GooD MorninG..!

@JMac The same happens to me. That's why I have stopped sleeping at any other time in the day except the night. Though I did once or twice sleep during the afternoon intentionally to gather "inspiration" (because when I wake up, my mind works quite differently, it feels as if I got high and then became normal again).

@123 Good morning.

@ACuriousMind And that is the time my mind comes up with totally weird ideas. I mean, this occasionally happened when I used to read/study in the afternoon. And my mind used to apply classical mechanics to find the double bond equivalent of an organic compound (weird stuff, you get it, right?), and then when my senses returned, I would have totally forgotten how I did it. I guess I might even have discovered the theory of everything in those half asleep moments (jk).

@FakeMod yo, what happened to your rep?

@ACuriousMind one thing i heard that works is drinking coffee right before taking a nap

two people told me it makes them wide awake when they get up

@satan29 I deleted my account during the last few months of JEE, and resurrected it after JEE was over. I have done this deal of deletion and recreation quite a lot of times in the past :)

@SirCumference How can one sleep after drinking a coffee? It make me "wide awake" even before I sleep.

ah, i see. cool :)

@FakeMod i think ya gotta nap before it kicks in

i guess if you're extremely tired then it might be viable

@SirCumference Hmm... I will try your coffee idea, if I need to sleep and, also, wake up awake.

theoi.com/Ouranios/MousaOurania.html

Urania doesn't have a lot of informations

How am I supposed to make a sacrifice to the muse of astronomy for general relativity powers

I can find some rituals, but they're all modern ones

neosalexandria.org/rituals/other-rituals/ritual-for-the-muses

I only want authentic ancient greek ones

Only elements are 1) lyre-ruling 2) golden headband 3) some hymn

Fairly short list

How can we create uniform vector field???

I know E of infinite sheet and between two oppositely charged plates. But why it is uniform everywhere in space. Because E depend on r.

Simple enough

just write down a uniform vector field

And compute its divergence

although it's not guaranteed that it will be a proper EM field

There may be no combination of electric and magnetic fields that give such a field

@Slereah :P GooD Answer. Physics way how E between two oppositely charged plates not depend on r

How can we explain bus accelerate because there is no external force which act on the bus??? Like newton's third law involve second object causes acceleration.

Another example is rocket acceleration moving upward. There is no external force act on it.

A rocket is simple enough, it ejects a bit of its mass in one direction

By conservation of momentum, this means that the rocket itself must accelerate in the other direction

I understand conservation of momentum is a explaination of this example. How can we describe this motions using newton's three laws

The simplest case is to consider it as two point masses

Where is external force

One representing the rocket, the other representing the gas

You're roughly doing an elastic collision

Ookay. pls explain

The gas molecule will bump into the rocket, and go the other way

The gas momentum is transferred to the rocket

What about bus?

Bus is a bit more complex

Engines and wheels

Although the rough idea is still the same, ie gas expansion is converted into motion

But this is not following external force idea

It is, if you consider the gas and the bus as two different systems

A bus is the motion of many different parts which you can consider as independent systems

and then it's just a transfer of such motion from one part to the other

Ookay Thanks.

It's fundamentally the same idea as throwing a ball at something

ok. got it

yo

afternoon

yo

@Charlie hi

Do you have a mood to discuss physics. I am struggling to understand conservative force idea

Quick question about a procedure just done in P&S, we want to isolate the interacting vacuum $|\Omega\rangle$ and we find an expression for it in terms of the free vacuum $|0\rangle$, does this imply that we are treating the interacting and free theories as existing in the same Hilbert space?

I could understand why this would be necessary if we don't fully know what the interacting Hilbert spaces are.

:P i am asking a question to you. You are also here to ask question

:O

@123 What do you not understand about conservative forces?

@Charlie Yes, and this is nonsense by Haag's theorem. Physicists usually don't care and it seems to work anyway :P

Ah I seee

$\vec{F}\cdot\vec{ds}$ here we take $\vec{s}$ in the direction of force.

right.

Does the Millenium prize regarding the Yang-Mills mass gap relate to rigorously constructing interacting QFTs?

I've read a bit about it but don't really know enough to say

@123 $\vec ds$ is not the direction of the force, it is the direction of travel through the force vector field

It is dot product. dot product means in we take the angle between two vectors and project it any one of the vector.

oh sorry you said "in" the direction of force, not "is" the direction of force

well the answer is still no

why???

i know inner product . but the idea is same

What you're doing is, at a given point, taking the the force vector $\vec F$ at that point, and taking its dot product with $\vec s$, the vector pointing in the direct of travel at that point

Yes at every point. but this is the same meaning what i said about every point

Then when you integrate along a path, you are effectively doing this at every point along the path and adding up the resulting dot products

When you said "we take $\vec s$ in the direction of force", that was incorrect, $\vec s$ is a vector pointing in the direction along which you are travelling through the force

Yes. I understand what integration does. i am confirming it is same result which we are getting from 2D.

Sorry my bad english.

I'm not sure what you mean "from 2D"

That procedure works in all dimensions

$\Re^2$

yes i meant to say that idea is same for all dimensions.

Oh no I know what you meant by 2D, but not "the same result which we are getting from 2D", that phrase doesn't make sense to me

The dot product is defined basically the same way in $\Bbb R^2$ and $\Bbb R^3$, so the procedure is the same

@Charlie means. In dot product in $Real^2$ or $Real^2$ we take one vector projection to another . right

yes

How do write Real in latex

\Bbb R^n

$\Bbb R$

$\Bbb R^n$

Thanks

As per this dot product analogy. we are just looking Force and arc length vector in the direction of force. right

Hi @Az

I don't know what you mean by arc length vector in the direction of force

Hi @Azmuth

@123 Hi :)

Arc length mean $\vec{s}$

What you're doing is drawing a line through $\Bbb R^3$, then at each point along this line you're taking the dot product of the tangent vector to that line with the force vector at that point.

tangent vector

I can not see upload option. I wanted to share a picture

There's an upload button just to the right of the text box

If this the case we in this $\vec{F}\cdot\vec{s}$ we are just interested in distance curve of tangent vector only in the direction of force. Here i want to confirm one thing.

Upload button was there yesterday. But today i have two buttons MathJax and send

oh

rip

It is difficult to tell without showing the pictures.

How can upload?

I'm not sure sry

See this image click it

ok I see it

The point is that both of those lines result in the same work done against gravity, since it is a conservative force

If curve go downward from starting point let which give negative work as we move upward it is positive by the same amount. It cancels out every point below from starting point. Same is true for ending point curve.

This is correct or not?

Yes

The point is that for every bit of energy "spent" going against the gravitational force the same amount is "gained" when going with the gravitational force. The net result is that both lines do the same amount of work

this is a feature of conservative vector fields, i.e. those for which $\nabla \times V=0$.

The curve between stating and endpoint give the same result as straight curve. Because this is the direction of force. and we are looking for curve tangent vector in the direction force. It gives the same result.

This is also true?

I'm not sure what your second and third sentence mean, but the first one is correct

Any two lines that start and end at the same point that travel through a conservative vector field will do the same work

@Charlie I want to say we have to segments joining starting and endpoint. One is straight and other is curve. Now i conclude what happened to work above and below the endpoints. Now i am figuring out between endpoints

As long as the starting point and the ending point are the same, all possible curves do the exact same amount of work against the field

The straight line segment is also the direction of force. If we take dot product between these two is same result which is given by the curved one with the force. Because we are taking dot product here.

@Charlie Yes because any thing below and above the endpoints cancel out completely and between curve by dot product we emphasize that just take the component of curve in the direction of force.

Is that true?

If I've understood what you've said correctly, then yes :P

:P what you don't understand from my bad english

:O

Your english is fine, it's just trying to understand your reasoning behind what you believe to be true

The picture i shared i was talking about this. In this picture i have divided curve in three parts.

1. Curve below the starting point

2. Curve above the endpoint

3. Curve between endpoints

Points 2 and 3 actually are equivalent but still

First I want to digest your point no 1

the i want to move your point no 2. This is also needed understanding.

The point is that any line that isn't directly from $A$ to $B$ adds up in such a way that it cancels itself out until it is equivalent to the line directly from $A$ to $B$.

If we take $\vec{F}\cdot\vec{s}$ Anything below and above endpoints give the result zero. Is that true?

Note that in the picture you linked, it doesn't matter if you first travel in the opposite direction, you end up having to go back up through the force which cancels out the energy "gained" by travelling with the field

@Charlie Ahaaa. you are right because we talking dot product between them.

$\vec F\cdot \vec s$ is not equal to zero at all points below the starting point and above the end points

@Charlie Whyyyyy????

This is equivalent to saying that $A+B=A+5+B-5$.

Because the field isn't equal to zero below the starting point and above the ending point, what we're saying is that they cancel out, not that they are both equal to zero

No it should be zero let say when we go down from starting point (bottom one i am taking starting point) it give us some scalar number then we move back to starting point up it give us another number which should additive inverse of same number when giong below. the sum to be zero...

Yes, what you've said there is correct

Thanks otherwise my mind blows up.

The total $\vec F\cdot d\vec s$ moving down and then back up again is zero, but at each point $\vec F \cdot d\vec s$ is not $=0$, which is what it sounded like you were saying

The big thing about work being independent of path is that it doesn't really matter how you end up in your final position. The potential energy just depends on where you are in the field, not the path you took to get there.

I did not want to say at each point.

I think you understand, it was just a miscommunication

yeah

@Charlie yes yes. thanks i was just saying any curve below and above endpoints add up to zero

Well, a special property of conservative vector fields is that the net work done by any closed loop is zero

The curve between the endpoints is same as we take it as straight line joining the two point this is we give this task to dot product to do this. right.

@Charlie This is why is zero because it is sum up to zero.

If I understand what you're saying, yes

hahaha.... :P

Now i am curve only between enpoints.

In a closed loop, if you add up the dot product of the tangent vector to the curve and the force field vector at each point you get zero

For a closed loop

in a conservative field

In close loop i understand. what if it is not close loop. The just go below and achieve the same height.

In my picture when you only see curve below starting point.

which goes down first then up and stop it where it has same height as starting point.

In a field like gravity that is just in the vertical direction, your horizontal position doesn't change the work done; but other conservative fields can depend on your horizontal position as well.

In this picture we end up at the same "height" above the ground, but the net work is not zero

@Charlie Yes what about this

The point is that the gravitational field is (assumed to be) homogenous

Yeah that's a good example of not only a horizontal field where horizontal distance does matter.

OoooKay...

what about my link

whould it be zero or not by same height

Yes, because the field is homogenous

Good example. Because in your example it is seen it is not zero.

But in my example it is not a closed loop.

But note that if the curve were closed in my example it would be zero

Yes, it doesn't matter that the loop isn't closed in your example because the field is completely homogenous, in my image the field isn't homogenous

we can say that if field is homogenous by achieving the same height also give result zero.

Well, homogenity has a direction associated with it

It's also worth noting that there are still a set of equipotential lines in that example where there are several locations where potential is the same and the loop isn't closed; but the field is more complicated so those potential lines aren't just horizontal.

^

Thanks @Charlie . Now i have few more questions. Why we take dot Force with curved tangent vector. Is there ant physics way interpretation or benefit?

We take the dot product of the force with the direction of travel because that is by definition how we calculate work done

@JMac Hi. Yes you are right. what is benefit of equipotential surface?

If you moved side to side in your example, the dot product is 0 because it doesn't move in the direction of force.

@Charlie Problem is that. We can feel displacement, time, mass, velocity , force, acceleration. How we feel energy work. KE PE.

@123 It shows a set of locations where the potential energy is the same even though position is different in a conservative field hyperphysics.phy-astr.gsu.edu/hbase/electric/equipot.html It's the red dashed lines in those pictures. Gravity is basically like a rotated version of that constant electric field example

I'm not sure what you mean by "we can feel" those things

Feel means we can measure distance it is real life experience also for time, force etc.. What about KE, PE, Work done

We can only measure changes in energy, we cannot directly measure the absolute "energy" of a system

you can see there are lot of lot of questions about energy in this forum problem is same, this idea doesn't have any real life experience people actually don't feel it visualize it in their brain.

Is there any intuitive way of defining or understanding KE PE work-done???

I don't like the word intuitive because I don't know what you consider intuitive, but energy isn't something you can visualise afaik, it's a fairly abstract concept

afaik = as far as I know *

Yeah energy seems "intuitive" enough to me; but it's still abstract as well.

What happened to the system. If system has more energy or less energy any explanation. So we can create comparative model of energy.

Again we can only measure differences in energy, not absolute values

This guys videos are incredible

We can say a particle has "more" kinetic energy if it has a greater value of $\frac{1}{2}mv^2$

Those god damn drawings of lines make some sense now...

@Charlie Energy is abstract that's why people will struggle to understand it and have a lot of lot of question just about this topic.

sure

@bolbteppa Thanks

I have seen this about 3 years ago.

Yo EVry 1 here

What is the difference if system has low and high energy?

@Azmuth Yo

What do you mean "difference", what kind of difference are you expecting?

@123 High energy atomic/molecular systems are unstable and seek to stablize by discharge of energy

Yeah, note that that is again talking about relative energy

Any example related to mechanics?

An object with high potential energy seeks to lower it by converting it to other forms...

@Charlie :P

The basic way to think about energy I've usually found good is "the capacity to perform work". That capacity allows systems to perform work on each other and change their kinetic and potential energies. The changes in kinetic energy can look complicated on the molecular scale and smaller (like heat); but that's basically what is going on.

This one on pions and kaons is stunning, drawing in the hidden curve explains everything

@JMac Ah, yes, that's a better explaination.

I have seen all the videos at this channel.

Beside math. Is there any mechanics example at which we can say two objects having same mass but one has high energy and another has lower energy?

Yes, two particles, one travelling at greater velocity than the other

You are again talking about relative energy, which is perfectly reasonable to define, what isn't easy to define is "absolute" energy because that requires you to define a "zero" energy

Another example: One mass is sitting still on the floor. The other mass is sitting still on the shelf several feet above. The higher mass has more energy in the presence of gravity.

@Charlie hahaha.. You are right. So i can say that energy is another kind physics which written in different formulation (like LM). but there is transformation between NM and LM

Everything is relative! Time, Life, soul, position and stupidity..... Well I'm not sure about Stupidity...

I'm not sure what you mean "energy is another kind of physics"

Lagrangian and Newtonian mechanics both have a concept of energy

@Azmuth :O

A (quantum) mechanics example is mirror nuclei (particles with same overall number of protons and neutrons) almost have the same mass and the energy spectra look almost the same but the one with more protons has higher energies due to the higher coulomb repulsion :p

I almost got an heartattack with that ping sound.... speakers were at 150% :P

@Azmuth Lower the sound ;P

@Azmuth I don't think that's how %'s work?

Linux Does!

aslimixer plugin to amplify sounds! :)

Oh that's not your speakers though.

@Charlie It means energy is just mathematical doesn't have any real life experience?

Mine, but with a different name

Well, energy is a physical thing, mathematicians don't ever talk about energy so it's definitely physical

@Azmuth Well your speakers often have their own sound control; you generally can't set that past 100 because it physically can't go higher. You can crank the gain on the computer side though.

@Charlie I protest.

@JMac crank the gain on the computer side though That's what ASLIMIXER does

@Charlie :P because students ask questions we we can't experience energy. they can experience distance ,time etc..

@Azmuth Mute it

Sure but that doesn't make it a purely mathematical thing :P

@JMac It increases tho!

@123 Done!

can anyone check if colab.research.google.com is down now or the link is not opening for me.

No. @Charlie gave me many many good ideas. your participation is also needed. :P

@123 I have a lot of practical experience with energy. I took an entire course called "Energy Management" which was basically about tracking energy through different processes and looking at ways to maximize the amount of useful work from the energy. So things like processing plants, you can trace where all the energy comes from and goes, and how much of it actually gets used for your main goal(s).

@Charlie Forget it. Pls explain why curl is zero in conservative force?

@JMac Great so can have discussion on this?

Because a conservative vector field can always be written as the gradient of some potential scalar field, $\vec V=\nabla f$, it is then a mathematical identity that $\nabla \times (\nabla f)=0$

Sure, what exactly do you want to know? I feel like it's something I just got more and more intuition about the more I studied physics.

"The curl of a gradient field is always zero"

yep, it's a scalar field

with straight lines

the scalar fields can have curved lines

for example that surrounding a central potential

I see...

@JMac My questions are same. we can clearly explain the object at earth surface or top of table by gravity distance etc.. this is same explanation which we give in chapters before work and energy. how do we connect students the same example with the idea of energy?

temperature is generally a simple measurement of (molecular) energy.

@Charlie You are right . Now if see if we already took dot product between two vector there is no perpendicular component remain. Thanks

...which reminds me of neumaiers "thermal interpretation of QM"

A Good paper ML ft. Ransom after a long time

Does my explanation correct about curl? @Charlie

The cross and dot product are definitely different operations, I'm not sure what you meant

If I can write a vector field $\vec V$ as the gradient field of some scalar field $f$, it is a mathematical fact that $\nabla \times \nabla f=0$

@123 The way I found was good was just getting comfortable with the basics of energy you see in physics books and have some understanding of the calculations. Then after that you can look at things like thermodynamics and heat transfer which highly involve looking at how energy flows between systems and the effects.

you said curl of gradient is zero

i take gradient as dot product and curl as cross product.

Gradient is not the same as the dot product

You can't take the curl of a dot product

Vector algebra is important while learning about Maxwell's eqnX :)

@Charlie You written earleir $\del\cross(\del\cdot\pi)$

\nabla is the correct symbol

and for cross product is small x

Note $\nabla f\neq \nabla\cdot f$, the second operation is not defined

for cross product the symbol is \times

Ahh ok ok i used it several times my fault thanks

@Charlie Yes this is not same.

The cross product is defined $$\times : V\rightarrow V,$$ the dot product is defined $$\cdot: V\times V\rightarrow \Bbb R,$$ and the gradient is defined as $$\text{grad}:C^2(\Bbb R)\rightarrow V.$$

OK i saw again your expression you did not used dot product.

Yes sorry mistakenly i wrote that.

short notes on Vector Calculus

Is it possible to give an example for KE, PE and work to students.

more speed means more KE . But speed can be visualize energy don't

Not sure what you mean

$\vec{P} = m\vec{v}$ and $KE = \frac{1}{2}mv^2$

@123 Use Graphs for visualisation, students will understand it better.

Don't get hung up trying to visualise energy

momentum can be defined as quantity of motion. Like the more mass object has needed more force to stop same is for velocity.

I have given explanation about momentum friend.

Not all energy is really easy to "see". Consider a pressurized container. From the outside, it looks pretty normal and not high energy; but then if you open a hole in it; suddenly all that stored energy escapes. You can't really visualize the energy until it does something else. I'm not sure if trying to visualize it like motion helps.

but energy also has mass and velocity terms in different way.

explaining some other quantity of same object at the same time. What is this quantity in physics way?

@JMac Problem is that energy is derived from work. The question is same then what is work.

@Charlie visualisation questions do frequently appear in JEE

How does one ask a question about visualisation

Like it is the motion only in the direction of force.

@Charlie graphs... They are pretty tricky and hard

@Charlie Let me give you an example frnds.

Sure but that's not a visualisation of energy itself

which is what it sounds like you're asking for @123

@123 $W= \sum_i \vec F. \vec s_i$

Note for conservative forces, $\oint \vec F. d \vec s = 0$

@Charlie These kinds of graph questions - studyadda.com/question-bank/jee-main-advanced/physics/…

If we pull a box on floor at some angle amount of force which parallel to the direction displacement is working to move the box (say x-axis) and perpendicular (y-axis) component of force balanced by box weight.

sorry wrong link

@Charlie this one

@123 check that link above^

@Azmuth You are right . But i am talking about any curve.

But these aren't asking you to visualise energy itself though, which isn't possible, they're asking you about how energy changes in certain systems :P

@Charlie But that greatly helps in improving understanding.

oh sure, but it's still not visualising energy itself

Does anyone here plays Among us?

somewhat very close to visualisation

@123 Work is essentially the quantifiable change in an objects position in a conservative field and/or a change in it's relative motion; which can be quantified described and equated using energy.

If take an example of projectile the only force acting on it is gravity.