The bending of starlight is twice the Newtonian prediction

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Summary:
Why does GR predict starlight bending twice Newtonian?
When light is viewed from an accelerating frame of reference it appears to bend at a rate of a/c (the acceleration a being at right angles to c). By the Principle of Equivalence the bending of a beam of light in a uniform gravitational field ought to be g/c rad/sec. In principle you could use this expression to calculate the total bending of light as it passed a massive object by fairly straightforward integration. I call this the 'Newtonian' prediction.
I am told that GR predicts the bending of starlight round a massive object to be twice this 'Newtonian' prediction. Does this mean that the bending of light in a uniform gravitational field is 2g/c in apparent contradiction of the Principle of Equivalence? If not, then what feature of GR causes light to bend by twice this amount when it passes through the non-uniform gravitational field of a star but not when it passes through a uniform field?
 

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  • #2
Ibix
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The equivalence principle only applies locally, so local measurements made by a hovering observer will show what you expect. However, the deflection of star light is not a local effect - the light has travelled in from infinity, passed near the Sun, and climbed back up to Earth. So you should not expect equivalence principle based arguments to work.

The simplifications you make to the Einstein field equations in order to recover Newtonian gravity involve neglecting what you might call spatial curvature, keeping only the time-time equation. So, roughly speaking, the solution is that local measurements will measure a tiny "Newtonian" deflection, but if you simply add them together you neglect the spatial curvature. A line of little labs along the light path won't quite fit together the way Euclid says they would - and that failure to fit together is the extra half of the deflection.
 
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  • #3
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The equivalence principle only applies locally, so local measurements made by a hovering observer will show what you expect. However, the deflection of star light is not a local effect - the light has travelled in from infinity, passed near the Sun, and climbed back up to Earth. So you should not expect equivalence principle based arguments to work.

The simplifications you make to the Einstein field equations in order to recover Newtonian gravity involve neglecting what you might call spatial curvature, keeping only the time-time equation. So, roughly speaking, the solution is that local measurements will measure a tiny "Newtonian" deflection, but if you simply add them together you neglect the spatial curvature. A line of little labs along the light path won't quite fit together the way Euclid says they would - and that failure to fit together is the extra half of the deflection.
Thanks for this reply. It is basically what I understood to be the case. What bothers me is why there is no spatial distortion in the uniform case when there is obviously a temporal distortion. I understand that at the bottom of a potential well time is dilated according to depth of the well. So why is there no equivalent spatial distortion?
The Schwartzchild metric around a spherical uncharged non rotating mass has equal potential factors in both the spatial and temporal terms. At distances much greater than the SR the gravitational field becomes essentially uniform. So why does the spatial term cease to matter?
Is it the case that as the line of sight of the star moves further and further away from the star, the bending approaches the Newtonian prediction? If so, why are we told that the GR prediction is always precisely twice the Newtonian one?
 
  • #4
PeroK
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Thanks for this reply. It is basically what I understood to be the case. What bothers me is why there is no spatial distortion in the uniform case when there is obviously a temporal distortion. I understand that at the bottom of a potential well time is dilated according to depth of the well. So why is there no equivalent spatial distortion?
The Schwartzchild metric around a spherical uncharged non rotating mass has equal potential factors in both the spatial and temporal terms. At distances much greater than the SR the gravitational field becomes essentially uniform. So why does the spatial term cease to matter?
Is it the case that as the line of sight of the star moves further and further away from the star, the bending approaches the Newtonian prediction? If so, why are we told that the GR prediction is always precisely twice the Newtonian one?
To get the answer you do the mathematics. Quantitative questions in physics are settled by calculations.
 
  • #5
anuttarasammyak
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I could find in the open text
https://web.mit.edu/6.055/old/S2009/notes/bending-of-light.pdf
"Newton’s theory is the limit of general relativity that considers only time curvature; general relativity itself also calculates the space curvature." You may get more detail by reading it.

[edit]
As one approaches to sun
time pass slowly
periphery/radius becomes longer
These double effects matter.
 
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  • #6
vanhees71
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Thanks for this reply. It is basically what I understood to be the case. What bothers me is why there is no spatial distortion in the uniform case when there is obviously a temporal distortion. I understand that at the bottom of a potential well time is dilated according to depth of the well. So why is there no equivalent spatial distortion?
The Schwartzchild metric around a spherical uncharged non rotating mass has equal potential factors in both the spatial and temporal terms. At distances much greater than the SR the gravitational field becomes essentially uniform. So why does the spatial term cease to matter?
Is it the case that as the line of sight of the star moves further and further away from the star, the bending approaches the Newtonian prediction? If so, why are we told that the GR prediction is always precisely twice the Newtonian one?
The answer is that you have to solve the Einstein field equations first. You either get the Schwarzschild solution when assuming a spherical symmetric solution (which is, outside of the matter, always the static vacuum Schwarzschild solution, which is known as Birkhoff's theorem) or you insist on the non-relativistic limit and solve for the linearized Einstein field equations, which gives not only a temporal but also a spatial component of the metric, ##g_{00}=1-2M/R## and ##g_{rr}=-(1+2M/R)##, which are both at the same order of magnitude, and in solving the geodesic equation for lightlike geodesics you cannot neglect the spatial part. The result is twice the result when (wrongly) neglecting the spatial perturbation.
 
  • #7
Ibix
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What bothers me is why there is no spatial distortion in the uniform case when there is obviously a temporal distortion. I understand that at the bottom of a potential well time is dilated according to depth of the well. So why is there no equivalent spatial distortion?
I'm not sure what metric you have in mind for the uniform case.

The Schwartzchild metric around a spherical uncharged non rotating mass has equal potential factors in both the spatial and temporal terms.
No - one is the inverse of the other. And it isn't really the metric you need to worry about here so much as the connection coefficients, which describe how tangent spaces (the formal concept more or less equivalent to the locally flat bits of spacetime) fit together. These are built from derivatives of the metric (which is why the inverse makes a difference) and it is from these that the field equations are built (and Poisson's equation falls out of these).
Is it the case that as the line of sight of the star moves further and further away from the star, the bending approaches the Newtonian prediction?
I don't think so, but I'd need to double check the maths.
 
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What you are saying, Ibix, makes a lot of sense. I am afraid the actual maths is totally beyond me but I can get a hint from your comment about the derivatives why the spatial and temporal components could differ.
Thinking a bit further about my final question, I suspect that the GR prediction will always be twice the Newtonian one. If it is the case it could, loosely speaking, be ascribed to the fact that as the light passes the star, the direction of the gravitational field through which it passes rotates through 180 degrees. The effect of this would be to 'drag' the light beam round with it. (Of course, this is only a manner of speaking - the real reason is because of the way the 'locally flat bits of space time fit together'.) The implication here is that when a light beam passes though a small region of space in which the direction of the gravitational field changes, then the light beam will bend. In a totally uniform field, there is no change in direction hence the only bending is due to the EP.
Suppose you arranged for a light beam to pass through a region of space which contained a gravitational field of constant magnitude but which rotated at a constant rate from 0 to 180 deg. How would the rate of bending vary along the trajectory, I wonder? My instinct suggests that the rate of bending would actually be constant; at the extremities it would be due to my 'drag' effect (i.e. the spatial distortion) but in the middle it would be due to the Newtonian (temporal) effect. Both effects would contribute equally to the total bending. What do you think?
 
  • #9
PeroK
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I am afraid the actual maths is totally beyond me but ...
You can't discuss physics at an I level (i.e. intermediate, undergraduate level) without mathematics.
 
  • #10
Dale
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Summary:: Why does GR predict starlight bending twice Newtonian?

Does this mean that the bending of light in a uniform gravitational field is 2g/c in apparent contradiction of the Principle of Equivalence?
No. In a uniform gravitational field the bending is the usual Newtonian value you mention. The equivalence principle holds throughout a uniform field.

However, the gravitational field of the sun is non-uniform. That is “curvature” in GR. The equivalence principle only holds locally near the sun, over regions small enough to consider uniform. Not over the whole trajectory. The factor of two accounts for the non-uniformity over the whole trajectory.

If not, then what feature of GR causes light to bend by twice this amount when it passes through the non-uniform gravitational field of a star but not when it passes through a uniform field?
Curvature. Many people think that in GR all gravitational effects are due to curvature. But in actuality curvature is specifically the effects of tidal gravity or non uniform gravity.
 
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  • #11
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You can't discuss physics at an I level (i.e. intermediate, undergraduate level) without mathematics.
I agree, of course, but mathematics and understanding go hand in hand. If you rely on either to the exclusion of the other you can go seriously astray.
Can you guide me through the mathematics of the scenario I suggested in my last post?
 
  • #12
vanhees71
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No. In a uniform gravitational field the bending is the usual Newtonian value you mention. The equivalence principle holds throughout a uniform field.

However, the gravitational field of the sun is non-uniform. That is “curvature” in GR. The equivalence principle only holds locally near the sun, over regions small enough to consider uniform. Not over the whole trajectory. The factor of two accounts for the non-uniformity over the whole trajectory.

Curvature. Many people think that in GR all gravitational effects are due to curvature. But in actuality curvature is specifically the effects of tidal gravity or non uniform gravity.
But you can do a further approximation, setting ##2M/r \simeq 2 M/R=\text{const}##, though I don't think it makes sense in the light-bending on a star, because there you consider the unbound (hyperbola-like) orbit of the "photon".

Mathematically, you still have both deviations from Minkowski in the metric, i.e., in both components ##g_{00}##, ##g_{11}##. I've never done the calculation for light bending in this approximation but isn't there the same factor of 2 for the deflection angle compared to the naive approximation ##g_{11}=-1##?
 
  • #13
PeroK
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I agree, of course, but mathematics and understanding go hand in hand. If you rely on either to the exclusion of the other you can go seriously astray.
Can you guide me through the mathematics of the scenario I suggested in my last post?
The calculation is non-trivial. It's in Hartle's book, culminating in the equation $$\delta \phi = \frac{4GM}{c^2b}$$ where ##\delta \phi## is the angular deflection and ##b## is the impact parameter.

We'd need to do the equivalent calculation for a particle in a Newtonian inverse-square force, travelling at the speed of light. I don't think I've got that in my notes anywhere!
 
  • #14
PeroK
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PS The classical inverse square deflection can be found here, equation (4.20):

https://www.damtp.cam.ac.uk/user/tong/relativity/four.pdf

$$\delta \phi = 2 \tan^{-1}(\frac{GM}{bc^2})$$ If we take the case that ##\frac{GM}{bc^2}## is small (the Hartle equation was explicitly for this case) and use the Taylor expansion for ##\tan^{-1}##, we get the :
$$\delta \phi = \frac{2GM}{bc^2}$$ which is half that calculated from GR.

Note also that these are approximations for the particular case of a small deflection. The maths, as it often does, gets messy.
 
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PS The classical calculation is also here (although he just quotes the GR result).

http://www.inference.org.uk/sanjoy/teaching/approximation/light-bending.pdf
Thanks for this. I had seen this reference before. The trouble with dimensional analysis is that it fails to give the very parameter we want, namely the coefficient!
I know that you think that once you have done the maths there is nothing more to be said, but I am sure there must be a way of explaining why the spatial distortions produced by a massive object cause light to bend. I take it that you don't like my idea of the rotating gravitational field 'dragging' the light round. Is there any way of using Flamm's paraboloid? Would a 'great circle' route on the paraboloid be an accurate way of plotting the path of a light beam under the influence of spatial distortions only (i.e. the EP effect would be in addition.)
 
  • #17
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I know that you think that once you have done the maths there is nothing more to be said, but I am sure there must be a way of explaining why the spatial distortions produced by a massive object cause light to bend. I take it that you don't like my idea of the rotating gravitational field 'dragging' the light round. Is there any way of using Flamm's paraboloid? Would a 'great circle' route on the paraboloid be an accurate way of plotting the path of a light beam under the influence of spatial distortions only (i.e. the EP effect would be in addition.)
Non-mathematical physics is a dead-end. You might be able to persuade yourself of this, that or the other, but you have no sure way to know where your reasoning will lead you astray.

Once you have a mathematical basis, there is everything to be said.

In any case, the notes I have linked to are undergraduate physics. Physics depends on mathematics and not woolly thinking. Phrases like "spatial distortions", "rotating gravitational field", "dragging light" are all so imprecise as to be meaningless.
 
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  • #18
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I have to say "show where a factor of 2 comes from without using mathematics" is a pretty tall order.
 
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Non-mathematical physics is a dead-end. You might be able to persuade yourself of this, that or the other, but you have no sure way to know where your reasoning will lead you astray.

Once you have a mathematical basis, there is everything to be said.

In any case, the notes I have linked to are undergraduate physics. Physics depends on mathematics and not woolly thinking. Phrases like "spatial distortions", "rotating gravitational field", "dragging light" are all so imprecise as to be meaningless.
Ok I think you have made your point. Thanks anyway.
 
  • #20
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but I am sure there must be a way of explaining why the spatial distortions produced by a massive object cause light to bend.
It's better not to think of light has bending, but rather moving in a straight line through curved spacetime.

Imagine that you are standing at the Earth's equator and start walking in s straight line due north. I do the same thing at the same time, except that my starting point is a few meters to your left. You will find that although our paths were initially parallel, my path is gradually turning towards you so that we collide at the north pole. You can explain this either by saying that you are walking in a straight line while my path is bending or by saying that we are both moving in a straight line but the surface of the earth is curved so that initially parallel paths draw closer and intersect.
Someone walking in a straight line on a curved two-dimensional space (the surface of the earth) is not a great analogy for a flash of light moving in a straight line through a curved four-dimensional spacetime, but it's as good of an explanation as we can do without the math.
I take it that you don't like my idea of the rotating gravitational field 'dragging' the light round. Is there any way of using Flamm's paraboloid? Would a 'great circle' route on the paraboloid be an accurate way of plotting the path of a light beam under the influence of spatial distortions only (i.e. the EP effect would be in addition.)
This one is trickier than it seems, because in GR the notion of "gravitational field" is very limited.

Something analogous to a classical gravitational field is only found in a few (important and frequently discussed - the spacetime around a massive body like the sun is one) situations, and none of these involve the sort of rotation that you're considering. For more general cases, in principle we specify the distribution of stress and energy, insert this in the Einstein field equations (64 coupled non-linear partial differential equations), solve for the metric tensor, and then calculate the paths of things (objects in free fall, flashes of light) as they move through the spacetime described by that metric tensor - a "gravitational field" that we consider to be exerting a force on objects just isn't anywhere in the solution.

Flamm's paraboloid is applicable to only one particular situation: a non-moving isolated massive spherically symmetric object like the sun. It won't be much help trying to visualize anything else.
 
  • #21
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I would be happy to stick to the case of a non-moving isolated massive spherically symmetric object like the sun!
The attached file shows Flamm's paraboloid on which I have traced a great circle route (black) and a 'straight line' route (red). Would you not agree that it is appropriate to say that the former is the 'shortest' route from A to B and that it does, at least qualitatively, describe the sort of path to be expected on the basis of the spatial distortions around such an object?
 

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  • #22
PeroK
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Would you not agree that it is appropriate to say that the former is the 'shortest' route from A to B and that it does, at least qualitatively, describe the sort of path to be expected on the basis of the spatial distortions around such an object?
In GR, the geodesics followed by massive particles are the longest (not the shortest) path between two points. That said, the null geodesics that light follows have zero length.
 
  • #23
PeterDonis
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Flamm's paraboloid
Is, unfortunately, not a good tool to use for your intended purpose. All it shows is one particular aspect of "space curvature" in one particular system of coordinates. It does help with some intuitions (for example, intuitions about purely radial motion and "climbing out of the gravity well"), but not with others (such as motion with a non-radial component, as you are considering here).

Would you not agree that it is appropriate to say that the former is the 'shortest' route from A to B and that it does, at least qualitatively, describe the sort of path to be expected on the basis of the spatial distortions around such an object?
No and no. The problem is that freely falling objects (light, planets in orbit, spacecraft launched by humans) travel on geodesics of spacetime, not space. The geodesics of the Flamm paraboloid are simply the wrong ones to be looking at.
 
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  • #24
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at least qualitatively, describe the sort of path to be expected....
You may want to step back from general relativity for a moment and consider how "distance" (the standard term is "interval") works in the flat spacetime of special relativity. This not a digression because any small patch on the surface of the paraboloid (more generally, any sufficiently small region of spacetime) can be treated as flat so you'll need it to understand it to understand what the paraboloid is.
 
  • #25
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I would like to thank all the above contributors for their views. I may not yet have a perfect understanding of the issue I raised but at least some errors of fact have been exposed so thank you.
 
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