The h Bar

heather

12:22 AM

hmm... what's this symbol that looks like a cursive r, in the first equation? detexify doesn't recognize it.

(it's in the denominator of the second fraction)

ah, nvm, found it:

12

Q: Script-r Symbol

The near-ubiquitous undergraduate reference on electrodynamics, "Introduction to Electrodynamics" by David Griffiths make extensive use of a script-r symbol, defined to be \vector{\scriptr} \equiv \vector{r} - \vector{r}^{\prime} and looks like: Although the question of how to produce such a s...

skullpetrol

Nice work Sherlock :-)

heather

ick, now I have to figure out how to format it in newcommand

so i take it back, my request for help still stands =P

0celo7

1:11 AM

$\mathfrak{r}$

$\mathscr{r}$

$\mathcal{SUCKS}\,\,\mathcal{TO}\,\,\mathcal{BE}\,\,\mathcal{YOU}$

@heather I have a problem for you to think about.

A long-term goal.

Physics Meta

0

Q: Why can't I post something on Physics Exchange for 2 days?!

WHY WHY WHY WHY WHY WHY WHY WHY WHY CAN'T I DO IT WHY why can't I do it that's stupid I want to do more

discussion

heather

1:46 AM

@0celo7 okay, sure.

Maks

2:06 AM

Hey, would someone like to discuss a hypothetical universe ?

What would happen it the second law of thermodynamics was inverted

Energy would from more concentrated to less concentrated

0celo7

@heather Ok, are you here?

heather

@0celo7 yep

0celo7

@heather Let $g:[0,1]\to\Bbb R$ be a fixed continuous function. Show that there exists a unique continuous function $f:[0,1]\to\Bbb R$ such that $$f(x)-\int_0^xf(x-t)e^{-x^2}\,dt=g(x).$$

$x\in[0,1]$ of course. @heather Does that make sense?

heather

i'm still deciphering it =)

i think i'll need to learn a bit more to even understand the problem.

0celo7

Hmm? What is confusing?

heather

2:18 AM

well, i'm not sure what is meant by the notation $g:[0,1]$

I've seen it before, but I can't remember what it means.

0celo7

$f:X\to Y$ is a function from the set $X$ to the set $Y$.

heather

ah, okay

okay, so then "fixed continuous function" is a phrase I'm not familiar with.

0celo7

Omit "fixed" then.

$g$ is given is what I'm trying to say.

@heather Do you not know what continuous means?

heather

i think so...

there's no point where the function is undefined.

nothing like $\frac{1}{x}$ (because when x=0, the function is undefined) for example

0celo7

wat

heather

2:23 AM

okay, I'm wrong

0celo7

$x\mapsto 1/x$ is not a function on $[0,1]$

heather

no, no, I know it isn't

I'm saying 1/x on $\mathbb{R}$, sorry - I was just trying to give an example.

0celo7

Hmm. I guess the best way to explain it is that a continuous function is one whose graph is "connected"

@heather It's not a function there either. $0\in\Bbb R$

heather

wait, what?

I'm so confused then

I know 0 is in $\mathbb{R}$

0celo7

$f(x)=1/x$ is not a function on $\Bbb R$

it is not defined at the origin

heather

2:25 AM

yes, that's what I'm saying

it's a non-continuous function.

or is that not how continuous function is defined?

0celo7

But it is continuous on its domain.

@heather It's one for which the graph is connected.

heather

@0celo7 where there is no number you plug in for which the output is undefined?

0celo7

@heather Please rephrase

heather

brb

heather

2:39 AM

sorry, my mom wanted me to do some stuff

@0celo7 okay, um...if you have a function $f:X\to Y$ and you input a $x\in X$, the output will be a $y\in Y$ (never a $y\notin Y$)

dmckee

It's funny how a person can be blasé about one class of risks and get the heeby-jeebies about another one that is objectively no more substantial. This ...

0

Q: How to make a small 100kHz pulsed laser with 1 uJ energy per impulse?

I want to make small (3x2cm) pulsed (100kHz) laser with 905 nm and 1 uJ energy output per impulse(~10 ns). It can be done with laser driver + diode. But I found small laser drivers only up to 70kHz. The size is small so I dont think that it is can be made with resonator, am I right? May be the...

optics laser fiber-optics light-emitting-diodes

makes me shudder ... scary class 4 blindificator. Eeek!

But i have worked with potentially lethal (slowly and nastily lethal, at that) radioactive sources more than once and take very matter of fact approach to dealing with them.

heather

i have to go for now. i will try to read more about your problem @0celo7

0celo7

2:58 AM

@heather You will need Banach space theory to solve it.