# Non-relativistic deflection

1. ### bobie

682
Hi,
I am trying to understand the difference between GR and classical deflection of light by the Sun.I found this
I do not now how to apply a 'weight sin3' so I worked out my own equation in which;

R = 2.5c (7.5*10^10 cm) point of impact, x distance of light from R in light-seconds, r (=√x2 +R2) distance of light from center of sun and then distance sun-earth = 200c (http://www.wolframalpha.com/input/?i=integrate+y=+2.5^3/+(sqrt(x^2+2.5^2))^3+from+0+to+200) = 2.4998
and distance star-sun -∞ (http://www.wolframalpha.com/input/?i=integrate+y=+2.5^3/+(sqrt(x^2+2.5^2))^3+from+-+infinity+to+0) = 2.5 (R/c)
the non relativistic pull should then be $\frac{GM}{R^2} = 23600 *2*2.5 \frac{2*R}{c} → \frac{2 GM}{Rc} ≈ 118 000 cm/s^2$

Can you check if it is right and clarify a couple of points:

- do observed light beams actually 'graze' the Sun? what is the usual distance of the observed beams? can you give me a link where to find details of recent observations? I read that deflection has been adjusted to 1.67'',but, at what distance from the center of the sun?
- In order to get the global pull by the sun we must multiply gsun (the force at x=0 (R)) by the integral and by 2 because the pull is exerted from -∞ to 200. This seems the obvious classical procedure, but BvU (I am not sure about mfb) considers this factor as the relativitic γ-factor, which would give 4 GM.
what is correct?

Last edited: Jul 25, 2014
2. ### Simon Bridge

15,259

What do you mean by "classical"?
In Newtonian mechanics - light is not deflected by the Sun because light has no mass.

The deflection of starlight by the Sun, or by other bodies, is well documented.
As for the grazing path: as the Sun traverses the sky it sometimes blocks the stars. It folows that there must be a time when the star appears just beyond the limb of the Sun as viewed from the Earth ... so the answer is "yes" light rays can graze the Sun.
Accessible source:
http://cosmictimes.gsfc.nasa.gov/teachers/guide/1919/guide/gravity_bends_starlight.html

The link between GR and Newtonian gravitation is better understood from the maths:
http://preposterousuniverse.com/grnotes/grtinypdf.pdf
http://csep10.phys.utk.edu/astr162/lect/cosmology/gravity.html

3. ### bobie

682
http://arxiv.org/abs/physics/0508030
-In 1803 Soldner gave a first extimate of the deflection, was at the time anything but Newtonian physics? Are you saying that he quoted thread treats the issue wrt to GR, does it? It seems it is integrating the Newtonian acceleration
That is above my math, Simon, I found this article https://www.dropbox.com/s/50rswg7ft72g90u/Light Deflection SM.pdf
that says (5.6) the that the formula is modified by the factor $\frac{1+\gamma}{2}$
where γ is 1 in GR and 0 in Newtonian gravity. I read that half bending comes from curvature of space and 1/2 from curvature of time.
If it is so, it is quite simple

-If you click on the quote you go to the thread
- Do you know an article that gives details of recent experiments of light grazing the sun?
do you think I should consider R = 7*10^10cm?

Last edited: Jul 25, 2014
4. ### Simon Bridge

15,259
Well OK.
By Newton's second law, the acceleration of gravity does not depend on the mass of the accelerating object - which would imply that light would be deflected by massive objects. iirc the "no mass" objection comes from trying to work out the gravity of the light - and considering the 3rd law. There is also a problem with there being a force of gravity where there is no mass, since the force depends on the mass of both objects.

If you ignore these, you can do calculations for the deflection of light by gravity from Newtonian physics and then compare with experiment.

Note: GR has to confirm the Newtonian results.
The analysis in the thread is extremely simplistic - I'd have done it as a central force problem for a small test mass travelling at the speed of light in the Newtonian framework.

I don't know of any recent "grazing light" observations, but "gravitational lensing" is standard in astrophysics these days and uses the same physics.

My undertstanding is that the Newtonian trajectory close to the Sun is flatter than the GR trajectory, but there is close agreement for greater distances.

It is not clear to me what you are trying to understand.
You should be aware that GR replaces Newtonian gravitation because there are some things that don't work in Newton. Therefore, trying to use your understanding of Newton to understand GR won't get you far. You wll also need to improve your maths.

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5. ### bobie

682
Hi Simon, I hope you got to the thread alright.
- if the analysis in that thread is non-relativistic.I thought so , as the result is 2GM... instead of 4GM (as in GR)
- I cannot understand why BvU says that 2* GM is what Einstein predicted, as he predicted 2*2GM.
- I'd like someone to confirm that the GR applies the same laws and parameter of Newtonian gravitation and simply multiplies it by a factor of 2 (1+γ(=1)/2) (did you read the article I quoted?)
and lastly I would like to find an article that describes in detail a recent observation of deflection with precise figures.

I have other minor queries , if you have time
Thanks.

6. ### Mr-R

75
It is non-relativistic. This is done by assuming that light is made of capsules with very small mass. Then the mass drops out anyway. Quasi-Newtonian its called I think. Or just by using the equivalence principle, therefore mass of the photon doesn't matter. You just use Newton's law of gravitation and cook up some stuff which allow light to get attracted to mass in Newtonian gravitation.

Probably he meant the other factor of two by GR.

Not really. GR is a geometrical theory of gravity. You need to use differential geometry and tensor calculus in order to use GR. See the PDF files Simon provided you with. The article used Einstein field equations to derive that. Not just by multiplying by that factor.

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7. ### bobie

682
Great, Mr R
that file is too heavy for me, what about the simpler one I quoted?
it uses tensor calculus or whatever, (derived formula: 5.5) but just to plug in there the Newtonian value (2*2GM/RC2) , which then is multiplied by 1+γ/2 (5.6), where γ is 1 in Gr and 0 in non-GR, producing 4GM/RC^2 *1/2 = 2GM/RC^2, did I get it wrong?

another thing I'd like to know is why, once you get the force 118000-128000 cm/s^2 (GM/Rc), you divide that value by C and get GM/Rc^2 , to determine the angle, shouldn't it be divided by C/R since light is deflected in its orbit ,which has radius R and not C ?

I look forward to a link about real experiments, the Eddington data are not reliable.
In the meanwhile, do you know how close can a photon graze the Sun without being altered?, does extreme heat affect light?

Thanks Mr R, your help has been invaluable.

Last edited: Jul 27, 2014
8. ### Mr-R

75
I actually meant the article you provided also used Tensor calculus to derive the terms which are used in the deflection calculation.

Are you referring to your calculation? I am a little bit lost. I think it is uausally done by calculating the change of speed of light and dividing that by its its vacuum speed c. $Sin(Δv/v)\approxθ\approxΔv/v$. v=c

Did not find anything yet. Just call me Rashid, I am a second year undergrad :tongue:. Started learning GR on my own with some help from PF!.
Okey, I started like this. I was curious about the light deflection so I did what you did. Let me calculate one for you and see if that helps clear some things

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9. ### Mr-R

75
I hope this helps. My last post said Sin. make it tan and I used the change of velocity here to get the angle.

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Last edited: Jul 27, 2014
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10. ### bobie

682
Hi Rashid, I think I found the right person that can understand me.
I am referring to any calculation, I just worked out an equation to find the total F acting on the photon, and I found GM/Rc, like anybody else, using the laws I would use for an asteroid or an electron. I meant to go on by the same laws , finding what angle corresponds to 128000 cm/s2, if the orbit is 7c and speed is C.
I do not understand the change of speed of light, never heard about it, can you explain?

11. ### Mr-R

75
But bobie, GM/Rc has the units of velocity (m/s) how is it the force?

I see, you want the F that when divided by will get you the angle. I think you mixed up some stuff in your calculation because your "Force" (GM/Rc) is actually the velocity component that deviated from the straight path the photon was travelling. and you divide that by the straight path (C) velocity to get the angle.

Velocity change?
Since in this quasi-Newtonian mumbo jumbo we considered the photons to be some sort of capsules that has mass. Then it has a variable velocity.

Plus: what do you mean by c? as in orbit is 7c? :shy:

12. ### bobie

682
I have little experience, Rashid, you must be patient with me.
We have Fg = GM/R2, which is the acceleration a point-mass gets in 1 sec at distance R from the sun, right?
If you integrate this value to find the acceleration it gets in 200 sec, whatever units you get , it's always acceleration (Fg) isnt' it? we merge the formula together but the gist is
Fg in 1 sec = 27500, Fg in 200 sec = Fg * (R/C = 2.5) = 128000.
I am not an expert in units, but the logic is that, right?

I made my example using c as unit, so R is 7c = 7(.5)*1010 cm and the orbit of the photon grazing the sun as radius 7c.
In Newtonian orbits , both for planets and electrons, we find the required acceleration to make an objet bend its trajectory by 1 radiant multiplying momentum/velocity by the ratio velocity/radius of the orbit (v*v/r= v2/r) I was trying to apply the same formulas I use for other orbits. Is it a total mess?

13. ### Mr-R

75
Dont worry. Me too, thats why I am taking some time to reply and understand :tongue:

yes

Integrate with respect to what? for example, if you integrated acceleration with respect to time then you get velocity.

I am afraid that I lost you here again bobie.

Most likely a mess bobie. BUT You should try again slowly and with no numbers. Numbers are calculated at the last stage. You will get it eventually!
I am really sorry for not being able to help. Where are the experts when we need them?! :rofl:

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682
My logic is naive, but rationalal. If you earn money in a variable way (≈ 7$per hour) and integrate over time to get the total, in a month 7*200 makes 1400, do you get money or velocity? But now I am beginning to understand why you talk about velocity v/c , is that so? so you find acceleration per second 27500, integrate and get velocity 128 000? then you divide it by C and you get what? what are now the units of v/C ? it should be a dimensionless number for what I know, why should it be an angle? Thanks a lot, you have been very helpful, perhaps an expert would not have understood me at all! P.S it's August, most people are on holidays/vacation, I suppose Last edited: Jul 27, 2014 15. ### Mr-R 75 money you earned. But there above, you integrated acceleration which is, in this money analogy,$7 per hour per hour. When you integrate you get \$7 per hour.

I think you are getting at it. So, you know that $tan(θ)≈θ$ for small θ, right? Now if you take a look at the paper I attached earlier, you will see that $tan(o/a)=tan(v/c)≈v/c=θ$

You are welcome bobie:thumbs:
Just forget your numbers. Do it again and calculate the numbers at the last moment.

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16. ### bobie

682
OK, let me understand:
the earth (v=106cm/s) gets roughly F = 1 c/s2. We integrate over 200 s and we get v=200, what does that mean?
what is here $tan(o/a)=tan(v/c)≈v/c=θ$
if 200/10^6 =θ, what is that?
and what happens when the angle is not small, for example near a BH?

Last edited: Jul 28, 2014
17. ### bobie

682
Could someone please confirm or correct that?
For a small angle it is sinθ ≈ tanθ , and not ≈θ ?

In this case v/c = 0.000004233, tan-1 = θ =0.0002425 and sin θ = 0.000004223

can you also explain what is θ for the deflection of the earth by the Sun in 200 sec:
tanθ = 200/106 = 0.0002
Is this angle small enough to consider tan-1
Thanks

### Staff: Mentor

Look at the series expansions.
##sin(x)=x-\frac{1}{6}x^3+....##
##tan(x)=x+\frac{1}{3}x^3+...##

If ##x## is small than ##x^3## will be even smaller and the terms further to the right will be yet smaller than that. So you can approximate either function to just ##x## any time that ##x## is small enough that an error of ##x^3## is acceptable.

19. ### bobie

682
Thanks Nugatory,
I am trying to understand the rationale of the formula, in the first place:

the ray is moving tangentially at 3*1010 cm/s
it gets a normal velocity 127 700 cm/s , right?

when we divide v/c = 127 700/3*1010 = tanθ = 0.000004256 we are dividing the legs of a rectangle and we get tan θ, which should be the angle of deflection, right?

Why do you need to approximate the tangent to the angle or to its sine,
why don't you simply do tan-1 θ ?
Is that really so? it doesn't make sense, apparently, as the pull is perpendicular to the motion

Last edited: Aug 4, 2014
20. ### DrStupid

676
In the usual hyperbolic path of deflected light there is only a single point where the pull is perpendicular to the motion.