The h Bar - 2018-06-16

« first day (2781 days earlier) ← previous day next day → last day (2142 days later) »

Anonymous

4:46 AM

@pleaseunban0celo7 Yo

5:34 AM

pretty please

I now just realized something cool. It is unbelievable that I was so stupid back then(several years ago) I spent 3 months thinking, and a year or two after about it and I could not see it

Hopefully, I finish this little thing in the way, and then embrace my destiny and start doing cool things

I think the time I spent thinking about this stuff was actually well spent

I can't believe, I was too stupid to realize this, but I had been pushed in the right direction 3 or four times. Once in NorCal, once in SoCal, Once SoFr, and Once in Germ . . .

This is it!!!!

It all makes sense now!!!

This is the holy grail

!!

6:20 AM

Ok so I just figured things out (at least the easy stuff aka academics and work)

6 hours later…

12:22 PM

user image

The board is back

What, and you haven't yet scribbled some GR stuff on it?

Well I just put it up :p

I shall use it to finish that bloody idea on thin-shell wormholes

Also maybe that stupid idea about the $S^7$ spacetime

But that is for later

Also I probably need to buy some pens back because it's been a while since I used it

I couldn't put it up in any rented place

3 hours later…

3:02 PM

@Mithrandir24601 Does anyone know how he died?

Novalium Company

3:54 PM

Guys, I finally understand derivatives, how to solve them and what they are, but what is the purpose of the derivatives? Why would I want to know the slope of a tangent line on a function graph?

The derivative of position is velocity

The second derivative of position is acceleration

One of Newton's laws says that $F = ma$ governs all of existence, so derivatives are related to the meaning of life

1 hour later…

5:08 PM

in Mathematics, 11 mins ago, by geocalc33

Given the function $ \Phi(s,x)=\zeta(s)^{1/\text{log}(x)} $ how do you solve for $s$

2 hours later…

6:39 PM

@NovaliumCompany They tell you the instantaneous rate of change of a function

7:35 PM

8

Q: Why is the derivative important?

Derivatives, both ordinary and partial, appear often in my mathematics courses. However, my teachers have never really given a good example of why the derivative is useful. My questions: Other than the usual instantaneous rate of change, what are some common uses of the derivative? What does t...

derivatives soft-question applications motivation

8:15 PM

@NovaliumCompany Calculus (roughly) is the study of change, in the same way geometry is the study of shapes and algebra is the study of operations. The tools you gain from it are very powerful when working with changing quantities

For example, suppose you know an object's position as a function of time. If you wanted to know its velocity as a function of time, you can just differentiate the position.

Velocity and so on are common physical examples of the derivative being used, but the derivatives come up in tons of different places

Novalium Company

8:53 PM

So with the derivative, I can know the velocity at any given time on a position(x)-time(t) graph?

Hmm, can't I just divide the position x by the time? Or maybe that will give me the average velocity, not the exact one at the given time?

@NovaliumCompany Yes, that will give you the average instead.

Novalium Company

Quick question: The second derivative is just taking the derivative of the first derivative?

Yup

Novalium Company

Wow, I love derivatives. Also I tried to look at integrals and it just says that it's the reverse of derivatives and using integrals you can find the area under a curve between to points on the horizontal axis. Well, can integrals relate somehow with the pos-time graph or there is some other usage for integrals in physics?

@NovaliumCompany Yep. By the way, you can think of "instantaneous velocity" at a point as finding the velocity over the shortest time period possible

Novalium Company

9:02 PM

Yep, makes sense.

@NovaliumCompany Well, the integral of velocity is position

I.e. distance per unit time, except we make the period of time infinitely small

@ACuriousMind *distance

Position is the antiderivative

@NovaliumCompany Basically, there's two concepts you should know: the antiderivative (sometimes called the "indefinite integral", though this is a bad name), and the integral (sometimes called the "definite integral")

...why is "indefinite integral" a bad name? That's precisely what it is - an integral where one of the bounds is variable, i.e. indefinite!

@ACuriousMind The antiderivative is the general solution to the ODE, $dy/dx = f(x)$. The integral is defined very differently from derivatives

Using an integral where one of the bounds is variable will get you a particular solution, but you can never recover the constant lost via differentiation

That's why it gets you "distance", not "position"

Novalium Company

And after all, why would I want to know the area under a curve? How does this relate to anything?

9:06 PM

@NovaliumCompany Ah, good question. Loosely speaking, integrals are like "better multiplication" under some circumstances

Think about how multiplying a number by another number gets you the area of a rectangle. Well, integration lets you do that when one of those "numbers" is a changing function, and the other is its variable

In that sense, you can use integration as a type of "multiplication" when one of your numbers is changing with respect to the other

@NovaliumCompany Whenever the graphed function is a "rate" of a flow something (like flow rate of water through a pipe, or electric current, etc.), its integral over some time period - the area under the curve - is the total amount of something that was transported during that time by the flow

@ACuriousMind I'm hesitant to introduce it as an opposite to derivatives. That's kind of true, but falls apart in cases like multivariable calculus, where not all integrands are rates of changes of something

E.g. there's no fundamental theorem of calculus for surface integrals

@SirCumference Uhhhh...I'm not convinced that's a helpful way to think about it - multiplication is a commutative binary operation, integration is not.

@SirCumference Sure there is, it's called Stokes' theorem.

Novalium Company

@ACuriousMind Why would you divide by time? Shouldn't it be the integral times the time?

@NovaliumCompany Who said to divide by time?

9:10 PM

@ACuriousMind Eh, all right, but another example is how gradient theorem only works on conservative fields

Novalium Company

@ACuriousMind "its integral over some time period"

@NovaliumCompany Oh, lol, that's not what that means

@ACuriousMind It gives a sense of when to use it for physical applications. It's like using the distributive property of multiplication to add up a bunch of products, except we are changing the terms

When one says "the integral over X" - where $X$ is an interval - that has nothing to do with division, it means that you are basically computing the area under the curve in that interval.

It's a loose explanation but should be helpful on when to use it

Novalium Company

9:12 PM

I thought when people say over, they mean division. I've heard people say, Velocity is Distance over Time.

Integration is not quite the opposite of differentiation, because the latter doesn't care about functional values whereas the former does. Differentiation only cares about how a function changes. So you can never recover the actual functional values with integration, only another function with the same curve.

Again, this is why the distinction between position and distance is important

@SirCumference So what? Being a conservative field is the 1-d-higher equivalent of "having an antiderivative", and the fundamental theorem likewise only makes sense for functions with an antiderivative. What changes is that "having an antiderivative" is a stronger condition on tensor fields than on functions of a single variable.

@NovaliumCompany One word can have multiple meanings :P

Novalium Company

Meh, ok :)

@ACuriousMind My point is that the fundamental theorem of calculus only relates the two when the integrand is a derivative, or higher dimensional analog. Integration is not reliant on the definition of the derivative.

Saying they are "opposites" is misleading

Fundamental theorem of calculus

The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function. The first part of the theorem, sometimes called the first fundamental theorem of calculus, states that one of the antiderivatives (also called indefinite integral), say F, of some function f may be obtained as the integral of f with a variable bound of integration. This implies the existence of antiderivatives for continuous functions. Conversely, the second part of the theorem, sometimes called the second fundamental theorem of calculus, states that...

"The fundamental theorem of calculus relates differentiation and integration, showing that these two operations are essentially inverses of one another. "

9:16 PM

@bolbteppa "essentially"

@SirCumference I don't think anyone ever implied that the integral is "reliant on the definition of the derivative".

But that doesn't mean they're not opposites :P

You cannot "undo" a derivative because it is nonunique

You can recreate the original curve though

That's why the theorem is framed in terms of definite integrals

The problem with thinking of them as inverses is about the functions you can apply each to, e.g. you can integrate continuous functions, can't differentiate all continuous functions

This thinking motivated distributional derivatives

The fundamental theorem boils down to "the sum of all the changes in a function equals the net change". It cannot give you the original functional values

I think I'm restating my point

@SirCumference I'm not sure why you are willing to accept "integration is like multiplication", which I don't really agree with, but are getting hung up on "integration is like the inverse of differentiation". The latter is true in a much more precise sense (i.e. the fundamental theorems) than the former.

Sure, it's not exactly an inverse, but no one ever claimed that.

9:19 PM

@ACuriousMind I'm not religiously accepting the former, it's just more indicative of when integrals are physically relevant than "the area under the curve"

Novalium Company

So guys, If I have a velocity-time graph, using integrals, I can know the position at any given point in time? And using derivatives, I can know the acceleration at any given point in time?

I'm just saying that treating them as "inverses" will confuse a lot of students on the difference between "definite" integrals and antiderivatives, which becomes more important later on

@NovaliumCompany You can know the distance covered at any point in time. What @SirCumference was saying is that you cannot recover "position", since the velocity does not contain information about the starting point.

@ACuriousMind Yup

Novalium Company

Yep, makes sense. So thanks then :)

9:21 PM

Sorry for sounding a bit pedantic, but the fundamental theorem becomes a lot more beautiful imo when you think of things in that way

Novalium Company

What will happend if I take the integral of a position-time graph?

The only way the fundamental theorem becomes beautiful is when you realise it's just a special case of Stokes' theorem :P

@NovaliumCompany You get a useless quantity :P

@ACuriousMind I don't know the generalized Stokes' theorem :( But I still think "summing all the changes equals the net change" is pretty beautiful

Novalium Company

Haaha.

I thought so :D

What about taking the derivative of acceleration-time graph? Jerk?

@NovaliumCompany I'm sure some engineer somewhere is interested in jerk, but generally it's not a very useful quantity, either

9:24 PM

I mean technically it is, but you won't really worry about it

Novalium Company

Hmm, ok then. Thanks so much guys, I'll be going to bed now :)

That's because the equations of motion are second order, i.e. involve position and its first and second derivative. Other (anti-)derivatives don't feature, so they don't play any important physical role

I actually have to go back to work, adios

Mithrandir24601

9:44 PM

@SirCumference All I know is that he died peacefully and it wasn't that unexpected as he'd been getting much worse over the past 5 years or so

10:25 PM

So, apparently a headline-news black hole has either 20 solar masses or 20-million solar masses. Turns out the latter looks consistent with the researchers' report, but jeeze.

« first day (2781 days earlier) ← previous day next day → last day (2142 days later) »

Transcript for

Jun '1816

The h Bar

General chat for Physics SE (physics.stackexchange.com). For M...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

site design / logo © 2024 Stack Exchange Inc; legal

mobile