DrRocket

This entire discussion was predicated on special relativity. That implies no gravity. Mass in general relativity is a whole new can of worms. http://en.wikipedia....eral_relativity

1. There is no meaning in mathematics to the term "reference frame". It is a concept from physics. I think you are referring to what in linear algebra is called a set of basis vectors. 2. Given two sets of basis vectors x1, x2, .., xn and y1,y2,...yn a change of basis is a matrix T for which the ith column is the coordinates of xi in terms of y1, y2, .., yn 3. So to solve the vector equation u= A.v in terms of matrices you need to express everything in terms of a single basis. The transformation T is actually the identity linear transformation, though not the identity matrix. If it takes the "left frame" to the "right frame" then your equation in matrix terms, referred to the "right frame" is u= ATv This reduces the problem to solving u= B.v in a single reference frame. Unless the vector space is 1-dimensional there are infinitely many such matrices B for any given u and v. 4. If with respect to a fixed basis, and given u and v you want to find a linear transformation such that u=Bv you can simply send v to u and extend linearly. There are infinitely many ways to do this in dimension 2 and above. One simple way is this: If u has coordinates (u1, u2, ..., un) and v has coordinates (v1, v2, ..., vn) B can be a diagonal matrix with diagonal elements u1/v1, u2/v2, ... , un/vn

integrate twice

1. Since the equation is non-linear, what makes you think that the solution is expressible as the sum of "the general homogeneous solution" plus an "inhomogeneous solution" ? When the differential operator is linear, any two solutions of the inhomogeneous equation differ by a solution of the homogeneous equation. Not so for non-linear equations. 2. [math] x(t) = e^x - \dfrac{x^2}{2} + cos(t) [/math] fails to express x explicitly as a function of t. 3. [math] \dfrac {d}{dt}( e^x - \dfrac{x^2}{2} + cos(t)) = e^{x(t)} \frac {dx}{dt} -x \frac {dx}{dt} - sin(t) [/math] [math] \dfrac {d^2}{dt^2}( e^x - \dfrac{x^2}{2} + cos(t))[/math][math] = e^{x(t)} (\frac {dx}{dt})^2 + e^{x(t)} \frac {d^2x}{dt^2} - (\frac {dx}{dt})^2 -x \frac {d^2x}{dt^2} - cos(t) [/math] This does not appear to help very much. 4. The technique you are applying in your [math]x''(t) + 1 = e^{x(t)}; \lambda^2+1 = 0 \Rightarrow \lambda_{1} = i, \lambda_{2} = -i ; x(t) = D_{1}cos(t) + D_{2}sin(t) \because x(0) = 1 , x'(0) = 0 [/math] is intended for homogeneous linear equations. But this is all screwed up. First because [math]x''+1=0[/math] is not homogeneous, and second because it does not help with [math]x''(t) + 1 = e^{x(t)}[/math] The solutions to [math]x"=-1[/math] are [math] x(t) = - \frac {t^2}{2} - at + c [/math], so you can see that you have misapplied the characteristic equation.

Sorry, brain fart. None of the techniques discussed are going to work, because this ODE is non-linear. [math] \dfrac {d^2 \ x}{dt^2} - e^x = -1[/math] Given the OP's statement that a solution in terms of a general solution to the homogeneous equation plus a particular solution to the inhomogeneous equation is expected (which applies to a linear ODE), I suspect that the equation written is not the equation of the real problem. Should it be [math] \dfrac {d^2 \ x}{dt^2} = e^t -1[/math] ? If so, then integrate twice.

http://www.scienceforums.net/topic/33180-cosmo-basics/

Your interests are tailor-made for biomedical engineering. There are even programs that can result in q combined MD/PhD.

Integrate twice.

For 1 & 2 all that you need is the definition of linear independence. For 3 you need the definitions of "span" and of "vector space". The statement "let W be a subspace of V containing S' means W contains S and W is a subspace of V.

What is synonymous are the statements "a is congruent to b mod n" and "a=b (mod n)". Congruence is not equality. However, when one is working with equivalence classes, one can correctly state that two equivalence classes are equal. It is often the case in mathematics that one is working with equivalence classes, and equality is equality of equivalence classes. Sometimes the fact that it is equivalence classes that are the objects of interest is understood but not explicitly stated -- as, for, instance in the theory of measure and integration where "equals" often means "equals almost everywhere" -- equivalence meaning equality except possibly on a set of measure zero. Whether the use of equivalence classes is explicitly stated or just understood depends on the intended audience. It is made explicit in introductory tests. It is often implicit when one is addressing experts.

The first, when done by real scientists, is a speculation based on insight and vision. The second is often a fantasy based on delusion and hallucination. Caveat: When one is merely talking about a different approach but not disputing the validity of established science, as with Lagrangian or Hamiltonian mechanics versus the standard Newtonian F=ma approach, then a new approach may have merit.

Those two statements are synonymous.

Let's be blunt. The problem is not really speculation in the true sense. Theoretical research, by its very nature is speculation, informed and responsible speculation. String theory, quantum loop gravity, and Lisi's E8 theory are all speculative. They ae also legitimate research avenues. Without speculation there would be no research, and no progress. What is objectionable are the "relativity is wrong because it violates my intuition", "I don't understand quantum mechanics so it must be wrong" or "Faraday's law is wrong" diatribes which have zero basis in either theory or empirical data. In short "speculative' is being used as a euphemism for "wacko". Everybody understands this, even wackos. Uinderstandably, wackos do not like to be clearly identified. Gee, that's unfortunate. The policy of limiting objectionable theories to the "Speculations" forum is appropriate and necessary. What is important is not what happens in "Speculations", but what, as a result, does not happen elsewhere. I hope that you and other mods keep up the good work.

Old people tend to have greater wealth than young people. Old people tend to die more frequently than younger people. Ergo, wealth will kill you. I am offering a similar service to yours , but more comprehensive as I will accept all items of value, and do not limit benefits to only a single threat to life. Radiation is only one of the items covered. Contribute and live longer. I am old enough to feel little burden from additional wealth, so you can contribute with a clear conscience. Credit cards not accepted.

The problem arises with what one means by [math]z^{\frac {1}{2}}[/math]. the convention that you learn in high school algebra is that it is the positive square root if [math]z>0[/math]. In elementary calculus you learn that [math]x^y = e^{y\ log \ x}[/math] when [math] x>0 [/math] This also works for complex numbers [math]z^y = e^{y\ log \ z}[/math], but one must be very careful with what is meant by [math]log \ z [/math]. [math] log \ z = log (|z|) + i \ Arg (z) [/math] but [math] Arg (z)[/math] is only defined modulo [math]2 \pi [/math] and as one winds around the origin [math]Arg[/math] is either ambiguous or discontinuous. Mathematicians handle this by either "picking a branch of the logarithm' or by working on a Riemann surface instead of the complex plane. In "picking a branch of the logarithm" one deletes from consideration some infinite line segment starting at the origin -- deleting the negative real axis corresponds to the usual convention of elementary algebra and calculus. The net result is that the simple aglebraic rules, like [math] x^a x^b = x^{a+b}[/math], [math](xy)^a = x^ay^a[/math] , [math](x^a)^b = x^{ab}[/math] do not hold in general when dealing with complex numbers.

Seems reasonable. So what do I find in the ad banner attached to this very thread, but a link to a company selling potassium iodide !!! http://www.nutrabio.com/Products/potassium_iodide.htm?gclid=CMXNwPCO3qcCFQImbAodNT7R-Q

No it means that the set of all linear combinations of elements of T does not include all of Mnn. One presumes that by Mnn you mean all nxn matrices, so the problem is to show that the space spanned by commutators is not all linear operators on n-space. To do that you might want to look for an invariant shared by commutators that is not a property of all operators.

I have only seen that occur in one instance in which a student insisted in proceeding to the dissertation defense over the advice of his advisor -- essentially suicide. Under ordinary circumstances that sort of thing should be prevented by the advisor. Oddly, that one instance that I can recall occured in a chemistry department -- I think not far from you . There was a potential for a similar incident in the Physics Department at the state university in the state in which I now live.. A baseless, and mathless, "theory of everything" was, and is, being promoted by a student with ties to the university president. There was no danger of a physics PhD being awarded for this, but the internal university politics were delicate. A thesis defense debacle was a possibility. Fortunately the student transferred to the Philosophy Department. I have not seen the science world being "very cut-throat", only very rigorous. I can recall, shortly before I graduated, giving a talk at a special session of the Annual Meeting of the American Mathematical Society to a room full of "big-shots". They were very cordial. They were equally cordial when one interrupted another speaker, a faculty member at Purdue, and showed why what he was saying was wrong. That speaker was a bit embarrassed for the moment, but told me personally that the guy who revealed the error was absolutely correct. There was no animosity on the part of either of them, and I think they are cordial to this day. My dealings with both of them were also always cordial. What might be interpreted as cut-throat is intolerance for wackos. That is something else entirely, something understandable and necessary since the internet has provided a soap box for purveyors of nonsense. That sort of intolerance, consistent with off-the-wall "theories' being relegated to "Speculations", is a good thing.

Good luck. I have yet to see a good popularization of meanungful mathematics. Physics popularizations generally rely on dumbing down the math. That pretty clearly won't work for your task. Talks that I have heard either lacked content or were understandable only by a small minority -- mathematicians and a few physicsts.

huh ?

I think you need to complete some of your planned studies of topology and group theory, plus some analysis on manifolds before you tackle Lie Algebras. A Lie group is a manifold that is also a group and for which the group operation is smooth. The Lie algebra of a Lie group is the tangent space at the identity with a non-associative multiplication that comes from viewing the tangent space as differential operators. if this seems a bit abstract, I am, not surprised, but it will make more sense after you have learned more mathematics. Don't get ahead of yourself. I am not at all sure that Lie algebras are an appropriate tool for your end application. I suspect not.

Haven't you studied the "quadratic formula" ? You should be able to find the roots of a quadratic equation with no calculator at all.

[math]A^2= \-I[/math] Note: [math](det \ A)^2=det(-I)=(-1)^n \ \Rightarrow \ n[/math] is even. Note also that the conclusuin that n is even and [math] det \ A \ = \ \pm 1 [/math] would not hold over the complex field, so the proof must take advantage of the fact that we are working over the reals. Sketch of proof: 1. Using Jordan canonical form A can be assumed to be block diagonal, with 2x2 blocks. This reduces the problem to the case n=2. 2. So now [math] A \in (O2)[/math]. [math]A[/math] is thus a composition of a rotation and, if [math]det \ A \ = \ -1[/math], an orientation changing transformation, which can be taken to be . [math] C= \left ( \begin{array} {cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right )[/math] Note that [math]C[/math] does not commute with rotations, but yet that any element in [math]O2[/math] can be written as [math]B[/math] or [math]BC[/math] where [math]B[/math] is some rotation. 3. Now examine the possibilities [math]A= BC[/math] where [math]B[/math] is a rotation and show that [math](BC)^2 \ne \ -I[/math] . 4. That the shows that [math] A [/math] is a rotation through [math]\frac {\pi}{2}[/math] or [math]\frac {3 \pi}{2}[/math] and hence has determinant 1.

Let [math] A= \left ( \begin{array} {cc} 1 & 0 \\ 0 & -1 \\ \end{array} \right )[/math] Then [math]A^2 = I[/math] and det[math]A=-1[/math]

Thanks anyway for spending some of your time and lending your expertise.