# Non-relativistic deflection

by bobie
Tags: deflection, nonrelativistic
PF Gold
P: 636
Hi,
I am trying to understand the difference between GR and classical deflection of light by the Sun.I found this
 Quote by BvU They want us to integrate dx / c from -∞ to +∞ with a weight sin3Θ (Hint: x = R/tanθ ; change to dθ ) My guess is that the integral gives 2R/c. Contrary to what Mentor claims: the 2 is still important. Spectacularly so in 1919 when Einstein was proved right...
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...+from+0+to+200) = 2.4998
and distance star-sun -∞ (http://www.wolframalpha.com/input/?i...+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?

 Homework Sci Advisor HW Helper Thanks P: 13,135 Please provide a link to the thread in question. 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/tea...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/l...y/gravity.html
PF Gold
P: 636
 Quote by Simon Bridge Please provide a link to the thread in question. In Newtonian mechanics - light is not deflected by the Sun because light has no mass.
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 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
 Quote by BvU ...
- 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?

 Homework Sci Advisor HW Helper Thanks P: 13,135 Non-relativistic deflection 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.
PF Gold
P: 636
 Quote by Simon Bridge It is not clear to me what you are trying to understand.
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.
P: 74
 Quote by bobie - if the analysis in that thread is non-relativistic.I thought so , as the result is 2GM... instead of 4GM (as in GR)
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.

 Quote by bobie - I cannot understand why BvU says that 2* GM is what Einstein predicted, as he predicted 2*2GM.
Probably he meant the other factor of two by GR.

 Quote by bobie - 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?)
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.

 Quote by bobie and lastly I would like to find an article that describes in detail a recent observation of deflection with precise figures.
PF Gold
P: 636
 Quote by Mr-R 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.
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.
P: 74
 Quote by bobie Great, Mr R that file is too heavy for me, what about the simpler one I quoted? https://www.dropbox.com/s/50rswg7ft7...ction%20SM.pdf it uses tensor calculus or whatever, (derived formula: 5.5) but just to plug in there the Newtonian value (2*2GM/RC^2) , 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?
I actually meant the article you provided also used Tensor calculus to derive the terms which are used in the deflection calculation.

 Quote by bobie 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 ?
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

 Quote by bobie 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.
Did not find anything yet. Just call me Rashid, I am a second year undergrad . 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
 P: 74 I hope this helps. My last post said Sin. make it tan and I used the change of velocity here to get the angle. Attached Thumbnails
PF Gold
P: 636
 Quote by Mr-R 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
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?
P: 74
 Quote by bobie 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?
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?
PF Gold
P: 636
 Quote by Mr-R But bobie, GM/Rc has the units of velocity (m/s) how is it the force? Plus: what do you mean by c? as in orbit is 7c?
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?
P: 74
 Quote by bobie I have little experience, Rashid, you must be patient with me.
Dont worry. Me too, thats why I am taking some time to reply and understand

 Quote by bobie We have Fg = GM/R2, which is the acceleration a point-mass gets in 1 sec at distance R from the sun, right?
yes

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

 Quote by bobie 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 am afraid that I lost you here again bobie.

 Quote by bobie 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?
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?!
PF Gold
P: 636
 Quote by Mr-R Integrate with respect to what? for example, if you integrated acceleration with respect to time then you get velocity.
My logic is naive, but rationalal. If you earn money in a variable way (≈ 7$per hour) and integrate over time to get the total, in a month 7*200 makes 1400, do you get money or velocity? But now I am beginning to understand why you talk about velocity v/c , is that so? so you find acceleration per second 27500, integrate and get velocity 128 000? then you divide it by C and you get what? what are now the units of v/C ? it should be a dimensionless number for what I know, why should it be an angle? Thanks a lot, you have been very helpful, perhaps an expert would not have understood me at all! P.S it's August, most people are on holidays/vacation, I suppose P: 74  Quote by bobie My logic is naive, but rationalal. If you earn money in a variable way (≈ 7$ per hour) and integrate over time to get the total, in a month 7*200 makes 1400, do you get money or velocity?
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.

 Quote by bobie 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?
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 $tan(o/a)=tan(v/c)≈v/c=θ$

 Quote by bobie Thanks a lot, you have been very helpful, perhaps an expert would not have understood me at all!
You are welcome bobie
Just forget your numbers. Do it again and calculate the numbers at the last moment.
PF Gold
P: 636
 Quote by Mr-R money you earned. But there above, you integrated acceleration which is,
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?
 Quote by Mr-R , you know that $tan(θ)≈θ$ for small θ, right? [/itex]
and what happens when the angle is not small, for example near a BH?
PF Gold
P: 636
 Quote by Mr-R , 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 $tan(o/a)=tan(v/c)≈v/c=θ$ .
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
Mentor
P: 3,975
 Quote by bobie Could someone please confirm or correct that? For a small angle it is sinθ ≈ tanθ , and not ≈θ ?
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.

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