1. Mar 2, 2008

### mitesh9

I recently read Einstein's book explaining SR and GR for general readers.
Though it was simple, at one place, while explaining proof of GR, he wrote about bending of light near sun to be collective effect of newtonian gravity and spacetime curvature (half-half). However, GR gravity (spacetime curvature) is itself revised version of newtonian gravity, and hence must replace newtonian theory, in which case, the effect of sun on bending of light can only be due to newtonian gravity or spacetime curvature, but not both.

Can anybody throw some light on this?

2. Mar 2, 2008

### Staff: Mentor

Welcome to PF mitesh9

I think you misread the paragraph. The bending of light is not half due to Newtonian gravity and half due to GR spacetime curvature. As you mention they are two different theories must each be considered separately.

The paragraph probably actually said that the prediction of Newtonian gravity was half of the prediction of GR. So this became the first experimental method of distinguishing between the two theories. If they measured the bending to be one value it would support Einstein's theory, if they measured half of that it would support Newton's, if they measured any other value it would refute both. They measured the GR value.

3. Mar 2, 2008

### DocZaius

I am surprised to hear that Newtonian theory would predict any light bending at all. It doesn't seem to me that the theory is equipped to deal with something as foreign to it as light.

4. Mar 2, 2008

### Stingray

The idea is to notice that the acceleration of a small particle in a Newtonian gravitational field doesn't depend on its mass. If you make believe that light is composed of a bunch of non-interacting particles moving at c, Newtonian mechanics can therefore be used to determine their deflection in a gravitational field. This should probably be thought of more like a reasonable guess than an actual prediction.

5. Mar 2, 2008

### Benzoate

I do not think General relativity is self contradictory. I think it doesn't explain the whole picture of the universe. The theory only describes certain characteristics of the macroscopic portion of space. In addition , it is an extension of newton universal law of gravity describing more in detail why objects attract each other and how gravitation occurs. Scientific theories and physical laws are always getting revised when more new discoveries are made about the universe and other physical phenomena. Sometime in the future , there will probably be a theory about how the curvature of space and subatomic particles are interrelated.

6. Mar 2, 2008

### yuiop

In relativity energy and mass are interchangeable so a photon can be regarded as having mass equivalent to its energy (hv/c^2). If the bending is calculated using Newtonian gravity theory (acting on the mass equivalent of the photon) then the result is half that predicted by GR.

I think it was Galileo that discovered that all objects fall at the same rate, whatever their mass and with a little imagination this can be extended to particles with no mass. ;) This is not quite true, but for small particles the acceleration of the larger body (sun) towards the small particle is negligible and the mass of the small particle becomes unimportant as pointed out by Stingray.

7. Mar 2, 2008

### shalayka

The OP is not mistaken. However, it is of interest to note that Einstein was referring to the curvature of space, and not spacetime.

From 'Relativity, The Special and the General Theory' by Albert Einstein. Section 'The Experimental Confirmation of the General Theory of Relativity', part b) Deflection of light by a gravitational field:

"It may be added that, according to the theory, half of this deflection is produced by the Newtonian field of attraction of the sun, and the other half by the geometrical modification ("curvature") of space caused by the sun."

I found this paragraph to be incredibly misleading as well. The only advice that I have is to disregard it completely and to focus entirely on learning the Schwarzschild solution as a better way to gain insight into this aspect of GR.

Last edited: Mar 2, 2008
8. Mar 2, 2008

### Staff: Mentor

Then perhaps it is a bad translation of the original German?

9. Mar 2, 2008

### shalayka

Considering this is the 15th edition from 1952 (and the kajillionth printing), I doubt it.

10. Mar 2, 2008

### pmb_phy

You're assuming that Einstein held that gravity and spacetime curvature are the same things. In reality Einstein never held that opinion, which corresponds to a different definition of the term "gravity." I.e. Einstein held the view that the presence of a gravitational field is dictated by the inertial acceleration of a single point particle, e.g. drop an object and it accelerates to the ground due to the presence of a gravitational field. Others held the view that the presence of a gravitational field is determined by the presence of tidal gradients. I.e. drop two particles in the Earth's gravitational field and there will be a relative acceleration between the two particles. The relativistic term of tidal gradients is "spacetime curvature." According to Einstein's interpretation oone can have a gravitational field in the absense of spacetime curvature. A perfect example is a uniform gravitational field. In the frame of reference in which such a field is present a particle dropped in such a field will accelerate but there is no spacetime curvature for such a field. If you recall from your reading of Einstein's book you'll see that Einstein said that if one is in an globally inertial frame of reference and there is a box (e.g. an elevator car) in that frame which is accelerating then observers in the boxes frame of reference there is a gravitational field present. A beam of light is seen by the box observers is seen to be deflected by this field. Observers in the inertial frame see the beam going in a straight line. Thus one *creates* a gravitational field when one changes their frame of reference from an inertial frame to a non-inertial frame.

I disagree. He was quite correct when he made that statement. Einstein's first calculation of the angle through which beam of light deflect by the Sun did not take into account the spatial curvature (which makes the spacetime curvature non-zero). When Einstein finished his GR theory he used the correct model where space was curved then he got the correct answer. Since the final value is twice his initial value then it is clear that half of the deflection is due to the acceleration due to gravity whereas the second half was due to the fact the space around the Sun is non-Euclidean.

Best wishes

Pete

11. Mar 3, 2008

### mitesh9

Hi everyone,

Thanks for ur suggestions and explanations...

U may say that I misinterpreted the explanation of Einstein, but what I have read translates to what I mentioned (so no misreading).

The exact phrase we (copy-pasted from pdf file) "It may be added that, according to the theory, half of this deflection is produced by the Newtonian field of attraction of the sun, and the other half by the geometrical modification (" curvature ") of space caused by the sun."

Eventhough I consider all the intuitiveness of the GR, and genius of Einstein, I am somehow not convinced about the GR.

Further, The point GR tries to explain with spacetime curvature and geodesic trajectory of moving objects are inherently contradicting. For instance, If the GR prediction of bending of light is double then that predicted by Einstein, The same doubled deflection should hold true for all objects in motion which are affected by spacetime curvature of sun (say planets), but the motion of the planets are very satisfactorily described by newtonian theory. Here, one can argue that, the deflection is double in case of photon due to it's speed, yet. I think, faster the object, it should be harder to deflect it. But GR says, faster the object, its easier to deflect. It is in contradiction with the principle of inertia, and yet, GR equates inertial mass to be the same as gravitational mass.

Regards,

mitesh patel

Last edited: Mar 3, 2008
12. Mar 3, 2008

### mitesh9

I think E=mc^2 may not be correct for photons, 'coz, the only energy it possess is kinetic energy and hence it should be E=(1/2)mv^2 according to kinetic energy formula, in which case, the mass of the photon will automatically be double then relativistic mass and hence, the newtonian gravitation calculated with this (putting E=hv in above equation, and putting this mass in newtonian formula) will actually be equal to deflection calculated by GR.

Regards,

Mitesh Patel

13. Mar 3, 2008

### Ich

The translation is quite correct, but they left out parentheses which give a hint to what Einstein intended to say:
While it is still misleading, it becomes clear when you look at how the result is derived:
1. A simplification of GR introducing special assumptions (weak fields, low coordinate speeds etc.) gives a description where space itself is flat and all gravitational interactions result from the time-component (R00) of curvature, which can be identified in this limit with the Newtonian potential. This simplified theory gives the same results as Newtonian theory, and this is what he means when he talks about the (Newtonian) field of attraction. He actually talks about a certain simplification of GR.
2. Because light is moving very fast, the assumption of low speeds is too stringent. One has to include some additional terms in the expansion. Those terms correspond to the curvature of space (R0i, Ri0), which cannot be identified with whatever feature of Newtonian gravity. This is what he means when he talks about the (“curvature”) of space.

So it's just two different levels of simplification of the same theory, GR, where the oversimplified theory (corresponding to Newtonian theory) gives only half the expected value.

Last edited: Mar 3, 2008
14. Mar 3, 2008

### 1effect

No one sait it was. The correct formula is $$E=hf$$

This is even worse.

15. Mar 3, 2008

### mitesh9

Can u please explain what is 'even worse'?

E=hf is known, but it may not be the only way to calculate the energy of a photon. Please correct me if I'm wrong.

16. Mar 3, 2008

### 1effect

$$E=\frac{mv^2}{2}$$ does not apply because m=0 for the photon. Yet, the photon has energy.

17. Mar 3, 2008

### Stingray

GR correctly reproduces the Newtonian limit as well as the full amount of light deflection that's been observed. There is no contradiction. Don't take vague statements in a popular book too seriously.

Also, none of this has anything to do with the energy of photons. The correct way to understand the propagation of light in classical physics (of which GR is a part) is to solve Maxwell's equations. In doing so, you find that a compact wavepacket tends to propagate along null geodesics in vacuum regions of weakly curved spacetimes. The same result is obtained for light of any reasonable intensity. This is nothing more than a generalization of the statement that light tends to move at c in our everyday experience. Its speed in vacuum does not depend on energy density.

18. Mar 3, 2008

### mitesh9

But we use the formula $$E=\frac{mv^2}{2}$$ to back calculate the velocity of radiation in case of nuclear reactions.

Further, there is no difference in mathematical treatment of mass while calculating the effect of mass (say gravity), may it be rest or relativistic, what counts is the resultant mass (which may be rest-mass + mass due to energy), 'coz photons may never be at rest.

19. Mar 3, 2008

### Stingray

No you don't. That applies only for slowly moving masses. It has nothing to do with light. Don't try to assign gravitational or inertial masses to an electromagnetic field. It just won't get you anything useful unless you're extremely careful.

20. Mar 3, 2008

### mitesh9

You may be right sir! Thank you for your suggestions...I'll be careful next time.