Is a star's size changed due to gravitational deflection (lensing)?

[Forreal]

TL;DR Summary

Does the mass of a star have any effect on it’s apparent size, due to gravitational deflection (or lensing)?

Using the Sun as example, we know from the Eddington experiment that the Sun deflects light.
Does light from other stars get deflected towards or away from the centre of mass, or is it a mix?
Does light coming from the Sun get deflected towards or away from the centre of mass?
Would this make an object appear smaller, larger or the same size?

 

[Dr_Nate]

Well, what have you found out about light deflection so far?

 

[Ibix]

Qualitatively, light behaves just like anything else in a gravitational field - it's deflected towards the mass. You can work out the optical effect by drawing a diagram of a spherical star, drawing a straight line from a point on the limb of the star to your eye, and then thinking how I would have to throw a ball from that point to reach your eye. Do I aim straight at you, above, or below?

That's really it for stars. Life gets more complicated near objects like spinning black holes because our intuitive feel for how gravity works isn't right and you need to do the maths properly.

 

[Forreal]

Well, what have you found out about light deflection so far?

Caused by gravity and proportional to mass.
Apparently not interchangeable with the term gravitational lensing.

Qualitatively, light behaves just like anything else in a gravitational field - it's deflected towards the mass. You can work out the optical effect by drawing a diagram of a spherical star, drawing a straight line from a point on the limb of the star to your eye, and then thinking how I would have to throw a ball from that point to reach your eye. Do I aim straight at you, above, or below?

Well you have to aim above, to counter-act gravity, right? If you were aiming in a straight line ( not taking gravity into account), the ball would end up below.
I was thinking along the lines of placing a bar magnet on a CRT.
Doing so would make an image look smaller in both cases?

 

[jbriggs444]

I was thinking along the lines of placing a bar magnet on a CRT.
Doing so would make an image look smaller in both cases?

You need to clarify your thinking here.

With a CRT/screan, you have a point source emitting a collimated beam. The magnet deflects the beam and affects the position where the beam lands. With the star/eye you have a source of non-negligible size (else you would not be asking about image size) emitting light in all directions. The only light that you see is the light that happens to strike your eye.

The emitted light may be deflected on the way to your eye.

The relevant bit for image size is: "From what angle will the deflected light then strike my eye?" The departure angle from the star is somewhat important because it tells you how the trajectory has to start. But what matters for image size is how that trajectory ends. [Or, more accurately, what matters is how the trajectories differ for light originating on the two opposite edges of the star]

 

[Forreal]

I'm getting a bit muddled.
Could someone please just let me know if the answer is larger or smaller?
Then I'll know if I've got everything mixed up and need to go brush up on optical lensing + other stuff, or not.

 

[jbriggs444]

I'm getting a bit muddled.
Could someone please just let me know if the answer is larger or smaller?
Then I'll know if I've got everything mixed up and need to go brush up on optical lensing + other stuff, or not.

Draw a diagram tracing a ray of light from the right edge of the star to the viewer's eye. Repeat for the left edge. The light rays should each follow a path roughly like one half of a hyperbolic orbit, concave inward toward the star.

Now trace two straight dotted lines back from the eye, each tangent to the path on which one of the two rays arrived. Draw in a dotted image of the star where the straight dotted lines end near the position of the real star. That dotted image is the virtual image that is seen.

Is it larger or smaller than the real star?

 

[Forreal]

Is it larger or smaller than the real star?

Smaller.
Right?

"So the statement gravitational deflection makes the Sun appear smaller than if there were no gravitational deflection" is true?

 

[jbriggs444]

Smaller.
Right?

"So the statement gravitational deflection makes the Sun appear smaller than if there were no gravitational deflection" is true?

Did you draw the picture?

 

[jbriggs444]

Tried my hand with mspaint. Black is original star. Blue is projected sight lines and virtual image.

 

[Forreal]

Did you draw the picture?

Well, I tried.
I think I mixed up concave and convex.

Tried my hand with mspaint. Black is original star. Blue is projected sight lines and virtual image.

Thank you! Very appreciated.

 

[sophiecentaur]

Tried my hand with mspaint. Black is original star. Blue is projected sight lines and virtual image.View attachment 256643

It's always good when someone actually bothers to draw a relevant diagram.

Compared with the effect on light from a distant star, I think any effect would be seriously minuscule. A parcel of light beams from a distant star, grazing the star's significant gravitational field would all be affected by a similar amount, producing a somewhat coherent virtual image. In your diagram, the light from one spot is emitted over a big range of angles and the virtual image you have drawn (blue) would be in a different place for every 'individual' ray from the surface. A family of rays from near the edge would all emerge in different directions because the field and path length will vary from ray to ray. (there is no equivalent 'grazing') That would not produce a coherent image; I suggest it could blur the perceived circumference but not give a noticeably bigger image.

 

[jbriggs444]

In your diagram, the light from one spot is emitted over a big range of angles and the virtual image you have drawn (blue) would be in a different place for every 'individual' ray from the surface.

This scattering feature is present for every object that we see. The light sprays out from illuminated spots in all directions. Only a tiny fraction of that light happens to shine in the direction of the pupils in our eyes to be gathered and focused at a point on the retina that corresponds to the single point on the object. This does not prevent the real image on the retina from being sharp.

A family of rays from near the edge would all emerge in different directions because the field and path length will vary from ray to ray. (there is no equivalent 'grazing') That would not produce a coherent image; I suggest it could blur the perceived circumference but not give a noticeably bigger image.

The rightmost rays visible to the eye are, naturally, the rays that left the rightmost edge of the star (*). As drawn, actually a tiny bit back from there due to the deflection. These are the rays that left the sun at a grazing angle. [If the angle were not grazing then there would be a viable trajectory even further right and intersecting the surface of the sun even farther back]

(*) Yes, we both know that this will be on the left edge of the inverted image on the retina.

This diagram differs from those that one would typically use when trying to present geometric optics using ray tracing through lenses. In those drawings, a typical feature is a trace of two different trajectories (e.g. through each rim of a concave lens) from a single point on the object to a single point on a real image. This drawing is just portraying sight lines -- one ray trace from one point on the object to the viewer's eye and another ray trace from another point on the object to the viewers eye. That is enough to determine the subtended angle. Subtended angle is the relevant feature, of course.

 

[sophiecentaur]

This scattering feature is present for every object that we see.

But the gravitational effect will depend on the mass of the object involved but it will be masked by the effect of an atmosphere, I imagine. This thread has become muddled because there are very different situations when 'self bending' and with normal gravitational lensing. I guess it would be possible to do a model to calculate the effect on the Sun.

I still say that the situation for a wide, nearby object is significantly different compared with the lensing effect of distant objects. The optics with a distributed source with varying angles will be less coherent than for a distant point for which the angles are very much the same.
Gravitational lensing will not magnify so much as re-positioning and increasing luminosity. There can be no magnifying effect for very distant stars because they will always be point sources. The Sun and nearby stars would be candidates for this effect. Otoh, distant galaxies could possibly have their dimensions / shapes changed.(Millions of times the angular dimension of their constituent stars).

This diagram differs from those that one would typically use when trying to present geometric optics . . . .

But it's not bad for showing the effect of deflection from each elemental point source. It's only necessary to thing a bit 'inside out' as when describing rainbow and halo formation. I think it implies that there would be a brighter ring round the outside due to light being bent over the horizon. But the outer parts of the Sun's disc are actually not as bright as the centre because of the effect of its atmosphere. So it's complicated. Any effect has to satisfy energy conservation.

PS I managed to get a result with a second hand Coronado PST solar telescope yesterday. I could see a flare that must have been at least 40 thousand km high. Makes you think dunnit?

 

[davenn]

I could see a flare

a prominence

yes, there was a decent one there

 

[sophiecentaur]

Hah. Whatever. It was still pretty big.
I must buy an observer’s book of the Sun. Then I can give you a fair fight.