# Interpretation of double bending / deflection

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1. Jun 7, 2015

### SpiderET

Probably the most famous experiment confirming GR is bending of light by Sun. Here are the best explanations to this topic what I have found:
http://www.mathpages.com/rr/s8-09/8-09.htm
http://mathpages.com/rr/s6-03/6-03.htm

Especially the graph of GR double bending vs Newtonian bending is interesting and the text under it (have bolded part of it):
According to the calculation of 1911, the rate of deflection is a maximum at the point of closest approach to the gravitating body (i.e., where x = 0 and y = R), and the calculation of 1915 gives the same rate of deflection at that point. However, the 1915 calculation, accounting for the spatial as well as temporal curvature, shows that there are actually two points of maximum rate of deflection, at the locations x = ±R/2. The integrated area under the 1911 curve is 2, whereas the integrated area under the 1915 curve is 4, but this plot shows that the relationship between the two is not as simple as one might think based on the fact that the latter happens to give twice the total deflection of the former (to the first order in m/r in the small-deflection limit).

So here comes my question: If I understand it right, there is no "fixed" curvature caused by Sun, which is causing this double deflection compared to Newtonian deflection value, but it is a function of spatial curvature and time dilation and it is heavily depending on the speed of the particle.

An example of my understanding (please correct me if Im wrong):
If we have photon bended by Sun, we have double bending compared to Newtonian calculation. If we would have an low invariant mass neutrino with speed 99,9999% of c, then we would still have nearly double bending compared to Newtonian calculation. But if we would have a neutrino with 50% of c, then the bending would be almost the same as in Newtonian calculation.

Last edited: Jun 7, 2015
2. Jun 7, 2015

### Mentz114

In the second reference the deflection is calculated by solving some equations of motion. The author asys

If this is true then the curvature of the spacetime must be included in the EOMs.

But, the metric is 1+1, so it cannot have the full curvature of the Schwarzschild solution (I conjecture). Is that what you mean ?

Incidentally, that article is very readable and tells the story well. But it leaves no room to doubt the results in view of modern observations in radio wavelengths.

3. Jun 7, 2015

### SpiderET

Why should be there any doubts for results in wavelengths?

4. Jun 7, 2015

### pervect

Staff Emeritus
I haven't read the linked explanations in great detail, but my quick impression was that they were based too much on Newtonian physics for me to feel they had much applicability to GR.

A class of explanations I'd find more convincing (I don't have a web reference, you'd probably find this class of explanations in Taylor & WHeeler's "Exploring Black Holes") would involve using Hamilton's principle, very loosely the principle of maximal aging, to find the path of bodies other than light (i.e. bodies moving at less than c). Since light doesn't have a rest mass and doesn't have a proper time, you can either take the limit of a massive particle approaching "c", or replace Hamilton's principle with Fermat's principle (loosely speaking again, the principle of minimum optical path length) to find the curvature of light.

In any case, I'd mostly agree with your conclusions. If we ignore spatial curvature (a precise statement of what this means is to set the PPN parameter gamma to zero, this parameter being a measure of spatial curvature as per the PPN formalism http://en.wikipedia.org/wiki/Parameterized_post-Newtonian_formalism), we'd get the traditional values for light deflection which are half of the observed values. The only metric coefficient we'd have in this case would be the metric coefficient for time dilation. If we include the spatial parts of the metric by setting the PPN parameter gamma to it's correct relativistic value of one, we get double the deflection for particles moving near "c". The PPN parameter beta has no effect on light deflection, IIRC. Also, for sufficiently slow moving particles, there is no extra deflection as you state. However, I don't think that 50% of c would be slow enough to avoid extra deflection. Offhand I'd expect 1.5x the deflection, though I haven't worked this out in detail.

5. Jun 7, 2015

### SpiderET

Thanks a lot for your opinion.

Regarding the deflection of 1,5x my rough guess would be that you would need around 87% of c to get that, because the relativity effects are rather in logarithmic than linear scale, but unfortunanely Im not able to calculate that precisely.

6. Jun 7, 2015

### Mentz114

You misunderstand what I wrote. I was referring to this ( from the second RR reference). Have you read it ?

7. Jun 7, 2015

### SpiderET

Yes, I have read this part and it seems perfectly OK to me.
Maybe we have some misundestanding of the meaning of your sentence: But it leaves no room to doubt the results in view of modern observations in radio wavelengths.

If the sentence would be: "It leaves no room to doubt the results in view of modern observations in radio wavelengths." then it would be perfectly in line what is written in the linked reference.

8. Jun 7, 2015

### Mentz114

I admit it was not well put.

9. Jun 7, 2015

### pervect

Staff Emeritus
If you want to actually work out the equations of motion and the resulting angle of deflection for a massive particle and see how it behaves vs velocity, you can find the required equations online in "Orbits in Strongly Curved Space-time", http://www.fourmilab.ch/gravitation/orbits/.

Compare these equations to the Newtonian results using the "Effective potential" approach, http://en.wikipedia.org/w/index.php?title=Effective_potential&oldid=654422579

The basic idea for the newtonian version is that E^2 = -gmM/r + (m/2) v^2, where v^2 = (dr/dt)^2 + (r d$\phi$/dt)^2,i.e. the totl energy (which is constant) is the sum of the kinetic energy and potential energy. The other constant of motion, the Newtonian angluar momentum L is m r^2 (d$\phi$/dt) The fourmilab page gives the correct relativistic equations of energy and angular momentum - the first is a bit different, the second is pretty much the same. Note that the fourmilab page uses geometric units so that G=1, and ~E is E/m

If you don't actually use the real relativistic equations of motion, relying on your Newtonian intuition is very likely to get you into trouble.

10. Jun 8, 2015

### SpiderET

I will take a look on the links, thanks for help.