[yor_on]

It's tricky Pete, rereading it I confuse myself I want to put it into my own terms, and then I would say that gravity is observer dependent. Then again, I'm always wanting to define it locally. when I say that I agree to that 'gravity must exist everywhere' I'm using what I call a 'global definition', as opposite to a local, it makes it clearer for me defining it such, a local definition of a geodesic versus one where I speak about a whole 'SpaceTime's' gravity.

You write "Einstein did not interpret gravity as a curvature of space-time, rather that space-time curvature is a manifestation of gravity." giving me a feeling that you then may consider it a 'force' defining a SpaceTime. That would then be from what I call a 'global definition', not local.

Alternatively I could read it as a statement that something is needed to create gravity, mass (energy), and so define a SpaceTime. And it is true that the elevator example only can be used ignoring tidal forces, as you otherwise would be able to differ between the 'gravity' you find in a uniformly constantly accelerating rocket, relative on Earth. And so restrict the equivalence principle.

The question then becomes what one consider tidal forces as? Myself I think of it as 'gravity' too and so consider it irrelevant for a wider definition of the equivalence between gravity, and a uniform constant acceleration? Either there is a equivalence, as I think now and then this exception (tidal forces) is something solvable, or we have a situation in where Earths gravity, as it involve tidal forces, can't be applicable to a uniform constant acceleration of that elevator.

I guess I'm using it in its wider sense, if I now would try to define it.
=

If a geodesic is something without 'forces' acting on it, in a free fall finding no gravity acting on you, no friction or resistance retarding your uniform motion, then that should include tidal forces? If I exclude tidal forces from that definition, I come to a definition in where I have to assume that this ultimately will retard my uniform motion, as I think. But I have to admit that I found geodesics to be one of the most difficult assumptions to make, as from my 'global definition' gravity and tidal forces exist everywhere matter is. And if we then include the way binary stars act gravitationally, as well as other tidal forces, then there is a 'friction' to a geodesic too. And all mass have a gravity acting on them, even in a geodesic, the gravity created by its own mass though.

I've been thinking about it actually, now and then, wondering if there is some better definition I can make, the one I'm leaning too is one in where a geodesic will be defined by gravity, including tidal forces, and it won't matter if matter 'spagettifies' under its influence. The directions 'they' take, under and after, such a event should still be geodesics, to make sense for me. So you might be able to see it as 'straight lines', without resistance, although for the poor bast* getting split , as well as for any other observers, might want to define a 'force' to it.
=

But that is what I call a global definition. From a local point of view a ideal geodesic must be free from 'friction', in a constant uniform motion (or uniform gravitational acceleration as it comes down to the same. No 'local gravity'), to fit the idea I have of it.
==

If you look at

"Can gravitation and inertia be identical?  This question leads directly to the General Theory of Relativity.  Is it not possible for me to regard the earth as free from rotation, if I conceive of the centrifugal force, which acts on all bodies at rest relatively to the earth, as being a "real" gravitational field of gravitation, or part of such a field?  If this idea can be carried out, then we shall have proved in very truth the identity of gravitation and inertia. For the same property which is regarded as inertia from the point of view of a system not taking part of the rotation can be interpreted as gravitation when considered with respect to a system that shares this rotation.

According to Newton, this interpretation is impossible, because in Newton's theory there is no "real" field of the "Coriolis-field" type.  But perhaps Newton's law of field could be replaced by another that fits in with the field which holds with respect to a "rotating" system of co-ordinates?  My conviction of the identity of inertial and gravitational mass aroused within me the feeling of absolute confidence in the correctness of this interpretation."

then I feel I have gotten the idea right That's how I think of 'tidal forces'.

[yor_on]

"I've been thinking about it actually, now and then, wondering if there is some better definition I can make, the one I'm leaning too is one in where a geodesic will be defined by gravity, including tidal forces, and it won't matter if matter 'spagettifies' under its influence. The directions 'they' take, under and after, such a event should still be geodesics, to make sense for me. So you might be able to see it as 'straight lines', without resistance, although for the poor bast* getting split , as well as for any other observers, might want to define a 'force' to it."

That one has to do, for me that is, with what a 'frame of reference' can be 'minimized' too. I assume that locally a 'frame of reference' can be defined to some scale, stopping at Planck scale. and if one look at a 'spagettifiezation' from a very local perspective then each particle will find its own geodesic. On the other hand you need a acceleration to 'split' those particles from each other, as there is a force binding particle to particle. so I'm of two minds you might say, the acceleration happening is not a geodesic, but what you had and what you get after should still be a geodesic. But geodesics are a very difficult idea to encompass, as it should be definable from 'test particles' but as you gather those into pieces of matter it seems to become a lot more complicated to me.
=

It also depends on how you define something being 'at rest' with each other. In a piece of matter, is its particles 'at rest' with each other? That depends on how I would like to define a 'frame of reference' does it not? From a view in where you have a decided minimized scale to what a 'frame of reference' might be, only test particles make sense, and they better be 'point like' to make me happy. On the other hand we have something being 'at rest' with each other in a uniform motion, and from that we can also find some defining it as you also can be 'at rest' in a acceleration, equivalent to some other accelerating object in a equivalent space. Thats also why I find your definition of two objects in a uniform gravity (field), presenting us with different definitions so interesting. I'm assuming them to be uniformly moving though, for this.

[yor_on]

Let us put it this way. Assume that you have a test particle passing a event horizon, it being 'indivisible' for this (point like). should I define it as it undergoing a acceleration meeting tidal forces, or should I define it as a 'straightest path' in a 'bent/distorted' SpaceTime?

[Pmb (OP)]

yor_on - I'm unble to follow your posts. They're too confusing for me. For some bizzare reason you keep changing the background color and you posts multilple posts one after another. Why do you do this?

[Pmb (OP)]

I'm curious about something. Do you folks understand that by the term "curvature" as it's used in "spacetime curvature" refers to what's known as intrinsic curvature? E.g. the surface of a sphere has intrinsic curvature but that of a cylinder extrinsic curvature.

See
http://mathworld.wolfram.com/ExtrinsicCurvature.html
http://mathworld.wolfram.com/IntrinsicCurvature.html

Einstein also used the term curvature to refer to acceleration as well, e.g. in his text Relativity: The Special and General theory

[yor_on]

I do not change the background colors?
Could it be your browser?

And yeah, I definitely wrote too much there
But I got two questions. The one about you using, if I got it right that is, equivalent observers, in a uniform gravitational field, finding them to have different accelerations. How would you go about describing it without the mathematics?

And the one about a 'test particle' meeting tidal forces, as passing some event horizon. It's about how one would define that particle? To have a acceleration at some point, or following a geodesic at all times? I think I read somewhere that Einstein defined it as long you don't find forces acting (free fall) on you, you're in a geodesic?

[Pmb (OP)]

And yeah, I definitely wrote too much there

If you don't mind me making a suggestion than you might want to consider shortening your post to more readable lengths. When a post is too long you read I skip over it no matter how much I'm interested in what the writer has to say. And I am interested in what you have to say. You have a reasonably good understanding of general relativity. I'd like to help you make it better. But I will not read multiple posts each of which is too long to begin with. I recommend that you try to shorten what you want to say and say it in fewer words. I.e. be more efficient in what you're writing. Think of this forum as you would a journal. In a journal article you only have so much space to get across what you want to and editors want you to say what you have to say using as little words as possible.

Also don't write posts one after each other. If there's something you want to add to your post and someone hasn't posted after you last one then use the edit function and edit that last post to say everything you want to in what would have been the next post. I've had this discussion with many other posters. We all want to hear what you have to say. We're just not going to spend what little time we have on the internet reading such long posts. Okay? Take all I’ve said here as a compliment, not as a criticism, okay my friend?

But I got two questions. The one about you using, if I got it right that is, equivalent observers, in a uniform gravitational field, finding them to have different accelerations. How would you go about describing it without the mathematics?

The equivalence of a uniform gravitational field and a uniformly accelerating frame of reference means that you cannot tell whether you’re in a uniformly accelerating frame of reference or a uniform gravitational field merely by observing nature, i.e. all particle experimental outcomes will be the same regardless of which frame of reference that you’re in. Toss a ball in a vacuum. Then the trajectory will be exactly the same in each frame and depend only on the initial conditions. Experiments in electromagnetism will be the same regardless of which frame you’re in. Etc.

And the one about a 'test particle' meeting tidal forces, as passing some event horizon.

Where did I say anything about such a thing?

[yor_on]

No problem Pete

How do you think of a intrinsic curvature? As a way to describe a space without referring to a embedding? If that is correct, how would you describe the space we and Earth traverse? It's a very interesting subject, but seems very tricky mathematically. As I get it it is a way to describe a curvature without referring to what the curvature may exist in?

Or, possibly giving you a same value/description no matter how you embed it, dimension-wise?
=

If you take a cylinder and then fold it out you get a 'plane' (a flat rectangular piece). So drawing a triangle on the outside would give you a same triangle when folded back, measuring by the triangles interior angles. That's called a extrinsic type of curvature. A intrinsic curvature would then be measuring the interior angles of a triangle, but on a ball, finding it to give you more than 180 degrees (as the triangles 'legs' bends over the ball). So telling you that you have a 'curved space'. Gaussian curvature. And all as I get it.

The first is a Euclidean 'flat' geometry. The one we use to define a triangle, and the one you can define by scaling the universe down to a 'locally flat space'. The other is a closer representation of our 'SpaceTime' as I get it. If our SpaceTime has a 'overall' curvature, taking you back to the point of origin without you ever deviating from a 'straight' path, you now should have defined both a extrinsic as well as intrinsic shape to our universe. But if you only are able to define intrinsic properties, what is the universe's shape? And does it need to have one?

Shape of the Universe. You might also say that it depends on if we got the Big Bang right, and so the homogeneous and isotropic universe we want
.

But the point still is that a intrinsic curvature does not have to relate to any shape at all as I get it, it's a intrinsic property we define, we can't place us outside our universe to find which way it is shaped, if it now is, in any special way. And all as I get it Pete

[yor_on]

Not you, it's me thinking about uniform motion, accelerations, and geodesics. I started to think about tidal forces, and wondering what it would mean if one used some particle, defined to to not be 'breakable' into smaller parts, passing a event horizon.
=

To me that 'test particle' should be in a geodesic at all times? And that matter break up under tidal forces being a measure of its particles finding new geodesics, as they get acted on by gravity, and acts. Maybe one also could see it as a question if gravity could be seen as a 'force' here?

[Pmb (OP)]

How do you think of a intrinsic curvature?

Itrinsic curvature is curvature that can be measured by observers who only have the ability to make measurements within the surface and they find that in their space geodesic which start out parallel do not remain parallel. E.g. the surface of a cylinder has zero intrinsic curvature.

See
http://mathworld.wolfram.com/IntrinsicCurvature.html
http://mathworld.wolfram.com/ExtrinsicCurvature.html

If you take a cylinder and then fold it out you get a 'plane' (a flat rectangular piece). So drawing a triangle on the outside would give you a same triangle when folded back, measuring by the triangles interior angles. That's called a extrinsic type of curvature.

That is wrong. The surface of the cylinder has zero intrinsic curvature for just the reasons you state

Gaussian curvature - Please note that Gaussian curvature is not defined for spaces of dimension other than two. So it can't be defined for spacetime, which has four dimensions.

Euclidean geometry is defined by the metric by the way so you need to be careful when using this term. E.g. even in a flat spacetime with no curvature the spacetime still can't be considered Euclidean unless the metric is defined in the same way it is in regular geometry, i.e. as

ds² = (cdt)² + dz² + dy² + dz²

If time is imaginary then this metric can be used in flat spacetime. Some authors still use imaginary time, e.g. Richard Mould does in his text Basic Relativity.

The way you used the term is not the correct way.

[Pmb (OP)]

Not you, it's me thinking about uniform motion, accelerations, and geodesics. I started to think about tidal forces, and wondering what it would mean if one used some particle, defined to to not be 'breakable' into smaller parts, passing a event horizon.
=

To me that 'test particle' should be in a geodesic at all times? And that matter break up under tidal forces being a measure of its particles finding new geodesics, as they get acted on by gravity, and acts. Maybe one also could see it as a question if gravity could be seen as a 'force' here?

I'm curioius yor_on. What do you think tidal forces are and how do they relate to spacetime curvature?

Request - Next time please append additional posts to your last post if nobody has posted since.

[yor_on]

= quote=

"If you take a cylinder and then fold it out you get a 'plane' (a flat rectangular piece). So drawing a triangle on the outside would give you a same triangle when folded back, measuring by the triangles interior angles. That's called a extrinsic type of curvature."

That is wrong. The surface of the cylinder has zero intrinsic curvature for just the reasons you state.

== End of quote


So when they write.

"Gaussian curvature is however in fact an intrinsic property of the surface, meaning it does not depend on the particular embedding of the surface; intuitively, this means that ants living on the surface could determine the Gaussian curvature. For example, an ant living on a sphere could measure the sum of the interior angles of a triangle and determine that it was greater than 180 degrees, implying that the space it inhabited had positive curvature.

On the other hand, an ant living on a cylinder would not detect any such departure from Euclidean geometry; in particular the ant could not detect that the two surfaces have different mean curvatures (see below), which is a purely extrinsic type of curvature."

You read this is a expression describing a two dimensional surface, right? It is a tricky subject, and one I haven't used myself.

(As for 'Euclidean geometry' I was referring to the type we used, and still use in school, before we get into SpaceTime and modern physics, topology, etc, suspecting that was how Euclid thought of it too?)
==

You know, rereading it it becomes even weirder. If we take the cylinder and assume it to describe some universe it seems to state that you can have a universe shaped as a cylinder and find it to be the same shape as when 'flattened out' measuring intrinsically. And then you say, if I got you right? That this only can be true in two dimensions. So what would I find, practically, measuring the same in three?

It also assumes 'ideal surfaces' it seems? How would I be able to be to measure anything, if I was part of that 'ideal surface'? To assume a 'ant' is either to introduce a third dimension, or mixing this subject with scales. But it's not about scales at all, well, as I read it? To me it's pure mathematical concepts?

Hopefully you will have a analogue, without the mathematics, that makes sense. Because practically, and as I think, we're part of the dimensions we measure, so when we find 'space' to 'bend' we should be 'bending' too, as a integral part of  those 'dimensions'?

[yor_on]

Why not answer the question Pete?
It can't be that strange?

If a 'undividable test particle' meets tidal forces, will it continuously be in a geodesic?
As for how I think of tidal forces, it's gravity, what else would it be?
=

As for relating it to a SpaceTime curvature, I better admit that I think it is 'space' myself
How would you define a 'ball shaped' universe as one proposition? Think Einstein used it himself to describe something having a 'straight path', still finding it ending at the origin?

[Pmb (OP)]

Why not answer the question Pete?

Which question are you referring to? I didn't realize that you asked me a question that I left unanswered. As I keep saying your posts are very confusing to me. You keep spreading them out over multiple posts and that makes things difficult for everyone here, not just me. And you know that since it's been pointed out to you on more than one occasion yet you still keep doing it.

My question to you is why not make your posts easier to understand and don't spread them out over multiple contiguous posts?
It can't be that strange?

If a 'undividable test particle' meets tidal forces, will it continuously be in a geodesic?

Only if it has no spin.

As for how I think of tidal forces, it's gravity, what else would it be?

I don't know. Some of the things I think are obvious turn out to be very different from what they really are with some people. I just want to be 100% correct that we're on the same page on this one point. Do you not wish to answer it? If not then why didn't you answer it?

As for relating it to a SpaceTime curvature, I better admit that I think it is 'space' myself

I'm trying to determine precisely what your understanding of general relativity are. If you're not going to answer the questions I ask then please state so now so that I won't waste my time asking questions that won't be answered.

The questions I answered have very straight forward responses. They're very simple to answer if you have a solid understanding of GR. Until you answer the questions I've already asked then I'm going to consider my discussion about curved spacetime with you as being over.

[yor_on]

You're free to do so Pete.

[Pmb (OP)]

You're free to do so Pete.

As are you.

It might seem as if I'm intentionally being hard to get along with but that's not quite right. You asked me why I didn't answer your question. The reason is that the way you post and spread it all out over multiple posts is so confusing that it's hard to follow. I've explained that to you on more than one occasion so this isn't new to you. You not only didn't make any changes to your posting style to be clearer but you made no attempt to explain why you can't do so. So don't be surprised if I choose not to respond to your questions. Especially when you make no attempt to answer my questions. Nope. That's not how the world works my friend.

[captcass]

Relativity is based upon the fact that, although we know the universe has an invariant rate of time, we cannot see it that way because it takes time for light to travel, creating Lorentz transformations, and energy density slows the rate of time. We see the illusion, not the reality. We see objects moving "through" space, but we know they are densities evolving forward in the quantum continuum. Although we see dimentionality, all three dimensions are being evolved forward simultaneously by time, in the forward direction of time. As time has no depth, neither do the dimensional spatial events being evolved forward. The evolution appears to turn this way and that way towards densities because time slows in densities. The denser the "mass", the greater its resistance to evolution...

So we are not seeing stellar systems revolving around MECOs "through space". We are seeing spatial densities evolving forward at different rates within the evolving continuum.  We see stellar systems evolving forward faster the farther they are from the steep dilation gradient at the center as the continuum evolves forward.

[Galileo1564]

Rest in Peace PMB.

We lost one of the few people who understood gravity as Einstein did.

"The interpretation of gravity as a curvature in space-time is an interpretation Einstein did not agree with."
-Peter M. Brown

https://arxiv.org/abs/physics/0204044 [nofollow]