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

[Hurkyl]

By the way are you suggesting that when we assume that geodesics are straight lines in space-time it must be true that we are talking about a flat space-time?

If by "straight line" you mean something that we would consider to be a line in Euclidean geometry, then yes.

Being flat roughly means that:

(1) Go three steps forward. Turn right 90 degrees. Go three steps forward. Turn left 90 degrees.

has exactly the same result as

(2) Turn right 90 degrees. Go three steps forward. Turn left 90 degrees. Go three steps forward.

and this is clearly the case where all Euclidean lines are geodesics for our manifold.


The semantics problem is that geodesics are, by definition, "straight" lines. It's just that "straightness" is determined by the metric on our manifold.

 

[MeJennifer]

If by "straight line" you mean something that we would consider to be a line in Euclidean geometry, then yes.

No, of course not, one can have a straight line in non Euclidean geometry as well.

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), that means when you take the coordinates of every worldline and draw lines with the same coordinates in an ordinary 4D euclidean space with 3 axes representing space and one representing time, the geodesic worldlines would have to look like ordinary straight lines in this representation, if all geodesics are indeed "straight" in terms of the coordinate system.

Right, so then why do you say:

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.

 

[JesseM]

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), that means when you take the coordinates of every worldline and draw lines with the same coordinates in an ordinary 4D euclidean space with 3 axes representing space and one representing time, the geodesic worldlines would have to look like ordinary straight lines in this representation, if all geodesics are indeed "straight" in terms of the coordinate system.

Right, so then why do you say:

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.

What are you confused about? I'm just saying that if you lay out a coordinate system on curved spacetime, then no matter what coordinate system you use, it is not in general going to work out that when you project the coordinates of all worldlines into 4D euclidean cartesian coordinates, the geodesics will always be projected into straight lines in this 4D euclidean space (which is what it would mean for a worldline to be 'straight' in terms of the coordinate system). My argument about the repeatedly-crossing orbits (which are both geodesics) is one way of seeing why it can't work out this way--how could two straight lines in a 4D euclidean space repeatedly cross each other?

 

[MeJennifer]

What are you confused about?
I'm just saying that if you lay out a coordinate system on curved spacetime, then no matter what coordinate system you use, it is not in general going to work out that when you project the coordinates of all worldlines into 4D euclidean cartesian coordinates, the geodesics will always be projected into straight lines in this 4D euclidean space (which is what it would mean for a worldline to be 'straight' in terms of the coordinate system).

I know you think that, you said that before. But you have not explained why

My argument about the repeatedly-crossing orbits (which are both geodesics) is one way of seeing why it can't work out this way--how could two straight lines in a 4D euclidean space repeatedly cross each other?

Right, I got that to!

Then I wrote:

Well straight lines can cross in curved space-time right?

Take the geometry on a sphere like the earth, two straight lines can cross right?

So perhaps I miss something, I am just trying to find out were your "straight lines cannot cross" thing comes from.

If we take a coordinate system of space-time where straight lines are geodesics, then why do you say we cannot have geodesics that are straight lines in some cases?

You give an example of where lines would cross.
But I don't understand why you see that as a problem?

After all, just because two lines cross does not mean they are not straight!

 

[JesseM]

Well straight lines can cross in curved space-time right?

Take the geometry on a sphere like the earth, two straight lines can cross right?

So perhaps I miss something, I am just trying to find out were your "straight lines cannot cross" thing comes from.

Your example involves lines that are "straight" in the sense of being geodesics, but I thought I already made clear I was talking about lines that are straight in the coordinate sense. There is no coordinate system you could use on the surface of a sphere that would have the property that all geodesics were straight in the coordinate sense.

If we take a coordinate system of space-time where straight lines are geodesics

This question doesn't make any sense to me, since whether worldlines are geodesics has nothing to do with what coordinate system you pick, it has to do with the metric.

Again, all that a "straight line" in the coordinate sense means is that the position coordinates are changing at a constant rate with respect to the time coordinate; or if you're talking about the surface of a sphere, it would mean that in terms of the x and y coordinates you assign to different points on the sphere, the x coordinate is changing at a constant rate with respect to the y-coordinate. 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.

You give an example of where lines would cross.
But I don't understand why you see that as a problem?

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.

 

[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

<br /> \frac{d^2x^a}{d\tau^2} + \Gamma^{a}{}_{bc}\frac{dx^b}{d\tau}\frac{dx^c}{d\tau} = 0<br />

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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\put(15,40){\line(1,0){10}}<br /> \put(25,40){\vector(1,0){2}}<br /> \end{picture}<br /> \[<br />There is a 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.