I The bending of starlight is twice the Newtonian prediction

[PeterDonis]

The paper which I linked gives such terminology:
obtained by the application of the equivalence principle and special relativistic considerations alone (the Newtonian value)!

That's the "quasi-Newtonian" calculation. There is no separate "Newtonian" vs. "SR" calculation that both give the same answer; there is just the one "quasi-Newtonian" calculation that gives an answer half as large as the GR calculation.

 

[PeterDonis]

For $g_{11}$ and other it gives 1, but for $g_{00}$ it gives above-mentioned part $1-2GM/r$.

This part of the paper (in the right column of p. 1195, the paragraph that starts "We note that in the Newtonian theory...") looks like handwaving to me. I don't see a reference given to the "Newtonian theory" to which this refers. The paper seems to have in mind a metric that uses $g_{00}$ from the first-order approximation to the Schwarzschild metric but ignores $g_{11}$ from that same approximation. But that isn't a valid global metric; it's invalid math.

A proper analysis using the equivalence principle does not write down a global metric at all. It just calculates the deflection of a transverse light ray as observed in an accelerated frame in a small patch of spacetime that is approximated as flat.

 

[exponent137]

OK, $g_{00}$ for SR should be -1, otherwise this is not SR approximation.

The paper seems to have in mind a metric that uses $g_{00}$ from the first-order approximation to the Schwarzschild metric but ignores $g_{11}$ from that same approximation. But that isn't a valid global metric; it's invalid math.

Maybe this approximation is valid at low speeds, for instance for a rest observer, in such case impact of $g_{00}$ is larger than of $g_{rr}$? Thus, that this is not valid for bending of a ray, but it is valid for other GR phenomena?

In principle, the terminology is not important for me, I only wish to link my abovementioned simple calculation with the calculation of $-g_{00}=1-2GM/r$.

 

[PeterDonis]

Maybe this approximation is valid at low speeds

There is a Newtonian approximation in which spatial curvature can be ignored, and yes, it is valid at low speeds (all speeds much less than $c$). This is the approximation used, for example, to model the orbits of the planets and make predictions about things like perihelion precession. But obviously that can't apply to a light ray.

I only wish to link my abovementioned simple calculation with the calculation of $-g_{00}=1-2GM/r$.

For a light ray, there is no "simple calculation" that gives this $g_{00}$ and gives $g_{11} = 1$.

 

[exponent137]

But obviously that can't apply to a light ray.For a light ray, there is no "simple calculation" that gives this $g_{00}$ and gives $g_{11} = 1$.

This is "my" simple calculation:
Newtonian version can be easily calculated if we integrate projection of gravitational acceleration on the path of a ray, we obtain such change of velocity which gives such angle 0,89 arc seconds.
It is probably calculated in one of the links given above.
Maybe it is not simple enough for others, but subjectively it gives me enough to imagine this problem. It is not important if it is physically wrong, it gives some picture.
More formal version is described with above $g_{00}$ (and other g's =1). I only wish to link this formal version with this "simple" calculation?

One step forward is that "There is a Newtonian approximation in which spatial curvature can be ignored, and yes, it is valid at low speeds. "

Maybe this is already done. (A corrects versions which gives factor 2, are already given in the paper and elsewhere.)

 

[PeterDonis]

This is "my" simple calculation:
Newtonian version can be easily calculated if we integrate projection of gravitational acceleration on the path of a ray, we obtain such change of velocity which gives such angle 0,89 arc seconds.

Okay, now back this up with math. You will find that you can do it in Newtonian physics (if you assume a light ray moves at $c$ in Newtonian absolute space but is subject to "acceleration due to gravity"), but Newtonian physics does not have a spacetime metric. See further comments below.

More formal version is described with above $g_{00}$ (and other g's =1).

No, it isn't, because, as noted above, in Newtonian physics there is no spacetime metric, so there is no such thing as $g_{00}$ or any other $g$. Once you use a spacetime metric, you are using relativity, and the only way to use relativity globally for this problem is to use general relativity, in which case, as we've seen, both $g_{00}$ and $g_{11}$ are no longer equal to $1$.

One step forward is that "There is a Newtonian approximation in which spatial curvature can be ignored, and yes, it is valid at low speeds. "

That's not a step forward at all for light, because, as has already been said, light does not move at low speeds and this approximation is not valid for light. And there is no other approximation that has the property you are looking for, that $g_{00}$ has a correction factor in it (is not equal to $1$), but $g_{11}$ does not (is equal to $1$). In other words, the "simple calculation" you are talking about does not exist.

 

[PeroK]

OK, $g_{00}$ for SR should be -1, otherwise this is not SR approximation.

Maybe this approximation is valid at low speeds, for instance for a rest observer, in such case impact of $g_{00}$ is larger than of $g_{rr}$? Thus, that this is not valid for bending of a ray, but it is valid for other GR phenomena?

In principle, the terminology is not important for me, I only wish to link my abovementioned simple calculation with the calculation of $-g_{00}=1-2GM/r$.

I would say that there is no gravity in SR, and no bending of light (*). In the sense that there is nothing in the theory to tell you how to calculate such a thing. You use either Newtonian gravity or GR.

(*) by massive objects.

 

[PeterDonis]

I would say that there is no gravity in SR

Yes, in the sense that there is no spacetime curvature.

and no bending of light.

No. An observer at rest in an accelerating rocket in flat spacetime will see light rays bend as they travel across the rocket (i.e., perpendicular to the direction of acceleration). This is the basis of the equivalence principle prediction that someone at rest in a gravitational field (and experiencing proper acceleration to stay there) will also see such bending locally.

 

[PeterDonis]

In the sense that there is nothing in the theory to tell you how to calculate such a thing.

A purely SR model would indeed not predict global bending of light in the sense that GR does.

 

[PeroK]

No. An observer at rest in an accelerating rocket in flat spacetime will see light rays bend as they travel across the rocket (i.e., perpendicular to the direction of acceleration). This is the basis of the equivalence principle prediction that someone at rest in a gravitational field (and experiencing proper acceleration to stay there) will also see such bending locally.

Fair enough, you can add the equivalence principle to SR and say something about gravity.

 

[PeterDonis]

you can add the equivalence principle to SR and say something about gravity.

Yes, you can. But "bending of light" is not the same as "gravity"; you don't even need to "add the equivalence principle to SR" to predict that light bending will be observed in an accelerating rocket in flat spacetime.

 

[Sagittarius A-Star]

Fair enough, you can add the equivalence principle to SR and say something about gravity.

A. Einstein did it in 1911 with SR and got the wrong result by a factor of 2 for the bending of light near the sun (see §4 in the link). Later, he did a new calculation, based on GR.
http://www.relativitycalculator.com...ation_on_the_Propagation_of_Light_English.pdf

 

[PeroK]

Yes, you can. But "bending of light" is not the same as "gravity"; you don't even need to "add the equivalence principle to SR" to predict that light bending will be observed in an accelerating rocket in flat spacetime.

We're not talking about accelerating rockets, we're talking about starlight being "bent" by the Sun. That's why we need a theory of gravity.

 

[PeterDonis]

We're not talking about accelerating rockets, we're talking about starlight being "bent" by the Sun. That's why we need a theory of gravity.

That still doesn't make it correct to say that, because there is no gravity in SR, there is no bending of light in SR. There is. And it occurs in accelerating rockets; you need the proper acceleration in order to derive the SR result. Which means you also need the proper acceleration to invoke the equivalence principle and make use of the SR result in the presence of gravity; locally a free-falling observer even in a curved spacetime will not see any bending of light.

 

[PeterDonis]

A. Einstein did it in 1911 with SR and got the wrong result by a factor of 2 for the bending of light near the sun (see §4 in the link).

The calculations in general in that paper are not really just doing it "with SR", because the concept of "uniform gravitational field" that Einstein uses is not SR. (It also turns out to be quite a bit more complicated than Einstein realized when he wrote that paper.) But even if we set that aside, section 4 extends the results to non-uniform gravitational fields like that of the Sun, which is certainly not SR.

 

[PeroK]

This calculation is not really just doing it "with SR", because the concept of "uniform gravitational field" that Einstein uses is not SR. (It also turns out to be quite a bit more complicated than Einstein realized when he wrote that paper.)

That, it seems to me, is precisely the point that @Sagittarius A-Star was making.

 

[Sagittarius A-Star]

That, it seems to me, is precisely the point that @Sagittarius A-Star was making.

Einstein applied in 1911 the equivalence principle globally. That was his error. Here is Einstein's GR calculation from 1916, see equation (74):
https://en.wikisource.org/wiki/The_Foundation_of_the_Generalised_Theory_of_Relativity

 

[Sagittarius A-Star]

The calculations in general in that paper are not really just doing it "with SR", because the concept of "uniform gravitational field" that Einstein uses is not SR.

I thought, SR is the theory of flat spacetime. So Einstein should have assumed, a "uniform gravitational field" only locally to apply the equations of SR, but he did it globally.

 

[PeroK]

I was wrong about there being no bending of light. It can also happen in a prism:

 

[Sagittarius A-Star]

I found an explanation for the factor 2 between Einstein's calculations from 1911 and 1916:

Doubling the Deflection
...
Intuitively, Einstein’s 1911 prediction was only half of the correct value because he did not account for the cumulative effect of spatial curvature over a sequence of small regions of spacetime, within each of which the principle of equivalence applies. This can be understood from the figure below, which depicts a ray of light passing through a sequence of “Einsteinian elevators” near the Sun.

By evaluating the absolute spacelike intervals Δs along the top and bottom of each “elevator car” we find that the walls which are parallel in terms of the local coordinates of the car are not parallel in terms of the global coordinates (except for the central car, where the 1911 and 1915 calculations do predict the same rate of deflection).
...

Source:
https://www.mathpages.com/rr/s8-09/8-09.htm

 

[Nugatory]

I found an explanation for the factor 2 between Einstein's calculations from 1911 and 1916:

And that is also the explanation provided by @Ibix in #2, the very first reply to the original post, four pages back.

 

[PeterDonis]

That, it seems to me, is precisely the point that @Sagittarius A-Star was making.

Saying "he did it with SR", which is what @Sagittarius A-Star said in the post I responded to, seems to me to be a strange way of saying "he didn't really do it with SR", which is what I said.

 

[PeterDonis]

I thought, SR is the theory of flat spacetime. So Einstein should have assumed, a "uniform gravitational field" only locally to apply the equations of SR, but he did it globally.

Yes, it is true that assuming there could be a "uniform gravitational field" globally was a problem with Einstein's 1911 calculation.

I think it's worth expanding some on why it is a problem. Take a step back and consider the flat spacetime of SR in a global inertial frame (which of course always exists in flat spacetime). It is obvious that in this global frame, there is no bending of light. In order to have bending of light in flat spacetime, you must be at rest in a uniformly accelerated frame (at least if we are talking about vacuum--with a material medium present you can, of course, have light bending in an inertial frame in flat spacetime, as @PeroK's example of a prism shows). But there is no global uniformly accelerated frame in flat spacetime. (Of course there isn't one in any curved spacetime either, but in 1911 Einstein hadn't quite gotten to curved spacetime conceptually.) We know that now because we know about Rindler coordinates (which were not discovered until decades later, IIRC), and we know that they only cover a portion of flat spacetime, not all of it.

So in extending his 1911 calculation globally, Einstein was in fact doing invalid math; he just didn't realize it.

 

[vanhees71]

The point is that when using the linearized Einstein field equations, which is for sure legitimate for our Sun, that it's not sufficient to take into account the perturbation of $g_{00}=1+h_{00}$ as it is for "slow" massive "test particles" (planets), because the naive "photon" is massless, and thus you have to take into account also the spatial corrections. For the radially symmetric vacuum solution in the usual spherical coordinates what you get is the corresponding approximation of the Schwarzschild solution,
$$\mathrm{d} s^2 \simeq (1-r_S/r)^2 \mathrm{d} t^2 - (1+r_S/r)^2 \mathrm{d} r^2 - r^2 (\mathrm{d} \vartheta^2 + \sin^2 \vartheta \mathrm{d} \varphi^2),$$
valid for trajectories with $r\ll r_S$.

 

[Sagittarius A-Star]

Yes, it is true that assuming there could be a "uniform gravitational field" globally was a problem with Einstein's 1911 calculation

I must correct by above posting #78. Einstein did not assume globally a "uniform gravitational field", but along the path of the light a gravitational field, of which the magnitude of gravitational acceleration depends on the distance to the sun. So he applied the equivalence principle locally, but assumed a wrong global model for the gravitational field, because in his model, it's direction is every the same.

 

[PeterDonis]

Einstein did not assume globally a "uniform gravitational field", but along the path of the light a gravitational field, of which the magnitude of gravitational acceleration depends on the distance to the sun.

Can you point out the specific place in the 1911 paper where Einstein makes this assumption (the one I've bolded in the above quote?

 

[Sagittarius A-Star]

Can you point out the specific place in the 1911 paper where Einstein makes this assumption (the one I've bolded in the above quote?

The distance to the sun is $r$ in the integral under equation (4):
http://www.relativitycalculator.com...ation_on_the_Propagation_of_Light_English.pdf

This integral is explained from the equivalence principle after equation (1) in

Now, we can read directly off the figure that the direction of the wave front changes by an amount equal to –∂c/∂y per unit of distance along the direction of the wave. Note that the total deflection is extremely small, so to the first order of approximation we can consider just the x component of its motion as depicted below.

The partial of c with respect to y is

To compute the total deflection of the wave front along the entire path, Einstein simply integrated this over the path.
...
(See below for a comment on how Einstein actually performed this integration.)

Source:
https://www.mathpages.com/rr/s8-09/8-09.htm

 

[PeterDonis]

The distance to the sun is $r$ in the integral under equation (4)

Ah, got it.

 

[Sagittarius A-Star]

Ah, got it.

I must also correct my posting #85. He makes no wrong assumption about the direction of the global gravitation field at a certain location. The only error is the missing spatial curvature.

 

[H_A_Landman]

I don't think so, but I'd need to double check the maths.

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?

It is not the case that in the weak-field limit, bending gets Newtonian. In that limit (of say the Schwarzschild metric) there is still a spatial term that ranges from 0% to 100% as large as the temporal term, depending on speed.

To get Newtonian gravity, you have to take the weak-field AND low-speed limits. The spatial term goes to zero at v=0, but equals the temporal term at v=c. So light cannot ever bend by just the Newtonian amount; because it travels at the speed of light, the spatial term is always at max, and the bending is exactly twice what Newton predicts.

At intermediate speeds, the spatial term is always smaller than the temporal term. Thus it is NEVER correct to simplify "gravity is due to the curvature of space-time" to "gravity is due to the curvature of space", but it is often correct to simplify it to "gravity is due to the curvature of time". For example, the gravity holding your butt in your chair is at least 99.9999% due to the curvature of time.