The h Bar

12:06 AM

This may be a bit out of scope, but what if SE had a study group type feature? Like say people wanted to study GR at an introductory level. Or all the people with homework-type questions

Cows

Nailed presentation/demo of software. Will be adding new features tonight. Then pick up moola on the 14th

Slapped that back end like a boss

I think the moment he saw me demo some skills on the spot he was like Jesus Christ lol.

Emilio Pisanty

14

A: Recent feature changes to Stack Exchange

July 2018 2018-07-12: The moderator-only post issue indicator has a new design that is smaller and uses icons. 2018-07-11: The top bar's sticky behavior has been rolled out to all network sites except Ask Ubuntu (due to clashes with the Ubuntu top bar). 2018-07-05: Favorite Tags renamed Watch...

> 2018-07-04: The algorithm for calculating the overall Stack Exchange reputation for Stack Exchange chat was changed to prevent users from simply signing up for 20 Stack Exchange accounts to gain enough rep to chat.

wait - that loophole was open until now?

danielunderwood

12:27 AM

Anyone doing that must really want to get into chat

danielunderwood

1:00 AM

I think I just heard the pun to end all puns. In one of Brian Greene's books, he talks about a hypothetical lineland and a line creature Kaluza K. Line

Maybe I'm just dumb, but I lost it

David Z

1:18 AM

@EmilioPisanty Apparently so :P

enumaris

1:33 AM

@Slereah lol, I am in fact not "a french"

Mikhail

3:06 AM

Yo, maybe you guys know this. Is there some way to make a laser fiber more broad band? What I really want to do is use a higher power laser, but increase the temporal bandwidth by a dozen nano meters. Aka go from 850 +/- 2nm to 850 +/- 15 nm.

John Rennie

3:38 AM

@danielunderwood people use chat rooms for that. It seems to work pretty well.

Sir Cumference

Just noticed SE implemented a network-wide sticky topbar

Neat

user400188

5:02 AM

In LP Lightman's book, "Problem book in relativity and gravitation" it is asked in the first question on page 17 to express $u^0$ in terms of $|\mathbf{v}|$. $\mathbf{u}$ here is the 4-velocity and $\mathbf{v}$ is the 3-velocity. Its supposed to be an easy question, but in the worked solution at the back of the problem book, they imply that $\frac{dt}{d\tau}=\gamma$. I have not seen this expression anywhere in the problem book or in the main book "Spacetime and geometry" by Sean Carrol.

For context, me and a group of friends are studying out of these two books in preparation for an honors course which most of us will do next year.

The main thing I want to know is, why is the rate of change of with with respect to the proper time $\tau$ equal to the Lorentz factor?

John Rennie

5:18 AM

@user400188 $dt/d\tau = \gamma$ is a standard result isn't it?

I'm sure I've derived this in an answer on the main site. Let me have a look ...

Yes, have a look at:

55

Q: What is time dilation really?

Please will someone explain what time dilation really is and how it occurs. There are lots of questions and answers going into how to calculate time dilation, but none that give an intuitive feel for how it happens.

user400188

Thanks @JohnRennie We are just starting the textbooks (chapter one in both) so we will probaly be introduced to the standard result later on in them.

user400188

5:35 AM

I just got to the bottom of tha page where $\tau=\frac{t}{\gamma}$. This means $\frac{dt}{d\tau}=\gamma\frac{dt}{dt}=\gamma$. Thanks again

John Rennie

@user400188 cool :-)

Novalium Company

Morning.

user400188

One thing that's bugging me, shouldn't $\tau=\frac{t}{|\gamma|}$? (From $c^2\tau^2=t^2c^2(1-\frac{v^2}{c^2})=t^2c^2\frac{1}{\gamma^2}$)

John Rennie

We always take $\gamma$ to be the positive root.

A negative root would mean t and $\tau$ were flowing in opposite directions, which would be ... weird :-)

user400188

6:08 AM

Does anyone know why the norm of a four velocity is $\pm c$? In my textbooks the sign convention is the negative one and $c=1$. I have read that its always timelike (always negative) but I can't seem to prove it. The best I can do is:
$\begin{align}
u\cdot u&=\eta^\mu_mu\frac{dx^\mu}{d\tau}\frac{dx_\mu}{d\tau}\\
&=\gamma^2(-1+(v^j)^2)
end{align}$

Where $v^j=(v_x v_y v_z)$

darn, found the last mistake in the edit 1 second before it was too late (missing a single "\")

Semiclassical

in the rest frame of a massive particle (i.e. not a photon) the four-velocity is $v^\mu=(c,0,0,0)$

so in its rest frame, one has $v^\mu v_\mu=\pm c^2$ depending on your convention

but $v^\mu v_\mu$ is a relativistic invariant

John Rennie

@user400188 I don't think the calculation is hard.

Calculate $u = (dt/d\tau, dx/d\tau, dy/d\tau, dz/d\tau)$ then use the fact that $u_\alpha = \eta_{\alpha\beta}u^\beta$

Semiclassical

the two of our remarks together give another proof of $dt=\gamma\,d\tau$

user400188

6:32 AM

I can see how this all falls out:

$u=(dt/d\tau\ dx/d\tau\ dy/d\tau\ dz/d\tau)=\gamma(1\ \partial x/\partial\tau\ \partial y/\partial\tau\ \partial z/\partial\tau)$

$u\cdot u=\eta_{\alpha\beta}u^\beta u_\alpha=\gamma^2(-1+v_x^2+v_y^2+v_z^2)$

then if $v_j=0$, $\gamma=1$ and $u\cdot u=-1$

the only problem is, the question I need to use this in asks me to write $u^0$ in terms of the other three components $u^j$, where j=1,2,3

The solution should hold for any velocity, not just a velocity of zero.

Semiclassical

by definition, what is $\gamma^2$?

user400188

$\gamma=\frac{1}{\sqrt{1+\frac{v^2}{c^2}}}$

$v$ here is a vector consisting of the x,y and z components.

Semiclassical

not quite...

(if $v=c$, then that formula would give $\gamma=1/\sqrt{1+1}=1/\sqrt{2}$)

user400188

$\gamma=\frac{dt}{d\tau}$?

yeah thats what I was thinking. It would seem (by the definitions I was using) that the solution only held for $v=0$. Which isn't what the question was asking for

Semiclassical

You're not remembering the definition of gamma correctly.

user400188

6:39 AM

oh its a minus not a plus

Semiclassical

Yeah

@user400188 also, the expression $-1+v_x^2+v_y^2+v_z^2$ is a bit off in terms of units

unless you're working in units where $c=1$

user400188

I'm missing the factor of $c$ in all my responces becuase no textbook I've ever read has had $c$ equal to anything other than one.

GPhys

@dmckee precisely!

Semiclassical

fair enough. but then it's also $c=1$ in your statement of $\gamma$

anyways. can you see how this gives $u\cdot u =-1$?

user400188

i placed it there because its one of the few formulas where I know where it goes, and not yet, I'm still committing thought to paper.

Unfortunately Ive got a stray square root floating around. $u\cdot u=-1\cdot \frac{1-v^2}{\sqrt{1-v^2}}=-\sqrt{1-v^2}$

Semiclassical

6:47 AM

$\gamma^2$, not $\gamma$

user400188

and that solves it. Thanks @Semiclassical :)

Semiclassical

np

Novalium Company

Guys, I want to make a drone that can fly around my house (without crashing into walls of course). What I was thinking is using a programmable ic and give it instructions where to go (to go forward put more voltage through propeller 3 and 4) and so on... Is this a right way to do it?

John Rennie

@GPhys I wonder how the density of the plasma at the (visible) surface of a star compares to the density at recombination ...

@b_jonas it would be on topic on the Chemistry SE. Or I can answer here if you want.

Novalium Company

Guys, when we have an FM car radio receiver, how does it tune to listen to exactly one frequency, I mean, the antenna receives so many frequency, how can a circuit remove all of them an listen to a desired one?

John Rennie

7:01 AM

@NovaliumCompany the receiver uses a resonant circuit that only produces a large signal when the radio wave frequency matches the frequency of the resonant circuit.

Novalium Company

So it's like a variable filter than can pass only the desired frequency?

John Rennie

In the old days this would have been a capacitor and inductor in parallel. These days radios use a phase locked loop:

Phase-locked loop

A phase-locked loop or phase lock loop abbreviated as PLL is a control system that generates an output signal whose phase is related to the phase of an input signal. There are several different types; the simplest is an electronic circuit consisting of a variable frequency oscillator and a phase detector in a feedback loop. The oscillator generates a periodic signal, and the phase detector compares the phase of that signal with the phase of the input periodic signal, adjusting the oscillator to keep the phases matched. Keeping the input and output phase in lock step also implies keeping the input...

@NovaliumCompany basically yes, though I think it's more accurate to say it only amplifies the desired frequency.

Novalium Company

Well, we know that in FM the frequency changes, how does the resonant circuit still picks up the desired frequency?

John Rennie

The resonant circuit has a resonance width that is tuned to match the range of frequencies used in radio tranmissions.

So it passes/amplifies a narrow range of frequencies centred on the transmission frequancy.

Novalium Company

Got it, thanks. So when radio companies buy frequencies to transmit their channel, their frequency must have let's say +-10Hertz of free space around the main freuqency?

John Rennie

7:06 AM

Have a look at:

71

Q: Why doesn't the motion of a car affect the frequency of radio stations?

When we go in a car and tune to an FM radio station, why doesn't our motion disturb the frequency? Like the Doppler effect?

waves electromagnetic-radiation doppler-effect

Novalium Company

Ok, but if I buy the license to stream at 50MHz for example, a span of frequencies around mine, must be free?

John Rennie

Yes, but the permissible frequencies are chosen to be widely enough spaced to allow a span of frequencies.

So when you buy the licence you automatically get a range of frequencies.

You aren't allowed to choose just any old frequency to buy.

Novalium Company

So that's why when for example tune to 20.10MHz, the next on is let's say for example at 20.50MHz, they need unoccupied frequencies to stream undirsturbed?

John Rennie

Yes

Novalium Company

@JohnRennie Ok thanks. How do you guys know so much about so many things.... ahh gods of knowledge :D

John Rennie

7:13 AM

@NovaliumCompany it just happens when you get old :-)

When you've spent 40 years listening to people tell you this stuff some of it sticks :-)

Novalium Company

I hope one day I'll know as much as you guys do. :)

John Rennie

Give it 40 years ...

Novalium Company

Well, indeed I've learned more here for a few months, than in school for a few years.

@JohnRennie Do you have some spare time? I'm reading "Practical Electronics for Inventors" and I've got to the AC part and there are some equations... with calculus.... I'm having hard time understanding them.

John Rennie

@NovaliumCompany I need to work for about 15 minutes. If you can wait that long I'll be happy to help.

Novalium Company

Of course, thanks. Tell me when you are free.

user351417

7:37 AM

Is there anyone who knows something about BAs in Physics? I'm applying for an undergrad physics major this fall, and a couple of the places I'm interested in don't offer BS degrees for physics: only BAs.

user351417

I'm wondering if BAs give you the appropriate foundation needed for grad physics and subsequent research.

John Rennie

@Chair in the UK BA and BSc are the same thing.

@NovaliumCompany free now :-)

Novalium Company

Ok, taleri.files.wordpress.com/2014/02/… go to page 115. The RMS Voltage and current kinda confuse me also the calculus.

John Rennie

@NovaliumCompany equation 2.28?

Novalium Company

Yep.

Am I suppossed to make some sense of it, or just memorize it?

John Rennie

7:47 AM

If you graph V^2 as a function of time it looks like this (diagram incoming ...):

@NovaliumCompany OK so far? (it's a slightly wonky curve - sorry abut that)

Novalium Company

When V(t) = sin(t)? or?

John Rennie

@NovaliumCompany $V = sin(\omega t + \phi)$ for some frequency $\omega$ and phase offset $\phi$

Novalium Company

I haven't heard of that sorry.

John Rennie

AC is normally a sine wave

Novalium Company

Well, V(t) = Vpeak * sin(2pi * f * t)?

John Rennie

7:53 AM

Yes

Novalium Company

I'm just following the book.

John Rennie

We use the symbol $\omega$ to mean $2 \pi f$.

Novalium Company

Oh, that's convenient.

John Rennie

($\omega$ is called the angular frequency)

Anonymous

mornin

Novalium Company

7:54 AM

@Blue mornin'

John Rennie

Hi :-)

Novalium Company

Ok so V(t) = Vpeak * sin(w * t)?

John Rennie

Yes, which means $V^2 = V_{peak}^2 \sin^2(\omega t)$ and that's the graph I drew

oops, I've just realised I mislabelled the y axis ...

Anonymous

@NovaliumCompany Yes, but keep in mind at at $t=0$ it is not necessary that $V$ will be $0$. That's why they include a tiny $\phi$ in there

Anonymous

$V = sin(\omega t + \phi)$

Novalium Company

7:56 AM

@Blue when we have a sine wave, in t = 0, the V should be 0 too?

Anonymous

@NovaliumCompany Not necessarily. Play around with this a bit: desmos.com/calculator/56nxbpsjvy

John Rennie

Anonymous

Shift the $c$ slider and see how it shifts

Anonymous

And especially how the initial value at $t=0$ changes

Anonymous

Yes, like in JR's picture ^ !

Novalium Company

7:59 AM

If $\phi$ = 0, then we don't need it?

Sorry I don't understand the point of the phi.

Anonymous

$\phi$ will be $0$ if your initial voltage is $0$

Anonymous

But your initial voltage need not always be $0$

Anonymous

It can begin at any value in between $-V_{\text{peak}}$ and $+V_{\text{peak}}$

John Rennie

@NovaliumCompany we can choose any time we want for our zero. That is, suppose we're timing the siganl with a stopwatch we can reset the stopwatch to zero at any point we want. It doesn't have to be when $V=0$.

We often choose $V=0$ when $t=0$ because it's simplest, but we don't have to. That's why in general there is a $\phi$ in the equation.

Novalium Company

I think I got it.

John Rennie

8:03 AM

But as you say if we choose to set $t=0$ when $V=0$ we can just ignore $\phi$.

But anyway, you were asking what that equation meant ...

Novalium Company

@JohnRennie Shoudn't your wave cover the negative side as well?

Anonymous

Lol, I just realized I have been sleeping for more than 12 hours. It's nearly 2pm and I just woke up. I'm getting lazier everyday :D

Anonymous

@NovaliumCompany He drew the graph for $V^2$

John Rennie

@NovaliumCompany no, because it's V^2 not V

Novalium Company

You mean $V(t)^2$?

John Rennie

8:04 AM

Yes

Novalium Company

So why if we square it, the negative side is gone?

Anonymous

$(-5)^2=+25$

Novalium Company

I'm so stupid sorry :D

Got it.

John Rennie

Have you got a spreadsheet to hand? You could graph sin(x) and sin^2(x) and see how they compare

Anonymous

Just use Desmos, lol

Novalium Company

8:06 AM

Yep, I saw it (using Desmos)

John Rennie

@Blue Learning to use a spreadsheet is a useful skill for you young fellows :-)

Novalium Company

As @Blue mentioned, getting lazier everyday... meaning... Desmos :D

Ok so what's next?

John Rennie

@NovaliumCompany are we OK to move on now?

Novalium Company

Yep

John Rennie

What we're trying to do is find the constant voltage that is on average the same as our AC voltage.

The trouble is that because the AC voltage swings both positive and negative it averages to zero. But if we square it then it swings between zero and $V^2_{peak}$ and we can then calculate an average.

Novalium Company

8:09 AM

Got it.

John Rennie

So if ou start with our graph of V^2:

11 mins ago, by John Rennie

We have to try and average it.

Novalium Company

$\frac{Vpeak^2}{2}$? No? :D

John Rennie

If we integrate the function that calculates the area under the graph:

Integrating gives us the area I've shaded in pink. Are you happy with this because this is a key point?

Novalium Company

I get it but why would we want the area under the curve?

John Rennie

Well, suppose we draw our average velocity on the same graph:

The green line is our average constant voltage (I just realised I missed off the "squared" - it's V^2rms)

Now suppose we integrate our $V^2_{rms}$, that gives us the area below it

Novalium Company

8:14 AM

Sorry I don't understand what's the point of finding the area under the Vpeak^2 curve?

John Rennie

The area under the constant average V^2 is the green area.

Now what we do is choose $V^2_{rms}$ so the two areas are equal.

Novalium Company

So the green area that you've marked is the same as the pink area above?

John Rennie

Yes

Novalium Company

Got it.

So that's how you average sine waves?

John Rennie

Oh, hang on, the green area is the same as the pink area in this graph:

Novalium Company

8:18 AM

Yep, that's what I meant :)

So this is how you find the average sine wave?

John Rennie

OK, there's a a good reason why we chose to average V^2

It's because if we connect some resistor to our voltage then the power dissipated in the resistor is given by $P = V^2/R$.

So power is proportional to $V^2$

Novalium Company

Integrating V^2 will gives us Vrms^2 (which is the averaged voltage)?

user228700

Hi, everyone :-)

John Rennie

So by choosing our $V_{rms}$ to give the same area as the AC wave we make the power generated by our constant Vrms the same as the power we get from the sine wave

@KaumudiH morning :-)

Anonymous

@NovaliumCompany Just adding a bit: Area under $V_{\text{peak}}^2$ = Area under $(V_{\text{rms}})^2$. That's the method we use to find the $V_{\text{rms}}$ voltage. RMS stands for "root mean square" i.e. $$V_{\text{rms}} = \sqrt{\frac{1}{T}\int_{0}^{T} V(t)^2 dt}$$ where $V(t)=A\sin(\omega t + \phi)$. In statistics terms are read backwards i.e. "square" $\to$ "mean" $\to$ "root".

Anonymous

8:21 AM

Notice that to get the expression of $V_{rms}$ you first square $V(t)$ followed by taking its mean, followed by taking the "root".

John Rennie

@NovaliumCompany that is, suppose I connect the voltage to my kettle to boil some water. Then the sine wave and my constant $V_{rms}$ both boil the kettle in the same time.

Novalium Company

So Vrms is bascially the averaged V?

John Rennie

Yes

It's the averaged V that gives the same electrical power

Novalium Company

And we can average V(t) by squaring it and integrating it?

(which again, is equal to Vrms^2)

John Rennie

Yes

Novalium Company

8:24 AM

We take the square root to get rid of the ^2 part?

John Rennie

Yes.

Novalium Company

Got it, and what's the point of the 1/T (frequency)?

Anonymous

@NovaliumCompany There's a very subtle difference. The statistical average which you have learnt in school would translate to $V_{\text{mean}} = \frac{1}{T}\int_0^T V(t)dt$, but that is not very useful in electric circuits.

John Rennie

@NovaliumCompany when you integrate the integral is done over some time $T$

Anonymous

We specifically use "root mean square" averages in the context of electrical circuits. Not the ordinary "mean". That is, you find the average voltage which gives you the same "electrical power" per unit time, as the fluctuating sine voltage, JR mentioned.

John Rennie

8:25 AM

That is you take the area under the curve from $t=0$ to $t=T$

And obviously this area increases as you increase the upper limit $T$. What we want is the area per unit time. So we integrate from time $t=0$ to time $t=T$ then we divide by the time $T$ to give the area per unit time.

Anonymous

Oh, hi @KaumudiH

Anonymous

How's it been?

Novalium Company

I mean, before the integral, there is some thing? $\frac{1}{T}$
(Which I guess if frequency if T is period)

Anonymous

@NovaliumCompany "Area under the curve per unit time"

Novalium Company

So the T stands for time, not period?

John Rennie

8:34 AM

@NovaliumCompany correct

Anonymous

Yes. $T$ can be any finite value of time

Novalium Company

I understand T (time) in the integral part, but why before the integral it's 1 over time?

Anonymous

Unitary method :P

Anonymous

If in 500 years the area under the curve is $A$

Anonymous

What is the area for 1 year?

Novalium Company

8:36 AM

$\frac{A}{500}$?

John Rennie

$T$ is the time we do the integral for.

Anonymous

@NovaliumCompany Right. So that is $\frac{1}{500}\times A$

Anonymous

Which is why you see the $\frac{1}{T}$

John Rennie

So for the green line, which is constant, the integral i.e. the area is equal to $V^2_{rms} T$. OK so far?

Novalium Company

@JohnRennie I understand the T in the integral part, the integration is done of some period of time from time 0 to time T?

John Rennie

8:38 AM

But that means our area depends on the value of $T$ that we choose

Novalium Company

Yep?

The larged the T, more area....

John Rennie

If we equate the two areas we get:

$$ V^2_{rms} T = \int_0^T V^2(t) dt $$

And we don't want that factor of $T$ on the left so we divide both sides by $T$:

Novalium Company

$V^2_{rms} = \int_0^T V^2(t) dt$ seems fine to me?

John Rennie

$$ V^2_{rms} = \frac{1}{T} \int_0^T V^2(t) dt $$

Novalium Company

I know, but why the large T (not in the integral)

John Rennie

8:41 AM

We've said we want to set the two areas equal. Yes?

Novalium Company

Ohhh, wait a second, lemme thing, I think I'm getting it.

Vrms is a function of T?

John Rennie

No, $V_{rms}$ is a constant

Anonymous

@NovaliumCompany No, you're missing a $T$ on the left hand side. Look at the equation right above yours

Novalium Company

I know mine is wrong, but I won't understand why we multiply Vrms^2 by T (time)?

John Rennie

The area of a rectangle is base times height. Yes?

Novalium Company

8:43 AM

Yes

John Rennie

The height is $V^2_{rms}$

and the base is $T$

7 mins ago, by John Rennie

Novalium Company

I think I get it but my mind is overheating :D

The integral of $V(t)^2$ alone should give us the area under the curve?

John Rennie

The integral of $V^2(t)$ gives us the area under the red curve, yes.

And the area under the green line is $V^2_{rms} T$

And what we're saying is that the two areas must be equal

So $$ V^2_{rms} T = \int_0^T V^2(t) dt $$

Novalium Company

Ok so the integral of $V^2(t)$ is equal to the area of the red curve and is equal to $Vrms^2$, which is not the area under the green box it's just the height and we multiply it by T (which is the base) to get the area under the green box

John Rennie

>Ok so the integral of V2(t) is equal to the area of the red curve and is equal to Vrms^2

Huh?

The two areas are equal

The area under the red curve is equal to the area under the green curve

Anonymous

8:53 AM

I think Novalium means that only. His sentence structure is a bit weird :P

Anonymous

@NovaliumCompany The area under $V^2(t)$ (red curve) for time $T$ is indeed given by the integral which see on the right hand side of that equation ^

Anonymous

And you're right that $V_{\text{rms}}^2$ by itself is not the area under the green curve for time $T$. You need to multiply it with the time to get the area under the curve

Novalium Company

Oh I think I start to get it.

My brain blocked for a few mins

So Vrms is basically the averaged V(t)?

Anonymous

@NovaliumCompany Sort of. But please avoid saying just "averaged". Either say "averaged such that electrical power is preserved" or "root mean square average"

Anonymous

There are too many types of "averages". Be specific

Novalium Company

8:59 AM

Yep, I think I understand the equation now, but I'm not sure :D

Anonymous

Try to solve some problems maybe

Novalium Company

Where?

Anonymous

School textbook?

Novalium Company

Oh wait, my book provides :D

@Blue I'm still 9th going to 10th grade and we probably won't study these stuff.

I don't think we'll even cover calculus.

Anonymous

I think you should download the NCERT grade 12 physics book and practice problems from there

Anonymous

9:03 AM

It's difficult to learn these things without practice

Anonymous

See the chapter on Alternating currents

Novalium Company

I have some problems to solve in the book (Practial electronics for inventors)

But I'll give it a look :)

@Blue and @JohnRennie thanks so much for all the help! I'll head for lunch now, see you :--)

Anonymous

Bye!

John Rennie

Enjoy lunch (it's flippin' hours until my lunch! :-)

user351417

@JohnRennie Ah okee... You familiar with the system in the US by any chance? Of course, it'll probably be pretty different for different universities (a quick look at the course requirements for a couple of places showed me that), but what about the way in which people 'look at' people with BAs? Like do you ever feel "Aw, poor guy only has a BA; he probably doesn't know enough physics"?

John Rennie

9:08 AM

I don't know about the US I'm afraid.

In the UK traditionally the older universities call a science degree a BA while the newer ones call it a BSc.

In the US does the same university offer both BA and BSc courses in physics?

Anonymous

@Chair There might be something of relevance here: physicsforums.com/threads/ba-vs-bs.193362

Anonymous

"Aw, poor guy only has a BA; he probably doesn't know enough physics"?....that's a funny way to judge people tho :P

John Rennie

If Princeton only offers a BA in physics I think that's pretty convincing evidence that a BA is as good as a BSc :-)

user351417

@Blue That physicsforums thing is the one that made me consider the way in which people perceive BAs, and I thought of asking here to see if that's actually true :) I was actually shocked by the fact that UChicago only offers BAs too. They have Fermilab and stuff; it seems like there's lots of important HEP stuff going on there.

user351417

So @JohnRennie you haven't seen people kind of looking down on BA people?

John Rennie

9:17 AM

@Chair not in the UK. I have a BA no a BSc because Cambridge gives BAs.

And an MA not an MSc come to that.

In the UK it is a meaningless distinction.

Anonymous

@Chair From that thread at least it seems that you should focus on getting into a good school with decent research facilities rather than worrying about the name of the degree. But then, I'm not merkan ;)

user351417

Cooilio. Thanks! It seems like a lot of the older places give BAs only: even Rice (apparently) user to give only BAs for a long time. However, I saw something on the website of University of Iowa (which offers both) that BAs in physics are meant for people going into management, so whatever.

Balarka Sen

PhD is literally Doctor of Philosophy

low-IQ degree

Who even does philosophy??

Only people who don't understand science

Anonymous

Everything is philosophy dude

user351417

Philosophy SE actually has some really fun, cool stuff, although that's not a good measure of whether philosophy's a legitimate discipline.

John Rennie

9:23 AM

@Chair if a university offers both a BA and a BSc in physics then I guess there's a difference. But that applies only to that particular university and its courses.

Anonymous

@Chair I see you're from India. ~~Gave~~ Took the SAT this year?

Secret

> I do not understand science

user351417

@Blue yep, I took it in march. I'm probably applying to the US though.

user351417

In India, there isn't so much emphasis upon research these days. It's primarily engineering.

user351417

Like it's usually assumed that if you're into physics/math, you're thinking of engineering.

user351417

9:32 AM

@JohnRennie Great, that clears things up. It makes sense since the list of courses needed for a physics BS is pretty different for almost all places.

Anonymous

@Chair Good! I think that's a good choice :) Just one advice: look up the faculty profiles and choose universities which have good faculties in the areas of your interest. For instance, some uni might have a good condensed matter group but you might be interested in astronomy instead.

Anonymous

@Chair That's pretty well known in this part of the Internet :P

user351417

@Blue I was leaning towards UChicago because they have Fermilab next door and I'm leaning towards HEP, though I'm not completely sure yet. Also, their essay prompts are glorious and I want to write those. Is it actually worth choosing a university because of 1 or 2 particular faculty members you're interested in? So you actually get the freedom to choose exactly who you'll do some research under or whatever?

user351417

But UChicago's pretty seriously hard to get into, so whatever.

user351417

9:47 AM

I think it's got an acceptance rate of 8% or something.

Balarka Sen

the feedback loop goes both ways. students are more interested in getting an socio-economically rewarding job than where mainstream research will lead you, therefore most students are going to engineering and medical sciences than mainstream science and the fraction of the population who are interested in mainstream science and can afford - not just economically - to study abroad are leaving the country,

therefore there are less interest in mainstream science in the country, and less government funding, therefore there aren't as many economically rewarding jobs in mainstream science, therefore....

Anonymous

@Chair "Is it actually worth choosing a university because of 1 or 2 particular faculty members you're interested in?" - absolutely. Recommendations from top faculties in your areas of interest will help you a lot of you're looking forward to pursuing graduate studies.

John Rennie

@Chair I'm not sure it makes that much difference. As long as you choose a good university I'm not sure which university you choose makes that much difference at degree level. Where it really matters is for your PhD.

Anonymous

But well, another factor to keep in mind is the cost of living and tuition there. Make sure you get good amount of funding in terms of scholarships, etc. During my time only a few schools offered 80% or more financial aid.

Balarka Sen

the reason everyone thinks being interested in math or physics automatically correlates with apting for an engineering career is because the indian society, backward like a perfect example of a third world country as it is, has integrated the idea that it needs to survive purely by spinning the feedback loop around itself

but there are no perpetual motion machines. the laws of physics applies to everything universally; this wheel will just drain out the energy it's running on

Anonymous

9:54 AM

@BalarkaSen That's a perfect description. But things are gradually changing

Balarka Sen

hence my last line

user351417

Yeah, strangely enough, a decent number of people in my batch seem to be less focused on engineering/medicine.

Anonymous

Fwiw I think the IISERs and IISc are doing excellent nowadays

user351417

IISc seems to be diluting the emphasis on pure science.

Anonymous

@Chair Why do you think so ?

user351417

9:56 AM

They introduced an undergrad course (though it's only 120 people or something), and if I'm not mistaken, there's something related to engineering being set up aswell.

user351417

Yep, just searched for "iisc engineering", and found this: iisc.ernet.in/ug/engg.html

Anonymous

Expanding the scope of courses doesn't necessarily mean that they're diluting pure science courses

Balarka Sen

It's suspicious though isn't it

user351417

I@Blue I dunno, I think that if they're no longer only pure science, that's the same as diluting the emphasis on the research.

Anonymous

Research doesn't only mean research in pure science

Balarka Sen

9:58 AM

It's as if not enough people are applying to IISc so they are slightly broadening their scope and goal of being a national institute for mainstream science to include a bit of engineering and applied stuff :p

Anonymous

There can be research in engineering too

user351417

@Blue that's a good point. I tend to misuse the word 'research'.

Anonymous

In fact I like an interdisciplinary approach to science and engineering

Anonymous

They have a great math faculty too

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