Undergrad What is the Principle of Equivalence and how was it determined?

[loislane]

So how do you call the acceleration(and associated forces) related to noninertial frames that is indeed measured by accelerometers when making the distinction with inertial frames (defined as those in which an accelerometer reads zero)?

 

[PeterDonis]

inertial frames (defined as those in which an accelerometer reads zero)?

No, that's not how "inertial frames" are defined. Inertial frames are defined as those in which an accelerometer at rest in the frame reads zero. For example, an accelerometer attached to you, floating inside a spaceship in deep space, will read zero, because you are at rest in an inertial frame. If the spaceship's engines fire, an accelerometer attached to the spaceship will no longer read zero, but that doesn't mean the spaceship disappears from the inertial frame (in which you are still at rest, with your accelerometer still reading zero).

So how do you call the acceleration(and associated forces) related to noninertial frames that is indeed measured by accelerometers

I have no idea what forces you are referring to here. But there have already been plenty of specific examples given in this thread of forces that can be measured by accelerometers, vs. "forces" that can't.

 

[stevendaryl]

this looks mainly semantic but I would like to get clear on this, accelerometers detect 1 g on the Earth's surface, I guess it is up to you to decide if that is due to the Earth's gravitational field, or to some other more contrived cause that avoids calling it gravity, or inertial force due to the Earth's being a noninertial frame, or to reactions forces, it all looks a bit arbitrary to me.

This is a very important point for you to get clear about. The accelerometer is NOT measuring the force of gravity. If you drop an accelerometer, the reading goes to zero (until it hits the ground). But the force of gravity isn't zero. So the accelerometer isn't measuring the force of gravity. That's that. It's not a matter of opinion, it doesn't depend on how you look at, it's not arbitrary.

Mathematically, we can state things this way:

Let \vec{F}_g be the force of gravity. Let \vec{F}_{up} be the force pushing up on the accelerometer. Let M be the mass of the accelerometer.

Then the total force acting on the accelerometer is \vec{F}_{total} = \vec{F}_g + \vec{F}_{up}. The coordinate acceleration (relative to an Earth-centered coordinate system) is given by \vec{a} = \vec{F}_{total}/M. The accelerometer does NOT measure this acceleration. Instead, it only measures the non-gravitation part:

\vec{a}_{non-grav} = \vec{F}_{up}/M

That's what people mean when they say that an accelerometer doesn't measure gravity. It measures only the non-gravitational part of the acceleration.

Now, getting back to your claim that an accelerometer measures the acceleration due to gravity: It doesn't, in general. However, there is a special case in which it does: If an object is at REST in the gravitational field, that means:

\vec{a} = 0 \Rightarrow \vec{F}_{total} = 0
\Rightarrow \vec{F}_g + \vec{F}_{up} = 0
\Rightarrow \vec{F}_{up} = - \vec{F}_g
\Rightarrow \vec{a}_{non-grav} = - \vec{F}_{g}/M
\Rightarrow a_{non-grav} = g (because a_g = -M g)

So it's always the case that an accelerometer measures the non-gravitational part of the acceleration. But if we choose the nongravitational part so that the total acceleration is zero, then (and only then) will an accelerometer measure the acceleration due to gravity.

Saying that it's measuring gravity is sort of like saying that you're weighing a pig when you're actually weighing a rock that happens to weigh the same as the pig.

 

[Dale]

this looks mainly semantic but I would like to get clear on this, accelerometers detect 1 g on the Earth's surface, I guess it is up to you to decide if that is due to the Earth's gravitational field, or to some other more contrived cause that avoids calling it gravity, or inertial force due to the Earth's being a noninertial frame, or to reactions forces, it all looks a bit arbitrary to me.

An accelerometer at rest on the Earth detects 1 g UPWARDS. So it definitely is not detecting gravity which is downwards. This is not a semantic difference. Physically the upwards pointing contact force is detected by the accelerometer. Physically the downwards pointing gravitational force is not detected by the accelerometer.

The only semantic component is in deciding if you even want to use the word "force" to describe gravity and inertial forces.

 

[Dale]

But if we choose the nongravitational part so that the total acceleration is zero, then (and only then) will an accelerometer measure the acceleration due to gravity.

Of course, as your math showed, you mean "measure the opposite of the acceleration due to gravity"

 

[loislane]

No, that's not how "inertial frames" are defined. Inertial frames are defined as those in which an accelerometer at rest in the frame reads zero.

Ok, that is what I was defining, I thought the "at rest" bit was implied by the way accelerators work(basically a damped mass on a spring like a dynamometer that is displaced proportionally to any force on the device casing).

I have no idea what forces you are referring to here.

Well, for instance tidal forces on free-falling objects(think Shoemaker-Levy 9 comet) . Are those not associated to noninertial frames and detectable by accelerometers?, but then I guess not since from your posts I infer you consider such objects in free-fall as being at rest in an inertial frame(accelerators read zero right?).

 

[stevendaryl]

Well, for instance tidal forces on free-falling objects(think Shoemaker-Levy 9 comet) . Are those not associated to noninertial frames and detectable by accelerometers?

No, and no. Tidal forces are measurable, physical quantities (like proper acceleration) that don't depend on what frame or coordinate system they are measured in. Basically, tidal forces are due to the gravitational field being different in one place than in another. If you are closer to the Earth, then gravity pulls harder than if you are farther away from the Earth. Although the force of gravity isn't directly measurable at a point, differences from point to point are measurable.

Imagine having a "dumbbell" with two masses connect by a strong bar. If you drop the dumbbell, so that it is allowed to fall freely, then the force of gravity appears to vanish; the dumbbell would appear to be floating in gravity-free space. But if the two ends are accelerated DIFFERENTLY by gravity, then the bar will be stressed--either compressed or stretched slightly by the differences. That's a measurable effect, and that's the tidal force. It's called that, because it's the source of the tides on the Earth. Both the center of the Earth and the oceans are pulled by the Moon's gravity. But the oceans, being slightly closer to the Moon (when the Moon is overhead), are pulled a little more than the ground underneath the ocean. That difference is the tidal force, and causes the oceans to rise relative to the rest of the Earth.

No, an accelerometer does not measure tidal forces, although a pair of accelerometers connected by a rigid bar could (in theory) measure it by comparing the accelerations of the two ends.

 

[loislane]

No, and no. Tidal forces are measurable, physical quantities (like proper acceleration) that don't depend on what frame or coordinate system they are measured in. Basically, tidal forces are due to the gravitational field being different in one place than in another. If you are closer to the Earth, then gravity pulls harder than if you are farther away from the Earth. Although the force of gravity isn't directly measurable at a point, differences from point to point are measurable.

Imagine having a "dumbbell" with two masses connect by a strong bar. If you drop the dumbbell, so that it is allowed to fall freely, then the force of gravity appears to vanish; the dumbbell would appear to be floating in gravity-free space. But if the two ends are accelerated DIFFERENTLY by gravity, then the bar will be stressed--either compressed or stretched slightly by the differences. That's a measurable effect, and that's the tidal force. It's called that, because it's the source of the tides on the Earth. Both the center of the Earth and the oceans are pulled by the Moon's gravity. But the oceans, being slightly closer to the Moon (when the Moon is overhead), are pulled a little more than the ground underneath the ocean. That difference is the tidal force, and causes the oceans to rise relative to the rest of the Earth.

No, an accelerometer does not measure tidal forces, although a pair of accelerometers connected by a rigid bar could (in theory) measure it by comparing the accelerations of the two ends.

I actually used the plural(accelerometers), but anyway could you explain how is it that a pair of accelerometers attached to different points of a free-falling object can give different readings if both are supposed to read zero on account of being on an inertial frame?

Also in a more mathematical view, ultimately according to GR tidal forces are what the Riemann curvature tensor represents, and as far as I know a nonzero curvature cannot be made to vanish at a point(unlike the christoffel symbols), so the tidal force can be measured at one point theoretically so it would suffice one accelerometer.

Of course you might mean that no real free-falling object is actually inertial, only some infinitesimal idealized test object is, but then you cannot attach an accelerometer to such a test object to check it.

 

[Dale]

I actually used the plural(accelerometers), but anyway could you explain how is it that a pair of accelerometers attached to different points of a free-falling object can give different readings if both are supposed to read zero on account of being on an inertial frame?

There is no such thing as an inertial frame if there is significant curvature to spacetime. Tidal forces indicate spacetime curvature.

 

[Dale]

Also in a more mathematical view, ultimately according to GR tidal forces are what the Riemann curvature tensor represents, and as far as I know a nonzero curvature cannot be made to vanish at a point(unlike the christoffel symbols), so the tidal force can be measured at one point theoretically so it would suffice one accelerometer.

Tidal forces have a length scale, so they can be made arbitrarily small by making your accelerometer arbitrarily small.

 

[stevendaryl]

I actually used the plural(accelerometers), but anyway could you explain how is it that a pair of accelerometers attached to different points of a free-falling object can give different readings if both are supposed to read zero on account of being on an inertial frame

Please don't use the phrase "being on an inertial frame". It's better to say "traveling inertially" or "in freefall". Every object is in every frame, so it doesn't make sense to say that something is in an inertial frame or in a noninertial frame. You could say that an object is at REST in some particular frame.

So, yes, if an accelerometer is in freefall (that means it has no non-gravitational forces acting on it) then it will measure zero--no acceleration. However, if you connect two accelerometers by a steel bar, then they are no longer in freefall. The bar itself exerts a nongravitational force. If one accelerometer is closer to the Earth than the other, then the closer one is pulled up by the bar (and so measures a nonzero upward acceleration) and the far one is pulled down by the bar (and so measures a nonzero downward acceleration). So the accelerations are different (and nonzero).

Also in a more mathematical view, ultimately according to GR tidal forces are what the Riemann curvature tensor represents, and as far as I know a nonzero curvature cannot be made to vanish at a point(unlike the christoffel symbols), so the tidal force can be measured at one point theoretically so it would suffice one accelerometer.

The definition of the Riemann curvature tensor is, exactly as I said, in terms of differences in freefall accelerations for nearby point masses. To actually get the curvature at a single point means taking the limit as the distance between the point masses goes to zero (basically, it's like a derivative of the acceleration due to gravity). In terms of my two dumbbells, you could make a device to measure (approximately) curvature at a point by taking two accelerometers at the end of a bar, and taking the limit as the length of the bar goes to zero (to get the full tensor, you would need to have several such devices oriented in different directions). A single accelerometer cannot measure curvature.

 

[loislane]

There is no such thing as an inertial frame if there is significant curvature to spacetime. Tidal forces indicate spacetime curvature.

Tidal forces have a length scale, so they can be made arbitrarily small by making your accelerometer arbitrarily small.

As I wrote in the not quoted last paragraph of my post there is no testable accelerometer that is arbitrarily small, so assertions of that type are out of the scope of physical science.

So, yes, if an accelerometer is in freefall (that means it has no non-gravitational forces acting on it) then it will measure zero--no acceleration. However, if you connect two accelerometers by a steel bar, then they are no longer in freefall. The bar itself exerts a nongravitational force. If one accelerometer is closer to the Earth than the other, then the closer one is pulled up by the bar (and so measures a nonzero upward acceleration) and the far one is pulled down by the bar (and so measures a nonzero downward acceleration). So the accelerations are different (and nonzero).

Excuse me but a physical accelerometer has every part of its structure (damped mass in a spring and encasing) connected in an equivalent way as your description of the two accelerometers and therefore each individual constituents of an accelerometer wouldn't be in free fall by your own explanation but the accelerometer as a whole itself wouldn, so I'm not sure if you are talking about physics or about some idealized metaphysics when making that distinction.
On the other hand as Dalespam says if there is significant curvature(that is, enough to produce tidal forces between the parts of an object) then it can't be called an object at rest in an inertial frame:

There is no such thing as an inertial frame

The definition of the Riemann curvature tensor is, exactly as I said, in terms of differences in freefall accelerations for nearby point masses. To actually get the curvature at a single point means taking the limit as the distance between the point masses goes to zero (basically, it's like a derivative of the acceleration due to gravity). In terms of my two dumbbells, you could make a device to measure (approximately) curvature at a point by taking two accelerometers at the end of a bar, and taking the limit as the length of the bar goes to zero (to get the full tensor, you would need to have several such devices oriented in different directions). A single accelerometer cannot measure curvature.

Following your reasoning, and Dalespam's about taking the limit I can agree that theoretically(since as explained above there is no way to check it directly in an idealized "point-accelerometer") a single measure would give just a component of the Riemann curvature tensor one would need more measurements in different orientations, but we are all again saying that since the curvature tensor for this setup with tidal forces doesn't vanish at any point, not even on the infinitesimal limit of lenth, one cannot talk about a free-falling object orbiting the Earth as at rest in an inertial frame, not even at the infinitesimal limit of considering the object a point mass.

 

[stevendaryl]

Excuse me but a physical accelerometer has every part of its structure (damped mass in a spring and encasing) connected in an equivalent way as your description of the two accelerometers and therefore each individual constituents of an accelerometer would be in free fall by your own explanation.

The pieces would be in freefall if there were no non-gravitational forces acting on them. The rod connecting the two accelerometers IS a non-gravitational force. So when the rod is present, the accelerometers are NOT in freefall.

You can think of an accelerometer as just a cubic box, and in the center of the box is a mass--a metal ball, for instance--connected to the 6 walls by 6 identical springs. You measure acceleration by measuring the displacement of the ball from the center.

Now, what I'm proposing is that you attack a metal rod to the outside of the box, perpendicular to one of the sides. The accelerometer will read a nonzero value if you pull or push on the rod.

Now, you attach a second accelerometer to the other end of the rod. Again, it will measure a nonzero value if you pull or push on the rod.

If there is no gravity, or if the gravitational force is constant and the accelerometers are in freefall, then both accelerometers will read zero. But near the Earth, the gravitational field is nonuniform--the gravitational pull is weaker the farther you are from the center of the Earth. So if one accelerometer is above the other, then the bottom one will be pulled stronger by gravity.

If you have two accelerometers dropping in a nonuniform gravitational field, connected by a rigid rod, then let

g_U = the acceleration due to gravity at the upper box
g_L = the acceleration due to gravity at the lower box
M = the mass of each box
T = the tension in the rod (let's ignore the mass of the rod for simplicity; assume it is a strong, thin wire)

Then T = m \dfrac{g_L - g_U}{2}

(note: if the gravity is uniform, then g_U = g_L so T=0)

Accelerometers only measure the non-gravitational part of the acceleration, which is due to the tension. So:

The lower box will be pulled up by the tension, so will measure an acceleration of + \dfrac{T}{M}.
The upper box will be pulled down by the tension, so will measure an acceleration of - \dfrac{T}{M}.

The accelerations measured by the two boxes are different (they have opposite signs), so the pair detects nonzero tidal forces.

...we are all again saying that since the curvature tensor for this setup with tidal forces doesn't vanish at any point, not even on the infinitesimal limit of lenth, one cannot talk about a free-falling object orbiting the Earth as at rest in an inertial frame, not even at the infinitesimal limit of considering the object a point mass.

It depends on exactly what you are measuring. Some experiments are sensitive to the curvature, and some are not.

There is an epsilon-delta type statement of the equivalence principle that holds even in non-uniform gravitational fields:

You tell me the accuracy \epsilon_l and \epsilon_t that you will be measuring distances and times.
I tell you a length scale \delta L and a time scale \delta T such that if you confine your measurements to a box with sides \delta L or smaller, and you perform an experiment that takes less than time \delta T, you will be unable to distinguish between the case in which that box is in a gravitational field and the case in which the box is in outer space, far from any gravitational sources.

Computing curvature at a point requires arbitrarily accurate measurements. If the measurements have a limited accuracy, then curvature is not measurable.

 

[loislane]

The pieces would be in freefall if there were no non-gravitational forces acting on them. The rod connecting the two accelerometers IS a non-gravitational force. So when the rod is present, the accelerometers are NOT in freefall.

Sorry, I wrote that backwards, it is edited now.

 

[loislane]

Computing curvature at a point requires arbitrarily accurate measurements. If the measurements have a limited accuracy, then curvature is not measurable.

Agreed: limited accuracy plus curvature small enough lead to no direct measurability. That is why I was speaking theoretically,i.e. in principle, meaning: if Riemann geometry is right, which I wouldn't have much doubt about. Also I would put Riemannian geometry before a heuristic physical principle.

 

[Dale]

As I wrote in the not quoted last paragraph of my post there is no testable accelerometer that is arbitrarily small, so assertions of that type are out of the scope of physical science.

I think that you are confusing the term "arbitrarily small" with the term "infinitely small". Both mathematically and experimentally the idea of "arbitrarily small" does not mean infinitely small, it means that for a given required accuracy and curvature there is a length scale below which curvature is negligible. So "arbitrarily small" accelerometers are definitely within the scope of physical science whereas "infinitely small" ones would not.

The "arbitrary" part is the arbitrary nature of setting the "epsilon" as stevendaryl mentioned above.

The other thing that puts it firmly within the realm of physical science is that for any finite size and curvature you can calculate the deviation from flatness that you would expect to measure experimentally.

 

[loislane]

I think that you are confusing the term "arbitrarily small" with the term "infinitely small". Both mathematically and experimentally the idea of "arbitrarily small" does not mean infinitely small, it means that for a given required accuracy and curvature there is a length scale below which curvature is negligible. So "arbitrarily small" accelerometers are definitely within the scope of physical science whereas "infinitely small" ones would not.

The "arbitrary" part is the arbitrary nature of setting the "epsilon" as stevendaryl mentioned above.

I'm not confusing anything with "infinitely small". There's the mathematical notion from calculus of infinitesimals that you might be garbling with "infinity", that uses epsilon in proofs, but that has nothing to do with what I meant in that particular reply. I was obviously referring to the impossibility of physically building an accelerometer that is "arbitrarily small", i.e. as small as one wishes, like as small as an atom, or a proton or the Plack length or smaller(arbitrarily, but finite nothing to do with infinitely). That's what made your assertion not empirically testable.

The other thing that puts it firmly within the realm of physical science is that for any finite size and curvature you can calculate the deviation from flatness that you would expect to measure experimentally.

That's correct.

 

[PeterDonis]

I was obviously referring to the impossibility of physically building an accelerometer that is "arbitrarily small", i.e. as small as one wishes

DaleSpam didn't say you had to be able to build an accelerometer of any size down to the Planck length. He just said you had to be able to build one that is small enough that tidal effects within it are negligible. Obviously what size that is depends on what is producing the tidal effects. For the Earth's gravity, an ordinary-sized accelerometer, such as the one inside your smartphone, is quite sufficient.

 

[loislane]

DaleSpam didn't say you had to be able to build an accelerometer of any size down to the Planck length. He just said you had to be able to build one that is small enough that tidal effects within it are negligible. Obviously what size that is depends on what is producing the tidal effects. For the Earth's gravity, an ordinary-sized accelerometer, such as the one inside your smartphone, is quite sufficient.

If that is what he meant then I'm fine with it. But the bottomline is that one thing is what one can measure and other what the math says, in this case the Riemannian geometry behind the GR theory when it refers to the Riemann tensor representing tidal forces, and the point here is that regardless of its practical measurability, the curvature doesn't vanish regardless of if you are using an accelerometer that is insensitive enough not to measure the tidal effects of the curvature.
And this seems to be at odds with an interpretation of the equivalence principle that equates(even for an infinitesimally small space and time region) a free-falling object accelerating towards the Earth with an object at rest in an inertial frame which would have to have an exactly vanishing Riemann curvature.

 

[PeterDonis]

the curvature doesn't vanish regardless of if you are using an accelerometer that is insensitive enough not to measure the tidal effects of the curvature.

In principle this is true, yes.

this seems to be at odds with an interpretation of the equivalence principle that equates(even for an infinitesimally small space and time region) a free-falling object accelerating towards the Earth with an object at rest in an inertial frame which would have to have an exactly vanishing Riemann curvature.

The EP doesn't "equate" those two things in all respects. It just says that, for a region of spacetime that is small enough that the effects of curvature can be ignored, the two things can be treated the same, so you can understand a lot of the physics of "gravitational fields" by looking at the physics in a local inertial frame. For example, the gravitational redshift and gravitational time dilation can be understood purely in terms of the behavior of accelerated clocks in a local inertial frame, without any reference to spacetime curvature.

This kind of procedure is by no means limited to relativity; it's done all the time in physics. In order to understand some particular piece of physics, we ignore other pieces of physics that are, for practical purposes, negligible in the specific circumstances under consideration. If we always had to solve the entire equations of physics to understand one little problem, we'd never get anywhere.

 

[loislane]

In principle this is true, yes.
The EP doesn't "equate" those two things in all respects. It just says that, for a region of spacetime that is small enough that the effects of curvature can be ignored, the two things can be treated the same, so you can understand a lot of the physics of "gravitational fields" by looking at the physics in a local inertial frame. For example, the gravitational redshift and gravitational time dilation can be understood purely in terms of the behavior of accelerated clocks in a local inertial frame, without any reference to spacetime curvature.

This kind of procedure is by no means limited to relativity; it's done all the time in physics. In order to understand some particular piece of physics, we ignore other pieces of physics that are, for practical purposes, negligible in the specific circumstances under consideration. If we always had to solve the entire equations of physics to understand one little problem, we'd never get anywhere.

So then you are saying the EP is just an approximation heuristic like it is indeed often used in physics?
I would say that since it is called a principle it is meant to be exact, besides if it is not an exact statement many things in GR wouldn't make much sense, like background independence, but I'm not a physicist so I can't assure what is meant by the EP.

 

[PeterDonis]

So then you are saying the EP is just an approximation heuristic like it is indeed often used in physics?

In the sense that in principle curvature is never exactly zero, you could call the EP an approximation, yes. Any local inertial frame, in which we assume that the EP holds and use the special relativistic laws of physics to make predictions, is in principle an approximation.

However, the EP is not a "heuristic", because the predictions you get when you use it to justify applying the SR laws of physics in a local inertial frame are not heuristic predictions. They are exact predictions using the laws of SR. The only approximation involved is the assumption that, to the accuracy of measurement, curvature effects are negligible, so we expect the predictions using the laws of SR in a local inertial frame to be correct to the accuracy of measurement. Empirically, we find that this is certainly true.

I would say that since it is called a principle it is meant to be exact

I believe the EP can be formulated mathematically as an exact statement, using the concept of "jet space" and "jet bundles". However, it's been quite a while since I looked into this, so I don't have a good reference handy.

if it is not an exact statement many things in GR wouldn't make much sense, like background independence

I'm not sure how any of this relates to background independence.

 

[Dale]

I was obviously referring to the impossibility of physically building an accelerometer that is "arbitrarily small", i.e. as small as one wishes, like as small as an atom, or a proton or the Plack length or smaller(arbitrarily, but finite nothing to do with infinitely). That's what made your assertion not empirically testable.

It still seems like you don't understand what is meant by "arbitrarily small" in this context. Or perhaps you are conflating present technological limitations with fundamental theoretical limitations. Yes, at any point there will always be some limit to what physical regimes we can experimentally probe. That doesn't mean that the theory itself is somehow not empirically testable, nor that statements made within the theory are non physical or non scientific.

If you wish to continue this line of assertions, then please produce a professional reference that explains and supports your characterization of the "arbitrarily small" concept as "out of the scope of physical science" or whatever.

 

[stevendaryl]

So then you are saying the EP is just an approximation heuristic like it is indeed often used in physics?
I would say that since it is called a principle it is meant to be exact, besides if it is not an exact statement many things in GR wouldn't make much sense, like background independence, but I'm not a physicist so I can't assure what is meant by the EP.

Well, the EP may be a heuristic, but it is the basis for a very precise scientific claim, which is that gravity only affects particles and fields through the metric tensor (and its derivatives). There is no separate "force" term for gravity in the equations of motion for particles and fields. What is commonly thought of as the "force" of gravity is actually seen to be the Christoffel coefficients used to describe motion in curvilinear coordinate systems. The equivalence principle tells us that gravity enters into the equations of motion in exactly the same way that fictitious forces such as centrifugal and coriolis forces do. The EP basically tells us that gravity is a manifestation of curved spacetime.

That doesn't uniquely determine General Relativity, because there are other theories of gravity that also satisfy the EP, incuding Newtonian gravity.

 

[loislane]

Well, the EP may be a heuristic, but it is the basis for a very precise scientific claim, which is that gravity only affects particles and fields through the metric tensor (and its derivatives).

What I'm after here is for someone to mathematically justify that the EP serves as the basis for such claim(not entering into the claim itself). Considering not only the metric and it first derivatives, but also the second derivatives necessary to define the Riemann curvature. I just don't see how the EP says anything about the second derivatives of the metric.

What is commonly thought of as the "force" of gravity is actually seen to be the Christoffel coefficients used to describe motion in curvilinear coordinate systems. The equivalence principle tells us that gravity enters into the equations of motion in exactly the same way that fictitious forces such as centrifugal and coriolis forces do.

And this I can accept as the mathematical essence of the EP, I think it only reaches up to the Christoffel symbols, the first derivatives of the metric.

The EP basically tells us that gravity is a manifestation of curved spacetime.

And above I'm suggesting that it says less than that.

That doesn't uniquely determine General Relativity, because there are other theories of gravity that also satisfy the EP, incuding Newtonian gravity.

This would seem to confirm my point since in Newtonian gravity the EP cannot certainly say that gravity is a manifestation of curvature(Newtonian space is euclidean).

 

[stevendaryl]

What I'm after here is for someone to mathematically justify that the EP serves as the basis for such claim(not entering into the claim itself). Considering not only the metric and it first derivatives, but also the second derivatives necessary to define the Riemann curvature. I just don't see how the EP says anything about the second derivatives of the metric.

It doesn't. To apply the EP, you never need anything other than the first derivatives of the metric. In the book Gravitation by Misner, Thorne and Wheeler, the two ideas of GR are summarized by:

1. Gravity tells matter how to move.
2. Matter tells spacetime how to curve.

Point 1, how gravity affects matter, pretty much only involves the first derivatives of the metric tensor--the Christoffel coefficients. If you write down the equations of motion for flat spacetime, then incorporating gravity means replacing ordinary partial derivatives by covariant derivatives (which means sticking Christoffel coefficients in at various places). The second derivatives don't come into play.

On the other hand, point 2, how matter affects gravity, must involve second derivatives (or higher-order derivatives, although these don't appear in GR). The reason for this is that first derivatives of the metric can always be made to vanish locally by choosing an appropriate coordinate system. Obviously, you can't make matter disappear through a coordinate change. So if matter is to affect gravity, then it must be through higher-order derivatives.

This would seem to confirm my point since in Newtonian gravity the EP cannot certainly say that gravity is a manifestation of curvature(Newtonian space is euclidean).

That's not true. If you describe gravity in a coordinate-independent way, you are led to a geometric view of Newtonian gravity, as well. That's not the way it's usually presented, but Newtonian gravity is completely equivalent to a geometric theory in which gravity is a manifestation of spacetime curvature. Going back to my two points about GR, point number 1 is the SAME in both Newtonian gravity (expressed in geometric terms) and GR. It's point number 2 that differs between the two theories. In Newtonian gravity, it's mass (a scalar quantity) that affects gravity, while in GR it's the energy-momentum tensor, which is a tensor.

 

[loislane]

It doesn't. To apply the EP, you never need anything other than the first derivatives of the metric. In the book Gravitation by Misner, Thorne and Wheeler, the two ideas of GR are summarized by:

1. Gravity tells matter how to move.
2. Matter tells spacetime how to curve.

Point 1, how gravity affects matter, pretty much only involves the first derivatives of the metric tensor--the Christoffel coefficients. If you write down the equations of motion for flat spacetime, then incorporating gravity means replacing ordinary partial derivatives by covariant derivatives (which means sticking Christoffel coefficients in at various places). The second derivatives don't come into play.

On the other hand, point 2, how matter affects gravity, must involve second derivatives (or higher-order derivatives, although these don't appear in GR). The reason for this is that first derivatives of the metric can always be made to vanish locally by choosing an appropriate coordinate system. Obviously, you can't make matter disappear through a coordinate change. So if matter is to affect gravity, then it must be through higher-order derivatives.

I don't have that book but I would doubt that any book other than some basic popularization could use such hand-waving arguments. They simply aren't mathematically sustainable. Gravity, that is represented by Riemann curvature in GR cannot afffect matter in a mathematically different way from how matter affects gravity as long as you are talking about the same Riemann curvature in both statements. So the second derivatives are necessarily required if curvature(gravity) is involved.

According to your reasoning about how gravity(curvature according to GR) affects matter you make the presence of curvature depend on coordinate changes(which you claim correctly that can't be done with matter). That is not mathematically sound when one understands the invariance of the curvature tensor.
When you say:"incorporating gravity means replacing ordinary partial derivatives by covariant derivatives (which means sticking Christoffel coefficients in at various places). The second derivatives don't come into play", it can hardly be more misleading because computing the Riemann curvature, that is gravity in the GR sense involves always second derivatives of the metric(derivatives of the christoffels) not simply sticking Christoffel coefficients.
Your point 1 could amount to what the EP says but that then doesn't include gravity as Riemann curvature.

 

[Dale]

What I'm after here is for someone to mathematically justify that the EP serves as the basis for such claim(not entering into the claim itself). Considering not only the metric and it first derivatives, but also the second derivatives necessary to define the Riemann curvature. I just don't see how the EP says anything about the second derivatives of the metric.

I am not sure what stevendaryl means above, but a good discussion of the EP can be found in chapter 4 here:
http://www.preposterousuniverse.com/grnotes/

 

[stevendaryl]

I don't have that book but I would doubt that any book other than some basic popularization could use such hand-waving arguments.

What argument? The two points I quoted is just a summary. It's not an argument.

They simply aren't mathematically sustainable. Gravity, that is represented by Riemann curvature in GR cannot afffect matter in a mathematically different way from how matter affects gravity as long as you are talking about the same Riemann curvature in both statements.

Let's take Newtonian gravity, for a simpler example than GR.

If you let \Phi be the gravitational potential, then gravity affects matter through the equation:

m \dfrac{d^2 x^j}{dt^2} = - m \dfrac{\partial \Phi}{\partial x^j}

Note: this involves the first derivative of \Phi

Matter affects gravity through the equation:

\sum_j \dfrac{\partial^2 \Phi}{(\partial x^j)^2} = 4 \pi G \rho

where \rho is the mass density. That involves the second derivative of \rho.

GR modifies both of these equations: Instead of the first equation, we have:

m \dfrac{d^2 x^\mu}{d \tau^2} = - m \Gamma^\mu_{\nu \sigma} \dfrac{dx^\nu}{d\tau} \dfrac{dx^\sigma}{d\tau}

where \Gamma^\mu_{\nu \sigma} is constructed from the first derivatives of the metric tensor.

Instead of the second equation, we have:

G_{\mu \nu} = 4 \pi T_{\mu \nu}

where G_{\mu \nu} is constructed from the second derivatives of the metric tensor, and T_{\mu \nu} is the energy-momentum tensor.

So what you're saying is mathematically unsustainable is a feature of both GR and Newtonian gravity. Which is a kind way of saying that you're completely wrong about this.

So the second derivatives are necessarily required if curvature(gravity) is involved.

Curvature does not (directly) influence the motion of particles.

According to your reasoning about how gravity(curvature according to GR) affects matter you make the presence of curvature depend on coordinate changes(which you claim correctly that can't be done with matter).

No, I said that the Christoffel coefficients are affected by coordinate changes. Curvature is covariant under coordinate changes.

That is not mathematically sound when one understands the invariance of the curvature tensor.

Look, there is nothing wrong with asking questions when you don't understand something. But don't pretend that you understand something when you don't.

When you say:"incorporating gravity means replacing ordinary partial derivatives by covariant derivatives (which means sticking Christoffel coefficients in at various places). The second derivatives don't come into play", it can hardly be more misleading because computing the Riemann curvature, that is gravity in the GR sense involves always second derivatives of the metric(derivatives of the christoffels) not simply sticking Christoffel coefficients.

I've explained this as best I can: If you want to compute the path of a particle under the influence of gravity, then you use the geodesic equation. That doesn't involve curvature, it only involves the Christoffel coefficients, which are constructed from the first derivatives of the metric tensor. If you want to compute the effect of matter, momentum and energy on gravity, then the appropriate equation is the field equations, which does involve the curvature tensor (and therefore, second derivatives of the metric tensor).

Your point 1 could amount to what the EP says but that then doesn't include gravity as Riemann curvature.

That's right. The EP basically says that test particles move on geodesics. The EP by itself doesn't say anything about curvature.

The EP is basically about the effect of gravity on the equations of motion of particles and fields. That does not involve curvature. Curvature is involved when you look at the other half of the story: how do particles and fields affect gravity.

 

[stevendaryl]

I've explained this as best I can: If you want to compute the path of a particle under the influence of gravity, then you use the geodesic equation. That doesn't involve curvature, it only involves the Christoffel coefficients, which are constructed from the first derivatives of the metric tensor. If you want to compute the effect of matter, momentum and energy on gravity, then the appropriate equation is the field equations, which does involve the curvature tensor (and therefore, second derivatives of the metric tensor)

There is a sense in which the two questions: how does gravity affect matter and fields, and how does matter and fields affect gravity, aren't actually separate in GR. I believe that it is possible to derive the prediction that test particles follow geodesics from the field equations.