admin 4 posts
Forum: 396 –
Topic: Welcome
When you choose “SVG-Edit → Save Image” in the SVG Edit window, it will save the SVG code back to your post. When you’re done with that, just “Save reply” and the SVG will be magically-rendered.
You can also create Tikzpictures:
\ begin{tikzpicture}[every tqft/.style={fill=orange,fill opacity=0.25},
tqft/every boundary component/.style={fill=yellow,fill opacity=.5},
tqft/every upper boundary component/.style={draw,thin,blue},
tqft/every lower boundary component/.style={draw,dashed,thin,blue}]
\usetikzlibrary{tqft}
\begin{scope}[tqft/every incoming boundary component/.style={fill=green, fill opacity= .25}]
\pic[tqft/pair of pants,draw,name=a,at={(0,0)}];
\pic[tqft/reverse pair of pants,draw,anchor=incoming boundary 2,name=e,at={(8,0)}];
\end{scope}
\begin{scope}[tqft/every outgoing boundary component/.style={draw,thin,blue,fill=yellow}]
\pic[tqft/reverse pair of pants,draw,anchor=incoming boundary 1,name=b,at=(a-outgoing boundary 2)];
\end{scope}
\begin{scope}[tqft/every incoming boundary component/.style={fill=green, fill opacity= .25}]
\pic[tqft/cylinder,draw,anchor=outgoing boundary 1,name=d,at=(b-incoming boundary 2)];
\end{scope}
\path (b-incoming boundary 2) ++(1.5,0) node[font=\Huge] {\(=\)};
\begin{scope}[tqft/every outgoing boundary component/.style={draw,thin,blue,fill=yellow}]
\pic[tqft/cylinder,draw,anchor=incoming boundary 1,name=c,at=(a-outgoing boundary 1)];
\pic[tqft/pair of pants,draw,anchor=incoming boundary 1,name=f,at=(e-outgoing boundary 1)];
\end{scope}
\end{tikzpicture}produces
admin 4 posts
edited 5 years ago
Forum: 396 –
Topic: Welcome
Since we’re stuck here in quarantine, it would be a good idea to have a place to discuss the subject of this course online.
You need to sign up (by clicking on the link at the upper left), before you can post.
Equations are pretty easy to type. If you type
The $N$-point tree-level S-Matrix is
$$
S_N(p_1,\dots,p_N) = \frac{8\pi i}{\alpha'} g_s^{2(N-2)}
\int |z_{1 2}|^2 |z_{1 3}|^2 |z_{2 3}|^2\langle O_1(z_1,\overline{z}_1)
\dots O_N(z_N,\overline{z}_N)\rangle d^2 z_4\dots d^2 z_N.
$$
where the $O_i$ are $h=\tilde{h}=1$ conformal primaries in the theory of 26 free scalars.You get
The N N
S N ( p 1 , … , p N ) = 8 π i α ′ g s 2 ( N − 2 ) ∫ | z 1 2 | 2 | z 1 3 | 2 | z 2 3 | 2 ⟨ O 1 ( z 1 , z ¯ 1 ) … O N ( z N , z ¯ N ) ⟩ d 2 z 4 … d 2 z N .
S_N(p_1,\dots,p_N) = \frac{8\pi i}{\alpha'} g_s^{2(N-2)}
\int |z_{1 2}|^2 |z_{1 3}|^2 |z_{2 3}|^2\langle O_1(z_1,\overline{z}_1)
\dots O_N(z_N,\overline{z}_N)\rangle d^2 z_4\dots d^2 z_N.
where the O i O_i h = h ˜ = 1 h=\tilde{h}=1
and
The 4-point tachyon amplitude is
\[
S_4(p_1,p_2,p_3,p_4) = \frac{8\pi i}{\alpha'} g_s^4 (2\pi)^{26}
\delta^{(26)}\bigl(\sum p_i\bigr)2\pi \frac{\Gamma\bigl(-\tfrac{\alpha' s}{4}-1\bigr)
\Gamma\bigl(-\tfrac{\alpha' t}{4}-1\bigr)\Gamma\bigl(-\tfrac{\alpha' u}{4}-1\bigr)
}{\Gamma\bigl(\tfrac{\alpha' u}{4}+2\bigr)\Gamma\bigl(\tfrac{\alpha' t}{4}+2\bigr)
\Gamma\bigl(\tfrac{\alpha' s}{4}+2\bigr)}
\]produces
The 4-point tachyon amplitude is
(1) S 4 ( p 1 , p 2 , p 3 , p 4 ) = 8 π i α ′ g s 4 ( 2 π ) 26 δ ( 26 ) ( ∑ p i ) 2 π Γ ( − α ′ s 4 − 1 ) Γ ( − α ′ t 4 − 1 ) Γ ( − α ′ u 4 − 1 ) Γ ( α ′ u 4 + 2 ) Γ ( α ′ t 4 + 2 ) Γ ( α ′ s 4 + 2 )
S_4(p_1,p_2,p_3,p_4) = \frac{8\pi i}{\alpha'} g_s^4 (2\pi)^{26} \delta^{(26)}\bigl(\sum p_i\bigr)2\pi \frac{\Gamma\bigl(-\tfrac{\alpha' s}{4}-1\bigr)\Gamma\bigl(-\tfrac{\alpha' t}{4}-1\bigr)\Gamma\bigl(-\tfrac{\alpha' u}{4}-1\bigr)}{\Gamma\bigl(\tfrac{\alpha' u}{4}+2\bigr)\Gamma\bigl(\tfrac{\alpha' t}{4}+2\bigr)\Gamma\bigl(\tfrac{\alpha' s}{4}+2\bigr)}
The full equation syntax may be a little daunting, but you probably won’t need much more than the above. For the rest, there’s Markdown .
Oh, and you can draw pictures, using the built-in drawing tool.
Layer 1
p
1
p_1
p
2
p_2
p
3
p_3
p
4
p_4
s
→
4
n
α
′
s\to \frac{4n}{\alpha'}
p
1
p_1
p
2
p_2
p
3
p_3
p
4
p_4
G
(
p
2
)
G(p^2)
=
∫
d
26
p
(
2
π
)
26
={\int\frac{d^{26}p}{(2\pi)^{26}}}
(Just click on the “Create SVG Graphic”) button.
admin 4 posts
Forum: 373 –
Topic: Welcome
Here’s a place to discuss Physics 373. You need to sign up (by clicking on the link at the upper left), before you can post.
Equations are pretty easy to type. If you type
The position, $x(t)$, is
$$
x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 .
$$
Eliminating $t$,
$$
v^2(t) = \sqrt{ v_0^2 + 2 a ( x(t) - x_0 ) }
$$You get
The position, x ( t ) x(t)
x ( t ) = x 0 + v 0 t + 1 2 a t 2 .
x(t) = x_0 + v_0 t + \frac{1}{2} a t^2 .
Eliminating t t
v 2 ( t ) = v 0 2 + 2 a ( x ( t ) − x 0 )
v^2(t) = \sqrt{ v_0^2 + 2 a ( x(t) - x_0 ) }
and
The angular momentum
\[
\vec{L} = \vec{r} \times \vec{p}
\]produces
The angular momentum
(1) L → = r → × p →
\vec{L} = \vec{r} \times \vec{p}
The full equation syntax may be a little daunting, but you probably won’t need much more than the above. For the rest, there’s Markdown .
Oh, and you can draw pictures, using the built-in drawing tool.
Layer 1
v
⇀
0
\vec{v}_0
x
=
0
x=0
x
=
L
x=L
(Just click on the “Create SVG Graphic”) button.
admin 4 posts
edited 8 years ago
Forum: 362L –
Topic: Fourier Transform Conventions
Here are our conventions for Fourier Transforms
In d d
f ( x ⇀ ) = ∫ d d p ⇀ ( 2 π ) d g ( p ⇀ ) e i p ⇀ ⋅ x ⇀ g ( p ⇀ ) = ∫ d d x ⇀ f ( x ⇀ ) e − i p ⇀ ⋅ x ⇀
\begin{aligned}
f(\vec{x}) &= \displaystyle{\int\frac{ d^d\vec{p} }{ {(2\pi)}^d } g(\vec{p}) e^{i\vec{p}\cdot\vec{x}}}\\
g(\vec{p}) &= \displaystyle{\int d^d\vec{x} f(\vec{x}) e^{-i \vec{p}\cdot \vec{x}}}
\end{aligned}
Normalization
∫ d d x ⇀ | f ( x ⇀ ) | 2 = 1 ⇔ ∫ d d p ⇀ ( 2 π ) d | g ( p ⇀ ) | 2 = 1
\int d^d\vec{x} |f(\vec{x})|^2 = 1\quad \xLeftrightarrow{\qquad}\quad \int \frac{d^d\vec{p}}{{(2\pi)}^d} |g(\vec{p})|^2 = 1
The δ \delta 1 1
δ ( d ) ( x ⇀ ) = ∫ d d p ⇀ ( 2 π ) d e i p ⇀ ⋅ x ⇀ ( 2 π ) d δ ( d ) ( p ⇀ ) = ∫ d d x ⇀ e − i p ⇀ ⋅ x ⇀
\begin{split}
\delta^{(d)}(\vec{x}) &= \displaystyle{\int \frac{d^d\vec{p}}{{(2\pi)}^d} e^{i\vec{p}\cdot\vec{x}}}\\
{(2\pi)}^d \delta^{(d)}(\vec{p}) &= \displaystyle{\int d^d\vec{x} e^{-i\vec{p}\cdot\vec{x}}}
\end{split}
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