# Freefall and Acceleration: A Discussion

• CJames

#### CJames

Just a little question.

I'm a little curious. In the classical sense, a reference frame in freefall would, of course, be said to accelerate. However, no forces are felt within a freefalling reference frame. Am I correct, then, in assuming a freefalling reference frame to be an inertial reference frame? Am I also correct in stating that a reference frame on the surface of Earth is an accelerating reference frame, since a force of gravity is felt toward the ground? (Even though in the classical sense, this reference frame is "at rest.")

Don't limit this thread to an answer to my question. I would like a continuous discussion. Please? LOL.

Good day everybody.

Umm, really intersting point !
Well i can't say lot, but i have an idea.
When you are on earth, the frame of refference is said to be 'inertial' because the frame itself is not under any force (other than centrifugal, which we (somehow) omit).
Now the gravitational force on objects inside this frame are not acting on the frame itself, but on the objects inside it, therefore you (as a person inside this frame) feel gravitational force.
Since you feel gravitational force, you can also know that (for example) as you get higher the gravitational force will become smaller.
But, if you are in frame of refference that is falling in the gravitational field of the earth, the frame itself has the force acting on it (meaning that if the origin point of the frame is moving under gravitational force), now here you will face a problem (and i think this is why it is not taken to be inertial).

In this frame you do not feel any force acting on you (omit air resistance), now suppose there is a ball 1 KM above you, it is supposed to stay at this distance from you (since there is not force acting on it as you see it).
But since the gravitational force acting on the ball will be smaller than this acting on you, then the ball will start to get farther and farther from you.
So the 1st law of Newton doesn't hold righ here, therefore the frame of refference is NOT inertial .

Just assume for a second gravity is not a force that acts between two bodies of mass,but only spacetime.since gravity curves spacetime,then also assume spacetime has energy that gravity attracts.so what if when a body of mass attracts this energy from spacetime,and pulls it towards it at light speed or more.then on a planet like the earth,what sends you into acceleration is your attraction to the spacetime energy that is in motion towards the earth.this could be why you feel no forces in freefall.so as the Earth spins,the spacetime energy is being pulled in,while being forced into a spiral motion at the same time.looks like a galaxy in motion.since your in the planets gravitational field your body has the combined field of the entire planet in the field,as matter adds each others field to each other as a mass grows.this is why objects fall at the same rate,they attract spacetimes energy with the entire planets field by itself.so in freefall no force is felt because your in acceleration by spacetimes motion,so of course how would you be able to feel space moving!

Last edited:
In the strictest sense of the definition, a frame in free-fall, that is if you know it is in free-fall, is not an inertial frame of refrense, also the Earth is not an inertial frame of refrence, even though it is usually referred as one.

MrCan,
In the strictest sense of the definition, a frame in free-fall, that is if you know it is in free-fall, is not an inertial frame of refrense, also the Earth is not an inertial frame of refrence, even though it is usually referred as one.
I'm curious as to what, exactly, they are then. If niether is inertial, are they both accelerating reference frames? If so, why?

Technically, of course, no body anywhere in the universe is free of a gravitational force, so from that standpoint, if a freefalling object and a standing object are both accelerating reference frames, there is no such thing as an inertial reference frame.

So I guess it's all more about approximation. Which is more inertial? Does that make sense? I'm just wondering in which case you apply SR, and in which you use GR. It seems to me that from the standing reference point, spacetime is curved. Yet spacetime must also be curved in freefall. So what is the difference? There has to be one, right?

matter has no interaction with matter itself only spacetime.so matter is just a big energy magnet sucking energy toward it.spacetime curves from this pull on the energy.then the energy is converted into magnetism for TIME.and gravity between to suns or masses is the motion of spacetime moving around,though,and between both masses,the attraction between both masses and the motion of spacetime energy forces a object into motion.so the bigger the current of spacetime energy,times the gravitational attraction to spacetime give a mass a better chance of being forced into motion

Any frame of reference with massive bodies affecting it is not truly an inertial frame of reference. There are tidal effects that tend to attenuate the observer frame, and radial effects causing differential increase of acceleration, nearest the mass (unless that frame is a singularity or no masses exist at all). Think of squaring the section in circular coordinates between r and r+dr, between [the] and [the]+d[the], where the mass is at the origin.

Welcome Loren.

Any frame of reference with massive bodies affecting it is not truly an inertial frame of reference.
I understand that now. However, there is no object in the universe that is truly free of gravitational pull or tidal forces. Earth orbits the sun, the sun orbits the center of the galaxy, the galaxy is being pulled on by other galaxies. So, in effect, a completely inertial reference frame doesn't actually exist. It is an idealization, an estimation of reality.

Anyway, what I'm curious about, is what a gravitational field "looks like" from the perspectives of the two reference frames I gave. How is the standing frame's perception of spacetime different from the falling frame's perceptions? In both, it seems, spacetime must be curved. But something is different, because one frame feels the pull of gravity, and the other only feels a slight tidal pull stronger at the "bottom" of the frame than at the "top". (Reference frames don't have bottoms and tops but work with me.)

Another question, what about a reference frame that is viewing Earth as it passes by it? What is the perception of spacetime there? (The spacetime surrounding Earth.)

re - "Just a little question. I'm a little curious. In the classical sense, a reference frame in freefall would, of course, be said to accelerate. However, no forces are felt within a freefalling reference frame. Am I correct, then, in assuming a freefalling reference frame to be an inertial reference frame?"

Yes.

re - "Am I also correct in stating that a reference frame on the surface of Earth is an accelerating reference frame, since a force of gravity is felt toward the ground? (Even though in the classical sense, this reference frame is "at rest.")"

It is a non-inertial frame. However when saying whether a frame is accelerating you have to say accelerating with respect to what? With respect to Earth? No. With respect to an inertial frame which is located where you are? Yes. With respect to an inertial frame whis is at rest 10 light years from you? No.

Pete

Don't limit this thread to an answer to my question. I would like a continuous discussion. Please? LOL.

Good day everybody.

__________________

Greetings !

Elementary dear CJames !
You can not possibly discuss your reference frame
as an observer standing on the Earth as that of
a body that is not affected by any forces (an
isolated system all by itself) IF you then
intend to consider something that IS affected by
the space-time curvature due to the presense of the
Earth. You simply can't consider something in
one reference frame in relation to another while
the contents of each system is different. The
trick is to define all the bodies and forces
in ONE system that you need and THEN pick reference
frames (easier not to make mistakes this way).

Live long and prosper.

Welcome drag,

The
trick is to define all the bodies and forces
in ONE system that you need and THEN pick reference
frames
Actually, I don't think that is the case, since the measurements of the bodies and forces are different from the point of view of each reference frame. I don't believe, by relativity, you are allowed to define anything without first choosing a reference frame.

You simply can't consider something in
one reference frame in relation to another while
the contents of each system is different.
I'm not entirely sure what you meant by this. I'm just trying to discuss the effects of spacetime within each reference frame, not necessarily use one reference frame to figure out what happens in another.

Take care and have a great day.

Thanks for your response pmb. Does anybody have anything to add to this? Is he right? (I rarely take one unknown source alone as a definite answer, nothing against you pmb.)

pmb points out that acceleration is relative (as the elevator equivalence model was supposed to resolve). Free fall is accelerative relative to an outside gravitating body such as Earth. Such an "inertial" frame is not even approximately so classically for the vast majority of geodesics.

Inertial frames, I believe, should be defined in terms of approaching the zero mass case. Whether one can determine this preferred inertial frame in spacetime is moot. (CMB?)

Also, think of the Earth represented by an equivalent black hole gravitating mass, with enough charge to maintain avoidance of collapse of your Earth-bound frame. The nature of a massive body insures an eventually singular frame with the inertial approaching the noninertial (or vice versa).

Greetings !
Originally posted by CJames
Actually, I don't think that is the case,
since the measurements of the bodies and
forces are different from the point of view
of each reference frame. I don't believe, by
relativity, you are allowed to define anything
without first choosing a reference frame.
When you define ANY physical system you must
consider ALL the bodies that are in it and
ALL the qualities of these bodies. The values
of the forces will of course be different.
Originally posted by CJames
I'm not entirely sure what you meant by this.
I'm just trying to discuss the effects of
spacetime within each reference frame, not
necessarily use one reference frame to
figure out what happens in another.
What could I POSSIBLY talk about ?!
Of course that the space-time curvature due to
the presense of the Earth is "felt" by bodies
in both reference frames equally. But, you
CAN'T just consider that the Earth exits in
the falling reference frame - creating the
space-time curvature and not existing in the
standing reference frame in terms of supporing -
pushing back on your standing body on its surface
(Not considering the Earth that you're standing on !).

If you consider the falling reference frame to be
at rest then you will see the standing frame
accelerating towards you. An object on the surface
of the Earth is at rest relative to the Earth.

Live long and prosper.

LBooda:
Free fall is accelerative relative to an outside gravitating body such as Earth.
Thanks, I can understand that. I'm just trying to get a general idea of what's going on in GR. Like, from what reference frame is the general idea of a bowling ball curving a trampoline employed? (Yeah, I know, it's way more complicated than that.) Is that what spacetime looks like from the reference frame of Earth, the reference frame of an inertial reference frame stationary w/ respect to Earth, or the reference frame of anything within Earth's gravitational field regardless of motion?

Hi drag,
But, you
CAN'T just consider that the Earth exits in
the falling reference frame - creating the
space-time curvature and not existing in the
standing reference frame in terms of supporing
Oops! Sorry, I think we have some confusion here. I missed it in your first post. I didn't realize you were under the impression that I was saying Earth doesn't exist from a certain reference frame. Yes, of course Earth exists in both. My point was that in freefall, an accelerometer will yield no measurement, while on the surface of Earth it will find an acceleration of 1g.

But now I can understand that if an elecator suddenly falls, it isn't exactly equivalent to if Earth's gravitational field had disappeared. There would be a measureable difference in force from the top to the bottom of the elevator. So it's not exactly equivalent unless you consider the elecator to be a point. My question now is what does this imply?

CJames - Its easy to see the true meaning of gravity in General Relativity if you keep three things in mind

(1) tidal forces and spacetime curvature are exactly the same things. The former in the language of Newtonian Physics and the later in the language of differential geometry.

(2) The gravitational field can always be transformed away at least at one point in spacetime - physically that means that there always exists a frame of referance in which a point partilce (with no internal structure) is at rest at the origin.

(3) Mass-energy is the source of gravity where "source of gravity" is to be interpreted in the same way one would interpret the statement "charge is the source of a EM field"

re - "I was saying Earth doesn't exist from a certain reference frame. Yes, of course Earth exists in both. My point was that in freefall, an accelerometer will yield no measurement, while on the surface of Earth it will find an acceleration of 1g."

This isn't quite correct. There is a distinction between gravity and spacetime curvature. The former can be measured with a gravity gradiometer and the later with an accelerometer. So using an accelerometer you can easily see that the gravitational field can be transformed away - just change to a frame of referance in free-fall. However the tidal forces can't be transformed away.

If that seems confusing then think in terms of a two equal charges. At the point in space midway between the two charges the electric field is zero - but only at that one point. However that does not mean that the field is uniform anywhere and if you place charged body with a non-zero spatial extent then there will be stresses imposed in the body due to the electric field gradient.

re - "So it's not exactly equivalent unless you consider the elecator to be a point. My question now is what does this imply?"

The equivalence principle states that a *uniform* gravitational field is equivalent to a uniformly accelerating frame of referance. It does say that you can't tell the difference between gravity and an accelerating frame of referance. In fact Einstein was very clear on this point and stated this explicity.

But what your example means is that spacetime curvature cannot be transformed away but that the gravitational field can. I.e. the gravitational force can always be transformed away but not gravitational tidal forces.

Pete

Thank you very much for your response pmb.

(1) tidal forces and spacetime curvature are exactly the same things. The former in the language of Newtonian Physics and the later in the language of differential geometry.
So, for example, an elevator in freefall and one at "rest" will measure the exact same tidal force? Is that correct? Another question (and I appologize for asking so many). What if freefall is approaching c, will special relativity effects alter this measurement, thus altering the measurement of spacetime curvature?

The equivalence principle states that a *uniform* gravitational field is equivalent to a uniformly accelerating frame of referance.
Does "uniform gravitational field" imply a gravitational field with no tidal forces?

But what your example means is that spacetime curvature cannot be transformed away but that the gravitational field can.
Cool.

So what differences are there in rotating reference frames and accelerating reference frames, in terms of spacetime curvature, compared to the curvature created by a large mass? For example, in a rotating reference frame, as you climb toward the center you will begin to feel a force in the direction opposite (I'm faily sure) of the frame's rotation, something that obviously doesn't occur via a mass-generated field. (This is due, in the classical sense, to your inertia.)

Hi CJames

re - "Thank you very much for your response pmb." - You're most welcome!

re - "So, for example, an elevator in freefall and one at "rest" will measure the exact same tidal force? Is that correct?"

Depends on the velocity. Tidal forces are velocity dependant - at least in General Relativity. The effect is probably too small to measure in practice.

re - "Another question (and I appologize for asking so many)."

Nah! Don't think twice about asking anything. If I didn't want to post then I wouldn't be here. Rememeber the most important thing about learning - When you can explain it to someone else such that they understand then you can rest assured that you yourself understand it. So the more I respond the better I get! At least that's the idea! :-)

I tutored all through collge and it showed me the benefits of explaing things.

re - "What if freefall is approaching c, will special relativity effects alter this measurement, thus altering the measurement of spacetime curvature?"

Consider what happens when a body is in free-fall and is heading to a black hole. *As measured from a remote observer* - The speed will increase at first. After a certain period the body will slow down and the speed will keep decreasing and never even get to the event horizon! The tidal forces will change as measured by an observer falling with the body. But this is wicked complicated to calculate and I've never gotten around to it ... yet! So I don't know the exact functional relation between the tidal forces and the speed of the body.

re - "Does "uniform gravitational field" imply a gravitational field with no tidal forces?"

Yes!

See journal referances to this effect in my web site

http://www.geocities.com/physics_world/uniform_field.htm

re - "Cool." - Yep! My sentiments exactly.

I thought I was nuts when I started to learn GR. People are always saying that gravity is a curvature in spacetime. Then they say that spacetime curvature is the same thing as tidal forces. Sorry. There's something wrong there. Well it turned out that the world was wrong and I was right! :-)

Well not the entire world. Einstein agrees with me, or to me exact - I agree with Einstein since it was he who came up with all of this. And then there is Dr. John Stachel, Boston University, GR expert, former editor of the Einstein papers project, and world renoun Einstein historian. I correspond with him and he confirmed this. I happened to have caught Kip Thorne (Black Hole Guru) at a lecture at Harvard and he too confirmed this.

I wrote a paper on this. See

http://arxiv.org/abs/physics/0204044

re - "So what differences are there in rotating reference frames and accelerating reference frames, in terms of spacetime curvature, compared to the curvature created by a large mass?"

If you're in an inertial frame of referance far from any gravitating body then there is no gravitational field. The spacetime is flat. There is no way to curve spacetime by changing your frame of referance. So now change to a frame rotating with respect to the first. No spacetime curvature - but gravitational field - where "gravitational field" is understood to mean "gravitaitonal field". At least according to Einstein.

re - "For example, in a rotating reference frame, as you climb toward the center you will begin to feel a force in the direction opposite (I'm faily sure) of the frame's rotation, something that obviously doesn't occur via a mass-generated field. (This is due, in the classical sense, to your inertia.)"

Well I don't know about that. Perhaps there is a configuration of matter which will do that. I don't know. But don't think exclusively a gravitational field as a point with a 1/r^2 force law.

In fact here's an example of a gravitational field generated by mass and yet there is no spacetime curvature. In fact here are three such examples

www.geocities.com/physics_world/grav_cavity.htm
www.geocities.com/physics_world/domain_wall.htm
www.geocities.com/physics_world/cosmic_string.htm

Pete

Thanks a lot and I will read those but first...I think I'm going to lose my mind! If spacetime curvature isn't equivalent to gravity...how does GR explain gravity? Don't tell me it doesn't! That would just ruin my whole day. I thought that GR described gravity purely in terms of geometry!?

Originally posted by CJames
Thanks a lot and I will read those but first...I think I'm going to lose my mind! If spacetime curvature isn't equivalent to gravity...how does GR explain gravity? Don't tell me it doesn't! That would just ruin my whole day. I thought that GR described gravity purely in terms of geometry!?

Yes. GR does explain gravity! :-)

And yes. It does, to some extent, explain gravity in terms of geometry. Think of gravity as *something that 'can' curve spacetime" rather than "gravity 'is' a curvature in spacetime'.

Suppose you're in a spaceship out in deep space (i.e. in an inertial frame in flat spacetime). Call this frame of referance O. There is a beam of light which enters the window on the port side and leaves on the starboard side which is directly across the cabin from port. The light travels in a straight line. Same thing would happen with a stone which was moving in stead of the light passing through the windows. If the spacehip was moving at constant velocity relative O then the light and stone would still travel in a straight line at constant velocity - straight line in spacetime- just a different straight line in space. However if the spaceship was accelerating then in that accelerating frame of referance, call it O', the rock and the light beam would no longer travel in a straight line but a curved line. And that would be true regardless of the mass of the stone too - i.e. the path the stone takes does not depend on the mass of the stone - only on it's velocity.

Now consider what this looks like mathetically. You start with one coordinate system and you change that coordinate system. Consider a straight line in a Cartesian coordinate system. The equation of the line is

y = ax + b

No switch to polar coordinates. The equation of the line is not as simple. Mattter of fact if you plot r and theta in a diagram which has the axes at right angles then this path becomes curved.

Now go back to relativity. Here too I am changing coordinate systems. But now I'm not changing the coordinates to polar but keeping the *spatial* coordinates Cartesian. But this is a change in coordinates of space*time*. And this ends up being a curved line in Cartesian coordinates and the motion of the particle is identical to the motion of the particle in an uniform gravitaitonal field. The geometry of general relativity is Riemman geometry (or is it psuedo-Riemman? I forgot) and Riemman geometry is the geometry of curvilinear coordinates. It so happens that the most general kind of geometry will have curvature - but it need not!

Listen to how Einstein explains this in his famous general
relativity paper

g_uv is the metric tensor - think of that as the quantities which define the gravitational field - i.e. think of them as a set of 10 gravitational potentials

The case of the ordinary theory of relativity arises out of the case here considered, if it is possible, by reason of the particular relations of the gab in a finite region, to choose the system of reference in the finite region in such a way that the g_uv assume the constant values [n_uv]. We shall find hereafter that the choice of such co-ordinates is, in general, not possible for a finite region. ... the quantities g_uv are to be regarded from the physical standpoint as the quantities which describe the gravitational field in relation to the chosen system of reference. For, if we now assume the special theory of relativity to apply to a certain four-dimensional region with co-ordinates properly chosen, then the gab have the values [n_uv]. A free material point then moves, relatively to this system, with uniform motion in a straight line. Then if we introduce new space-time coordinates x0, x1, x2, x3, by means of any substitution we choose, the g_uv in this new system will no longer be constants but functions of space and time. At the same time the motion will present itself in the new coordinates as a curvilinear non-uniform motion, and the law of this motion will be independent of the nature of the moving particle. We shall interpret this motion as a motion under the influence of a gravitational field. We thus find the occurrence of a gravitational field connected with a space-time variability of the g_uv. So, too, in the general case, when we are no longer able by a suitable choice of co-ordinates to apply the special theory of relativity to a finite region, we shall hold fast to the view that the gab describe the gravitational field.

So this means "gravity is inertial accelertation" basically. Now the later part refers to a finite region. If metric can't have the values n_uv in a finite region then that means the spacetime is curved and the field vanishes at most at one point in spacetime.

The curvilinear don't have a *length* which changes upono a change in coordinates. The straight line is a geodesic in spacetime (distance means something else so be careful when you interpret the work "lenght" in my comment here). If two particles are moving on geodescics and these geodescisc start out parallel but do not remain parallel then the spacetime is curved.

re - "However, there is no object in the universe that is truly free of gravitational pull or tidal forces."

Except for point particles. I don't see how a single photon can be subject to tidal forces myself.

re - "Anyway, what I'm curious about, is what a gravitational field "looks like" from the perspectives of the two reference frames I gave. How is the standing frame's perception of spacetime different from the falling frame's perceptions?"

Suppose the standing man extends his arms out to his sides and drops two stones. The stones will not just accelerate down but will accelerate towards each other. If he drops one from his head and feet at the same time then they wil both accelerate downwards, the one at his feet accelerating faster.

However according to the man in free-fall the ones dropped to the sides are accelerating inward towards him and the ones drop along head and feet will accelerate away from him. So the astronaut is being crushed from the sides but stretched from head to toe.

re - "In both, it seems, spacetime must be curved."

Yes. That's because the gravitational field of the Earth generates tidal forces - it's not a uniform field. Therefore it curves spacetime.

re - "Another question, what about a reference frame that is viewing Earth as it passes by it? What is the perception of spacetime there? "

Picture it in your head. Picture the two stones being dropped (Make believe the Earth is invisible so you don't get distracted by the Earth). What do you see the stones doing?

re - "Any frame of reference with massive bodies affecting it is not truly an inertial frame of reference."

True. But one can always find a *locally* inertial frame. Even if that frame is as small as the nucleas of an atom.

Pete

But something is different, because one frame feels the pull of gravity, and the other only feels a slight tidal pull stronger at the "bottom" of the frame than at the "top". (Reference frames don't have bottoms and tops but work with me.)
Yup. Quite true. And that is why one needs to be careful.

Wow, thank you very much for this pmb. So, what you're saying is that there a two ways of explaining gravity in GR. If it is a uniform gravitational field it can be explained in terms of simple accelerating reference frames.

If it is non-uniform then some type of spacetime curvature is necessary in explaining what is going on. Let me see if I have the idea correct. Curved spacetime will cause a worldline within that spacetime to curve as well (if it is to follow a geodesic, the shortest distance between two points in spacetime). As anybody who has taken precalculus or possibly even algebra should know, a curve on a graph of space and time corresponds with acceleration or deceleration. In relativity we realize that acceleration and deceleration are one and the same. Therefore, a curved line in spacetime corresponds to acceleration. Since the curvature will be different at each point in curved spacetime, it therefore makes sense to state that the acceleration is different at each point in spacetime. Hence the effect of tidal forces (differential acceleration) is a direct reflection of the curvature of spacetime.

Any misconceptions in there or have I caught the drift? If you have anything to add please do.

To me then it seems like a rotating frame of reference would experience spacetime distortion as well. For one thing, the pull is stronger toward the feet than it is toward the head. This distortion would look very different from the type of distortion generated by a massive body, but should still follow the rules of GR. Correct?

And finally, since a uniform gravitational field has no tidal forces there is therefore no spacetime curvature and no time dilation (within that reference frame.)

Hi CJames

re - "Wow, thank you very much for this pmb."

My pleasure. But I prefer to be called Pete. More friendly that way. :-)

re - "So, what you're saying is that there a two ways of explaining gravity in GR."

Not according to Einstein. Then again you could say that there's Einstein's way and there's Max Von Laue's way since he was the one to come up with this "gravity = spacetime curvature" meaning - and Einstein did not agree with this definition.

A gravitational field exists if a test particle undergoes inertial acceleration (inertial acceleration is an acceleration which does not depend on the mass of the object).

If there are tidal forces present then the same is true. But in that case the gravitational field doesn't completely vanish for an observer in free-fall.

re - "Curved spacetime will cause a worldline within that spacetime to curve as well"

No. Curved spacetime means that if there are two geodesics which start out parallel then they do not remain parallel - hence the term "geodesic deviation"

Do *not* confuse "worldline" with "geodesic" since they mean different things. A worldline is the path in spacetime taken by an object. A geodesic is the worldline of an object which is subject only to the force of gravity.

re - "geodesic, the shortest distance between two points in spacetime"

No. A geodesic is not defined as the shortest distance between two points. If you read that somewhere then it was wrong. A geodesic is a path of 'extremal length'. Are you familiar with this term?

See www.geocities.com/physics_world/geodesic.htm

I derived the geodesic equation in that page by three different methods just for clarity.

re - "In relativity we realize that acceleration and deceleration are one and the same."

I would disagree with that statement. "deceleration" means to "slow down/speed decreases" where as "acceleration" means "to change velocity." For example: If a particle is moving a constant speed in a cirlce then its not decelerating since the speed is not changing. However the particle is accelerating since the velocity is changing since velocity is a vector quantity and thus has a magnitude and direction and the direction of motion is constantly changing.

re - "Therefore, a curved line in spacetime corresponds to acceleration. Since the curvature will be different at each point in curved spacetime, it therefore makes sense to state that the acceleration is different at each point in spacetime. Hence the effect of tidal forces (differential acceleration) is a direct reflection of the curvature of spacetime."

I disagree. The curvature of the line being different at different places does not mean that the object has a non-constant acceleration. The acceleration that is being referred to in GR is what is called "local acceleration." The difference is difficult to explain without getting into details. Plus I'm not sure that the curve

z = (1/c)gt^2

has a constant curvature. Don't think of the curvature of the line as being related in any way shape or form to spacetime curvature. They are *very* different.

re - "To me then it seems like a rotating frame of reference would experience spacetime distortion as well."

No. A rotating frame of referance is obtain by taking a in inertial frame of referance in flat spacetime and change to a rotating frame. Spacetime curvature is an inherent property of spacetime and it either exists or it doesn't. You can't get spacetime curvaure by changing coordinates.

re - "For one thing, the pull is stronger toward the feet than it is toward the head."

Actually what is happening is that you have the obsever at the outer rim accelerating at a different rate than an observer near the center. So a particle moving along inertially seems to move faster when it's at different places. However tidal forces are present only when there is a "relative" acceleration between two particles in free-fall. Or if there is geodesic deviation. Think, not in terms of space but think it terms of space*time*.

re - "And finally, since a uniform gravitational field has no tidal forces there is therefore no spacetime curvature and no time dilation (within that reference frame.)"

No. In a uniform g- field there can be redshift. I made a page on that too.

www.geocities.com/physics_world/red_shift.htm

This is something which is totally misunderstood in general relativity. Even by the best of the best of them. However Einstein's very first derivation of redshift actually was done in a uniform gravitational field!

But if someone tells you that redshift implies spacetime is curved tell them they are wrong and then refer them to Einsteins derivations on it.

Pete

re - "To me then it seems like a rotating frame of reference would experience spacetime distortion as well."

No. A rotating frame of referance is obtain by taking a in inertial frame of referance in flat spacetime and change to a rotating frame. Spacetime curvature is an inherent property of spacetime and it either exists or it doesn't. You can't get spacetime curvaure by changing coordinates.

Kinda-sorta.

The curvature tensor will remain 0 when changing coordinates, but the metric will be different.

Hurkyl

Originally posted by Hurkyl
Kinda-sorta.

The curvature tensor will remain 0 when changing coordinates, but the metric will be different.

Hurkyl

What do you mean "Kinda-sorta"?

I said two things about the rotating frame

(1) There is a gravitational field in it. I.e. the metric is not the Minkoski metric.

(2) There is no spacetime curvature. I.e. the Riemman tensor (i.e. the 'curvature tensor') vanishes.

Mathematically - The reason for the relative existence of the g-field in GR has to do with the (a) the criteria for the field existing and (b) the nature of the metric

(a) Criteria - While the criteria for curvature is the vanishing of the curvature tensor such is not the case for the gravitational field tensor - i.e. the metric. We only require that the components of the metric not be constants when expressed in Cartesian coordinates (at least locally)

(b) By its very nature the metric never vanishes - unlike the curvature tensor.

Pmb

I feel so stupid right now. Okay, I'll take another stab at this Pete. You can call me Carter, haven't been signing as that like usual, sorry.

re - "So, what you're saying is that there a two ways of explaining gravity in GR."

Not according to Einstein.
Well, I didn't mean two ways of defining gravity. What I was trying to say is that there are two different forms of gravity, that which involves spacetime curvature and that which does not. Please let me be right about that?

No. Curved spacetime means that if there are two geodesics which start out parallel then they do not remain parallel - hence the term "geodesic deviation"
I'm glad you point out these things to me, but I would like to know if I have the general idea. If two worldlines without any other forces acting on them pass through curved spacetime they will separate in this same way, correct?

Do *not* confuse "worldline" with "geodesic" since they mean different things. A worldline is the path in spacetime taken by an object. A geodesic is the worldline of an object which is subject only to the force of gravity.
Forgive me. I'm working with what I know here. I wasn't confusing terms. I was assuming when I said worldline that the object was experiencing no other forces, and would hence follow a geodesic. I realize I'm mixing terms here, it's just that I want to get a visual picture in my mind. For me a geodesic isn't enough, I have to imagine something traveling through the field. That's just how my feeble mind works...

No. A geodesic is not defined as the shortest distance between two points.
Well, for instance a geodesic is the shortest distance between two points on the surface of Earth. If a geodesic is defined as "the straightest possible line in a curved space" then I assumed it is also the shortest distance between two points in a curved space (two events in curved spacetime?). [?] I'm sorry I don't know what extremal length is. I'm still struggling through high school calculus.

EDIT: Alright, I'm starting to think of extreme examples such as a geodesic that travels through the center of a neutron star, in which case it could in fact be more distant than simply traveling around the neutron star.

I would disagree with that statement. "deceleration" means to "slow down/speed decreases" where as "acceleration" means "to change velocity."
You're right, I favored more flavorful wording over clarity. It's more appropriate for me to say deceleration is a form of acceleration.

I disagree. The curvature of the line being different at different places does not mean that the object has a non-constant acceleration. The acceleration that is being referred to in GR is what is called "local acceleration." The difference is difficult to explain without getting into details.
What I'm trying to say is that each geodesic (switching terms altert) will curve a different amount. I assume you would find the local acceleration by finding the second derivative at any given point along the geodesic? (Or something similar to that. Just let me know if I have the right idea or not.)

Don't think of the curvature of the line as being related in any way shape or form to spacetime curvature. They are *very* different.
But it's related in the sense that the "shape" of the geodesic is dependant upon the curvature of spacetime?

No. In a uniform g- field there can be redshift.
So a clock above my head in a uniform g-field will tick faster, and one below will tick slower. But in a uniform gravitational field the distance from me won't change how large this time dilation is? Or will it? (Of course I'm still talking about within the accelerating reference frame, not about objects outside this reference frame, where SR comes into play too.)

EDIT: Nevermind, scratch that. Light from a source farther above my head would take much longer to reach it, we would therefore accelerate toward it longer than a nearby source, and the clock would therefore tick quite a bit faster.

Thanks for the discussion, I hope I'm getting something out of it. --Carter.

Last edited:
I'm going to add this. Imagine two parallel worldlines following two geodesics. The lines in curved space will separate away from each other. Hence, from the perspective of one worldline, the other is curved and is therefore accelerating away. If these two worldlines correspond to the top and bottom of an elevator, we can see that a tidal force is taking place. (Of course, in reality, the top and bottom of the elevator are attatched too rigidly to accelerate away from each other, and will in reality only "try" to accelerate away from each other. But is the basic concept correct?)

My mind works in examples and I appologize for that.

Last edited:
Hi Carter
What I was trying to say is that there are two different forms of gravity, that which involves spacetime curvature and that which does not.
I guess you could say that if you want. It's like saying there are two kinds of surfaces. Those which are flat and those which aren't. To me there are just surfaces, but sure, why not! :-)

If two worldlines without any other forces acting on them pass through curved spacetime they will separate in this same way, correct?
In general they will deviate. They may come together or they may seperate..

That's just how my feeble mind works...
Give yourself some credit Carter. After all this is the most dificult thing one can try to imagine right? I mean after all it took Einstein years to figure it out and he said it was incredibly hard! So take pride in the fact that you can even begin to understand it.

Well, for instance a geodesic is the shortest distance between two points on the surface of Earth.
Depends on the geodesic. Take any two points on the sphere. Run a plan through these two points and the center of the sphere. The plane intersects the sphere in a curve which is called a greate cirlce. If you think about it you have two choices on how to get to one of the points on the cirlce to the other depending on which way you walk. One curve is, in general, shorter than the other. Each is a geodesic.

What I'm trying to say is that each geodesic (switching terms altert) will curve a different amount. I assume you would find the local acceleration by finding the second derivative at any given point along the geodesic? (Or something similar to that. Just let me know if I have the right idea or not.)
The derivatives are with respect to proper-time but the curves are plotted in a spacetime diagram with respect to coordinate time.

But it's related in the sense that the "shape" of the geodesic is dependant upon the curvature of spacetime?
Not neccesarily. I see no way to determine the presence of curvature if all you know about is one geodesic. There may be, I simply don't know.

So a clock above my head in a uniform g-field will tick faster, and one below will tick slower.
Yes. The difference is related to the difference in gravitational potential. If two clocks are at a different gravitational potential they will run at a different rate.

But in a uniform gravitational field the distance from me won't change how large this time dilation is? Or will it?
Sure it will. See above comment.

Pete

In a freefalling elevator, the air within the elevator
isn't accelerating, so there will be a measurable pressure difference
in the elevator proportional to the acceleration; if so why is it equivelant to being inertial if an experiment can prove it is not?

mich

In a freefalling elevator, the air within the elevator
isn't accelerating

Sure it is; the air is in freefall just like the observer inside.

What do you mean "Kinda-sorta"?

I mean that while you're exactly correct, I don't think it was really about what CJames was asking.

Switching into a rotating frame changes the geometry of space. The rotating observer will see the universe differently than the nonrotating observer because in his frame, space-time has a different geometry. Space-time, in both pictures, has zero curvature, but the geometry is different, thus switching frames will introduce a "space-time distortion".

Hurkyl

Hi Hurkyl

Switching into a rotating frame changes the geometry of space. The rotating observer will see the universe differently than the nonrotating observer because in his frame, space-time has a different geometry. Space-time, in both pictures, has zero curvature, but the geometry is different, thus switching frames will introduce a "space-time distortion".

And I wonder about the rest of what you say. To me the "geometry" is an intrinsic property of a manifold once a metric is attached to it. When you change frames all you're really doing is chaning the coordinate system. You're not chaning the geometry of spacetime itself. The metric is a geometric object and when you change coordinates wha you're doing is merely changing the components - not altering the geometric entity itself. E.g. Think of the analogy with vectors since they too are geometric objects like the metric. If you have a Cartesian vector and you change coordinates then you change the components of the vector but you don't change the geometric object itself.

Pmb

Originally posted by mich
In a freefalling elevator, the air within the elevator
isn't accelerating, so there will be a measurable pressure difference
in the elevator proportional to the acceleration; if so why is it equivelant to being inertial if an experiment can prove it is not?

mich

Inside a freely falling elevator the air *is* accelerating. It's accelerating doward with the elevator - that's if the elevator is sealed such that it's isolated.

Pmb

I guess I should say the geometry of the reference frame rather than the geometry of space-time!

Originally posted by Hurkyl
I guess I should say the geometry of the reference frame rather than the geometry of space-time!

But what do you mean by this? Let's consider Euclinean geometry for a moment. Specifically E^2 - a plane for example. In cartesian coordinates the metric is given by

dL*2 = dx^2 + dy^2

I know change coordinates to polar coordinates

dL^2 = dr^2 + r^2 dtheta^2

If I understand you correctly this situation is analogous to the one in GR with regard to the rotating frame. I start with a flat space - I change coordinates - I still have a flat space. The metric looks different because of the curvilinear coordinates. What is the "distortion" you're referring to?

Perhaps you're referring to the curvilinear coordinates and its the curvilinear coordinates that's the distortion? Is so how is this different/same for a uniform g-field/accelerating frame of referance?

Thanks

Pete

The alternate coordinates don't even have to be that exotic. Spinning your reference frame amounts to a helical twist:

t' = t
x' = x cos &omega t - y sin &omega t
y' = x sin &omega t + y cos &omega t
z' = z

The twist preserves locally Minowski frames located on its axis, and is a smooth deformation from the identity map.

Is so how is this different/same for a uniform g-field/accelerating frame of referance?

Here the distortion is that (via a change of coordinates) one reference frame sees a uniform nonzero classical gravitational field and another reference frame sees a Minowski metric everywhere.