Xerxes

And what, pray, is this this called "reality"? Surely not snooker balls on trampolines?

You have a problem with mathematics? How else would you describe the Theory of General Relativity?

Actually, curvature is defined as the second derivative of the metric field. So if the metric field is constant the curvature field is zero - elementary calculus. Non-zero otherwise

I ask again - where is your argument that says this is the case? I have one, although to say it is hand waving would be to flatter it!

Why? Or Rather how? Your argument is...........

As this question was posted in the Linear Algebra and Group Theory section it seems reasonable to assume it was referring to the theorem that the order of any group is an integer multiple of that of any of it's subgroups (proof is easy - try it; hint - use the definition of cosets) Whether it applies in so-called daily life, I have no idea. Should we care?

I wonder if it might be helpful to point out the difference between the ordinal numbers and the cardinal numbers, since we seem to restricting the discussion to the natural numbers \(\mathbb{N}\) So an ordinal number roughly speaking describes the position of an element in an ordered set. In contrast a cardinal number describes the size of a set, ordered or otherwise. Notice these are quite different concepts. Now, by construction the natural numbers are ordered (there is a theorem that any set can be ordered - the proof is hellacious and not relevant here). So it is fairly easy to see that, for any subset of \(\mathbb{N}\) (it's ordered recall) if there exists a largest ordinal \(n\) then this corresponds to the cardinality of our subset and it must be finite. Otherwise the gloves are off. The largest non-finite ordinal, by an arbitrary convention is denoted as \(\omega\). This is still an ordinal., and can in no way denote the cardinality of a non finite subset of \(\mathbb{N}\) e.g \(\mathbb{N}\) itself. For this we use the arbitrary symbol \(\aleph_0\). Any help?

Wrong formatting, sorry

In fact in science "points of view" have no value. None. Whatever. The so-called "curvature of spacetime" arises because in the presence of mass/energy the most useful/appropriate coordinates to describe the geometry of the 4-dimensional spacetime manifold are curvilinear, not rectilinear. "Spacetime curvature" is an unhelpful pop-science term. It is easy to see that in the (theoretical) absence of a gravitational source, or if you prefer, at infinite distance from one, the rectilinear (i.e. quasi-Cartesian) coordinates will suffice. Yes, as far as is known - but note that, in this context, gravitation is considered a form of mass/energy. That is gravitation causes itself!! For this reason, mathematicians call the General Theory of Relativity non-lineaar

Hi Function. If you were submitting your thesis to a UK university board, there are no "rules". There are, however, conventions, not only in theses but generally. 1. You can refer to yourself however you like - but John Maynard Octavious Smith, Jr. would be considered pretentious 2. In attributions or thanks, it is usually sufficient to use the suffix title only - Prof., Dr., etc. But be careful - in a clinical context, do not refer to someone as Mr. unless they are a surgeon. 3. "Trailing" qualifications (MD, FRCS, PhD, BSc etc) are not normally used in this context 4. If in doubt, ask the people concerned what they would prefer. That seems the simplest course

!!!!! +1

Good God, did I really say this? It appears that I did - I cannot imagine what I was thinking, as it is quite clearly nuts. Apologies

No, I don't think this is quite correct. If [math]n \in \mathbb{N}[/math] and [math]\mathbb{N}[/math] is infinite but countable i.e has cardinality [math]\aleph_0[/math], then the vector spaces [math]\mathbb{R}^n[/math] and [math]\mathbb{C}^n[/math] are necessarily infinite dimensional vector spaces.

Then if you know better than those who try to guide you, you do not need to ask the question, right? Just a reminder: the image of your first function is [math]h(x)[/math]. Then you claim it's preimage is [math]h^{-1}(x)[/math]. So [math]h(x)=x[/math]. Agreed? And so on.......

No, this forum doesn't quite work like that. First if this is homework, you are in the wrong sub-forum - try "homework". Second, nobody will do your homework for you, although they may give hints if you show what you have tried. If it is not homework, you are wasting everybody's time with a pointless question

No, never! I agree in general, I think. Would you accept the following........ .....a coordinate set is referred to as an inertial reference frame iff for a body considered in inertial motion relative to one coordinate set, there exists a global orthogonal transformation that brings our body the rest relative to these new coordinates. This seems to be the case in the Special Theory, as far as I understand it. Of course, in the General Theory, global coordinates do not exist, so the term "inertial reference frame", by the above ad hoc definition, is not appropriate.

If you are using spatial coordinates here. Yes Now you are using spacetime coordinates. Is it not the case that, given spatial coordinates [math]x,\,y,\,z[/math] and time-like coordinate [math]ct[/math] then under some arbitrary transformation that [math]x^2+y^2+z^2 \ne x'^2+y'^2+z'^2[/math] [math](ct)^2 \ne(ct')^2[/math] and yet [math]x^2+y^2+z^2-(ct)^2=x'^2+y'^2+z'^2-(ct')^2[/math].

The cube has 8 vertices, events in spacetime are points in a spacetime 4-manifold i.e. uniquely specified by 4 Real numbers Hmm. Is this true when I choose coordinates such that, say, the spatial coordinate [math]x^4 = ct[/math]? Not being a physicist I had always understood that a "point" in spacetime was an "event" as used in common parlance.

So why are you on this site if you have nothing to share? With this attitude, as far as I am concerned, I would be disinclined to answer any questions from you. You are abusing the forum and insulting its members

Neither. If I say for some arbitrary [math]a \in \mathbb{R}[/math] that [math]f(a) \in \mathbb{R}[/math] I am entitled to ask how the image varies as the argument varies. In other words, [math]a[/math] is a variable. For reasons I gave in another post, I can write this as [math]\frac{df}{da}[/math] it being understood this is unambiguous shorthand for [math]\frac{df(a)}{a}[/math]. You seem to believe that [math]x[/math] is the only possible label I can attach to an arbitrary Real number - it's not Of course not. [math]\pi[/math] is a constant. You cannot differentiate with respect to a constant Interesting - I don't like it one bit. But let's not go there......

Irrelevant link. [math]a^a=b \Rightarrow a=\log_a b =(^a\sqrt{b})[/math]. Of course this generalizes - [math]x^y=z \Rightarrow y=log_xz\,\,\ne \,\,(^y\sqrt{z})[/math]

Ha! So I am fired, in the nicest possible way! *wink* Do not feel bad, wtf. Differential geometry is a hard subject, as you would see if you had all 5 volumes of Michael Spivak's work. I do not pretend to have his depth of knowledge - I merely took a college course. Moreover his reputation as a teacher is extremely high, whereas mine is ....... (do NOT insert comment here!) Regarding applications, all I can say is that I am neither an engineer nor a physicist, so as far as bridge bolts etc. you would need to ask somebody else. On the other hand, it is not possible to study differential geometry without at some point encountering tensor fields, especially metric fields and the curvature fields that arise from them. These are the principle objects of interest in the General Theory of Gravitation. If I offered to give guidance on this subject, it would be strictly as an outsider, an amateur.

This is not quite correct. The transformation [math]A[/math] that you refer to has of course a matrix representation. But if you mean to imply that the matrix for a contravariant transformation is the inverse of that for a covariant transformation, you'd be wrong - the 2 matrices are mutual transposes. Now it is true that there are matrices/transformations where the transpose IS the inverse - these are called unitary transformations - in general coordinate transformations are not of this type. Specifically, the Lorentz transformation is not unitary. Maybe, but distance is a scalar quantity, and for any physically meaningful transformation, one requires that scalars be invariant. Doubly so in the case of distance, since again, physically meaningful transformations are compelled to preserve the metric. The terms co- and contravariant, although out-moded, refer to vector-like entities ONLY

Yes, but at no pint did I assert that a continuous function needs to be differentiable. Rather I asserted the converse - a differentiable function must be continuous. Maybe I did not make myself clear. I said that the [math]C^k[/math] property for a function "subsumes" the [math]C^0[/math] property. If we attach the obvious meaning to the [math]C[/math] in [math]C^k[/math] we will say that a [math]C^0[/math] function is continuous to order zero, a [math]C^1[/math] function is continuous to order one..... a [math]C^k[/math] function is continuous to order [math]k[/math] I am sorry if my language was not sufficiently clear.

So, in spite of a sudden lack of interest, I will continue talking to myself, as I hate loose ends. Recall I gave you in post#27 that, for set open sets in [math]U \cap U'[/math] we will have the coordinate transformations [math]x'^j=x'^j(x^k)[/math]. Notice I am here treating the [math]x'^j[/math] as functions, and the [math]x^k[/math] as arguments Suppose some point [math]m \in U \cap U'[/math] and a vector space [math]T_mM[/math] defined over this point. Recall also I said in post#41 that for any [math]v \in T_mM[/math] that [math]v(x^j)=\alpha^j[/math] which are called the components of [math]v= \alpha^j \frac{\partial}{\partial x^j}[/math]. Likewise I must have that [math]v=\alpha'^k \frac{\partial}{\partial x'^k}[/math]. We may assume these are equal, since our vector [math]v[/math] is a Real Thing So that [math]\alpha^j=v(x^j)[/math] and [math]\alpha'^k=v(x'^k)[/math], we must have that [math]\alpha'^k= \alpha^j\frac{\partial x'^k}{\partial x^j}[/math]. This is the transformation law for the components of a tangent vector, also known (by virtue of the above) as a type (1,0) tensor. It is no work at at to extract the transformation laws for higher rank tensors, and very little to extract those for type (0,n) tensors. PS I do wish that members would not ask questions where either they they are not equipped to understand the answers, or have no real interest in the subject they raise