Recent content by anuttarasammyak

I see. “The moving inertial frame against the original IFR” would eliminate those cases.

I do not think so. In the text, "(absolute) velocity of the moving frame" is better.

An example of my point. Say light is passing on element ##dx^\mu##, i.e., $$ g_{\mu\nu}dx^\mu dx^\nu = 0 $$ After variation of metric $$ \bar{g}_{\mu\nu}dx^\mu dx^\nu \neq 0 $$ The light changes its course and ##dx^\mu## means something different from the light path. ##dx^\mu## is a 4-vector in...

Yes. For $$g_{\mu\nu}= g_{\mu\alpha} g_{\beta\nu} g^{\alpha\beta},\ g^{\mu\alpha} g_{\alpha\nu}=\delta^\mu_\nu $$ symbolically I write $$ 1=1*1*1,\ 1*1=1 $$ and for all barred things $$ 1+\epsilon=(1+\epsilon)(1+\epsilon)(1-\epsilon),\ (1+\epsilon)*(1-\epsilon)=1 $$...

$$\bar{g}_{\mu\nu} - g_{\mu\nu} = \delta g _{\mu\nu} = - \delta \bar{g}_{\mu\nu} $$ ##\bar{g}_{\mu\nu}## is a tensor in the world where metric is ##\bar{g}_{\mu\nu}## itself. ##g_{\mu\nu}## is a tensor in the world where metric is ##g_{\mu\nu}## itself. Their difference ##\delta g _{\mu\nu} =...

I agree that index raise- lower operation should be done using one same, old or new, metric for all the indexes of all the entities in equations.

My bad. I withdraw it.

To me the point is two metrics: original and varied. In varied metric world what kind of things or relations remain and what does not ?

Thanks. I will restate it to confirm my understanding. Here I try not to refer controversial (at least to me) tensor and concentrate on what I found in making use of this example. The metric $$ \bar{g}_{\mu\nu}=mg_{\mu\nu} $$ With $$ \bar{g}_{\mu\alpha} \bar{g}^{\alpha\nu}...

First of all, thanks @JimWhoKnew So in general cases which could include ##\delta g^{\mu\nu}##, do we need further information or condition to say it is a tensor or not, though the answer of #32 was affirmative ? I agree that the sum of tensors is a tensor. And sum of no tensors could be...

I am sorry I don't understand "the same signature". If A is g itself, I think that B = A = g. A=mg, B=1/m g where m is a real number ##\neq 0## satisfy the relation. I am sorry not to be smart enough to relate A ##\neq## B with the discussion.

Going back to basics, Dirac's textbook defines a tensor as follows: $$ T^{\alpha'\beta'}_{\gamma'}=x^{\alpha'}_{,\lambda}x^{\beta'}_{,\mu}x^{\nu}_{,\gamma'}T^{\lambda\mu}_{\nu} \ \ \ (3.6)$$ How exactly should I perform coordinate transformations to investigate the quantity I'm considering? Or...

I will summarize what I have written. The index-raising and -lowering relation for the metric tensor $$ g_{\mu\nu}=g_{\mu\alpha}g_{\nu\beta}g^{\alpha\beta} ,$$ holds for a varied metric tensor as $$\bar{g}_{\mu\nu}=\bar{g}_{\mu\alpha}\bar{g}_{\nu\beta}\bar{g}^{\alpha\beta} $$ where $$...

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 ?