# Sun bending light

1. Nov 30, 2005

### michael879

wasnt the fact that the sun bends light coming from distant stars proof that gravity warps space-time? Im sure im wrong but can someone explain to me how they know it wasnt just refraction of the light as it passed the sun's "atmosphere"? Light bends around earth due to our atmosphere and refraction..

2. Nov 30, 2005

### mathman

The famous 1919 experiment (Eddington) compared star positions passing the sun (observed during an eclipse) against star positions at night. The atmospheric effect would be the same in both cases.

The important thing to remember is that Newton's thoery also predicted a bending. However, GR predicted bending 2X that of Newton. 1919 results bore out GR.

3. Nov 30, 2005

### michael879

no I dont mean bending do to our atmosphere, I mean, the sun shoots stuff into space, itll have something resembling an atmosphere that would bend light that comes close to it right?

4. Nov 30, 2005

### RandallB

?? What was/is the Newtonian prediction based on ?
Mass of the ‘zero mass’ photons?
'assuming' a mass based on the photon Energy?

On the optical bending of light through the (Hydrogen/Helium??) atmosphere of the sun. I don’t think the light of the star was close enough to have any significant refraction due to the sun’s atmosphere.

5. Nov 30, 2005

### Janus

Staff Emeritus
This experiment was just the first in a series of test that provided evidence for the bending of light by gravity.

The reason it was considered strong evidence was that it not only showed that the light was bent, but that it was bent by the amount predicted by GR. It would have been an extremely unlikely coincidence that any atmospheric refraction by the Sun would be exactly that needed to match that prediction.

Another fact is that refraction also produces an effect know as chomatic aberration, where the different color components that make up the light are bent by different angles. Gravity bending doesn't produce this effect.

6. Nov 30, 2005

### michael879

he means that if you take the debroglie mass of light, the amount the sun bends it is different from the amount predicted by F = GMm/r^2. Thanks janus, twas the answer I was looking for. Newtonian gravity would give a different force for each wavelength too right? cause the photon mass is based on the frequency.

7. Nov 30, 2005

### SpaceTiger

Staff Emeritus
The classical calculation of light being bent by the sun is usually given as:

$$\alpha=\frac{2GM}{rc^2}$$,

where M is the mass of the lensing object and r is the radius at which the light passes. This is a factor of two less than the GR prediction, but they're both independent of frequency.

I think it's debatable, however, whether Newtonian gravity really predicts any bending of light.

8. Nov 30, 2005

### michael879

I thought the classical way of finding the bending of light was by finding the mass of light with whatever debroglies thing is (I forget but I know frequency is in there), and then using that mass to find the gravitational attraction with F = GMm/r^2. and is it rly off by exactly a factor of 2? that seems like a huge coincidence....

9. Dec 1, 2005

### Staff: Mentor

Also, the observed amount of bending varies with the angular distance of the source from the sun (relative to the earth), in a way that agrees with the predictions of general relativity, and disagrees strongly with what solar atmospheric refraction would predict. I've read that this data now extends out to 90 degrees from the sun or more. That is, we can measure the deflection of electromagnetic radiation from a source that is directly overhead, when the sun is at the horizon! I think this is done with radio waves from pulsars, not light, but they're both electromagnetic radiation.

10. Dec 1, 2005

### pervect

Staff Emeritus
If one goes back to pure Newtonian theory (no SR as well as no GR), the acceleration of a falling body is independent of its mass or speed. One can then imagine light as being just like a bullet. The exact mass doesn't even matter to how fast it falls.

Of course this idea doesn't agree with experiment, but we didn't know that until it was tested and Einstein's ideas were confirmed.

11. Dec 1, 2005

### da_willem

Can someone with more acquaintance with GR explain, in words using plausibility arguments or using some formulae, explain why the factor 2 appears?

12. Dec 1, 2005

### Garth

Take the sun warping space-time by its mass.

In a frame of reference co-moving with the Sun we can slice through (foliate) space-time dividing it into hyper-surfaces of space separated by time.

Each such space-like hyper-surface is also curved by the Sun's mass.

A ray of light passing close to the Sun on this surface follows a straight line, a (null) geodesic, along the surface. This ray is 'bent' because the surface it travels along is curved.

The deflection of a ray just grazing the Sun's surface at distance R from its centre, is determined by the curvature of 'space' to be
$$\alpha = \frac{2GM}{Rc^2}$$ radians as stated by ST.

However there is a further effect due to the dilation of time.
This extra effect is caused by space-time being curved rather than just space.

Instead of there being a 'curvature of time' as well as a 'curvature of space' the time component of space-time curvature reveals itself as a time dilation between clocks at different levels in the gravitational field as determined by the equivalence principle, or equivalently by the SR treatment of Newtonian gravity. [Edit I originally put this down the other way round; why do I do that?]

The time dilation component adds an extra
$$\alpha = \frac{2GM}{Rc^2}$$ radians

making a total of

$$\alpha = \frac{4GM}{Rc^2}$$ radians = 1.75" arc for the Sun, as is observed.

I hope this helps.

Garth

Last edited: Dec 1, 2005
13. Dec 1, 2005

### SpaceTiger

Staff Emeritus
Calculating an effective mass for light doesn't really do anything for you because the deflection angle doesn't depend on the mass of the object being deflected. One way you can roughly derive the classical deflection angle is with the impulse approximation. Imagine you have a massive object with momentum, $p=mv$, coming towards a more massive object in a way such that the distance of closest approach is $r$. We'll assume that the total deflection angle is small, so we can simply calculate the total impulse provided to the moving body along a straight line path. We could use calculus to get it precisely, but it's easier to say that:

$$\Delta p=F\Delta t\simeq\frac{GMm}{r^2}\times\frac{2r}{v}=\frac{2GMm}{rv}$$

The deflection angle is just the added momentum (above) divided by its original momentum. This is because the impulse was perpendicular to the direction of motion. Anyway, this gives:

$$\alpha=\frac{\Delta p}{p}\simeq\frac{2GM}{rv^2}$$

Since the result is independent of mass, it's not unreasonable to expect that light would experience a similar deviation, so naively, we can just substitude $c$ for $v$ and get the result I gave above:

$$\alpha=\frac{2GM}{rc^2}$$

Keep in mind that this is not the correct result. The correct one is relativistic and gives a factor of two larger, this is just an explanation of how a classical physicist might have approached the problem.

It's not a coincidence, really. It's common for results obtained by classical analysis to be similar to their relativistic equivalents. After all, relativity approaches classical physics in certain limits.

Last edited: Dec 1, 2005
14. Dec 1, 2005

### Garth

If we write the general Schwarzschild metric in its standard form expanded in the Robertson Post Newtonian Approxmation parameters $\alpha, \beta, \gamma$ we get:

$$d\tau^2=(1-2\alpha\frac{MG}{rc^2}+...)dt^2 - (1+2\gamma\frac{MG}{rc^2}+...)dr^2 -r^2d\theta^2-r^2sin^2\thetad\phi^2$$.

Where $\alpha$ = unity in GR, is a measure of how much the measured Newtonian constant differs from the G entered into the metric, and $\gamma$ is a measure of the curvature of space by unit mass, then the deflection of light is given by:

$$\delta=(\frac{4MG}{rc^2})(\frac{1+\gamma}{2})$$ .

You can see that if $\gamma$ = 0, i.e. there is no curvature of space, then the deflection caused by the equivalence principle, treating light as if it were simply falling under Newtonian gravity, is half the full GR prediction, which is obtained when $\gamma$ = 1.

Garth

N.B. note my edit in my previous post.

Last edited: Dec 2, 2005
15. Dec 2, 2005

### pervect

Staff Emeritus
In GR, the path that light traverses is determined by solving the geodesic equations.

Do you know what a Christoffel symbol is? I could write the equations for geodesic motion out in terms of the Christoffel symbols if that would be helpful. But I suspect it may not be :-(.

I can say, though, that the equations of motions are ordinary differential equations. The Christoffel symbols are expressions that can be calculated from the metric. There are 64 of them, though they exhibit a high degree of symmetry, they are usually denoted as $\Gamma^a{}_{bc}.$, where a,b,c take on the values 0,1,2,3.

Solving the geodesic equations gives the equations of a geodesic path. I.e. in a coordinate system [t,x,y,z] one winds up with 4 functions that define a path

t(tau), x(tau), y(tau), z(tau), where tau can be "proper time" (for any physical body), or a so-called "affine parameter".

The closest I can come to explaning this in words is that the time-part of the curvature of space-time is the only part that's important at low velocities, and that this can be interpreted as "forces". These forces are one specific set of Christoffel symbols.

For high velocities, the space-part of the curvature becomes equally important as the time part, and light deflects both due to "forces" and to the "curvature" of space (the purely spatial part of the space-time curvature).

This explanation plays a bit fast and lose with the defintion of "curvature". When I say "curvature" here, what I really mean are the Christoffel symbols (not any of the various curvature tensors).

16. Dec 2, 2005

### da_willem

It does, thanks, pervect as well!

17. Dec 3, 2005

### RandallB

This is for an observation with light passing at 2.1 x 10 6 km from the sun. Or about a full diameter of the sun/moon away from the edge of the eclipse (Well away from any significant sun atmosphere)
I assume without SR, there were no “Classical Newtonian” predictions made by anyone to ‘compete’ with Einstein’s prediction.
Did anyone at the time actually propose the Classical SR solution, that 1/2 the GR prediction should be expected?
Or are these just given by more modern exercises in understanding what the classical would have expected?
Interesting and excellent view of the classical SR view.
But what looks like a huge coincidence, of a factor of two difference to GR, that michael879 pointed out has me concerned.
The assumption was to find the deflection from a “straight line path” valid for very small angles which this is. However looking at this deflection as coming from the straight line of the tangent to Perigee of the hyperbolic (escape orbit), may only be providing half the solution. From this perspective it seems the solution is only providing the deflection angle from that tangent as it turns toward the observer on earth. To reach that trajectory the incoming path must follow a bend as well to reach Perigee. The hyperbola is symmetric so the incoming angle would be identical to the outgoing angle, for a total deflection of double as calculated. That resulting in the “classic SR” matching GR prediction exactly not just 1/2!

Is there some detail that can be followed in the full calculus to show if the “total impulse” here is accounting for all of the deflection, both incoming and outgoing.
Or, just the outgoing deflection off the assumed straight line tangent?
RB

18. Dec 3, 2005

### Garth

Yes - Einstein! In 1915; he corrected his prediction just in time for the 1919 ecilpse, if he hadn't the history of 20th Century physics might have been completely different.....

Garth

19. Dec 3, 2005

### Garth

No, you are correct to say, "The hyperbola is symmetric so the incoming angle would be identical to the outgoing angle", the full 'SR' derivation of 1/2 the GR prediction takes into account the incoming and outgoing deviations from a straight line, that is where ST's factor of 2 comes from.

NB - also the angle of 1.75" arc is the deviation of a ray just grazing the Sun, the actual stars observed were further away, as you rightly said, and their deflections were proportionally less.

Garth

Last edited: Dec 3, 2005
20. Dec 3, 2005

### SpaceTiger

Staff Emeritus
If you're unsure about my hand-wavy result, you can just do the integral directly. It becomes (replacing the old $r$ with $r_0$):

$$\Delta p=\int_{-\infty}^{\infty}Fdt = \int_{-\infty}^{\infty}\frac{GMm}{r^2}\frac{r_0}{r}\frac{dx}{v}$$

where x is the distance along the light's path.

$$\int_{-\infty}^{\infty}\frac{GMm}{r^2}\frac{r_0}{r}\frac{dx}{v}=\frac{GMmr_0}{v}\int_{-\infty}^{\infty}\frac{1}{r^3}dx$$

$$\frac{GMmr_0}{v}\int_{-\infty}^{\infty}\frac{1}{r^3}dx=\frac{GMmr_0}{v}\int_{-\infty}^{\infty}\frac{1}{(x^2+r_0^2)^{3/2}}dx=\frac{GMmr_0}{v}\times\frac{2}{r_0^2}$$

$$\Delta p = \frac{2GMm}{r_0v}$$

Same as before