Succumbing to LaTeX

Update: The post below was written back when CV was on its own. Here on the new Discover site, the way to put something into Latex is to start with

$latex

and end with a simple

This stands in marked contrast with the previous system, explained below.

——————————————————-

For a long time I was reluctant to joint the many other sciencey blogs that had integrated equations by providing support for LaTeX, the technical typesetting system that nearly every physicist and mathematician uses. Possible reasons for this attitude include:

We felt it was important to remain accessible to a wide range of readership, and feared that the appearance of equations would put people off (and tempt us into being unnecessarily technical).
It sounded like work.

You can decide for yourself which is more true. The good thing is, there is no wrong answer!

But right now I am uninspired to blog because my brain is preoccupied with real science stuff. So I thought of posting about some of the fun ideas in quantum mechanics I’ve been learning about. But there’s really no way to do it without equations. So for that reason, and in belated honor of Donald Knuth’s birthday, I went and installed the LatexRenderer plugin. (Amazingly, InMotion Hosting already had LaTeX installed on our server. Yay for them!)

So now it’s easy to include equations; they should even be available in comments. All you have to do is type [tex], then your LaTeX commands, then [/tex]. So for example

[tex]R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi G T_{\mu\nu}[/tex]

should produce

$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi G T_{\mu\nu}$ .

There are a million online tutorials; try this list of commands to get you started. Use comments to this post to try it out. (Sadly, no preview, so be careful, and this post will remain open for playing around.) One thing I’ve noticed: don’t use linebreaks within the formulas, just put everything on the same line. And use “\displaystyle” if you want the look of a set-off (rather than in-line) equation.

But now I should get back to work. So to keep you thinking, here are a couple of equations from the stuff I’m thinking about and hopefully will explain soon:

$\displaystyle \langle\langle \hat{\mathcal O}\rangle\rangle =\lim_{t \rightarrow \infty}\frac{1}{t}\int^t_0 \langle \psi_s|\hat{\mathcal O}|\psi_s\rangle ds = \rm{tr}(\hat\rho \hat{\mathcal O})\,,$

$\displaystyle \hat\rho = \frac{1}{Z} \exp{\left(-\beta \hat{H} - \sum^n_{i=2} \mu_i \hat{F}_i\right)}\,.$

Kind of beautiful, in an austere way, don’t you think?

January 22nd, 2008 8:27 PM
in Cosmic Variance | 200 comments | RSS feed | Trackback >

200 Responses to “Succumbing to LaTeX”

1. Sean Says:
January 22nd, 2008 at 8:28 pm

Okay, here goes nothing:
$\displaystyle \frac{d}{dx}e^x = e^x$ .
2. Sean Says:
January 22nd, 2008 at 8:29 pm

Success!
3. Sam Says:
January 22nd, 2008 at 8:47 pm

[ tex] e_i\pi=-1 [ /tex]
4. Big Vlad Says:
January 22nd, 2008 at 8:52 pm

$latex e^{\PI i} = -1 [\tex]
5. Sean Says:
January 22nd, 2008 at 8:52 pm

Shouldn’t have any spaces between the “[” and the “tex” or “\tex”.
6. Sean Says:
January 22nd, 2008 at 8:53 pm

And it’s \pi, not \PI.
7. Big Vlad Says:
January 22nd, 2008 at 8:53 pm
$e^{\PI i} = -1$
8. Julianne Says:
January 22nd, 2008 at 8:55 pm

This has to be the most fabulously geeky comment thread, EVAH!
9. Neil B. Says:
January 22nd, 2008 at 8:56 pm

Hi, I am still looking for good (and hopefully free) program to convert Word or maybe WPerfect to LaTex. The one I downloaded (Word2tex) seems unable to perform, likely because I just don’t get how to use all those wacky little files. I mean, I just want to rev up a simple click-to-start program (like any other Windows product) to take the Word doc and turn it into a LaTex document (and look at it to be sure it is right) without hassle, any help please? tx
10. Sean Says:
January 22nd, 2008 at 9:03 pm

Julianne, don’t you mean $\ \epsilon\nu\alpha\hbar$ ?
11. HB Says:
January 22nd, 2008 at 9:09 pm
$\zeta(1/2+\pi^{-e}+i 10^{500} )=0$
12. Bob Munck Says:
January 22nd, 2008 at 9:22 pm

LaTeX. That’s so 20th century. The modern, up-to-date way to support equations is MathML; it’s what all the kids are using these days.

To be honest, we tried for weeks to get MathML to work on one of the Space Elevator forums, and never succeeded. Your LaTeX renderer is somewhat of a kludge, but one can’t argue with the fact that it works.

Did you know that one of the programming languages used by NASA for the Space Shuttle software, HAL/S, supported multiple-line statements containing 2-D equations?
13. Christina Pikas Says:
January 22nd, 2008 at 9:31 pm

Not sure if it’s important, but it doesn’t seem to work in the feed. Reading the feed in bloglines, I just got the markup, not the equation.
14. Sam Gralla Says:
January 22nd, 2008 at 9:39 pm
$\nabla^c \nabla_c h_{ab} - 2 R^c{}_{ab}{}^d h_{cd} = - 16 \pi M \int_\gamma \delta_4(x,z(\tau))\,u^a(\tau)u^b(\tau)\,d\tau,$
$\nabla^b h_{ab}=O(\mu^2)$
15. Sam Gralla Says:
January 22nd, 2008 at 9:41 pm

Oh yeah, first try. I’m a winner.
16. Sean Says:
January 22nd, 2008 at 9:48 pm

Christina, really? It shows up okay for me, both in bloglines and in Google reader. Which feed are you using? Do you usually see images? (Because that’s all they are.) Do you have some weird background color?
17. tacitus Says:
January 22nd, 2008 at 10:00 pm

Kind of beautiful, in an austere way, don’t you think?

…and utter gobbledygook to physics and maths illiterates like me

Have fun.
18. Eugene Says:
January 22nd, 2008 at 10:01 pm

I see your density matrices and raise you a unitary operator, stuff that I am thinking about

$latex i\frac{\partial}{\partial t} U(t,t’) = H_I(t)U(t,t’)[\tex]
19. Freiddie Says:
January 22nd, 2008 at 10:05 pm

$latex \epsilon_123=1[\tex]
Does it work?
Does the first equation with the R_mu_nu in your blog post have anything to do with general relativity?
And I haven’t got a clue what those “austerely beautiful” equations are, or mean.
20. Freiddie Says:
January 22nd, 2008 at 10:06 pm

$latex \epsilon_{123}=1[\tex] Trying again
21. Jerry Boetje Says:
January 22nd, 2008 at 10:09 pm

Bob, I remember HAL from about an eon ago. Developed at Draper Labs in Cambridge and named for the gentleman who developed much of the Apollo navigation code. You could write a Kalman filter in 2 lines of code using the sub/superscript capability. Very cool. Thanks for reminding me of those days.
22. Sean Says:
January 22nd, 2008 at 10:11 pm

Forward slashes on the closing /tex, guys.

Precision counts!
$\epsilon_{123}=1$
And yes, the first example is Einstein’s equation of general relativity.
23. Helge Says:
January 22nd, 2008 at 10:12 pm

$latex det\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = -1 {\tex]
sucks.
24. Helge Says:
January 22nd, 2008 at 10:13 pm

$latex det\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = -1 [\tex]
sucks.
25. Jim Says:
January 22nd, 2008 at 10:15 pm

Sean, this is fabulous! I’ve been looking for a host that has $latex \LaTeX[\tex] installed. InMotion may be exactly what I’ve been looking for and may prompt a server switch in the very near future.
26. Jim Says:
January 22nd, 2008 at 10:16 pm

Huh, I think I got the ending tag wrong (it’s counterintuitive). It should have been $\LaTeX$ .
27. Brad Holden Says:
January 22nd, 2008 at 10:23 pm
$i_{775} - z_{850} > 1.5$
Its hard not to type the $, \( or \[
28. astropixie Says:
January 22nd, 2008 at 10:25 pm

“For a long time I was reluctant to joint the many other sciencey blogs that had integrated equations by providing support for LaTeX”

heck, my fellow grad student’s advisor remains reluctant to join the many other sciencey folks that had integrated equations using anything other than FORTRAN!!
29. seriously Says:
January 22nd, 2008 at 10:29 pm

$latex W^{h^{e^{e^{e^{e^{e^{e^{e^{e^{e^{e^e}}}}}}}}}}}[\tex]

$latex W_{h_{o_{o_{o_{o_{o_{o_{o_{o_{o_{o_o}}}}}}}}}}}[\tex]
30. seriously Says:
January 22nd, 2008 at 10:30 pm
$W^{h^{e^{e^{e^{e^{e^{e^{e^{e^{e^{e^e}}}}}}}}}}}$ $W_{h_{o_{o_{o_{o_{o_{o_{o_{o_{o_{o_o}}}}}}}}}}}$
31. Freiddie Says:
January 22nd, 2008 at 10:35 pm
$\epsilon_123=1$
32. Freiddie Says:
January 22nd, 2008 at 10:35 pm

$latex \epsilon_{123}=1[\tex] Now this should work
33. Freiddie Says:
January 22nd, 2008 at 10:35 pm

$\epsilon_{123}=1$ Now this must work!
34. Bob Munck Says:
January 22nd, 2008 at 10:52 pm

Bob, I remember HAL from about an eon ago. Developed at Draper Labs in Cambridge and named for the gentleman who developed much of the Apollo navigation code.

Really? I thought HAL/S stood for Houston Aerospace Language/Shuttle. It was supported by a Fresh Pond company named Intermetrics (a rival of my own SofTech). I had a little bit to do with re-targeting it to a French CPU, the Metra 4 (I think) when I was working for ESA on Spacelab.
35. Name (required) Says:
January 22nd, 2008 at 11:28 pm

$\langle \langle \hat{\mathcal{O}} \rangle \rangle = \lim_{t \rightarrow \infty} \frac{1}{t} \int_0^t \langle \psi_s \left| \hat{\cal{O}} \right| \psi_s \rangle ds = \mbox{Tr} \left( \hat{\rho} \hat{\mathcal{O}} \right) ,$ $\hat{\rho} = \frac{1}{Z} \mbox{exp} \left( \beta \hat{H} + \sum_{i=2}^n \mu_i \hat{F}_i \right) .$
36. Name (required) Says:
January 22nd, 2008 at 11:30 pm

Hmm, seems to evaluate as inline LaTeX for some reason. Maybe it’s the html tags?
$\langle \langle \hat{\mathcal{O}} \rangle \rangle = \lim_{t \rightarrow \infty} \frac{1}{t} \int_0^t \langle \psi_s \left| \hat{\cal{O}} \right| \psi_s \rangle ds = \mbox{Tr} \left( \hat{\rho} \hat{\mathcal{O}} \right) ,$ $\hat{\rho} = \frac{1}{Z} \mbox{exp} \left( \beta \hat{H} + \sum_{i=2}^n \mu_i \hat{F}_i \right) .$
37. Name (required) Says:
January 22nd, 2008 at 11:32 pm

No, still no good. Excuse the debugging,
$\frac{1}{2}$ $\int_0^{\infty}$
38. Name (required) Says:
January 22nd, 2008 at 11:34 pm
$\frac{0}{0}$
39. efp Says:
January 22nd, 2008 at 11:35 pm
$\pi \approx \ln 6^{\ln 5^{\ln 4^{\ln 3^{\ln 2}}}}$
40. Brendon Brewer Says:
January 22nd, 2008 at 11:37 pm

Hmm, looks like Sean’s thinking about the long-term time averages of certain quantities in quantum statistical mechanics. I wonder why?
41. Sam Says:
January 22nd, 2008 at 11:40 pm
$e_{\pi i}=-1$
second try…
42. Name (required) Says:
January 22nd, 2008 at 11:40 pm
$\begin{displaymath} \int_0^\infty \frac{1}{x^x} dx \end{displaymath}$
$latex \begin{displaystyle} \frac{1}{\pi} \int_0^\pi \cos \left( n t – x \sin t \right) dt \end{displaymath}{/tex]
43. Sam Says:
January 22nd, 2008 at 11:41 pm
$e^{\pi i}=-1$
third try…
44. Sean Says:
January 22nd, 2008 at 11:50 pm

It’s really just “\displaystyle”; e.g.

(tex)\sum_{n=0}^\infty(/tex)

(with square brackets instead of parentheses) gives
$\sum_{n=0}^\infty$
while

(tex)\displaystyle \sum_{n=0}^\infty(/tex)

gives
$\displaystyle \sum_{n=0}^\infty$
45. Name (required) Says:
January 22nd, 2008 at 11:56 pm

Oh well.

Hmm, looks like Sean’s thinking about the long-term time averages of certain quantities in quantum statistical mechanics. I wonder why?

Apparently he’s relating the long-term time average of a pure-state expectation value in the limit $t\rightarrow \infty$ , with the instantaneous exp. value of a mixed state
$\langle \langle \hat{\mathcal{O}} \rangle \rangle = \lim_{t \rightarrow \infty} \frac{1}{t} \int_0^t \langle \psi_s \left| \hat{\cal{O}} \right| \psi_s \rangle ds = \mbox{Tr} \left( \hat{\rho} \hat{\mathcal{O}} \right) ,$
where $\rho$ looks like a partition function, with n-1 funny-looking terms added to the hamiltonian
$\hat{\rho} = \frac{1}{Z} \mbox{exp} \left( \beta \hat{H} + \sum_{i=2}^n \mu_i \hat{F}_i \right) .$
The last time Sean blogged thermodynamics, it was about Boltzmann brains. Maybe these are quantum Boltzmann brains.
46. Name (required) Says:
January 22nd, 2008 at 11:58 pm
$\displaystyle{ \langle \langle \hat{\mathcal{O}} \rangle \rangle = \lim_{t \rightarrow \infty} \frac{1}{t} \int_0^t \langle \psi_s \left| \hat{\cal{O}} \right| \psi_s \rangle ds = \mbox{Tr} \left( \hat{\rho} \hat{\mathcal{O}} \right) , }$
$latex \displaystyle{ \hat{\rho} = \frac{1}{Z} \mbox{exp} \left( \beta \hat{H} + \sum_{i=2}^n \mu_i \hat{F}_i \right) .}
47. Name (required) Says:
January 22nd, 2008 at 11:59 pm

Eureka, it worked!
$\displaystyle \langle \langle \hat{\mathcal{O}} \rangle \rangle = \lim_{t \rightarrow \infty} \frac{1}{t} \int_0^t \langle \psi_s \left| \hat{\cal{O}} \right| \psi_s \rangle ds = \mbox{Tr} \left( \hat{\rho} \hat{\mathcal{O}} \right) ,$ $\displaystyle \hat{\rho} = \frac{1}{Z} \mbox{exp} \left( \beta \hat{H} + \sum_{i=2}^n \mu_i \hat{F}_i \right) .$
48. TomR Says:
January 22nd, 2008 at 11:59 pm

The electic field of a charge falls of like
exp(-r/lambda)/r^2, where lambda is the so-called Debye screening length of the plasma.

The electric field of a charge falls off like $e^{-r/\lambda}/r^2$ where $\lambda$ is the so-called Debye screening length of the plasma
49. TomR Says:
January 23rd, 2008 at 12:02 am

The electic field of a charge falls of like
exp(-r/lambda)/r^2, where lambda is the so-called Debye screening length of the plasma.

The electric field of a charge falls off like $\dfrac{e^{-r/\lambda}}{r^2}$ where $\lambda$ is the so-called Debye screening length of the plasma
50. Eric Says:
January 23rd, 2008 at 12:16 am
$S=-\frac{T}{2}\int d^2\zeta \sqrt{-\gamma} \gamma^{ab} h_{ab}$
51. Brendon Brewer Says:
January 23rd, 2008 at 12:21 am

Apparently he’s relating the long-term time average of a pure-state expectation value in the limit t\rightarrow \infty, with the instantaneous exp. value of a mixed state

Hmm, that sounds like an ergodic theorem. I don’t like ergodic theorems because they pretty much have zero relevance except in algorithm design (MCMC and stuff).
52. Carl Brannen Says:
January 23rd, 2008 at 12:21 am

The essence of the pure density operator formalism: States are operators instead of vectors. Example: $|a\rangle \to \rho_a = |a\rangle \langle a|$ . Example: $\psi(x) \to \rho(x,x') = \psi^*(x')\psi(x)$

Let $\hat{S}$ be an operator that squares to 1. Then $(1\pm\hat{S})/2$ is (an opeartor and is also) an eigenstate of $\hat{S}$ with eigenvalue $\pm 1$ . Example, spin 1/2 in the X direction has the operator:
$\hat{X} = \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$
and therefore the density matrix state corresponding to it is:
$\rho_X = 0.5\left(\begin{array}{cc}1&1\\1&1\end{array}\right)$

If it passes this, that’s pretty good…
53. Carl Brannen Says:
January 23rd, 2008 at 12:24 am

Hmm. ampersands sort of don’t work in the definition of arrays, they get an extra “amp;” .
54. Aristotle Pagaltzis Says:
January 23rd, 2008 at 1:58 am

Regarding preview, may I recommend the AJAX Comment Preview plugin? It’s the best preview function I’ve seen anywhere, and easy to set up.
55. Jason Says:
January 23rd, 2008 at 4:14 am

Very nifty.

$latex \begin{displaymath}
\mathbf{A} =
\left( \begin{array}{ccccc}
1 & 0 & 0 & 0 &\ddots \\
d & -2d -a^2 & d & 0 & \ddots \\
0 & d & -2d -a^2 & d & \ddots \\
\ddots & \ddots & \ddots & \ddots & \ddots \\
\hdots & 0 & 0 & 0 & 1
\end{array} \right).
\end{displaymath}$
56. Jason Says:
January 23rd, 2008 at 4:15 am

Well, it worked in my term paper… ah, well.
57. Christian Says:
January 23rd, 2008 at 4:32 am
$\lim_{3\rightarrow4}\sqrt{3}=2$
58. tca Says:
January 23rd, 2008 at 4:45 am

You could simply use the Emacs Muse package for the “all in one” Emacs. It supports latex2png.

Testing LaTeX in wordpress:

[ tex]
R=\left(
\begin{matrix}
2q_0^2-1+2q_1^2&2q_1q_2-2q_0q_3&2q_0q_2+2q_1q_3\cr 2q_1q_2+2q_0q_3&2q_0^2-1+2q_2^2&2q_2q_3-2q_0q_1\cr
2q_1q_3-2q_0q_2&2q_0q_1+2q_2q_3&2q_0^2-1+2q_3^2
\end{matrix}
\right)
59. tca Says:
January 23rd, 2008 at 4:50 am

Did not work… Sorry!
60. LaTeX « Os Manos DALTON Says:
January 23rd, 2008 at 4:59 am

[...] 23, 2008 Ora aqui está qualquer coisa de muito importante que acabei de descobrir através do Cosmic Variance: como escrever equações neste blog, e apenas em Wordpress, usando LaTeX… Fiquei [...]
61. Mark Says:
January 23rd, 2008 at 5:23 am

A limerick:
$\int_1^\sqrt{3} z^2\,dz \cross cos(\frac{3\pi}{9} = \ln\sqrt[3]{3}$
62. Mark Says:
January 23rd, 2008 at 5:25 am

2nd try:

A limerick:
$\int_1^{\sqrt{3}} z^2\,dz \cross cos(\frac{3\pi}{9}) = \ln\sqrt[3]{3}$
63. Mark Says:
January 23rd, 2008 at 5:27 am

And of course I get the formatting right but screw up the formula.
$\int_1^{\sqrt[3]{3}} z^2\,dz cos(\frac{3\pi}{9}) = \ln\sqrt[3]{e}$
64. feg Says:
January 23rd, 2008 at 5:40 am
$e^{i*pi}+1=0$
65. andy Says:
January 23rd, 2008 at 7:02 am

Hmm… not sure I know how this works, but…
$T = \left(\dfrac{L_\ast \left(1-A\right)}{ 16\pi\sigmaD^2 }\right)^\frac{1}{4}$
66. andy Says:
January 23rd, 2008 at 7:03 am

Try again…
$T = \left(\dfrac{L_\ast \left(1-A\right)}{ 16\pi \sigma D^2 }\right)^\frac{1}{4}$
67. Joerg Says:
January 23rd, 2008 at 8:14 am
$\partial_t\vec{v}+[\nabla\cdot\vec{v}]=-\frac{1}{\rho}\nabla\vec{p}-\vec{g}-\nabla² \vec{v}²$
68. Joerg Says:
January 23rd, 2008 at 8:15 am

Wow that wasn’t right
$\partial_t\vec{v}+[\nabla\cdot\vec{v}]=-\frac{1}{\rho}\nabla\vec{p}-\vec{g}-\nu\nabla^2 \vec{v}^2$
69. Count Iblis Says:
January 23rd, 2008 at 9:39 am
$\displaystyle\int_{0}^{\infty}\frac{x^{-p}}{1+x}dx=\frac{\pi}{\sin\left(\pi p\right)}$
70. Frank Oswalt Says:
January 23rd, 2008 at 10:01 am
$\frac{Love - 0}{No Limit}$
71. Erick Says:
January 23rd, 2008 at 10:07 am

How can I use Latex on Windows? I’ve downloaded the Windows thing but when the interface comes up, there’s nothing to do. You know, to start a new document and start typing with those sciencey fonts..what am I missing?
72. Navneeth Says:
January 23rd, 2008 at 10:09 am

Let $latex \epsilon
73. Navneeth Says:
January 23rd, 2008 at 10:11 am

$latex Let \epsilon
74. citrine Says:
January 23rd, 2008 at 10:21 am

Formatting LaTEX commands seeems cumbersome to me. I’d think typing the Standard Model Lagrangian would be a nightmare! Embarrassing confession – I typed two dissertations using the Word equations editor. I’m waiting for the point and click version of LaTEX – essentially a more versatile version of Word E.E.
75. efp Says:
January 23rd, 2008 at 10:31 am

Re #71 & #74:

http://www.mackichan.com

It’s commercial, but its &ltbad french accent&gt very nice &lt/bad french accent&gt

If anyone knows of a open source/cross-platform equivalent, I’m all ears. But don’t tell my Lyx, it doesn’t come close.
76. efp Says:
January 23rd, 2008 at 10:32 am

Damn, what this blog needs is a preview button…
77. Christina Says:
January 23rd, 2008 at 11:50 am

wrt feed (sorry slow to get back to this), on bloglines, I’m sub’d to: http://blogs.discovermagazine.com/cosmicvariance/?feed=rss2
I do see the pictures, have just a white background… or actually gray for that one (every other item is gray). I can put a screenshot somewhere if that would be helpful.
78. Sean Says:
January 23rd, 2008 at 11:56 am

Frank Oswalt wins.

Christina, it seems to work with

http://blogs.discovermagazine.com/cosmicvariance/feed/rss

Try switching to that one to see if it works.

Sorry about the lack of preview; we used to have one, but it broke for some unknown reason. The AJAX preview is also seemingly incompatible with our theme.
79. Dave Bacon Says:
January 23rd, 2008 at 12:27 pm

I see no equations using IE 7.0. There is a javascript error: Line 56, ‘document.getElementById(..) is null or not an object’
80. Sean Says:
January 23rd, 2008 at 12:34 pm

Sorry, I think that was my fault — tried to mess with a preview plugin, and mistakenly deactivated the latex plugin. Should be okay now.
81. chemicalscum Says:
January 23rd, 2008 at 12:35 pm

I see no equations using IE 7.0. There is a javascript error: Line 56, ‘document.getElementById(..) is null or not an object’

I see no equations either on both FF 2.0.0.11 and IE6 on Windows. Haven’t yet tried it on a Linux box.
82. chemicalscum Says:
January 23rd, 2008 at 12:36 pm

Yep it’s working for me now.
83. Navneeth Says:
January 23rd, 2008 at 12:41 pm
$\alpha\beta\gamma$
84. chemicalscum Says:
January 23rd, 2008 at 12:50 pm

OK lets give it a try see if I can get some Dirac Notation:

$latex % define new matrix element, \ME command
\newcommand{\ME}[3]{\ensuremath{\left \langle \left. #1
\right. \right| #2 \left| \left. #3 \right. \right \rangle}}

\ME{a}{M}{b}$
85. chemicalscum Says:
January 23rd, 2008 at 12:51 pm

Oh well I’ll try again later.
86. bswift Says:
January 23rd, 2008 at 1:47 pm

oh, good times…
87. mollishka Says:
January 23rd, 2008 at 2:58 pm
$\displaystyle \theta_{E}^{2} = \frac{4GM}{c^2}\frac{D_{LS}}{D_{L}D_{S}}$
88. MedallionOfFerret Says:
January 23rd, 2008 at 4:38 pm

Emulation is the sincerest form of flattery:
$\displaystyle \langle\langle \hat{\mathcal O}\rangle\rangle =\lim_{t \rightarrow \infty}\frac{1}{t}\int^t_0 \langle \psi_s|\hat{\mathcal O}|\psi_s\rangle ds = \rm{tr}(\hat\rho \hat{\mathcal O})\,,$ $\displaystyle \hat\rho = \frac{1}{Z} \exp{\left(-\beta \hat{H} - \sum^n_{i=2} \mu_i \hat{F}_i\right)}\,.$
89. MedallionOfFerret Says:
January 23rd, 2008 at 4:38 pm

I also do “The rain in Spain falls mainly on the plain” quite well.
90. andy Says:
January 23rd, 2008 at 6:34 pm
$\left[\dfrac{abc}{def}\right]$
91. andy Says:
January 23rd, 2008 at 6:35 pm

Er… oops…
92. andy Says:
January 23rd, 2008 at 6:37 pm
$\left\[\dfrac{abc}{def}\right\]$
93. andy Says:
January 23rd, 2008 at 6:39 pm

Hmmm… doesn’t like square brackets…
$\dfrac{abc}{def}$
$\left \[ \dfrac{abc}{def} \right \]$
94. Count Iblis Says:
January 23rd, 2008 at 6:45 pm
$\left[A,B\right]=AB - BA$
95. Count Iblis Says:
January 23rd, 2008 at 6:47 pm

So, I think it may be the dfrac command, Andy
96. Hal S Says:
January 23rd, 2008 at 7:47 pm
$P_{\a\v\g}$
97. Hal S Says:
January 23rd, 2008 at 7:50 pm

$latex P_{avg}[\tex]
98. Hal S Says:
January 23rd, 2008 at 7:51 pm
$P_{avg}$
99. Hal S Says:
January 23rd, 2008 at 8:05 pm
$P_{avg}=\frac{\Deltam}{\Deltat}c^2=\frac{\Deltap}{\Deltax}c^2$
100. Hal S Says:
January 23rd, 2008 at 8:08 pm

still practicing
$\drac{\Delta t}$
101. Hal S Says:
January 23rd, 2008 at 8:11 pm
$P_{avg}=\dfrac{\Delta m}{\Delta t} c^2=\dfrac{\Delta p}{\Delta x} c^2$
102. Hal S Says:
January 23rd, 2008 at 8:12 pm

cool
103. Lawrence B. Crowell Says:
January 23rd, 2008 at 8:19 pm

$\partial_a\partial^a\phi~=~-m^2\phi~+~g^2\phi^3 F_{ab}F^{ab}$
with the action
$latex
S~\simeq~ \int d^4x \Big(\frac{1}{2}|\nabla\phi|^2~+~\phi^2\big(\sqrt{-g}\kappa R~+~\frac{1}{2}m^2~-~\frac{g^2}{4}\phi^2 F_{ab}F^{ab}\big)\Big),$
104. Freiddie Says:
January 23rd, 2008 at 9:46 pm

I’m seeing all sorts of familiar and unfamiliar maths here! Hey, is that a Lie bracket?
105. Blake Stacey Says:
January 23rd, 2008 at 11:29 pm

Dang it, now I feel really guilty I didn’t finish my next SUSY QM post today. I find it’s actually harder to write the prose in between the equations than the equations themselves. . . .
$\{Q, Q^\dag\} = \mathcal{H},$ $\{Q, Q\} = \{Q^\dag, Q^\dag\} = Z.$
106. agm Says:
January 24th, 2008 at 4:51 am

Sean, I vote for Christian (#57) winning the thread. For the obvious reason.
107. James Gallagher Says:
January 24th, 2008 at 6:21 am
$i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H\left|\Psi(t)\right>$
OK, but you do need a previewer.

LatexRender is cool in its quirky way, slightly naive, limited and won’t be the future, but hey, it’s fun.

You wouldn’t get the superstring guys using it though.
108. James Gallagher Says:
January 24th, 2008 at 7:02 am

For the pedantic, let’s make that Hamiltonian explicitly time dependent
$i\hbar\frac{\partial}{\partial t}\left|\Psi(t)\right>=H(t)\left|\Psi(t)\right>$
109. aatish Says:
January 24th, 2008 at 8:33 am

$\sum_{n=1}^{\infty}n = -\frac{1}{12}$ , therefore physics is crazy!

I discovered latex on wordpress recently, so I’ve been going latex crazy on my website.
110. Hal S Says:
January 24th, 2008 at 11:09 am

Electrical theory
111. Ghiret Says:
January 24th, 2008 at 12:15 pm

So, as you say in your book, covariant derivative of $V ^\mu$ is:
$\nabla _{\nu}V^\mu = \partial _\nu V^\mu+\Gamma^\mu^{\sigma\nu}V^\sigma$
I think, but I suppose I don’t remenber index placement really.

(anyway, I was just trying latex )
112. Ghiret Says:
January 24th, 2008 at 12:17 pm

Mmm, $\sigma\mu$ should be down… :S
113. James Nightshade Says:
January 24th, 2008 at 12:35 pm
$(x+y)^n=\displaystyle \sum_{k=0}^n \left ( n \atop k \right ) x^{n-k}y^k$
114. Blake Stacey Says:
January 24th, 2008 at 2:05 pm

Now, the real question is, can we use LaTeX to write a Hamiltonian for the Quantum of Solace?
115. benji Says:
January 24th, 2008 at 2:28 pm

Quadratic thingy, try 1
$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
116. benji Says:
January 24th, 2008 at 2:29 pm

Cool!
117. andy.s Says:
January 24th, 2008 at 2:40 pm
$-\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi = i\hbar\frac{\partial}{\partial t}\Psi$
118. andy.s Says:
January 24th, 2008 at 2:45 pm

Whoo hoo! Worked first time!

Previews are for wussies!
119. Chris Says:
January 24th, 2008 at 6:24 pm

You guys better worry whether Steven Hawking was right in his introduction to _A Brief History of Time_. He repeated advice given to him that for every equation used in the text, he would decrease his readership by 50%. By my estimates, the readership of this blog will soon be approximating the Planck Length.
120. Count Iblis Says:
January 24th, 2008 at 8:21 pm
$\displaystyle x^3 + p x + q = 0 \Longrightarrow$ $\displaystyle \sqrt[3]{\sqrt{\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2} - \frac{q}{2}} - \sqrt[3]{\sqrt{\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2} + \frac{q}{2}}$
121. Count Iblis Says:
January 24th, 2008 at 8:24 pm

I meant to say:
$\displaystyle x^3 + p x + q = 0 \Longrightarrow$ $\displaystyle x = \sqrt[3]{\sqrt{\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2} - \frac{q}{2}} - \sqrt[3]{\sqrt{\left(\frac{p}{3}\right)^3+\left(\frac{q}{2}\right)^2} + \frac{q}{2}}$
122. Brian Says:
January 24th, 2008 at 9:34 pm
$\displaystyle \oint_s \vec{B} \cdot \da$
here goes…
123. chris Says:
January 25th, 2008 at 3:45 am
$\begin{tabular}{ll}1 &2\\1&2\end{tabular}$
124. chris Says:
January 25th, 2008 at 3:48 am

$latex \begin{tabular}{ll} 1 2\\ 1 2\end{tabular}
125. chris Says:
January 25th, 2008 at 3:48 am

$latex \begin{tabular}{ll} 1 2\\ 1 2\end{tabular}[\tex]
126. chris Says:
January 25th, 2008 at 3:49 am
$\left( \begin{tabular}{ll} 1 2\\ 1 2\end{tabular}\right)$
127. chris Says:
January 25th, 2008 at 3:50 am
$\begin{tabular}{ll} 1 \left( \begin{tabular}{ll}1 2 \\ 1 2 \end{tabular}\right) \\ 1 2\end{tabular}$
128. chris Says:
January 25th, 2008 at 3:51 am
$\left(\begin{tabular}{ll} 1 \left( \begin{tabular}{ll}1 2 \\ 1 2 \end{tabular}\right) \\ 1 2\end{tabular}\right)$
129. chris Says:
January 25th, 2008 at 3:53 am

< ![CDATA[latex]\texbf{feel free to axe the above ones if things have become cluttered.}[/latex]>
130. chris Says:
January 25th, 2008 at 3:53 am
$\texbf{feel free to axe the above ones if things have become cluttered.}$
131. chris Says:
January 25th, 2008 at 3:59 am
$\textbf{\sigma}=\right\{ \left( \begin{tabular}{ll}0 1// 1 0\end{tabular}\right) ,\left( \begin{tabular}{ll}0 -i// i 0\end{tabular}\right) , \left( \begin{tabular}{ll}1 0// -1 0\end{tabular}\right) \right\}$
132. chris Says:
January 25th, 2008 at 4:00 am

That would have been a lot more impressive had it worked.
133. chris Says:
January 25th, 2008 at 4:05 am
$\textbf{\sigma}=\right( \left( \begin{tabular}{ll}0 1\\ 1 0\end{tabular}\right) ,\left( \begin{tabular}{ll}0 -i \\ i 0\end{tabular}\right) , \left( \begin{tabular}{ll}1 0\\ -1 0\end{tabular}\right) \right)$
134. chris Says:
January 25th, 2008 at 4:06 am

Okay, so much for that. Sorry about the big white splotches, you might want to get rid of them. Congrats on getting the latex!
135. Seth Says:
January 25th, 2008 at 9:03 am

Can’t resist..
$\sigma(p(P)p(P) \to Y + X) = \int_0^1dx_1\int_0^1dx_2f_1(x_1)f_2(x_2)\sigma(q_1(x_1P)q_2(x_2P) \to Y)$
136. Seth Says:
January 25th, 2008 at 9:04 am

Err… that was:

\sigma(p(P)p(P) \to Y + X) = \int_0^1dx_1\int_0^1dx_2f_1(x_1)f_2(x_2)\sigma(q_1(x_1P)q_2(x_2P) \to Y)

What’s wrong?
137. Neil B. Says:
January 25th, 2008 at 1:05 pm

It looks like everyone forgot or overlooked my question at #9, LMK if you have an answer, tx
138. Brian Drell Says:
January 25th, 2008 at 11:00 pm

$latex E=\frac{n^2\pi^2\hbar^2}{2mL^2}[\tex]
139. glenn Says:
January 26th, 2008 at 5:44 am
$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi G T_{\mu\nu}$
hehe. I’m plagiarizing!
140. James Gallagher Says:
January 26th, 2008 at 8:25 am

I’ll try your formula Seth:

$\sigma(p(P)p(P) \to Y + X) = \int_0^1dx_1\int_0^1dx_2f_1(x_1)f_2(x_2)\sigma(q_1(x_1P)q_2(x_2P) \to Y$ latex

“Potentially dangerous”, wow, these days you just don’t know where you are safe.

Chris, I like your recent contributions, ala G Lissi, you’ve created a Theory of Nothing
141. James Gallagher Says:
January 26th, 2008 at 8:26 am

I’ll try your formula Seth:
$\sigma(p(P)p(P) \to Y + X) = \int_0^1dx_1\int_0^1dx_2f_1(x_1)f_2(x_2)\sigma(q_1(x_1P)q_2(x_2P) \to Y$
“Potentially dangerous”, wow, these days you just don’t know where you are safe.

Chris, I like your recent contributions, ala G Lissi, you’ve created a Theory of Nothing
142. Count Iblis Says:
January 26th, 2008 at 8:53 am

Let’s give Seth’s formula another try:

$latex \sigma(p(P)p(P) \rightarrow Y + X) = \int_0^1dx_1\int_0^1dx_2f_1(x_1)f_2(x_2)\sigma(q_1(x_1P)q_2(x_2P) \rightarrow Y[\tex]
143. Count Iblis Says:
January 26th, 2008 at 8:54 am

OOps, made a trivial error, let’s try again:
$\sigma(p(P)p(P) \rightarrow Y + X) = \int_0^1dx_1\int_0^1dx_2f_1(x_1)f_2(x_2)\sigma(q_1(x_1P)q_2(x_2P) \rightarrow Y$
144. Count Iblis Says:
January 26th, 2008 at 8:57 am

The software they use at Physicsforums does not complain, so perhaps here we need to care about the delimeters being closed properly… let’s try again:
$\sigma(p(P)p(P) \to Y + X) = \int_0^1dx_1\int_0^1dx_2f_1(x_1)f_2(x_2)\sigma(q_1(x_1P)q_2(x_2P) \to Y)$
145. Freiddie Says:
January 26th, 2008 at 12:23 pm

I tried parsing the formula in Wikipedia – works fine,
but I’m not sure it’ll work here.
This is what you have been trying, right?
\sigma(p(P)p(P) \rightarrow Y + X) = \int_0^1dx_1\int_0^1dx_2f_1(x_1)f_2(x_2)\sigma(q_1(x_1P)q_2(x_2P) \rightarrow Y
Now, here’s the formula (may not work…):
$\sigma(p(P)p(P) \rightarrow Y + X) = \int_0^1dx_1\int_0^1dx_2f_1(x_1)f_2(x_2)\sigma(q_1(x_1P)q_2(x_2P) \rightarrow Y$
Does it work?
146. Freiddie Says:
January 26th, 2008 at 12:25 pm
$\sigma(p(P)p(P) \rightarrow Y + X)$ [latex] = \int_0^1dx_1[/latex] $\int_0^1dx_2f_1(x_1)f_2(x_2)$ [latex]\sigma(q_1(x_1P)q_2(x_2P)[/latex] $\rightarrow Y$
147. Freiddie Says:
January 26th, 2008 at 12:26 pm

Hey, it works! (I broke the entire formula into multiple pieces of [ tex ] and [ /tex ].
148. C Says:
January 26th, 2008 at 2:11 pm

$latex \psset{unit=0.5cm}
\begin{pspicture}(-4,-0.5)(4,8)
\psgrid[subgriddiv=0,griddots=5,gridlabels=7pt](-4,-0.5)(4,8)
\psline[linewidth=1pt]{->}(-4,0)(+4,0)
\psline[linewidth=1pt]{->}(0,-0.5)(0,8)
\psplot[plotstyle=curve,linewidth=0.5pt]{-4}{0.9}{10 x exp}% postscript function
\rput[l](1,7.5){$10^x$}
\psplot[plotstyle=curve,linecolor=red,linewidth=0.5pt]{-4}{3}{2 x exp}% postscript function
\rput[l](2.2,7.5){\color{blue}$e^x$}
\psplot[plotstyle=curve,linecolor=blue,linewidth=0.5pt]{-4}{2.05}{2.7183 x exp}% postscript function
\rput[l](3.2,7.5){\color{red}$2^x$}
\rput(4,8.5){\color{white}change\normalcolor}
\rput(-4,-1){\color{white}bounding box\normalcolor}
\end{pspicture}
$
149. C Says:
January 26th, 2008 at 2:13 pm

Not quite working properly…maybe some packages need to be installed? For example, see http://sixthform.info/steve/wordpress/?p=24. Also, it would be nice to have the javascript option, where clicking on the equation brings up the latex code that generated it.

\psset{unit=0.5cm}
\begin{pspicture}(-4,-0.5)(4,8)
\psgrid[subgriddiv=0,griddots=5,gridlabels=7pt](-4,-0.5)(4,8)
\psline[linewidth=1pt]{->}(-4,0)(+4,0)
\psline[linewidth=1pt]{->}(0,-0.5)(0,8)
\psplot[plotstyle=curve,linewidth=0.5pt]{-4}{0.9}{10 x exp}% postscript function
\rput[l](1,7.5){$10^x$}
\psplot[plotstyle=curve,linecolor=red,linewidth=0.5pt]{-4}{3}{2 x exp}% postscript function
\rput[l](2.2,7.5){\color{blue}$e^x$}
\psplot[plotstyle=curve,linecolor=blue,linewidth=0.5pt]{-4}{2.05}{2.7183 x exp}% postscript function
\rput[l](3.2,7.5){\color{red}$2^x$}
\rput(4,8.5){\color{white}change\normalcolor}
\rput(-4,-1){\color{white}bounding box\normalcolor}
\end{pspicture}
150. Count Iblis Says:
January 26th, 2008 at 4:54 pm

C, isn’t there anything to render postscript directly? Then we could all just compile the latex code (including possible eps figures) to postscript and copy and paste that on this blog
151. Freiddie Says:
January 26th, 2008 at 5:23 pm

If you want to see the code (provided the equation is rendered correctly), simply look at the tooltip that appears when you move your mouse over the equation.
152. Plato Says:
January 27th, 2008 at 12:21 am

In Clifford’s Sandbox here are some tools for people to use. Carl helps quite a bit there. And, someone else mentioned mouse overs help too.
153. Plato Says:
January 27th, 2008 at 12:42 am

Also see Latex Spoken here for further examination
154. Jason Says:
January 27th, 2008 at 2:22 pm

Excellent! Now I can submit the proof that girls are evil:
$\mathrm{girls} = \mathrm{time}\times\mathrm{money}$
$\mathrm{time} = \mathrm{money}$
$\mathrm{girls} = \mathrm{money}^2$
$\mathrm{money} = \sqrt{\mathrm{evil}}$
$\mathrm{girls} = \sqrt{\mathrm{evil}^2} = \mathrm{evil}$
155. Sean Says:
January 27th, 2008 at 7:11 pm

Shouldn’t that be plus or minus evil?
156. James Gallagher Says:
January 27th, 2008 at 8:04 pm

Nope, it’s +evil, he made a slight error in the formatting, it should be
$\mathrm{girls} = \sqrt{\mathrm{evil}}^2$
157. Hal S Says:
January 28th, 2008 at 9:52 am
$\langle \rangle$
158. Eugene Says:
January 28th, 2008 at 1:10 pm

$latex U=0[\tex]
159. Eugene Says:
January 28th, 2008 at 1:11 pm

Bah
$U=0$
160. Hal S Says:
January 29th, 2008 at 7:24 am
$\dfrac{h \nu}{e^{\frac{h \nu}{k T}}-1}$
161. Hal S Says:
January 29th, 2008 at 7:44 am
$1…$
162. Hal S Says:
January 29th, 2008 at 8:54 am
$\int \dfrac{\delta E}{T}\$
163. Seth Says:
January 29th, 2008 at 6:16 pm

So I guess it was just the length of my formula that made it dangerous. Not an ideal feature, but then again that’s probably not a formula you’d write out in a blog entry. (It’s really more comprehensible when explained in words anyway.)
164. andy.s Says:
February 4th, 2008 at 10:11 am

I must say, Sean, the Google Ads sidebar comes up with some interesting associations from the repeated use of the word “Latex”.
165. fishcake Says:
February 4th, 2008 at 3:58 pm
$\oint$
166. andy.s Says:
February 14th, 2008 at 2:50 pm
$\pi^+ = \frac{\bar{u}u - \bar{d}d}{\sqrt{2}}$
167. nuno Says:
February 15th, 2008 at 11:07 am
$-\frac{\hbar}{2m}\psi+V(x)\psi=i\hbar\frac{dH}{dt}$
168. nuno Says:
February 15th, 2008 at 11:09 am
$-\frac{\hbar^2}{2m}\nabla^2\psi+V(x)\psi=i\hbar\frac{dH}{dt}$
169. Traums Says:
February 23rd, 2008 at 7:49 am
$19 = 1 x 2^4 + 0 x 2^3 + 0 x 2^2 + 1 x 2^1 + 1 x 2^0$
170. Traums Says:
February 23rd, 2008 at 7:50 am
$19 = 1 * 2^4 + 0 * 2^3 + 0 * 2^2 + 1 * 2^1 + 1 * 2^0$
171. andy.s Says:
July 8th, 2008 at 2:56 pm

test post:
$|\Psi(t^\prime)\rangle = U(t, t^\prime) |\Psi(t)\rangle$
meaning the wave function at a future t’ is some operator times the wave function at t – the operator depends on t and t’.

Obviously,
$U(t, t) = I$
So in the infinitesimal region around t, U is probably something like,
$U(t, t+\delta t) = I + \delta t \Omega + (\delta t)^2 (whatever)$

That makes
$|\Psi(t+\delta t)\rangle = (I + \delta t \Omega + (\delta t)^2)(whatever) ) |\Psi(t)\rangle$

Getting the derivative gives you:
$|\dot{\Psi}\rangle = \Omega |\Psi\rangle$

Big whoop right? We had some operator U that we didn’t know anything about, now we have some operator $\Omega$ that we don’t know anything about. But hit it on the left with $\langle \Psi |$ and you get:
$\langle \Psi |\dot{\Psi}\rangle = \langle \Psi |\Omega |\Psi\rangle$
Adding the complex conjugate,
$\langle \Psi |\dot{\Psi}\rangle + \langle \dot{\Psi} |\Psi\rangle = \frac{\partial}{\partial t}\langle \Psi |\Psi\rangle = 0 = \langle \Psi |\Omega +\Omega^\dagger|\Psi\rangle$
Which implies that $\Omega^\dagger = -\Omega$ .
Cool. So $\Omega$ is anti-hermitian.

But we like Hermitian operators, so let’s multiply both sides by i. This gets us
$i|\dot{\Psi)}\rangle = i \Omega |\Psi)\rangle$
That makes the operator on the right side Hermitian. That’s where the “i” in the Schroedinger equation comes from.

Next is the units. $|\Psi)\rangle$ always has some weird units like length^-1/2, and naturally $|\dot{\Psi)}\rangle$ would be length^-1/2 s^-1. That means that $\Omega$ will have units of s^-1 or frequency.
Lucky we called it $\Omega$ then.

So $i\Omega$ is Hermitian and has units of frequency, so it must be some kind of frequency observable. Well, the energy levels of atoms are always associated with absorption and emission frequencies so we can get the operator to have energy units by multiplying both sides by $\hbar$ to get:
$i\hbar|\dot{\Psi)}\rangle = i \hbar\Omega |\Psi)\rangle$
Which is just the Schroedinger equation, with $i\hbar\Omega$ identified with the Hamiltonian.
172. andy.s Says:
July 8th, 2008 at 3:02 pm

next try:
$|\Psi(t^\prime)\rangle = U(t, t^\prime) |\Psi(t)\rangle$
$U(t, t) = I$
$U(t, t+\delta t) = I + \delta t \Omega + (\delta t)^2 (whatever)$
$|\Psi(t+\delta t)\rangle = (I + \delta t \Omega + (\delta t)^2(whatever) ) |\Psi(t)\rangle$
$|\dot{\Psi}\rangle = \Omega |\Psi\rangle$
$\langle \Psi |\dot{\Psi}\rangle = \langle \Psi |\Omega |\Psi\rangle$
$\langle \Psi |\dot{\Psi}\rangle + \langle \dot{\Psi} |\Psi\rangle = \frac{\partial}{\partial t}\langle \Psi |\Psi\rangle = 0 = \langle \Psi |\Omega +\Omega^\dagger|\Psi\rangle$

Which implies that $\Omega^\dagger = -\Omega$ .
$i|\dot{\Psi}\rangle = i \Omega |\Psi\rangle$
Next is the units. $|\Psi\rangle$ always has some weird units like length^-1/2, and naturally $|\dot{\Psi}\rangle$ would be length^-1/2 s^-1. That means that $\Omega$ will have units of s^-1 or frequency.
Lucky we called it $\Omega$ then.
$i\hbar|\dot{\Psi}\rangle = i \hbar\Omega |\Psi\rangle$
173. andy.s Says:
July 8th, 2008 at 3:05 pm
$U(t, t+\delta t) = I + \delta t \Omega + (\delta t)^2 (whatever)$ $|\Psi(t+\delta t)\rangle = (I + \delta t \Omega + (\delta t)^2(whatever) ) |\Psi(t)\rangle$
174. Proteus Says:
July 8th, 2008 at 5:36 pm
$G(k,i\omega)=\int\frac{-1}{\pi}\frac{G^{''}(k,\omega_1)}{i\omega-\omega_1}$
175. Proteus Says:
July 8th, 2008 at 5:39 pm

$latex G(k,i\omega)=-\int\frac{1}{\pi}\frac{Im(G(k\omega_1))}{i\omega-\omega_1}[\tex]

i think the blog needs a preview function; probably true in general. soon our productivity can $latex \mathrm{Productivity}\rightarrow 0[\tex].
176. Mike M Says:
July 8th, 2008 at 6:12 pm

${\rm H}_2{\rm SO}_4$ , Professor. And $\sqrt{\pi}/$ latex to your good lady wife.
177. Mike M Says:
July 8th, 2008 at 6:13 pm

${\rm H}_2{\rm SO}_4$ , Professor. And $\sqrt{\pi}$ to your good lady wife.
178. Wearing LaTeX « Standard Deviation Says:
July 11th, 2008 at 8:49 pm

[...] Filed under: General — Tags: General — theoreticalperson @ 1:49 am Ever since Cosmic Variance enabled LaTeX, I have to admit I’ve been extremely jealous. It is almost contradictory to [...]
179. A Student Says:
October 29th, 2008 at 8:20 pm
$\mid Z_{1} \mid$
180. A Student Says:
October 29th, 2008 at 8:21 pm
$|Z_{1}|$
181. A Student Says:
October 30th, 2008 at 8:09 pm

$latex P(A \cup B)[\tex]
182. A Student Says:
October 30th, 2008 at 8:10 pm
$P(A \cup B)$
183. A Student Says:
October 30th, 2008 at 8:11 pm
$P(A\cupB)$
184. A Student Says:
October 30th, 2008 at 8:12 pm
$P(A\cup B)$
185. Brad Says:
December 31st, 2008 at 1:14 pm
$i \hbar \frac{\partial \boldsymbol{\Psi}}{\partial t} = \Big[ - \frac{\hbar^2}{2m} \nabla^2 + V \Big] \boldsymbol{\Psi}$
It worked in Wikipedia:Sandbox, let’s see if it works here.
186. Brad Says:
December 31st, 2008 at 1:15 pm

Oh, right…
$i \hbar \frac{\partial \boldsymbol{\Psi}}{\partial t} = \Big[ - \frac{\hbar^2}{2m} \nabla^2 + V \Big] \boldsymbol{\Psi} \displaystyle$
187. Brad Says:
December 31st, 2008 at 1:15 pm

Oh well, it looks squished.
188. andy.s Says:
January 8th, 2009 at 7:01 pm
$\LaTeX$
189. andy.s Says:
January 17th, 2009 at 4:36 pm

I look forward to it Sean. But I bet you’d sell more if you explained it all to your dog.

Also: re entropy.

Consider a muon. Before it decays is structureless and just like a stable particle in terms of properties, but when it decays there is then more complexity in the universe than before etc – how can entropy be coherently defined for such entities

Not sure if entropy would increase in this reaction. Muon decay goes like $\mu^- \to \nu_\mu + W^- \to \nu_\mu + e^- + \bar{\nu_e}$ . That reaction should be reversible, and in chemistry, IIRC, a reversible reaction does not increase entropy. I don’t know, maybe it’s different with particles.
190. notaquant Says:
February 3rd, 2009 at 1:45 pm
$\frac{\partial\psi}{\partial t}(t,x)=-\frac{\partial^2\psi}{\partial x^2}(t,x)+V(t,x)\psi(t,x)$
191. asdfhgh Says:
March 29th, 2009 at 4:04 am

[tex]| \psi \rangle[/tex]
192. asdfhgh Says:
March 29th, 2009 at 4:05 am
$| \psi \rangle$
193. hmm Says:
April 4th, 2009 at 12:11 pm

[tex]R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi G T_{\mu\nu}[/tex]
194. notaquant Says:
May 10th, 2009 at 10:52 am

$latex
i\hbar\frac{\partial\psi}{\partial t}(t,x)=-\frac{\hbar^2}{2m}\nabla^2\psi(t,x)+V(t,x)\psi(t,x)
$
195. notaquant Says:
May 10th, 2009 at 10:53 am

[tex]
i\hbar\frac{\partial\psi}{\partial t}(t,x)=-\frac{\hbar^2}{2m}\nabla^2\psi(t,x)+V(t,x)\psi(t,x)
[/tex]
196. Successful Researcher: How to Become One Says:
May 26th, 2009 at 10:40 am

Just a test: $E=m c^2$
197. macho Says:
July 2nd, 2009 at 5:17 pm

adding a link:
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