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.

 

[SpaceTiger]

How is the photon or light’s 'mass' h\nu/c^2 helpful.

It isn't, really, but it can be used as a justification for the application of Newton's Laws to the problem. Without giving the light a mass, you effectively have action without reaction (the Newtonian gravitational force would have to be set to zero).

By the way, I made a mistake in my latex code in an earlier post (forgot to include the \frac). Here's the correct expression for the semilatus rectum:

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

 

[RandallB]

It isn't, really, but it can be used as a justification for the application of Newton's Laws to the problem.

Yes, I accept the idea of using an “implied mass” for the light photon to apply existing laws. Especially when the formulas don’t use the mass of the object in motion in-order to describe their path of motion, it’s the same for all values. That is angles of deflection here are not going to change based on different mass or different frequency of the traveling object – no prism effect.

Where I’m having trouble is using that assumption and deriving the 2nd half of the GR deflection that comes from time dilation. It makes sense that it should provide some addition push, force, bend, change - however you want to describe the causal effect that gives an additional bend beyond the bend caused by following the GR curve of space. Both being needed to reach the correct GR total.

Other than the statement that the measure of that defection caused by the time dilation component happens to be exactly the same as the space curve component; I’ve found nothing that will start me off on defining or deriving that number by using the “the SR treatment of Newtonian gravity” or any other. So I’m a little lost on this one.

PS: If the system still has the edit open on your old post - It might be worth it for other readers to edit post #26 for that “\frac”

 

[Wizardsblade]

I was wondering when I read this, light does not follow normal Newtonian physics (it has constant velocity.) So when light is pulled towards the sun it can not accelerate as a normal particle would. So would this effect the equations for the space part of GR's equation? Or does this effect have no significant on lights arc?

I'm also curious about the derivation of the time factor.

 

[Garth]

I was wondering when I read this, light does not follow normal Newtonian physics (it has constant velocity.) So when light is pulled towards the sun it can not accelerate as a normal particle would. So would this effect the equations for the space part of GR's equation? Or does this effect have no significant on lights arc?
I'm also curious about the derivation of the time factor.

As light is attracted towards a mass and its gravitational field the photon does not speed up in the normal sense, instead its path is deflected as discussed above and its energy increases, (as measured by apparatus of standard mass), that is, its frequency increases - it is blue shifted. As it moves away from a mass, and its gravitational field, its energy decreases and it is red shifted.

Garth

 

[RandallB]

and its energy increases, (as measured by apparatus of standard mass), that is, its frequency increases - it is blue shifted. As it moves away from a mass, and its gravitational field, its energy decreases and it is red shifted.

Is this the key to figuring out the additional time dilation angle to be added to the GR curve in space (only 1/2 the needed angle)? Similar to the 90⁰ Doppler effect of red shift ( I don’t recall - what’s the correct term for this?) when a high speed light emitting source passes at it’s closest point to an observer.

While object (light) follows the curve space path of GR as it descends deeper into the gravity well and experiences a Blue Shift . Can we use that apparent increase in energy as an “apparent increase” in mass (that goes away as it departs the sun) to calculate an additional displacement for the higher mass to account for the angular change caused by the time dilation?

 

[Garth]

Is this the key to figuring out the additional time dilation angle to be added to the GR curve in space (only 1/2 the needed angle)? Similar to the 90⁰ Doppler effect of red shift ( I don’t recall - what’s the correct term for this?) when a high speed light emitting source passes at it’s closest point to an observer.
While object (light) follows the curve space path of GR as it descends deeper into the gravity well and experiences a Blue Shift . Can we use that apparent increase in energy as an “apparent increase” in mass (that goes away as it departs the sun) to calculate an additional displacement for the higher mass to account for the angular change caused by the time dilation?

I think you are following a false trail here.

The photon travels along a straight line - a geodesic - and as it is light traveling at the speed of light that geodesic is a null-geodesic because the photons's proper time along its path is zero.

That straight line appears 'curved', that is it is deflected, because the space-time hyper-surface along which it is traveling is curved by the presence of the Sun's mass and this curvature is that which gives the Sun its gravitational field.

The photon has no rest mass, the change of its energy does not alter its path as all objects or photons fall at the same rate in GR, whatever their mass or energy - that is the essence of the Equivalence Principle and verified in experiments by Galileo onwards. (Cannon balls and feathers fall at the same rate in a vacuum.)

In order to calculate the amount of deflection you have to solve the equation of motion of the photon across the curved space-time surface, it is an involved calculation that however may be broken down into two components.

One component is due to the photon 'falling' towards the Sun, which is half the total, and this can be identified with the effect of time dilation at a series of momentarily stationary frames of reference along the path through which the light ray passes. The other half component is due to the curvature of space on its own.

I hope this helps.


Garth

 

[lightgrav]

I was wondering when I read this, light does not follow normal Newtonian physics (it has constant velocity.) So when light is pulled towards the sun it can not accelerate as a normal particle would. So would this effect the equations for the space part of GR's equation? Or does this effect have no significant on lights arc?
I'm also curious about the derivation of the time factor.

Using the corpuscular theory of light that was fashionable in Newton's day,
theorists would have expected light to speed up as it "fell" toward the Sun.
Unlike SpaceTiger, who pulled the "v" outside the integral in his impulse approximation, Newton would've treated it as a ponderous object which would travel faster near the Sun, and so be deflected less than you might expect based on the constant-speed straight-line path.
It is remarkable that the actual trajectory of a massive object takes it just enough closer than the straight-line path, so that the angular deflection infinitely far from the Sun is the same as derived in the impulse approximation (for small deflections). This is a coincidence, only being true for 1/r potentials.

The EASY way to get the time-factor is to consider the (scalar) Gravitational Potential (/c^2) as an increase in the index of refraction.

 

[SpaceTiger]

Unlike SpaceTiger, who pulled the "v" outside the integral in his impulse approximation, Newton would've treated it as a ponderous object which would travel faster near the Sun, and so be deflected less than you might expect based on the constant-speed straight-line path.

These results are only valid in the small-angle (nearly constant velocity) limit. The deflection angle at infinity is not equal to \frac{2GM}{rv^2} for a body whose kinetic energy is comparable to its potential energy at closest approach.

so that the angular deflection infinitely far from the Sun is the same as derived in the impulse approximation (for small deflections). This is a coincidence, only being true for 1/r potentials.

I don't feel like solving the necessary differential equations to check this, but I'm pretty sure that's incorrect. For the 1/r^2 case, as one approaches the limit I mentioned above, the total impulse given to a passing body will be dominated by the time during which it's closest to the central mass. If the force law were steeper (such as 1/r^3), this would be even more true and I would expect the above approximation to be even more accurate. Clearly, it will break down for some force laws (like F~const.), but the result almost certainly isn't a coincidence of 1/r^2.

 

[RandallB]

I think you are following a false trail here.
…. ….. …. to calculate the amount of deflection .. to solve the equation of motion …. may be broken down into two components.
One component is due to the photon 'falling' towards the Sun, which is half the total, and this can be identified with the effect of time dilation at a series of momentarily stationary frames of reference along the path through which the light ray passes.

The other half component is due to the curvature of space on its own.
I hope this helps.
Garth

It’s your trail from an earlier post I’m trying to pick up.

So far this one thread has given more about Einstein’s bending starlight prediction then anything I’ve ever found on the net. Most do little better than recite a news headline about Einstein being the predictor of star locations.

I thing most will get the idea about the curve in space:
From your post #12

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.

Even without a detailed derivation here it compares well with Swartzchild formula AND SpaceTiger gave excellent Newtonian examples in a couple different ways.
Even Einstein only had this much before 1916. The key to GR is his additional part of the prediction to include time dilation.
Again from your post #12

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

Of the two options to “reveal” the time dilation component 1) as determined by the equivalence principle, or 2) equivalently by the SR treatment of Newtonian gravity.
It’s by using the “SR treatment of Newtonian gravity” to derive this last half of the total deflection that I believe will be the most understandable explanation.

That’s the only piece missing, to have this thread hold a complete explanation of Einstein’s GR calculation of starlight deflection by the sun.
Can you or someone show a derivation to build this last half, or reference that does? It would make this thread complete.

 

[RandallB]

Just following up on your post #12

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
making a total of
\alpha = \frac{4GM}{Rc^2} radians = 1.75" arc for the Sun

Do you remember how to derive these time dilation parts or a reference where it came from?
I was hoping the “equivalently by the SR treatment of Newtonian gravity” part would be even simpler than the impulse version of the first half (non-time dilation portion) of the answer.

 

[Garth]

You will find the splitting of the time dilation and space curvature components of light deflection in C. M. Will, Was Einstein Right?: Putting General Relativity to the Test, Basic Books (1993). (This is a popular account of tests of general relativity.)

The mathematics of the time dilation part as a "SR" treatment of Newtonian gravity has already been given by ST above.


The space-time deflection of light in terms of the Robertson parameters for a general metric is given by
\theta =\frac{4G_mM}R\left( \frac{1+\gamma }2\right)
where \gamma is the amount of space (only) curvature per unit mass, and in GR \gamma = 1.

By elimination you can see that the other component (also 1) is that caused by the dilation of time.

I hope this helps.

Garth

 

[RandallB]

The mathematics of the time dilation part as a "SR" treatment of Newtonian gravity has already been given by ST above.

Sorry I don’t see that.
I assume by ST you mean Space Tiger not “Space-Time”
He did provide very valuable examples of the Newtonian equivalents for finding the sun’s bending of light.

One used the impulse approximation for any object applied to gravity in the Newtonian limit. Very effective and intuitive once you understand it.https://www.physicsforums.com/newreply.php?do=newreply&p=857454

Another using any in-falling mass following a hyperbolic orbit he diagramed the angle of the semilatus rectum to arrive at the same answer. A little harder to follow but it works.

As both of these approaches give the same answer for objects of any small mass, it is no stretch at all to apply the same conclusions to the “no-mass” photon, by considering no mass as included in “any small mass”.
And these results match perfectly with Einstein’s GR predictions of 1911 to 1915 and are all only half the total correct solution. (Not until 1916 did Einstein’s final adjustments to GR include the affect of time dilation in it for the total solution we accept today.)

BUT, I don’t see where any of these have derived anything to support that final half of the light bending solution for time-dilation. And I consider this thread incomplete without it. As I said earlier I find most web references to simple state the end result equation with no explanation as to how they get there. What is need is the explanation or derivation of how the time dilation was built into the final equation, not just that final equation shown in two parts.

The editorial I find on your referenced book says it will “explain the implications of this highly complex theory without using any mathematics beyond geometry”. Explaining implications doesn’t sound like it is focused on providing a derivation of the original. But I’ll try to get hold of a copy and see what’s there, if I can extract a reasonable explanation of the time dilation half of the solution is I will post it here. I just figured it would be better to come from someone that already knew and understood more about that time dilation calculation than I do.

 

[Garth]

First there is only one effect in GR, the deflection of the light ray as a consequence of space-time curvature. To do it properly you have to solve the equation of motion of a null-geodesic grazing the Sun. There is no easy alternative, see
Weinberg "Gravitation & Cosmology" page 188 ff. or
MTW "Gravitation" page 184-5 (a simpler derivation) or
Wald "General Relativity" pages 144-8.

The result using the generalised metric in the Eddington & Robertson expansion is as above:
\theta =\frac{4G_mM}R\left( \frac{1+\gamma }2\right)
where the Robertson parameter \gamma is the amount of space (only) curvature per unit mass, and in GR \gamma = 1.

So in GR half the effect is due to the curvature of space alone. But what is the other half due to? It must be the non-space part of space-time curvature i.e. time dilation.

That is my argument.

By the Equivalence Principle time dilation is the measure of the relative acceleration of two momentarily stationary freely falling particles at different altitudes in a gravitational well.

This acceleration is the normal Newtonian gravitational acceleration, therefore the calculation of a photon 'falling towards' the Sun is equivalent to the calculation of the time dilation effect of the metric at different points along the light path.

Furthermore, the Newtonian calculation of a photon 'falling towards' the Sun makes no allowance for the curvature of space (alone) as space is Euclidean in Newtonian physics.

Garth

 

[RandallB]

in GR half the effect is due to the curvature of space alone. But what is the other half due to? It must be the non-space part of space-time curvature i.e. time dilation.
That is my argument.
This acceleration is the normal Newtonian gravitational acceleration, therefore the calculation of a photon 'falling towards' the Sun is equivalent to the calculation of the time dilation effect ...
Furthermore, the Newtonian calculation of a photon ... makes no allowance for the curvature of space

Whoa – I think I see the flaw in your argument that has you running up the wrong trail!

If I understand, your arguing that Space Tigers explanations match up to the GR time-dilation components and not the Space Curve part of GR.
That sounds absurd (To borrow from J. Bell), STiger never used anything involving time changes.
And of course what he has shown us relates to the GR curve in space not in time. First of all we are not dealing with a photon that is “falling”! That implies accelerating into the sun due to it’s gravity. These are “traveling by” the sun well above escape speed and being affected by the suns gravity (gravity well) and crossing altitudes of changing time.

Take a good look at the formulas Tiger has given us, not based on time they are based on any mass following normal Newtonian orbits, including escape orbits.
Now reduce those speeds but keep the same fundament formulas; reduced till you get a circular orbit based on the perigee we are investigating. With this orbit there is no time levels being crossed but the Newtonian is giving us an orbit. How could GR ever correlate this with a curve in time! It cannot, the altitudes are not changing! The GR explanation for a circular orbit is the curve in space, the gravity well not GR time dilation. You can not just take those same Newtonian fundamental formulas that tie directly to the GR curve in space at low speeds and somehow claim they transform into time dilation at high speeds. They are comparable to the curve in space not time. So of course “the Newtonian calculation makes no allowance for the curvature of space” it is the equivalent of it. How could a Newtonian calc use a GR curve anyway?

I have no doubt the additional bend is due to light traveling by and crossing different altitudes of time. And the correct number for that part of it is:
\alpha = \frac{2GM}{Rc^2} radians

But we have yet to see how arriving at that part of the solution can be described or was derived.

Maybe someone else has some ideas or knows.

 

[Garth]

If I understand, your arguing that Space Tigers explanations match up to the GR time-dilation components and not the Space Curve part of GR.
That sounds absurd (To borrow from J. Bell), STiger never used anything involving time changes.
And of course what he has shown us relates to the GR curve in space not in time.

You are confusing the orbit curving in space (Newtonian) and the curvature of space. GR explains the curvature of an orbit, such as the circular orbit of the Earth around the Sun as a 'straight line' - that is a geodesic - along a hyper-surface of space-time that itself is "curved". For such curvature you do not have to embed the hypersurface in a higher, fifth dimension, although it might help to visualise it, curvature is described by its intrinsic geometrical properties.

If the Earth's elliptical orbit around the Sun does not look much like a straight line that is because you are not viewing it in 4D space-time. To scale the Earth follows a spiral through space time where the radius is 1 AU and the pitch is I light year, about 10⁵ AUs.

First of all we are not dealing with a photon that is “falling”! That implies accelerating into the sun due to it’s gravity.

In the Newtonian/SR treatment yes we are - the photon is traveling so fast that it is deflected, "accelerating into the sun", by only a small amount.

I have no doubt the additional bend is due to light traveling by and crossing different altitudes of time. And the correct number for that part of it is:
\alpha = \frac{2GM}{Rc^2} radians
But we have yet to see how arriving at that part of the solution can be described or was derived.

As I said above

To do it properly you have to solve the equation of motion of a null-geodesic grazing the Sun. There is no easy alternative, see
Weinberg "Gravitation & Cosmology" page 188 ff. or
MTW "Gravitation" page 184-5 (a simpler derivation) or
Wald "General Relativity" pages 144-8.

Garth

 

[RandallB]

You are confusing the orbit curving in space (Newtonian) and the curvature of space. GR explains the curvature of an orbit, such as the circular orbit of the Earth around the Sun as a 'straight line' - that is a geodesic - along a hyper-surface of space-time that itself is "curved". For such curvature you do not have to embed the hypersurface in a higher, fifth dimension, although it might help to visualise it, curvature is described by its intrinsic geometrical properties.

Now you’re just grabbing at clichés – no one needs the 5th dimension hyper-surface here stick w/ GR.

You’ve only assumed that GR curve doesn’t fit with those STiger explanations.
And give no rational explanation to claim they should be compared with time dilation other than to say “It must be the non-space part of space-time curvature” based on an assumption.

The correlation between STiger explanations and the GR curve in space I gave in post #42 is simple and direct.

If all you have is 5-D and hand waving to try to hold onto comparing these Newtonian explanations to GR time dilation, let someone else weigh in.
Or at least take some time to think it through.

 

[Garth]

For such curvature you do not have to embed the hypersurface in a higher, fifth dimension, although it might help to visualise it, curvature is described by its intrinsic geometrical properties

Now you’re just grabbing at clichés – no one needs the 5th dimension hyper-surface here stick w/ GR.

You obviously do not read my posts so there is no point in trying.

Garth

 

[RandallB]

You obviously do not read my posts so there is no point in trying.
Garth

No your obviousy not even trying to see the point, but at least maybe other can learn.

 

[exponent137]

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

I search, but I do not find derivation and explanation for this time dilatition component.
Space time curvature component can be easily calculated. I can show and Tiger Space showed (not to zero precision).
But how to understand this time dilatation part?

 

[exponent137]

I tried another model: Space is not curved, but light moves faster in one height than in the other (because of time dilatation). Besides, it moves faster in one direction as in another (dilatation of lenght). Light is curved on lens. So it can be curved also on such dilatated space (because of sun).

I tried to calculate something and calculation with dilatation seem right. But I cannot calculate with length dilatation. I tried to calculate with small graviational fields, this means bending of light because of sun.

Is something wrong with my model?

Best regards