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

[MeJennifer]

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. My example was just an illustration of a curved space (a mathematical space) where straight lines cross.

You can't generally find a coordinate system where all geodesics are straight lines in the coordinate sense in curved spacetime or on curved 2D surfaces like the surface of a sphere. Because if two lines cross more than once, it cannot be the case that each line's x-coordinate is changing at a constant rate with respect to its y-coordinate (in the case of a 2D space) or that each line's space coordinates are changing at a constant rate with respect to the time coordinate. You can see that this must be true if you project the coordinates of the two lines onto a cartesian coordinate system in a euclidean space of the appropriate dimension, where "position coordinates/x-coordinate changing at a constant rate with respect to time coordinate/y-coordinate" always means a straight line in this euclidean space, and obviously two straight lines cannot cross more than once in euclidean space.

In Euclidean space yes, but space-time is curved and hence space-time is not Euclidean, not Euclidean as to the metric and not Euclidean as to the Euclidean postulates.
So why make this constraint that it must be Euclidean?

 

[JesseM]

Yes and I am talking about that as well. My example was just an illustration of a curved space (a mathematical space) where straight lines cross.

But these lines (geodesics on a sphere) cannot all be "straight" in the coordinate sense, no matter what coordinate system you use!

In Euclidean space yes, but space-time is curved and hence space-time is not Euclidean, not Euclidean as to the metric and not Euclidean as to the Euclidean postulates.
So why make this constraint that it must be Euclidean?

I don't think you're understanding my argument. If a line is "straight" in the coordinate sense, it is a necessary consequence of this that when a line with the same coordinates is drawn in a cartesian coordinate system in a euclidean space of the right dimension, the result will be a "straight line" in the ordinary euclidean sense. In other words:

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?

If you agree with this, then what I'm saying is we can see that it's impossible for two lines that are straight in the coordinate sense to cross multiple times, since "straight in the coordinate sense" is logically equivalent to "straight in the euclidean sense when the coordinates of the line are projected into euclidean space", and we know that straight lines in euclidean space can't cross multiple times.

 

[MeJennifer]

Ok let me narrow this problem down:

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?

 

[JesseM]

Ok let me narrow this problem down:

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?

As long as we are using a grid-like coordinate system where the coordinates represent distances from the origin along different paths--they can be bent paths in a distorted grid laid out on a curved space/spacetime, of course--then if we take "straight lines" to mean straight in the coordinate sense that the rate of change in one coordinate will be constant as you vary another coordinate (for example, dx/dt and dy/dt are constants for the line), then no, it should not in general be possible to find a coordinate system where all geodesics will be straight lines in the coordinate sense. If you disagree with this, please address the question in my last post:

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?

 

[MeJennifer]

As long as we are using a grid-like coordinate system where the coordinates represent distances from the origin along different paths--they can be bent paths in a distorted grid laid out on a curved space/spacetime, of course--then if we take "straight lines" to mean straight in the coordinate sense that the rate of change in one coordinate will be constant as you vary another coordinate (for example, dx/dt and dy/dt are constants for the line), then no, it should not in general be possible to find a coordinate system where all geodesics will be straight lines in the coordinate sense.

Seems incorrect to me.

For instance when you talk about two objects in orbit, "crossing" each other more than once.
Think about a double helix for instance!

 

[JesseM]

Seems incorrect to me.

Then please address this question of mine:

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?

For instance when you talk about two objects in orbit, "crossing" each other more than once.
Think about a double helix for instance!

The strands of a double helix never actually cross. And if you displaced them so that they did, then it would not be possible to come up with a coordinate system where both strands were straight in a coordinate sense.

 

[pervect]

Ok let me narrow this problem down:

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?

Let's look at the geodesic equations:

http://en.wikipedia.org/wiki/Solving_the_geodesic_equations

$$\frac{d^2x^a}{d\tau^2} + \Gamma^{a}{}_{bc}\frac{dx^b}{d\tau}\frac{dx^c}{d\tau} = 0$$

A geodesic will solve these equations.

Now, in a curved space-time, it is possible to make the Christoffel symbols (those are the $\Gamma^a{}_{bc}$ zero at anyone particular point, but not everywhere.

If you define "a straight line" as

t = k1*tau
x = k2*tau
y = k3*tau
z = k4*tau

then a straight line solves the geodesic equation when all the Christoffel symbols are zero.

In addition, if all possible straight lines through a point satisfy the geodesic equation, then all the Chrsitoffel symbols must be zero.

If some straight lines through a point satisfy the geodesic equation, and not others, that means that there exist directions in which the solution to the geodesic equation are not a straight line. (There is exactly one geodesic through a point in every direction. If the straight line in that direction isn't a solution to the geodesic equation, then there must be some other solution in that direction).

Conclusion: in a curved space-time, you can make all the geodesics at (near) a single point "straight lines" but you can't make all the geodesics at all points "straight lines" . It is possible to make all the Christoffel symbols zero at a single point, but not possible to make all of them zero everywhere.

 

[MeJennifer]

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?

 

[pervect]

A more concrete example might help. Let's take a point near the Earth, or some other large mass.

At anyone particular point, by following a geodesic for that point, we can set up a locally, "almost inertial" coordinate system.

The Christoffel symbols at that point would be zero.

An orbiting space-station would be an example. The station is going to be in zero g at the center of the space-station.

Tidal forces from the Earth will be present, though, meaning that someone who is not at the center of the space-station will have to accelerate to keep himself stationary with respect to the space-station.

So here we see again the significance of the "one point" rule - the center of mass of the space-station is the one point where the Christoffel symbols are zero, and there are no tidal forces. At other points, we have non-zero Christoffel symbols, and non-zero tidal forces.

Now we know that there are "stretching" radial tidal forces. If we work out the details, we would see that there are also "comprssive" tidal forces (I'm not going to do that here unless someone requests it).

If we consider a 2-d slice of the space-station, we can make a diagram of the tidal forces.

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possibility (I think it is a probability) that there will be some particular directions that don't experience tidal forces from this diagram, but I think I'd better work out the details before I make any claims, so I'm editing this section of the post.

 

[pervect]

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?

Please define "a straight line" for me so that I'm sure we are using the same defintions.

I'm a little leery that we may have a misunderstanding due to the usage of words.

Because you are asking me to repeat something I've already talked about, I'm doubly leery of semantic issues - i.e. what your concept of a "straight line" is, and if it is the same as mine.

 

[JesseM]

So are you saying that in a non Euclidean coordinate system

What does "non-Euclidean coordinate system" mean? Euclidean vs. non-Euclidean depends on the metric, not the coordinate system.

modeling space-time there exist a wordline on a geodesic that is not straight?

"Straight" in the coordinate sense that I was discussing before? (constant rate of change of one coordinate with respect to another for the worldline's description in this coordinate system)

 

[MeJennifer]

What does "non-Euclidean coordinate system" mean? Euclidean vs. non-Euclidean depends on the metric, not the coordinate system.

"Euclidean" is used for two things, one pertains to the metric and then there is the more classical one that pertains to the adherence to the Euclidean postulates.

 

[MeJennifer]

At anyone particular point, by following a geodesic for that point, we can set up a locally, "almost inertial" coordinate system.

True.

The Christoffel symbols at that point would be zero.

Ok.

An orbiting space-station would be an example. The station is going to be in zero g at the center of the space-station.

Ok.

Tidal forces from the Earth will be present, though, meaning that someone who is not at the center of the space-station will have to accelerate to keep himself stationary with respect to the space-station.

True, but his worldline is different than that of the center of the space station!

Of course, if we were to "plot out" a flat (!) coordinate system from the center of the spacetime we attribute "forces" acting on all those who are not in the center. But clearly those "forces" are fictional, they are simply the result of the curvatures involved.

So here we see again the significance of the "one point" rule - the center of mass of the space-station is the one point where the Christoffel symbols are zero, and there are no tidal forces. At other points, we have non-zero Christoffel symbols, and non-zero tidal forces.

No disagreement here.

 

[JesseM]

"Euclidean" is used for two things, one pertains to the metric and then there is the more classical one that pertains to the adherence to the Euclidean postulates.

But they are mathematically equivalent, are they not? With a euclidean metric all the euclidean postulates will be obeyed, and with a non-euclidean metric they won't--it doesn't depend on the coordinate system you use, since euclid's postulates don't involve assigning coordinates to the objects they describe (that wasn't done until the invention of algebraic geometry).

 

[pervect]

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?

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?

 

[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:

$$u^a \, \nabla_a u^b = 0$$

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.