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Nereid
2010-Nov-19, 08:36 PM
Has the GR gravitational deflection of light, by the Sun, been observed in any region of the electromagnetic (EM) spectrum blue-ward of the Lyman limit (i.e. EUV, x-ray, gamma-ray)? If so, where are such observations published?

Would it be possible for Chandra, or XMM-Newton, to detect it, if either observed the Sun eclipsing (occulting?) the Crab nebula (or some other very bright x-ray point source)? I suspect it might, just - their astrometric resolution is just good enough - except that they may not be equipped to stare at the Sun.

What is the (wavelength-dependent) deflection of EM radiation, of distant sources, by the Sun near its limb, due to the inner corona? I think in all wavebands it would be far, far too small to detect, except possibly at some microwave/radio frequencies.

Amber Robot
2010-Nov-19, 10:19 PM
Would it be possible for Chandra, or XMM-Newton, to detect it, if either observed the Sun eclipsing (occulting?) the Crab nebula (or some other very bright x-ray point source)?

Both of those satellites have sun avoidance angles. Maybe if it were the last observation you'd ever do with the satellite they might point in that direction. But I'm guessing that your pointing wouldn't be accurate enough to do the measurement you want because I'm sure the startrackers they have aren't built for looking at the sun.

macaw
2010-Nov-20, 12:13 AM
Has the GR gravitational deflection of light, by the Sun, been observed in any region of the electromagnetic (EM) spectrum blue-ward of the Lyman limit (i.e. EUV, x-ray, gamma-ray)? If so, where are such observations published?

Would it be possible for Chandra, or XMM-Newton, to detect it, if either observed the Sun eclipsing (occulting?) the Crab nebula (or some other very bright x-ray point source)? I suspect it might, just - their astrometric resolution is just good enough - except that they may not be equipped to stare at the Sun.

What is the (wavelength-dependent) deflection of EM radiation, of distant sources, by the Sun near its limb, due to the inner corona? I think in all wavebands it would be far, far too small to detect, except possibly at some microwave/radio frequencies.

Hi Nereid

I only know about some research in the very high frequency (20GHz) spectrum, see here (http://en.wikipedia.org/wiki/Gravitational_lens#Search_for_gravitational_lenses ).

StupendousMan
2010-Nov-20, 03:12 PM
In order to measure the gravitational deflection of light by the Sun, one needs to have a set of sources which are close enough to the Sun to suffer significant deflection, and bright enough to have very precisely measured positions.

Even if one could point X-ray or gamma-ray telescopes towards the Sun (which, as another poster has pointed out, one can't), I doubt that there are any fields rich enough in bright X-ray sources to create the sort of astrometric net required to measure the deflection.

Nereid
2010-Nov-20, 03:47 PM
In order to measure the gravitational deflection of light by the Sun, one needs to have a set of sources which are close enough to the Sun to suffer significant deflection, and bright enough to have very precisely measured positions.

Even if one could point X-ray or gamma-ray telescopes towards the Sun (which, as another poster has pointed out, one can't), I doubt that there are any fields rich enough in bright X-ray sources to create the sort of astrometric net required to measure the deflection.
Thanks.

You could certainly do this with gamma-ray sources, especially high energy/hard ones (the Sun is pretty dim in those wavebands); however, you'd need a pretty big instrument (gamma ray fluxes, except from GRBs, are really very low), and one with an angular resolution (far) better than any current instrument or facility.

In at least some x-ray bands, it may be possible to sidestep an astrometric net to some extent; if the centre of the Sun could be located sufficiently precisely - from its x-ray emission alone - then there are point sources bright enough that timing information should give precise positions wrt the Sun. Of course, as has been pointed out, existing space-based x-ray telescopes are not easily pointed at the Sun! :clap:

Nereid
2010-Nov-20, 03:48 PM
Hi Nereid

I only know about some research in the very high frequency (20GHz) spectrum, see here (http://en.wikipedia.org/wiki/Gravitational_lens#Search_for_gravitational_lenses ).
Thanks.

macaw
2010-Nov-20, 04:11 PM
Thanks.

Your chances are much better if you abandon the idea of using the Sun as the source of bending. See here (http://www.sciencedaily.com/releases/2010/01/100119172846.htm).

ngc3314
2010-Nov-20, 08:19 PM
Your chances are much better if you abandon the idea of using the Sun as the source of bending. See here (http://www.sciencedaily.com/releases/2010/01/100119172846.htm).

Come to think of it, ROSAT and Chandra have detected the lensed images at least in the "original" lensed QSO system 0957+561.

macaw
2010-Nov-20, 09:27 PM
Come to think of it, ROSAT and Chandra have detected the lensed images at least in the "original" lensed QSO system 0957+561.

Black holes seem to be a lot better bet than the Sun when it comes to lensing.

a1call
2010-Nov-20, 10:34 PM
Distortion from the Solar Atmosphere (http://en.wikipedia.org/wiki/Sun#Atmosphere) would probably be an obstacle for making precise optical observations.

StupendousMan
2010-Nov-21, 04:24 AM
Black holes seem to be a lot better bet than the Sun when it comes to lensing.

An object with one solar mass located halfway between us and a target will do a MUCH better job at lensing then an object with one solar mass which is only 0.000000000000000000000000000000001% of the way from us to the target.

StupendousMan
2010-Nov-21, 04:26 AM
Thanks.


In at least some x-ray bands, it may be possible to sidestep an astrometric net to some extent; if the centre of the Sun could be located sufficiently precisely - from its x-ray emission alone - then there are point sources bright enough that timing information should give precise positions wrt the Sun. Of course, as has been pointed out, existing space-based x-ray telescopes are not easily pointed at the Sun! :clap:

First, the Sun is not a symmetric source in X-rays, so it would be pretty hard to define its center; especially as the X-ray appearance changes over time, both due to rotation and changes in the solar X-ray luminosity at different locations.

Second, I don't understand what you mean by "timing information." Could you explain?

macaw
2010-Nov-21, 05:43 AM
An object with one solar mass located halfway between us and a target will do a MUCH better job at lensing then an object with one solar mass which is only 0.000000000000000000000000000000001% of the way from us to the target.

True but I don't see though what this has to do with my comment re black holes.Besides, there are quite a few black holes that have masses much bigger than the Sun.

ngc3314
2010-Nov-21, 01:08 PM
True but I don't see though what this has to do with my comment re black holes.Besides, there are quite a few black holes that have masses much bigger than the Sun.

They are all so distant from us that lensing by a black hole has yet to be detected. On the other hand, the intense X-ay emission from accreting black holes makes them really good background sources to be lensed.

Jeff Root
2010-Nov-21, 02:07 PM
An object with one solar mass located halfway between us and a
target will do a MUCH better job at lensing then an object with one
solar mass which is only 0.000000000000000000000000000000001%
of the way from us to the target.
If the solar mass is the Sun, at the distance between the Sun and the
Earth, the target in your example would be 1.5E30 light-years away.

-- Jeff, in Minneapolis

macaw
2010-Nov-21, 03:37 PM
They are all so distant from us that lensing by a black hole has yet to be detected. On the other hand, the intense X-ay emission from accreting black holes makes them really good background sources to be lensed.


Thank you

grapes
2010-Nov-21, 06:06 PM
If the solar mass is the Sun, at the distance between the Sun and the
Earth, the target in your example would be 1.5E30 light-years away.I think they got carried away with the zeroes, or exaggerated some. I guess the point is that a gravitational lense has a "focal length" and the sun is a weak gravitational lense--the gravitational "bend" is given as 1.75 seconds of arc (http://www.mathpages.com/rr/s6-03/6-03.htm), which for a sun of radius of 700,000 km would focus at a distance of 700,000 * 360 * 60 * 60 / 1.75, about 5 x 1011km, 100 times the orbit of Pluto.
,

StupendousMan
2010-Nov-21, 11:23 PM
True but I don't see though what this has to do with my comment re black holes.Besides, there are quite a few black holes that have masses much bigger than the Sun.

Sorry, I was too brief. What I was trying to communicate was the fact that it's not only the mass of an object which is important for bending light rays: it's also the location of the lensing object relative to the observer and to the target object. The optimal situation is when the lensing object is exactly halfway between the observer and the target: in that situation, the degree to which the light rays are bent will be a maximum.

We are so close to the Sun, and so far away from most astronomical sources, that the bending of the target's light rays by the Sun is very difficult to detect.

Jeff Root
2010-Nov-22, 01:24 AM
The optimal situation is when the lensing object is exactly halfway
between the observer and the target: in that situation, the degree to
which the light rays are bent will be a maximum.
Is that also true of lensing by ordinary optics? I suspect that it is,
but has no relevance for astronomical imaging.

-- Jeff, in Minneapolis

macaw
2010-Nov-22, 01:42 AM
We are so close to the Sun, and so far away from most astronomical sources, that the bending of the target's light rays by the Sun is very difficult to detect.

Precisely, this (and other reasons such as much larger mass) is why I was advocating using the lensing effect of black holes INSTEAD of the lensing effect of the sun. this is why I didn't understand the relevance of your post.


The optimal situation is when the lensing object is exactly halfway between the observer and the target: in that situation, the degree to which the light rays are bent will be a maximum.

I had to go back to the GR books to verify the above, it turns that it is not true. The books tend to to the calculations for the case when the bending body is situated half way between the source and the observer because that makes the calculation easier. The deflection angle is directly proportional with the mass and inversely proportional with the distance to the center of the bending body. There is no dependency on the relative position between source, bending body and observer.

ngc3314
2010-Nov-22, 02:01 PM
The optimal situation is when the lensing object is exactly halfway between the observer and the target: in that situation, the degree to which the light rays are bent will be a maximum.

I had to go back to the GR books to verify the above, it turns that it is not true. The books tend to to the calculations for the case when the bending body is situated half way between the source and the observer because that makes the calculation easier. The deflection angle is directly proportional with the mass and inversely proportional with the distance to the center of the bending body. There is no dependency on the relative position between source, bending body and observer.

Both statements can be correct. The deflection angle is dependent only on lens properties and impact parameter, but that's not what we measure or select objects on. The optimal observability comes from the largest apparent separation between lensed images (or angular radius of Einstein ring), which happens (for fixed lens mass and in the Euclidean approximation) for a lens halfway to the source. (In cosmological applications, the appropriate definitions of distance for each segment need to be used.) The connection between deflection angle and image separation comes because the lens properties and our distance select what impact parameter applies to the rays which reach the observer.

ngc3314
2010-Nov-22, 02:10 PM
Has the GR gravitational deflection of light, by the Sun, been observed in any region of the electromagnetic (EM) spectrum blue-ward of the Lyman limit (i.e. EUV, x-ray, gamma-ray)? If so, where are such observations published?


It's not the most ironclad result, but there is now a report of detection in the time domain. The blazar PKS 1830-211 is lensed - in the radio there are two images with a complete fainter Einstein ring. This field is at low galactic latitude, so optical data lagged well behind - HST eventually showed the spiral lens galaxy. Now this paper reports a statistically defensible detection of a time delay in gamma rays, so they can se an effect of the lensing even with the poor angular resolution of Fermi:

First evidence for a gravitational lensing-induced echo in gamma rays with Fermi LAT (http://arxiv.org/abs/1011.4498), by A. Barnacka, J-F.Glicenstein, and Y. Moudden (Submitted on 19 Nov 2010).



Aims. This article shows the first evidence for gravitational lensing phenomena in high energy gamma-rays. This evidence comes from the observation of a gravitational lens induced echo in the light curve of the distant blazar PKS 1830-211. Methods. Traditional methods for the estimation of time delays in gravitational lensing systems rely on the cross-correlation of the light curves of the individual images. In this paper, we use 300 MeV-30 GeV photons detected by the Fermi-LAT instrument. The Fermi-LAT instrument cannot separate the images of known lenses. The observed light curve is thus the superposition of individual image light curves. The Fermi-LAT instrument has the advantage of providing long, evenly spaced, time series. In addition, the photon noise level is very low. This allows to use directly Fourier transform methods. Results. A time delay between the two compact images of PKS 1830-211 has been searched for both by the autocorrelation method and the “double power spectrum” method. The double power spectrum shows a 3-sigma evidence for a time delay of 27.5 +/- 1.3 days, consistent with the result from Lovell et al. (1998). The relative uncertainty on the time delay estimation is reduced from 20% to 5%.

macaw
2010-Nov-22, 04:10 PM
Both statements can be correct. The deflection angle is dependent only on lens properties and impact parameter, but that's not what we measure or select objects on. The optimal observability comes from the largest apparent separation between lensed images (or angular radius of Einstein ring), which happens (for fixed lens mass and in the Euclidean approximation) for a lens halfway to the source.

Wouldn't that happen for the case of the lens being actually very close to the source rather than being half-way between the source and the observer?

StupendousMan
2010-Nov-22, 08:59 PM
For those who have a little time and mathematical aptitude, there's a pretty good description of the physics of gravitational lensing in this doctoral dissertation by Robert Schmidt:

http://www.ari.uni-heidelberg.de/mitarbeiter/rschmidt/papers/thesis_html/thesis_publish.html

I recommend especially the section called "The Point Mass Lens."

There's a simple simple simple introduction to the topic in one of my class lectures, at

http://spiff.rit.edu/classes/phys240/lectures/grav_lens/grav_lens.html

but it clearly doesn't tell one enough to answer the questions in this thread. It might be enough for someone to do a quick order-of-magnitude computation of the appearance of some particular situation, though.


Perhaps others can tell us their favorite sources for details on gravitational lensing?

caveman1917
2010-Nov-22, 09:12 PM
Wouldn't that happen for the case of the lens being actually very close to the source rather than being half-way between the source and the observer?

Wouldn't it be for the lens being very close to the observer, rather than the source or half-way?

ETA: nope, it's indeed half-way.

macaw
2010-Nov-23, 02:01 AM
Wouldn't it be for the lens being very close to the observer, rather than the source or half-way?

ETA: nope, it's indeed half-way.

I don't think so, I think that :

-the angle depends only on the ratio M/R (there is no disagreement on that)

-the linear displacement tends to be larger as the lense gets closer to the source (there is debate on this issue).

StupendousMan
2010-Nov-23, 02:45 AM
I
-the linear displacement tends to be larger as the lense gets closer to the source (there is debate on this issue).

Maybe the problem here is that we might be using similar words to describe different things. "The linear displacement" -- what might that mean?

Consider the following way to describe lensing. Start with a source of light, far from the observer. Set up a coordinate system so that the source lies at the origin. Let the observer be located far away, along the positive z-axis. In the absence of any lens, the observer will look back along the z-axis and see the source at the origin of an x-y plane, projected onto the sky.

Now, place a lensing object somewhere along the z-axis, between the observer and the source. The gravity of the lens bends the light rays from the source, so the observer now sees image(s) of the source which are shifted somewhat away from the origin.

The question is: for some particular lensing object, where should we place it to cause the images of the source -- as perceived by the observer -- to be as far from the origin of the x-y plane as possible?

Does this help to clarify the issue we are discussing?

Jeff Root
2010-Nov-23, 02:55 AM
Diagrams! I want diagrams!

-- Jeff, in Minneapolis

macaw
2010-Nov-23, 03:21 AM
Maybe the problem here is that we might be using similar words to describe different things. "The linear displacement" -- what might that mean?

Distance_between_lens_and_observer*Deflection_angl e (i.e. theta)




Consider the following way to describe lensing. Start with a source of light, far from the observer. Set up a coordinate system so that the source lies at the origin. Let the observer be located far away, along the positive z-axis. In the absence of any lens, the observer will look back along the z-axis and see the source at the origin of an x-y plane, projected onto the sky.

Now, place a lensing object somewhere along the z-axis, between the observer and the source. The gravity of the lens bends the light rays from the source, so the observer now sees image(s) of the source which are shifted somewhat away from the origin.

The question is: for some particular lensing object, where should we place it to cause the images of the source -- as perceived by the observer -- to be as far from the origin of the x-y plane as possible?

Does this help to clarify the issue we are discussing?

I understand GR lensing extremely well, I have derived the formulas many times, there is no no distance observer-lens dependency in the determination of the deflection angle .

StupendousMan
2010-Nov-23, 03:49 AM
I am trying to understand the reason we are failing to agree on the situation in which lensing is most noticeable.

Suppose we work in the situation I described two messages ago: source of light at origin, observer far away along the z-axis. You are given a point mass to act as the lens. You are free to place this mass anywhere along the z-axis.

My claim is that you will make the lensing most obvious to the observer if you place the lens halfway between source and observer.

Your claim is that the lens should be placed close to the source -- is that right? Could you confirm this?

Is it possible that there is a misunderstanding of the meaning of the deflection angle "theta"? There are two angles one might discuss in this situation:

a) the angle between the line of sight from observer to source, and the line of sight from observer to lensed image

b) the angle between the line of sight from observer to source, and the direction of a light ray which has been bent by the maximum deflection angle, but which does NOT reach the observer.

Perhaps you are considering possibility b), while I've been considering possibility a).

macaw
2010-Nov-23, 04:37 AM
I am trying to understand the reason we are failing to agree on the situation in which lensing is most noticeable.

Suppose we work in the situation I described two messages ago: source of light at origin, observer far away along the z-axis. You are given a point mass to act as the lens. You are free to place this mass anywhere along the z-axis.

My claim is that you will make the lensing most obvious to the observer if you place the lens halfway between source and observer.

Can you produce a mathematical formalism that proves the above? It would help a lot in clarifying your claim.


Your claim is that the lens should be placed close to the source -- is that right? Could you confirm this?

Yes.




Is it possible that there is a misunderstanding of the meaning of the deflection angle "theta"? There are two angles one might discuss in this situation:

a) the angle between the line of sight from observer to source, and the line of sight from observer to lensed image

b) the angle between the line of sight from observer to source, and the direction of a light ray which has been bent by the maximum deflection angle, but which does NOT reach the observer.

Perhaps you are considering possibility b), while I've been considering possibility a).

I am considering case a) as well, this is the definition of the deflection angle in any thextbook as in:

theta=4GM/(Rc^2)

As you can see, theta does not depend on the position of the lens between the source and the observer.

caveman1917
2010-Nov-23, 09:49 AM
Are we talking about the same theta here? I though theta is the angle of deflection along the lens (though i may be wrong in this, and in that case i'll have to reconsider my point, but i don't have time to do that now - have to go work in 15min). So angle 'a' as in StupendousMan's post would be ((pi - theta) / 2) for a half-way lens.

ngc3314
2010-Nov-23, 01:00 PM
I am considering case a) as well, this is the definition of the deflection angle in any thextbook as in:

theta=4GM/(Rc^2)

As you can see, theta does not depend on the position of the lens between the source and the observer.

Yes, we can see that (it is a standard result, and no one is quibbling about that form - only how it applies to specific lensing cases as we actually observe them), and we can also see that this angle is not the case (a) listed by StupendousMan. His case (a) is the angular offset we observe between the source position if it could be seen free of lensing effects and the one we observe (the angular radius of the Einstein ring in the symmetric case). If I read it correctly now, I have misinterpreted what you were calling the linear offset - from your formula above, am I right that it is the extrapolated linear difference between the point where the original and deflected rays pass the observer's position? I thought earlier it was the impact parameter at the lens.

StupendousMan
2010-Nov-23, 01:57 PM
Perhaps diagrams will help. I have been concentrating on the angle by which the images of a source appear to be deflected from the true position of the source, as seen by some distant observer. In other words, the angle "theta_E" in the diagram below.

http://spiff.rit.edu/classes/phys240/lectures/grav_lens/lens_diagram_B.gif

I suspect that you are concentrating on some other angle. Perhaps it is the angle by which rays from infinity are deflected as they move past a lensing object, like this:

http://spiff.rit.edu/classes/phys240/lectures/grav_lens/lens_diagram_A.gif

If this is the case, then we have just been talking about two different aspects of lensing. Yes, you are right, the angle "theta" shown in the second diagram doesn't depend on the distance to the observer. But I'm also right: if we place a lensing object in the wrong place, even through the angle "theta" may be very large, the deflected rays won't go to the observer, and so, despite the strong lensing effect, we on Earth won't see any lensed images.

macaw
2010-Nov-23, 04:13 PM
Yes, we can see that (it is a standard result, and no one is quibbling about that form - only how it applies to specific lensing cases as we actually observe them), and we can also see that this angle is not the case (a) listed by StupendousMan. His case (a) is the angular offset we observe between the source position if it could be seen free of lensing effects and the one we observe (the angular radius of the Einstein ring in the symmetric case). If I read it correctly now, I have misinterpreted what you were calling the linear offset - from your formula above, am I right that it is the extrapolated linear difference between the point where the original and deflected rays pass the observer's position.

Yes, this is correct.

macaw
2010-Nov-23, 04:18 PM
Perhaps diagrams will help. I have been concentrating on the angle by which the images of a source appear to be deflected from the true position of the source, as seen by some distant observer. In other words, the angle "theta_E" in the diagram below.

http://spiff.rit.edu/classes/phys240/lectures/grav_lens/lens_diagram_B.gif

I suspect that you are concentrating on some other angle. Perhaps it is the angle by which rays from infinity are deflected as they move past a lensing object, like this:

http://spiff.rit.edu/classes/phys240/lectures/grav_lens/lens_diagram_A.gif

If this is the case, then we have just been talking about two different aspects of lensing. Yes, you are right, the angle "theta" shown in the second diagram doesn't depend on the distance to the observer. But I'm also right: if we place a lensing object in the wrong place, even through the angle "theta" may be very large, the deflected rays won't go to the observer, and so, despite the strong lensing effect, we on Earth won't see any lensed images.

Excellent, thank you, this couple of pictures solves the issue.I was talking about the second picture, I understand now that you were talking about the first one. If the lens is not midway between the observer and the source, the observer will not get to see the source.Thank you very much for your help.

caveman1917
2010-Nov-23, 09:26 PM
I had been thinking about the 'blue case', won't our angle theta-E grow larger when we position the observer closer to the lens?
Though we might have R get smaller than the photon sphere radius, which wouldn't work.

13958

ETA: how do i make it appear as big as it should be? I just worked on the same file from StupendousMan and re-attached

Jeff Root
2010-Nov-23, 09:58 PM
Lemme just mention that for a "point" light source (such as a star) and
a "point" lensing mass (such as another star, or a black hole), the light
would be spread out in the image, not concentrated, so the brightness
of the image would be reduced. If the light source is spread out (such
as a galaxy), or the lensing mass is spread out (such as a galaxy or a
cluster), the light can be concentrated by the lens, producing an image
brighter than if it were not lensed.

-- Jeff, in Minneapolis

StupendousMan
2010-Nov-24, 12:20 AM
Actually, no, that's pretty much the opposite of the real situation: if a star is lensed by another point mass, so that the lensed images of the star have a larger apparent area than the original star ... then the images are brighter than the original star. That's the whole point of microlensing: lensing can cause the apparent brightness of a star to rise considerably for a short time (until the lensing object moves out of the line of sight).

Counterintuitive, perhaps, but true, and proved by many sets of observations.

Jeff Root
2010-Nov-24, 12:45 AM
Okay, you are right if the "point mass" is so far away that the
light passing around it forms a circle so small that it is essentially
also a point on the detector. I wasn't thinking of the point mass
being so far away that the circle could not be resolved, but of
course that would always be the case. With the circle unresolved,
all the light from the full circle would land on the same point on
the detector.

-- Jeff, in Minneapolis

StupendousMan
2010-Nov-24, 12:26 PM
No, that wasn't what I meant. Gravitational lensing increases the amount of light reaching the observer. It does not simply take the same amount of light emitted by the source and spread it out over a larger area.

I suggest that you do a little reading on "gravitational microlensing" and the surveys by the MACHO and PLANET collaborations (among others).

Jeff Root
2010-Nov-24, 03:13 PM
What you are saying is what I said. Although I distinctly
remember saying explicitly that the total amount of light
would be greater with lensing, and I don't see that in my
posts, so maybe it disappeared while editing.

The total amount of light reaching a detector is greater with
lensing. For a nearby object, with angles comparable to those
in your diagrams above, which I had in mind, the lensed light
would be spread out in a circle, reducing its surface brightness
compared to the unlensed image. For actual star distances,
where the light rays are essentially parallel, and the star is
unresolved, no circle is seen, and all the light is concentrated
into a small spot, the same size as that of the image of the
unlensed star, resulting in a greater surface brightness.

-- Jeff, in Minneapolis

ngc3314
2010-Nov-24, 08:47 PM
For a nearby object, with angles comparable to those in your diagrams above, which I had in mind, the lensed light would be spread out in a circle, reducing its surface brightness compared to the unlensed image.

Gravitational lensing actually preserves surface brightness (locally within the object image), so that the geometric distortion is connected to the flux amplification. This comes from Liouville's theorem (which is in fact pointed out in the Wikipedia article (http://en.wikipedia.org/wiki/Gravitational_lensing_formalism)). Early on (around 1980), this caused some confusion because of applying point-source formalism in situations in which it really made no sense.

Nereid
2010-Dec-01, 10:10 AM
First, the Sun is not a symmetric source in X-rays, so it would be pretty hard to define its center; especially as the X-ray appearance changes over time, both due to rotation and changes in the solar X-ray luminosity at different locations.

Second, I don't understand what you mean by "timing information." Could you explain?
Sorry, I missed this question.

It is, as you point out, moot, because the sky position of the geometric centre of the Sun cannot be estimated based on the Sun's x-ray emission. However, here's what I had in mind (a rough sketch only):

The Sun moves at approx 1" (angular distance) every 24 seconds (of time), wrt distant objects such as x-ray bright quasars or the Crab.

These x-ray bright sources are point sources, at the ~1" resolution of something like Chandra.

If the Sun and one of these sources - undergoing an occultation by the Sun - were bright enough that integration times of ~1 second would be sufficient to determine their relative positions, in a Chandra-like telescope's FOV*, then a sequence of 1 second integrations would produce estimates of the angular distance between the two sources, as a function of time.

That estimated angular separation would be a straight line on a chart/graph/plot (against time) if there were no bending; deviations from a straight line could be used to estimate the amount of bending.

* the geometric distortion across this field of view is assumed to be stable and well-characterised

StupendousMan
2010-Dec-01, 01:27 PM
Sorry, I missed this question.

If the Sun and one of these sources - undergoing an occultation by the Sun - were bright enough that integration times of ~1 second would be sufficient to determine their relative positions, in a Chandra-like telescope's FOV*, then a sequence of 1 second integrations would produce estimates of the angular distance between the two sources, as a function of time.


I think you're making the problem harder than it needs to be. You don't have to measure the position of the Sun at all if you have 3 or 4 or more objects in the field of view, scattered roughly symmetrically around the Sun; just measure the relative positions of these distant sources as a function of time.

Still, this discussion seems a little bit silly. Astronomers have measured gravitational deflection of optical light by the Sun, and radio waves by the Sun. There's really no big reason to try it with X-rays, especially given that the design of current X-ray satellites precludes it.

Jeff Root
2010-Dec-01, 05:00 PM
The Sun moves at approx 1" (angular distance) every 24 seconds
(of time), wrt distant objects such as x-ray bright quasars or the Crab.
Chances are that I'm the one messing up, but I get (as a crude
approximation) 1' (one minute of angle) every 4 seconds of time.
That is the same as 1 degree every 4 minutes of time.

-- Jeff, in Minneapolis

Amber Robot
2010-Dec-01, 07:33 PM
Chances are that I'm the one messing up, but I get (as a crude
approximation) 1' (one minute of angle) every 4 seconds of time.
That is the same as 1 degree every 4 minutes of time.

-- Jeff, in Minneapolis

I'm assuming that Nereid's calculation went something like this: 365 days per year * 24 hours per day * 60 minutes per hour * 60 seconds per minute = 3.15x10^7 seconds per year. The Earth goes around the sun 360 degrees per year, so that's 360 degrees * 60 arcminutes * 60 arcseconds = 1.296x10^6 arcseconds.

Dividing the two gives you 24.3 seconds per arcsecond. This is of course an average because the Earth's orbit is not circular, so the actual speed will change throughout the year.

caveman1917
2010-Dec-01, 09:36 PM
Chances are that I'm the one messing up, but I get (as a crude
approximation) 1' (one minute of angle) every 4 seconds of time.
That is the same as 1 degree every 4 minutes of time.

-- Jeff, in Minneapolis

Nereid's calculation was with respect to the distant stars ("wrt distant objects"). I'm guessing your calculation is with respect to the horizon (ie the apparent movement throughout the day)?

Jeff Root
2010-Dec-01, 11:35 PM
Hoo boy! Yes, I envisioned the Sun moving across the sky due
to Earth's rotation, and figured that the annual motion and other
factors would be minor adjustments to that. Wrong, wrong, all
wrong! Entirely the other way around!

-- Emily Litella, in Minneapolis

Nereid
2010-Dec-02, 05:38 PM
I'm assuming that Nereid's calculation went something like this: 365 days per year * 24 hours per day * 60 minutes per hour * 60 seconds per minute = 3.15x10^7 seconds per year. The Earth goes around the sun 360 degrees per year, so that's 360 degrees * 60 arcminutes * 60 arcseconds = 1.296x10^6 arcseconds.

Dividing the two gives you 24.3 seconds per arcsecond. This is of course an average because the Earth's orbit is not circular, so the actual speed will change throughout the year.
That's it, but I was a bit lazier than this ... the Earth goes round the Sun in ~365 days, which is 360 more or less. A day has 24 hours, and the number of seconds (time) in an hour (time) is the same as the number of second (angle) in a degree (angle), so the Sun moves 1" in ~24 seconds (time) :)

All I needed for the purpose I had then was to know that the relative (angular) separation of the Sun and a distant object would change quite slowly, wrt the integration time needed to estimate the position of the bright, distant object to within its resolution (~1" is what I assumed).

Also, as to what I was trying to do: by the time I got back to this thread, a better solution had already been given, even if an x-ray telescope with a 1" resolution could be pointed at the Sun, and even if the position of the Sun could be determined to within ~1" from its x-ray emission alone; namely, x-ray observations of lensed quasars. You see, I was - and still am - interested in good confirmations of the GR-predicted 'bending' of light, in as many regions of the electromagnetic spectrum as possible. Now gamma-rays are out (unless, perhaps, a really gamma-ray bright lensed quasar were to be observed, being eclipsed by the Moon, say), as are EUV (I may be wrong about this). For UV (blue-ward of the atmosphere's cutoff) there are possibly observations by GALEX, FUSE, or a now-defunct UV system on the Hubble (I can't think of its name just now); for IR - near, mid, far - there's lots to choose from, with lensed quasars again being the likely best bet. Ditto microwave and radio out to the ionosphere's plasma frequency cutoff (though VLBI is surely needed for low frequencies), with the exception of interplanetary spacecraft-as-sources (e.g. Cassini) for whatever channel they work on.

Amber Robot
2010-Dec-02, 08:00 PM
For UV (blue-ward of the atmosphere's cutoff) there are possibly observations by GALEX, FUSE, or a now-defunct UV system on the Hubble (I can't think of its name just now);

None of these would be pointed toward the sun.

I'm not sure which "defunct" UV system on HST you are referring to -- there's STIS and COS, which are both functional, and ACS and WFC3 have some UV capability.

ngc3314
2010-Dec-23, 11:39 PM
Also, as to what I was trying to do: by the time I got back to this thread, a better solution had already been given, even if an x-ray telescope with a 1" resolution could be pointed at the Sun, and even if the position of the Sun could be determined to within ~1" from its x-ray emission alone; namely, x-ray observations of lensed quasars. You see, I was - and still am - interested in good confirmations of the GR-predicted 'bending' of light, in as many regions of the electromagnetic spectrum as possible.

Just saw this and it reminded me of the thread:

X-ray microlensing in the quadruply lensed quasar Q2237+0305 (http://arxiv.org/pdf/1012.4807v1)

They concentrate on the temporal analysis, but their Fig. 1 shows that Chandra's PSF is good enough to separate at least 3 of the images in the Einstein Cross lens 2237+03, as well as get light curves allowing them to separate microlensing effects from the multiply-delayed intrinsic quasar variations. (As to the phenomenon itself, though, ROSAT separated the image components of 0957+561, and the Einstein observatory missed that only because the fainter image seems to have blended with emission from the foreground galaxy group; at least the latter is what I surmise from reading a meeting abstract...)