A few comments on Gravity

I have him on ignore, so I don't feel the need to respond any longer. I'm glad y'all are keeping him in check. Or trying...

 

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That, to say the least, is a desperately patronizing comment.

I am SO grateful for that liberty.

It is for others to judge whether my contributions here have been "useful". But I stand by my knowledge of differential geometry, and I say EMPHATICALLY you have no such knowledge, and should not lecture others on the General Theory

 

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I sometimes forget what I had for breakfast yesterday, but I don't remember posting anything in this tread about what I believe... At least as far as any comments I have made relative to your beliefs... Well other than context itself is important and you don't seem to recognize context as anything other than your own misinterpretation...

 

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More in anger that in "love" I copy this post of mine own from another forum.

(To some extent, derived from the late and much missed Hanno Rund)

Let us see if we can derive a metric for an arbitrary smooth (\(C^{\infty}\)) manifold \(M\)

I will first assume that we have a vector space \(T_pM\) defined at the point \(p \in M\), and further assume that the norm - the length - \(|V|\) of \(V=A^j \frac{\partial}{\partial x^j}\) is given by

\(|V|^2=\sum\nolimits(A^j)^2\) just as in the Euclidean case - that is the norm is a quadratic in its components.

Now we know that each point \(p \in M\) is uniquely specified by the set \(\{x^j\}\) of coordinate functions, so I make two further assumptions (possibly three):

Here, as in the Euclidean case, from the vector norm - its quadratic form - I can extract a bilinear form - an inner product (and vice versa)

However we are not in a Euclidean space, so I make a further assumption, or rather stipulation

Since for an arbitrary manifold, with or with a metric, the inner product of, say, \(V \in T_pM\) with \(W \in T_qM\) can have no meaning whatever (this comes from the Hausdorff property of our manifold viewed as a topological space), I stipulate that whatever I may have to say about a metric will refer to a specified point and that point only.

So to work....

Given all the above, I define a real-valued function \(G(x^j,A^j)\).

I make one further stipulation - that this function be homogeneous linear in the first degree in the arguments \(A^j\). That is, for any real number \(\alpha\) that

\(G(x^j,\alpha A^j)= \alpha G(x^j,A^j)\)

So I now define a curve in \(M\) by \(\gamma:[0,1] \to M\) with \(t\in \gamma\) (a Real number, of course) so that \(\gamma(t) = p_t \equiv x^j(t)\), since \(p = \{x^1,x^2,....,x^n\}\)

I now suppose that, to an arbitrary displacement \(dx^j\) at tex]p[/tex], there corresponds a vector tangent to a curve that "passes through" this point, and accordingly, for the length of arc \(s\) between say \(p_1\) and \(p_2\), whose coordinates functions are parametized by \(t_1\) and \(t_2\) respectively

\(\displaystyle = \int\nolimits_{p_1}^{p_2}G(x^j,dx^j)=\int\nolimits_{t_1}^{t_2}G(x^j,\frac{dx^j}{dt})dt\)

For legibility I write \(\frac{dx}{dt}\equiv \dot{x}\) - this is standard notation in this sort of problem.

By virtue of the fact that I have declared my function to be positively linear homogeneous of degree 1, I now refer you to

Euler's homogeneous function theorem

from which we may infer that

\(\displaystyle\frac{\partial G(x^j,\dot{x}^j)}{\partial \dot{x}^j}\dot{x}^j= G(x^j,\dot{x}^j)\) and

\(\displaystyle\frac{\partial^2G(x^j,\dot{x}^j)}{\partial \dot{x}^j\partial \dot{x}^k}\dot{x}^k = 0\)

Let's drop superscripts when they merely appear on arguments of our function - it simplifies things a lot and leads to no errors (I promise!).

Now by a very slight abuse of the product rule from elementary calculus I write

\(\displaystyle\frac{\partial G^2(x,\dot{x})}{\partial \dot{x}^j}=2[G(x,\dot{x})\frac{\partial G(x,\dot{x})}{\partial \dot{x}^j}]\)

from which it follows that

\(\displaystyle \frac{1}{2}\frac{\partial^2 G^2(x,\dot{x})}{\partial \dot{x}^j \partial \dot{x}^k}\)

\( \displaystyle =\frac{\partial G(x, \dot{x})}{\partial \dot{x}^j}\frac{\partial G (x,\dot{x})}{\partial \dot{x}^k}\) \(\displaystyle +G(x,\dot{x})\frac{\partial^2 G^2(x,\dot{x})}{\partial \dot{x}^j \partial \dot{x}^k}\)

But from Euler's theorem the las term on the RHS is zero, while the first is just

\(\displaystyle G^2(x,\dot{x}) = \frac{1}{2}\frac{\partial^2 G^2(x,\dot{x})}{\partial \dot{x}^j \partial \dot{x}^k}\dot{x}^y \dot{x}^k\)

I now DEFINE the gadget \(\displaystyle g_{jk}(x,\dot{x}) = \frac{1}{2}\frac{\partial^2 G^2(x,\dot{x})}{\partial \dot{x}^j \partial \dot{x}^k}\), without making claims as to its true nature, which of course we will need to come back to.

But for now notice the above (but one) can be written as \(g_{jk}(x,\dot{x})\dot{x}^j \dot{x}^k\), so Farsight's homework is......

Prove that \(\dot{x}^i\) are the components of a type (1,0) tensor (Hint: use the chain rule to show the properties of a coordinate transformation on the scalar components of such a tensor)

 

No it is not it is not recognized as 'akin' to that at all. You would know this if you do the math, oh that's right you can't do the math, which means you can't do the physics.

You are a combination of arrogance and ignorance. That my dear fellow is a horrible combination. At least your misplaced arrogance shields you from realizing what a laughing stock you are.

 

Is it safe to say that this thread has run its (inevitable) course and has exceeded its life expectancy?

 

Very. The thought of reading "popscience trash" wave-offs again makes me ill.