The Equivalence Principle as a Starting point of GR

[fog37]

Hello World,

I have understood the following: in SR, time intervals and space intervals (distances, lengths) are relative and inertial reference frame dependent. Space and time is not absolute anymore. However, acceleration is still absolute: different inertial frames see the same acceleration.

When introduced to GR, we run into the equivalence principle:
an accelerated frame of reference in the absence of a gravitational field is indistinguishable from an inertial (unaccelerated) frame in the presence of a homogeneous gravitational field: the local effects of gravity are the same as those of being in an accelerating reference frame.

I understand this, as well as the example of the light beam curved downward trajectory from an accelerated spaceship frame while an inertial observer on Earth concludes that the light path is curved by gravity (but doesn't the Earth observer see the path as straight and notices the upward accelerating spaceship?)

Why is the equivalence principle so useful? Why is it so powerful and novel to state that the local effects of gravity are the same as those of being in an accelerating reference frame?

In classical physics, gravity is a force acting on massive objects. Light, which is radiation, should not be affected and curved by mass, i.e. gravity...What does the equivalence principle add to the discussion? Is the mentioned equivalence the equivalence between inertial mass and gravitational mass? How does that equality bring us to the idea that mass-energy curves spacetime and light truly follows this geodesic paths?

Thank you!

 

[AVentura]

(but doesn't the Earth observer see the path as straight and notices the upward accelerating spaceship?)

you have mixed the scenarios together here. There is either gravity or acceleration. One observer at a time.

 

[PeterDonis]

acceleration is still absolute: different inertial frames see the same acceleration.

No. What you are describing is coordinate acceleration, which is not the same in all inertial frames.

What is the same in all frames is proper acceleration, i.e., acceleration measured by an accelerometer or felt by an observer as weight.

an accelerated frame of reference in the absence of a gravitational field is indistinguishable from an inertial (unaccelerated) frame in the presence of a homogeneous gravitational field

No, this is not what the EP says. The EP does not say that an accelerated frame is indistinguishable from an inertial frame. That would be obviously false.

What the EP says is that an accelerated frame of reference in the absence of gravity, such as a frame attached to a rocket accelerating in flat spacetime by firing its engines, is indistinguishable from an accelerated frame in the presence of gravity, such as a frame attached to an observer standing at rest on a planet with the same proper acceleration as an observer inside the accelerating rocket.

the light beam curved downward trajectory from an accelerated spaceship frame while an inertial observer on Earth concludes that the light path is curved by gravity (but doesn't the Earth observer see the path as straight and notices the upward accelerating spaceship?)

No, you have it backwards, just as you have the EP wrong above. The inertial observer--the free-falling observer--sees the light path as straight. The accelerated observer--such as one standing on the surface of a planet or the one inside the accelerating spaceship--sees the light path as curved.

Why is the equivalence principle so useful? Why is it so powerful and novel to state that the local effects of gravity are the same as those of being in an accelerating reference frame?

Because this property picks out a fundamental difference between gravity and all other interactions--no other interaction obeys the EP. And this property of gravity allows us to interpret it as a manifestation of spacetime geometry rather than a "force".

Also, the EP allows us to clearly separate the "removable" effects of gravity--the "acceleration due to gravity"--from the non-removable ones, the ones that you can't make disappear by changing your reference frame. The latter effects--tidal gravity, or spacetime curvature--are the ones that then must be determined by the distribution of matter and energy via the Einstein Field Equation.

Light, which is radiation, should not be affected and curved by mass, i.e. gravity

This is not necessarily required in Newtonian physics. It is perfectly possible to model light as being affected by gravity in Newtonian physics. However, the predicted deflection of light paths will not be the same as the GR prediction, and we know experimentally that the GR prediction is correct. So this is still a case where the Newtonian prediction is known to be wrong.

Is the mentioned equivalence the equivalence between inertial mass and gravitational mass?

That's one way of looking at it, yes.

How does that equality bring us to the idea that mass-energy curves spacetime and light truly follows this geodesic paths?

See above. If you want more detail than that, you are basically asking for a course in GR, which is beyond the scope of a PF discussion.

 

[fog37]

Thanks PeterDonis. I am processing.

So gravity is a force in Newtonian physics but according to GR, which is the best theory we have about gravity, it is not a force anymore (but originates from the curvature of spacetime). I know physicists are trying to unify the four forces of nature (weak nuclear force, strong nuclear force, gravity, electromagnetic force) with the weak nuclear and electromagnetic one already being unified...But what happens with gravity now since it is not a force anymore and is explained/tied to this geometrical framework?

Thanks for any insight.

 

[PeterDonis]

what happens with gravity now since it is not a force anymore and is explained/tied to this geometrical framework?

It depends on where the geometrical framework comes from. A big part of the search for a theory of quantum gravity is to figure that out, and in the process figure out how (or whether) gravity gets unified with the other fundamental interactions. At this point I'm aware of basically two leading candidates:

(1) A quantum field theory of gravity (which might arise from string theory or might arise from some other foundation). We already know there is a QFT of a massless spin-2 field whose field equation is the Einstein Field Equation, and therefore can give rise to effects that, at the classical level, look like curved spacetime geometry. However, this QFT is not renormalizable and is not considered a viable fundamental theory in itself of which GR could be the classical limit. But this QFT, or something like it, might possibly be an effective field theory that arises from some more fundamental field theory (such as string theory) in an appropriate approximation.

(2) Some very different underlying theory, that doesn't look like a "geometry" or a "field" at all, could give rise, in an appropriate approximation, to something that looks like a classical curved spacetime geometry, plus fields that look like the other fundamental particles and interactions that we know from the current Standard Model of particle physics. Loop quantum gravity is an example of this approach.

 

[fog37]

Really cool. Thanks.

I just looked up Einstein equation, which is a tensor equation where the involved tensors have 2 indices (hence each tensor has 9 components). However, the equation is truly10 coupled, nonlinear PDEs.

Why 10 equations and not 9? I would think 9 equations, each relating the components of the various tensors...

 

[Nugatory]

I just looked up Einstein equation, which is a tensor equation where the involved tensors have 2 indices (hence each tensor has 9 components)

16, not 9. This is a four-dimensional spacetime so the indices run from zero to three. However, the tensors we’re working with (Ricci, Einstein, metric, stress-energy) are symmetrical so only ten independent components.

 

[pervect]

Really cool. Thanks.

I just looked up Einstein equation, which is a tensor equation where the involved tensors have 2 indices (hence each tensor has 9 components). However, the equation is truly10 coupled, nonlinear PDEs.

Why 10 equations and not 9? I would think 9 equations, each relating the components of the various tensors...

The Einstein tensor and the stress-energy tensor do have two indices, but because they're space-time tensors, that's 4x4 = 16 components.

The tensors are symmetric, i.e. $T^{ab} = T^{ba}$, similarly $G^{ab}=G^{ba}$. There are 4 diagaonal components $T^{ii}$. There are 12 off diagoanl components $T^{ab}$ with a not equal to b. But by symmetry, $T^{ab} = T^{ba}$, so the off diagonal components only give six different equations. The diagonal components give 4, for a total of 10.

 

[fog37]

Thank you.

Just to rewind for a minute, in Newtonian physics, the inertial mass and the gravitational mass were suspected to be equivalent but nobody was sure why these two different concepts, both called mass, had exactly the same value. Inertial mass represents the inertia of an object to changes its status of motion whereas gravitational mass has to do with a different phenomenon, gravity, and instead promotes the object's acceleration and it is also called mass.

In GR, Einstein does his two thought experiments (elevator in free fall and elevator pulled upward at constant acceleration) and concludes that the two masses must be indeed the same thing. Honestly, I am still trying to close the loop on why the two masses are obviously the same thing in reference to those two experiments. But I am close :)

 

[AVentura]

I don't think it is at all obvious that they are the same mass. The theory reveals the consequences if they are. The predictions are almost fantastical but apparently true (like black holes).

 

[PeterDonis]

am still trying to close the loop on why the two masses are obviously the same thing in reference to those two experiments.

Because in the case of no gravity (floating in free space or accelerating in a elevator being pulled), the response of objects inside the elevator is determined by their inertial mass. But in the case of gravity (free-falling in a gravitational field, or elevator suspended at rest in the field), the response of objects inside the elevator is determined by their gravitational mass. So since the responses are the same in both cases, the masses must be the same.

 

[fog37]

Haha! Great. Significant help!

Thank you!

 

[fog37]

Thank you!

• Spacetime is a four-dimensional, smooth, manifold (a manifold of dimension N can (always?) be embedded in another manifold of higher dimension N+1) which means that it is four-dimensional space that locally behaves like a "flat" space in which Euclidean geometry works perfectly. A flat space has zero curvature everywhere. However, if there is a nonzero mass or energy at a certain spacetime point the curvature must become nonzero. Can we still apply the smoothness property for the manifold in that case and consider the local geometry as Euclidean?

• As mentioned, the entire four-dimensional space of SR is flat, which should mean zero curvature anywhere. Einstein worries about the effects of mass in GR. But in SR, objects have mass and travel at very high speed acquiring a lot of energy. Shouldn't the energy of moving objects in SR distort spacetime rendering it curved instead of flat?

 

[PeterDonis]

a manifold of dimension N can (always?) be embedded in another manifold of higher dimension N+1

Not quite. A manifold of dimension N can always be embedded in another manifold of dimension greater than N, but that dimension will not always be N + 1.

which means that it is four-dimensional space that locally behaves like a "flat" space in which Euclidean geometry works perfectly

No, it locally behaves like a flat spacetime in which Minkowskian geometry works perfectly. Minkowskian geometry is not the same as Euclidean geometry.

Shouldn't the energy of moving objects in SR distort spacetime

No, because the "motion" you are talking about in SR is frame-dependent, and so is the "energy". But the action of any piece of matter as a source of spacetime curvature in GR is not frame-dependent.

In SR, spacetime is assumed to be flat. That means all objects are assumed to be negligible as sources of gravity. In other words, SR is an approximation to GR. That approximation does not work in cases where there are objects present that are not negligible as sources of gravity.

 

[fog37]

Cool. Thanks. I see how masses in SR are not huge but often their speeds are large (large fractions of $c$). That must not be enough to curve spacetime...

I am also looking into the difference between Euclidean (only spacelike) and Minkowski geometries.

 

[jbriggs444]

but often their speeds are large (large fractions of ccc). That must not be enough to curve spacetime...

If you are considering a single object that is moving then that motion does not curve [otherwise-flat] space-time at all. The curvature of space-time is an invariant property. The motion of a single object amounts to nothing more than a choice of coordinates. By definition, coordinate choice does not affect invariant properties.

If you were considering two objects then their relative motion is an invariant fact. That can result in space-time curvature.

 

[fog37]

Ahhh, yes I was considering as single object with mass $m$ and moving at high speed. Didn't know it would not curve spacetime at all alone, even in virtue of its large kinetic energy and mass...thanks. I will look into it.

 

[Nugatory]

Ahhh, yes I was considering as single object with mass mmm and moving at high speed. Didn't know it would not curve spacetime at all alone, even in virtue of its large kinetic energy and mass...thanks.

Any kinetic energy that can be made to disappear just by changing frames cannot contribute to curvature, because the curvature will be the same in all frames. In the most extreme case, consider the frame in which your single object is at rest - zero kinetic energy, zero curvature. And if the curvature is zero in that frame it will be zero in all frames.

 

[fog37]

I guess that is true also in GR, with a single huge mass: no curving of spacetime...

 

[Nugatory]

I guess that is true also in GR, with a single huge mass: no curving of spacetime...

No, a single huge mass does curve spacetime - this is the Schwarzschild solution, which predicts black holes and gravitational lensings and explains the anomalous precssion of Mercury's orbit.

There is no frame where you can make the rest mass of an object disappear (or even change - the rest mass is an invariant).

 

[Mister T]

In GR, Einstein does his two thought experiments (elevator in free fall and elevator pulled upward at constant acceleration) and concludes that the two masses must be indeed the same thing.

That doesn't sound right to me. The equivalence of gravitational and inertial mass is part of Newtonian physics and was well known by Einstein and his contemporaries. It's a consequence of the fact that particles free fall with the same acceleration, regardless of their mass. Something asserted by Galileo well before Newton formalized it with his second law of motion and his law of universal gravitation.

At the time that Einstein originated that thought experiment, his point was to establish the equivalence of gravity and acceleration, which would of course be possible only if inertial and gravitational mass are equivalent.

 

[cianfa72]

Because in the case of no gravity (floating in free space or accelerating in a elevator being pulled), the response of objects inside the elevator is determined by their inertial mass. But in the case of gravity (free-falling in a gravitational field, or elevator suspended at rest in the field), the response of objects inside the elevator is determined by their gravitational mass. So since the responses are the same in both cases, the masses must be the same.

Please, could you elaborate a bit around about what you mean with "response of objects inside the elevator" in both cases (no-gravity, gravity) ? thanks

 

[Nugatory]

Please, could you elaborate a bit around about what you mean with "response of objects inside the elevator" in both cases (no-gravity, gravity) ? thanks

Suspend the object from the ceiling of the elevator with a spring. The spring stretches by the same amount in both cases, telling us that the same force is required to hold the object in place.

 

[cianfa72]

Suspend the object from the ceiling of the elevator with a spring. The spring stretches by the same amount in both cases, telling us that the same force is required to hold the object in place.

ok, this kind of spring attached one side to the ceiling of the elevator does not "implement" an accelerometer I believe; to measure the object proper acceleration we need to attach an accelerometer (possibly built itself with a spring inside its own casing) to it, right ?

 

[jbriggs444]

ok, this kind of spring attached one side to the ceiling of the elevator does not "implement" an accelerometer I believe; to measure the object proper acceleration we need to attach an accelerometer (possibly built itself with a spring inside its own casing) to it, right ?

If you weld an accelerometer case around the spring, and bolt it to the elevator ceiling, it still reads the same.

 

[cianfa72]

If you weld an accelerometer case around the spring, and bolt it to the elevator ceiling, it still reads the same.

in this sentence it stands for "the spring" bolt to the elevator ceiling ?

 

[jbriggs444]

in this sentence it stands for "the spring" bolt to the elevator ceiling ?

The idea is that if you bolt the case to the ceiling, it does not matter whether you attach the spring to the case or to the ceiling.

 

[PeterDonis]

could you elaborate a bit around about what you mean with "response of objects inside the elevator"

The motion of the objects relative to the elevator (for example, what happens if you drop an object and let it move freely), and the proper acceleration the objects feel.

 

[cianfa72]

The idea is that if you bolt the case to the ceiling, it does not matter whether you attach the spring to the case or to the ceiling.

Excuse me, are you talking about the spring in the experiment proposed by Nurgatory or just a spring the accelerometer could be built of ?

 

[jbriggs444]

Excuse me, are you talking about the spring in the experiment proposed by Nurgatory or just a spring the accelerometer could be built of ?

Either or both. What difference do you imagine?

 

[cianfa72]

Either or both. What difference do you imagine?

Sorry, I've not a precise idea about as an accelerometer is made. My point was just that to measure the proper acceleration felt by an object you have to attach the accelerometer case to it

 

[jbriggs444]

Sorry, I've not a precise idea about as an accelerometer is made. My point was just that to measure the proper acceleration felt by an object you have to attach the accelerometer case to it

The case around an accelerometer is not essential. In principle, a [single axis] accelerometer is nothing more than a known mass on a spring and a scale against which the spring's length can be assessed. You attach the measuring scale and the free end of the spring to the object whose acceleration you wish to measure. You then see how long the spring is when the mass has settled into a position.

Putting a case around the whole assembly keeps contaminants out and keeps anyone from "putting their fingers on the scale"

 

[cianfa72]

The idea is that if you bolt the case to the ceiling, it does not matter whether you attach the spring to the case or to the ceiling.

Thus basically the accelerometer bolted to the elevator ceiling and the object attached to the spring 'share' the same proper acceleration measured by the accelerometer (actually the spring inside it) against its measuring scale

 

[Mister T]

Please, could you elaborate a bit around about what you mean with "response of objects inside the elevator" in both cases (no-gravity, gravity) ? thanks

With gravity, you release a ball and it accelerates towards the floor. Without gravity, you release a ball and the floor accelerates towards the ball. This is an "equivalence".

 

[epovo]

No. What you are describing is coordinate acceleration, which is not the same in all inertial frames.

Do you mean that each inertial frame may have its spatial coordinates rotated in its own way, giving different coordinate values but the same magnitude to the observed acceleration vector?