[watsisname]

Yeah, I thought you referred to crossing event horizon instead of falling to singularity (which, I should note, I suspect is a mathematical artefact from the theory rather than a physical object at the "bottom" of the black hole).

It is, yeah.  General relativity is a classical theory.  It treats space-time as continuous and infinitely differentiable.  A singularity in GR even has a very precise definition, beyond "infinite density".  If you find a place on a space-time manifold where all geodesics (space-like, time-like, and null) and all time-like paths with bound acceleration (any allowed observer), end after finite proper time, and the manifold cannot be smoothly extended beyond that location, then you have found a singularity.  Classically, gravitational collapse is guaranteed to produce such a region anytime mass falls within its own Schwarzschild radius and forms a trapped surface, as proved by various singularity theorems (e.g. by Penrose).

Of course, the correct theory of gravitation might not be classical.  Perhaps some future theory of quantum gravitation may totally rewrite our understanding of what happens in the depths of a black hole.  In practice, however, it makes no difference to the fate of a person falling into one.  Tidal forces become lethal before you'd be close enough to discern any difference from the classical description (unless you happen to be a human the size of a subatomic particle).  There's something disconcerting about that -- it means not only can you not report your discoveries about the 'singularity' to your fellows outside the horizon, you can't even make those discoveries in the first place.  Singularities are very well protected.

There are of course other possibilities -- metrics become pretty weird when you allow for nonzero values of charge/spin.  Or that 'cosmology with torsion' idea that was tossed around here a while back.  And then there's those theoreticians discussing black hole thermodynamics/holography, some saying that maybe there's this firewall sort of thing just above the event horizon.  I don't know, it sounds silly to me, but I sure don't understand it well enough.

I'm sure you're aware of the gravitational time dilation causing a time difference between surface and satellites, being substantial enough that GPS and other geolocation systems need to be specifically corrected to take it into account.

Certainly!  The gravitational time dilation effect was experimentally verified back in 1959 at Harvard, and it is critically important to consider for the proper functioning of GPS.  The system would very rapidly become useless if we didn't!  But this isn't 'curvature of time', it is again curvature of 'space time'.  We do not explain time dilation with 'time curvature' -- the concept makes no physical sense and does not help us to do calculations.

Your thought experiment examines the ability to measure the change in volume elements due to curvature (based on the idea that gravitational field is positive curvature in 4d-Riemannian geometry).  This underlying reasoning is correct -- volume in non-Euclidean geometry is different.  What I was trying to explain to you is that in practice, it is a terrible idea to try measuring curvature in this way (there are far better ways, and I gave an example).  Even excluding the complications of solid state physics, the effects of tidal forces are going to be overwhelming in comparison.  You would need an impossibly rigid structure to avoid this.  But yes, in answer to your question of an idealized sphere, so that this is purely a hypothetical thought experiment, you would expect a change in the enclosed volume due to the change in curvature.

[Herra Tohtori]

Ok, yeah, that makes sense. In other words the experiment is one of those that are valid on thought level, but in practice would require the arsenal of physics tools similar to massless wire, rigid rod (in this case a rigid shell), frictionless surface, ideal gas, and perfect black body...


So I guess instead of providing basis for an experiment, it can still be an useful way to visualize the effects of positive curvature on space-time, and what the phrase "curved space-time" functionally means. I suppose, fundamentally the problem is that when you build a geometric shape in one curvature, it will inevitably be distorted when it's brought to a different curvature.


But it's quite interesting - to consider that geometric shapes or regions of real space might have more volume than they externally appear to hold, depending on the properties of the space within them. It's basically the same thing as TARDIS... or Bag of Holding, for that matter. Just in much smaller extent in most situations.

[Phantom Hoover]

I saw a pretty cool demonstration on youtube of a device that was rigged to demonstrate the effect of curvature on world-lines on a 2D surface. Aha, found it:

[watsisname]

massless wire, rigid rod (in this case a rigid shell), frictionless surface, ideal gas, and perfect black body...

It's okay, as long as they don't take away my spherical cows.  They're so cute and lovable and maximally efficient methane farms.

So I guess instead of providing basis for an experiment, it can still be an useful way to visualize the effects of positive curvature on space-time, and what the phrase "curved space-time" functionally means. I suppose, fundamentally the problem is that when you build a geometric shape in one curvature, it will inevitably be distorted when it's brought to a different curvature.

That it is!  Another consequence of curvature we can visualize fairly easily is the separation or convergence of initially parallel light rays (this is a good way to test the average density of the universe).  Or we can consider the reverse situation to what we were looking at earlier -- instead of moving an object through changes in curvature, have changes of curvature move through the object.  Those are gravitational waves!  Or just watch what Phantom Hoover posted; that guy got it right.  (Thanks Phantom, that was cool!)

I should point out real quick that spatial volume can become a really tricky thing in extreme space-time curvature, even in the theoretical (instead of just practical) sense.  For instance, the surface area of an event horizon has a unique interpretation that everyone can agree on.  Not so for the volume enclosed by it!  That depends on the choice of coordinates.  One could even say that the volume within an event horizon is zero!

So, 3-volume is not an invariant quantity in GR.  There is however a 4-volume, akin to 'hypervolume', which is.

[Phantom Hoover]

One particularly bad problem with the standard 'bowling ball on a trampoline' illustration is that the trampoline's curvature is actually negative.

[watsisname]

No it isn't, it's positive.  Just check whether initially parallel rays come together or separate, or whether the sum of angles in a triangle is more or less than 180 degrees.

Curvature isn't "whether it bends up or down", like "bending downward is negative curvature".  If that was true then you can turn negative curvature into positive curvature just by inverting your perspective.  But that would be ridiculous, curvature should not depend on your point of view.  The curvature of a surface is an intrinsic property of that surface and does not depend on how it is embedded in a higher dimensional space.  That's one of the failings of our minds.  It's pretty hard to imagine a surface, or a manifold, without imagining what it is embedded in.  I can't visualize a 2D manifold without imagining 3D space around it.

So what does negative curvature look like?  Negative curvature is like the region immediately around a saddle point.  In one direction, it bends up, in the other direction it bends down.  Now try extending that, so that the neighborhood of every point on the surface has that property.  I don't know about you, but I can't visualize such a thing.  My brain can only extend it so far and then simply gives up.

Positive curvature on the other hand is extremely easy to visualize.  It looks like the surface of a sphere.  In fact, the surface of a sphere has the same positive curvature everywhere.  Try to draw two parallel lines on the surface of a sphere.  If you start them out parallel, then they will come together and intersect on the opposite side of the sphere.  Those are longitude lines!  They are also perfectly straight lines.  You can walk all the way around one without ever turning your step.  What about latitude lines, you might ask?  They are parallel everywhere, aren't they?  Yes, but they are in general not straight.   If you follow any longitude latitude line other than the equator, you will have to turn slightly at every step, or very sharply if you're near the pole.

And that's the idea behind geodesics -- locally straight lines.  Longitude lines are geodesics on the positively curved surface of the Earth.  Latitude lines are not geodesics.  You must accelerate (constantly change your direction) to follow one.  The secret to general relativity is to recognize that all objects that are in freefall follow geodesics.  What gravitation does is make those straight lines appear curved, because the underlying geometry of space-time is curved.  That curvature arises because of the presence of some mass nearby.  "Mass tells geometry how to curve, curvature tells matter how to move." 

The cover art and first chapter of the Gravitation textbook is all about this idea and is probably the most awesome description of it I've ever seen.  Far better than any rubber sheet analogy.  The rubber sheet analogy is only good if you recognize it only for its description of the effect of curvature on geodesics.  If you think of it as the curving of space, then it fails and only causes more confusion.  If you ask what causes objects to follow those paths ("don't you need gravity to explain the paths") then it fails again.  Ignore gravity in the rubber sheet analogy.  Just ask what happens to straight lines when you introduce that warp.  And remember that the rubber sheet is not analogous to space, it is analogous to space-time.

edit:  Mixed up lats and longs.  I do that sometimes.

[Phantom Hoover]

The curvature of the trampoline is definitely negative over a significant part of its area, though not right next to the bowling ball. And you can totally visualise negatively-curved surfaces, what you can't have is a surface that has constant negative curvature and has infinite extent in all directions.

[watsisname]

I'm still not sure I see any place on it with negative curvature, but perhaps it is my visualization of the rubber sheet depression itself that isn't right.

Added:  Ah, you are right, I was wrong.   My problem was thinking of the depression as being well fit by a spherical surface over most of the region.  Obviously that's not true everywhere, especially if the depression is more funnel shaped, like a tornado.  So it is possible to imagine broad regions of negative curvature, but as you point out with Hilbert's Theorem, not globally with constant negative curvature.  That's neat!

[Herra Tohtori]

The curvature of a funnel that most accurately simulates Newtonian gravitation would be one where elevation is relative to the square of distance, making it the approximate shape of the gravitational potential. That makes a ball rolling on the surface experience a component of gravitation that is aligned with the gradient of the surface, and that sort of makes the effect analogous to Newtonian gravity. Hence, good way to visualize effects of gravity, but I do agree it has fundamental problems with explaining the nature of gravity in general relativity.

One of which is that the curvature of the 2D shape that reproduces effects that resemble Newtonian gravity best is not necessarily analogous to the curvature of a planar sheet of 4-space.


Visualizing negatively curved surfaces is easy (at least objects with locally negative curvature). Negatively curved space is a bit of a different matter, though...


As said before I tend to visualize the effect of curvature instead of exactly trying to figure out how to make a 2D analogy of it or anything of the sort. Basically what I do is imagine that I can isolate an ideal geometric shape of the space, and then measure its properties, and compare those properties to what that shape should have in flat, euclidian geometry.

In a constant geometry you can use a triangle, measure the angles in the corners, and sum it together. But on a surface, it's sufficient to draw one triangle and that's it. In space, that only measures the curvature in the plane of the triangle, and it only returns the average curvature inside the triangle.

To get better results, you'd need a 3D shape instead of a 2D shape, which would allow you to measure the curvature in all three spatial dimensions at the same time.


Another solution, which works better for me, is to take a geometric shape of 4-space again (sphere works best because symmetry and simple definition), and compare its properties to its ideal, euclidian definition. For example, is the relationship between the diameter and circumference exactly pi?

However that again means we have to measure the diameter of the sphere in at least the three cardinal directions, and likewise the circumference, because we're not exactly using the entire sphere but instead circles aligned on the three spatial dimensions.

What's more elegant in my mind (and which spawned the question about whether making a measurement based on it is practical) is to look at the volume of the sphere.


Ideally, in euclidian space, the volume of any sphere regardless of its size is 4/3 pi r³.

The curvature of space affects that, and moreover the relationship between volume and radius becomes non-constant.

On my mind, I visualize positive curvature of space as something that causes the volume of the sphere to become more than 4/3 pi r³.

By contrast, negative curvature would reduce the volume compared to Euclidian norm.


The problem of trying to do an analogy of negative spatial curvature on the rubber sheet representing a world sheet is that the shape at which a region of rubber sheet has the smallest surface area is always flat.

In order to change the curvature of the sheet to any direction, positive or negative, you have to displace it from being flat. That means you're stretching the sheet, which basically always increases its surface area.



In my opinion, that more or less makes the "stretched" rubber sheet closer to being analogous to positive curvature of space, irrespective of the actual shape of the deformation that a weight causes on the sheet.

To have any hope of creating a visualization of negative curvature on a 2D sheet analogy is if you have some way of contracting the sheet instead of stretching it. You can do this by pre-stretching a sheet of rubber or spandex or whatever flexible material, then draw a co-ordinate grid on it.

You can't really visualize negative curvature as analogous to any deformation of the plane, but what you can do is to create an area where you reduce the stress of the sheet, by for example pinching it and then attaching tape on the underside to keep it on the "un-stretched" position. What you're doing is analogous to negative energy causing the space to "contract" from the euclidian standard.


Then you can basically look at the world lines passing the disturbance, and think of them as the routes that a photon would take when it passes that location.

If world lines curve towards the center of disturbance, you have positive curvature.

If the world line curves away from the disturbance, you have negative curvature. This doesn't necessarily depend on the actual shape of the rubber sheet in 3D space. Just concentrate on the coordinate grid, and see whether it contracts or expands relative to what we define as the "standard".



In this analogy, the "pre-stretched" sheet of rubber represents actually un-disturbed euclidian space with zero energy tensor. The "pinching" (or reducing the stress) is analogous to negative energy tensor, while stretching the sheet is analogous to positive energy tensor. It's still a bad analogy in the sense that we need to pre-stress the sheet in order to be able to reverse it by "negative" energy (pinching), but it seems necessary to avoid crumpling the sheet.

Technically it could be possible to use a fabric that doesn't need to be pre-stretched. A very good visualization tool could be material that expands or contracts significantly based on small temperature changes, which could allow creating a "cold spot" that causes the fabric to contract, and a "hot spot" to cause it to expand, and that would distort the world-lines (or co-ordinates) on the sheet so as to visualize the stress tensor's effects on space.


Oh and by the way, here's a fun little detail. A photon has the same acceleration due to gravity that we measure for everything else - approximately 9.81 m/s². They just travel so fast that they appear to travel straight, but they do fall. If you're doing very precise optical measurements you have to account for that - and if you're doing vertical beams of significant length, and you're measuring wavelength, you might also need to take red-shift or blue-shift into account (although this is not a practical concern, I can imagine it might be necessary in some experiments requiring extreme accuracy).

But in a falling elevator you would measure a photon traveling perfectly straight. As you would in a spaceship on orbit, or in any kind of free-fall situation.


In other words, curvature of space and time seems different depending on your reference frame. Photons traveling vertically separated would still diverge, though...


And every time I get into thinking this stuff I feel like my head's going to implode.

[watsisname]

A photon has the same acceleration due to gravity that we measure for everything else - approximately 9.81 m/s2. They just travel so fast that they appear to travel straight, but they do fall.

This is great for Newtonian thinking, but it actually doesn't work very well in General.  Oh, photons are affected by gravity, certainly, and their paths are deflected by it.  The equivalence principle teaches us this much.  But if you treat a photon as something moving at c but accelerating with g, you won't get the right answers.  E.g. for gravitational lensing the Newtonian calculation underestimates the deflection angle by a factor of 2.  It also fails spectacularly for describing what happens in extreme curvature.   You really need the geometric theory of gravitation to get the correct results.


So, for extreme curvature, consider a black hole like always.

If we think of photons in the Newtonian perspective, then a photon emitted radially outward from just inside the horizon will move outward, slow down, turn around, and come right back down.  Sounds correct, doesn't it?  It's just like the orbit of an object with less than the escape speed.  But that's not what happens!  If it was within the horizon, then it never gains altitude.  It gets dragged farther inward from the instant it was emitted.  A photon emitted from just barely inside the horizon does not and cannot make it to the horizon.

Similarly, photons emitted from a singularity don't just go up and come back down.  They don't leave the singularity at all.  You would not see a singularity even if you could put your nose right up next to it.

If a photon is emitted radially outward right on the horizon, then it just stays there, apparently motionless.  (This is actually not quite right either, because a photon is not infinitesimal -- it has wave-like properties and so it gets smeared out.  Classically though, treating a photon as a point, then it just hovers there on the horizon.)  We can say that the photon is moving at c, but space there is also moving inward at c.  Or, we can say that space-time is so strongly curved there that null geodesics directed normal to the horizon appear as points in space.  An event horizon is a null surface.

In this sense, an effective visualization of a gravitational well (especially that of a black hole) is to think of it not as being like a warped sheet, but like a waterfall.  Susskind gives a great presentation of this analogy in his lecture on The Black Hole War, see here about 8 minutes in.  Highly recommended.

So the correct description of motion in GR is that the photon, and everything else which is in free-fall, has exactly zero acceleration.  (This is probably the most awkward but most important principle in all of GR!)  There is no acceleration and so the paths are all geodesics, with the photon being the precise case of a null geodesic (ds²=0) because it moves at the speed of light.  The geodesics look curved because the geometry is curved. 

What acceleration actually is, is your departure from a geodesic.  Sitting or standing here on the surface of the Earth, you are accelerating!  Jump off of a cliff, that's when you cease to accelerate (though note quite, because air resistance).  Or more happily, go into orbit!  GR teaches us to totally rethink what acceleration actually means.


Some more detail on ds²=0:
ds is how we measure distances in space-time, aka a metric.
ds²=0 means the sum of the squares of the space and time distances is zero, which sounds really weird until you remember space-time is Minkowskian instead of Euclidean, so the definition of distance is different.  In Euclidian 4D space, distance is a logical extension of Pythagoras' Theorem: 
dx² = dx₀² + dx₁² + dx₂² + dx₃²,
where dx is the differential distance, and the subscripts 0 through 3 designate the coordinate axes.

But in Minkowskian space, time has opposite sign as space, so the measure of distance is really
dx² = dx₀² - dx₁² - dx₂² - dx₃²,
where dx₀ represents the time coordinate and we use +--- sign convention (time-like convention).  We could write it with the signs the other way (positive for space coordinates and negative for time) for a space-like convention.  Doesn't matter which we use as long as we are consistent.  We also generally write the metric as ds instead of dx, for space-time interval:
ds² = c²dt² - dx_r², with r being a dummy index to sum over the three spatial dimensions.

[Herra Tohtori]

A photon has the same acceleration due to gravity that we measure for everything else - approximately 9.81 m/s2. They just travel so fast that they appear to travel straight, but they do fall.

This is great for Newtonian thinking, but it actually doesn't work very well in General.  Oh, photons are affected by gravity, certainly, and their paths are deflected by it.  The equivalence principle teaches us this much.

Well, yeah. But the equivalency principle is a really, really awesome* thing and worth mentioning every time it's even remotely relevant!

The part about extreme curvatures is quite interesting - space dimensions become time-like inside event horizon and the only way to progress is towards singularity etc - but for practical purposes it's almost always necessary to convert things back into the "Newtonian" thinking, right? Especially in non-extreme cases, like consideration of how photons behave here on the surface of the Earth.


Oh, and I was being a bit inaccurate in my previous post. The curvature of space-time can of course be measured in free-fall by the divergence of parallel light rays, or tidal forces, or frame-dragging. The equivalency principle equalizes acceleration with homogenous gravity field (not that gravity is a field in GR anyway, but just to use the familiar term).

I'm not entirely sure what kind of curvature would be required to produce an ideal, homogenous "gravity field". In fact, technically it should be impossible to detect at all, right?


Although I'm not sure if what you wrote actually means that we would be able to detect an universe-wide homogenous gravity field based on photons acting differently than they should?



*Things that equivalency principle makes awesome: Photons have no (invariant) mass, but light - as a collection of photons - still has lots of properties that are equivalent to mass...

[watsisname]

Well, yeah. But the equivalency principle is a really, really awesome* thing and worth mentioning every time it's even remotely relevant!

Oh yes.  Equivalence principle is crucial, as we'll discover shortly below.

As for reverting back to Newtonian for practical purposes, you definitely do if you're only interested in low approximation or the weak field limit.  If the difference in your result due to GR's effects is greater than your intended precision, then you probably want to use GR.  Otherwise, using GR is a huge waste of computational time for no practical benefit.

The curvature of space-time can of course be measured in free-fall by the divergence of parallel light rays, or tidal forces, or frame-dragging. The equivalency principle equalizes acceleration with homogenous gravity field (not that gravity is a field in GR anyway, but just to use the familiar term).

Exactly.  Equivalence principle assumes that you take a reference frame that is sufficiently limited that the curvature becomes negligible over it.  It's the same idea as the surface of a sphere appearing flat if you are sufficiently small / close to it.  So the equivalence principle makes a real gravitational field look like a homogenous one, so that tidal forces, etc, vanish. 

Why would we care about this?  Why would we want to intentionally limit our perspective (I remember Luis had trouble with this idea earlier)?  Because it acts as a stepping stone for the development of general relativity from special.  It teaches us that we need to rethink what we thought we knew about how to do gravitational physics.  Furthermore, it is a very powerful tool for us later on, in distinguishing whether statements in relativity can or cannot be true.

In special we learn how to develop the Lorentz transformations once we recognize the constancy of the speed of light (and more fundamentally, the laws of physics) for all observers.  We learn of the Minkowskian nature of space-time and the associated coordinate transformations and tensors.  But we have a pretty hard time bringing gravitation into special relativity.  In fact, they are not compatible.  There's a leap of insight that we need, and that leap is facilitated by recognizing that the effects of a uniform gravitational field are indistinguishable from a uniformly accelerated reference frame (equivalence principle).  How can we explain this?  What we find is that a geometric interpretation of gravitation naturally provides the answer.  Indeed, the effect of gravitational redshift implies curvature!  (See Schild, 1960, 62, 67).

We'll now discover the power of the equivalence principle with your question about detection/existence of homogenous fields:

I'm not entirely sure what kind of curvature would be required to produce an ideal, homogenous "gravity field". In fact, technically it should be impossible to detect at all, right?

OK, so what kind of curvature would produce a homogenous gravity field?  Remember what we discussed earlier about the development of the curvature tensor from geodetic deviation.  Let's suppose we have some region of space-time operating under what we will call a homogenous gravitational field.  We will put two test masses in it, let them fall, and measure their separations over time.  What will we see?  Nothing.  No separation.  There are no tidal forces.  So we conclude that there is no curvature.  And no curvature means no gravitational field! 

Equivalently, we might notice that a simple coordinate transformation will remove the effect of uniform acceleration.  And if we can do that, we can again conclude that there is no gravity.

Does that sound nuts, or no?  Consider this instead.  Suppose we have a little box in orbit above the Earth.  Or in free-fall toward the Earth, doesn't matter, they're equivalent for our purposes.  There are two test particles in the box.  We watch their motions.  We see no geodetic separation.  As far as we can tell, the two particles just move in straight lines.  So we conclude there's no curvature, thus no gravity?  Earth's grav field doesn't exist?  No, we must remember what the equivalence principle tells us: if you use a frame of reference so limited that curvature appears flat, then you see no geodetic separation, no tidal forces, no gravity.  That's the key!  If you observe no separation of geodesics, then you conclude the space-time is flat to within precision of measurement and extent of your reference frame.

So, extend the reference frame.  Or improve our precision.  Then, for a real field, we'll start to detect those tidal forces.  We'll conclude that there is a gravitational field there.  Thus we can know that a gravitational field exists if, and only if, we can detect curvature.  We cannot detect curvature if things accelerate uniformly -- that's the equivalence principle.  Equivalently, we know it exists if we cannot globally remove it by a simple coordinate transformation.  I.e. we cannot remove the effects of the Earth's field everywhere just by changing coordinates.

Now, return to the concept of a globally homogenous field.  If we suppose the universe is permeated by such a field, then we find that this is fundamentally untestable.  There would be absolutely no experiment that could detect its effects.  No reference frame large enough, no precision high enough.

[Added]  Just in case there's any confusion between this and dark energy, no this does not mean dark energy is untestable.  I'm leaving out a little bit of detail, which is the importance of symmetries (in this case, isotropy).  Dark energy is homogenous in its energy distribution, but it is also isotropic.  It has curvature, and the same curvature everywhere.  It therefore affects the expansion rate of the universe.  The homogenous field Herra and I speak of above, which can be considered as being produced from a flat, infinitely extended mass distribution, is not isotropic.  It does not produce curvature, and it does not affect the expansion rate.
[/added]

It is also worthwhile to point out that this homogenous field is inconsistent with the formulation of general relativity.  There is no solution to the equations that preserves all the conditions of such a field.  (See Section 7.4 of Crowell's text on GR.)

Although I'm not sure if what you wrote actually means that we would be able to detect an universe-wide homogenous gravity field based on photons acting differently than they should?

"Acting differently than they should?"  How are photons supposed to act?  In GR, photons follow null geodesics.  How would a globally homogenous field cause them to do otherwise?

[Phantom Hoover]

Something I recently realised is that the best answer for the whole question of 'if you travel back in time do you end up floating in space because the earth moved away?' is that your time machine just stays in its inertial reference frame and starts orbiting the core of the earth.

[Beskargam]

Reading through this, I feel like a first year undergrad again; reading through science papers, recognizing a word or two, needing a dictionary for the rest, and lacking context for it all. So much to learn, so little time....

[Mongoose]

I at least know the general basics of some of what's being discussed, but I'm still left pulling a complete Keanu "woooooaaaaah" with most of it.

(Also I'm strongly reminded of why I never went any further with physics.  I loved most of the conceptual stuff, but the math that arose as you delved into it just destroyed me.)

[watsisname]

Yeah, the mathematics becomes kind of brickwallish in the difficulty curve, especially once you hit tensors.  I'm learning my way through tensors through some texts, but actually getting to the point of being comfortable with using them is a slow, arduous endeavor.  I'd like to get there before grad school, but we'll see.

@Herra, I was reflecting on what you were asking about the behavior of photons earlier, and wondered if you were thinking about what photons do in a uniform field as used in the equivalence principle in the following sense:  In a uniform field like that, we would not see the photons bend, right?  The field has no effect that we could measure.  But in a uniformly accelerated frame, we do see them bend.  Let me know if that's close to right and if this helps clarify or not:

What we mean by a uniformly accelerated frame is like having a uniform field permeating that frame, but it is not the same as a frame which is freely falling within a uniform field.  The uniformly accelerated frame is like being in a spaceship which is firing its engines.  It is also the same as being in a ship sitting on the surface of a planet.  If the acceleration of the ship in scenario (a) is the same as the surface gravity of the planet in scenario (b), then there's no way to distinguish the two. 

The other example of the equivalence principle is for freefall within a gravitational field, vs. not being subjected to a field at all.  You again can't tell the difference.

But, that's not quite right because there is of course an exception.  The exception is if your frame of reference is sufficiently large, and precision of measurement sufficiently fine, that you can detect tidal forces.  A real gravitational field produces tidal forces, whereas a uniformly accelerated frame (the rocket firing its engines, or a universally uniform field) does not produce them.  This is how you detect gravity.  You stop trying to care about what's going on around you.  Ignore the temptation to look out the gedanken experiment's window to see if there's a gravitational mass nearby.   Don't try to do physics with a global perspective.  Physics is simple only when examined locally.  Locally, you detect gravitational fields because they produce tidal forces.  You cannot distinguish gravity through a uniform acceleration.

(Incidentally, this has a lot to do with why we sometimes call gravity a fictitious force, just like Coriolis or centrifugal force.  In the local frame, the effect of these forces can be made to vanish by a coordinate transformation.  This happens any time you have a force which is proportional to mass, so that all test particles behave the same way.  Thinking of gravity as 'fictitious' is pretty mind bending.)