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

[Born2Perform]

A twin is on a spaceship, the other on the earth.
first twin accelerates and get distant form the earth, we know he is moving.
He dies in the spaceship and his son takes control of it.
Now, for his son the system is linear, he does not know the spaceship is moving, and he has not a possibility to discover it.

Halls of ivy said that:
The statement "he has not a way to discover the truth, that his ship is moving" is incomplete. Relative to what frame of reference? His ship is moving relative to some frames of reference, not moving, relative to others (in particular, a person is alway "not moving" relative to his own frame of reference). It simply isn't true to say it is "the truth, that his ship is moving".

It was good for a week, but i think there's a bug: Nobody can deny that the ship is moving, at least you can say it is or it is not relative to some frame of reference. But,
Have you canceled the acceleration? In a moment of time it happened, for any frame.
And, motion depends from frame to frame, but cannot a frame of reference be wrong? the ship is definitely moving, relative to nothing.

 

[MeJennifer]

the ship is definitely moving, relative to nothing.

No, it is not moving relative to nothing.
He is simply moving in space-time, and so is his uncle.

 

[Born2Perform]

No, it is not moving relative to nothing.
He is simply moving in space-time, and so is his uncle.

So you are according that the relativity principle is a fallacy?

 

[MeJennifer]

So you are according that the relativity principle is a fallacy?

How can something moving relative no nothing be related to relativity. It takes at least two to tango.

With regards to the "relativity principle", you don't need in GR.
It is sufficient to have a metric and a description on how space-time curvature and matter relate.

By the way what do you understand the relativity principle is?

 

[selfAdjoint]

The only motion that counts is the relative motion between the son and his uncle. Both of them are currently unaccelerated according to the statement of the problem, so they are inertial, and each of them is entitled to regard himself as being at rest, according to relativity. As the problem is stated, the spaceship has not turned around, so that if son and uncle communicate by radio, each of them will see the other's coordinates as Lorentz transformed relative to himself.

The son presumably has a record of how much time has passed on the ship since his father set out (we ignore the period of acceleration for simplicity, but if necessary it can be included at the cost of some complexity). Now the son sends this time length in a message to his uncle, and the uncle sends his own elapsed time to the son.

And both of them say "Wow, it took longer for him than it has taken for me!"

 

[MeJennifer]

The only motion that counts is the relative motion between the son and his uncle. Both of them are currently unaccelerated according to the statement of the problem, so they are inertial, and each of them is entitled to regard himself as being at rest, according to relativity. As the problem is stated, the spaceship has not turned around, so that if son and uncle communicate by radio, each of them will see the other's coordinates as Lorentz transformed relative to himself.

The son presumably has a record of how much time has passed on the ship since his father set out (we ignore the period of acceleration for simplicity, but if necessary it can be included at the cost of some complexity). Now the son sends this time length in a message to his uncle, and the uncle sends his own elapsed time to the son.

And both of them say "Wow, it took longer for him than it has taken for me!"

Well let's drill down a bit:

A and B are both in a spacecraft in an inertial frame. A and B are going to travel away from the spacecraft in opposite directions. Both are instructed to accelerate a for t seconds and then continue with a constant speed of s. Both A and B get an additional the instruction that as soon as their distance from the spacecraft is x to send a digitized message as to what their chronometer indicates as elapsed time. Their chronometers are synchronized, A leaves from the front and B leaves from the back. Both are instructed to accelerate a for t seconds and then continue with a constant speed of s.

Now are you claiming that when A and B get their respective messages the time will differ with the one they sent?

 

[Born2Perform]

Tell me this:

According with my brain, the spaceship is or isn't moving in spacetime?

If it is, and the son-uncle system is symmetrical, does not mean that there is an absolute motion, that there is a truth but in a symmetrical system we cannot discover it?
You cannot say no because relativity principle says no, here i put in discussion the principle itself.

 

[JesseM]

According with my brain, the spaceship is or isn't moving in spacetime?

The phrase "moving in spacetime" is not usually used in relativity, since from the perspective of 4D spacetime there is no movement in the usual sense, just various worldlines embedded in spacetime like lines on a 2D piece of paper or strings embedded in a 3D block of ice. There is no absolute sense in which either the ship or the Earth is moving, you can pick a frame where the Earth is moving while the ship is at rest or a frame where the opposite is true, the laws of physics will work the same way in both frames.

 

[MeJennifer]

According with my brain, the spaceship is or isn't moving in spacetime?

Any object is always moving in space-time, except for the limit condition of a space-time singularity.

The whole problem with SR is that it is a theory that is completely irrelevant to our existing space-time.
Our space-time has matter distribution, which is a map of reality!
To apply SR in the real universe would mean that you question this reality and are prepared to apply linear transformations, rotations, or even deformations of space-time to "make it work".

All this endless comparing of how frame X sees what frame Y does and vice versa is all due to simple coordinate transformations. There is no objective reality to it.
It is like wondering "how come in the English word "yes" is different in Hungarian and vice versa.

 

[HallsofIvy]

Tell me this:

According with my brain, the spaceship is or isn't moving in spacetime?

Perhaps you could explain what your brain means by "moving in spacetime". Since "spacetime" includes both space and time, an object moving in space corresponds to a single path in "spacetime"

If it is, and the son-uncle system is symmetrical, does not mean that there is an absolute motion, that there is a truth but in a symmetrical system we cannot discover it?
You cannot say no because relativity principle says no, here i put in discussion the principle itself.

I can say "no" because nothing you have said implies "absolute motion". Each party, son or uncle, sees the other moving relative to himself. If you are questioning the very concept of "relative motion", that is based on experimental evidence (and, except for electro-magnetic phenomena he did not know about, goes back to Galileo). If you want to "question" it, you will need to reference an experiment that contradicts it.

 

[Born2Perform]

i understand your rationale, but I'm convinced in something more fundamental:

-------
A body is accelerating.
is it or is it not moving?
-------

and don't say: "according to which frame of reference?"
it's not physics its pure logic, this is the core of my doubt.

 

[MeJennifer]

-------
A body is accelerating.
is it or is it not moving?
-------

and don't say: "according to which frame of reference?"
it's not physics its pure logic, this is the core of my doubt.

I know that you don't like to hear it but it does depend on the frame of reference.
But if you ask is a curved timeline always curved in space-time then the answer is yes.

 

[Born2Perform]

As hight school student i guess i will not undersant this for some years. i don't catch you rationale.

Last thing:
is there a probability that i could be right?
is possible that the sentence:

"A body is accelerating. It is moving."

could be right, or this possibility doesn't exists?

if it does not, i'll have my soul in peace until i'll reach university, if it could be i will think to it more deeply.
thanks.

 

[MeJennifer]

As hight school student i guess i will not undersant this for some years. i don't catch you rationale.

Last thing:
is there a probability that i could be right?
is possible that the sentence:

could be right, or this possibility doesn't exists?

if it does not, i'll have my soul in peace until i'll reach university, if it could be i will think to it more deeply.
thanks.

Well let me give you an example of a body that is accelerating but not moving in a particular frame of reference.
It is when you stand with your feet on the earth. You are in fact accelerating upwards! The (mostly) electro-magnetic forces accelerate you, and you can feel it as well.

 

[Born2Perform]

Well let me give you an example of a body that is accelerating but not moving in a particular frame of reference.
It is when you stand with your feet on the earth. You are in fact accelerating upwards! The (mostly) electro-magnetic forces accelerate you, and you can feel it as well.

I'm practising a force upwards, but accelerating?
and however forces that i can't feel are "feelable" with some machine.

saying

"A body is accelerating. It is moving."

is like to say 1+1=2, its Aristoteles logics!
i never have been more certain of something than now.

Really nobody of you feels to say that i could be right, also with a small, or ridiculous probability?

 

[JesseM]

saying

"A body is accelerating. It is moving."

is like to say 1+1=2, its Aristoteles logics!
i never have been more certain of something than now.

Really nobody of you feels to say that i could be right, also with a small, or ridiculous probability?

From the perspective of any given coordinate system, if an object is accelerating in that coordinate system (ie its coordinate velocity is changing with coordinate time), then of course it must also be moving in that coordinate system (its coordinate position is changing with coordinate time). And if you stick to inertial coordinate systems as is usually done in special relativity, then if something is accelerating in one inertial coordinate system, it is accelerating in all coordinate systems. However, if you allow non-inertial coordinate systems, something which is accelerating in one coordinate system may be at rest or moving at constant velocity in another.

 

[MeJennifer]

From the perspective of any given coordinate system, if an object is accelerating in that coordinate system (ie its coordinate velocity is changing with coordinate time), then of course it must also be moving in that coordinate system (its coordinate position is changing with coordinate time). And if you stick to inertial coordinate systems as is usually done in special relativity, then if something is accelerating in one inertial coordinate system, it is accelerating in all coordinate systems. However, if you allow non-inertial coordinate systems, something which is accelerating in one coordinate system may be at rest or moving at constant velocity in another.

Well with the proper coordinate systems one can make an elephant out of a monkey.
What matters, at least to me, is what is the significance to reality.

While inertial movements are indeed relative, it is different for movements that are not inertial.

For instance to deny that a rocket leaving our Earth is accelerating because all the objects in the whole universe could be accelerating in the opposite direction is just plain silly.
Think about four rockets leaving Earth in a plane of opposite directions, north, south, east and west. Now the rockets may not move because of the relativity principle?

Space-time is absolute, so if something moves by a (non gravitational) force, then it is a curved timeline in space-time.

 

[selfAdjoint]

Well let's drill down a bit:

A and B are both in a spacecraft in an inertial frame. A and B are going to travel away from the spacecraft in opposite directions. Both are instructed to accelerate a for t seconds and then continue with a constant speed of s. Both A and B get an additional the instruction that as soon as their distance from the spacecraft is x to send a digitized message as to what their chronometer indicates as elapsed time. Their chronometers are synchronized, A leaves from the front and B leaves from the back. Both are instructed to accelerate a for t seconds and then continue with a constant speed of s.

Now are you claiming that when A and B get their respective messages the time will differ with the one they sent?

Isn't this a little confused? First you say A and B are to send messages, and then you ask aboout the messages they receive. And you don't say whether x is before or after they stop accelerating. I am going to say after.

So both of A and B have relative speed s with respect to the ship and are currently inertial when they send or receive their messages. The ship is also inertial. So each of A & B sees ship length L_S as shorter, depending on s, and the ship time T_S will be seen by A and B to be longer also depending on s. Since A and B, although in different directions, have the same speed relative to the ship, they will experience the same transformation in seeing the ship data.

And what will the ship see from A's and B's messages? Exactly the same! L_A and L_B will both be less than L_S (assuming it's some common length like a meter stick that they agreed on back when both A and B were on the ship. And T_A and T_B will both be greater than T_S, with the same caveat.

The point is that all inertial systems are equal, and if one of a speed related pair sees the other's data treansformed, the second of the pair sees exactly the same transformation for the data from the first. To break that symmetry you have to accelerate. But your acceleration happened before they started comparing, so it doesn't factor in.

 

[HallsofIvy]

I'm practising a force upwards, but accelerating?
and however forces that i can't feel are "feelable" with some machine.

saying
a body is accelerating, it is moving
is like to say 1+1=2, its Aristoteles logics!
i never have been more certain of something than now.

Really nobody of you feels to say that i could be right, also with a small, or ridiculous probability?

Yes, that's true. And if you kept asserting that 2+ 2= 5, nobody would feel like saying you could be right "also with small, or ridiculous probability" either!
I might also point out that if you assert

saying
a body is accelerating, it is moving
is like to say 1+1=2, its Aristoteles logics!

then I can only include that you do not know "Aristoteles logics". Neither of those statements has anything to do with Aristotelian logic.

I can also conclude that you do not know what acceleration is. It is quite possible for a body to be moving with a non-zero (constant for simplicity) acceleration and yet, at a given instant, have 0 speed (i.e. not be moving) relative to some inertial coordinate system. Of course, it follows, from the definition of "inertial coordinate system" that an accelerating object can only momentarily motionless relative to a specific inertial coordinate system. However, it follows that such an object is, at any instant, motionless relative to some inertial coordinate system. And, of course, if we drop the "inertial", we can always assert that an object is motionless relative to its own coordinate system. Asserting that an accelerating object "must be moving" doesn't prove there exist "absolute" motion because acceleration itself must be relative to some coordinate system.

And, finally, just asserting that YOU don't understand relativity does not prove that it is wrong. All of relativity is based on experimental results- the only way to dis-prove a physical theory is to demonstrate experimental results it cannot explain. Can you do that?

 

[JesseM]

For instance to deny that a rocket leaving our Earth is accelerating because all the objects in the whole universe could be accelerating in the opposite direction is just plain silly.

Are you familiar with the concept of "diffeomorphism invariance" in GR? My understanding is diffeomorphism invariance means that because of the way the laws of GR are formulated, virtually any coordinate system you can think of (except perhaps for poorly-behaved ones where the same event is assigned multiple sets of coordinates or problems along those lines) is equally good, the same laws of physics will be obeyed in all of them (see http://www.advancedphysics.org/forum/archive/index.php/t-3532.html of how using GR, you can analyze the problem from the point of view of a coordinate system where the twin who accelerates (from the point of view of inertial coordinate systems) is at rest throughout the entire trip, and the G-forces she feels during the phase of the trip where inertial observers say she's accelerating are explained in this coordinate system by the creation of a "uniform gravitational field" (although this term should be taken with caution, as the page explains) during this phase of the trip:

We'll pick a frame of reference in which Stella is at rest the whole time! When she ignites her thrusters for the Turnaround, she is forced to assume that a uniform "gravitational" field suddenly permeates the universe; the field exactly cancels the force of her thrusters, so she stays motionless.

Not so Terence. The field causes him to accelerate, but he feels nothing new since he's in free-fall (or rather the Earth as a whole is). There's an enormous potential difference between him and Stella: remember, he's light-years from Stella, in a uniform "gravitational" field! Stella's at the bottom of the well, he's at the top (or they would be, if the well weren't bottomless and topless). So by uniform "gravitational" time dilation, he ages years during Stella's Turnaround.

...

You may be bothered by the Big Coincidence: how come the uniform "gravitational" field happens to spring up just as Stella engages her thrusters? You might as well ask children on a merry-go-round why centrifugal force suddenly appears when the carnival operator cranks up the engine. There's a reason such forces have had to endure the derisive prefix "pseudo" in so many books.

You may find uniform "gravitational" time dilation, the second assertion, a mite too convenient. Where did it come from? Is it just a fudge factor that Einstein introduced to resolve the twin paradox? Not at all. Einstein gave a couple of derivations for it, having nothing to do with the twin paradox. These arguments don't need the Principle of Equivalence. I won't repeat Einstein's arguments (chase down some of the references if you're curious), but I do have a bit more to say about this effect in the section titled Too Many Explanations.

If you still find the analysis from the perspective of this coordinate system "silly", despite the fact that the same laws of GR are being obeyed in it, can you explain whether "silly" is just an aesthetic opinion or whether it has some more rigorous physical meaning? After all, someone might also consider it "silly" to use an inertial frame where the whole Earth is moving at high speed while the rocket the twin is riding is at rest, but hopefully you'd agree all inertial frames are equally valid from the perspective of SR, and the situation with different non-inertial coordinate systems in GR seems analogous to this.

 

[MeJennifer]

"... she is forced to assume that a uniform "gravitational" field suddenly permeates the universe"

So you are not really sure that every time a space-shuttle takes off we could not instead have some uniform gravitational field permeating the universe instead? Sorry, but than one might as well believe that some invisible pink unicorn is suddenly deforming space-time.

Sorry, but I guess I am a simpleton, I simply don't believe that uniform "gravitational" (whatever the double quotes supposed to mean) fields can suddenly permeate the universe as soon as a rocket takes off. Do you believe that?

It is simply amazing to me what people want to offer not to "hurt Einstein's feelings", it is almost like religion.
Again inertial worldlines are relative, no problem with that whatsoever, but non-inertial worldlines, which are caused by non gravitational forces, are not.
But, again, I apologize, I am a simpleton.

hopefully you'd agree all inertial frames are equally valid from the perspective of SR, and the situation with different non-inertial coordinate systems in GR seems analogous to this.

I don't.

 

[JesseM]

"... she is forced to assume that a uniform "gravitational" field suddenly permeates the universe"

So you are not really sure that every time a space-shuttle takes off we could not instead have some uniform gravitational field permeating the universe instead? Sorry but than one might as well say that some invisible pink unicorn is suddenly deforming space-time.

All I can say is that if you adopt this alternate coordinate system, diffeomorphism invariance means the laws of physics will be no different in this system. Did you look at http://www.einstein-online.info/en/spotlights/background_independence/index.html explaining diffeomorphism-invariance and background independence?

Your argument seems to just be one of aesthetic distaste for the analysis from a particular coordinate system, but that is obviously not a rigorous way of deciding which coordinate systems are acceptable and which are not. A critic of special relativity might come up with a similar aesthetic argument against the equivalence of different inertial frames--they might say something like, "surely you don't expect me to believe that if I am flying away from the Earth in a rocket moving at constant velocity relative to the earth, it is actually the earth, the sun, in fact the entire galaxy which are moving away from the rocket while the rocket is standing still? Ridiculous!" How would you respond to such an argument? I think the only way to respond would be to point out that the criterion physicists used to decide when coordinate systems are "equivalent" is based on checking whether or not the laws of physics work the same way in the different coordinate systems--this is why in special relativity non-inertial coordinate systems are not equivalent to inertial ones, because if you tried to apply the usual non-tensor equations of SR that work in inertial systems like the rule for time dilation as a function of coordinate velocity, you'd get the wrong answers to physical questions like how much a clock will tick between two events on its worldline. Likewise, electromagnetic fields will obey the non-tensor form of Maxwell's laws in all inertial coordinate systems, but not in non-inertial ones. So this is why physicists say that all inertial coordinate systems are equally valid in SR, but that non-inertial ones are not equivalent to inertial ones. But when you go to GR, you find that the laws of physics expressed in tensor form will work just the same way in all coordinate systems, including non-inertial ones, so if you apply the same criterion you should conclude that there is no reason to prefer one coordinate system over another.

If you disagree with this reasoning, does that mean you disagree with the rule that if the laws of physics work the same way in different coordinate systems, these coordinate systems are equally valid? If you do reject that rule, then I don't think you have any basis for saying the argument of the SR skeptic I imagined above is wrong, and in fact I don't think you have any rigorous rule for deciding which coordinate systems are valid and which aren't, it would seem to be based on nothing but your own aesthetic preferences (if you do have a rigorous rule that differs from the one used by physicists, then please state it).

Sorry, but I guess I am a simpleton, I simply don't believe that uniform "gravitational" (whatever the double quotes supposed to mean)

If you read the rest of that page, they explain the reason for the scare quotes. A "uniform gravitational field" is distinguished from a regular gravitational field because it does not involve matter/energy curving spacetime, not all physicists would even use the term "gravitational field " to describe this.

fields can suddenly permeate the universe as soon as a rocket takes off. Do you believe that?

What does it mean to "believe" or "disbelieve" something which is not a physical fact, but which depends on your choice of coordinate systems? If the relative velocity between me and the Earth is 0.5c, do you "believe" that I am at rest while the Earth is moving away from me at 0.5c, as would be true in my rest frame? Do you "believe" that my current position along the x-axis is x=203.5 meters, as would be true in a coordinate system whose origin is 203.5 meters away from me and whose x-axis crosses my current position? These seem like meaningless questions to me, I think you are taking an overly concrete approach to coordinate-based statements. The only thing I "believe" that's relevant here is that all the laws of physics, when stated in terms of tensor equations, will obey exactly the same equations in a coordinate system where the traveling twin was at rest throughout the journey as they do in a coordinate system where the earth-twin was at rest between the time the traveling twin left and came back, and that this is the criterion that physicists use to decide whether different coordinate systems are equally valid.

hopefully you'd agree all inertial frames are equally valid from the perspective of SR, and the situation with different non-inertial coordinate systems in GR seems analogous to this.

I don't.

I think you may have misunderstood me--I was just saying "hopefully you'd agree" with the statement "all inertial frames are equally valid from the perspective of SR", I wasn't asking whether you agreed with the separate statement that "the situation with different non-inertial coordinate systems in GR seems analagous to this", that second part was my own assertion. It's analogous in the sense that exactly the same criterion is used to decide whether different coordinate systems are equally valid in both cases--namely, the criterion of whether the laws of physics obey the same equations in the different coordinate systems. Again, if you have an alternate criterion that's just as rigorous, please describe it; if you don't, then the only conclusion is that your judgements about the validity of different coordinate systems are based on your own personal aesthetic criteria.

 

[MeJennifer]

Ok so, let's take a coordinate system of space-time where a geodesic is a straight line. Now let's go back to the spaceship taking off from the landing platform, and accelerating away from earth.
Would you then agree or disagree that the principle of relativity holds in this case? Can we say who accelerated or is it all relative?

 

[selfAdjoint]

Ok so, let's take a coordinate system of space-time where a geodesic is a straight line. [/quote

So you're talking about a "flat" Minkowski space time.

Now let's go back to the spaceship taking off from the landing platform, and accelerating away from earth.
Would you then agree or disagree that the principle of relativity holds in this case? Can we say who accelerated or is it all relative?

Special relativity is what is called a "kinetic" theory; this means it has speed built into it (via c soming into the metric); that means you can solve "pure speed" problems straight out with just algebra and arithmetic. But accelerations are not so easy; they are not built into special relativity and in addition to the limits and calculus you always have to use for acceleration, you have also the formalism of "comoving frames" which is necessary to do relativitstic calculation for accelerating objects.

That said, the answer to your question is that accelerations are not relative; the people on the ship feel the acceleration and they are not inertial. There are length contractions and time dilations that apply to them (relative to the launching pad) but they are calulated using those comoving frames. In this case what the people on the ship see is not the same as what the people on the launching pad see.

 

[MeJennifer]

That said, the answer to your question is that accelerations are not relative; the people on the ship feel the acceleration and they are not inertial.

Well I agree with you.

So you're talking about a "flat" Minkowski space time.

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?

 

[JesseM]

Ok so, let's take a coordinate system of space-time where a geodesic is a straight line.

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.

Now let's go back to the spaceship taking off from the landing platform, and accelerating away from earth.
Would you then agree or disagree that the principle of relativity holds in this case? Can we say who accelerated or is it all relative?

It depends whether you mean "accelerated" to mean "non-geodesic path" or whether you're talking about coordinate acceleration. For example, would you say that an object falling towards the Earth is not accelerating?

 

[MeJennifer]

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.

Really, I don't see the problem actually.

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

 

[JesseM]

Really, I don't see the problem actually.

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

Orbits, for example. Suppose two satellites are orbiting the Earth in opposite directions, such that they pass next to each other every half-orbit--I don't see how you could come up with a coordinate system where both are straight lines, since the two worldlines should cross periodically.

 

[MeJennifer]

Orbits, for example. Suppose two satellites are orbiting the Earth in opposite directions, such that they pass next to each other every half-orbit--I don't see how you could come up with a coordinate system where both are straight lines, since the two worldlines should cross periodically.

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?

 

[JesseM]

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

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), 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.

 

[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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\multiput(62.37,28.83)(0.43,0.91){1}{\line(0,1){0.91}}<br /> \multiput(62.8,29.74)(0.39,0.93){1}{\line(0,1){0.93}}<br /> \multiput(63.19,30.67)(0.36,0.94){1}{\line(0,1){0.94}}<br /> \multiput(63.55,31.61)(0.32,0.96){1}{\line(0,1){0.96}}<br /> \multiput(63.87,32.56)(0.28,0.97){1}{\line(0,1){0.97}}<br /> \multiput(64.15,33.53)(0.24,0.98){1}{\line(0,1){0.98}}<br /> \multiput(64.39,34.51)(0.2,0.99){1}{\line(0,1){0.99}}<br /> \multiput(64.59,35.49)(0.16,0.99){1}{\line(0,1){0.99}}<br /> \multiput(64.75,36.49)(0.12,1){1}{\line(0,1){1}}<br /> \multiput(64.87,37.49)(0.08,1){1}{\line(0,1){1}}<br /> \multiput(64.95,38.49)(0.04,1.01){1}{\line(0,1){1.01}}<br /> \linethickness{0.3mm}<br /> \put(40,5){\line(0,1){10}}<br /> \put(40,5){\vector(0,-1){2}}<br /> \linethickness{0.3mm}<br /> \put(40,65){\line(0,1){10}}<br /> \put(40,75){\vector(0,1){2}}<br /> \linethickness{0.3mm}<br /> \put(55,40){\line(1,0){10}}<br /> \put(55,40){\vector(-1,0){2}}<br /> \linethickness{0.3mm}<br /> \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?