Xerxes

Senior Members
  • Content count

    246
  • Joined

  • Last visited

  • Days Won

    3

Xerxes last won the day on December 14 2015

Xerxes had the most liked content!

Community Reputation

45 Good

About Xerxes

  • Rank
    Meson
  • Birthday May 19

Profile Information

  • Gender
    Male
  • Location
    UK
  • College Major/Degree
    PhD@the best
  • Favorite Area of Science
    All
  • Biography
    I am a weird person
  • Occupation
    Trainee alcoholic
  1. 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
  2. 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
  3. !!!!! +1
  4. 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
  5. 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.
  6. 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.......
  7. 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
  8. 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.
  9. 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].
  10. 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.
  11. 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
  12. 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......
  13. 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]
  14. 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.
  15. 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