The h Bar

12:54 AM

@JohnRennie Hello, John, if you could spare a little time, could you take a brief look at this question? physics.stackexchange.com/questions/519489/…

Thank you very much!!

Sir Cumference

4:35 AM

Physics Meta

4:51 AM

-2

Q: New flag suggestion

More often than not the correct answer here is "Have you considered learning philosophy or art instead?" Now this is kind of blunt but i stress that it correct answer most of the time.

feature-request

John Rennie

6:58 AM

@frt132 I didn't understand what you are asking in that question. I'll be in the chat room for a few hours this morning if you want to discuss it here.

Aaron Stevens

8:14 AM

abc.net.au/news/science/2019-12-13/…

I agree with the mission to teach kids science, but I don't think their motivation is the best.

skullpetrol

8:32 AM

tsk tsk

dsm

8:57 AM

@JohnRennie as I recall, you're quite the GR fan -- chance you can help me understand some formality in that problem? I feel like it's very simple, but I'm fundamentally not understanding something. if I just follow definitions, it seems $\tilde{g}^{\mu\nu} \equiv \tilde{g}(e^\mu,e^\nu) \equiv g(e_\mu,e_\nu) = g_{\mu\nu}$. I don't know.

John Rennie

@dsm the problem is that I never formally learned differential geometry. I just picked up the bits and pieces I needed as I went along. So when people ask me questions about diff geo I generally have to admit I have no idea what the answer is. Sorry :-(

dsm

no worries, thanks for the response :) happy holidays

ACuriousMind

10:11 AM

@dsm Things with different free index positions can never be equal! You should have $\tilde{g}(e^\mu, e^\nu) = g(g^{\mu\rho}e_\rho, g^{\nu\sigma}e_\sigma) = g^{\mu\rho}g^{\nu\sigma}g(e_\rho, e_\sigma)$

dsm

10:38 AM

@ACuriousMind Hi there, thanks for the input. Sorry, still confused. In his definition of the non-degen bilinear form (ndbf) $\tilde{g}$ on $V^*$, it's equal to the ndbf $g$ on $V$ but with the original dual arguments $f,g$ now their vectors $\tilde{f},\tilde{g}$. Aren't your arguments of $g$ in that second equality still the dual vectors $e^\mu$ and $e^\nu$?

ACuriousMind

@dsm No - the $g^{\mu\rho}$ are numbers and the $e_\rho$ are vectors, so $g^{\mu\rho}e_\rho$ is a linear combination of vectors (the image of $e^\mu$ under duality), not a dual vector.

If your basis is not orthonormal, it is not necessarily true that the dual of $e^\mu$ is $e_\mu$.

dsm

Ahhh, I think I see what you're saying. Can you clarify "the image of $e^\mu$ under duality"? The image of the dual under duality? Does that mean it's $e^\mu$ represented as a vector? e.g. for finite dimensional, transposing the row dual into a column, hence living in the vector space?

Nevermind, that makes perfect sense. Thank you greatly

ACuriousMind

11:03 AM

You're welcome :)

Slereah

Also hello

dsm

@ACuriousMind Before I go, I'm also confused by one more thing on the same topic. I want to show that for a metric $g$ on $V$, the $(1,1)$ tensor associated to $g$ is the identity operator; i.e. $g_{i}$$^j$$ = \delta_i$$^j$. Following in your footsteps...

$g_{i}$$^j = g(e_i,e^j) = g(g_{i\mu}e^\mu,g^{j\nu}e_\nu) = g_{i\mu}g^{j\nu}g^\mu$$_\nu$

But then I'm stuck. Is there an identity I'm not seeing?

ACuriousMind

You have defined $g^{\mu\nu}$ above as the inverse of $g_{\mu\nu}$. What does that mean for $g_{\mu\nu}g^{\nu\rho}$?

dsm

that would be equal to $\delta_\mu$$^\rho$, right?

ACuriousMind

Indeed

dsm

11:18 AM

does that help me with $g_{i\mu}g^{j\nu}g^\mu$$_\nu$? the mixed indices are throwing me off

or perhaps my approach to showing that about the $(1,1)$ tensor of $g$ is not the way to go

ACuriousMind

@dsm Don't take the dual of both arguments of the mixed $g$

Just take the dual of one of them.

dsm

Ahh there it is, thanks! Slowly getting the hang of this -- should have done this self-study a long time ago. I'm off to dinner, cheers :)

Slereah

12:02 PM

Ryan hasn't been 'round in a while

I think I need some alternate strategy to get that thesis

I guess maybe I can try asking the librarians again

Explaining the situation better

dsm

1:12 PM

@ACuriousMind I confused myself thinking about that second result during dinner. The metric $g$ on $V$ is the mapping $g: V\times V \rightarrow \mathbb{R}$. With that, why does it make sense to feed in a vector and a dual into that metric in order to get the $(1,1)$ tensor $g_i$$^j$? I'm confused why $g(e_i,e^j)$ is a valid expression, appearing to be mapping $g:V\times V^*\rightarrow \mathbb{R}$.

Fine with the first question you helped me out on, as we had properly defined that $\tilde{g}$ metric on $V^*$.

dsm

1:27 PM

I want to justify it to myself by saying well, the second argument is rewritten as $e^j = g^{j\mu}e_\mu$, so all is well and we're feeding in vectors. But I guess that's the crux of my confusion. Writing that dual as the sum of vectors, it's still a dual, yes?

When I have some dual vector $f$ and I want to express the components of it in terms of its vector $\tilde{f}$, it's easy to show $f_\mu = g_{\mu\nu}\tilde{f}^\nu\equiv g_{\mu\nu}f^\nu$, but that's for the components of the dual. Am I making sense? New in this area of things.

ACuriousMind

@dsm If you wanted to be fully consistent, you would have to use a different symbol for the (1,1)-tensor, just like you did for $\tilde{g}$.

dsm

Ah, ok. For example, defining a metric $\bar{g}: V\times V^* \rightarrow \mathbb{R}$ to be $\bar{g}(\tilde{f},h) = g(\tilde{f},\tilde{h})$?

ACuriousMind

yep

dsm

Ahhh, I see. Thank you kind sir

Physics Meta

2:08 PM

0

Q: Should I edit closed or duplicate, but recently active questions?

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Artificial Intelligence

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