Register to reply

Implications of the statement Acceleration is not relative

by GregAshmore
Tags: implications, statement
Share this thread:
DaleSpam
#109
Feb15-13, 07:27 AM
Mentor
P: 16,961
Quote Quote by harrylin View Post
There are many objections to make against Einstein's induced gravitational fields. Consider such things as cause and effect, speed of gravitational wave, etc. I have never seen a paper that even attempts to address those issues, let alone solve them.
Probably because most people don't believe they are issues. The cause is obviously the choice of coordinates, and this meaning of "gravitational fields" does not mathematically produce "gravitational waves" in the usual sense of a wave equation with a characteristic propagation speed.
harrylin
#110
Feb15-13, 07:53 AM
P: 3,181
Quote Quote by DaleSpam View Post
Probably because most people don't believe they are issues. The cause is obviously the choice of coordinates, and this meaning of "gravitational fields" does not mathematically produce "gravitational waves" in the usual sense of a wave equation with a definite propagation speed.
A change of coordinates isn't a gravitational field. I don't follow Einstein's Machian explanation of invoking a physical, induced gravitational field (for enabling the interpretation that the rocket is constantly "truly in rest"): "all the stars that are in the universe, can be conceived as taking part in bringing forth the gravitational field [..] during the accelerated phases of the coordinate system K' ".
stevendaryl
#111
Feb15-13, 08:33 AM
Sci Advisor
P: 1,945
Quote Quote by harrylin View Post
A change of coordinates isn't a gravitational field.
It depends on how you define "gravitational field". If you drop an object, and it follows a trajectory [itex]h = -\frac{1}{2}g t^2[/itex] for some constant [itex]g[/itex], then many people would call [itex]g[/itex] the "acceleration due to gravity" or the "gravitational field". But it certainly is affected by a coordinate change. It can be made to vanish by careful choice of coordinates. If [itex]g=0[/itex] (no gravitational field), then a change of coordinates to accelerated coordinates can make [itex]g[/itex] nonzero.

The equivalence principle is about this notion of gravitational field: there is no difference (other than the variation of [itex]g[/itex] with location) between the effects of a [itex]g[/itex] due to gravity and a [itex]g[/itex] due to the use of accelerated cooridnates.
DaleSpam
#112
Feb15-13, 08:40 AM
Mentor
P: 16,961
Quote Quote by harrylin View Post
A change of coordinates isn't a gravitational field.
Remember that there is more than one possible meaning of the term "gravitational field" in GR. In the sense that Einsetin meant it (also my prefered sense) the "gravitational field" is the Christoffel symbols, and the Christoffel symbols do in fact change under a change of coordinates. So a change of coordinates does cause a gravitational field in the sense Einstein used the term.

[EDIT: stevendaryl made essentially the same point, but faster!]
harrylin
#113
Feb15-13, 10:02 AM
P: 3,181
Quote Quote by stevendaryl View Post
It depends on how you define "gravitational field". If you drop an object, and it follows a trajectory [itex]h = -\frac{1}{2}g t^2[/itex] for some constant [itex]g[/itex], then many people would call [itex]g[/itex] the "acceleration due to gravity" or the "gravitational field". But it certainly is affected by a coordinate change. It can be made to vanish by careful choice of coordinates. If [itex]g=0[/itex] (no gravitational field), then a change of coordinates to accelerated coordinates can make [itex]g[/itex] nonzero.

The equivalence principle is about this notion of gravitational field: there is no difference (other than the variation of [itex]g[/itex] with location) between the effects of a [itex]g[/itex] due to gravity and a [itex]g[/itex] due to the use of accelerated cooridnates.
A change in coordinates can of course produce a fictitious gravitational field; but if you don't consider this a real, physical field, then you side with the objector ("Critic") of early GR. Einstein countered that "the accelerated coordinate systems cannot be called upon as real causes for the field". As I elaborated, the issue here is with Einstein's Machian explanation of a gravitational field that is induced by the distant stars.
Quote Quote by DaleSpam View Post
[..] a change of coordinates does cause a gravitational field in the sense Einstein used the term. [..]
As you see here above, Einstein stated just the contrary.
DaleSpam
#114
Feb15-13, 10:35 AM
Mentor
P: 16,961
Quote Quote by harrylin View Post
A change in coordinates can of course produce a fictitious gravitational field; but if you don't consider this a real, physical field, then you side with the objector ("Critic") of early GR.
I don't worry too much about "real" or "fictitious" except where it is part of standard terminology (i.e. "real numbers" or "fictitious forces"), and I won't take either side of such a debate.

Quote Quote by harrylin View Post
Einstein countered that "the accelerated coordinate systems cannot be called upon as real causes for the field". As I elaborated, the issue here is with Einstein's Machian explanation of a gravitational field that is induced by the distant stars.

As you see here above, Einstein stated just the contrary.
In the same discussion he also said "the gravitational field in a space-time point is still not a quantity that is independent of coordinate choice; thus the gravitational field at a certain place does not correspond to something 'physically real', but in connection with other data it does. Therefore one can neither say, that the gravitational field in a certain place is something 'real', nor that it is 'merely fictitious'." He also said in the same discussion "Rather than distinguishing between 'real' and 'unreal' we want to more clearly distinguish between quantities that are inherent in the physical system as such (independent from the choice of coordinate system), and quantities that depend on the coordinate system." and "the distinction real - unreal is hardly helpful".

My point remains, that the "gravitational field" he refers to by "A gravitational field appears, that is directed towards the negative x-axis. Clock U1 is accelerated in free fall, until it has reached velocity v. An external force acts upon clock U2, preventing it from being set in motion by the gravitational field. When the clock U1 has reached velocity v the gravitational field disappears" is the field from the Christoffel symbols and it is entirely determined by the choice of coordinates.

I have not read all of Einsteins writings, but I believe that this is the meaning he usually attributes to the term. If you believe that he refers to something besides the Christoffel symbols then please be explicit about what mathematical term you think he intends and why.
Mentz114
#115
Feb15-13, 12:08 PM
PF Gold
P: 4,087
Quote Quote by harrylin View Post
What you here call "at rest" is also what Einstein calls "at rest"; your disagreement seems to be with the definition of acceleration that he used.
I don't know what you are talking about. The case I gave refers to proper acceleration which is unambiguous.

That's correct. In Langevin's "twin" scenario, the space capsule feels no force during the voyage.
I still don't know what you mean. You are moving the gaoalposts and and being slippery, because you're wrong.
GregAshmore
#116
Feb15-13, 05:56 PM
P: 221
Quote Quote by DaleSpam View Post
I did. I suspect that you didn't bother to read it, but stop acting as though the objection were summarily dismissed and no answer were given when you simply haven't bothered to read the answer given.
Yesterday I skimmed some posts, believing that I had "the gist" of the argument, without stopping to digest all the details. It's possible I missed your "gist", which would be sloppy of me. However, I don't think I missed what I was looking, for. I don't recall seeing it in posts prior to yesterday, either.

Before I say what I was looking for, I will accept a good deal of the criticism in recent posts. I have made some overly broad and not well thought out statements, resulting in errors. I realized that as I went through the mental exercise of constructing the spacetime diagram for the twin paradox, rather than accepting it as a finished product.

In my previous post, I asked for a SR solution of the problem in which the non-inertial rocket is always at rest. I don't recall anyone presenting such a solution. My sense is that my calls for such a solution have been rebuffed.

Every SR solution that I have seen has the rocket in motion. Certainly, in every SR spacetime diagram I have seen, the rocket is in motion; it is the reversing motion of the rocket which produces the knee in the rocket's worldline, and thus the difference in proper times. The article discussing acceleration in special relativity also speaks of the accelerating object as in motion. (http://math.ucr.edu/home/baez/physic...eleration.html)

I, in my rocket, claim that I am not in motion. I have been sitting in my rocket in the same position throughout the episode. You claim that I will be younger than my twin at the end of the episode. As proof, you present to me a diagram that shows me in motion. I reject it. I categorically deny that the diagram applies to me. The diagram shows me in motion; I have not moved.

I do not deny that my experience has been non-inertial. I do not deny that the experience of my twin has been inertial. I do not even deny that I have experienced "proper acceleration", because you have told me that I can experience proper acceleration while remaining motionless.

I deny that I have been in motion. Therefore, I insist on a solution that does not put me in motion.

What would a diagram of such a solution look like? There would have to be a point representing my position and time at the beginning of the episode, another point representing my position and time at the end of the episode, and a (possibly curvilinear) path that connects the two points. Whatever the shape of the path, the value of the position coordinate must not vary from the beginning to the end of the path, because I do not move.

It is maintained that SR can be used to solve this problem. I will attempt to draw the spacetime diagram. For convenience, I'll make my time and position axes orthogonal, in the usual manner with time positive toward the top of the page. And as usual I will set my starting position and time at the origin.

Because I do not move, my worldline must be parallel to the time axis; in this case it will be coincident with the time axis. I realize that this is the worldline of an inertial object. I am not inertial. It doesn't seem right that my worldline should be inertial, but that's how it must be, because I do not move.

Now to draw the earth. The earth is moving in my frame, and inertial in its own frame. I cannot find any worldline that will satisfy both conditions, and also meet me at the end of my worldline. I cannot complete the spacetime diagram.

Being unable to represent in a spacetime diagram the perfectly legitimate scenario of myself at rest throughout the episode, and the earth reversing, I conclude that SR is not suitable for solving the problem.

I may be wrong in my conclusion, as has been claimed. If so, I would like to see a SR solution which has me in one position throughout the episode.
ghwellsjr
#117
Feb15-13, 06:10 PM
PF Gold
P: 4,682
Quote Quote by GregAshmore View Post
I would like to see a SR solution which has me in one position throughout the episode.
How about this one:



from this thread.
GregAshmore
#118
Feb15-13, 06:14 PM
P: 221
Quote Quote by ghwellsjr View Post
How about this one:



from this thread.
I haven't read it yet. I was just coming on to see if I could save myself from the same mistake I've made before: seeing the thing in my head without checking it on paper. Some people learn slowly.
DaleSpam
#119
Feb15-13, 09:14 PM
Mentor
P: 16,961
Quote Quote by GregAshmore View Post
However, I don't think I missed what I was looking, for. I don't recall seeing it in posts prior to yesterday, either.
...
In my previous post, I asked for a SR solution of the problem in which the non-inertial rocket is always at rest. I don't recall anyone presenting such a solution.
...
What would a diagram of such a solution look like?
...
I may be wrong in my conclusion, as has been claimed. If so, I would like to see a SR solution which has me in one position throughout the episode
See post 80.

Quote Quote by GregAshmore View Post
Because I do not move, my worldline must be parallel to the time axis; in this case it will be coincident with the time axis. I realize that this is the worldline of an inertial object. I am not inertial. It doesn't seem right that my worldline should be inertial, but that's how it must be, because I do not move.
Don't forget, your frame is non-inertial so straight lines do not correspond to inertial worldlines. Inertial worldlines are geodesics, which is not the same thing as a straight line when you are using non-inertial coordinates.

Another, more accurate, way to say it is that your coordinate lines are bent. So lines of constant coordinates are not straight lines and straight lines don't have constant coordinates. A similar thing happens, e.g. in polar coordinates. Since the coordinates are curved the equation r=mθ+b does not represent a straight line.

Quote Quote by GregAshmore View Post
Being unable to represent in a spacetime diagram the perfectly legitimate scenario of myself at rest throughout the episode, and the earth reversing, I conclude that SR is not suitable for solving the problem.
Your inability to solve this reflects your own personal limitation, not a limitation of SR. It has been made abundantly clear to you that SR is not limited in this way.
GregAshmore
#120
Feb16-13, 07:56 AM
P: 221
Quote Quote by DaleSpam View Post
See post 80.
I have downloaded the paper.

Quote Quote by DaleSpam View Post
Don't forget, your frame is non-inertial so straight lines do not correspond to inertial worldlines. Inertial worldlines are geodesics, which is not the same thing as a straight line when you are using non-inertial coordinates.

Another, more accurate, way to say it is that your coordinate lines are bent. So lines of constant coordinates are not straight lines and straight lines don't have constant coordinates. A similar thing happens, e.g. in polar coordinates. Since the coordinates are curved the equation r=mθ+b does not represent a straight line.
I'll have to work on this to understand it.

Quote Quote by DaleSpam View Post
Your inability to solve this reflects your own personal limitation, not a limitation of SR.
I fully expect to find that you are correct in this.

Quote Quote by DaleSpam View Post
It has been made abundantly clear to you that SR is not limited in this way.
Well, no, it hasn't been made clear. It ought to be clear, I'm sure. The fact that it isn't clear is much more a factor of my response to what has been said than a factor of the content.

I have the impression that some of the contributors on this forum are teachers by trade, so what follows may be of interest. If not, no need to read further.

I've been trying to figure out why I have had so much trouble learning relativity. In particular, I have never had the experience of repeatedly thinking that I understand a subject, only to discover that I am profoundly wrong.

One reason, no doubt, is the bizarre premises that we are called on to accept. However, that was much more of a stumbling block at the beginning than it is now. At this point, I can "suspend disbelief" and treat the problem as an exercise in abstract logic. The "truth" or "reality" of the premises can be evaluated later.

So perhaps I'm just not good at abstract logic. Maybe. I'm sure I'm no Einstein, at any rate. But I'm not profoundly stupid, either. So how do I repeatedly find myself in the position of being profoundly wrong?

I had an "aha" moment on this a couple of weeks ago, which was reinforced and clarified last night. It has to do with my pattern of learning.

The entire subject of relativity is completely hands-off for me. I'll never see, much less operate, a particle accelerator. I learn new things all the time in my work, but in every case I can test my understanding of what ought to happen against what actually happens when I act on my understanding.

In addition, many aspects of relativity are hypothetical (hands-off) for everyone, at least in our lifetimes. We'll never travel at relativistic speeds. So none of us have the opportunity to directly test our understanding by experiment. (We have indirect experimental evidence to support what is predicted; that's not the same thing as making the prediction come to pass.)

I have made the mistake of thinking that "unable (in practice) to test by experiment" means "completely unable to verify". In my usual method of learning, I form a mental picture of what ought to happen, then I test it by experiment. Because I am unable to test, I have been in the habit of stopping after forming the mental picture.

In my work, I may do calculations after forming a mental image and before conducting an experiment. (I nearly always did when I designed machinery; I rarely do now that I work in a larger company and only write software.) The calculations are viewed as a means of avoiding failure in the experiment; they are never seen as a verification of anything. The calculations are never an end in themselves; they are a means of getting to the end, which is a functioning piece of equipment.

In relativity, for someone in my situation, the calculations are both the verification and the deliverable. That's what finally penetrated my thick skull last night. I can no more put something on this forum without verifying it by calculation than I can deliver an untested product to a customer.

Now maybe we'll see fewer dumb statements by me on the forum.

I did find this interesting, for perspective. Errors are never acceptable. But if even professionals have trouble, I should not be surprised if I have trouble, too.

From the paper referenced in post #80:
The path through this confusion existed already in Einstein’s original paper[9], and was popularised by Bondi in his work on ‘k-calculus’. It lies in the correct application of ‘radar time’ (referred to as ‘Marzke-Wheeler Coordinates’ in Pauri et al.[10]). This concept is not new. Indeed Bohm[1] and D’Inverno[2] both devote a whole chapter to k-calculus, and use ‘radar time’ (not under that name) to derive the hypersurfaces of simultaneity of inertial observers. However, both authors then apply this definition wrongly to the travelling twin.
stevendaryl
#121
Feb16-13, 08:53 AM
Sci Advisor
P: 1,945
Quote Quote by GregAshmore View Post
In my previous post, I asked for a SR solution of the problem in which the non-inertial rocket is always at rest. I don't recall anyone presenting such a solution. My sense is that my calls for such a solution have been rebuffed.
Once you understand how things work in inertial coordinates, then how they work in any other coordinate system (such as one in which the traveling twin is always at "rest") is just an application of calculus. So you are demanding that someone demonstrate a calculus exercise to you?

Okay, if it really will make you happy, I will post such a demonstration.
stevendaryl
#122
Feb16-13, 09:54 AM
Sci Advisor
P: 1,945
Accelerated Rocket in Inertial Coordinates

Let's choose the zero for [itex]x[/itex] and [itex]t[/itex] to make the mathematics as simple as possible.

So assume that one twin is at rest at some location [itex]x=L[/itex], in inertial coordinates. The other twin starts off at location [itex]x=L_1[/itex] at time [itex]t = -t_1[/itex], travels in the [itex]-x[/itex] direction until he reaches [itex]x=L_0[/itex] at time [itex]t=0[/itex], and then travels in the [itex]+x[/itex] direction until he reaches [itex]x=L[/itex] again at time [itex]t=+t_1[/itex]. The time origin is chosen so that his journey is symmetric in time about the point [itex]t=0[/itex]. The mathematical description of the traveling twin's path is:

[itex]x(t) = \sqrt{(L_0)^2 + c^2 t^2}[/itex]

Time [itex]t_1[/itex] is chosen so that [itex](t_1)^2 = \dfrac{L_1^2 - L_0^2}{c^2}[/itex]

The velocity of the traveling twin at any time is given by:

[itex]v(t) = \dfrac{c^2 t}{\sqrt{(L_0)^2 + c^2 t^2}}[/itex]

and the time-dilation factor [itex]\sqrt{1-\dfrac{v^2}{c^2}}[/itex] is given by:

[itex]\sqrt{1-\dfrac{v^2}{c^2}} = \dfrac{L_0}{\sqrt{(L_0)^2 + c^2 t^2}}[/itex]

The elapsed time for the traveling twin is given by:
[itex]\tau = \int \sqrt{1-\dfrac{v^2}{c^2}} dt[/itex]

I'm not going to do the integral, but you can see that the integrand is less than 1, so the result will certainly be less than [itex]\int dt = 2 t_1[/itex]. So the traveling twin will be younger.

I'm going to make another post where I describe this same situation from the point of view of the traveling twin.
stevendaryl
#123
Feb16-13, 12:06 PM
Sci Advisor
P: 1,945
Accelerated Rocket in Accelerated Coordinates

In inertial coordinates, the path of the traveling twin is given by:
[itex]x(t) = \sqrt{L_0^2 + c^2 t^2}[/itex]

Now, let's switch to a coordinate system in which the traveling twin is at rest, by making the transformation:

[itex]X = \sqrt{x^2 - c^2 t^2}[/itex]
[itex]T = \dfrac{L_0}{c} arctanh(\dfrac{ct}{x})[/itex]

where [itex]arctanh[/itex] is the inverse of the hyberbolic tangent function.

In terms of these coordinates, the path of the stay-at-home twin is given by:

[itex]X(T) = L_1\ sech(\dfrac{cT}{L_0})[/itex]

where [itex]sech[/itex] is the hyperbolic secant function.

To see that this is reasonable, you can look at the Taylor series expansion: [itex]sech(\theta) = 1 - \frac{1}{2} \theta^2 + \ldots[/itex]. So for small values of [itex]T[/itex], we have:

[itex]X(T) = L_1(1 - \frac{1}{2} \dfrac{c^2T^2}{L_0^2} + \ldots)[/itex]
[itex] = L_1 - \frac{1}{2} g_1 T^2 + \ldots[/itex]

where [itex]g_1 = \dfrac{c^2 L_1}{L_0^2}[/itex]

That is the path of an object that starts off at [itex]X = L_1[/itex] at time [itex]T=0[/itex] and falls at the acceleration rate of [itex]g_1[/itex].

So in these coordinates, the traveling twin is always at the location [itex]X = L_0[/itex], while the stay-at-home is at [itex]X = L_0[/itex] at some point (at some time prior to [itex]T=0[/itex], rises to [itex]X=L_1[/itex] at time [itex]T=0[/itex], and then falls back to [itex]X=L_0[/itex] at some later time.

In terms of the coordinates [itex]X,T[/itex], the elapsed time [itex]d\tau[/itex] for a traveling clock is given by:

[itex]d\tau = \sqrt{\dfrac{X^2}{L_0^2} - \dfrac{V^2}{c^2}} dT[/itex]

where [itex]V = \dfrac{dX}{dT}[/itex]

Note the difference with the formula for an inertial frame: [itex]d\tau[/itex] not only on the velocity, but on the position [itex]X[/itex].

For the traveling twin, who is always at [itex]X = L_0[/itex], his proper time is:
[itex]d\tau = dT[/itex]

For the stay-at-home twin, whose position [itex]X[/itex] is always greater than or equal to [itex]L_0[/itex], the first term [itex]\dfrac{X^2}{L_0^2} > 1[/itex]. If you work out the details, you will find that [itex]d\tau = \sqrt{\dfrac{X^2}{L_0^2} - \dfrac{V^2}{c^2}} dT > dT[/itex]

So in accelerated coordinates, it's also the case that the stay-at-home twin ages more than traveling twin.
PeterDonis
#124
Feb16-13, 12:16 PM
Physics
Sci Advisor
PF Gold
P: 6,039
Quote Quote by harrylin View Post
In Langevin's "twin" scenario, the space capsule feels no force during the voyage.
IIRC this is only because spacetime in Langevin's scenario has a different topology than standard Minkowski spacetime; it is spatially a cylinder in the x direction instead of an infinite line. The "traveling" twin goes around the cylinder whereas the "stay at home" twin does not. The two paths belong to different topological classes; you can't continuously deform one into the other.

In this kind of situation you can have multiple free-fall paths between the same pair of events with different proper times; but each free-fall path has maximal proper time compared to all non-free-fall paths within the same topological class. (There are other topological classes possible as well; for example, there could be another free-fall twin that went around the cylinder twice, etc.) For example, an accelerating "twin" that didn't go around the cylinder would have less elapsed proper time than the free-fall stay-at-home twin (but not necessarily less than the free-fall traveling twin); and an accelerating "twin" that *did* go around the cylinder would have less elapsed proper time than the free-fall traveling twin.
DaleSpam
#125
Feb16-13, 01:38 PM
Mentor
P: 16,961
Quote Quote by GregAshmore View Post
I'll have to work on this to understand it.
I think that this is probably one of the key topics of modern relativity.

The key concepts of relativity, both special and general, are geometrical (Minkowski geometry for SR and pseudo-Riemannian geometry for GR).

Just like you can take a piece of paper and draw geometrical figures and discuss many things, such as lengths and angles, without ever setting up a coordinate system. The same thing is possible in relativity. The "piece of paper" is spacetime which has the geometrical structure of a manifold. The "geometrical figures" are worldlines, events, vectors, etc. that represent the motion of objects, collisions, energy-momentum, etc.

In this geometrical approach, the twin scenario is simply a triangle, and the fact that the travelling twin is younger is simply the triangle inequality for Minkowski geometry. In a coordinate-independent sense, the traveler's worldline is bent, and that in turn implies that his worldline is necessarily shorter as a direct consequence of the Minkowski geometry.

Now, on top of that underlying geometry, you can optionally add coordinates. Coordinates are simply a mapping between points in the manifold (events in spacetime) and points in R4. The mapping must be smooth and invertible, but little else, so there is considerable freedom in choosing the mapping. It is possible to choose a mapping which maps straight lines in spacetime to straight lines in R4, such mappings are called inertial frames.

It is also possible to choose a mapping which maps bent lines in spacetime to straight lines in R4, a non-inertial frame. Such a mapping does nothing to alter the underlying geometry. The bent lines are still bent in a coordinate-independent geometrical sense, but because it simplifies the representation in R4 it can still be useful on occasion in order to simplify calculations.

Because the mapping is invertible, in many ways it doesn't matter if you are talking about points in the manifold or points in R4. So you talk about things being "at rest" based on R4, and things happening "simultaneously" based on R4, and many other things. However, it is occasionally important to remember the underlying geometry.

I hope this helps.

Quote Quote by GregAshmore View Post
Well, no, it hasn't been made clear. It ought to be clear, I'm sure.
It has, in fact, been made clear to you that SR can handle the twins paradox. What obviously hasn't been made clear to you is why. Your repetition of bald assertions that have already been contradicted is unhelpful. It wastes your time in repeating it and it wastes our time in repeating our responses. It also irritates those (maybe only me) who feel like their well-considered and helpful responses have been completely ignored by you.

You have been provided explanations and references addressing the topic, which clearly didn't "do it" for you. Read the explanations and references and point out the specific things that you don't understand or don't agree with. Then we can help clarify and make some progress.
TrickyDicky
#126
Feb16-13, 02:56 PM
P: 3,001
Quote Quote by DaleSpam View Post
Just like you can take a piece of paper and draw geometrical figures and discuss many things, such as lengths and angles, without ever setting up a coordinate system. The same thing is possible in relativity. The "piece of paper" is spacetime which has the geometrical structure of a manifold.
If this is so I wonder why rigorous definitions of manifolds are based on charts (that define local coordinate systems). IOW and simplifying: any object (that is of course also a topological space with certain topological features that make it well behaved) that can be "charted" is a manifold.
So the "piece of paper" must have the property that you can set up a coordinate system on it, that is its defining property if you want to call it a manifold. The fact that "you don't have to" is obviously relying on the fact that it is implicit in the definition, so I always have trouble understanding the insistence on banning charts.


Register to reply

Related Discussions
In binary can we have a value with deci centi mili or more lower valued prefix? Computers 14
Gravitational field vs. acceleration due to gravity Classical Physics 2
The terms absolute and relative Special & General Relativity 13
Universe not accidental: Is this Steinhardt statement rather pathetic ? If so, why? Cosmology 114
Checking Chadwick's statement about the mysterious neutral radiation ? Advanced Physics Homework 4