# Deflection of light

1. May 24, 2004

### kurious

When a photon is deflected by the sun will it be deflected by twice the Newonian expectation for any distance from the centre of the sun or will the difference between the deflection relativity gives and Newtonian gravity gives change with distance?

2. May 24, 2004

### marcus

I dont fully understand the question. I guess it is that i dont understand what deflection Newtonian gravity predicts.

However just to get it out on the table, for comparison, I will say what deflection angle ordinary GR predicts

If the closest approach is R (i.e. the ray of light comes within distance R of sun center) then the angle the ray is bent (expressed in radians) is

$$\frac{4GM}{c^2R}$$

It happens that

$$\frac{4GM}{c^2}$$

is around 6 kilometers, so if the closest approach of a grazing ray is 600,000 kilometers then to find the angle you just divide

6 km divided by 600,000 km is 1/100,000

so it gets bent by 1/100,000 of a radian

I may be telling you what you already know, but someone else reading the thread might want to know the formula for finding the angle.

what i dont get is what angle newtonian gravity would predict

Last edited: May 24, 2004
3. May 24, 2004

### kurious

what i dont get is what angle newtonian gravity would predict

--------------------------------------------------------------------------------
I've just read this a couple of times over the years.The Newtonian deflection gives 0.85 arcseconds at the surface of the sun and relativity 1,7 arc seconds.I have no idea how the Newtonian figure was calculated.Perhaps a photon was given its mass equivalent via E=mc^2.

I am going to use the equation you have given for the deflection to
try and gain some insight into how virtual photons might attract one
another.I've started another thread in the Quantum fields section
about qed and gravity.I was hoping to use Newton's laws to try and gain some insight by adjusting them for relativistic effects.

Last edited: May 24, 2004
4. May 24, 2004

### turin

Marcus,
Isn't the formula that you gave an approximation? In other words, isn't there some minimum value of R below which this formula is not valid? It appears to be directly related to the Schwarzschild radius, so that seems to suggest exactness as opposed to approximation (assuming a Schwarzschild geometry), but for some reason I thought that the formula was an approximation that is only valid for small deflections/large radii (but which would be, of course, valid at radii outside of the Sun). I'm not good with diff. eq.; the equation (for a geodesic in Schwarzschild geometry) is non-linear and involves special functions with which I am unfamiliar. I am interested in a better understanding of the solution that leads to the simple formula for deflection. Is it the fact that we are dealing with light that makes the formula exact instead of an approximation?

kurious,
Regarding the Newtonian vs. GR theory:

- Newtonian theory posits an ~r-2 central force, whereas GR posits a spherically symmetric geometry.

- Both lead to a diff. eq. of the (very general) form:

d2u/dφ2 = f(u)

where u = r-1, and GR's f(u) is obscenely complicated IMO, thus I find the "fu" appropriate in that case.

- In the Newtonian theory, the (I believe exact) hyperbolic solution is known (you can find it in evey text book on mechanics that I've seen) to be:

r = a( ε2 - 1 ) / { (+/-)1 + εcosφ }

where ε is the eccentricity of the orbit (>1 for a hyperbola) and can be found from the energy of the particle and distance of closest approach and a is some parameter.

From what I can tell, the angle between the assymptotes of the hyperbola is:

Δφ = π - 2α

where α = cos-1(ε-1). This can of course be related through α to ε, then through ε to the energy and impact parameter (b)* of the light ray, and thus can be compared to the GR deflection.

ε = √{ 1 + ( 2EL2 / [ G2M2m3 ] } = √{ 1 + [ b2c4 / ( G2M2 ) ] }

α = cos-1(ε-1) = cos-1(√{ G2M2 / [ G2M2 + b2c4 ] })

Δφ = π - 2α = π - 2 cos-1(√{ G2M2 / [ G2M2 + b2c4 ] })

For b >> GM/c2 (which is necessarily satisfied for an observed ray of light when considering what Marcus said), then:

Δφ ~ π - 2 cos-1( GM / [ bc2 ] )

This can be expressed as a complex natural logarithm which can in turn be expanded to lowest order. I don't have the motivation to type it out. I'm pretty sure that this approximation gives you:

2GM / [ bc2 ] + O(higher)

which demonstrates the factor of 2.

* The impact parameter, b, is the offset from a parallel trajectory that would be observed at an infinite distance that would collide directly with the target. In other words, it is how far off course the incident object begins its path. I would have to conduct further investigation to find its relationship to the distance of closest approach (R < b), but b is what would approximately be observed anyway, since the earth is approximately an infinite distance from the sun for these purposes. Also note that b ~ R in the approximation above.

Last edited: May 24, 2004
5. May 25, 2004

### kurious

TURIN: thanks for the info.
I'm going to assume that Marcus's equation is a good approximation at small distances.
I'm interested to find out if qed invalidates macroscopic force relations at
small distances or if it is the interaction of different fields that invalidate macroscopic force relations.

6. May 26, 2004

### pmb_phy

The Newtonian prediction is found by treating a photon as a particle moving at v = c. The deflection is off by a factor of two due to the failuer of Newtonian mechanics to take into account curved space. So half the deflection is due to curved space.

Pmb