# What curves space and bends light: Mass or Gravity?

1. Oct 29, 2015

### Doug Brown

my father who is an MBA not a PhD asked me:

Which is it more correct to say Mass curves space and bends light or Gravity curves space and bends light or Mass curves space and gravity bends light or some other phrasing?

Interesting distinction you're seeking

Gravity causes space-time curvature and the latter cause light to bend but Gravity has two sources, one of which has nothing to do with mass.

Gravity not mass is thus Supreme

First consider the Equivalence Principle

The Equivalence Principle says that there is an equivalence between a frame of reference in a gravitational field (the person in the elevator feels gravity pulling him onto the floor) and a frame of reference which is accelerating at the right rate (the person still feels pulled to the floor)

So you have the feeling by an observer of gravity in the two cases. 1) a nearby mass causes gravity or 2) a constant acceleration causes gravity without a nearby mass

The issue is if there is no nearby mass then one wonders what agency or agent is causing a constant acceleration of the reference frame? Clearly it takes a lot of power or energy to constantly accelerate a reference frame (an observing person) so one wonders if mass isn't doing it what is? (A rocket engine could be hurtling the observer through empty space-time far from any masses)

Assuming simply that there is constant acceleration (by whatever cause) then it would appear that it is the constant acceleration causing curvature of space which bends light

When objects fall or ppl are forced into elevator floors or light bends it is equivalent to all of those objects following curved space time lines

To conclude let's define gravity as equivalent to constant acceleration

You can have space-time curvature without mass

You cannot have space-time curvature without gravity (constant acceleration)

So I'd say it was constant acceleration or gravity which causes all these effects: curvature of space, bending of light, ppl being forced into elevator floors

Since they all can happen without mass but which then implies some unknown cause for constant acceleration this means that mass cannot be Primal Cause

Constant acceleration or Gravity is the Primal Cause for all of these phenomena and matter or mass just is a simple way of achieving Gravity but there is another way to cause Gravity without matter or mass and that is by externally applying a force which results in constant acceleration

Big question is if mass or matter isn't present then how in tarnations is constant acceleration being caused or forced ?

The answer is a rocket engine delivering propulsion or thrust.

If that answer is irrelevant then Gravity alone is Primal Cause but mass or matter isn't!

Do General Relativists or other physicists or interested observers of this web forum agree?

TO boil it down to its utter simplicity, choose one of the 4 following mutually exclusive and collectively exhaustive statements as being more correct than the others:

1. Mass curves space and bends light.
2. Mass curves space and gravity bends light.
3. Gravity curves space and mass bends light.
4. Gravity curves space and bends light.

2. Oct 29, 2015

### Staff: Mentor

No, gravity (more precisely, tidal gravity) is spacetime curvature. Spacetime curvature is caused by stress-energy. See below.

The "gravity" that you observe in an accelerating rocket in flat spacetime has no "source". Spacetime is flat; there is no stress-energy anywhere. That's why, in GR, we draw a distinction between "gravity", which is an ambiguous term that can have a number of meanings, and "tidal gravity", which refers specifically to spacetime curvature, and is what has a "source".

Obviously it's the rocket engine, or whatever other means of propulsion you are using. You even say as much later in your post.

This is a bad definition, because it (apparently) has misled you into thinking that you can curve spacetime by accelerating in a rocket. You can't.

I can't speak for others, but I don't. See above.

None of them are correct as you write them. A correct statement would be: stress-energy curves spacetime, and spacetime curvature (aka tidal gravity) bends light.

3. Oct 29, 2015

### Doug Brown

I stand corrected. Thank you. Physics Forum has functioned exactly as it was designed to do here since I thought I knew but in actuality didn't. Thank you Peter Donis for taking the time to weigh in. I also have available to me the similar titled threads suggested at the time I created the title for this. Of course, there are many additional details on those very related threads which apply here as well. I thank the Physics Forum members for generously contributing their time and expertise to questions like these

4. Oct 29, 2015

### phinds

Peter, I would appreciate it if you would explain how it is that a photon from a distant galaxy, passing by another galaxy and ending up at Earth as part of an Einstein Ring (and that thus clearly has been "bent") is experiencing tidal gravity. I always thought it was just getting bent by the gravitational field of the galaxy it's passing. I suspect my problem may be somewhat in my terminology but it may be a more fundamental misunderstanding, so please help me understand this. Thanks.

5. Oct 29, 2015

### Staff: Mentor

The Einstein Ring is not made up of just one photon. You can't observe tidal gravity by looking at just one trajectory. You have to look at all of the trajectories involved. Actually, it suffices to look at a group of nearby trajectories.

For example, consider the trajectories of a group of photons emitted from the distant galaxy at slightly different angles, which pass by the second galaxy at slightly different angles, and reach Earth from slightly different angles, and are seen as adjacent pieces of the Einstein Ring. If you look at what happens to all these trajectories, you will see that they must diverge from the first galaxy to the second galaxy, but must again converge from the second galaxy to Earth. So the second galaxy changed them from diverging to converging. That's tidal gravity. Local light bending, the kind you would derive just from the equivalence principle, won't do that: it would bend all the trajectories the same way, so it could not change a group of trajectories from diverging to converging.

Locally, you can view things that way. But the Einstein Ring is not a local effect. To account for it, you have to look at all the photon trajectories, not just one. When you look at them all, as above, you will see that different photons are bent at different angles, so the entire family of them changes from diverging to converging. That is tidal gravity, and without it, you can't account for the Einstein Ring. The local bending of light that you would derive from the equivalence principle is not enough by itself.

6. Oct 29, 2015

### phinds

Hm. I'm still a bit confused but I'll give it some thought. Thanks for that clarification.

7. Oct 29, 2015

### Doug Brown

wait, now I am confused by your use of saying the equivalence principle (EP) derives a form of light bending. As I told my father, I think the EP shows that gravity caused by a large nearby mass is equivalent to that felt in an accelerating frame (propelled by some engine in flat space). If there is no curvature in flat space then how can the EP "cause" bending which must result from or indicate the presence of curvature?

8. Oct 29, 2015

### Staff: Mentor

The light-bending that is observed on an accelerating spaceship happens because the beam of light is travelling in a straight line in flat spacetime while the acceleration of the spaceship is moving the floor and ceiling and walls of the spaceship increasingly away from the path of the light. It's the spaceship that is travelling on a curved path through spacetime, and it's following a curved path because its engines are pushing it off the straight-line path that inertia makes it want to follow.

In fact, there is no curved spacetime that will produce exactly the same effects as the gravity-like pseudo-force that you feel on an accelerating space spaceship (although some come very close) - there's no such solution to the Einstein field equations. To see the difference, imagine that you hold two objects one meter apart and drop them at the same time. On the spaceship, they will both hit the floor at the same time and at points also one meter apart. Try this on the surface of the earth and they will still hit the floor at the same time, but at points very slightly less than one meter apart. This happens because they are both falling towards the center of the earth, so their trajectories are not quite parallel - they converge to intersect at the center of the earth. That's an example of tidal gravity at work.

Last edited: Oct 29, 2015
9. Oct 29, 2015

### Staff: Mentor

And I pointed out that "gravity" is an ambiguous term. The "gravity" that the EP says is equivalent to what you feel in an accelerating frame is not the same as the "gravity" that is caused by a large nearby mass. The latter involves spacetime curvature; the former does not.

Even in a curved spacetime, such as that around a large mass, if you are in a small room, for example, sitting at rest on the surface of the large mass, you will not be able to see, locally, any effects of spacetime curvature (tidal gravity), so you won't be able to tell that the room is not inside an accelerating rocket in flat spacetime. That is really what the EP says: that what you feel standing at rest on the surface of a large mass is locally equivalent to what you would feel in an accelerating rocket. The "locally" qualifier is crucial.

This is incorrect. The path of light bends in an accelerating rocket in flat spacetime, with no spacetime curvature.

What does not happen in an accelerating rocket in flat spacetime is different light rays bending in different ways. But that is not a local effect. See my response to phinds.

10. Oct 29, 2015

### phinds

Aha ! I think I'm beginning to see the light (pun intended )

11. Oct 29, 2015

### Doug Brown

so if it is a strictly non-vanishing Energy-Momentum Tensor which causes the curvature of space and the existence of mass implies non-vanishing matrix elements of this tensor, then is it possible to have non-zero matrix elements of this tensor which have nothing to do with mass? Does each non-zero matrix element imply a mass or are there other sources of non-zero matrix elements which are not mass?

12. Oct 29, 2015

### PAllen

Unfortunately, there isn't a satisfactory answer in GR. I think, close to the best you can say, is:

If the universe is asymptotically flat (i.e. like SR at great distance from 'everything' - to put it poetically), AND if there are limits on exotic matter, then all the regions with non-zero stress energy tensor correspond to some positive total mass.

Unfortunately, remove any of the caveats (e.g. asymptotically flat, or unlimited violation of what are called energy conditions - which are added to the Einstein field equations, not part of them), and you are left with nothing. Of particular significance is that our universe is definitively NOT asymptotically flat. Thus, what Peter said in terms of the stress-energy is accurate, but attempts to make it more accessible by substituting 'mass' will fail in classical GR as currently understood.

13. Oct 29, 2015

### Doug Brown

then again our universe does not have any empty space regions, at least in the sense of vacuum polarization. If at every point in space there are energy-time uncertainty principle "energy violations" (that is Energy from nothing Delta E ) which happen for infinitesimally small times (Delta T) such that the Energy Time Uncertainty Principle is satisfied, then the empty vacuum is not empty. Quantum Fluctuations (or Wheeler's Foam) of this off-mass-shell (virtual particle) type fill the vacuum and hence every point in space has this "background" "energy" in it at all points in time but for fleeting amounts of time but overall more or less continuously

Therefore, all space has off-mass-shell virtual particle mass in it even if only of the quantum fluctuating off-mass-shell type, and hence all space exhibits a corresponding relative amount of curvature though at presumably a much smaller scale (both in amount and in time) than compared with the curvature that accompanies the space around real on-shell mass

But all of this curvature due to quantum fluctuations must be negligible compared to the curvature resulting from actual on-mass-shell mass

Is this off-mass-shell curvature from quantum fluctuations what you mean by "our universe is definitively NOT asymptotically flat"?

14. Oct 29, 2015

### Staff: Mentor

The stress-energy tensor includes more than just mass (or energy) density. It also includes momentum, pressure, and other stresses. But all of those things are normally found "attached" to something that has mass (or energy--EM radiation has a nonzero stress-energy tensor, but it has no rest mass). So I'm not sure what you mean by elements that "have nothing to do with mass".

This is not really a good description of the nonzero energy density due to quantum fluctuations of the vacuum. But yes, quantum fluctuations of the vacuum do have a nonzero stress-energy tensor. It has some properties that are different from those of the stress-energy tensor of ordinary matter or radiation; the key one is that energy density due to quantum fluctuations of the vacuum causes the expansion of the universe to accelerate, whereas energy density due to ordinary matter or radiation causes the expansion of the universe to decelerate.

No. Our universe would definitely not be asymptotically flat even if there were no energy density due to quantum vacuum fluctuations. "Asymptotically flat" means basically "an isolated region of matter and energy surrounded by empty space". Our universe does not meet that description even if we leave out the energy due to quantum vacuum fluctuations.

15. Oct 29, 2015

### Staff: Mentor

No. General relativity is a classical theory, no quantum effects, so in GR the vacuum really is a proper classical vacuum. And it's not asymptotically flat - you cannot find a region where the curvature is arbitrarily small by travelling an arbitrarily large distance away from where you are now.

Bringing quantum fluctuations and the like into the picture requires a theory of quantum gravity, but you can't go there until after you know GR like the back of your hand.

16. Oct 29, 2015

### Staff: Mentor

This isn't quite true. We can model quantum vacuum fluctuations (or more precisely an expectation value for the energy due to them) using a classical stress-energy tensor, and we can use observational data to assign an estimated value to the components of that tensor. That's what is done in the standard $\Lambda$ CDM cosmological model.

What we can't do without a theory of quantum gravity is explain why the value we get from the observational data, and which we plug into the cosmological model, is 120 or more orders of magnitude smaller than the value we think is implied by our current understanding of quantum field theory.

John Baez' article on this topic is worth a read:

http://math.ucr.edu/home/baez/vacuum.html

17. Oct 29, 2015

### Staff: Mentor

To see how curvature is involved, consider longitude lines going from the south pole to the north pole. At the south pole they are diverging, at each point along the line they go due north without turning (geodesics), at the equator they are parallel, and at the north pole they are converging. That can only happen because the sphere is curved.

18. Oct 29, 2015

### phinds

That's not what I'm having a problem with at all. My problem is tidal gravity vs non-tidal gravity and why Einstein Rings are caused by tidal gravity instead of just the gravitational field caused by a galaxy's mass. Curvature isn't my problem at all as far as I'm aware.

19. Oct 29, 2015

### Staff: Mentor

Try this: imagine an accelerating rocket with a bunch of light rays passing through it. The light rays are diverging from a common source before they pass through the rocket. Is it possible for the light rays to look like they are converging, to an observer at rest in the rocket, because of the "bending" of light seen by that observer? (The correct answer is no. Can you see why?)

Yes, curvature is your problem, because curvature--spacetime curvature--is tidal gravity. They're the same thing.

20. Oct 30, 2015

### A.T.

Non-tidal gravity means a uniform acceleration field. The gravitational field caused by a galaxy isn`t uniform.