The h Bar

vzn

12:39 AM

@Cows new job? congrats! :) whats that about?

Jake Rose

12:58 AM

Could somebody give me a hand with a double integral

$\int_{x=-a}^{x=a} \int_{y=x^2}^{y=(1-x^2)^(1/2)} dydx$

bolbteppa

1:25 AM

@JakeRose is that $\int_{x=-a}^{x=a} \int_{y=x^2}^{y = \sqrt{1-x^2}} dy dx$? Have you learned how to do double integrals over non-rectangular domains?

tutorial.math.lamar.edu/Classes/CalcIII/DIGeneralRegion.aspx

Jake Rose

Yeah

Im just strugglong with the integral of the top limit

once its subbed in and all that jazz you have to integrate it you see

bolbteppa

$\int_{x=-a}^{x=a} \int_{y=x^2}^{y = \sqrt{1-x^2}} dy dx = \int_{x=-a}^{x=a} y|^{\sqrt{1-x^2}}_{x^2}dx = \int_{x=-a}^{x=a} (\sqrt{1-x^2} - x^2)dx = \int \cos^2 (\theta) d \theta - \frac{x^3}{3}|^a_{-a} = \dots$ etc right?

Jake Rose

yeah thats sort of where I got up to

bolbteppa

I see, so what's wrong with doing the rest

Jake Rose

Its just going wrong somewhere unfortunately

actually nevermind i think ive got it

bolbteppa

1:41 AM

It's probably a mess because the limits are $a,-a$ not $1,-1$

HsMjstyMstdn

If I already have an x double-dot to show it to be an acceleration, do I still need to put an arrow above it to show it's a vector or is that just too much ?

Jake Rose

You need the arrow really

bolbteppa

If you are talking about a vector, then you can either write $\ddot{\overline{x}}$ or in more advanced notation $\ddot{x}^i$ but if you write $\ddot{x}$ people will think it's a scalar unless the context is very clear

Jake Rose

Hey bolba

Im getting a acsin that cancels

When in the answer Ive got it shouldnt

Nevermind

I havent got the right answer tbf

But I at least have some of the right terms

bolbteppa

If you're comparing to some online calculator, it could easily give a bad answer tbh, best try to work the logic of it out carefully, do similar examples

e.g. $\cos^2(x) = \frac{1+\cos(2x)}{2}$ etc

and it's probably a mess because of those $a$'s

Jake Rose

1:59 AM

Nah I have a answer sheet with the final (ish) answers

Im not being crazy right

but $[x^3/3] with limits a and -a is \frac{2}{3}a^3$

I forgot to go out of latex

My sheet says 1/3 instead of 2/3

Maybe its wrong

bolbteppa

$\frac{x^3}{3}|^{x=a}_{x=-a} = \frac{a^3}{3} - \frac{(-a)^3}{3} = \frac{a^3}{3} - \frac{-a^3}{3} = \frac{2a^3}{3}$

gian

can someone explain to me how to go from the non-relativistic to the relativistic lagrangian of a free particle?

bolbteppa

@gian you don't, you go the other way

gian

2:15 AM

well yes i understand that the non-relativistic is just a limit of the relativistic. i'm just confused about the -m constant and why we integrate with respect to proper time

Vivek

@DavidZ I have improved my question which was closed by you... Can you please review it? I want it to be reopened.

bolbteppa

Just as non-relativistically you find the minimum path in space, relativistically you find the minimum path in spacetime, and the arc length in spacetime is $ds^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 = c^2 d \tau^2 - 0$ in certain frames

gian

by certain frames you mean the rest frame?

bolbteppa

In the rest frame of a particle, where we have $dx = dy = dz = 0$, then $ds^2 = c^2 d \tau^2$

gian

right. what about the $-m$?

or i guess $-mc$

bolbteppa

2:30 AM

When you take the non-relativistic limit you want it to reduce to the usual Lagrangian, try it with $S = \alpha \int ds$

gian

thanks

bolbteppa

A more physical argument is, since we can always scale actions by arbitrary constants anyway, because $S = \alpha \int ds = \alpha \int c d \tau = \alpha \int c \sqrt{1-v^2/c^2}dt$, the minimum is clearly $0$ i.e. when $v = c$ so that $S = 0$, while the maximum is clearly a straight line in spacetime, so that $S = c \tau$, however when you put in the minus, the minimum is a straight line while the max is 0, so now you do the non-rel limit on $S = - \alpha \int c \sqrt{1-v^2/c^2}dt$

Eran Medan

3:16 AM

Hey everyone! I asked this noob question (https://physics.stackexchange.com/questions/390315/time-dilation-questions-ive-been-trying-to-figure-out-for-years-and-was-too-afr?noredirect=1#comment876975_390315) on SE physics and was recommended to ask around here...

I am determined to really understand special relativity before I die (and then general relativity) not just the concepts but also the math and formal physics behind it

I think I somewhat understood (or have enough material) for #1

e.g. that the mirror thought experiment's math works no matter how much the clock is rotated (due to space contraction etc...)

but I'm still struggling with the rest :(

Sir Cumference

4:03 AM

Electric Potential near Electric Dipole | video in HINDI

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Mesentery

Hello! Physics.SE. In, 1100/10.2 by using significant figure rules, the answer is 110. But in, 1100 m/s divided by 10.2 m/s the answer is 108. The actual answer is 107.8431373. Can anybody please explain that why 110 turned 108 just by using units? Thanks!