The h Bar

12:22 AM

@tpg2114 Well if you know how to program, you can do anything, right?

12:47 AM

@dmckee I look forward to the day when that ancient formula, which you saw through immediately, has no purchase anywhere. Note the irony of the "time is priceless" phrase.

Stan Shunpike

2:21 AM

What is a good metric to begin with for calculating curvature? I am using these formulas

$$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho{}_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma}-\Gamma^\rho{}_{\nu\lambda‌}\Gamma^\lambda{}_{\mu\sigma}$$

$$\Gamma^l{}_{jk} = \tfrac{1}{2} g^{lr} \left \{\partial _k g_{rj} + \partial _j g_{rk} - \partial _r g_{jk} \right \}$$

According to Wald, these should be sufficient to calculate the curvature given a metric. I would like to calculate some metric that models actual curvature. I was thinking Schwarschild, but Im not really sure how to approach this. Wald makes it sound like "plug and play"

bolbteppa

@StanShunpike cartesian coordinates, polar coordinates, spherical coordinates, cylindrical coordinates, ellipsoidal coordinates, Schwarzschild metric,

Alfred Centauri

2:50 AM

@StanShunpike it's a subtle thing, of course, but there's a distinction between the geometry (metric) and line element (of the metric on some coordinate system). For example, "Schwarzschild" might mean metric (solution, geometry) or coordinates which are distinctly different. The Schwarzschild metric (geometry) can be expressed in Schwarzschild coordinates or, e.g., Kruskal–Szekeres coordinates. The geometry (solution) is invariant and independent of the coordinates.

Stan Shunpike

@AlfredCentauri I meant metric. But an interesting point.

Stan Shunpike

3:25 AM

Okay, once I calculate the Christoffel symbols, I need to plug them into the equation. But for $(t,r,\theta,\phi$, there is one 4x4 matrix each upper index of the Christoffel symbols. And then, for instance, I have to take the partial derivatives of those terms....so doesnt that mean...something like...4*4*16 = 256 versions of the equatiosn!?!?!?!?!?

Alfred Centauri

3:58 AM

@StanShunpike There are just 20 independent quantities due to the symmetries of the curvature tensor. 10 of the DOF are in the Einstein tensor - the remaining are in the Weyl tensor.

Stan Shunpike

4:18 AM

@AlfredCentauri so out of 256, 20 are independent?

I'm not sure where i got 256 from. MTW mention that number in their book

David Z

@Jimnosperm yeah, but that's my point. If a question says "Do [X] for me" it's off topic. So I think the corresponding "Point me to a book that shows how to do [X]" should also be off topic.

Stan Shunpike

Once I have the Christoffel symbols and I plug this into the curvature tensor formula, what does this give me? Obviously it gives me the curvature tensor. But I mean, what do I then use it for?

Like I don't really get how this tells me anything about curvature yet. This is just a tensor. How does this tell me about the difference between an initial vector and a vector parallel transported around a patch on my manifold?

Do I have to plug stuff into the tensor to measure curvature?

Stan Shunpike

5:45 AM

@0celo7 ^ any words of wisdom on this?

@DavidZ Most homework questions I have seen seem like they could be thought through better and rephrased to be conceptual questions. Is that your impression?

David Z

6:36 AM

Yeah, most of them could, it's just a matter of how much of a change would be required

Even the questions that copy and paste a homework question and show no effort whatsoever could be rephrased into conceptual questions. History suggests that's unlikely to happen though.

Stan Shunpike

Precisely. I think it's tricky, but there really is no value in just asking someone else to do your work. I mean, that's not a site I would find interesting. The word parasitic seems strong but it wouldn't be a good use of ppls time.

Like i mean, i don't know if I'd call having someone else doing your homework parasitic....but it kinda reminds me of it because its just one person taking

David Z

@Jimnosperm just looked over this and my thoughts are: (1) When in doubt, just flag it. Worst case is that your flag gets declined, which nobody really cares about. (2) I think the answer is arguing that cables transmit signals via free-space EM waves, not electronic oscillations, by acting as a waveguide. Which is wrong (right?), and not well enough explained to be useful, but I can see how it qualifies as an answer.

I first saw this problem on the Google Labs Aptitude Test. A professor and I filled a blackboard without getting anywhere. Have fun.

Stan Shunpike

@DavidZ very very true lol

@Danu if you scroll a few lines up, I had a question about the curvature tensor. Ping me if you got any thoughts about it.

David Z

6:55 AM

@StanShunpike I think Wikipedia seems to have an explanation

Physics Meta

0

Q: Why should one answer a no-answer question?

There are over 6,500 questions with no answer here, and over 8,000 practically unanswered. I tried one, but I have seen that nobody even takes a look at old questions. Do you know why? How does the date influence the usefulness of a question/answer? You have no interest that most questions be pr...

discussion

Stan Shunpike

7:09 AM

@DavidZ So the idea is that, by taking the limit as the area of the patch goes to zero, the limit is the Riemann curvature tensor. And so if we are given a vector field $Z$, then $R(X,Y)Z$ yields how much Z has changed by? (To be clear, I have read that section 10x but I didn't know what it said)

David Z

You don't need a vector field, just a manifold, and one vector $Z$ at a particular point in the manifold.

Stan Shunpike

And when I plug that into the Riemann tensor, that yields the curvature?

David Z

If you parallel-transport $Z$ around a small rectangle, then when you get back to the original point, it won't be pointing in the same direction. The Riemann tensor tells you how the direction changes, in the limit as the size of the rectangle goes to zero.

@StanShunpike well curvature is going to be some contraction of the Riemann tensor.

Or you could consider the tensor itself the curvature, in a slightly different sense. The word "curvature" is not really a specific technical term.

There are different kinds of curvature.

Curvature of Riemannian manifolds

In mathematics, specifically differential geometry, the infinitesimal geometry of Riemannian manifolds with dimension at least 3 is too complicated to be described by a single number at a given point. Riemann introduced an abstract and rigorous way to define it, now known as the curvature tensor. Similar notions have found applications everywhere in differential geometry. For a more elementary discussion see the article on curvature which discusses the curvature of curves and surfaces in 2 and 3 dimensions, as well as Differential geometry of surfaces. The curvature of a pseudo-Riemannian manifold...

Stan Shunpike

@DavidZ aren't we just talking about intrinsic curvature? I didn't know there were multiple kinds of intrinsic curvature. I have heard of Gaussian curvature, sectional curvature, and scalar curvature.

Is curvature an umbrella term?

David Z

There's intrinstic and extrinsic, of course, but that's not what I mean. Yes, I'm talking about different kinds of intrinsic curvature. It is an umbrella term, as far as I know.

There is not one thing that everybody means when they say "curvature"

Stan Shunpike

7:19 AM

Damn, that's confusing. So the Riemann tensor itself doesn't yield any information about curvature but requires contractions? If we contract once, we get Ricci curvature and if we contract twice, we get scalar curvature, right? So are those the types of curvature physicists mean when they say curvature?

David Z

Well, if you mean the Riemann tensor when you say "curvature" then it does yield information about curvature ;-)

Stan Shunpike

LOL wait, I thought u just said "curvature is going to be some contraction of the Riemann tensor"

So that implies its doesn't yield info about curvature, right?

Not without doing something further to it

David Z

Well, that was when I thought you were talking about one of the simpler kinds of curvature

but in general, any of these things can be considered curvature

The question of whether the Riemann tensor yields information about curvature is meaningless unless you are precise about what you mean by "curvature"

Stan Shunpike

Well, I am considering the Schwarzchild metric. What kinds of curvature does that mean I am interested in?

David Z

I dunno, it depends on what you want to do with it/them

Stan Shunpike

7:29 AM

How about approximating the different gravitational fields of objects in our solar system? For instance, the curvature around the sun.

What kind of curvature does that involve?

David Z

"curvature around the sun" still doesn't specify anything

But I guess you're looking for a curvature that relates to gravitational potential or gravitational force?

That I don't know offhand

Stan Shunpike

Okay. I was very stuck and not making much progress until I found en.m.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution which allowed me to understand physicspages.com/2013/12/25/…

Now that I understand how to compute the Christoffel symbols and plug them in, I need to figure out what doing all that accomplishes. You helped me see part of that since now I understand applying the curvature tensor to $Z$ gives me the difference between the original and parallel transported vectors.

But I now need to understand exactly what kinds of curvature this means I am measuring and what kinds of physical information this gives me about gravitational systems that fit the Schwarzchild metric.

David Z

Yeah, as far as I know it really depends on your application.

Someone who works with GR could tell you more.

Stan Shunpike

By application, do you mean the particular physical characteristics of the system I am studying?

David Z

No, I mean what you're trying to do. What you're applying the Riemann tensor (and Christoffel symbols and metric) to.

Stan Shunpike

7:42 AM

Isn't this all just to calculate the gravitational field? What else could I do or be interested in?

David Z

8:06 AM

@StanShunpike Gravitational waves, time dilation, black hole entropy and firewalls, inflation, cosmological expansion, wormholes, etc. etc. etc.

Geodesy

Geodesy (/dʒiːˈɒdɨsi/), — also known as geodetics or geodetics engineering — a branch of applied mathematics and earth sciences, is the scientific discipline that deals with the measurement and representation of the Earth, including its gravitational field, in a three-dimensional time-varying space. Geodesists also study geodynamical phenomena such as crustal motion, tides, and polar motion. For this they design global and national control networks, using space and terrestrial techniques while relying on datums and coordinate systems. == Definition == Geodesy — from the Greek word γεωδαισία or...

If it were only about calculating gravitational fields, the study of gravity would have ended 80 years ago. But in reality, it's a huge research area.

Oh, and then there's also quantum field theory in curved spacetime.

Danu

2:02 PM

Could this be a typo?

I'd expect $[(x_1,x_2)-(y_1,y_2)]+i[(x_1,y_2)+(y_1,x_2)]$

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