I The bending of starlight is twice the Newtonian prediction

[vanhees71]

This is a very bad attitude for a teacher. You should not teach anything that is really wrong. Of course, you have to make simplifications and omit topics that cannot be treated with the level of mathematics your students understand and can handle. It is very well justified to teach classical Newtonian mechanics and classical electromagnetics. The latter subject is already problematic, because in principle you need vector calculus for it, but at least in electrostatic you can get pretty far by starting from Coulomb's Law and describe the electrostatic forces etc.

What's really bad is if the teaching provides even a qualitatively wrong picture, e.g., by insisting on teaching Bohr's model of the atom (cementing the idea of electrons moving on orbits around the nucleus as if it were a tiny planetary system).

Concerning general relativity it's very hard to teach right without quite advanced math (tensor calculus on non-Euclidean manifolds). You can get some of the flavor by carefully (!) referring to the equivalence principle, but as this thread shows there's always the danger to overstate this qualitative picture.

Treating the propagation of light in GR space time as if it were light going through a dielectric medium is also dangerous, and I'd not refer to this. To understand the bending of light in a gravitational field (i.e., geometrically spoken wave propagation in curved space-time) you can refer to the geodesic motion of a massless particle, but it's in fact the solution of what's called the eikonal approximation of an electromagnetic wave in this curved space-time, it describes the change of the wave vector of a locally plane em. wave. It's the description of light in terms of geometric optics.

In this and related context one should not talk about "photons" as if they were point-like particles, although that's the picture you find very often even in university-level textbooks, but that's another sin of bad physics didactics, because it provides again an even qualitatively wrong picture about electromagnetic waves/light.

 

[PeterDonis]

when it comes to teaching physics, almost nothing that we teach is strictly true

If by "strictly true" you mean "a model that is exactly correct", then yes, I agree. None of our scientific models are exactly correct.

However, the onus is still on you as a teacher to correctly describe the models themselves (as well as the experimental facts and how well the predictions of the models match the experimental facts). If you can't correctly describe the models, because you don't think your students will be able to understand a correct description (or because you're not sure you understand a correct description yourself), then IMO you should not describe the models at all; you should just describe the experimental facts, and perhaps a very simple description of an underlying model (I will give an example of that below), and point students who are interested in learning more at a reference that can help them learn a correct description of the underlying models for themselves.

My job is two-fold; to give my non-scientific students a basic understanding of how the world works and why things do what they do

There's a very simple way to do that for the topic given in the title of this thread: the bending of light is twice the Newtonian prediction because of spacetime curvature.

Most students, who as you say will not be studying physics at the university level, will probably be satisfied with that answer and you won't need to go into any more detail. The ones who will be curious for more detail are covered by my next comment below.

and secondly, to enthuse those students with the required ability and curiosity to take the subject further so that you have some undergrads to fill your lecture halls!

Students who have that ability and curiosity will be able to folllow up references themselves. They won't need to be told things in class that they then have to unlearn later.

 

[Nugatory]

Is it that you do not agree that the pulses of light received by Tom are time dilated? Or do you think that the ruler at the bottom of the shaft changes length as Tom climbs out? or do you think that Tom's conclusion is misguided?

We agree that if a source at rest (using a particular coordinate system) emits flashes of light at intervals $\Delta T_S$, a receiver at rest (using the same coordinate system) higher in that gravity well will receive them at intervals $\Delta T_R$, that these are intervals measured using the same clock and therefore are proper times, and that $\Delta T_R \gt \Delta T_S$. This is the phenomenon that is often casually referred to as gravitational time dilation; it is predicted by general relativity and confirmed by (among other tests) the Pound-Rebka experiment.

We agree that the ruler does not change length. This follows directly from the definition of the meter as the distance that light travels in $1/c$ seconds and the fact that the clock in the middle of the ruler receives a flash from one side or the other every $1/c$ seconds.

We agree that Tom can calculate the quantity $1/\Delta T_R$ and that this wil be something less than $c$. If Tom does not already know that this is an example of what is properly called a "coordinate speed", it is high time that he learn about coordinate speeds and their limitations.

Where things go off the rails is trying to assign any physical significance to this (or any other - that's one of the limitations) coordinate speed. In particular thinking that the difference between that coordinate speed and $c$ suggests that there is something different about the spacetime deeper in the well, as does your analogy with boggy and hilly ground, is just plain wrong.

I will concede that the bogs and hills analogy is consistent (at a hand-waving math-free level) with the observed deflection of starlight, but just because it works in that one case doesn't make it good pedagogy.
It would be like teaching that Apollo's chariot pulls the sun from east to west every day - a fine explanation if we care only about the motion of the sun through the sky, but one that fails for the motion of other heavenly bodies and must be unlearned before the student can make further progress.

 

[pervect]

Please. What is it that you object to in the scenario I have described?

The soldiers in the swamp are presumably minimizing travel time, rather than maximizing proper time. Other than that, it seems as good as any other popularization, though I'm not quite sure how badly it might fail if a student relies on it as their sole understanding of the issues.

A truly correct understanding of the physics would require understanding space-time as a Lorentzian geometry, and the principle of maximal aging. Which would be more ambitious, though it has the advantage of actually being correct.

It's definitely better than the bowling ball on the trampoline analogy, but that's not a high bar.

 

[PeroK]

@J O Linton

Just to add my tuppence ha'penny worth:

The book I referenced previously Gravity, an Introduction to Einstein's GR, by James Hartle is probably the theoretical minimum for GR. Even so, he doesn't cover the geometry of a spherical star and Schwarzschild coordinates until chapter 9.

I found that you need a lot of experience of mathematical physics and a certain maturity of thought before you can digest GR. The first time I looked at that book I was way beyond A-level (I found A-level physics a piece of cake even after 30 years break from maths and physics) - but I still backed off GR at the first attempt. When I came back two years later I was older and wiser, and it was much more accessible.

The other issue is that you cannot really teach GR if you are floundering in the dark. There are a great set of lectures by Professor Scott Hughes of MIT on YouTube - that's a graduate physics course and starts with a re-representation of SR in geometric terms - starting with no assumed knowledge of GR. Unless you understand GR at that level, it's difficult to teach. At the very least, you would need to have worked through Hartle's book and/or the MIT lectures to teach GR in any meaningful way - you can't just make stuff up yourself!

My final point is that even if you do soldier on and try to teach GR, you really ought to take note of what has been said in this thread.

 

[berkeman]

Instead pretend that I am an intelligent sixth former studying physics at A level and explain exactly what is wrong with the argument

I would just like to say that when it comes to teaching physics, almost nothing that we teach is strictly true.

But the majority of my students will not study physics at university level. My job is two-fold; to give my non-scientific students a basic understanding of how the world works and why things do what they do; and secondly, to enthuse those students with the required ability and curiosity to take the subject further so that you have some undergrads to fill your lecture halls!

Yeah, I Googled your username, and found your home page:

http://www.jolinton.co.uk/index.html

I am a retired physics teacher

So who are your students now? Are you referring to the folks who read your blog and previously published articles?

In any case, it does appear that your articles at your website are more "popular press" in nature, rather than scientifically rigorous.

 

[stevendaryl]

I am sorry if I did not make myself clear and I am sorry if I forgot to mention that the ruler was horizontal (but I did say that it was at right angles to the GF). I agree that Tom at the top of the shaft and Dick at the bottom will measure the same value for the speed of light but it remains the case that the pulses of light which Tom receives from the device at the bottom will be time dilated and he can therefore legitimately conclude that from his point of view light travels slower at the bottom of the shaft then at the top. It is a trivial matter to deduce quantitatively that the refraction (i.e. bending) of a horizontal light beam is equal to g/c in perfect agreement with the EP. I am baffled as to why everyone is having so much difficulty with this argument.

Well, the description of gravitational time dilation in terms of light traveling slower in a gravity well is not really correct. The equivalence principle shows this: The same dilation occurs aboard an accelerating rocket.

 

[exponent137]

Summary:: Why does GR predict starlight bending twice Newtonian?

Maybe this paper will help you at this:
A simple calculation of the deflection of light in a Schwarzschild gravitational field
https://www.researchgate.net/publication/238984034_A_simple_calculation_of_the_deflection_of_light_in_a_Schwarzschild_gravitational_field
It helped me.

But, I do not understand enough why Newtonian (or classical) and SR calculations give the same result? 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. But SR calculation contains part $(1-\frac{2GM}{r})dt^2$, why it has the same effect?

 

[PeterDonis]

I do not understand enough why Newtonian (or classical) and SR calculations give the same result?

What "SR" calculation are you talking about?

SR calculation contains part $(1-\frac{2GM}{r})dt^2$, why it has the same effect?

Any calculation with such a factor in it is a GR calculation, not an SR calculation. The GR calculation leads to a result twice as large as the "quasi-Newtonian" calculation.

 

[exponent137]

What "SR" calculation are you talking about?Any calculation with such a factor in it is a GR calculation, not an SR calculation. The GR calculation leads to a result twice as large as the "quasi-Newtonian" calculation.

The paper which I linked gives such terminology:
obtained by the application of the equivalence principle and special relativistic considerations alone (the Newtonian value)!
For $g_{11}$ and other it gives 1, but for $g_{00}$ it gives above-mentioned part $1-2GM/r$.

 

[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.

 

[haushofer]

That mathpages treatment ("doubling the deflection") is really insightful. It shows that the total deflection is doubled w.r.t. Newton, but not necessarily at every point of the photon's path; that depends on your coordinates.