Sun Bending Light: Gravity Warps Space-Time?

[michael879]

wasnt the fact that the sun bends light coming from distant stars proof that gravity warps space-time? I am sure I am 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..

 

[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 theory also predicted a bending. However, GR predicted bending 2X that of Newton. 1919 results bore out GR.

 

[michael879]

no I don't 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?

 

[RandallB]

The important thing to remember is that Newton's theory also predicted a bending. However, GR predicted bending 2X that of Newton.

?? 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.

 

[Janus]

wasnt the fact that the sun bends light coming from distant stars proof that gravity warps space-time? I am sure I am 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..

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.

 

[michael879]

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

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.

 

[SpaceTiger]

Newtonian gravity would give a different force for each wavelength too right? cause the photon mass is based on the frequency.

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.

 

[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...

 

[jtbell]

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.

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.

 

[pervect]

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.

 

[da_willem]

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...

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

 

[Garth]

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

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

 

[SpaceTiger]

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.

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.

and is it rly off by exactly a factor of 2? that seems like a huge coincidence...

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.

 

[Garth]

and is it rly off by exactly a factor of 2? that seems like a huge coincidence...

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.

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.

 

[pervect]

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

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).

 

[da_willem]

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

It does, thanks, pervect as well!

 

[RandallB]

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

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?

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}

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

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

 

[Garth]

Did anyone at the time actually propose the Classical SR solution, that 1/2 the GR prediction should be expected?
RB

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

 

[Garth]

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!...Or, just the outgoing deflection off the assumed straight line tangent?
RB

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

 

[SpaceTiger]

That resulting in the “classic SR” matching GR prediction exactly not just 1/2!

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

 

[RandallB]

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

It seems unlikely that in 1915 with his focus on GR that Einstein would have been making a prediction of the deflection at eclipse based on SR. Rather I’d assume he had made a prediction based on the GR 'curvature of space' formula that you gave, but (absent minded professor?) forgot to include SR where:

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

Do you have a reference as to when and how he corrected his 1915 prediction – another paper published before 1919 I’d assume.

Also, for that required time dilation component, do you know or have a reference on how that formula is derived “by the SR treatment of Newtonian gravity” to match the “equivalence principle”?

And Yes it would be questionable how histroy would have treated Einstien if that correction was made after the observation rather than before.

PS: I think my error was by ‘3.14’ in converting radians when trying to find the distance for your 1.75” arc-s.

 

[RandallB]

you can just do the integral directly.

I’m not real current on that but I’ll work on it – to do so I think I need to make sure I’m clear on working with a hyperbola.
From the integration where:

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

Does this come from a function of a hyperbola that somehow gives something like:

r⁶ = (x² + r₀²)³

Not sure how to get there.

 

[SpaceTiger]

I’m not real current on that but I’ll work on it – to do so I think I need to make sure I’m clear on working with a hyperbola.
From the integration where:
\int_{-\infty}^{\infty}\frac{1}{r^3}dx = \int_{-\infty}^{\infty}\frac{1}{(x^2+r_0^2)^{3/2}}dx
Does this come from a function of a hyperbola that somehow gives something like:
r⁶ = (x² + r₀²)³
Not sure how to get there.

No, we're working in the impulse approximation, so it's just a straight line. Draw a diagram with a point (representing the large mass) and a line (representing the particle/light) passing near the point. Draw a perpendicular from the line to the point and label it r_0. At a particular time, the particle/light will be at a distance x from the intersection of this perpendicular with its path. The distance of the particle from the massive body is then given by the pythagorean theorem:

r=\sqrt{r_0^2+x^2}

 

[RandallB]

r=\sqrt{r_0^2+x^2}

Got it – just substituting that for "r" gives:
r³ = (x² + r₀²)^3/2
I should have noticed :
r⁶ = (x² + r₀²)³
Was the same as the Pythagorean :
r² = x² + r₀²

Simple and classic – just never used “Impulse” before.


Still cannot find a copy of, or even an account of, the actual predictions & prediction revisions Einstein made about the eclipse prior to 1919. Wouldn’t think that would be so hard to find.

 

[robphy]

http://relativity.livingreviews.org/open?pubNo=lrr-1998-12&page=node2.html
Gravitational Lensing in Astronomy
Joachim Wambsganss

In the year 1911 - more than a century later - Albert Einstein [50]directly addressed the influence of gravity on light (``Über den Einfluß der Schwerkraft auf die Ausbreitung des Lichtes'' (``On the Influence of Gravity on the Propagation of Light''). At this time, the General Theory of Relativity was not fully developed. This is the reason why Einstein obtained - unaware of the earlier result - the same value for the deflection angle as Soldner had calculated with Newtonian physics. In this paper, Einstein found \tilde \alpha = 2GM_{sun}/c^2R_{sun}=0.83{\rm\ arcsec} for the deflection angle of a ray grazing the sun (here M_{sun} and R_{sun} are the mass and the radius of the sun, c and G are the velocity of light and the gravitational constant, respectively)...

-snip-

See [9] to view a facsimile of a letter Einstein wrote to G.E. Hale on October 14, 1913. In the letter, Einstein asked Hale whether it would be possible to determine the light deflection at the solar limb during the day. However, there was a ``wrong'' value of the deflection angle in a sketch Einstein included in the letter.

-snip-

With the completion of the General Theory of Relativity, Einstein was the first to derive the correct deflection angle \tilde\alpha of a light ray passing at a distance r from an object of mass M as
\tilde \alpha = \frac{4GM}{c^2} \displaystyle\frac{1}{r}
where G is the constant of gravity and c is the velocity of light. The additional factor of two (compared to the ``Newtonian'' value) reflects the spatial curvature (which is missed if photons are just treated as particles). With the solar values for radius and mass Einstein obtained [51, 52]:
\tilde \alpha_{sun} = \frac{4GM_{sun}}{c^2} \displaystyle\frac{1}{R_{sun}}=1.74{\rm\ arcsec}

[9] American Institute of Physics, ``A. Einstein, Images and Impact. World Fame I'', (1996), [Part of Online Exhibition]: cited on 14 September 1998, http://www.aip.org/history/einstein/ae24.htm

[50] Einstein, A., ``Über den Einfluß der Schwerkraft auf die Ausbreitung des Lichtes'', Annalen der Physik, 35, 898, (1911).

[51] Einstein, A., ``Die Grundlage der allgemeinen Relativitätstheorie'', Annalen der Physik, 49, 769, (1916).

lists pre-GR and GR references.

 

[SpaceTiger]

The impulse approximation is quite handy for calculating atomic cross sections (that's where I learned it), but it can just as easily be applied to gravity in the Newtonian limit. I find it more intuitive than a lot of the other options.

Of course, there are other ways of getting the same result. As you've noted, an infalling mass with E > 0 will follow a hyperbolic orbit. The general solution for the two-body problem takes the form:

r=\frac{p}{1+e cos(\theta)}

where p is the semilatus rectum, given by:

p=\frac{h^2}{GM}={v^2r_0^2}{GM}

The semilatus rectum can also be thought of as the horizontal distance from the primary mass to the particle's orbit (\theta=\frac{\pi}{2}). If you draw a diagram of this, you can convince yourself that this forms a right triangle with r_0 and that the angle that it deviates from the horizontal axis is just:

\alpha'\simeq\frac{r_0GM}{v^2r_0^2}=\frac{GM}{v^2r_0}

In this case, we did only consider one half of the orbit, so you need to multiply by two to get the full angle of deflection:

\alpha=\frac{2GM}{v^2r_0}

 

[RandallB]

The semilatus rectum - one half of the orbit,

Thanks Ilike this one too. More for me to look up.

Thanks ROBPHY too excelent ref:
RB

 

[RandallB]

I feel like I understand most all the parts bending making up the bending of light by the sun. Actually the GR particle path is easier than the Newtonian part that comes up short by 1/2 . The only part I’m not clear on is the Time Dilation part added for GR:

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

Does anyone have a resource or a hint on deriving this part from
1) equivalence principle
and also by using
2) the SR treatment of Newtonian gravity

 

[Garth]

I feel like I understand most all the parts bending making up the bending of light by the sun. Actually the GR particle path is easier than the Newtonian part that comes up short by 1/2 . The only part I’m not clear on is the Time Dilation part added for GR: Does anyone have a resource or a hint on deriving this part from
1) equivalence principle
and also by using
2) the SR treatment of Newtonian gravity

The equivalence principle simply implies that photons and light 'fall' at the same rate, with Newtonian acceleration - this is the SR treatment of a photon with 'mass' h\nu/c^2. That is ST's calculation.

For the full GR prediction you have to also allow for the curvature of space on top, which doubles it.

Garth

 

[RandallB]

The equivalence principle simply implies that photons and light 'fall' at the same rate, with Newtonian acceleration - this is the SR treatment of a photon with 'mass' h\nu/c^2. That is ST's calculation.
For the full GR prediction you have to also allow for the curvature of space on top, which doubles it.
Garth

Yes we have one:
\alpha=\frac{2GM}{rc^2}
from the space curve of GR.

But to get the second
\alpha=\frac{2GM}{rc^2}
for the time dilation you pointed out, is the question.

How is the photon or light’s 'mass' h\nu/c^2 helpful.
We don’t know, nor should care what the ‘mass’ is for the moving object (photon).
We’re just looking at time dilation.
How do we use “the SR treatment of Newtonian gravity” you mentioned to derive this other half of the total GR deflection to account for the time dilation.