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

[MeJennifer]

Going back to my previous example, it should be obvious, I hope, that the space-station in Earth orbit is following a geodesic, and that the worldline describing its orbit is not "straight" in the sense that dx/dt = dy/dt = dz/dt = constant. In fact, the orbit is in genreal an ellipse (for the example I had in mind, a circle).
Was that really your question, or did I misunderstand something?

This whole discussion is about space-time not about space.

The relevant question was:

So are you saying that in a non Euclidean coordinate system modeling space-time there exist a wordline on a geodesic that is not straight?

We are not talking about if an observer "sees" the line straight or curved.

 

[JesseM]

Going back to my previous example, it should be obvious, I hope, that the space-station in Earth orbit is following a geodesic, and that the worldline describing its orbit is not "straight" in the sense that dx/dt = dy/dt = dz/dt = constant. In fact, the orbit is in genreal an ellipse (for the example I had in mind, a circle).
Was that really your question, or did I misunderstand something?

This whole discussion is about space-time not about space.

If something is moving in an elliptical path in terms of your spatial coordinates, there is no way that dx/dt, dy/dt and dz/dt (which involve time as well as space) can all be constant.

 

[MeJennifer]

If something is moving in an elliptical path in terms of your spatial coordinates, there is no way that dx/dt, dy/dt and dz/dt (which involve time as well as space) can all be constant.

But you are thinking in terms of space not in terms of space-time.

For instance one can make an elipse looks like some kind of a helix (a helix + some time deformations on it) in a simple model of spacetime.
So are you then saying that the lines of a helix cannot be straight in curved space-time?

 

[JesseM]

So for instance are you saying that the lines of a helix cannot be straight in curved space-time?

"Straight" in the coordinate sense rather than the geodesic sense? In the coordinate sense I suppose you could pick a distorted coordinate system where the helix was straight, but then the coordinate positions it was covering would no longer describe a circle, in terms of this coordinate system its spatial path would just be a straight line or a point. If the spatial coordinates that a path travels through describe an ellipse in terms of that coordinate system (meaning that when you look at every (x,y,t) corresponding to a point on the worldline, you find that the x and y coordinates always satisfy some ellipse equation like (x - h)^2 / a^2 + (y - k)^2 / b^2 = 1), then there is no way that dx/dt and dy/dt can be constant in terms of that same coordinate system.

As with the helix, you could always pick some new coordinate system where that particular worldline is straight, in which case the spatial path will be a line or a point rather than an ellipse in that coordinate system. But for a general curved spacetime, you won't be able to find a coordinate system where all geodesics are straight in the coordinate sense. I gave a simple proof involving orbits that repeatedly cross--if you still don't understand/agree with this proof, I ask again that you address my question in post #41--and pervect gave what looks like a more general proof in post #42, although I don't have the knowledge of tensor mathematics to follow it.

 

[pervect]

Then I don't understand your question. So I'll have to ask again - what do you mean by a straight line, if you don't mean

dx/dt = dy/dt = dz/dt = constant?

I thought that's the definition we had agreed on? There's some communication problem here...

A few notes which may or may not help:

Straight lines in Euclidean space don't involve time at all, and minimize the distance between two points. You can work out using the calculus of variations the fact that the shortest distance between two points in an Euclidean space is a straight line.

Geodesics in space-time do involve both space and time, and maximize the Lorentz interval between two points.

The similarity is the fact that some sort of "distance intergal" (regular distance in Euclidean geometry, Lorentz interval in relativity) is made to have a locally extereme value. This implies that the trajectory must satisfy the Euler-Lagrange equations.

http://en.wikipedia.org/wiki/Euler-Lagrange_equations

The Euler-Lagrange equation, developed by Leonhard Euler and Joseph-Louis Lagrange in the 1750s, is the major formula of the calculus of variations. It provides a way to solve for functions which extremize a given cost functional. It is widely used to solve optimization problems, and in conjunction with the action principle to calculate trajectories. It is analogous to the result from calculus that when a smooth function attains its extreme values its derivative vanishes.

The "distance" intergal in my terminology is the "cost functional" referred to in the Wikipidea article above. A functional is anything that assigns a value to a function. One specifies a trajectory (technically, a function), and gets some sort of number in return (the distance, time, or the "cost functional"). One seeks to find the optimum trajectory which causes the returned number (functional, etc.) to have a maximum or minimum value.

The use of the Euler-Lagrange equations to show that the shortest path between two points as a straight line is undoubtedly overkill, but is a good place for "jumping off" into generalizations of geodesics.

 

[MeJennifer]

Then I don't understand your question. So I'll have to ask again - what do you mean by a straight line, if you don't mean

dx/dt = dy/dt = dz/dt = constant?

I thought that's the definition we had agreed on? There's some communication problem here...

Sorry pervect but it seems to me that you can only take those derivatives in a flat space-time. Once it becomes curved we have to take the curvature in consideration as well and for each event!

Clearly space-time is not uniformly curved, and that is what I don't understand when JesseM speaks about coordinates. Each space-time event has its own measure of curvature, isn't that correct?

 

[Hurkyl]

Sorry pervect but it seems to me that you can only take those derivatives in a flat space-time.

There is no metric involved. x is a scalar function, d/dt is a tangent vector, and dx/dt is well-defined.

 

[MeJennifer]

There is no metric involved. x is a scalar function, d/dt is a tangent vector, and dx/dt is well-defined.

What do you mean no metric involved, we are talking about the worldline of something influenced by a mass.
We have to take into account not only the curvature of space but also the curvature of time due to this mass!

 

[JesseM]

What do you mean no metric involved, we are talking about the worldline of something influenced by a mass.
We have to take into account not only the curvature of space but also the curvature of time due to this mass!

If you first construct the coordinate system and then want to predict the behavior of worldlines in your chosen coordinate system, then physical considerations like the curvature of spacetime would be involved in your prediction. But once you have already figured out the parametrization of the worldline in your coordinate system, getting some function x(t) to tell you how the x-coordinate changes as a function of the t-coordinate, then no additional physical considerations are needed to figure out dx/dt, it's just a straightforward derivative. If the worldline can be described by x(t) = 2t in your chosen coordinate system, that automatically means that dx/dt = 2 in this coordinate system. dx/dt just means the rate the x coordinate changes as a function of the t-coordinate...if you add 1 to the t-coordinate the x-coordinate always increases by 2, in this example. Like I said all along, a "straight line in coordinate terms" just means that the spatial coordinates change at a constant rate as you vary the time-coordinate, nothing more.

This means, for example, that if you have already found the functions x(t), y(t) and z(t) for a worldline, and found that the relation between the spatial coordinates satisfies the equations of some ellipse, then this is enough to tell you that dx/dt, dy/dt and dz/dt can't all be constant, so the worldline cannot be "straight" in the coordinate sense.

 

[MeJennifer]

Ok I give up! I am either crazy or not understood, or both!

I simply don't understand how we can get from a question concerning space-time, not space, but space-time into a discussion about "dx/dt = dy/dt = dz/dt = constant is not valid for an elipse".
The elipse is not in space-time, there is no elipse in space-time!
In space-time we have a helix (plus some time distortions) on a local area of curved space-time.

Remember the original question:

If we model space-time in a coordinate system which is not constrained to be Euclidean (Euclidean as in validating the axioms of Euclid), then do you claim that no such system can exist where all geodesics can be represented as straight lines?

To me it seems there is no problem whatsoever to have a system like that.

 

[JesseM]

Ok I give up! I am either crazy or not understood, or both!

No, you're just not understanding the argument.

I simply don't understand how we can get from a question concerning space-time, not space, but space-time into a discussion about "dx/dt = dy/dt = dz/dt = constant is not valid for an elipse".
The elipse is not in space-time, there is no elipse in space-time!
In space-time we have a helix (plus some time distortions) on a local area of curved space-time.

I agree, we have something like a helix in spacetime. The point is that if you know the function for a worldline in space and time, you can also figure out the set of points in space alone[/i] that the path crosses through, at different times. You can think of this as the "shadow" of the entire 4D worldline on a single 3D plane. For example, the "shadow" of a verticle helix would just be a circle, and the spatial coordinates of an event on the helix at any any time would always lie on that circle. As an example, the helix might be described by the following equations:

x(t) = 2*cos(t / 5 seconds) meters
y(t) = 2*sin(t / 5 seconds) meters
z(t) = 0

Now, if we let t take any possible value, would you agree that these equations are a coordinate description of a "helix" in 4D spacetime? And would you agree that no matter what t you choose, the z coordinate at that t will always be 0, and the x and y coordinates will always lie on a circle on the xy plane which satisfies the equation x^2 + y^2 = 4 meters^2? For example, at t=0 seconds, the object following this path will be at position x = 2 meters and y = 0 meters; at t = 5 * pi/2 seconds, the object will be at position x = 0 meters and y = 2 meters; at t = 15 * pi/2 seconds, the object will be at position x = 0 meters and y = -2 meters; and at t = 10 * pi seconds, the object will have returned to position x = 2 meters and y = 0 meters. So no matter what t you choose, from -infinity to +infinity, corresponding to different points along the vertical dimension of the helix, the shadow of the object onto the xy plane will always be on the circle x^2 + y^2 = 4.

The point here is that the "shadow" of a path that satisfies dx/dt = constant and dy/dt = constant and dz/dt = constant will always be either a straight line or a point (in the coordinate sense, not the geodesic sense), never a circle or an ellipse or any other curve. Can you see why this is true? (hint: try integrating dx/dt = C to find allowable functions for x(t), and likewise with dy/dt and dz/dt) Notice, for example, that dx/dt and dy/dt are not satisfied by the equations for the 4D helix I give above. It should be obvious that a path whose x-coordinate cycles back and forth over a finite range, like a circle or an ellipse, cannot have a constant dx/dt, since constant dx/dt means the x-coordinate is decreasing or increasing at a constant rate as t increases. Likewise for dy/dt and dz/dt. So now can you see the relevance of looking just at the spatial path--the "shadow" of the entire worldline in spacetime--to answering the question of whether or not dx/dt and dy/dt and dz/dt (which involve both space and time) are constant?

 

[MeJennifer]

No, you're just not understanding the argument.

Well if you remember, I brought up the issue because you made the claim that some geodesics in space-time (note: not space and time, but space-time) cannot be represented as straight lines. I disagreed with that, space-time is curved (locally) in quite a complex way. It may be curved globally as well, but that is not particularly relevant to the issue at hand.

And the whole reason that particular discussion came up was in relation to the alleged equivalence between an inertial frame and a non-inertial frame (acceleration by a non-gravitational force). You made the claim that GR is simply an extension to SR in that respect and that the acceleration is relative. I claimed that that was not the case. After that it seems that there was some sort of confusion with regards to a straight line and a geodesic. Is that your recollection of the history as well Jesse? Please correct me if I misunderstood something.

Anyway...

I am talking about wordlines in space-time.

In other words I don't care about space and time. This whole argument is about space-time not about an observer dependent view on things.

So, ok now you agree with me it is some sort of helix.

Do you agree that there might be surfaces where a helix could be considered straight?

 

[JesseM]

Well if you remember, I brought up the issue because you made the claim that some geodesics in space-time (note: not space and time, but space-time)

I don't understand the distinction you're making between a "geodesic in space and time" and "a geodesic in spacetime", I would understand these terms to be synonymous. All geodesics must have extremal values of some measure of "distance"--what exactly would a geodesic in space and time be, if not a path with an extremal value of the proper time like a geodesic in spacetime?

cannot be represented as straight lines. I disagreed with that, space-time is curved (locally) in quite a complex way.

What does the "complexity" of the curvature of spacetime have to do with it? Look, I've already made a simple argument involving two orbits that cross repeatedly to show why you can't find a coordinate system where all geodesics in spacetime will have the property of constant dx/dt, dy/dt and dz/dt; if you disagree with this argument, then please address the following questions from post #41, which I've already asked three times in a row and you keep ignoring:

straight line in coordinate sense <-> projection of that line's coordinates into euclidean space results in straight line in euclidean sense

non-straight line in coordinate sense <-> projection of that line's coordinates into euclidean space results in non-straight line in euclidean sense

Do you agree that these pairs of statements are logically equivalent, so that if one of a pair is true the other must be true as well, and if one is false the other must be false as well?

For example, can you think of a function y(x) representing the coordinates of a line in some 2D space such that dy/dx is constant (meaning the line is 'straight' in the coordinate sense) but when you plot this function y(x) in cartesian coordinates in 2D euclidean space, the result is a curve that is not straight in the euclidean sense? I would say this is impossible, do you disagree?

And the whole reason that particular discussion came up was in relation to the alleged equivalence between an inertial frame and a non-inertial frame (acceleration by a non-gravitational force). You made the claim that GR is simply an extension to SR in that respect and that the acceleration is relative. I claimed that that was not the case. After that it seems that I or you was confused and switching straight line and geodesic cleared everything up. Is that your recollection of the history as well Jesse? Please correct me if I misunderstood something.

I was confused by your use of terminology, because "straight vs. curved worldline" is not the usual way of describing geodesic vs. non-geodesic paths. I agree that it is not relative whether a path is a geodesic or not, but I also say that a coordinate system where a non-geodesic path is represented as a straight line in terms of the coordinates, or a geodesic path is represented as a non-straight line in coordinate terms, is just as valid as one where the opposite is true (and as I've been saying, it's not possible to find a coordinate system where all geodesics are straight lines in coordinate terms). This is one sense in which GR extends SR--where SR said that all inertial coordinate systems are equally valid, GR says that all smooth coordinate systems are equally valid, because the laws of physics obey the same equations in all of them.

Anyway...

I am talking about wordlines in space-time.

In other words I don't care about space and time. This whole argument is about space-time not about an observer dependent view on things.

I thought the argument was about coordinate systems, which do depend on arbitrary choices made by the observer. You had agreed several times that we were only using the word "straight" in the coordinate sense, not in the sense of being a geodesic. If you're using it in some other way, please clarify, because my argument is just that you can't find a coordinate system where all geodesics are straight in the coordinate sense.

Also, I still don't understand the distinction you're making between a line in "space and time" and a line "in spacetime"--in terms of coordinate systems, I would understand both to mean that each point on the line has 3 space coordinates and 1 time coordinate, and that the line can be parametrized by functions x(t), y(t) and z(t). And once you have found the functions that parametrize the line in your chosen coordinate system, the line will be straight "in the coordinate sense" if and only if dx/dt, dy/dt and dz/dt are all constant.

So, ok now you agree with me it is some sort of helix.

Do you agree that there might be surfaces where a helix could be considered straight?

By "surfaces", do you mean coordinate systems? If we are talking about a line that has the equation of a helix in that coordinate system, then it is impossible that in that coordinate system the derivatives dx/dt, dy/dt and dz/dt will all be constant. On the other hand, a worldline that might come out as a helix in a more "natural" coordinate system, like an object moving in a circular orbit as seen in an inertial coordinate system, could be turned into a straight line in another coordinate system which is very distorted relative to the more natural one. But the fact that it is a straight line means that it is not a helix in terms of this new coordinate system--the equations x(t), y(t) and z(t) will not be the equations of a helix.

 

[MeJennifer]

By "surfaces", do you mean coordinate systems?

No.
Curved space-time is not an ordinary coordinate system Jesse. How could it be, since it has local curvatures! Every wordline of a mass particle creates a curvature in space-time.
But these curvatures are not a global property of space-time. So when you talk about coordinate systems, I fail to see the relevance for curved space-time.

Some of those curvatures create a surface, such that, if a wordline traverses it, it appears (in space-time not in observer space!) it is traversing a kind of helix.

So for instance a worldline of a space station "trapped" in an orbit of a planet has a continuous "realignment" with the worldline of that planet by curvature. Such realignment appears on the form of a helix, actually not exactly a helix because in addition to x, y, and z, time is also curved.
Do you understand what I am saying?

 

[Rach3]

Geodesics are properties of test particles in the system, not massive objects. Parallel transport works on what is already there and curved, not what is being transported which is just a tangent vector. And I don't think you can even talk about massive point particles in GR.

 

[pervect]

MeJennifer, perhaps you are looking for the defintion:

<br /> u^a \, \nabla_a u^b = 0<br />

which defines a geodesic as a curve which "parallel transports" its tangent vector along itself.
Here u^a is the "tangent vector" to the curve, aka the 4-velocity.

This is another, perfectly acceptable defintion of a geodesic, but it requires that one know what "parallel transport" is. The \nabla_a symbol is the covariant derivative (the generalization of the flat space derivative).

To quote Wald

Intuitively, geodesics are lines that "curve as little as possible"; they are the "straightest possible lines" one can draw in a curved geometry.

Wald then presents the same equation I did.

 

[JesseM]

No.
Curved space-time is not an ordinary coordinate system Jesse.

I don't know what you mean by "not an ordinary coordinate sytem"--it's not any sort of coordinate system, any more than photons or galaxies are coordinate systems! Curved spacetime is a metric on a 4D manifold, the coordinate system is something totally separate which you draw over the manifold to give names to different points on it.

How could it be, since it has local curvatures!

A coordinate system cannot have curvature, nor can it be flat. It's just a function that assigns 4 numbers--the x-coordinate, the y-coordinate, the z-coordinate, and the t-coordinate--to each event in spacetime. That's all! Nothing to do with the curvature in itself, although once you have a coordinate system you can talk about the value of the metric tensor at a given set of coordinates.

But these curvatures are not a global property of space-time. So when you talk about coordinate systems, I fail to see the relevance for curved space-time.

Again, whether the curvature is "local" or "global" has to do with the metric. Nothing to do with what coordinate system you choose to place on spacetime.

Some of those curvatures create a surface, such that, if a wordline traverses it, it appears (in space-time not in observer space!) it is traversing a kind of helix.

Are you suggesting there is some "objective" sense in which it is traversing a helix as opposed to a straight line or some other shape, independent of your arbitrary choice of coordinate system? If so, I don't understand what that would mean, unless you can define particular shapes of worldlines like "helix" purely in terms of proper distances and proper times (which only depend on the metric, not the coordinate system), as opposed to defining it in coordinate terms. But even if you could, this is not relevant to the question of whether a particular worldline is "straight" in coordinate terms, which is what we were supposed to be talking about.

So for instance a worldline of a space station "trapped" in an orbit of a planet has a continuous "realignment" with the worldline of that planet by curvature. Such realignment appears on the form of a helix, actually not exactly a helix because in addition to x, y, and z, time is also curved.

Coordinates themselves cannot be curved, curvature is defined in terms of the metric, although can look at the way the metric tensor varies at different points in spacetime corresponding to different sets of coordinates in whatever coordinate system you're using.

Do you understand what I am saying?

Not completely, you seem to be confused about the difference between objective physical statements about the curvature of spacetime and coordinate-dependent statements.

 

[MeJennifer]

Are you suggesting there is some "objective" sense in which it is traversing a helix as opposed to a straight line or some other shape, independent of your arbitrary choice of coordinate system?

An orbit is a kind of helix (with a "twist" because time is warped as well) in space-time.

If so, I don't understand what that would mean, unless you can define particular shapes of worldlines like "helix" purely in terms of proper distances and proper times (which only depend on the metric, not the coordinate system), as opposed to defining it in coordinate terms.

There would be no point, since proper time and proper distance are observer dependent quantities, irrelevant to space-time.

But even if you could, this is not relevant to the question of whether a particular worldline is "straight" in coordinate terms, which is what we were supposed to be talking about.

Well as I said I do not see an issue with that.

 

[JesseM]

An orbit is a kind of helix (with a "twist" because time is warped as well) in space-time.

In a the typical type of coordinate system this might be true, but one could find a coordinate system where the orbit was a straight line or some other shape. If you believe there is some coordinate-independent way in which an orbit is objectively helix-shaped, you need to define it.

There would be no point, since proper time and proper distance are observer dependent quantities, irrelevant to space-time.

No, they are observer-independent measures of "distance" in spacetime, as I understand it the metric is based on one or the other, at least in infinitesimal form (for example, in flat spacetime the metric would have the signature ds^2 = dx^2 + dy^2 + dz^2 - c^2 dt^2 in a locally inertial coordinate system, which is just the infinitesimal form of proper distance). In flat spacetime the proper time between two events with timelike separation, or the proper distance between two points with spacelike separation, is the same in all coordinate systems; in curved spacetime you should still be able to talk about the proper time along the unique geodesic between any two events with timelike separation, and that will be coordinate-independent, I'm not sure if there is any coordinate-independent notion of distance between events with spacelike separation in curved spacetime though.

But even if you could, this is not relevant to the question of whether a particular worldline is "straight" in coordinate terms, which is what we were supposed to be talking about.

Well as I said I do not see an issue with that.

What does that mean? Do you understand now that a coordinate system is nothing more than a function which assigns 4 coordinates--x, y, z, and t--to each point in spacetime, so that once you have this you can find the functions x(t), y(t), z(t) for the set of events that any particular worldline passes through? Do you understand that "straight in coordinate terms" means that x, y and z change at a constant rate with respect to t, meaning that dx/dt and dy/dt and dz/dt for these functions x(t), y(t) and z(t) must be constant? And given these definitions, do you understand why it is impossible that two worldlines which are "straight in coordinate terms" could cross more than once in the region the coordinate system was defined? (assuming each worldline is continuous in coordinate terms so there's no disappearing off one boundary of the region the coordinate system is defined and reappearing on the other)

 

[MeJennifer]

What does that mean?

I suppose it means that when I am talking about space-time you want to talk about coordinates systems.
Why can't you talk about space-time without a notion of a coordinate system?

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.

 

[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!)