# Gravity as a force and as a curvature

1. Feb 11, 2014

### shounakbhatta

Hello All,

As far as the Newtonian mechanics and Einstein's GR is concerned, I am a little bit confused in the following things:

(a) Concerning the bending of light due to gravity: Some lectures and opinions show that light bends due to the force of gravity as shown in the event of a solar eclipse, light bends and takes a different route before reaching the observer. Another argument shows that as photons are mass less, hence how can they be affected by gravity? The curvature of spacetime allows the photons to take the curved path. WHICH ONE IS CORRECT?

In case we use F= G m1m2/r^2 and use photons, substituting m=0 for mass less photons, we get F=0. From that can we tell that for photons there is no force?

(b) Concerning the red shift and blue shift: It shows that photons coming out of gravitation zone, looses energy and hence the wavelength becomes longer and causes red shift and vice-verse causing the blue shift. Here the energy of the photon is affected by gravity while IN THE ABOVE CASE IT IS NOT. AM I GETTING SOMETHING WRONG?

Thanks.

2. Feb 11, 2014

### tiny-tim

hello shounakbhatta!
no, light does not bend

light does not take a bent path

light always takes a straight path through space-time

but the straight path isn't always what you might think it is

(for example, if you look "down" on the solar system from a long way "above", then the path that the light takes looks bent to you)

mass has nothing to do with it …

the light follows a straight path through space-time because it has no reason to do otherwise

3. Feb 11, 2014

### dauto

You're making two mistakes.
1st, Even in Newtonian mechanics it is NOT correct to assume that a massless particle does not accelerate. True, you get F=0. Try plugging that into Newton's 2nd law to find the acceleration. What did you get?

2nd, In GR it is NOT correct to assume that gravity only acts on massive objects.

Last edited: Feb 11, 2014
4. Feb 11, 2014

### .Scott

Photons have mass, but they have no rest mass.
But even if there is a particle with no mass (or negative mass) it would still be affected by gravity. Although the gravitational force applied to an object is proportional to its mass, so is the amount of force required to accelerate it.

5. Feb 11, 2014

### Staff: Mentor

First, this argument is based on Newtonian gravity which has been shown experimentally to be incorrect regarding light and gravity. Second, in the context of Newtonian gravity the argument is incorrect. In Newtonian gravity the acceleration of an object is independent of its mass, so a massless object accelerates at the same rate as a massive object. True, the force is 0, but by F=ma it is clear that it takes 0 force to accelerate an object with 0 mass, so you have to determine the acceleration by some other means than a calculation of the force.

6. Feb 11, 2014

### shounakbhatta

Hello,

Tiny Tim, thanks for your answer. I have one question:photons have relativistic mass?

Now what we get an explanation during solar eclipse, is that the star light bends as the massive object like sun causes a bend, a dip in the spacetime curvature. Is that so? Please correct me as it might seem rudimentary questions at the initial level.

7. Feb 12, 2014

### Ibix

In general relativity, the source of gravity is the stress-energy tensor, which you can think of as a 4x4 matrix. One element of that is the rest mass, the "m" in Newton's equations. That is zero for a photon, but not all of the components are zero. Some people describe this as .Scott did, as the photon having "relativistic mass"; others don't like the terminology since it's also used for $\gamma m$ and $\gamma^3m$ at least.

To the extent that I'm qualified to have an opinion, I prefer not to use the term "relativistic mass" at all, and I seem to be in the majority (at least round here). But, whatever language you use, photons have non-zero components in their stress-energy tensors. Just not the classical mass element of it.

On the curvature of spacetime, "a dip" is probably not the best way to describe it. But it is often depicted that way. Don't read too much into that, though.

8. Feb 12, 2014

### A.T.

The often shown "dip" is just the spatial curvature, which accounts for half of the light bending. See the lower picture here:
http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

The other half of light bending is due to gravitational time dilation (or "time curvature"): Light at different distances from the mass propagates at different speeds when measured with the same clock, so a wave front propagating tangentially changes direction. This is similar to light bending in a medium with varying refractive index:
http://www.nature.com/nchina/2013/131106/full/nchina.2013.113.html

Last edited by a moderator: May 6, 2017
9. Feb 12, 2014

### tiny-tim

hello shounakbhatta!

(just got up :zzz:)
some people call the energy of a particle its relativistic mass, so in that sense, yes
i prefer to call it an increase in curvature …

space everywhere has some curvature (there's no place with zero gravity! )

the stronger the gravity, the stronger the curvature, and the more "parallel" lines deviate from each other (much like "parallel" lines of longitude on the earth) … this is also known as geodesic deviation

(technically, i shouldn't have been saying "straight", i should have been saying "geodesic" … but "straight" is easier! )

10. Feb 12, 2014

### pervect

Staff Emeritus
The GR analysis better fits observations. For instance, famously, one of the early tests of GR was a test of "extra' light deflection.

I suspect that the issue is that you don't fully understand the GR paradigm of "curved space-time". Which is rather understandable, unless you've studied it - and it's not generally taught until college. AT has some nice diagrams that can help, if you are familiar enough with special relativity to understand space-time diagrams. If you are not familiar enough with SR to understand space-time diagrams, it seems unlikely that you can understand the significance of the space-time diagrams being drawn on a curved surface (which is what curved space-time means, operationally). So you'd need to study SR to the level at which you understood space-time diagrams first, before the GR analysis will make any sense to you.

This is an attempt at a purely Newtonian analysis, it has nothing to do with GR. In GR, it's not "mass" that causes gravity - as I'm sure other posters have mentioned, it's the stress-energy tensor. In terms of the Newtonian analysis, it's a bit unclear to me how the "expected" Newtonian light prediction (the one that was half the GR value) was calculated. My personal answer would be that the Newtonian formulation of light deflection is known to be wrong anyway, so I'm not too interested in the details of how it was formulated.

In the Newtonian analysis, photons do gain or loose energy as it falls. In the GR analysis, the "energy-at-infinity" of the photon is a constant as it falls.

With the GR defintion of "energy-at-infinity" in the GR analysis it's the clocks that measure the photon's frequency that are affected and which cause the red and blue shift.

There is another meaning of "energy" in GR, the locally measured energy. This does red and blue shift as the photon falls, and is probably more simlar to the Newtonian notion of energy with which you are familiar. But the local energy is not a conserved quantity, while the "energy-at-infinity" is conserved (well, there's some fine print - it's conserved for those space-times in which it can be defined, like static space-times).

11. Feb 12, 2014

### Maxila

I believe for many people some confusion is because space-time is often referenced in a way that may indicate it to be some kind of independent entity or a structure of substance, rather than it simply being coordinates, location(s). Einstein specifically said in the last appendix he added to his book “Relativity The Special and General Theory” that, "Space-time does not claim existence on its own, but only as a structural quality of the field." [Page 155 under Appendix 5. Year 1952].

In the presence of a gravitational field space-time coordinates are curved, that is literately what is meant when you read, or see someone say, “light travels a straight path in curved space-time. It means light follows the curved coordinate path defined by space-time and there is no other path it can follow, that is the straightest (shortest) possible route, unlike the curve of a road that may ‘cut-off’ and a distance traveled can be shortened. That also shows why a photons lack of mass is irrelevant to why gravity effects its path (in addition to gravity effecting the path of any mass, it effects the very fabric, or space-time coordinates, of space itself.

Last edited: Feb 12, 2014
12. Feb 12, 2014

### WannabeNewton

This is terribly incorrect I'm afraid. Space-time has a geometric structure and kinematics/dynamics on space-time have invariant geometric forms all of which are several strata above coordinate systems. In fact coordinates are only computational tools, nothing more, and in principle never have to be introduced; we can talk about general relativity and space-time physics without ever mentioning coordinates-it will be quixotic and highly inefficient at times but that doesn't mean it can't be done in principle. I can refer you to general relativity textbooks that rarely if ever introduce coordinates-most of the calculations in the texts will be done using the coordinate-free abstract index and/or index-free calculus and most definitions will be set in this geometric language. It's far more elegant albeit far less practical, sometimes intractable, for down to Earth problems such as Perihelion precession or ultra-relativistic gravitational scattering.

Last edited: Feb 12, 2014
13. Feb 12, 2014

### Maxila

I still see defining a structure of space, a manifold, topological space, being paramount to the locations that define them, and in agreement with the quote of Einstein "Space-time does not claim existence on its own, but only as a structural quality of the field." However I stand corrected in implying space-time is a coordinate system and redact that implication.

14. Feb 12, 2014

### shounakbhatta

Thank you very much for the wonderful answers. Pervect, it was great to read your explanations. Well, tiny tim, can I conclude, that in a curved spacetime, light follows the geodesic. For us it 'appears to bend'. Right?

Also, taking the classical concept of a field affecting the photos, gravitational fields, never bends or draw photons inward. Right?

15. Feb 13, 2014

### A.T.

However, this classical prediction for "light particles" is only half of what is actually observed. The other half is due to the spatial-distortion, which is only present in General Relativity. So the observational test of Newtonian "light particles" vs. General Relativity is the amount of light bending, not it's existence.

16. Feb 13, 2014

### tiny-tim

hi shounakbhatta!
right!

the geodesics 'appear' to bend, in the sense that geodesics that are initially parallel will get closer, then get further again

but they aren't actually bent, any more than lines of longitude on the earth are bent!
i think this is a semantic issue

you can separate the field equations into two parts, which happen to have equal effects: a gravitational part and a curvature part …

on that view, there's a gravitational force on the photon, and it has an effect additional to that of curvature​

but if you regard gravity as not being a force, but as part of the general structure of space-time, then of course the question is meaningless (ie, no gravitational field anyway): everything is governed by curvature

17. Feb 13, 2014

### Staff: Mentor

What are you describing here? Are you talking about separating the Christoffel symbols into two groups somehow or are you talking about something else?

18. Feb 13, 2014

### tiny-tim

i was thinking about the way the general relativity equations can be regarded as a correction to the special relativity equations (which give only half the correct result, as i believe einstein originally obtained before GR) …

perhaps i'm misremembering my history of physics, but i thought there was a pre-GR way of applying SR to curved space-time: if so, you could say that that is the curvature contribution, and the GR "correction" is the gravitational contribution

19. Feb 13, 2014

### Staff: Mentor

Hmm, OK, I am not familiar with that approach, but then I never spent much effort investigating historical models.

20. Feb 13, 2014

### Staff: Mentor

I haven't seen anything like this in what I've read of pre-GR work in relativity (not that I have read everything there is to read by any means). Also, trying to apply SR to curved spacetime seems inconsistent to me, because curved spacetime breaks a basic assumption underlying SR, the assumption that free-falling observers who start out at rest relative to each other will stay at rest relative to each other.

21. Feb 14, 2014

### tiny-tim

ah, i did a google search and decided i must have been thinking of this, from http://en.wikipedia.org/wiki/Histor...e_development_of_the_Einstein_field_equations

When Einstein realized that general covariance was actually tenable, he quickly completed the development of the field equations that are named after him. However, he made a now-famous mistake. The field equations he published in October 1915 were

$R_{{\mu \nu }}=T_{{\mu \nu }}$,,

where $R_{{\mu \nu }}$ is the Ricci tensor, and $T_{{\mu \nu }}$ the energy-momentum tensor. This predicted the non-Newtonian perihelion precession of Mercury, and so had Einstein very excited. However, it was soon realized that they were inconsistent with the local conservation of energy-momentum unless the universe had a constant density of mass-energy-momentum. In other words, air, rock and even a vacuum should all have the same density. This inconsistency with observation sent Einstein back to the drawing board. However, the solution was all but obvious, and in November 1915 Einstein published the actual Einstein field equations:

$R_{{\mu \nu }}-{1 \over 2}Rg_{{\mu \nu }}=T_{{\mu \nu }}$,

where R is the Ricci scalar and $g_{{\mu \nu }}$ the metric tensor. With the publication of the field equations, the issue became one of solving them for various cases and interpreting the solutions. This and experimental verification have dominated general relativity research ever since.​

… but both parts are due to curvature, so it seems you can't separate out the gravity

22. Feb 14, 2014

### pervect

Staff Emeritus
I view the extra light deflection in GR as arising from the spatial part of the curvature, but I'm not sure there is general agreement about this

One of the issues is what does "the spatial part of the curvature" even mean? To make sense of this, you need to slice space-time into space+time, for generality one needs to specify that the slicing is done in the usual means by the time-like Killing vector.

Given that slicing, there are two approaches I'm aware of for arguing that the extra light deflection is due to the spatial part of the curvature

One is to use the Riemann tensor, and it's Bel decomposition into the electrogravitic, magnetogravitic, and topogravitic parts. Then one can show that the topogravitic local deflection is equal to the electrogravitic local deflection. I find this rather elegant, but I'm not sure I've seen it published anywhere. In this approach, The topogravitic part of the Bel decomposition of the Riemann tensor is what one means by "spatial curvature".

Another way is to use the non-tensor PPN parameter gamma, and note that it can be interpreted as "the amount of spatial curvature per unit mass". This description doesn't really define what "spatial curvature" is, but it is published (in MTW, Wiki also uses this).

23. Feb 22, 2014

### shounakbhatta

Thanks everybody for the explanations.

24. Feb 22, 2014

### stevendaryl

Staff Emeritus
I would think that "SR in curved spacetime" would just be General Relativity without the field equations. That is, GR in a fixed, background geometry. So it would take into account the effects of curvature on equations of motion (mostly, replacing partial derivatives by covariant derivatives), but would not take into account the effect of mass/energy on curvature.

25. Feb 22, 2014

### Staff: Mentor

First of all, such a theory would be inconsistent, because the equations of motion for matter are connected to the Einstein Field Equation; you can't separate them. (This is because the EFE obeys the Bianchi identities, and the Bianchi identities applied to the stress-energy tensor give equations of motion for the matter that the stress-energy tensor describes.)

Second, such a theory wouldn't be SR, because as soon as you allow curved spacetime to affect the motion of objects, you break a basic assumption of SR, as I said in my previous post: if you have curved spacetime, you have free-falling objects that start out at rest relative to each other but don't stay at rest relative to each other, and SR can't handle that.

You can certainly do SR in curvilinear *coordinates*, which requires you to deal with covariant derivatives instead of partial derivatives; but spacetime itself must still be flat if you're doing SR (i.e., you can have nonzero connection coefficients, but the Riemann curvature tensor must still be zero).