[Mika]

15 sig figs?  How goddamn accurate were your equipment and measurements?!  That's just about the most amazing accuracy I've ever heard of (the most accurately determined value in physics, the g-factor of the electron, is known to something like 20 sig figs).  I think I managed six or seven in my spectroscopy lab.  For the absolute value sum, what kind of error were you calculating?  The sum of squares is mathematically the only way to do it for different factors correlated in the way I described.

15 significant figures? No, it didn't mean that and tried to avoid it by saying units; and the effect of using that rule is actually the opposite. For example, if the calculated result is 6.340658, using this rule causes one to compare the last two significant figures to the result. Since 58>15, one rounds 6.340658 ~ 6.34066, and yet again 66 > 15 -> ~ 6.3407. Since 07 < 15, the result would be 6.3407. Usually, I only managed to get one or two significant figures after the roundings. I know the rounding rule is arbitrary, but I suppose it was used because the equipment was really antiquated and they didn't want that the students would think they could measure something accurately with those things. On the other hand, I don't blame them for that, and made us understand that any sort of rounding rule just depends on what kind of effect is sought after.

For the use of summed absolute values instead of summed squared values, take a look at here. There is a table that describes the use of Ra and Rq, both of them are used to describe surface texture roughness. Ra is used by metal industry, Rq is used by optical industry. What's the difference? Rq is affected more by surface spikes due to the square in the equation, while Ra isn't usually that sensitive to them. So for your local mechanical work shop, it doesn't matter if there is a big local spike on the surface if there is only one a few them; they are likely going to get neutered in any case when the piece fixed with another one. Not so in Optics, as these spikes will cause scattering effects and they need to be reduced. I thought this analogue might be the best way to describe the difference between using summed absolute values and summed squares in error estimation.

A shift of 3", while noticeable, isn't necessarily that big.  Most systems studied in astronomy are at least several arcminutes wide, so a few arcseconds is at least an order of magnitude smaller, sometimes two (compared to the angular diameter of a big galaxy cluster, for example).  As for aberrations, maybe; I don't know.  They would be described differently than optical ones, however, because a gravitational lens' power increases the smaller the impact parameter is, the opposite of what an optical lens will have.

Following will be some stupid questions by a person who doesn't have background in Astronomy, so bear with me:
- Thinking the above from the perspective of Earth, does this mean that there will be systematic ~3" uncertainty in the absolute direction of all measured galaxies, and that it applies equally to everything that we see as the Sun is located closer to the edge than center of Milky Way whose gravitation will direct the light?
- And seen from the perspective of the photon that is starting from a star in a galaxy far away, the closer it is to the center of the galaxy, the more its path will be bend? For me that would mean that there is less photons arriving on the telescope from the center than from the edge, and that the images of the galaxies have a dimmer center than they actually should have?

[Astronomiya]

15 significant figures? No, it didn't mean that and tried to avoid it by saying units; and the effect of using that rule is actually the opposite. For example, if the calculated result is 6.340658, using this rule causes one to compare the last two significant figures to the result. Since 58>15, one rounds 6.340658 ~ 6.34066, and yet again 66 > 15 -> ~ 6.3407. Since 07 < 15, the result would be 6.3407. Usually, I only managed to get one or two significant figures after the roundings. I know the rounding rule is arbitrary, but I suppose it was used because the equipment was really antiquated and they didn't want that the students would think they could measure something accurately with those things. On the other hand, I don't blame them for that, and made us understand that any sort of rounding rule just depends on what kind of effect is sought after.

I'd never heard of that rounding rule before.  In my line of work, sig figs for the measured value are given until you hit the error; so, for example, if you measure some distance to be 156846 km with an error of 1000 km, the final result is 157000+-1000 km.  Careful error estimation that takes into account all the systematic errors (basically, errors related to the equipment used) should give a good idea of the actual accuracy of a measurement.

For the use of summed absolute values instead of summed squared values, take a look at here. There is a table that describes the use of Ra and Rq, both of them are used to describe surface texture roughness. Ra is used by metal industry, Rq is used by optical industry. What's the difference? Rq is affected more by surface spikes due to the square in the equation, while Ra isn't usually that sensitive to them. So for your local mechanical work shop, it doesn't matter if there is a big local spike on the surface if there is only one a few them; they are likely going to get neutered in any case when the piece fixed with another one. Not so in Optics, as these spikes will cause scattering effects and they need to be reduced. I thought this analogue might be the best way to describe the difference between using summed absolute values and summed squares in error estimation.

I think we may have something of a language barrier here.  I wouldn't call what you posted any kind of error measurement; in fact, I would expect there to be an error associated with the roughness measurement, so the result would given like R_q = 5+-1 microinch or something.

Following will be some stupid questions by a person who doesn't have background in Astronomy, so bear with me:
- Thinking the above from the perspective of Earth, does this mean that there will be systematic ~3" uncertainty in the absolute direction of all measured galaxies, and that it applies equally to everything that we see as the Sun is located closer to the edge than center of Milky Way whose gravitation will direct the light?
- And seen from the perspective of the photon that is starting from a star in a galaxy far away, the closer it is to the center of the galaxy, the more its path will be bend? For me that would mean that there is less photons arriving on the telescope from the center than from the edge, and that the images of the galaxies have a dimmer center than they actually should have?

I don't have time to answer these right now; I'll get to them tonight.

[watsisname]

I'll try to take a stab at these.  I will warn that I am not 100% certain of the accuracy of my answers, so if someone finds I made a mistake then please correct me.

- Thinking the above from the perspective of Earth, does this mean that there will be systematic ~3" uncertainty in the absolute direction of all measured galaxies, and that it applies equally to everything that we see as the Sun is located closer to the edge than center of Milky Way whose gravitation will direct the light?

The deflection shouldn't be equal in all directions, and it should be able to be accounted for if we need to (so I wouldn't call it an uncertainty, exactly).  Let's look at two example scenarios and see if this helps make things more clear:

Consider a galaxy which is located in the opposite side of the sky from the center of our own galaxy.  (Pefectly 180° away).  Then the light from that galaxy, en route to our eyes, is traveling in a straight line from one galaxy to the other and thus it experiences no deflection angle at all.  However, it will experience a redshift as it leaves the source galaxy's gravitational field, followed by a blueshift as it enters our gravitational field.  If the galaxies' masses are similar then you might expect the shifts to mostly cancel each other out.  (And of course there's also a redshift from the expansion of space during the light's transit -- the more distant the galaxy, the greater the redshift).

Now consider a galaxy that is on a line (in reality, not visually) 90° away from the center of our own.  Then the light from that galaxy that was originally directed towards us will be deflected toward the center of our galaxy, therefore missing us.  Instead we see the light that was originally heading slightly outward, meaning that the galaxy appears to be a little more than 90° away from our galactic center than it is in reality.

That's one of the weird things about gravitational lensing; the image of the background object being lensed is displaced away from the foreground object that is acting as the lens.  I guess in that sense it is like looking through a convex lens. 

- And seen from the perspective of the photon that is starting from a star in a galaxy far away, the closer it is to the center of the galaxy, the more its path will be bend? For me that would mean that there is less photons arriving on the telescope from the center than from the edge, and that the images of the galaxies have a dimmer center than they actually should have?

Well if the star is in line with the center of the source galaxy then there's no deflection at all, since the light ray going from the star to our eyes is traveling perfectly in line with the gravitational field gradient.  Imagine throwing a ball straight upward -- ignoring the rotation of the earth, the ball's trajectory remains vertical.

However, the image of a star that was on the "side" of the galaxy would be deflected, and it'd appear shifted away from the galaxy's center.  So this would mean that galaxies appear slightly larger than they really are, though again I'm quite sure the effect is fairly minimal.  (I hesitate to throw numbers out there since I'm not to well versed in GR.) 

As for the apparent brightness, I don't think this would be affected in any way -- we're still seeing the same number of light rays from all parts of the galaxy, just some of them were originally heading in slightly different directions as others.


@ Astronomiya:  Nice to see someone working with improving exoplanet detection.   My central interest is exoplanets, though I still have several years of education to go before I can contribute to the field.

[Astronomiya]

Mika, sorry it's taken me so long to get back to this thread.  However, it looks like watsisname answered your questions (I can't find fault with his answers, that's for sure).  I agree with him that the apparent brightness of the galaxy shouldn't be affected in most cases, be it the center or edges.

[Mika]

I'd never heard of that rounding rule before.  In my line of work, sig figs for the measured value are given until you hit the error; so, for example, if you measure some distance to be 156846 km with an error of 1000 km, the final result is 157000+-1000 km.  Careful error estimation that takes into account all the systematic errors (basically, errors related to the equipment used) should give a good idea of the actual accuracy of a measurement.

You are not the first one who hasn't heard about it  . Most of us thought it was idiotic back then, but on the hindsight it actually taught more about the error estimation than using standard techniques by the book and top of the line equipment. And boy did they make us think about what went wrong! I recall measuring a gravitational constant of 7.3*10^-11 [m^3/kg/s^2 ], but at least I got the decade right. During that particular measurement we had to measure a distance inside a room using a clothesline...

And what about that effective focal length of a convex mirror? The one with the focal length of  0.4 m,  +/- 0.3 m 
I had rather sarcastic comments for the assistant in that measurement report.

The 15 number rounding rule only applied on the result of differential error calculation itself, and there was even more of that which I forgot. Assume that after substitution of deltas and relevant terms to the differential error equation one gets an error of 5.658861234769. Now the least accurate term in the error equation determined the number of decimals allowed. Say that one participant of those terms had accuracy of 3 decimals, so we start to think 5.658|861234769. Since 58 > 15, -> 5.658 ~ 5.66 and since again 66>15, -> 5.66 ~ 5.7, and yet again since 57>15, the end result will be 6. I'm not sure if the last rounding was necessary. So instead of that 3 decimals that we thought to be reasonable, I ended up with either one decimal, or no decimals at all. The reason for this operation was that at least we were not downplaying the error 

Say that the calculated measurement result itself was 126.4575. The error determines the number of decimals we put there and luckily the result itself would get rounded with normal rounding rules. So the result would have been 126 +/-6.

I think we may have something of a language barrier here.  I wouldn't call what you posted any kind of error measurement; in fact, I would expect there to be an error associated with the roughness measurement, so the result would given like Rq = 5+-1 microinch or something.

This was an analogy to the differences in the error estimation using absolute sums instead of squared sums. When using squared sums in the error estimation, you will weigh more single terms with higher uncertainities. The effect will be greater when there are more terms in the error equation. Then one needs to think if this kind of effect is desirable / realistic or not. I'm not sure, but I think this has something to do with the probability distribution of the error too, for example say a distance tolerance of +/-0.1 m uniformly distributed is different from +/-0.1 m normally distributed tolerance.

Now back to Astronomy in the next post.

[Mika]

Well, after writing some stuff, I found out that perhaps it is better to do that tomorrow, brain doesn't seem to be the sharpest around 01:23.

The thought I had went somehow like this: the light bending equation assumes b>>M and thus can't be applied when the source is close to the axis between two galaxies. There is no bending if the source lies on the axis between two galaxies, but the effect will be greatest if the source is slightly offset from the axis? There's something I don't understand here, but perhaps tomorrow will be a better day. Is there a some sort of maximum bending angle if one doesn't do the approximation of b>>M?

Bah, I need to review my GR notes. Some time has already passed since I attended those lectures, and there has been nothing comparable to that in my daily work.

[watsisname]

Ahh, I think I see where some of the confusion is coming from.  There is a difference when we're considering a lightsource within a galaxy or a lightsource behind a galaxy.  The equation Astronomiya had provided was (I believe) for the latter case, while my previous post was mainly dealing with the former.

If the lightsource, say a quasar, is directly behind a (assume symmetrical) foreground galaxy, then the image of the quasar will become a perfect ring around the foreground galaxy.  If the alignment isn't perfect then you get some kind of offset pair of images, and then things can get increasingly complex from there.

Maybe a good way of conceptualizing this is to try it on paper.  Draw a pair of galaxies and try to figure out (roughly) where the lightray paths go to reach your eyes.  All you have to remember is that they get curved toward the center of a gravitational field; the stronger the field the stronger the curvature.  Example:

[Astronomiya]

The 15 number rounding rule only applied on the result of differential error calculation itself, and there was even more of that which I forgot. Assume that after substitution of deltas and relevant terms to the differential error equation one gets an error of 5.658861234769. Now the least accurate term in the error equation determined the number of decimals allowed. Say that one participant of those terms had accuracy of 3 decimals, so we start to think 5.658|861234769. Since 58 > 15, -> 5.658 ~ 5.66 and since again 66>15, -> 5.66 ~ 5.7, and yet again since 57>15, the end result will be 6. I'm not sure if the last rounding was necessary. So instead of that 3 decimals that we thought to be reasonable, I ended up with either one decimal, or no decimals at all. The reason for this operation was that at least we were not downplaying the error 

Say that the calculated measurement result itself was 126.4575. The error determines the number of decimals we put there and luckily the result itself would get rounded with normal rounding rules. So the result would have been 126 +/-6.

That's kinda weird.  I was taught that only the first two digits in the error made any sense whatever, so you could have an error of 5.6, or 56, say, but not 56.0.  For the rest of your post, I guess I'll go check out a statistics book or something.  That would be a useful learning experience anyway.

Now, on to GR...

Well, after writing some stuff, I found out that perhaps it is better to do that tomorrow, brain doesn't seem to be the sharpest around 01:23.

The thought I had went somehow like this: the light bending equation assumes b>>M and thus can't be applied when the source is close to the axis between two galaxies. There is no bending if the source lies on the axis between two galaxies, but the effect will be greatest if the source is slightly offset from the axis? There's something I don't understand here, but perhaps tomorrow will be a better day. Is there a some sort of maximum bending angle if one doesn't do the approximation of b>>M?

Bah, I need to review my GR notes. Some time has already passed since I attended those lectures, and there has been nothing comparable to that in my daily work.

Remember what M means in this case.  In MLT (i.e., G = c = 1; I was using these before somewhat, and I should have mentioned it then, sorry) units, the Schwarzschild metric (which I've been using up to this point), reads

ds² = -(1 - 2M/r)dt² + (1 - 2M/r)^-1dr² + r²dΩ²

2M is the Schwarzschild radius in these units, which for the 1E12 solar masses in the Milky Way, is about a third of a light year.  So really, as long as we have spherical symmetry in the problem, we really don't have to worry about the approximation becoming invalid.  However, the Milky Way is rather obviously not spherically symmetric, and especially for light traveling within the galactic plane, I would not expect any serious deflection unless it passed very close to a black hole or other such object inside the galaxy itself.

There can be lensing even if the object is directly behind the gravitational lens; see Einstein's Cross as an example, and the picture watsisname just posted.  It's true that light rays on a direct line from the source through the lensing object to the observer will not be bent.  However, since luminous objects tend to emit light in all directions, some of it will not be heading straight for our eyes, but would instead have missed us completely were it not for the presence of the gravitational lens.  Does this clear things up a bit, Mika?

For the maximum bending angle, it depends on what kind of object we're dealing with.  Most things, like stars and planets, have a maximum deflection capability given by using their radius as b, so the photons are just skimming their surface (for these spherical objects, the Schwarzschild metric applies).  For the extreme case, a black hole, the angle can be made infinite, if the photon comes in on a path that puts it in the photon sphere, the point where photons have circular orbits (I think it's at r = 3M, again in MLT units).  Assuming you can send a photon off precisely enough, you can cause it to complete any number of orbits you want before it's either swallowed or ejected.

The equation Astronomiya had provided was (I believe) for the latter case, while my previous post was mainly dealing with the former.

Correct.  Sorry for any confusion I've caused, guys.  I should have been more clear.

[Mika]

It seems we have partially identified the issue, I know the gravitational lensing effect in a sense of imaging effect, i.e. a galaxy between us and a quasar images the quasar to us. Since we only see the last known direction of those photons, we get a ring.

My problem is trying to understand the effect of the gravitational potential of a galaxy on the photon paths when the source was in the same galaxy to begin with, not a gravitational lensing effect in the usual sense where some distant target is imaged. I recall that there is usually (?) a massive black hole in the middle of the galaxy, or at least some kind of source of massive gravitational potential. If it is so that the photons experience more bending closer to the gravitational potential, I would think that there is radial distortion on the position of the stars from the center of the galaxy that displaces stars towards the edge of the galaxy, and the closer their actual position is to the center, the greater the effect. I think I would need to put a drawing of this one out here, but on the other hand I wouldn't like to register in to yet again one more net service to provide the drawing.

Yes, I think now too that the brightness shouldn't be affected since the stars should be taken as point sources and then by definition any direction has equal weight.

[watsisname]

I recall that there is usually (?) a massive black hole in the middle of the galaxy, or at least some kind of source of massive gravitational potential.

  Indeed.  Our own galaxy has one of about 4 million solar masses, and observations suggest that most massive galaxies contain them as well.  Interestingly, the mass of the hole appears to be directly related to the mass of the host galaxies' bulge, for reasons we don't quite understand yet.  But anyways,

If it is so that the photons experience more bending closer to the gravitational potential, I would think that there is radial distortion on the position of the stars from the center of the galaxy that displaces stars towards the edge of the galaxy, and the closer their actual position is to the center, the greater the effect.

I think Astronomiya is in a better position to answer this than I am, though I'm fairly certain that any distortion due to a supermassive black hole should be negligible for objects beyond a few light years of it.  Most of the extreme gravity of a black hole is contained in a very small region of space.  IIRC our own galaxy's SM-black hole would fit within the orbit of mercury, and stellar-mass black holes are on the order of kilometers across.  Now if on the other hand you're specifically looking at an object within a few radii of a black hole, then there's definitely distortion in the image.

I think I would need to put a drawing of this one out here, but on the other hand I wouldn't like to register in to yet again one more net service to provide the drawing.

You can put it as an attachment, or just upload it to imageshack.

[Mika]

The time goes fast, almost too fast. I'll try to get the picture done tomorrow. If I don't get it uploaded here tomorrow, expect something new from me earliest on next Saturday.

Sorry about cutting off from the good discussion for a couple of days.

[Astronomiya]

It seems we have partially identified the issue, I know the gravitational lensing effect in a sense of imaging effect, i.e. a galaxy between us and a quasar images the quasar to us. Since we only see the last known direction of those photons, we get a ring.

My problem is trying to understand the effect of the gravitational potential of a galaxy on the photon paths when the source was in the same galaxy to begin with, not a gravitational lensing effect in the usual sense where some distant target is imaged. I recall that there is usually (?) a massive black hole in the middle of the galaxy, or at least some kind of source of massive gravitational potential. If it is so that the photons experience more bending closer to the gravitational potential, I would think that there is radial distortion on the position of the stars from the center of the galaxy that displaces stars towards the edge of the galaxy, and the closer their actual position is to the center, the greater the effect. I think I would need to put a drawing of this one out here, but on the other hand I wouldn't like to register in to yet again one more net service to provide the drawing.

Yes, I think now too that the brightness shouldn't be affected since the stars should be taken as point sources and then by definition any direction has equal weight.

Yes, the Milky Way does have a supermassive black hole at its center, which has a radius of about .08 AU (an AU, or astronomical unit, is the mean Earth-Sun distance), well within Mercury's orbit of about .39 AU (semi-major axis).  For future reference, remember that the Schwarzschild radius of a black hole is given by

R_s = 2GM/c^2

In applying our approximation, two orders of magnitude is sufficient (more than, usually) for a "much greater" or "much less" approximation like the deflection angle one to hold; so, we can apply it to photons that pass no closer than 100 times R_s, which for the Milky Way's black hole, is about 8 AU, or out near Saturn's orbit in our solar system.  The deflection of a photon that gets that close is approximately a whopping two radians, or about 120 degrees.  However, this is of course absurdly close for a photon to get; what about b = 1 light year?  That means the deflection angle is instead .1", which is tiny.  Therefore, we can conclude that the central black hole has almost no lensing influence except in its immediate vicinity.  Since the effect from stars is even less (the Sun has a mere 1.7" deflection for a surface-skimming photon), we can safely conclude that when observing targets in our own galaxy, gravitational lensing can be safely ignored in just about every situation.

One way of thinking about it is remembering that GR, despite being at its heart nonlinear (specifically in the metric*), in the weak field limit is linear, because it must give back Newtonian gravity in that situation, which IS linear.  Then you can think about simply summing up the contributions from each mass a given photon encounters, because the principle of superposition applies.  In this manner, it should become clear that given the vastness of space between the stars, virtually no lensing takes place unless it passes very close to something, which is very unlikely (just think about all the stars you can see at night in the Milky Way, and how their light is almost totally unobstructed, passing directly from them to your eyes).  If that light does pass very close to something massive, the deflection is noticeable and must be accounted for.

Now you might ask, why do we see significant lensing when galaxy clusters are involved, but not star clusters, despite even vaster distances being involved?  Well, it turns out that while stars are amazingly far apart compared to their sizes, galaxies are not.  Consider the distance between us and Andromeda, the nearest other massive galaxy to us.  This is about 2.5 million light years, or 25 Milky Way diameters, discounting the dark matter halo.  25 solar diameters doesn't even get you to Mercury if you start from the Sun!  This is to say nothing of getting to Alpha Centauri, which is about 40 million or so solar diameters away.  So we can see that on the galactic scale, clusters are much denser than star clusters are, and thus are going to have a much greater lensing effect.

*Here are the Einstein field equations:  R_ab + (Λ - R/2)g_ab = 8πG/c⁴*T_ab

They look superficially linear in the metric g_ab at first, until you remember that R_ab and T_ab, the Riemann curvature and stress-energy tensor respectively, both depend on the Christoffel symbols, which depend on derivatives of the metric.  Happy happy fun times.

[Mika]

I had no idea ImageShack could be used without registration!


[Galaxy NGC4414 and Earth images are originally by NASA, both from Wikipedia.]

I thought the thread was in dire need of the image of the scenario. I think Astronomiya may have partially answered my questions, but I still need to make this sure.
Above you see a telescope seeing a Galaxy, FOV being limited by the blue lines and white is the optical "straight" axis. Needless to say, all images are greatly exaggerated to visualize the effect. Sorry about the bad quality, but it's getting really late and I want to get this done and this will have to do, for now. When I get back on Saturday, I'll have more time to draw better figures.

Rightmost is a concept figure of a galaxy behaving as a GRIN lens, where greatest bending is happening closest to the optical axis.
Leftmost figure displays the effect of such bending. Earth based observer will see equi-distant star images on his telescope, while in reality, stars were not at all equidistant in the galaxy, they just appear so. This would be barrel distortion if I didn't make a mistake here between the source and the image.

The blackhole in the middle will distort the close by approaching photon paths. But orbiting the black hole, there is a large amount of mass in celestial objects. I might need to repeat a question now, but doesn't the mass around the black hole sum up too, to extend the effect of bending? I understand now that the distances between stars is so great that their gravitational fields should have a neglible effect by Newtons equation, but isn't it still possible to centralize a mass inside a spherical volume of radius r in the middle point of that sphere and then sum up the masses?

I, again, may need to think this with rested brain. It is 02.27 and I need to go to sleep. You'll hear from me earliest on Saturday. What I didn't know was that galaxies themselves aren't that far apart on their scale and I need to think about that too.

[Astronomiya]

The blackhole in the middle will distort the close by approaching photon paths. But orbiting the black hole, there is a large amount of mass in celestial objects. I might need to repeat a question now, but doesn't the mass around the black hole sum up too, to extend the effect of bending? I understand now that the distances between stars is so great that their gravitational fields should have a neglible effect by Newtons equation, but isn't it still possible to centralize a mass inside a spherical volume of radius r in the middle point of that sphere and then sum up the masses?

In the weak field limit, yes, as long as there is spherical symmetry.  A disk is much more complicated, and I'm not sure how to get exact effects here (there appear to be several papers on the subject, but the math in them requires a lot of effort to go through, and I unfortunately don't have the time to really go through them).  In any case, your diagram is, so far as I can tell, correct.  Even though our galaxy is a barred spiral, and thus the bulge is NOT spherically symmetric (not even close), I'm going to be lazy and call it close enough.  Imagine that we all live in Andromeda instead if you like.  In any case, the central bulge has a mass of about 2E10 solar masses, and a radius of some 10,000 light years (for our idealized spherical one; in actuality, each lobe of the bar is about 14,000 light years long).  Using the deflection formula I posted way back when, we find that a photon encountering the edge of the bulge will be deflected about .3", which really isn't much, though measurable (the Hipparcos satellite was able to obtain precision on the order of half a milliarcsecond).

I hope I'm not confusing you too badly; GR is already confusing enough without my help (this thread is actually helping me understand this stuff better too).

[watsisname]

If I might interject a question regarding black holes, this has been bugging me for a while:

My understanding is that with GR, an outside observer will never see an object actually cross an event horizon (due to whatever process you want to label it as -- infinite redshift, infinite time dilation, etc).  Wouldn't this also mean that, again for the outside observer, a black hole would be perpetually be in a state of forming/collapsing, since the matter that collapsed create the black hole in the first place would be stuck on the event horizon for all eternity?  If correct, is there any practical difference between this and the classical description of a black hole?

[Astronomiya]

If I might interject a question regarding black holes, this has been bugging me for a while:

My understanding is that with GR, an outside observer will never see an object actually cross an event horizon (due to whatever process you want to label it as -- infinite redshift, infinite time dilation, etc).  Wouldn't this also mean that, again for the outside observer, a black hole would be perpetually be in a state of forming/collapsing, since the matter that collapsed create the black hole in the first place would be stuck on the event horizon for all eternity?  If correct, is there any practical difference between this and the classical description of a black hole?

Not really.  If you run the calculation, you find that it all redshifts to infinity or close enough quite quickly (I think on the order of the fall-in time or thereabouts, actually).  So I suppose you could kinda think of it like all the matter that collapses to form the black hole becomes its event horizon - and if Leonard Susskind and company are correct, at the very least this is exactly what happens to the information contained in the matter that falls into a black hole.  From the infalling matter's point of view, it just keeps right on going to the singularity in finite time, and all its information content goes with it.  From outside, it appears as if the information has smeared itself completely evenly over the event horizon.

Incidentally, this is an excellent website when it comes to visualizing what happens near and inside black holes.

[Mika]

Hello, returned from my trip to Germany yesterday.

First of all, I managed to find something related to the topic we are talking about in the James Webb telescope design report. On page 18, Chapter 4.4 "Anisotropy measures and weak lensing".  It is mentioned there that the distortion introduced by gravity is countered by taking images of thousands of galaxies and then averaging the effect. Is there more about this somewhere, that would be publicly available (= free)? The effect is called "weak lensing" there, and luckily there seems to be something in the Wikipedia too. The general gravitational lens formalism can be found from there too. Unfortunately, that seems to consider the cases where a cluster of galaxies is imaging a distant target on Earth, and what I'm wondering about is different from that.

I also tried to use a GRIN lens to check some effects of the radial distortion, but realized that the GRIN lens models are usually given as n(r) = n₀ + n₁r + n₂r², where n(r) is the index of refraction given by the distance from the optical axis. n₁ and n₂ are coefficients that determine the index of refraction when stepping outside the optical axis (r>0). The problem with these models is that the bending effect is greatest on the edge (r = r_max), instead of next to optical axis. I would need to reverse the behavior of the GRIN model, and that would mean programming a new surface model - I'm not yet going to do that.

What I'm thinking now is that how does the distortion angle due to gravity behave as a function of distance from the optical axis in the case I drew in the figure. 3 arcseconds at the edge, but what happens close to the optical axis? "Close" meaning 10 or 5 percent of the radius of the galaxy - is that within the bulge volume?

[Astronomiya]

I also tried to use a GRIN lens to check some effects of the radial distortion, but realized that the GRIN lens models are usually given as n(r) = n₀ + n₁r + n₂r², where n(r) is the index of refraction given by the distance from the optical axis. n₁ and n₂ are coefficients that determine the index of refraction when stepping outside the optical axis (r>0). The problem with these models is that the bending effect is greatest on the edge (r = r_max), instead of next to optical axis. I would need to reverse the behavior of the GRIN model, and that would mean programming a new surface model - I'm not yet going to do that.

What I'm thinking now is that how does the distortion angle due to gravity behave as a function of distance from the optical axis in the case I drew in the figure. 3 arcseconds at the edge, but what happens close to the optical axis? "Close" meaning 10 or 5 percent of the radius of the galaxy - is that within the bulge volume?

In actuality, using that straight GRIN model might be the way to go; if you read my last post answering you, you'll find that I calculated the bulge had a full lensing power somewhat under an arcsecond, which is a good deal less than the galaxy's full lensing power.  Remember that as you go farther and farther out, you add more and more mass; you in fact end up adding a considerable amount (the bulge has a mass of about 2E10 solar masses, or maybe 5% of the galaxy's total mass).  So the statement that a gravitational lens's power increases the closer you get to it is only true as long as the deflected light is still "seeing," as it were, all of the lens's mass.  In the case of a galaxy, this is of course not true at all, unless the impact parameter is larger than the radius of the galaxy.

In answer to your question about closeness, for a spiral galaxy, yes you are within the bulge at 5-10% of its diameter; elliptical galaxies do not have one.

And paydirt for a freely available paper on gravitational lensing due to spiral galaxies:  http://iopscience.iop.org/0004-637X/486/2/681/pdf/0004-637X_486_2_681.pdf

[Mika]

The article you posted was a pretty fun read, they actually didn't create their own space time metric, but instead just constructed an different sort of model, whose justification is that it happens to work.

I know that the mass of the galaxy is decreasing, but so does the impact parameter b, which made me think earlier if there was some kind of maximum between these two at some radial distance from the center. That might not be the case if the distances between stars is so great. It probably wouldn't be a simple GRIN lens type curving either.

One more thing, I understand that there is no dust outside the disk of a galaxy, but how about on the disk? It is probably not like rings of Saturnus, but I would expect that there would be more within the plane of the disk?

[Astronomiya]

The article you posted was a pretty fun read, they actually didn't create their own space time metric, but instead just constructed an different sort of model, whose justification is that it happens to work.

Welcome to science; the justification is pretty much always "it happens to work."

I know that the mass of the galaxy is decreasing, but so does the impact parameter b, which made me think earlier if there was some kind of maximum between these two at some radial distance from the center. That might not be the case if the distances between stars is so great. It probably wouldn't be a simple GRIN lens type curving either.

There might be; I don't know enough about galactic mass distributions to say (you might have some luck hitting up ADS with an abstract search).  I share your doubt that a simple GRIN model would be the correct one.

One more thing, I understand that there is no dust outside the disk of a galaxy, but how about on the disk? It is probably not like rings of Saturnus, but I would expect that there would be more within the plane of the disk?

Oh, there's plenty of dust and free gas in the galactic plane in the form of the interstellar medium, but in total it only accounts for about 15% of the normal matter in the galaxy, the rest being in stars.  Of that 15%, only 1% is dust; the rest is mostly hydrogen.  Also keep in mind that visible matter only accounts for maybe 20% of the galaxy's total mass.  This means that for the purposes of gravitational lensing, the ISM can almost be ignored entirely, and the dust component certainly can be.