Does Newtonian gravity bend light?

[smithpa9]

Looking for help interpreting a proof of G.R. please. . .

One of the earliest proofs of G.R. was the deflection of light by a gravitational field, first shown in 1919 during a solar eclipse.

I think Einstein predicted approximately 1.7 seconds of arc deflection during that eclipse. But in his book, Relativity (written for the general reader), Einstein says that "half of this deflection is produced by the Newtonian field of attraction of the Sun, and the other half by the geometrical curvature of space"


1) Does that mean that classical Newtonian gravitational theory would have predicted roughly 0.8 seconds of arc deflection of light, and G.R. predicts 1.7?

But wasn't light considered massless to Newton?

If so, how does Newton's gravity theory predict the attraction of light?

2) If the answer is 'no' to #1, then what does Einstein mean by the quote above?

thanks !

Paul

 

[robphy]

http://en.wikipedia.org/wiki/Bending_of_starlight
search the page for "Soldner"

Possibly useful
http://www.inference.phy.cam.ac.uk/sanjoy/teaching/approximation/light-bending.pdf
(I haven't read through it.)

 

[Zeit]

Hi smithpa9,

In Classical Mechanics, the formula for gravitational acceleration is :

F = Gmm'/r²

According to Newton's second law, F = ma, we find :

a = Gm'/r²

This equation does not imply the attracted object's mass. So, even if a photon is massless, according to this last equation, light should be curved by an astronomical object. After calculation, we find that light should be curved.

GR introduces some corrections (relativistic effects). Compare GR light curvature to classical light curvature and you find a quotient equal at 2.

 

[actionintegral]

Hi Zeit,

I think there is a problem.

In Classical Mechanics, the formula for coulomb acceleration is :

F = m'qq'/r²

According to Newton's second law, F = ma, we find :

a = m'q'/r²

This equation does not imply the attracted object's mass. So, even if a neutron is chargeless, according to this last equation, neutrons should be curved by an electrically charged object.

----------> I don't think a massless photon would be affected by Newtonian gravitation under your argument

 

[selfAdjoint]

Wasn't it Laplace who calculated that (using Newton's gravitation and his corpuscular definition of light) a sufficiently massive body would have an escape speed greater than c?

 

[robphy]

http://en.wikipedia.org/wiki/Dark_star

 

[Zeit]

Hi

Hi Zeit,

I think there is a problem.

In Classical Mechanics, the formula for coulomb acceleration is :

F = m'qq'/r²

According to Newton's second law, F = ma, we find :

a = m'q'/r²

This equation does not imply the attracted object's mass. So, even if a neutron is chargeless, according to this last equation, neutrons should be curved by an electrically charged object.

----------> I don't think a massless photon would be affected by Newtonian gravitation under your argument

Coulomb's formula is :

F = (1/4πε)(qq'/r²)

Where q and q' are charges (coulomb :C) of two motionless particles , ε is the electric constant ε≈8.854 C/Vm (coulomb per volt-meter).

Because Newton's second law, we found :

a = F/m = (1/4πε)(qq'/r²)(1/m)

I have never seen this equation somewhere, but the dimension r/t² is good.

However, gravitational field is Gm'/r² and electric field is (1/4πε)(q'/r²). So, the electric field is not, at the contrary of the gravitational field, an acceleration. But the effect is the same : it describes the force applied in any point in space.

Wasn't it Laplace who calculated that (using Newton's gravitation and his corpuscular definition of light) a sufficiently massive body would have an escape speed greater than c?

Yes, I think so (robphy alredy gave us a link). But, the Dark Star concept was different than the concept of black hole of course.

Zeit

EDIT : You can look at this, especially at "Parallels between electrostatics and gravity" : http://en.wikipedia.org/wiki/Electric_field"

 

[smithpa9]

Zeit - In the Newtonian formula for Gravitational force you gave,

F = Gmm'/r²

So, for my question about if light bends when close to the Sun under Newtonian gravity theory, I think m = mass of Sun (or photon of light) and m' = mass of photon of light (or Sun).

If either one of these masses is zero, then the right hand size of the equation = 0, hence F=0, hence no Force of gravity between the two, hence no bending of light.

What's wrong with my logic?

 

[smithpa9]

OK, I've thought about this a little more. . . let's see if this is correct:

Prior to SPECIAL relativity, photons were considered massless, always. So everyone before Einstein would have believed that light does NOT bend near a massive object, like the Sun.

Nowdays, photons are thought to have a rest-mass of zero, but of course you never find one "resting." At the speed of light, photons have some mass due to their speed, hence energy and the equivalence of mass and energy (E=mc^2). (True?) so, they would bend due to gravity near the Sun, and would do so approximately 0.8 arc-seconds.

Then, as part of Einstein's discussion of GENERAL relativity, he postulated that light passing near the Sun would bend EVEN FURTHER due to the additional effect of the curvature of space, which added another 0.9 arc-seconds of bending, for a total of 1.7. ?

In sum:

1) under purely pre-Relativity mechanics, light would NOT be thought to bend near the Sun. Photons are massless, space-time is flat, hence no bending of light for any reason.

2) considering only SPECIAL RELATIVITY, photons have energy and hence mass-equivalence, and are therefore affected by gravity. Photons have mass, spacetime is still flat however, hence, SOME bending of light near the Sun.

3) considering SPECIAL and GENERAL RELATIVITY, photons have mass, spacetime is curved, hence MORE bending of light near the Sun, for both reasons.

Does this sound right?

 

[Mute]

OK, I've thought about this a little more. . . let's see if this is correct:

Prior to SPECIAL relativity, photons were considered massless, always. So everyone before Einstein would have believed that light does NOT bend near a massive object, like the Sun.

Nowdays, photons are thought to have a rest-mass of zero, but of course you never find one "resting." At the speed of light, photons have some mass due to their speed, hence energy and the equivalence of mass and energy (E=mc^2). (True?) so, they would bend due to gravity near the Sun, and would do so approximately 0.8 arc-seconds.

...

2) considering only SPECIAL RELATIVITY, photons have energy and hence mass-equivalence, and are therefore affected by gravity. Photons have mass, spacetime is still flat however, hence, SOME bending of light near the Sun.

3) considering SPECIAL and GENERAL RELATIVITY, photons have mass, spacetime is curved, hence MORE bending of light near the Sun, for both reasons.

Does this sound right?

No. E = mc^2 relates the rest energy of an object to its rest mass, i.e., it relates energy in the reference frame in which the object is at rest to the invariant mass of the object. There is no such thing as a rest frame for a photon - it has no rest energy, and no rest mass.

See https://www.physicsforums.com/showthread.php?t=125613"

Additionally, things with mass cannot travel at c. In the same vein, anything massless must travel at c.

 

[Ich]

F = Gmm'/r²

So, for my question about if light bends when close to the Sun under Newtonian gravity theory, I think m = mass of Sun (or photon of light) and m' = mass of photon of light (or Sun).

If either one of these masses is zero, then the right hand size of the equation = 0, hence F=0, hence no Force of gravity between the two, hence no bending of light.

What's wrong with my logic?

With a=F/m, a stays constant, regardless of m'. In the limit m'->0, a is still the same. At m'=0, a is mathematically defined by L'Hôpital's rule, if you prefer not to cancel it earlier.

 

[Zeit]

Hi everybody,

If either one of these masses is zero, then the right hand size of the equation = 0, hence F=0, hence no Force of gravity between the two, hence no bending of light.

No, it means that you haven't to exert force to bend light. As Ich noted, a = F/m and it's a "constant". Because F = Gmm'/r² (m' is Sun's mass here), g = Gm'/r² and is independent from the attracted object's mass, even if it's a photon, you or Earth. Because when m increases, F does too ; when m decreases, F does too. However, it suggests that we don't need the force concept to describe gravitation, only the gravitational field concept. It's closer to space-time curvature than to Newton's force.

Here a link (in French... I haven't found one in English about our subject) :
http://home.tiscali.be/jp.delavallee/Gravitation/DeviationLumiere/deviation_lumiere.htm"

Nowdays, photons are thought to have a rest-mass of zero, but of course you never find one "resting." At the speed of light, photons have some mass due to their speed, hence energy and the equivalence of mass and energy (E=mc^2). (True?) so, they would bend due to gravity near the Sun, and would do so approximately 0.8 arc-seconds.

The entire equation for energy in special relativity is :

E² = (pc)² + (mc²)²

where, usually, p = ymv (y is Lorentz factor).

If the object is motionless, (pc)² = 0, so we have E = mc². If the object is massless, (mc²)² = 0, so E = pc.

But, using the "normal" formula for momentum, E = 0. We used de Broglie formula for momentum : p = h/λ , where h is Planck constant and λ is wavelenght.

You can't use m = E/c²

Zeit

PS : Here another link about deflection of light : http://www.mathpages.com/rr/s6-03/6-03.htm"

 

[Jorrie]

The easiest way to understand GRs bending of light is that half of it comes from the equivalence principle (essentially 'time curvature') and the other half from space curvature. Newton's gravity only bends light if the 'corpuscle theory' of light is used. In the 'wave theory' there is no bending under Newton's gravity.

 

[actionintegral]

Newton's gravity only bends light if the 'corpuscle theory' of light is used.

Does the corpuscle theory of light assign a mass to these corpuscles?

 

[vanesch]

Zeit - In the Newtonian formula for Gravitational force you gave,

F = Gmm'/r²

So, for my question about if light bends when close to the Sun under Newtonian gravity theory, I think m = mass of Sun (or photon of light) and m' = mass of photon of light (or Sun).

If either one of these masses is zero, then the right hand size of the equation = 0, hence F=0, hence no Force of gravity between the two, hence no bending of light.

What's wrong with my logic?

I agree with Zeit here: in a Newtonian corpuscular light theory with massless light particles, light would be bend by gravity. Of course it's a different issue with light described by classical EM: there, one has in fact the choice (and it would lead, in any case, to some kind of inconsistency).


Let us consider the Newtonian case where a body A exerts a gravitational force on body B. We denote the (active) gravitational mass of body A by m2, the (passive) gravitational mass of body B by m3, then the force acting upon body B by body A is given by:

F = - G m2 m3 / R^2 with R the distance between both objects and the force F in the radial direction.

Now, the force F acting upon body B, accelerates body B, by Newton's equation: F = m1 a
Here, m1 is the INERTIAL MASS of body B.

Now, it is an experimental fact that for a given body, the inertial mass equals the passive gravitational mass, so we have that m1 = m3. In Newton's theory, this didn't need to be the case, because m3 is the "gravitational charge" which defines the gravity force, while m1 is the "inertial resistance". A priori, those two concepts have nothing to do with one another, but they turn out to be equal.

It also turns out that the gravitational force F acting on body B is equal and opposite to the gravitational force F' acting upon body A. In the last case, however, the "active" role is now played by B, and the passive role by A. So this also means that the active and passive gravitational mass for a body are equal. This, however, IS a consequence of the action=reaction principle.

So everything together:

for a given body, it turns out that active gravitational mass = passive gravitational mass = intertial mass. hence we call this quantity simply the mass of the body, without distinguishing between the 3 different functions of mass.

This is (a version of) the *equivalence principle*.

From the equivalence principle follows now an important concept:

A body B, with mass m, will undergo a gravitational force F which will be the sum of different contributions of different other bodies A1, A2, ...:

F = G m m1 / r1^2 1_r1 + G m m2 / r2^2 1_r2 + ...

and we can factor out G m:

F = G m (m1 / r1^2 1_r1 + m2 / r2^2 1_r2 +...)

here, the factor in () corresponds to the geometry and all the (active) contributions of all other bodies, but does not depend upon anything else but the place of body B (does not depend on its mass).

Applying Newton's law:

m . a = F = G m (m1 / r1^2 1_r1 + m2 / r2^2 1_r2 +...)

allows us to cancel m if m is non-zero (or, as was pointed out, to use l'Hopital's rule for the ratio m.a/F in the limit of m -> 0):

a = G (m1 / r1^2 1_r1 + m2 / r2^2 1_r2 +...)

So we see that the acceleration of body B is independent of any of its properties, but only of its LOCATION. So we can just as well say that this is a property of the LOCATION of B, and not of B itself.

ANY body that is at the said location, will undergo an acceleration a.

So also a massless particle. If we know the velocity of the massless particle, we can calculate its trajectory. It will be the same trajectory as a particle with mass, and the same velocity.

This reduces gravity already to a geometrical effect, even in Newtonian physics. I think it was Cartan who worked out a "Newtonian general relativity" by reformulating the above concept in entirely geometrical terms of curved spacetime. However, the theory is more complicated (and less accurate) than general relativity.

 

[hellfire]

Take a look to this post: https://www.physicsforums.com/showpost.php?p=842731&postcount=13

 

[Jorrie]

Does the corpuscle theory of light assign a mass to these corpuscles?

No, light-bending exercises simply assumes that the trajectory will be a hyperbola - like "tiny, high speed comets", e.g. Faber 1983: Differential Geometry and Relativity.

 

[smithpa9]

Wow! Thanks to everyone for some very complete and thorough answers. This was very enlightening (no pun intended).

:-)

 

[Creator]

It may be of interest to some of you posting here that the full (1.7 sec.of arc) relativistic light deflection can be (and was actually derived) without the calculations of general relativity by L.I. Schiff in 1959.

Simply using equivalence principle and special relativity considerations Schiff provided a very simple derivation whereby the full angular solar deflection (theta) was shown to be :

Theta = (4GM/Rc^2), where R is the radius of the sun. (Put in the figures to get the 1.7 ").

This was important because solar/starlight deflection was suppose to be one of 3 'crucial' tests of Gen Rel. , (gravitational doppler shift and the perihelion precession of Mercury's orbit being the other two).

Since grav. doppler shift and starlight deflection could both be derived outside of Gen Relativity (that is, without the goedesic or field eqns. of GR), Schff concluded that only the Mercurial orbital precession provided any real 'crucial' test of GR and that seemed very insufficient.

(See: "On Experimental Tests of General Relativity", L. I. Schiff, Amer. J. of Phys., 28, 340,(1960).

The point is: This lack of 'crucial' substantiation of GR became the motivation for Schiff to look for possible NEW experimental tests of GR, and was the precurser of his arriving (in very next year) of calculations showing that Gen. Relativistic precessional changes of a gyroscope in Earth orbit could fit the bill.
Thus was born the beginnings of an experiment later to be called Gravity Probe B.

Creator

 

[exponent137]

It may be of interest to some of you posting here that the full (1.7 sec.of arc) relativistic light deflection can be (and was actually derived) without the calculations of general relativity by L.I. Schiff in 1959.

Simply using equivalence principle and special relativity considerations Schiff provided a very simple derivation whereby the full angular solar deflection (theta) was shown to be :

Theta = (4GM/Rc^2), where R is the radius of the sun. (Put in the figures to get the 1.7 ").

I please for additional explanation.
(As first, special relativity enter in the game, because it limiting speed of light to c. Newton gravity gave it unlimited velocity.)
I calculated angle at sun classically and it is 2mG/(Rc2) = 0.729’’. It is calculated by supposition that light can be a little accelerated above speed of light.
But, what is (simply) additional reason in general relativity, that angle is duplicated? How equivalence principle is included in calculation?
It was also written that this factor is because of curvature of space. I please for more explanation.

I think also that all formulae in high physics should be explained graphically and simpler. For instance factor 2 above. Or principle of uncertainty is well explained by by two Gaussian curves, where one rises the other becomes thinner.

 

[actionintegral]

No, light-bending exercises simply assumes that the trajectory will be a hyperbola - like "tiny, high speed comets", e.g. Faber 1983: Differential Geometry and Relativity.

I meant that when people thought light were little corpuscles, before the wave picture was invented, did they assume that these corpuscles have mass? I am asking an historical question.

 

[exponent137]

I calculated this in elevator, which accelerate.
Classicaly
Phi=a l/c^2.
l is length of light ray in elevator. a is acceleration of elevator.
At the beginning light ray entered in the elevator at height Lx0 and exit at height Lx1.
But elevator touches lower Lx0 for x = ½ a l^2/c^2. Because it moves at that small time.
When I include that the distance from x point Lx is smaller due to time dilatation (I have larger velocity) and shortening of distance, I obtain Lx1 = Lx0 – ½ a L^2/c^2. Together with x this gives Lx1 = Lx0 –a L^2/c^2.
A think that this in homogenic field is simpler as calculation around sun.

 

[Jorrie]

Corpuscle theory of light

Does the corpuscle theory of light assign a mass to these corpuscles?

No, the corpuscles are massless. This does not change the fact that their paths are curved by gravity.

 

[actionintegral]

No, the corpuscles are massless. This does not change the fact that their paths are curved by gravity.

Thank you for sticking with me on this. Now how in the world did the people who believed te corpuscular theory of light (take Newton for example) come to the conclusion that light was affected by gravity if it had no mass?

 

[Jorrie]

Thank you for sticking with me on this. Now how in the world did the people who believed te corpuscular theory of light (take Newton for example) come to the conclusion that light was affected by gravity if it had no mass?

Pleasure. In response to your question: I don't think Newton believed that and neither did his theory predict it directly. It was actually Kepler's laws of planetary motion that were used to show that light would be bent by half as much as what GR predicts. But if I'm not mistaken, Newton's gravitational equations were developed to explain Kepler's laws.

In the modern era, any gravitation theory that is compatible with the weak equivalence principle predicts the "Newton half" of the deflection of light. The other half depends on how the theory handles space curvature. In Newton's theory there is no space curvature.

 

[exponent137]

Correction and question

I calculated this in elevator, which accelerate.
Classicaly
Phi=a l/c^2.


l is length of light ray in elevator. a is acceleration of elevator.
At the beginning light ray entered in the elevator at height Lx0 and exit at height Lx1.
But elevator touches lower Lx0 for x = ½ a l^2/c^2. Because it moves at that small time.
When I include that the distance from x point Lx is smaller due to time dilatation (I have larger velocity) and shortening of distance, I obtain Lx1 = Lx0 – ½ a L^2/c^2. Together with x this gives Lx1 = Lx0 –a L^2/c^2.
A think that this in homogenic field is simpler as calculation around sun.

Calculation in first approximation in Newton's physics gives that this angle theta is:
Phi= a L/c^2.
a is acceleration
L is horizontal length of elevator
C is speed of light.

or
or phi=dv/c=a dt/c=a dx/c^2=a L/c^2

Now I enter in special relativity. I expect (including deflection of light by sun) that this the first relativity approximation phi is twice larger. But, how? Calculation above is not correct? In attachement I should to calculate with velocity, not with distance.

Or simply, can someone glue or describe schiff's derivation, mentioned above?

I think that article is really "On Experimental Tests of the General Theory of Relativity"
American Journal of Physics -- April 1960 -- Volume 28, Issue 4, pp. 340-343

 

[pervect]

OK, I've thought about this a little more. . . let's see if this is correct:

Prior to SPECIAL relativity, photons were considered massless, always. So everyone before Einstein would have believed that light does NOT bend near a massive object, like the Sun.

The original idea was the corpuscular theory of light - the idea of the photon came in with quantum mechanics.

Prior to special relativity, I don't think people had assigned any definite mass to the photon, or to corpuscle of light. After special relativity, people realized that a photon had a zero invariant mass.

Nowdays, photons are thought to have a rest-mass of zero, but of course you never find one "resting." At the speed of light, photons have some mass due to their speed, hence energy and the equivalence of mass and energy (E=mc^2). (True?) so, they would bend due to gravity near the Sun, and would do so approximately 0.8 arc-seconds.

Look up "relativistic mass" and "invariant mass", in the wikipedia http://en.wikipedia.org/wiki/Mass, or see the sci.physics.faq "does mass change with velocity" http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html.

So one can say that a photon has zero invariant mass, while still saying that it has relativistic mass - if one really wants to. This is not the only way of looking at things, though. You have "imported" the old Newtonian idea that gravity is due to mass. General relatlvity actually says that space-time is curved by the stress-energy tensor, not by mass. The results of GR for the bending of light are different. So one can say the following.

In GR, energy (and momentum, and pressure) all have some effect on the curvature of space-time. This effect is formalized by the use of the stress-energy tensor.

Light, and small test particles, always follow geodesics in this curved space-time.

The end result of a full GR calculation is that light deflects twice as much as a quasi-Newtonian calculation. The quasi-Newtonian calculation assumes that light (or corpuscles) travels at 'c', and that they have an equal gravitational and inertial mass. The latter actually introduces a bit of GR, because pure Newtonian theory doesn't incorporate the equivalence principle yet.

 

[MeJennifer]

General relatlvity actually says that space-time is curved by the stress-energy tensor, not by mass.

I thought it is actually by the stress-energy tensor and the Weyl tensor, am I wrong?

 

[robphy]

I thought it is actually by the stress-energy tensor and the Weyl tensor, am I wrong?

The stress-tensor is algebraically related to the Ricci Tensor, and differentially related to the Weyl Tensor.

 

[MeJennifer]

The stress-tensor is algebraically related to the Ricci Tensor, and differentially related to the Weyl Tensor.

Ok, but as I understand it both tensors determine the curvature not just the stress-energy tensor. In other words the stress-energy tensor is not enough to completely determine the curvature of spacetime.

 

[pervect]

re: needing more than the stress-energy tensor

In GR, the stress-energy tensor is the density of energy and momentum. It is somewhat similar to the charge density in E&M. In E&M, one can interpret the "source" of electromagnetism as a scalar, charge, which has a density given by a rank-1 tensor, the four-current.

In GR, one can interpret the "source" of gravity as energy and momentum (a four-vector) which has a density given by a rank-2 tensor, the stress-energy tensor.

Thus the four-current, multipled by a vector representing a volume, gives the charge density (a scalar) contained in that volume in E&M. The stress-energy tensor, a rank 2 tensor, mutliplied by a vector representing a volume, gives the energy-momentum (a four-vector) contained in that volume.

In order to solve either E&M, one needs more than the distribution of charge. One also needs additional boundary conditions. E&M is linear, so that one can add any homogeneous solution of Maxwell's equations to an inhomogenoeous solution, and the result will still be a solution of Maxwell's equations. In E&M, it's usually more convenient to deal with the electromagnetic potential rather than the fields. In terms of the potential, one talks for instance about the solution to Poisson's equations (in a region where there is charge) or Laplace's equation (in a region that is charge free), but one needs boundary conditions, for instance Direchlett boundary conditions (see http://en.wikipedia.org/wiki/Boundary_condition) to specify a unique solution.

Gravity is not linear, so it's not as simple as it is for E&M, but one still needs both the stress-energy tensor AND the boundary conditions. Just knowing the source distribution is not enough. A common boundary conditions in E&M is "zero potential at infinity", in gravity a rougly similar assumption is "asymptotically flat space-time" which says that the metric "at infinity" is Minkowskian. The metric coefficeints g_ij obey second order differential equations in GR, so they are formally similar to the potentials in E&M, which also obey second order differential equations. Thus specifying the metric at infintiy is very similar to specfiying the E&M potential at infinity.

 

[MeJennifer]

Gravity is not linear, so it's not as simple as it is for E&M, but one still needs both the stress-energy tensor AND the boundary conditions.

And these boundary conditions are part of the Weyl tensor right?