# Thread: Gravitational deflection of light - measurements

1. Order of Kilopi
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The only undetermined system I can see is the one describing your views here. Can you spell out, simply and clearly, what you think is wrong and why? You seem to be setting up a strawman argument by claiming that the Hipparcos mission measureds deflection in a way that they did not. And, as has been said, your cryptic responses don't help.

I think it may be you who is not looking at the bigger picture, instead focusing on an incorrect mental model of how these measurements are made and what that means for cosmology. GR, including the bending of light rays, has far more evidence than just these measurements.

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Excuse me, but I have nothing to Hipparcos measurements.

And in cosmology we observe greater deflection than expected.
So perhaps we can accept also greater for sun - no?

a - b = 1.7
a + b = 8.3

Now we solve the system: a = 5 and b = 3.3
And what now do not agree with the observations?

You can have your cake and eat it too!

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GR's predictions were confirmed by measurements, most accurately at radio wavelengths. Hipparcos data showed the same result. I have no idea where you get the statement that we see a greater deflection - if we did then the predictions made by GR would not be hailed as a good piece of evidence for the utility of GR.

I think you have your answer about measuing differential deflections rather than absolute. Do you feel you have outstanding questions for which you want a mainstream answer?

4. Originally Posted by Hetman
Excuse me, but I have nothing to Hipparcos measurements.

And in cosmology we observe greater deflection than expected.
Yes, if there is dark matter we didn't know about before making the observation.
So perhaps we can accept also greater for sun - no?
Why should we?

a - b = 1.7
That is the difference between the absolute deflections of the stars a and b. It equals the difference between the observed spacing and what the spacing would be in the absence of any deflection.
a + b = 8.3
You have yet to give us an intelligible statement of what that equation represents, let alone how to measure it and find 8.3.

Now we solve the system: a = 5 and b = 3.3
And what now do not agree with the observations?
What observations?

You can have your cake and eat it too!

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It's nothing special - insert a number and you're done.

The deflection of 1.7'' was ever measured from different distance to the Sun, for example, Jupiter or at least from Mars?

6. Originally Posted by Hetman
Excuse me, but I have nothing to Hipparcos measurements.

And in cosmology we observe greater deflection than expected.
So perhaps we can accept also greater for sun - no?

a - b = 1.7
a + b = 8.3

Now we solve the system: a = 5 and b = 3.3
And what now do not agree with the observations?

You can have your cake and eat it too!
Hornblower mentioned in post #2 that the deflection at right angles is 4 millisecond seconds. So your equations would be more like

a - b = 1.7
a + b = 1.708

which results in a = 1.704 and b = .004

That's not much different than a = 1.7 and b = 0, and is sufficient for the accuracies of that day. Today we can make the much finer measurements.

7. Originally Posted by Hetman
It's nothing special - insert a number and you're done.

The deflection of 1.7'' was ever measured from different distance to the Sun, for example, Jupiter or at least from Mars?
Sure, you could plug in any old number for x and go through the motions of solving the simultaneous equations for a and b, regardless of whether or not x represents something you could measure on a photo of a star field. You still have not shown us what it is on that photo that you could measure to get a specific number for x. Please clarify. If you have no means of scanning and uploading a picture, look for someone to assist you.

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Originally Posted by Hetman
The deflection of 1.7'' was ever measured from different distance to the Sun, for example, Jupiter or at least from Mars?
There are reports of successful observations of the gravitational deflection due to Jupiter (well, due to its mass) in the literature (would anyone like some references?).

The distance of the Earth - and the Hipparcos satellite - from the Sun is (was) not constant; to the extent that the accuracy and precision of the observed deflections (mostly in the radio part of the electromagnetic spectrum, as Shaula said) is sufficient to detect this variation in distance, then the answer to your question is yes.

Various interplanetary missions have been used to test the Shapiro time delay (e.g. Cassini); whether any of these also measured (or attempted to measure) the gravitational deflection (due to the Sun), at various distances from it (significantly different from ~1 au), I do not know.

If GAIA meets its design specifications, the gravitational deflection due to all the solar system's planets (except Mercury; it won't observe that close to the Sun), several of its biggest moons, and perhaps its biggest dwarf planets will be detectable. It will certainly be sensitive enough to detect variations in the Sun's deflection, due to changes in GAIA's distance from the Sun.

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Originally Posted by grapes
Hornblower mentioned in post #2 that the deflection at right angles is 4 millisecond seconds. So your equations would be more like

a - b = 1.7
a + b = 1.708

which results in a = 1.704 and b = .004

That's not much different than a = 1.7 and b = 0, and is sufficient for the accuracies of that day. Today we can make the much finer measurements.
Then we get a system of equations:

a - b = 1.7
a + b = x

and one new:
c - b = 0.004

10. Originally Posted by Hetman
Then we get a system of equations:

a - b = 1.7
a + b = x

and one new:
c - b = 0.004
Once again, what would you measure while making an observation to get a number for your a + b? Until you answer that your other remarks remain a mystery to me.

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Originally Posted by Hornblower
Sure, you could plug in any old number for x and go through the motions of solving the simultaneous equations for a and b, regardless of whether or not x represents something you could measure on a photo of a star field. You still have not shown us what it is on that photo that you could measure to get a specific number for x. Please clarify. If you have no means of scanning and uploading a picture, look for someone to assist you.
OK. I found the solution to the problem.

Full deflection of light on the edge of the Sun indeed is about 8''.
Measurements from the small distance to the Sun give always less due to geometrical considerations.

Thank you for the fruitful discussion.

12. Full deflection of a light ray which grazes the photosphere of the sun
is about 1.7 arcsec, as one can see at

http://www.astro.ucla.edu/~wright/deflection-delay.html

and many other places.

13. Originally Posted by Hetman
OK. I found the solution to the problem.

Full deflection of light on the edge of the Sun indeed is about 8''.
Measurements from the small distance to the Sun give always less due to geometrical considerations.

Thank you for the fruitful discussion.
You have not answered my questions in my previous posts. For all we know your recent algebraic exercises could be using numbers obtained by blindly throwing darts at a chart. Unless you can show us how a number you use for a + b relates to actual measurements of star positions, I have no choice but to dismiss your assertions as without astrometric merit.

14. Originally Posted by Hetman
Then we get a system of equations:

a - b = 1.7
a + b = x

and one new:
c - b = 0.004
How are you going to solve that?

And why did you change 1.708 into x?

15. Originally Posted by Hetman
Then we get a system of equations:

a - b = 1.7
a + b = x

and one new:
c - b = 0.004
What is c?

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Originally Posted by grapes
How are you going to solve that?

And why did you change 1.708 into x?
a + b is unknown here, because a and b are unknown.
We know only difference: a - b and similarly c - b

and: 1.708 = (a-b) + 2(c-b); redundant equation.

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Originally Posted by Hornblower
What is c?
a, b, c - a real deflections in directions: 0, 180, 90 degrees, measured relative to the vector pointing towards the Sun.

One more question:

Deflection 1.7'' (ie: 4 m / r) is the deflection observed directly, or doubled?

I mean something like this - two images of one star (on two different photographs):

* a - the star during an eclipse
|
|
* b - the same star, but at night
----------- -> rim of the Sun

directly from the measurement gives:
a - b = 1.7'', or half of it: a - b = 0.85'' ?
Last edited by Hetman; 2012-Mar-24 at 05:30 PM.

18. Originally Posted by Hetman
a, b, c - a real deflections in directions: 0, 180, 90 degrees, measured relative to the vector pointing towards the Sun.

One more question:

Deflection 1.7'' (ie: 4 m / r) is the deflection observed directly, or doubled?

I mean something like this - two images of one star (on two different photographs):

* a - the star during an eclipse
|
|
* b - the same star, but at night
----------- -> rim of the Sun

directly from the measurement gives:
a - b = 1.7'', or half of it: a - b = 0.85'' ?
The only thing that is clear from your writing is that you are trying to do it the hard way. Let’s show how I would do it. I have held off until now to see how you would respond to my questions.

First things first. Astrometric work such as Eddington did in 1919 is done only on small fields that would fit on a single photographic plate. There is simply no way to measure momentary displacement of a star relative to another one far away in the sky with the necessary precision, using instruments that were available then.

I would photograph the eclipse with whatever exposure and development is needed to get measureable star images on the plate. It would be great if we could have the Pleiades in the background but that isn’t going to happen, so we take what we can get. Six months later I would photograph the same star field at the antisolar point and then compare the two plates on a precision measuring engine.

I would feel safe in assuming no significant deflection of the star images at or near the antisolar point. Not only is the gravitational effect very weak there in proportion to the effect at the grazing path during the eclipse, it is virtually parallel to the ray instead of crossways.

Using the antisolar plate as a reference, we can see the absolute deflection of each star away from the center of the Sun, and compare the amount with the predictions of Newton and Einstein. If the error bars are not too big, we can see which one is in better agreement.

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Okay, I found that the direct deflection, calculated from the comparison of two photographs, is whole: 1.7'', not half.
http://grupos.unican.es/glendama/Historical_intro.htm

http://claesjohnson.blogspot.com/201...y-gravity.html

1.75" x R/D

It is the correct scheme of deflection of light on spherical lens?

No. Ray bends symmetrically - the same on both sides of the Sun.
Thus, taking into account only this fact: the full deflection can not be less than 2 x 1.7''!

20. Originally Posted by Hetman
It is the correct scheme of deflection of light on spherical lens?

No. Ray bends symmetrically - the same on both sides of the Sun.
Thus, taking into account only this fact: the full deflection can not be less than 2 x 1.7''!
If you make the angle from a straight line 8.5 on both sides, then the total deflection is 17, which is to be the angle measured on the plates.

That line would be halfway between the orange dotted line and the yellow line, no?

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Originally Posted by grapes
If you make the angle from a straight line 8.5 on both sides, then the total deflection is 17, which is to be the angle measured on the plates.

That line would be halfway between the orange dotted line and the yellow line, no?
I know, but Eddington measured the offset 1.7'' directly on the photographs (previously was a link to the work of Eddington and there you will see it clearly).

When making a photo, during an eclipse, we do not record the same rays as in the night, but others - running phi / 2 above, then they are bend down phi, so we measure the shift phi / 2 compared to straight rays (without lens). There is a chord of the arc, not tangent.

22. From a remote star, the incoming rays are virtually parallel when they reach the Sun. In the absence of the Sun, a ray that is parallel to the incoming yellow ray and below it will be seen by the observer. The angle between it and the refracted ray will be equal to the angle marked in blue in post 49, which is equal to the angle through which the ray actually bends as it rounds the Sun. That is a detail that is neglected in this drawing, in which the "true position" of the star is about where Venus would be at superior conjunction. If we were looking at Venus instead of a distant star, yes indeed its angular change of position on the photo would be much less than the angle through which its grazing ray is deflected.
Last edited by Hornblower; 2012-Mar-25 at 01:06 AM. Reason: Fix a typo

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That's right, but with parallel rays, the result should be the same.
They then run with the same angle, but from another point (above), instead of with different angle and from the same point.

I see that it will be necessary to calculate the trajectory of rays, otherwise it can not be resolved.

24. Originally Posted by Hetman
That's right, but with parallel rays, the result should be the same.
They then run with the same angle, but from another point (above), instead of with different angle and from the same point.

I see that it will be necessary to calculate the trajectory of rays, otherwise it can not be resolved.
Here is a sketch.
http://img580.imageshack.us/img580/8127/docu0279.jpg

I brought two virtually parallel rays in from the right, coming from a star at a vast distance. The upper one grazes the Sun, is bent 1.7", and reaches the viewpoint. The long dashes to the left show where that upper ray would have gone had it not been deflected, and the long dashes to the right show where the ray appears to have come from to the observer. The lower solid line, from the same star, would have reached the viewpoint had the Sun not been there, and shows the direction to the true position of the star. The geometry of a diagonal line crossing two parallel lines shows that the angle of deflection as seen by the observer is equal to the angle of deflection actually undergone by the incoming ray.

I placed Venus on the upper ray, so it will be seen as occulting the star as viewed along the grazing ray. The short dashes show the ray that would reach the viewpoint from Venus in the absence of the Sun's gravity. In the undistorted view Venus would be above the star rather than occulting it. The apparent change of position for Venus as seen at the viewpoint is less than half of that of the star, even though the deflection of the incoming ray as it grazes the Sun is the same.

Clear as mud? If you have further questions or comments, please speak up.

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Originally Posted by Hornblower
Here is a sketch.
http://img580.imageshack.us/img580/8127/docu0279.jpg
In the undistorted view Venus would be above the star rather than occulting it. The apparent change of position for Venus as seen at the viewpoint is less than half of that of the star, even though the deflection of the incoming ray as it grazes the Sun is the same.

.
How do you figure, HB??
The equation for light deflection angle, Beta, is given by simply:

Beta = 4GM / (C^2)(R)...where R is the impact parameter.
From the eqn. there's no distance dependence to the light source; it is only dependent on the closest approach to the gravitational mass. So how can you alter the solar gravitational deflection for planets?

I could be wrong, but (concerning your sketch) it seems to me the sketch is not accurately portraying the star since you are using two rays parallel, rather than two rays from one POINT source for the star.

If you redo your sketch to show Venus occulation of the star in the undistorted view, ignoring of course the angular diameter of Venus, (using, say, only the center .01 " for simplicity). I think then you will see what I am talking about. In that case both the planet and star rays coincide (in undistorted view), and thus both must deflect by the same angle in the grav. distorted view..
Or have I missed something ??

G^2

26. Originally Posted by Gsquare
How do you figure, HB??
The equation for light deflection angle, Beta, is given by simply:

Beta = 4GM / (C^2)(R)...where R is the impact parameter.
From the eqn. there's no distance dependence to the light source; it is only dependent on the closest approach to the gravitational mass. So how can you alter the solar gravitational deflection for planets?
I did not assert any such thing, and the sketch shows no such thing. The latter part of my sentence in your boldface copy clearly says, "even though the deflection of the incoming ray as it grazes the Sun is the same." The chart shows the same thing. The grazing ray from Venus coincides with the one from the star, and it stays that way as it is deflected to the viewpoint.
I could be wrong, but (concerning your sketch) it seems to me the sketch is not accurately portraying the star since you are using two rays parallel, rather than two rays from one POINT source for the star.
They are drawn as parallel because the star is somewhere off to the right at least hundreds of thousands of times farther away than Venus. The lines would be out of parallel by at most less than 0.000,001", and much less for most stars. I deliberately omitted the actual star from the chart and showed only the approaching parallel rays, to drive that point home. For illustration I put Venus and a star in parentheses on the long dash extension from the viewpoint to indicate their deflected apparent positions.

If you redo your sketch to show Venus occulation of the star in the undistorted view, ignoring of course the angular diameter of Venus, (using, say, only the center .01 " for simplicity). I think then you will see what I am talking about. In that case both the planet and star rays coincide (in undistorted view), and thus both must deflect by the same angle in the grav. distorted view..
Or have I missed something ??

G^2
You missed the parallax effect that comes into play with Venus. Suppose I move Venus down to the lower solid line, which is the star ray that would go straight to the viewpoint if we could turn off the Sun's gravity. That would create an occultation in the undistorted view, but now the grazing ray from Venus would no longer coincide with the grazing ray from the star. It would come to the grazing point at a different angle, be deflected by the same 1.7", and thus proceed toward the viewpoint at a different angle. Now the star and Venus would be separated in the distorted view.
Last edited by Hornblower; 2012-Mar-26 at 03:42 AM. Reason: Fix a typo

27. In overnight hindsight I can see that my original sketch may have been hard to follow because of the visual clutter of superimposing the ray tracings of two objects at different distances behind the Sun. Let's try this one.

http://img717.imageshack.us/img717/2605/docu0280.jpg

In each case the solid ray is grazing the Sun and is bent 1.7".

The long dash ray is the undistorted line of sight from the viewpoint to the source in the absence of solar gravity.

For Venus and Jupiter the small angles v and s add up to 1.7", because the apex angle of the triangle is 180o - 1.7". The two angles are inversely proportional to the distances between the Sun and their respective corners, a good approximation for such small angles.

Thus angle s is smaller for Jupiter and approaches zero as we move the source out toward the distant stars and make the incoming rays more nearly parallel.

It follows that angle v is smallest for Venus, larger for Jupiter, and approaches 1.7" as we consider sources at extreme distances.

28. Order of Kilopi
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I wasn't confused by the first diagram, but I did fail to get the point.
Your second diagram makes the point clear by using progressivly
greater distances to Venus, Jupiter, and a distant star.

-- Jeff, in Minneapolis

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Originally Posted by Hornblower
Here is a sketch.
http://img580.imageshack.us/img580/8127/docu0279.jpg

I brought two virtually parallel rays in from the right, coming from a star at a vast distance. The upper one grazes the Sun, is bent 1.7", and reaches the viewpoint. The long dashes to the left show where that upper ray would have gone had it not been deflected, and the long dashes to the right show where the ray appears to have come from to the observer. The lower solid line, from the same star, would have reached the viewpoint had the Sun not been there, and shows the direction to the true position of the star. The geometry of a diagonal line crossing two parallel lines shows that the angle of deflection as seen by the observer is equal to the angle of deflection actually undergone by the incoming ray.

I placed Venus on the upper ray, so it will be seen as occulting the star as viewed along the grazing ray. The short dashes show the ray that would reach the viewpoint from Venus in the absence of the Sun's gravity. In the undistorted view Venus would be above the star rather than occulting it. The apparent change of position for Venus as seen at the viewpoint is less than half of that of the star, even though the deflection of the incoming ray as it grazes the Sun is the same.

Clear as mud? If you have further questions or comments, please speak up.
Thanks a lot.

But according to the drawing, the image of the star, as seen from Venus, should be shifted by a missing angle to full deflection for Venus, which is half of the 1.7'', right?

And now I have a new problem.

The deflection is an illusion, or whether there actually runs something along the arc instead of straight?

I checked this using the Shapiro delay (for straight ray).
Delay:

and in terms of distance: z(r) = -cdt

Direction of the wavefront tarns:
tan a = dz/dr, for small angles: tan a = ~ a

Thus, the rays run straight, only the wavefront rotates due to the diversity of delays.
Similarly, stellar aberration - here the wavefront turns by a fixed angle of ~ v/c, and we see the images moved forward.

Conclusion: we can not observe the stars below the edge of the Sun.
Last edited by Hetman; 2012-Mar-27 at 04:25 PM.

30. Well, that's that then. Now under ATM forum rules, with Hetman the OP.

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