Newton's Laws and The Bending of Light

RJ Emery · Post by **RJ Emery** » Sat Sep 09, 2006 1:33 pm

How is it that Newton's laws of gravitational attraction can be applied to the bending of light?

The following is an excerpt from Archives of the Universe, Marcia Bartusiak, editor, Chapter 36, "General Relativity and the Solar Eclipse Test", p. 303 and 629. Bold emphasis is mine:

Einstein himself suggested another test to confirm the space-time curvatures proposed in his theory: to photograph a field of stars at night, then for comparison photograph those same stars when they pass near the Sun s limb during a solar eclipse. A stellar beam of light passing right by the Sun would be gravitationally attracted to the Sun and get bent. Moreover, the attraction would be twice the bending calculated from Newton s laws alone.[8] The extra contribution, according to Einstein, comes from the warping of space-time near the Sun. Einstein calculated that a ray of starlight just grazing the Sun would get deflected by 1.7 arcseconds.

8. Several scientists used Newton's laws to contemplate gravity acting upon light. They include the British scientist john Michell in 1783 (see Chapter 42), the French astronomer Pierre-Simon Laplace in 1799, and the German mathematician and astronomer Johann Soldner (see "On the Deflection of a Light Ray from It Straight Motion Due to the Attraction of a World Body Which it Passes Closely. Berliner Astronomisches Jahrbuch an das Jahr 1804). Newton himself briefly considered the possibility as early as 1704.

(end excerpt)

rigelan · Post by **rigelan** » Sat Sep 09, 2006 4:42 pm

I didn't know that you could calculate bending of light from newton's gravitational laws.

edit: that is unless you assigned light a mass. Did we decide light was massless before or because of Einstein? I don't remember.

You could from his optics laws, if you assume there is a slight gas around a star with varying density and distances.

And of course we believe Einstein's laws of the bending of space around massive objects to account for light bending around stars

Qev · Post by **Qev** » Sat Sep 09, 2006 4:42 pm

That's interesting! I never thougth that Newtonian gravity would predict the curvature of light in a gravitational field. But apparently it does, just not to the degree that General Relativity does.

The acceleration due to gravity from a massive body is equal to:

a = Gm/r^2

...which makes absolutely no mention of the mass of the object being influenced by the gravitational field. Implying that even a 0-mass photon will still be deflected under Newtonian gravity.

I had no idea.

rigelan · Post by **rigelan** » Sat Sep 09, 2006 4:45 pm

qev: That equation comes from F=GMm/r^2, and only simplifies if we assume m (the mass of the object) is non-zero.

because

a = F/m = (GMm/r^2) / m

and you have to do some mathematical tricks to divide by 0, I think.

cosmo_uk · Post by **cosmo_uk** » Sat Sep 09, 2006 5:25 pm

correct there is no Newtonian way that light can be influenced by gravity unless it is assigned a mass. GR means that massive objects curve spacetime thus light follows this curved path and its path is therefore deviated

RJ Emery · Post by **RJ Emery** » Sat Sep 09, 2006 7:26 pm

Qev wrote:That's interesting! I never thougth that Newtonian gravity would predict the curvature of light in a gravitational field. But apparently it does, just not to the degree that General Relativity does.

The acceleration due to gravity from a massive body is equal to:

a = Gm/r^2

Where does this formula come from? How does this translate into deflection of a light beam?

astro_uk · Post by **astro_uk** » Sat Sep 09, 2006 8:42 pm

Hi all

I would guess that

a = Gm/r^2

comes from taking F=Gm(1)m(2)/r^2
Newtons gravity equation where m(2) is the mass of the smaller mass object (in this case the photon)

and combining it with F=m(2)a

You then get a=Gm(1)/r^2

i.e the accelaration due to the gravity of the larger mass when located a distance r away from the centre of the mass.

Im not sure if this does apply if m(2) =0 (i.e a massless photon), I would think it might, something to do with limits tending to zero etc.

I've just had a look at several websites, wikipedia included but they seem to disagree over this issue.

If anyone has a copy of Misner, Thorne and Wheeler I think it would solve the issue, If someone hasnt answered it by Monday I'll check the copy at work. Or ask the GR experts there.

RJ Emery · Post by **RJ Emery** » Sat Sep 09, 2006 9:11 pm

astro_uk wrote:If anyone has a copy of Misner, Thorne and Wheeler I think it would solve the issue, If someone hasnt answered it by Monday I'll check the copy at work. Or ask the GR experts there.

I, for one, would appreciate just that. If the forumula is applicable, I would like to know how it is derived, and how the deviation in arcseconds would then be calculated. Thanks.

Qev · Post by **Qev** » Sun Sep 10, 2006 5:31 pm

Cool, Astro. I'm honestly not sure if that equation does actually apply or no, so it'll be good to find out for sure.

ckam · Post by **ckam** » Mon Sep 11, 2006 11:08 am

so many posts, when it takes only few minutes to search for an answer

astro_uk · Post by **astro_uk** » Mon Sep 11, 2006 11:35 am

Hi All,

Well I've asked some people here, and their response is interesting. The answer seems to be that Newtonian physics can predict light bending if you want it to, the reason given above is correct that the accelaration felt by an object due to the mass of another object is independent of the smaller bodies mass.

I.e
a = GM/r^2

If you treat this as true then you get a certain amount of bending. The reason GR predicts twice as much is that GR acts on space/time, so you get some bending due to the way GR gravity bends space and you get some more due to the GR effects on time.

However I'm still confused, there is clearly something going on here, perhaps to do with the fact that even though photons dont have mass they do have momentum, and force is rate of change of momentum, so a massless photon can still feel a force.

Although p=hv / nu so if you changed the photons momentum to deflect it you would also introduce a gravitational redshift. Which I think is ok.

Not sure this really made anything clearer.

ckam · Post by **ckam** » Mon Sep 11, 2006 3:08 pm

imho, you should be able to get non-zero mass for photon in newtonian theory from hv = mc^2/2

aichip · Post by **aichip** » Mon Sep 11, 2006 9:03 pm

The point of this is that space itself is being curved by the mass. As a result, anything moving through it, whether it is massless or otherwise, will have to follow the minimum energy path, and that bends the light.

Imagine a sheet of curved material. Plot a straight line above a nearly flat section of it and measure the actual distance on the surface. Now plot another straight line over a more curved section and then measure the actual distance of the more curved surface.

You will find that the more curved surface will have a longer actual path, and so when mass is in spacetime, it curves spacetime slightly and makes the path longer. But it also means that anything freely falling along that curved space will take a different path than you might expect.

Dr. Skeptic · Post by **Dr. Skeptic** » Tue Sep 12, 2006 12:09 am

Wonderful.
But your analogy fails to explain why a "mass less" object's trajectory would be categorized as a "free fall" suggesting an influenced of gravity.

But you're close.

If observing the universe in the same "relativity" perspective as the photon traveling passed the Sun, no change in trajectory, effect of gravity or curvature of space/time would be seen or felt. A free falling object would be able to measure a relative change in trajectory in respect to the rest of the universe.

rigelan · Post by **rigelan** » Tue Sep 12, 2006 3:03 am

It reminds me of this example:

Shoot a proton through an electrically polarized field that is perpendicular to the path of the proton. The proton curves according to the potential of the field.

Shoot a neutron through the same field, it has no effect. The proton continues on its path. The potential doesn't exist because the field will only affect charged objects.

A charged object has no electric force or electric potential on a 'chargeless' object.

Likewise, it seems a massive object should have no gravitational effect upon a 'massless' one. the field shouldn't be created because the field exists to explain the interaction of two objects with mass.

rigelan · Post by **rigelan** » Tue Sep 12, 2006 3:05 am

But how would you explain it if you didn't know about relativity?

How could Newton have explained it?

rigelan · Post by **rigelan** » Tue Sep 12, 2006 3:28 am

So wait. . .

IF we assume that light has a mass, as newton might (any mass you like, it doesn't matter), then we should know the gravitational potential for the path around the sun or any other star.

And if we know the speed of light (I don't know how well this was known before the 20th century), then we could calculate a path for the light bending around a star.

According to current theory this doesn't make sense because light has no mass to speak of. But according to the people before the 20th century, this could make sense. But unfortunately, as was pointed out in the original post, what was predicted from this didn't match reality. Only when we got to Einstein, space-time, zero rest-mass light, field theory and so on dothe theories actually agree with the reality.

But I don't think we can make an argument about it using relativity-talk. This is more of an historical question.

dcmcp · Post by **dcmcp** » Tue Sep 12, 2006 5:29 am

rigelan wrote:But how would you explain it if you didn't know about relativity?

How could Newton have explained it?

and later

But I don't think we can make an argument about it using relativity-talk. This is more of an historical question.

You don't need to invoke relativity to get an (albeit incorrect) explanation of how light bends around stars. There is an explanation in the link posted previously (which I think you've read).

My own approach (in a Newtonian world) would be:

We can see that light bends around stars;
We know that photons have no mass (how? dunno, let's just go with the assumption for now)

From Newton:

F=GMm/r^2
and
F=ma
Therefore
a=F/m
=(GMm/r^2)/m
=(GM/r^2)(m/m)

So far so good, but we seem to have hit an obstacle: m/m = 0/0 - - which is undefined.
The way around this was alluded to by astro_uk in this post: limits!
The limit as m->0 of m/m is 1. This is a well-behaved limit because it is continuous (you get the same result if you approach 0 from above or below) and smooth. So we have

a=GM/r^2

which is independent of mass.
(This analysis is necessarily simplistic - because my maths isn't up to something more sophisticated)

So, in a Newtonian universe we can get gravitational bending of the paths of massless particles. The theory might not predict it, but it can be invoked after the event to explain it.

(This is the same conclusion Qev used earlier.)

...And if you assume light is a wave, not a particle, you're still in deep doggy doo-doo.

ckam · Post by **ckam** » Tue Sep 12, 2006 9:19 am

I would like to re-iterate my "newtonian" equation: hv (or any amount of photon kinetic energy) = mc^2/2.

It would seem that "mass less" particle would be bound to gain infinite velocity and always travel along straight lines in that theory. Then again, noone thought that light is made of particles at that time...

Dr. Skeptic · Post by **Dr. Skeptic** » Tue Sep 12, 2006 11:47 am

FYI: Doesn't a photon have a pseudomass measured in eV?

aichip · Post by **aichip** » Tue Sep 12, 2006 6:17 pm

Dr. Skeptic wrote:

Wonderful.
But your analogy fails to explain why a "mass less" object's trajectory would be categorized as a "free fall" suggesting an influenced of gravity.

Any object follows a minimum energy path. The object does not feel a force when it does this. It cannot. If it did, then it would be forced to follow a different path. You certainly do not feel the force of gravitation when you are freely falling, therefore you cannot. That is because it is uniformly present due to the curvature of the space around you.

Dr. Skeptic wrote:

If observing the universe in the same "relativity" perspective as the photon traveling passed the Sun, no change in trajectory, effect of gravity or curvature of space/time would be seen or felt.

This is untrue and at odds with observation. What makes you reach this conclusion?

Dr. Skeptic wrote:

A free falling object would be able to measure a relative change in trajectory in respect to the rest of the universe.

No, it cannot. It is being uniformly influenced and therefore sees or feels no difference. The only exception to this is when the curvature of the metric changes significantly over the extent of the object (like where the object extends enough that it has more than one orbital velocity). This is called tidal force.

The big difference in Einstein's concept of gravitation and Newton's in this case is that Newton only figured the field from the Sun's mass alone. That is why the numbers were off a little. Einstein takes into account that the energy of the Sun's gravitational field itself has a virtual mass and therefore also causes a further curvature of space. When the two fields are superimposed, the correct deflection is arrived at.

Once again, a freely falling oject feels no acceleration. It responds to it, but that is another matter entirely. A uniform application of force results in a net force on the whole object, but no differential across the object (and therefore, nothing to feel).

That explains very easily how gravitational lensing works. The light paths are curved because they are following the minimum energy paths through curved spacetime.

dcmcp · Post by **dcmcp** » Wed Sep 13, 2006 7:03 am

Hi ckam,

ckam wrote:I would like to re-iterate my "newtonian" equation: hv (or any amount of photon kinetic energy) = mc^2/2.

I don't like this approach. Not because it is any less valid than mine (or indeed the rest of the discussion). But for purely aesthetic reasons:

Classically KE=mv^2/2 and p=mv
So p=2KE/v - this gives a result that does not match the measured momentum of photons. (The measured momentum is actually p=E/c)
E=hv is a result from experiments with the photoelectric effect. The photoelectric effect was explained by a bloke with the name of Einstein (it was what he got the Nobel prize for). I feel that's just a little too close to Special Relativity (no, it's not a logical argument).
The photoelectric effect was one of the observations that led to quantum mechanics, because it demonstrated the electromagnetic energy was quantized. A result that could not be explained by classical Electromagnetism (Gauss, Maxwell et al).

Having said that, we can't afford to look to closely at any of the flaws or we quickly end up invoking non-Newtonian results...

It would seem that "mass less" particle would be bound to gain infinite velocity and always travel along straight lines in that theory....

That's why yez gotta use limits

The Gravitational force on a massless particle is zero
a=F/m=0/0=???

Then again, noone thought that light is made of particles at that time

At the time of Newton, it was not decided of light was made of particles or waves. Waves explained a lot of the observed behaviour (refraction, diffraction) but the big problem was "What is it that is waving?" Newton himself conducted experiments that confirmed wave-like behaviour, but could not answer this question.

It was not until the speed of light was measured and compared to a result from Maxwell's equations that it was realised that what was waving was electromagnetic fields. This finally killed the "corpuscular theory" of light - well, until the late 19th century when there were these inconvenient results which couldn't be entirely expained by waves....

+ + + + +

Dr. Skeptic wrote:FYI: Doesn't a photon have a pseudomass measured in eV?

Yeah, but this is a concept from SR.

ckam · Post by **ckam** » Wed Sep 13, 2006 9:57 am

dcmcp wrote:
ckam wrote:It would seem that "mass less" particle would be bound to gain infinite velocity and always travel along straight lines in that theory....
That's why yez gotta use limits
The Gravitational force on a massless particle is zero
a=F/m=0/0=???

I mean different thing. Suppose there's massless particle moving with 1m/s and 0 kinetic energy. Now you get your baseball bat and hit it. How much velocity would it gain?

Dr. Skeptic · Post by **Dr. Skeptic** » Wed Sep 13, 2006 12:36 pm

Dr. Skeptic wrote:Quote:
A free falling object would be able to measure a relative change in trajectory in respect to the rest of the universe.

No, it cannot. It is being uniformly influenced and therefore sees or feels no difference. The only exception to this is when the curvature of the metric changes significantly over the extent of the object (like where the object extends enough that it has more than one orbital velocity). This is called tidal force.

The moon is in free fall around the Earth, I can't detect a change in trajectory if I'm located on the Moon? I would observe a constant change in the location of the stars enough to know I wasn't moving in a linear path. A photon bending around a star would experience a true linear trajectory while an observer would see it otherwise.

ckam · Post by **ckam** » Wed Sep 13, 2006 1:29 pm

Dr. Skeptic wrote:I would observe a constant change in the location of the stars enough to know I wasn't moving in a linear path.

No it's not enough, there is a crazy chance that universal gravitational field makes stars to change location. You should use a pendulum instead

Starship Asterisk*

Newton's Laws and The Bending of Light

Newton's Laws and The Bending of Light

bent photon paths

free falling paths and force