View Full Version : Einstein vs. Newton

Cheap Astronomy

2009-Jul-01, 07:55 AM

Hi,

I've read that even Newton's law suggests a degree of gravitational lensing (at least in the context of the 1919 eclipse observation that confirmed GR) - but that GR predicted that the amount of bending of a light path would be exactly twice that predicted by Newton. See here (http://en.wikipedia.org/wiki/Tests_of_general_relativity#Deflection_of_light_by _the_Sun).

Can anyone talk me through why it's exactly twice (noting I'm not really a maths person).

Thanks

NEOWatcher

2009-Jul-01, 11:28 AM

Can anyone talk me through why it's exactly twice (noting I'm not really a maths person).

Can you point to me where in the article it says "exactly".

The article said Einstien said "half" but didn't qualify with the precision.

Now; maybe it is, and there might be a derivation of the computation that does it, but I don't see it mentioned.

grant hutchison

2009-Jul-01, 02:01 PM

It is an exact factor of two: 2m/r versus 4m/r. There's a moderately technical discussion and derivation here (http://www.mathpages.com/HOME/kmath115/kmath115.htm).

In handwaving terms (taken from Bernard Schutz's excellent book Gravity: From The Ground Up), the difference is because light rays "sample" both the spacial and temporal components of spacetime curvature, whereas Newtonian gravity reflects only the temporal curvature.

The planets of the solar system move relatively slowly, and as a result their orbits are determined almost entirely by "time curvature": that is, gravitational time dilation. Newtonian gravity is based on those motions, and so makes no allowance for spacial curvature. (In his book, Schutz builds a toy spacetime with flat space and curved time, and reproduces Newtonian gravity.)

Deviations from Newtonian gravity become more apparent as objects move faster: they begin to "see" the spacial curvature better. (Hence the anomalous motion of the perihelion of the planet Mercury when compared to simple Newtonian predictions.) At the limit, something moving at light-speed samples both curvatures equally, so ends up with twice the deflection that would be predicted by naive, time-based Newtonian gravity. Material objects that move at close to lightspeed will likewise be deflected at something close to twice the Newtonian prediction.

Grant Hutchison

blueshift

2009-Jul-01, 03:46 PM

Excellent post Grant! I'll have to consider buying Shutz's book. Thank you!

Ken G

2009-Jul-01, 04:42 PM

Yes, that "toy model" is quite an interesting way of looking at it. It seems to demystify some of the differences between Newtonian gravity and GR. Indeed, I find it interesting that you can get the identical Newtonian theory in a model where gravity is not a force-- sort of a nonrelativitistic equivalence principle. It's also a very nice example of the importance of keeping separate the predictions our theories make from the stories we tell about "what is happening".

grant hutchison

2009-Jul-01, 05:17 PM

Excellent post Grant!Thanks. :)

It needs one of those authorial disclaimers: credit for everything correct and useful goes to Schutz; blame for anything wrong or misleading comes to me.

Grant Hutchison

Ken G

2009-Jul-01, 06:19 PM

It needs one of those authorial disclaimers: credit for everything correct and useful goes to Schutz; blame for anything wrong or misleading comes to me.

With one important exception to the rule of giving all the credit to Schutz-- he didn't bring this result to our attention!

publius

2009-Jul-02, 05:16 AM

I'll present another toy model of time-only curvature, one that should be very familiar: the Rindler metric. For an acceleration in the z-direction, we have g_00 (g_tt) = (1 + gz/c^2)^2. The spatial part is flat, and there are no off diagonal terms.

But that is just a different coordinatization of flat space-time, one with curved coordinates, not actual invariant space-time curvature. The "gravity" there is entirely a coordinate effect. :) Well, if one wishes to define real gravity as space-time curvature. That is the modern view, although Einstein himself didn't seem to hold with that view. To him gravity was indeed in the Christoffel symbols, and he said as much, although he was criticized for that. He seemed to think it was silly, and I think I agree.

And, while the above stuff above space vs time curvature is a neat and clever trick, I'll have to throw a wet blanket and say that is just a coordinate dependent view. If we switch coordinates, we'll switch the balance of the time vs spatial curvature, and, more importantly as illustrated by Rindler, we can introduce coordinate curvature that gives us the psuedo force of gravity without any real spacetime curvature.

And another thing that Rindler illustrates. We get the Newtonian limit in the *weak field* limit of the time-curvature part. We will get deviations from Newton in the strong field, even with time-only curvature. In Newtonian terms, that translates into "small differences in potential".

Indeed this "time-only curvature" limit can be expressed as, in a sufficiently small local neighborhood,

g_00 = 1 + 2Phi/c^2, where Phi is the Newtonian potential. The spatial parts can be set to one. You may say, wait a minute, Rindler doesn't exactly look like that. Well, expand out the square:

g_00 = 1 + 2Phi/c^2 + (Phi/c^2)^2, where Phi = gz. (It's a rather neat trick, and a subject of some discussion of why that turns out to be exact all the way down for Schwarzschild).

At any rate, you can see that for Rindler, things are very different in the strong field. Indeed Newton would predict the horizon to be twice as far away as it actually is. It's that neat trick that it gives the correct result for Schwarzschild.

-Richard

Ken G

2009-Jul-02, 03:47 PM

But in another way of framing what Richard is saying is not so wet of a blanket after all-- we know that Newtonian physics does not get its gravity to work by curving spacetime. So in Richard's language, this means that Newtonian gravity is not a "real" gravity (no spacetime curvature), but could be viewed relativistically as purely a coordinate effect. Thus it really is a "nonrelativistic equivalence principle". It seems to me this might be saying that if you marry a local equivalence principle with the global behavior of Newton's gravity, but keep Galilean relativity, you get Newton's gravity, which is purely a local coordinate effect rather than a spacetime curvature, but since you're not doing relativity, you take that coordinate effect perfectly literally. Gravity is then a fictitious force like the centrifugal force, it just doesn't behave correctly if you look at it from a high speed.

This would help to know: Does Schutz's curvature involve c? The above g_00 metric loses all gravity when c goes to infinity, and Lorentz goes to Galilean, so that suggests you need a finite c to have gravity. But Newtonian gravity can assume c is infinite and still work, it does not refer directly to c in any way.

PraedSt

2009-Jul-02, 05:48 PM

Newton rules! :)

publius

2009-Jul-02, 06:13 PM

This would help to know: Does Schutz's curvature involve c? The above g_00 metric loses all gravity when c goes to infinity, and Lorentz goes to Galilean, so that suggests you need a finite c to have gravity. But Newtonian gravity can assume c is infinite and still work, it does not refer directly to c in any way.

Ah, it looks that way at first blush, but the 'c' will cancel out in the weak field limit, and you recover the Newtonian 'g' as the gradient of Phi. That happens locally, but I don't know what it would look like globally. That is, if we let c --> infinity, would we recover Newton globally? I don't know.

g_00 is about clock rates, and, "coordinate rest energy" if you like. Take E = mc^2. What does that mean when c -- > infinity?

Remeber the metric has a factor of d(ct) on the time part. While I'm not sure, what I think happens is the time part of the metric just blows away the spatial part. The time part is ct, or the spatial part is x/c. As c grows large, the time part is all that matters.

-Richard

Ken G

2009-Jul-02, 06:27 PM

Take E = mc^2. What does that mean when c -- > infinity? It means the rest energy is infinite, but it is kind of like a "renormalizable" infinity-- the kinetic energy is all that would matter, and it would be finite (and the usual Newtonian value). A person with infinite money wouldn't care about the "next dollar", but the next dollar on top of the infinite rest mass would still be able to control the physics of motion.

Remeber the metric has a factor of d(ct) on the time part. While I'm not sure, what I think happens is the time part of the metric just blows away the spatial part. The time part is ct, or the spatial part is x/c. As c grows large, the time part is all that matters. Yes, I think that must be true-- the limit of infinite c is the limit where all spacetime curvature is in the time component, so any theory that reduced to Newton in the low speed limit would continue to reduce to Newton at all speeds if c went infinite. That was basically the expectation prior to Michelson-Morley (not that the speed of light was infinite, but that the parameter in the metric was infinite, so all coordinate changes affected only the time component). This is even easier to see if we use the sensible convention of expressing the metric in terms of t and d/c!

What is new to me is that this means that an equivalence principle in the Newton/Galileo mechanics could still work for a low-speed version of gravity, or for any speed had Galilean relativity been right. That seems to be the conclusion here, I haven't seen how it would work out in detail-- the equivalence principle is something quite separate from Lorentzian relativity, and someone might have been able to suggest "gravity is not a force" even prior to Michelson-Morely. In other words, a form of GR could have preceded SR!

grant hutchison

2009-Jul-02, 06:59 PM

And another thing that Rindler illustrates. We get the Newtonian limit in the *weak field* limit of the time-curvature part. We will get deviations from Newton in the strong field, even with time-only curvature.Yeah, Schutz does a few dance steps there. He uses a binomial approximation on the way to his "time only" metric for Newton:

Δs² = -(1 – 2GM/c²r)(cΔt)² + (Δx)² + (Δy)² + (Δz)²

but then reveals that his binomial "approximation" actually produces the correct form for both weak and strong fields.

Grant Hutchison

publius

2009-Jul-02, 07:16 PM

Yeah, Schutz does a few dance steps there. He uses a binomial approximation on the way to his "time only" metric for Newton:

Δs² = -(1 – 2GM/c²r)(cΔt)² + (Δx)² + (Δy)² + (Δz)²

but then reveals that his binomial "approximation" actually produces the correct form for both weak and strong fields.

Grant Hutchison

I think that is that "neat trick" of Schwarzschild at work there. That trick shows up in various places. One of many examples is the clock rate of circular orbits. If you plug in the Newtonian orbital speed and ignore higher order terms, you get a simple result. But it turns out that simple result is exact all the way to the circular orbit limit at the photon sphere because the corrections to the orbital speed exactly cancel out the higher order terms. So deviations from Newton in one place exactly cancel out deviations in other places.

That is the neat trick of Schwarzschild and is the subject of some discussion. It's not magic, but, IIRC, a consquence of spherical symmetry in the EFE.

In the general case that special balance doesn't hold.

-Richard

Ken G

2009-Jul-02, 07:17 PM

I wonder why people are always so married to the idea that a metric is a distance, even when they are expressly talking about a time curvature? Let's be sensible and divide through by c2 and call the invariant a squared proper time tau2 (and reverse the signs):

Δtau² = (1 – 2GM/c²r)(Δt)² - (Δx/c)² - (Δy/c)² - (Δz/c)²

Then we can take the limit as c gets large as

Δtau² = (1 – 2GM/c²r)(Δt)²

Now what's interesting is that it looks like gravity goes away as c gets large, but of course we can do whatever we want with G/c2, like call it A, giving

Δtau² = (1 – 2AM/r)(Δt)²

Now we have something that looks like purely nonrelativistic time curvature, where if we let A have the value that G/c2 has in our universe, we get Newtonian gravity just like ours, but in a Galilean universe-- and here gravity isn't a force, it's a Rindler-like coordinate effect. Like publius, I don't know how this all extends globally, but it seems to make sense at first peek.

publius

2009-Jul-03, 12:33 AM

Thinking about all this some more, we have the following. It is a renormalizable infinity.

Our coordinate rest energy, at least in nice enough metrics is this:

E = mc^2*sqrt(g_00)

The general expression is some tensor indez gymnastics mess that reduces to the above when the metric has some relatively simple form. In stationary space-times, we're good to go with it I think.

Now, let's plug in the weak field limit for g_00 in terms of the Newtonian potential:

E = mc^2*sqrt(1 + 2Phi/c^2) = mc^2*(1 + Phi/c^2 + HOT)

E ~= mc^2 + m*Phi + HOT(which are increasing inverse powers of c^2)

If we let c go to infinity, the HOT goes to zero. E is still infinite, but the maginal difference above infinity is m*Phi, something that doesn't depend on c. That's where the Newtonian g will come from. The non-Newtonian behavior comes from the ratio of the values of E when it is finite in addition to the marginal difference. When c goes to infinity, the effect is to make the weak field limit be global. All fields are weak compared to the infinite baseline, and the only thing that governs the physics is the marginal difference, which is just m*Phi.

Also note that clock rates depend on the ratios of E at different points. In the infinite c limit, there is no time dilation! All clocks tick at the same rate. That's Newtonian, of course. But the Newtonian g comes from those marginal differences of energy "above infinity".

So it is thus apparently incorrect in constructing this "time curvature" picture to say the gravitational acceleration comes from time dilation! In the infinite c limit, there would be no time dilation.

By taking the infinite c limit, we are forcing the weak field limit global, and there are no differences in clock rates, but nonetheless, things still fall.

The above is all very rough, but I think its true.

-Richard

Powered by vBulletin® Version 4.2.3 Copyright © 2019 vBulletin Solutions, Inc. All rights reserved.