Can a frame of reference be wrong in determining motion in space-time?

[JesseM]

I suppose it means that when I am talking about space-time you want to talk about coordinates systems.

But I continually emphasized that I was talking about coordinate systems throughout this thread, and you never seemed to have a problem with this before. This whole issue of straight lines seems to have gotten started in post #26, where I said:

Apart from flat spacetime, I don't think it's necessarily going to be possible to find coordinate systems where all geodesics end up being straight lines.

This is clearly a statement about coordinate systems. You took issue with it, quoting this statement in post #27 and replying:

Really, I don't see the problem actually.

What kind of geodesics are you thinking of that would be a problem?

In post #28 I gave the examples of orbits which repeatedly cross, and then in post #29 you said you didn't see a problem because "straight lines can cross in curved space-time", and in post #30 I emphasized that I meant "straight" in coordinate terms:

But we were talking about straight lines in terms of the coordinate system, not just straight lines in terms of geodesics. If a worldline is straight in terms of the coordinates (meaning its position coordinate changes as a constant rate when you vary the time coordinate)

Then when you continued to take issue with my statement that paths which are "straight" cannot cross multiple times, I said again that I only meant "straight" in the coordinate sense, and in post #36 you replied:

I thought I already made clear I was talking about lines that are straight in the coordinate sense.

Yes and I am talking about that as well.

So are you saying now that you weren't talking about "straight" in the coordinate sense? Because I kept emphasizing again and again in subsequent posts that that is all I was talking about, how come you didn't ever say something like "I'm not talking about straightness in the coordinate sense, I'm not interested in talking about coordinate-dependent notions"? Will you at least agree now, in retrospect, that I made it pretty clear that this is what I was talking about all along, and that you failed to pick up on this or misunderstood the difference between coordinate-dependent notions and coordinate-independent notions like curvature?

Why can't you talk about space-time without a notion of a coordinate system?

You can, but the only way to do it in a physically meaningful way is to phrase all your comments in terms of coordinate-independent quantities, which is why I said you'd need to come up with a coordinate-independent way of judging whether a given worldline is helix-shaped or not (I don't have any opinion on whether this would be possible or not).

With regards to proper time and proper distance, they are the time and distance from a local frame of reference. Other frames of reference will measure different times and distances.

In flat spacetime, for an inertial worldline, the proper time is simply the time in the frame where that object is at rest (you are misusing the phrase 'local' though, in relativity that means the infinitesimal neighborhood of a single point in spacetime). It is still coordinate-independent, because all frames will agree on the proper time. And for curved worldlines in flat spacetime, or any worldline in curved spacetime, the proper time isn't based on any "reference frame" at all--it means the total time that would be elapsed by a clock traveling along that worldline between the two events you're talking about. Again, this is coordinate-independent, since all coordinate systems must get the same answer for the proper time between two events on a given worldline. For example, if you have a curved worldline in flat spacetime, and in a given inertial frame the two events on this worldline you want to know the proper time between happen at $$t_0$$ and $$t_1$$, and the velocity of the object on the worldline as a function of the time-coordinate in the inertial frame you're using is v(t), then the proper time would be the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$. This integral would have to come out the same in all inertial coordinate systems, even though each coordinate system will have a different pair of coordinate times $$t_0$$ and $$t_1$$ and a different function v(t) for coordinate velocity as a function of coordinate time. So, the proper time is coordinate-independent in this sense.

It is standard to call the proper time "coordinate independent" or just "invariant", if you say it is coordinate-dependent physicists will look at you funny. For example, on http://www.astro.ku.dk/~cramer/RelViz/text/geom_web/node2.html says:

Proper length is analogous to proper time. The difference is that proper length is the invariant interval of a spacelike path while proper time is the invariant interval of a timelike path.

Also, just as proper time can be defined along a general timelike path, whether inertial or non-inertial, so proper length can be defined along a general spacelike path, with the formula in curved spacetime given in terms of an integral involving the metric, which is given on the above wikipedia page.

And if you agree the notion of a "geodesic" between two points is a coordinate-independent one, note that geodesic just means the path that goes through both points which has the extremal value (usually the maximum value) of the proper time between those points! That's why, for example, in the twin paradox in flat spacetime the twin who moves inertially between the event of the other twin leaving and returning will always be older than the twin who moves non-inertially between these events, regardless of the path taken by the non-inertial twin--the inertial path is the unique geodesic between these points in flat spacetime, and thus has a greater value for the proper time than any other path.

 

[MeJennifer]

Ok, hopefully we get it right now:

In my view geodesics are straight lines in space-time. Even lines that go through warped space-time are still straight. So I am not talking about any particular coordinate system here, simply space time.

Now you claim that we cannot have any coordinate system where all geodesics are represented as a straight line.
Ok, that is your position, and it seems I will not be able to convince you of the contrary.

One of your objections is repeatedly crossing lines in two objects that are in orbit in opposite directions in space.

I tried to explain to you that when we look at those orbits in space-time (so not in space but in space-time) they do not cross at all and that they look like a (time deformed) helix.

Yes proper time and proper distance can be calculated but these properties are not observed except for an observer who is at rest to it.

 

[selfAdjoint]

Ok, hopefully we get it right now:

In my view geodesics are straight lines in space-time. Even lines that go through warped space-time are still straight. So I am not talking about any particular coordinate system here, simply space time.

Now you claim that we cannot have any coordinate system where all geodesics are represented as a straight line.
Ok, that is your position, and it seems I will not be able to convince you of the contrary.

One of your objections is repeatedly crossing lines in two objects that are in orbit in opposite directions in space.

I tried to explain to you that when we look at those orbits in space-time (so not in space but in space-time) they do not cross at all and that they look like a (time deformed) helix.

Look, MeJennifer, can you have straight lines on the surface of the Earth? Not chords which go through the interior, but straight lines restricted to lie on the surface? How on Earth ( ) can that be? If the staight line touches the surface at some point then it will head off into space as a tangent, won't it?

Well, exactly the same thing happens with the idea of straight lines in curved spacetime. Straight lines don't live in spacetime, but in the tangent Minkowski spaces to spacetime.

Geodesics on the surface of the Earth are not straight lines but great circles. Particular curved arcs. Likewise geodesics in curved spacetime are curved lines. They are the "straightest possible" curves between their endpoints, but that doesn't make them straight, it just makes them less sharply curved than other paths between those points.

 

[MeJennifer]

Look, MeJennifer, can you have straight lines on the surface of the Earth? Not chords which go through the interior, but straight lines restricted to lie on the surface? How on Earth ( ) can that be? If the staight line touches the surface at some point then it will head off into space as a tangent, won't it?

Well, exactly the same thing happens with the idea of straight lines in curved spacetime. Straight lines don't live in spacetime, but in the tangent Minkowski spaces to spacetime.

Geodesics on the surface of the Earth are not straight lines but great circles. Particular curved arcs. Likewise geodesics in curved spacetime are curved lines. They are the "straightest possible" curves between their endpoints, but that doesn't make them straight, it just makes them less sharply curved than other paths between those points.

Yes one can have straight lines on a surface. Space-time is a 4D manifold, not a 5 dimensional space. So where do you suppose your line is heading for if not on the surface?

Straight lines don't live in spacetime, but in the tangent Minkowski spaces to spacetime.

Sounds very confusing to me, what do you mean?

We can have a tangent vector on a point in the worldline and we can construct a plane of simultaneity on that, which is basically the view of 3D space for an observer on that worldline.
Kind of how you and I see the world

But there is no such thing as a tangent Minkowski space in space-time, unless I miss something.

Edited: Oh I see what you mean, but that is not relevant to whether something is a straight line in space-time.

 

[JesseM]

Ok, hopefully we get it right now:

In my view geodesics are straight lines in space-time. Even lines that go through warped space-time are still straight. So I am not talking about any particular coordinate system here, simply space time.

Fine, but then your definition of "straight line" has nothing to do with mine.

Now you claim that we cannot have any coordinate system where all geodesics are represented as a straight line.

I only make that claim if "straight line" is understood to mean "constant dx/dt, dy/dt and dz/dt", not if a geodesic is defined as a straight line. Do you agree that if we use my definition of "straight line", then it is impossible to find a coordinate system where all geodesics are straight lines? Obviously if we use your definition they will be, because you have simply defined geodesics as "straight lines"!

Ok, that is your position, and it seems I will not be able to convince you of the contrary.

Does "convince me of the contrary" mean convincing me to adopt your definition instead of mine (in which case I agree that geodesics are straight, by definition), or does it mean you want to convince me that even if we use my definition instead of yours, it is still wrong to claim there's no coordinate system where all geodesics are "straight lines"? In debates about science, you must understand that it's essential to define all your terms, my discussions with you often seem to founder on the fact that you use nonstandard terminology and have unclear or shifting definitions of what your own terms mean.

I tried to explain to you that when we look at those orbits in space-time (so not in space but in space-time) they do not cross at all

Er, when did you say that? I don't remember you ever saying they don't cross at all, that argument is clearly wrong. After all, I specifically said I was talking about two objects orbiting in opposite directions (one clockwise and the other counterclockwise, for example) which pass right next to each other (understood to mean infinitesimally close) with each orbit--in other words, each crossing happens at a unique point in spacetime. In this case, the two worldlines in spacetime will not look like a double helix, where the two strands always maintain the same distance--instead the two strands will repeatedly cross at a single point in spacetime, then move apart as they move to opposite sides of their orbits, then approach each other and cross again.

and that they look like a (time deformed) helix.

Not if they are orbiting in opposite directions and repeatedly crossing arbitrarily close to one another, as I have said. If they were orbiting in the same direction such that they were always on opposite sides of the same circular orbit, then their worldlines would look like a double helix (at least in a typical coordinate system--like I said before, I don't know what it means to say a worldline is shaped like a corkscrew or a straight line or any other shape without having a rigorous coordinate-independent notion of 'shape'.)

Yes proper time and proper distance can be calculated but these properties are not observed except for an observer who is at rest to it.

What is it that is being obeserved, exactly? All observers will see a clock moving along a particular worldline tick the same amount of ticks between two events on its own worldline, and that is how "proper time" is defined, not in terms of the time between those events in the observer's own frame.

Anyway, you are once again making up your own definitions of words instead of using the standard definitions that all physicists use. All physicists define quantities that are the same in all coordinate systems to be "coordinate-independent" or "invariant", they don't make a distinction between quantities that are "calculated" vs. quantities that are "observed". This doesn't seem like a well-defined distinction anyway, since as I pointed out above you can sometimes think of a quantity that is calculated in terms of something directly "observed", and the reverse would probably be true as well--can you name any coordinate-independent quantities which you think are purely "observed" rather than calculated? Also, do you agree that there is no way to define a "geodesic" without making reference to proper time or proper distance?

 

[jcsd]

Yes one can have straight lines on a surface. Space-time is a 4D manifold, not a 5 dimensional space. So where do you suppose your line is heading for if not on the surface?

In Euclidean space starightlines are geodesics because thye are locally minimize the distanc which is a general and defining property of geodesics. However Euclidean space has the parallel postulate (Minkowski space has the parallel psotulate, but spacetime in general does not), i.e. that parallel straight lines never intersect; in other spaces, for example the surface of a sphere, all geodesics intersect.

So geodesics share the length minimizing properties of startight lines in Euclidean spaces, but they do not generally share the other properties of staright lines in Euclidean spaces.

Sounds very confusing to me, what do you mean?

We can have a tangent vector on a point in the worldline and we can construct a plane of simultaneity of that which is basically the view of 3D space from an observer's perspective.
But there is no such thing as a tangent Minkowski space in space-time, unless I miss something.

Spacetimes are pseudo-Riemannian manifolds and as such each point is equipped with a tangent space.

 

[MeJennifer]

In Euclidean space starightlines are geodesics because thye are locally minimize the distanc which is a general and defining property of geodesics. However Euclidean space has the parallel postulate (Minkowski space has the parallel psotulate, but spacetime in general does not), i.e. that parallel straight lines never intersect; in other spaces, for example the surface of a sphere, all geodesics intersect.

So geodesics share the length minimizing properties of startight lines in Euclidean spaces, but they do not generally share the other properties of staright lines in Euclidean spaces.

I do not disagree with you on that at all. But that does not mean you cannot call a line on a surface straight.

Spacetimes are pseudo-Riemannian manifolds and as such each point is equipped with a tangent space.

Yes true, I am not sure what your point is?

 

[jcsd]

Vectors are not happy unless they have a space to live in with all their buddies. A tangent space is where all the tangent vectors at a point live.

 

[MeJennifer]

Vectors are not happy unless they have a space to live in with all their buddies. A tangent space is where all the tangent vectors at a point live.

I still do not see the point you are trying to make.

 

[jcsd]

A straight line has a meaning that is not synonymous with geodesic.

In the case of Lorentzian manifolds the tangent spaces are Minkowski spaces.

 

[MeJennifer]

A straight line has a meaning that is not synonymous with geodesic.

In the case of Lorentzian manifolds the tangent spaces are Minkowski spaces.

Ok correct, but again what is the relevance?

 

[JesseM]

Ok correct, but again what is the relevance?

The relevance is that you were calling geodesics "straight lines", and that you said:

But there is no such thing as a tangent Minkowski space in space-time, unless I miss something.

Calling geodesics straight lines is debatable I think--they don't have all the properties of straight lines in Euclidean space, but maybe you could argue that Euclidean straight lines are just a special case of a more general notion of "straight lines", I don't think this term has any set definition--but as I understand it the tangent space to any point in curved spacetime is definitely a Minkowski space, and I think that's what jcsd is saying.

 

[MeJennifer]

as I understand it the tangent space to any point in curved spacetime is definitely a Minkowski space, and I think that's what jcsd is saying.

He is correct.
But again it has no relevance on a straight lines in space-time.

Perhaps we should call it quits. The topic has been dragging on for a while.

 

[JesseM]

He is correct.

Are you admitting you were wrong in your statement about tangent spaces, then?

But again it has no relevance on a straight lines in space-time.

No, but it has relevance to your statement "there is no such thing as a tangent Minkowski space in space-time", doesn't it?

Perhaps we should call it quits. The topic has been dragging on for a while.

Up to you, although I would have been interested in seeing your response to my post #75. After all this time I still am not clear on whether you are actually disagreeing with the main point I have been making all along, namely that it is impossible to find a coordinate system where all geodesics are "straight lines" in the coordinate sense that I have defined (dx/dt, dy/dt and dz/dt constant), not according to any other definition.

 

[MeJennifer]

Ok, since you asked!

Fine, but then your definition of "straight line" has nothing to do with mine. I only make that claim if "straight line" is understood to mean "constant dx/dt, dy/dt and dz/dt", not if a geodesic is defined as a straight line. Do you agree that if we use my definition of "straight line", then it is impossible to find a coordinate system where all geodesics are straight lines?

Well my answer to that is how can that possibly refer to curved space-time?

Let's for the sake of argument consider this, then for a flat space-time there is no problem. If it is constant the line is straight, if not then it is curved.
But we are not talking about flat space-time but about space-time that can have local warpings of both the space and time components. Generally such warping takes place when we have a mass present.
So clearly if we simply take dx/dt, dy/dt and dz/dt and such a local warping is near it's path it is not going to work. So in other words, to determine if the wordline remains straight we have to follow the surface.
Does that make sense to you? Do you see that you cannot simply take dx/dt, dy/dt and dz/dt over a warped region?

In a simple world analogy, imagine skiing from one place to another in the mountains with hills and valleys in between. You could still ski in a straight line right? However a guy looking at it from within a plane is going to see you make funny curves.

Er, when did you say that? I don't remember you ever saying they don't cross at all, that argument is clearly wrong.

Ok, well they come close together.

After all, I specifically said I was talking about two objects orbiting in opposite directions (one clockwise and the other counterclockwise, for example) which pass right next to each other (understood to mean infinitesimally close) with each orbit--in other words, each crossing happens at a unique point in spacetime. In this case, the two worldlines in spacetime will not look like a double helix, where the two strands always maintain the same distance--instead the two strands will repeatedly cross at a single point in spacetime, then move apart as they move to opposite sides of their orbits, then approach each other and cross again. Not if they are orbiting in opposite directions and repeatedly crossing arbitrarily close to one another, as I have said.

Well they are simply two helices going in opposite directions.

like I said before, I don't know what it means to say a worldline is shaped like a corkscrew or a straight line or any other shape without having a rigorous coordinate-independent notion of 'shape'.)

Well that is your prerogative, but frankly I do not see the point. I am quite happy with one particular view. Must be a personal application of Ockham's razor.

 

[JesseM]

Well my answer to that how does that refer to curved space-time?

Because I'm talking about geodesics in curved spacetime, and a coordinate system on this curved spacetime (again, a coordinate system is the same thing in any spacetime, flat or curved--a function that assigns 3 space coordinates and 1 time coordinate to each point in the spacetime.)

But we are not talking about flat space-time but a space-time that can have local warping of both the space and time components. Generally such warping takes place when we have a mass present.
So clearly if we simply take dx/dt, dy/dt and dz/dt and such a local warping is in it's path it is not going to work.

I have no idea what you mean by "it's not going to work"--again, you seem to be totally confused about the difference between a metric and a coordinate system. A coordinate system is just a function that assigns each point in the spacetime coordinate x,y,z,and t, and once we have labeled every point in this way, we can look at a particular worldline and look at what its x coordinate is at a given t-coordinate, giving the function x(t) (for example, if the worldline passed through a point which the coordinate system labelled as x=5 meters, t=7 seconds, then x(7 seconds) = 5 meters.) Likewise with y(t) and z(t). Once we have these functions, dx/dt and dy/dt and dz/dt are simply the derivatives, we don't have to worry about the metric at all. That's just how coordinate systems work!

Does that make sense to you? Do you see that you cannot simply take dx/dt, dy/dt and dz/dt over a warped region?

No, you are fundamentally confused about the difference between coordinate systems and metrics. I promise you, any physicist would agree that if you have the parametrization of a worldline in a particular coordinate system like x(t), y(t), z(t), then dx/dt and dy/dt and dz/dt are just the ordinary derivatives of these functions (all that dx/dt means is 'the rate the x-coordinate of the worldline is changing as a function of its t coordinate', it has no physical significance beyond that), the metric doesn't enter into it (although you do need the metric if you first define the coordinate system and then want to predict the functions x(t), y(t) and z(t) for an object's position as a function of time in this coordinate system).

Ok, well they come close together.

Not just "close", the idea is that the distance is zero when they pass. This is an idealization, but it's no difference then the idealization used in the twin paradox, where you treat the moment they reunite as a single point in spacetime even though the distance between two flesh-and-blood twins couldn't be zero. Just imagine two orbiting spheres going in opposite directions which actually graze each other each time they pass, and then imagine taking the limit as the size of the spheres goes to zero, so they are two orbiting points which unite at a single point in spacetime each time they pass.

Well they are simply two helices going in opposite directions.

They are helices that cross paths repeatedly.

like I said before, I don't know what it means to say a worldline is shaped like a corkscrew or a straight line or any other shape without having a rigorous coordinate-independent notion of 'shape'.)

Well that is your prerogative, but frankly I do not see the point.

The point is that you can't approach physics like it was politics or some other highly subjective topic. All your terms should have precisely definable meanings in physics, if they don't then you're just relying on vague intuitions and as Pauli once said, such arguments are "not even wrong".

 

[Rach3]

So clearly if we simply take dx/dt, dy/dt and dz/dt and such a local warping is near it's path it is not going to work. So in other words, to determine if the wordline remains straight we have to follow the surface.
Does that make sense to you? Do you see that you cannot simply take dx/dt, dy/dt and dz/dt over a warped region?

In a simple world analogy, imagine skiing from one place to another in the mountains with hills and valleys in between. You could still ski in a straight line right? However a guy looking at it from within a plane is going to see you make funny curves.

That analogy (in blue) has nothing to do with your problem (in black). Derivatives are a local property of functions. You define a derivative at a single point on your snowy hill; the surrounding valleys and such are irrelevant.

 

[Rach3]

MeJennifer - I suggest there's no reason why you're making these arguments here. You are confused simply because you haven't yet learned the formalism of GR, which is differential geometry. There's no sense in asking, "how could this possibly work?", because it certainly does work, and everything is internally consistent; and you will only be able to understand how it is so when you learn the math. There's simply no shortcut. So if you're really interested in what geodesics are, or how to define derivatives on a manifold, it will take effort on your part, and a very thick textbook.

 

[MeJennifer]

That analogy (in blue) has nothing to do with your problem (in black). Derivatives are a local property of functions. You define a derivative at a single point on your snowy hill; the surrounding valleys and such are irrelevant.

The question is: can we have a coordinate system of space-time where all geodesics can be represented as straight lines.
To determine if a geodesic is straight that passes a warped region in space-time we would need a description of the curvature.
That's if why we cannot conclude that if dx/dt, dy/dt and dz/dt remains constant we have a straight line. Or in other words without a metric describing the warping at all places on the wordline we would not be able to conclude that.

 

[HallsofIvy]

What do you mean by "represented as straight lines"? Given any manifold, there exist a coordinate system in which geodesics, locally, are straight lines (that is, that the metric tensor, evaluated at that point, is Euclidean).

In a simple world analogy, imagine skiing from one place to another in the mountains with hills and valleys in between. You could still ski in a straight line right?

No, of course not! What do you mean by "straight line" here?

 

[selfAdjoint]

The question is: can we have a coordinate system of space-time where all geodesics can be represented as straight lines.
To determine if a geodesic is straight that passes a warped region in space-time we would need a description of the curvature.
That's if why we cannot conclude that if dx/dt, dy/dt and dz/dt remains constant we have a straight line. Or in other words without a metric describing the warping at all places on the wordline we would not be able to conclude that.

This is exactly right. I have bolded the key statement. And in (pseudo-)Riemannian geometry we have a metric tensor from which, in this case, we can derive a connection, a mathematical expression in the partial derivatives of the metric tensor, from which we get a covariant derivative which finally gives us a Riemannian or Curvature tensor, which describes the curvature. All of these derivations are straight computations from the metric; that is a feature of Riemannian geometry. Once we have this mathematical machinery that determines the curvature, we can do further mathematical derivations to find the form of the geodesic equations.

Now if the Riemann tensor is not identically zero then the form of the geodesic equations is not linear. Translation into English; if spacetime has nonzero curvature somewhere, then the geodesics there cannot be straight lines. This again is a straight computation and so it really only depends on the metric. In a Riemannian geometry the metric determines the curvature, and the geodesics for non zero curvature are not linear

 

[MeJennifer]

if spacetime has nonzero curvature somewhere, then the geodesics there cannot be straight lines.

Why not?
What else is a geodesic but a straight line on a curved surface?

 

[MeJennifer]

if spacetime has nonzero curvature somewhere, then the geodesics there cannot be straight lines.

Then what do you consider a straight line on a curved surface?
What else is a geodesic but a straight line on a curved surface?

 

[selfAdjoint]

Why not?
What else is a geodesic but a straight line on a curved surface?

A path which has minimum curvature between its endpoints.

 

[MeJennifer]

A path which has minimum curvature between its endpoints.

And you say that calling that a straight line is wrong?

 

[selfAdjoint]

And you say that calling that a straight line is wrong?

A straight liine is one where the local tangent vector to it at some point stays parallel to itself as you move along the line. In a curved geometry that can't happen; there are no mathematically straight lines in a curved Riemannian geometry that can serve for geodesics.

 

[MeJennifer]

A straight liine is one where the local tangent vector to it at some point stays parallel to itself as you move along the line. In a curved geometry that can't happen; there are no mathematically straight lines in a curved Riemannian geometry that can serve for geodesics.

You basically limit the concept of a straight line to Euclidean surfaces only.

But anyway I learned my lesson, no more straight line when I can use geodesic.

 

[nrqed]

You basically limit the concept of a straight line to Euclidean surfaces only.

But anyway I learned my lesson, no more straight line when I can use geodesic.

The whole discussion arose because you had decided to make you own definition of "straight lines" as being the same as geodesics, without making first the effort to learning what the rest of the physics community defines as a straight line. Sure, you can do that if you want, but then the price to pay is endless discussions like this. And this is only because you did not pause and ask, humbly, "I always thought that a straight line is the same as a geodesic. Can someone confirm this or correct me?"



SelfAdjoint has given the standard definitions of straight lines and geodesics. Two-dimensional beings living on on the surface of a sphere could realize they live on a cruved surface using local measurements only and that would be based on the fact that geodesics would *not* be straight lines. For example by using parallel transport of a vector along a short geodesic, turning by 90 degrees and so on until one is back to the starting point after having gone through a paralleliped. The vector would come out rotated which would be an indication that the geodesics were not straight lines (according to the definition used by the physics community). The amount of rotation would allow the calculation of the curvature.

What did *you* call a line for which the tangent vector remains parallel to itself?

You have to be willing to learn the terminology used by everyone before arguing that they are wrong about something.

Regards

 

[nrqed]

You basically limit the concept of a straight line to Euclidean surfaces only.

But anyway I learned my lesson, no more straight line when I can use geodesic.

He did not "limit" the concept of straight line to Euclidiean spaces only, he gave it a rigorous mathematical definition, which is the one used by everyone in the field. The fact that in a curved Riemannian geometry straight lines do not correspond to geodesics follow from their mathematical defintion.

One could as well argue that *you* had restricted (in a completely different way) the meaning of straight lines too! You had restricted them to be identical to geodesics! The main difference with SelfAdjoint's "limitation" is that his corresponds to what everyone uses in the field.

Regards

Patrick

 

[JesseM]

The question is: can we have a coordinate system of space-time where all geodesics can be represented as straight lines.
To determine if a geodesic is straight that passes a warped region in space-time we would need a description of the curvature.
That's if why we cannot conclude that if dx/dt, dy/dt and dz/dt remains constant we have a straight line.

Sigh. I made it very clear throughout this entire thread that when I said you couldn't find a coordinate system where all geodesics were "straight in the coordinate sense", I was defining the term "straight in the coordinate sense" to mean constant dx/dt, dy/dt, and dz/dt. I repeated this over and over again in many posts, just to make sure there was no confusion on this point. So if you are conceding that you cannot find a coordinate system where all geodesics have constant dx/dt, dy/dt and dz/dt, regardless of whether this would disqualify them from being "straight lines" under your preferred definition, then you are either admitting you were wrong all along in disagreeing with me, or admitting that you were not even paying a bare minimum of attention to what I was actually saying in my posts (since I repeated this definition in like every other post!)