "inner mass" (total mass at smaller radii) is large enough to create a Schwarzschild radius larger than the particle's radial coordinate, the particle is inside the (now adjusted) region, and will fall inwards.

This leads to the conclusion that the density distribution has a couple of conditions that is must conform to:
- Spherical symmetry;
- It must fall off to zero at large radii;
- There cannot be matter on the outer surface (as this would collapse inwards);
- The total mass must be such that the Schwarzschild radius is larger than the density distribution's maximum radius.
- At no radius smaller than the outer surface's, can the "inner mass" be such that the Schwarzschild radius of the "inner mass" is larger than or equal to that radius.

The last point indicates that the Schwarzschild radius' equation basically becomes an upper bound to the "inner mass". Note that mass scales to the third power compared to radius when assuming a constant density. This strongly implies (in a hand-wavy way) that if one uses a smooth enough density distribution, at large radii it must drop off at least as fast as a cube-root function, with the possible exception of the outer surface: the total mass needs to be large enough to create an event horizon. This creates a seemingly pathological density distribution: a dense core, decreasing density the larger the radius, but with a suddenly quite (energy-)dense shell as the outer surface. Note that the outer surface cannot contain matter; it must be composed entirely of massless particles.

As already pointed out earlier, the slightest "inner mass" mass increase will cause this outer surface of massless particles to fall inward. As this outer surface collapses, all matter it passes will collapse inwards as well (massive particles must fall inwards when on the edge of the Schwarzschild region). The inward-falling massless particles never see a mass-decrease of their "inner mass", because any massive particles that could do so start falling inward as well, as the Schwarzschild region shift with the outer surface. In effect, the entire structure starts collapsing, because the outer surface is falling inward forever, "sweeping up" anything along its path.

Conclusion: The required density distribution is either pathological, or meta-stable.

19. NotEinsteinRegistered Senior Member

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Because such objects cannot thus exist (for long) in practice, neutron stars in general will not have such density distributions. Neutron stars also don't seem to have the cube-root distribution, meaning the outer surface of photons must be even denser in order to generate the Schwarzschild radius at that point.

Let's see what happens if the "inner mass" decreases a bit instead. The mass decrease will cause the outer Schwarzschild radius to shrink. The outer surface will (counter-intuitively?) not move, because it can't expand outwards: it's right up to the Schwarzschild radius. As long as the total mass of the object is enough to generate an event horizon outside itself, reducing the mass will not enlarge the radius of the outer surface (it might affect its "thickness", but that's a whole other discussion). When this point is hit (which coincides with the lower and higher bounds on the Schwarzschild region's radius becoming equal), the outer surface will "poke out" of the event horizon (this is not a proper GR approach; it's merely speculation based on my inadequate knowledge of GR). At that point the entire outer surface of the object (remember, massless particles traveling at the speed of light) will escape, and start travelling to infinity. The remaining central object will have no event horizon (as it still complies to the density distribution that doesn't allow this per construction). We are left with "just" a celestial body.

Conclusion: A mass decrease sufficient to cause the event horizon to shrink to the outer surface fatally disrupts the density distribution required to generate any event horizon.

Under the assumption that my reasoning isn't too far off, that leaves us with the overall conclusion that:
- We need a density distribution that is very strange and unusual.
- Any particle dropping into the object causes it to fully collapse (unless it happens to annihilate fully at exact r = 1).
- Any photon dropping into the object causes it to fully collapse.
- A mass decrease sufficient to shrink the event horizon to the outer surface causes the entire outer surface to escape, leaving behind an object with no event horizon.

So you can't increase its mass, and decreasing it will eventually destroy the desired configuration. This means that I see no way for such an object to form in nature. In other words: neutron stars cannot generate these objects, at least not for very long (not even cosmologically; this is more millisecond work).

I invite RajeshTrivedi to provide a density distribution (and perhaps a way neutron star can generate it) that does work.

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Agreed
The BH consuming more matter and the EH getting bigger, and the possibility of HR evaporating the BH over time, the claim about hovering "forever" probably does not really happen.
But we are using the Schwarzchild metric as an idealisation. So, based on that model it could be there forever. Obviously though, the "thought experiment" is unrealistic for multiple reasons.

21. NotEinsteinRegistered Senior Member

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In fact, you can make a stronger claim: it's not just adding matter or Hawking radiation: if you take Birkhoff's theorem into account: https://en.wikipedia.org/wiki/Birkhoff's_theorem_(relativity)
"In general relativity, Birkhoff's theorem states that any spherically symmetric solution of the vacuum field equations must be static and asymptotically flat."
You can't have any change over time in the energy density at all, or you will necessarily disrupt the spherical symmetry, which opens a whole other can of worms.

22. NotEinsteinRegistered Senior Member

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Time for some small corrections before anybody else points them out.

"(the calculated radius will be equal to or smaller than this value)"

It's their only option if they ever want to have r = 1 in the future. It's either staying at r = 1, or set on a path where r will constantly decrease and eventually reach zero.

It's the upper bound of the Schwarzschild region that will grow, but it doesn't follow the lower bound will "grow" inwards. To see this, you have to use the modified definition of the Schwarzschild region I use later on. As the inner mass increases the tiniest bit, the critical mass for an event horizon will be reached at a slightly smaller radius, and all photons outside of that radius will start falling inward.

There's bound to be many more such oversights; I welcome you to hunt for them!

23. RajeshTrivediValued Senior Member

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This argument is qualitatively good and I am not objecting to that. But even when the matter is collapsing a photon can be produced at r = 0. There is no mention of this photon's worldline anywhere in the literature.

The closest I have come (BH WD and NS by Shapiro, freely available on net) has an example of photon produced at r = 2 M (Event Horizon). That does not help.

Moreover there are two metrics involved here. One is schwarzschild metric which is applicable external to surface and the other one is inside to the surface. The boundary matching leaves no scope for any stable object at or inside EH.


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# Simple example of a short exact sequence of groups

I am new to this and would like to understand $$0 \overset{a0}{\to} B \overset{a1}{\to} A \overset{a2}{\to} A/B \overset{a3}{\to} 0,$$ where $B \subset A$ and they are both Abelian groups. Also maybe it is helpful to know that every subgroup of an Abelian group is normal? This is in the context of a cochain (we were taught starting this direction).

So here is what I understand so far:

If it is a short exact sequence (s.e.s.) (is this given or is it just iff $a_0$ is injective AND $a_3$ is surjective?), then, with the identity element of an additive (for instance) Abelian group being $0$, it is such that,

$$\begin{eqnarray} ker\; a_0 \leq_{sg} 0 \;\Rightarrow ker\; a_0=0 \\ ---------------- {} \\ im\; a_1 = ker\; a_2 \text{ and } ker\; a_2 = B \\ H^2(A) = \frac{ker \; a_2}{im\; a_1} = \frac{B}{B} = 0 \\ ----------------- {} \\ a_2: A \to A/B \hspace{1cm} a_2(a_i) = a_i\,B \\ ker\; a_2 = \{a\in A : a_2 \circ a = 0\} = \{a \in A : a_2\circ a = B\} = B\\ im\; a_2 \cong A/B \\ ---------------- {} \\ a_3\; \text{ zero map } \Rightarrow a_3(a\,B) = B \hspace{1cm}\text{ as the zero of A/B is B ? } \\ H^3(A/B) = \frac{ker\; a_3}{im \; a_2} = B/(A/B) \end{eqnarray}$$

I'm not sure of these results, but I hope it is simple and clear enough to get some feedback. Thanks!

Edit: I forgot to mention that with this construction, $a_1$ induces an isomorphism from $B \to (\underset{s.g.}{{\cdot}}\subset A)$ and $a_2$ induces an isomorphism from $A/im\;B \to A/B$.

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I cannot tell what you are asking. What exactly is it that you want to understand? –  Mariano Suárez-Alvarez Sep 14 '12 at 19:52
Well I was wondering if what I wrote down (in the middle) is correct. Any other comments or feedback is appreciated. –  nate Sep 14 '12 at 19:56
Saying that a sequence is exact means that all the (co)homology groups vanish, i.e. $\operatorname{ker} a_{i+1}/\operatorname{im}a_i = 0$ for all $i.$ –  Andrew Sep 14 '12 at 19:57
All subgroups of abelian groups are normal subgroups. –  Thomas Andrews Sep 14 '12 at 20:18

• A priori, this is just a diagram, that is any arrow from $X$ to $Y$ merely denotes a homomorphism.
• It should be noted explicitly, but the linear structure suggests that it is a complex, that is the composition of any two arrows in straight succession is the 0 homomorphism; if there is $X\overset f\longrightarrow Y\overset g\longrightarrow Z$ in the diagram, that this states that $g\circ f=0$ holds.
• If desired, it should also be noted explicitly, that it is an exact sequence (or in general context: which joints are exact and which maybe not), that is $\ker g =\operatorname{im} f$ holds whenever we see $X\overset f\longrightarrow Y\overset g\longrightarrow Z$. So exactness is a claim about that diagram. Note however, that a short exact sequence $$0\overset \alpha \longrightarrow X\overset \beta \longrightarrow Y \overset \gamma \longrightarrow Z\overset \delta \longrightarrow 0$$ is exact iff the following holds

• exactness at $X$: Since $\alpha$ is the only homomorphism $0\to X$ and its image is $0$, we must have that $\ker \beta=0$, aka. $\beta$ is injective. (Formally, the kernel is an arrow; I speak of kernel, image etc. here as subobjects in the object manner throughout)

• exactness at $Z$: Similarly $\delta$ is the only homomorphism $Z\to 0$ and its kernel is all of $Z$; hence we need that $\operatorname{im} \gamma=Z$, aka. $\gamma$ is surjective.
• exactness at $Y$ is maybe the most interesting point: We need $\operatorname{im}\beta=\ker \gamma$.

What (co)homology does is to measure by how much a complex fails to be exact.

Your calculations are in principle correct, however there are a few probable typoes and e.g. $H^2(A)$ would not be used to denote ker/im of this complex that also depends on $B$ etc., but rather what can be computed from e.g. an injective resolution $$0\to A\to I_1\to I_2\to I_3\to\ldots$$ of $A$ by injectives.

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# Approximation of the FE feat. loose notation

1. Sep 19, 2007

### Spinny

Approximation of the FE feat. "loose notation"

I'm looking for a (professional) relativist to help me clarify something. I refer to the article General Relativity Resolves Galactic Rotation Without Exotic Dark Matter by Cooperstock and Tieu, available here: http://arxiv.org/abs/astro-ph/0507619v1.

They talk about the Einstein field equations to order $$G$$, and list them without any more details except a footnote stating that it's a "loose notation favored by many relativists".

I was wondering if anyone here are familiar with such an approximation scheme and could elaborate on how one goes about making such approximations, or perhaps refer me to some litterature on the subject.

Finally let me just specify that these approximations are not your standard linear approximations, and I can't seem to find any similar procedures in any textbooks (although that may just be because I'm not looking in the right books or at the right places in the books).

Thanks.

2. Sep 19, 2007

### pervect

Staff Emeritus
Cooperstock & Tiu's paper has been criticized by other authors, such as

This paper by Vogt and Letelier

While Cooperstock and Tiu continue to defend their paper, I personally think it's badly flawed, i.e. I agree with Vogt et al. Some other past threads on the general topic of this paper are:

https://www.physicsforums.com/showthread.php?t=103248
https://www.physicsforums.com/showthread.php?t=96935

As far as their notation goes, my impression was that they were basically doing a taylor series expansion of the metric in terms of some small paramter, so G(n) should be just a n'th order Taylor series approximation to the actual function that represents G. I'd have to look at the paper to refresh my memory. Unfortunatlely my browser isn't cooperating at the moment, perhaps a reboot will fix the problem.

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