The bending of starlight is twice the Newtonian prediction

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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 traveled 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
J O Linton
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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 traveled 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?
 
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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.
 
  • #8
J O Linton
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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
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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
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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
J O Linton
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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
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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, traveling at the speed of light. I don't think I've got that in my notes anywhere!
 
  • #14
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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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J O Linton
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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.)
 
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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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  • #19
J O Linton
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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
J O Linton
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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
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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
J O Linton
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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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  • #26
A.T.
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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?
As noted by @PeterDonis light follows a gedesic path in spacetime, not in space (on the paraboloid). What those geodesics on the paraboloid hint at qualitatively is the additional global bending effect, that comes on top of the local free fall (coordinate) acceleration, for which the time dimension is key.
 
  • #27
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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.
It's hard to visualize this, because the high-dimensionality. This book (free to download) makes a good job (Chapter 10-11, page 171 and following are about the additional bending due to the spatial distortion):
https://archive.org/details/L.EpsteinRelativityVisualizedelemTxt1994Insight/page/n183/mode/2up
 
  • #28
anuttarasammyak
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[edit]
As one approaches to sun
time pass slowly
periphery/radius becomes longer
These double effects matter.
I draw sketch to explain it.
 

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  • #29
J O Linton
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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 traveled 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.
Thank you. I think I have come to a satisfactory understanding of the issue now.

Firstly I don't think it is quite true to imply that the spatial effect is global which the temporal (EP) effect is local. All physical effects are local and the bending of light in any region of space near a gravitating mass is solely determined by the nature of space-time in the locality. As John Wheeler put it - matter tells space-time how to curve in the locality, space-time tells matter how to move in the locality (my additions). My mistake was to think that at a sufficient distance from a mass, the gravitational field would become effectively uniform. This is not the case.

It is clear to me now that the temporal bending is due to the slowing down of light (as seen by an external comoving observer) at points lower down in the potential well, an effect which it is easy to show is equal to g/c in agreement with the EP. The spatial component of the bending in any locality is due to the non-uniform nature of the field in the locality and I strongly suspect that it is equal (or proportional to) ##\sqrt{dg/dr}## . If this is true than, while g falls off as an inverse square, the spatial effect will fall off more slowly so at great distances the spatial effect will, in fact dominate, not the temporal one.

Regardless of whether my hunch is correct or not, it remains true that the bending of light in a local region near a spherically symmetrical mass will always require both effects to be taken into account. By considering a strictly uniform gravitational field, in which dg/dr = 0 the spatial component has been removed.

My suggestion about Flamm's paraboloid was dismissed by one contributor but I think he mistook the diagram for a potential well. In fact I believe that Flamm's paraboloid (which as you know is a surface whose metric is the same as the spatial component of the Schwartzchild metric) can be used to provide a legitimate (qualitative) explanation of the bending of light, particularly if the paraboloid is inverted so that there can be no confusion with the more familiar potential well diagram.

Essentially the idea is that the temporal component of the bending is due to the fact that light passing near the star travels more slowly (just like sound waves traveling over water or light waves in a graded index optical fibre); the spatial component is due to the fact that radial distances are 'stretched' closer to the star and therefore light which passes close to the star has further to travel. In traveling from a point A on one side of the star to a point B on the other, a light ray chooses the path with the shortest travel time. (The contributor who mentioned that in GR a geodesic maximises the proper time was confusing proper time with elapsed time. In any case the proper time of a light ray is always zero whichever way it travels.)

The attached diagrams should illustrate the way in which I shall in future explain to my students why the bending of starlight is twice the bending predicted by Newton and the EP.

(BTW Please do not think I am directing these comments at you, Ibix, as I am sure your understanding of GR is much better than mine. I have explained my ideas at length in the hope that other readers of this post will gain a better understanding.)
 

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  • #30
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I think I have come to a satisfactory understanding of the issue now.
Unfortunately, you have not. See below.

I don't think it is quite true to imply that the spatial effect is global which the temporal (EP) effect is local.
You are incorrect. If what you call the "spatial effect" were local, it would appear in the equivalence principle analysis, and it doesn't.

All physical effects are local
This is incorrect as well. The way different local patches fit together, which is where the "spatial effect" you refer to comes from, is global, not local. (Note that this means the terms "spatial effect" and "temporal effect" are not good terms; see further comments below.)

As John Wheeler put it - matter tells space-time how to curve in the locality, space-time tells matter how to move in the locality (my additions).
If you add your additions, you aren't stating it as John Wheeler put it. You should have stuck with the way John Wheeler did put it.

It is clear to me now that the temporal bending is due to the slowing down of light (as seen by an external comoving observer) at points lower down in the potential well, an effect which it is easy to show is equal to g/c in agreement with the EP. The spatial component of the bending in any locality is due to the non-uniform nature of the field in the locality and I strongly suspect that it is equal (or proportional to) . If this is true than, while g falls off as an inverse square, the spatial effect will fall off more slowly so at great distances the spatial effect will, in fact dominate, not the temporal one.
All of this is wrong. So is pretty much all of the rest of your post.

As I commented above, the terms "temporal effect" and "spatial effect" are not good terms. It is unfortunate that similar terminology is often used in heuristic or pop science descriptions of light bending. A much better way to look at it was described in earlier posts in this thread: you have a local effect, which can be understood using the equivalence principle, which predicts light bending, but which, if added up globally in the way one would naturally do it in flat spacetime, gives an answer only half as large as the GR prediction and the actual experimental result. The discrepancy is due to the fact that the spacetime around a massive object is curved, so the local patches along the light ray's worldline do not "add up" the way they would if the spacetime were flat. When the effects in each local patch are "added up" in the correct way for the actual curved spacetime, they produce a result (the GR prediction) which is twice as large, and which matches the experimental result. This is true regardless of how far away from the massive object the light ray's point of closest approach is; the overall magnitude of the bending obviously decreases as the point of closest approach gets further away, but the "split" between the "local" part and the "how things get added up globally" part remains the same.

The attached diagrams should illustrate the way in which I shall in future explain to my students why the bending of starlight is twice the bending predicted by Newton and the EP.
If you do this you will be giving your students false information. See above.
 
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  • #31
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Essentially the idea is that the temporal component of the bending is due to the fact that light passing near the star travels more slowly (just like sound waves traveling over water or light waves in a graded index optical fibre)
Except that this just isn't right. To see the problem we have to define exactly what "Travels more slowly" means, and that is harder than it looks. Say that we have two points A and B on the path of the light beam. We have a detector at point A and a detector at point B; the light flash passes A, a bit later it arrives at arrives at B; it seems obvious that if we divide the distance between A and B by the travel time between the two detections we'll have the speed of light... but both that distance and that time depend on our choice of coordinates.

First, what is the distance? That will be the spacetime interval between the point on A's worldline where the detection happened and that point on B's worldline which happened "at the same time", which is to say has the same time coordinate. Clearly this will depend on our simultaneity convention, we can make it come out to be pretty much any arbitrary number according to what "at the same time" means.

Second, what is the travel time? It's the difference between what A's clock read at its detection event and what B's clock reads at its detection event - but only if the two clocks are synchronized, meaning that they read the same thing at the same time, so again we require a simultaneity convention. It doesn't help to appeal to some third party, say us astronomers far from the sun and looking at our lab clock; that just means that we're choosing an arbitrary simultaneity convention between our lab clock and the two detectors.

There is a standard way out of this impasse: A and B are close enough that we can approximate the spacetime between them as locally flat and use Einstein clock synchronization and the methods of special relativity to define both the travel time and the travel distance. When we do this (and it would be somewhat perverse not to) we will always find that the distance divided by the time is ##c##.

(The contributor who mentioned that in GR a geodesic maximises the proper time was confusing proper time with elapsed time. In any case the proper time of a light ray is always zero whichever way it travels.)
I fear that the confusion is on your end. What you are calling an "elapsed time" is a proper time: It is the proper time between two readings of some remote observer's clock, when the readings are made at the same time (the simultaneity convention reenters the discussion) as the two events between which you are determining the elapsed time.
 
  • #32
J O Linton
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I stand by my statements.
Tom takes a metre ruler and a clock to the bottom of a mine shaft. He sets up a device consisting of a light beam pulsing back and forth along the ruler. The device emits pulses of light from each end A and B when the light beam is reflected. He places his clock midway along the rule and calibrates it so that it reads a time interval between pulses equal to ##\frac{1}{c}## He then carries his clock to the top of the mine shaft. Due to gravitational time dilation he now discovers that the time interval between pulses is now ##\gamma \frac{1}{c}##. The proper time between events A and B is, of course, zero. From Tom's point of view the elapsed time between these events is ##\gamma \frac{1}{c}## and the apparent speed with which the light travels from A to B is (from Tom's point of view) ##\frac {c}{\gamma}##.
 
  • #33
PeroK
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I hope your students are spared that confusion of ideas and terminology!
 
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  • #34
J O Linton
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Please. What is it that you object to in the scenario I have described?
 
  • #35
A.T.
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Firstly I don't think it is quite true to imply that the spatial effect is global which the temporal (EP) effect is local. All physical effects are local ...
But the total amount of bending after passing a large region is a cumulative effect. It's not local measurement.

... and the bending of light in any region of space near a gravitating mass is solely determined by the nature of space-time in the locality.
The bending relative to the local patch of space time is one thing. But the way how those spacetime patches are combined to form a larger region also affects the total amount of total bending, after passing that region.
 

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