Recent content by anuttarasammyak

Thank you @JimWhoKnew for the clear and easy-to-understand explanation for the "anomaly". I have preffered to regard "new" ##\bar{g}_{\mu\nu}## as "basic" then old ##g_{\mu\nu}## , ##\delta{g}_{\mu\nu}## and their reciprocals are not representations of the same coordinate free object...

Thanks. Even with this general caution, may we take it obvious that in tensor division of $$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} $$ all ## \bar{g}_{\mu\nu}## , ##g_{\mu\nu}## and ##\delta g_{\mu\nu}## are tensors in the world of metric undertaking variation ?

In the world of metric tensor undertaking variation where the metric tensor is $$ \bar{g}_{\mu\nu} := g_{\mu\nu}+\delta g_{\mu\nu} $$, Tensor indeces up-down will be done as $$ A_{\mu\nu}= \bar{g}_{\alpha\mu}\bar{g}_{\nu\beta}A^{\alpha\beta}$$ I consider the case that $$...

@haushofer it is a good chance to learn from your comments. May I say that in GR when tensor ##A_{abc..}^{def...}## is written arbitrary as $$A_{abc..}^{def...}=X_{abc..}^{def...}+(A_{abc..}^{def...}-X_{abc..}^{def...})$$ , are X and A-X also tensors ? In GR any linear combination of tensors...

@haushofer Thanks for the teaching. $$ g_{\mu\nu}+\delta g_{\mu\nu} =( g_{\alpha\nu}+\delta g_{\alpha\nu}) (g_{\mu\beta}+\delta g_{\mu\beta})( g^{\alpha\beta}+\delta g^{\alpha\beta} )$$ Is it an incorrect equation though the result coincides with the equation (*) ?

No, I don't know any case in tensors. That is one of the reasons that I think : I am afraid that you are against it in post #11, stating that one of them variation infinitesimal δgμν is a tensor. I should appreciate it if you could explain the validation of this minus-sign for tensor δgμν.

@haushofer The equation is the application of the rule for raising and lowering the indices in use of the (varied) metric tensor to the (varied) metric tensor itself. Then how can we distinguish which rule, (*) or (**) in OP, holds for each tensorial quantity of our interest ?

Under variation of metric tensor, ##g_{\mu\nu}+\delta g_{\mu\nu} ## is a tensor but its parts ##g_{\mu\nu}## and ##\delta g_{\mu\nu} ## are not tensors. Its indeces up-down relation is $$ g_{\mu\nu}+\delta g_{\mu\nu} =( g_{\alpha\nu}+\delta g_{\alpha\nu}) (g_{\mu\beta}+\delta g_{\mu\beta})(...

Why don't you start with the familiar case that all $$x^0,x^1,x^2,x^3$$ have dimension of length. All the metric tensor components, thus its determinant g also, are dimensionless. Then you may investigate more general case, e.g., cylindrical or polar type coordinates, if you wish.

@Kostic As for your question the explanation of L-L : seems a good complementary to Dirac.

I do not have academic background but I am involved in management of quantum computing research in a public institute. I observe most of condensed matter physics have something to do with QIS. I hope you will find your own field has some relation with QIS which would give you better...

Thanks @Ibix So in other words $$ \frac{1}{\bar{v}}=\sum_i \frac{r_i}{v_i} $$ where r_i is ratio of i-th segment's length to the whole length. Voltage is shared by the parallel resistors but there is no such sharing for segment runs. Here 1/2 or 1/N comes from taking average (of inverse speed)...

Your "arismetic mean of inverse" in reminds me of law of synthesizede resistance of parallel ressitors $$ \frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2} $$ They have difference of factor 2. I would like to understnd the situation difference clearly.

Just for a fun I observe $$ \bar{v}=\frac{\sqrt{v_1v_2}}{\frac{v_1+v_2}{2}}\sqrt{v_1v_2} \leq \sqrt{v_1v_2} $$

$$\bar{v}=\frac{2L}{L/v_1+L/v_2}=\frac{2v_1v_2}{v_1+v_2}$$ I am not sure what are harmonic mean and weighted average you say in this result.