Introducing LaTeX Math Typesetting

kingyof2thejring

[tex]mc^2(\frac{hc}{\lambda_0}-\frac{hc}{\lambda_1})=\frac{hc}{\lambda_0}\frac{hc}{\lambda_1}(1-cos\theta)[/tex]
[tex]hmc^3(\frac{1}{\lambda_0}-\frac{1}{\lambda_1})=\frac{h^2c^2}{\lambda_0\lambda_1}(1-cos\theta)[/tex]
[tex]\lambda_0\lambda_1(\frac{1}{\lambda_0}-\frac{1}{\lambda_1})=\frac{h^2c^2}{hmc^3}(1-cos\theta)[/tex]
[tex]\lambda_1-\lambda_0=\frac{h}{mc}(1-cos\theta)[/tex]

[tex]E_0=hf_0[/tex] [tex]f_0=\frac{c}{\lambda_0}[/tex]
[tex]E_1=hf_1[/tex] [tex]f_1=\frac{c}{\lambda_1}[/tex]

Divisionbyzer0

This would be better :

b2386

Hi all,

Here is my question:

In the [tex]E>U_0[/tex] potential barrier, there should be no reflected wave when the incident wave is at one of the transmisson resonances. Assuming that a beam of particles is incident at the first transmission resonance, [tex]E=U_0+(\frac{\pi^2 h^2}{2mL^2})[/tex], combine the continuity conditions to show that B=0. Here are the continuity conditions:

1st [tex]A+B=C+D[/tex]

2nd [tex]k(A- B)=k^{'}(C-D)[/tex]

3rd [tex]Ce^{ik^{'}L}+De^{-ik^{'}L}=Fe^{ikL}[/tex]

4th [tex]k^{'}(Ce^{ik^{'}L}-De^{-ik^{'}L})=kFe^{ikL}[/tex]

A couple more equations that we already know are [tex]k=\sqrt{\frac{2mE}{h^2}}[/tex] and [tex]k^'=\sqrt{\frac{2m(E-U_0)}{h^2}[/tex]

Here is my attempted solution:

I divided the 4th equation by K and then set equation 3 and 4 equal to each other. I then used the new equation to solve for C in terms of D giving me

[tex]C=De^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}[/tex] where I substituted [tex]\frac{\pi}{L}[/tex] in for k'.

I substituted this result into the first equation to now give me

[tex]A+B={De^{-2i \pi}\frac{k^{'}+k}{k^{'}-K} + D[/tex].

Solving for D gives me

[tex]\frac{A+B}{e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K} + 1} = D[/tex]

Now, plugging in our solutions for D and C into the 2nd equation

[tex]\frac{(A+B)e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}}{e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}+1}-\frac{A+B}{e^{-2i \pi}\frac{k^{'}+k}{k^{'}-K}+1}=\frac{k}{k^{'}}(A-B})[/tex]

At this point, it seems impossible to simplify the equation to a point where it is obvious that B = 0. Am I on the right track or is there an easier way?

Couperin

[tex]17(1-\frac{1}{17^2})^\frac{1}{2}[/tex]

Does anyone know of a website that allows me to use tex code and generate latex graphics, that isn't necessarily part of a forum like this one?

BTW, the guy who coded LaTeX into this forum is a bloody wizard.

EDIT: God! LaTeX is amazing! I've been checking the examples in this thread... what an invention!

Couperin

Found one here: http://at.org/~cola/tex2img/index.php

(sorry for double post, I can't find the EDIT button...)

actionintegral

[tex]\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
[tex]\frac{x-vt}{\sqrt{2}}[/tex]
[tex]\sqrt{2}[/tex]

Greg Bernhardt

For testing please use:
http://at.org/~cola/tex2img/index.php

Greg Bernhardt

Here is another latex preview tool suggested by Ranger

http://rogercortesi.com/eqn/index.php

Doc Al

FYI: Latex Command Window now available!

chroot

Test...

[latex]A = A'[/latex]

- Warren

chroot

Test

[latex]E = mc^{10}[/latex]

???

Greg Bernhardt

This page can be used for unlimited LaTeX previewing and testing. It's a little crude. Still working some details out.

http://www.physicsforums.com/mathjax/test/preview.html