Recent content by anuttarasammyak

This anti-symmetric equation has six non-zero solutions.

Thank you @Demystifier 1. I observe that a. ## \bar{g}_{\mu\nu}-g_{\mu\nu}## transforms as a covariant tensor under coordinate transformation b. ## \bar{g}^{\mu\nu}-g^{\mu\nu}## transforms as a contravariant tensor under coordinate transformation. c. There exists a tensor whose covariant...

I should appreciate it if you could teach me how to verify it.

Thank you, @Demystifier. In general, the difference of two tensors is again a tensor. However, in the context of variations of the metric tensor, the quantities $$ \bar{g}_{\mu\nu}-g_{\mu\nu}$$ $$ \bar{g}^{\mu\nu}-g^{\mu\nu}$$ do not transform as tensors. I have verified this.

You forgot to put double dollar signs at head and tail of the equalion lines. I put them. ------------ Has any one any idea to solve this equation ##J[1, x] = (x^2/10)*(J[1, x] + J[3, x])##, in which J are spherical Bessel function normally write as ##j_1 (x)## and ##j_3(x)## Methods 1 serial...

It is a tensor whose covariant and contravariant components are ##\delta g_{\mu\nu}## and ##-\delta g^{\mu\nu}##, respectively. Returning to the original question, does there exist such a quantity whose covariant and contravariant components are ##\delta g_{\mu\nu}## and ##\delta g^{\mu\nu}## ...

1. We can derive the equation $$ \delta g^{\mu\nu}=-g^{\mu\alpha}g^{\nu\beta}\delta g_{\alpha\beta} $$ by indeces up down operation. In the world whose metric tensor is g, $$ g^{\mu\nu}=g^{\mu\alpha}g^{\nu\beta}g_{\alpha\beta} $$ In another new world whose metric tensor is ##\bar{g}##, $$...

Now I know that $$ (\delta g)^{\mu\nu}=g^{\mu\alpha}g^{\nu\beta}(\delta g)_{\alpha\beta} $$ $$ \delta g^{\mu\nu}=-g^{\mu\alpha}g^{\nu\beta}\delta g_{\alpha\beta} $$ Here I have a question. Is $$(\delta g)_{\alpha\beta}=\delta g_{\alpha\beta}$$ $$(\delta g)^{\alpha\beta}=-\delta...

@PAllen @Demystifier Thanks. I went a bit easy on T .

To put it simply $$ \bar{\delta g}=-\delta \bar{g} $$ May I understand that ##\{\delta g, \bar{\delta g}\}=\{\delta g, -\delta \bar{g} \}## are tensor indeces up-down pair but ##\{\delta g, \delta \bar{g}\}## are not that pair which does not follow ordinary indeces up-down transformation but...

I have read the matrix representation in Landau and Lifshitz.

Yes. No charge, no current.

Yes, loop integral of electric field is generated with no regard to the wire. Free electrons in the conductor wire on the loop are required for generation of the force which are charge times electric field.

We are sharing the situation. I do not find any essential difference in the drawings.