/sci/ - Science & Math

[Anonymous No.5689366]

How do /sci/ remember gradient, divergenci, rotational... in other coordinates, like polar of spherical?

[Anonymous No.5689503]

bump

[Anonymous No.5689518]

Vector calculus laws (i.e. specializations of Stokes' theorem) applied to little boxes with constant-coordinate surfaces as faces.

[Anonymous No.5689519]

I don't, I prove them every time.

[Anonymous No.5689534]

Look them up

[Anonymous No.5689558]

So it seems that nobody knows how...

[Anonymous No.5689728]

>>5689366
I do not believe that you have to look for this everytime you need to use them, do you always recall it?
Or you can deduce it fast?

Like... how is the laplacian in polar coordinates

[Anonymous No.5690637]

bump

[metric !1tensorX8Q No.5690648]

I don't. Mostly I work in generalized coordinates anyway. 99 times out of 100 if I'm taking a gradient I'll use <span class="math">\partial_i[/spoiler], where i runs from 1-3, each representing a different coordinate in my coordinate system.

[Anonymous No.5690704]

Divergence, for example, is
<div class="math">\lim_{\rm volume \to 0}
\frac{\rm flux~of~\vec{V}~out~of~volume}{\rm volume}</div>so for the small volume between r and <span class="math">r+\Delta r[/spoiler] etc., the divergence is approximately
<div class="math">\frac{
\Delta_r (V_r (r \Delta \theta) (r \sin \theta \Delta \phi))
+ \Delta_\theta (V_\theta (\Delta r)(r \sin \theta \Delta \phi))
+ \Delta_\phi (V_\phi (\Delta r) (r \Delta \theta))
}{
(\Delta r)(r \Delta\theta)(r \sin \theta \Delta\phi)
}</div>where I'm using <span class="math">\Delta_r[/spoiler] etc. to indicate the difference is taken between equivalent points on the constant-r faces.

So pulling out the things that are constant, we end up with
<div class="math">
\frac{1}{r^2} \frac{\partial (r^2 V_r)}{\partial r}
+ \frac{1}{r \sin \theta} \frac{\partial (\sin \theta V_\theta)}{\partial \theta}
+ \frac{1}{r \sin \theta} \frac{\partial V_\phi}{\partial \phi}.
</div>

[Anonymous No.5690753]

>>5690704
Thats what I was looking for, but I do not understand how does your operator
\Delta_r
behave... could you explain it a little more detailed pls, or write a few steps of how do you end with the final formula?

[Anonymous No.5690758]

>>5690753
It's just the difference in the quantity between the opposite faces.

[Anonymous No.5690762]

>>5690758
ah, ok

[Anonymous No.5690765]

>>5690758
For example, the flux out through the face at <span class="math">r + \Delta r[/spoiler] would be
<div class="math">V_r (r \Delta \theta) (r \sin \theta \Delta \phi);</div>subtract the flux in through the opposite face, and you get the
<div class="math">\Delta_r (V_r (r \Delta \theta) (r \sin \theta \Delta \phi)).</div>

[Anonymous No.5690774]

>>5690758
I got it, but I still do not know
[latex]\Delta_r (V_r (r \Delta \theta) (r \sin \theta \Delta \phi))=
\frac{\partial V_r }{\partial r}r^2(\Delta \theta) (\sin \theta \Delta \phi) + V_r\frac{\partial r^2}{\partial r}(\Delta \theta) (\sin \theta \Delta \phi)[/latex]

[Anonymous No.5690775]

>>5690774
But this term doesnt appear in the final equation,

<span class="math"> V_r\frac{\partial r^2}{\partial r}(\Delta \theta) (\sin \theta \Delta \phi)[/spoiler]

[Anonymous No.5690776]

You have to find the scale factors (called h1, h2, h3) and use the scale factors to define new unit basis vectors (called e1, e2, e3 and in cartesian these are i, j, k) in whatever coordinate system you're working in. Since you mentioned only orthogonal coordinate systems (e1 x e2 = e3, much like how i x j = k) I'm going to assume that's all you need.

Wikipedia has insanely detailed documentation on this. As an example, I'll use cylindrical coordinates. There are three steps (p.s. hope you know suffix notation):

1) Find the scale factors for cylindrical coordinates. The equation for this is given by:

<span class="math">h_i=\sqrt{\left(\frac{\partial x_j}{\partial q_i}\right )^2}[/spoiler]

Where x_j is each coordinate in cartesian (x1, x2, x3) and q_i is each coordinate in cylindrical (rho, phi, x3). Clearly, the scale factor for the third coordinate is going to be 1; there is no reason to scale the same exact coordinate.

After you've completed this, you get 1, rho, and 1.

2) Find the unit basis vectors rho-hat, phi-hat, and x3-hat.

x3-hat is simply k-hat; again, no change. To find rho-hat and phi-hat:

<span class="math">e_i=\frac{\partial x_j}{\partial q_i}[/spoiler]

where e_i is the tangent vector of each coordinate in cylindrical. We want the unit vector for this so we'd divide by the magnitude of this vector and... holy shit, the scale factor is the magnitude of the tangent vector! Therefore:

<span class="math">\hat{e_i}=\frac{e_i}{h_i}[/spoiler]

This should come out to be http://en.wikipedia.org/wiki/Unit_vector#Cylindrical_coordinates (too lazy to type in latex)

3) Fortunately, the del operator is defined for cylindrical. Can you find it yourself based on the information above? Of course you can, try it out. Good luck.

[Anonymous No.5690783]

>>5690704
man this is genius! no more jacobian and shitty ways to remember!,
so fast...

[Anonymous No.5690801]

>>5690704
I've succesfully done the gradient (really, really easy if its done this way), but how whould you do the rotational and laplacian?

[metric !1tensorX8Q No.5690814]

>>5690783
The Jacobian is extremely easy to remember.

<div class="math">J^{i}_{~\bar{j}}=\partial_{\bar{j}} x^{i}</div>
The determinant relating the new coordinates to Cartesian coordinates gives the element volume:

<div class="math">dxdydz=\mathrm{det}(J^{i}_{~\bar{j}})dx^1 dx^2 dx^3</div>

[Anonymous No.5690820]

>>5690814
Yes, but it isnt usefull, you have to memorize the formulas for gradient, divergency etc... or look them up, because I think they are not as trivial as this geometric way to compute them... am I wrong?

[metric !1tensorX8Q No.5690826]

>>5690820
What, exactly, is there to memorize? It's about as simple an equation as I can think of.

[Anonymous No.5690835]

>>5690826
I'm not talking about the jacobian, i mean for expample, how would you write the divergency in spherical coordinates. Until today >>5690704
I had to look the formula up, how do you deal with that?

[Anonymous No.5690841]

>>5690801
For curl, you would look at the circulation around a small loop. For the Laplacian, once you have the divergence and gradient operators, you can work out the divergence of the gradient.

[Anonymous No.5690855]

>>5690841
I tried, but I not sure how should I write the curl. For example in spherical coordinates in the r direction

<span class="math">\frac{1}{(r\Delta \theta)(r \sin \theta \Delta \phi)}[\Delta_\phi(U_\phi r \sin \theta \Delta \phi) - \Delta_{\phi - \Delta \phi}(U_\phi r \sin \theta \Delta \phi) + \Delta_\theta(U_\theta r \Delta \theta) - \Delta_{\theta - \Delta \theta}(U_\theta r \Delta \theta][/spoiler]

Its close... but this may be wrong:
<span class="math">\Delta_{\phi + \Delta \phi}[/spoiler]

[metric !1tensorX8Q No.5690863]

>>5690835
In generalized coordinates the divergence of a vector <span class="math">v^i[/spoiler] is just <span class="math">\partial_i v^i=\frac{\partial v^i}{\partial x^i}[/spoiler], using the summation convention.

The "difficulty" comes from trying to express the gradient operator as a vector, rather than a one-form (i.e. upper index instead of a lower index). Then you have to invoke the metric tensor, complicating things. The metric in spherical coordinates is:

<div class="math">g_{ij}=\mathrm{diag}(1,r^2,r^2sin^2\theta )</div>
The inverse of which is:

<div class="math">g^{ij}=\mathrm{diag} (1,r^{-2},r^{-2}sin^{-2}\theta )</div>
The "vector divergence" is then just:

<div class="math">g^{ij}\partial_i v^j
In Cartesian coordinates vectors and one-forms are equivalent. But when you move to curvilinear coordinates or curved manifolds, they become separate entities related by the metric. The difficulty comes from trying to express everything as a vector instead of its natural formulation, a combination of vectors and one-forms.</div>

[Anonymous No.5690869]

>>5690863
but the you have to obtain the metric tensor and multiply them, isnt it quicker the geometric way if I'm working with vectors and not generalized coordinates?

[metric !1tensorX8Q No.5690870]

>>5690835
In generalized coordinates the divergence of a vector <span class="math">v^i[/spoiler] is just <span class="math">\partial_i v^i=\frac{\partial v^i}{\partial x^i}[/spoiler], using the summation convention.

The "difficulty" comes from trying to express the gradient operator as a vector, rather than a one-form (i.e. upper index instead of a lower index). Then you have to invoke the metric tensor, which complicates things. The metric in spherical coordinates is:

<div class="math">g_{ij}=\mathrm{diag}(1,r^2,r^2sin^2 \theta )</div>
The inverse of which is:

<div class="math">g^{ij}=\mathrm{diag} (1,r^{-2},r^{-2}sin^{-2}\theta )</div>
The "vector divergence" is then just:

<div class="math">g^{ij}\partial_i v^j</div>
In Cartesian coordinates vectors and one-forms are equivalent. But when you move to curvilinear coordinates or curved manifolds, they become separate entities related by the metric. The difficulty comes from trying to express everything as a vector instead of its natural formulation, a combination of vectors and one-forms.

[Anonymous No.5690871]

>>5690869
sorry It was stupid what I said, Its been I while since I studied generalized coordinates

[metric !1tensorX8Q No.5690873]

>>5690869
Not really. The metric of the most commonly used coordinate systems (Cartesian, spherical, cylindrical, Schwarzschild, etc.) are very easy to memorize, and deriving them isn't too hard either.

[Anonymous No.5690874]

>>5690870
Do you remember the other operators? divergency its easy... I dont recall the others...

[Anonymous No.5690877]

>>5690841
Could you give me an example for the curl?

[metric !1tensorX8Q No.5690896]

>>5690874
The Laplacian, for example, (i.e. <span class="math">\nabla^2[/spoiler] is just <span class="math">\nabla^2=\partial_i \partial^i=g^{ij} \partial_i \partial_j[/spoiler]. This becomes a very very useful notation when moving into Special Relativity, because the wave-operator (a Lorentz-invariant operator) is just:

<div class="math">\square=-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}+\nabla^2=\partial_\mu \partial^\mu = g^{\mu \nu} \partial_{\mu} \partial_{\nu}</div>
Curl, on the other hand, is a bit more complicated. The curl of two vectors is:

<div class="math">(v \times u)^i =\epsilon^{i}_{~jk} v^j u^k</div>
where <span class="math">\epsilon[/spoiler] is the Levi-Civita tensor, which you can find more information on at the wiki page: http://en.wikipedia.org/wiki/Levi-Civita_symbol

The Levi-Cevita tensor also generalizes to higher dimensions, so a "cross" product can be defined in four or higher dimensions.

[metric !1tensorX8Q No.5690899]

>>5690896
Fuck this site's terrible TeX. Should say:

<div class="math">-\frac{1}{c^2}\frac{\partia l^2}{\partial t^2}+\nabla^2=\partial_\mu \partial^\mu = g^{\mu \nu} \partial_{\mu} \partial_{\nu}</div>

[metric !1tensorX8Q No.5690901]

>>5690896
Fuck this site's terrible TeX. Should say:

<div class="math">-\frac{1}{c^2} \frac{\partial^2}{\partial t^2} + \nabla^2 = \partial_{\mu} \partial^{\mu} = g^{\mu \nu} \partial_{\mu} \partial_{\nu}</div>

[Anonymous No.5690904]

Fucking physicists on my /sci/.

[Anonymous No.5690906]

>>5690904
Excuse me sir?

[Anonymous No.5690911]

ITT: plebs not using combinations of the exterior derivative, hodge dual, and musical isomorphisms to easily find their vector operators in any coordinate system

[Anonymous No.5690912]

>>5690906
Mathematicians do not specify coordinates.

[Anonymous No.5690915]

>>5690912
and?

[Anonymous No.5690918]

>>5690915
So get out of my /sci/.

[metric !1tensorX8Q No.5690920]

>>5690912
What does that have to do with this thread? Also, I think you're forgetting that index-notation proved to be far more useful in the long run than index-free notation. "Abstract index-notation" was sort of a concession to this, even though the indices in relativity were always considered abstract to be begin with.

[Anonymous No.5690922]

>>5690918
LOL BUTTHURT MATHEMATICIAN DOESNT LIKE HOW PHYSICISTS HAVE A USEFUL NOTATION THAT ALLOWS FOR ACTUAL USE OF MATH IN THINGS THAT ARE COOL

"you know what sounds like a good idea? proving 10000 obvious things about open sets"- mathematician

"you know what sounds like a good idea? using math to describe theories about THE FUNDAMENTAL WORKINGS OF THE UNIVERSE"- physicist

[Anonymous No.5690924]

>>5690918
did you notice it's /sci/ and not /mat/

[Anonymous No.5690926]

>>5690924
good point

[Anonymous No.5690942]

>>5690855
Can somebody help with this?

[Anonymous No.5690949]

Sorry for the stupid question, but I felt like it didn't deserve it's own thread. Anyway:

Can someone explain spherical triangles to me? Why isn't the sum of angles 180 and how do you solve for anything then?

[Anonymous No.5691182]

>>5690877
I finally got it man,
(spherical, r direction)
<span class="math">\frac{1}{r \sin \theta \Delta \theta \Delta \phi}[\Delta_\theta (r \sin \theta \Delta \phi) V_\phi - \Delta_\phi (\Delta \theta) V_\theta ][/spoiler]

This will save me so much time until I master generalized coordinates. Its so easy to visualize this and just write the result

[Anonymous No.5691184]

>>5691182

<span class="math">\displaystyle \frac{1}{r^2 \sin \theta \Delta \theta \Delta \phi}[\Delta_\theta (r \sin \theta \Delta \phi) V_\phi - \Delta_\phi (\Delta \theta) V_\theta ][/spoiler]

[Anonymous No.5691253]

i would just like to say fuck the person who decided to swap the meanings of theta and phi in physics. asshole. nskes no goddamn sense.

[Anonymous No.5691255]

>>5691253
Yes. Taking physics and maths is tricky sometimes.