The h Bar

Sir Cumference

00:49

@Secret "$dd \over dt$"

dmckee

03:35

@Knight The thing that increase is the quantity $\gamma m$. This is sometimes called the "relativistic mass". But in the modern framing of the subject we simply don't assign a name to that quantity and don't frame explanations in terms of it.

Instead we use "mass" to mean only the invariant mass (what would be called "rest mass" in t he old nomenclature).

The big advantage of that approach is that "Invariant" means "the same in every frame", and it is also easy to compute in every frame: just find the square of the four-momentum

This is basically a shift in terminology to follow a shift in focus away from "that which is weird and unexpected if you know Newtonian physics" to "that which is clear to explain and easy to compute with on it's own terms".

Both systems give the same results, but I would argue that it is easier to confuse yourself with the old system. (In fact I have argued this is presentations at minor conferences...)

user8718165

Hello sir @dmckee

dmckee

Hello.

user8718165

@dmckee can I ask you a question please?

@dmckee sir, I have a question about forces.

dmckee

03:56

@user8718165 Don't ask about asking. Just ask.

user8718165

@dmckee sir, if an object is moving in +ve x direction, and I suddenly apply a force to it in the +ve y direction, only the object's $|\mathbf v|$ will change. I'm confused about this. Won't the object's velocity increase in the +ve y direction?

dmckee

Both the magnitude and direction of velocity will change. What will be unchanged is the x-component of the velocity.

And the change coms int he form of they-component changing.

user8718165

@dmckee yes sir. But, while teaching the concept of work, it's said that a perpendicular force can't change the object's KE... That's bugging me.

@dmckee Okay sir. Thanks

dmckee

Ah.

user8718165

@dmckee sorry, if I said something wrong.

dmckee

04:03

OK, if you applied the force in the y-direction for an infinitesimal time it would generate no work. But if you apply it for a finite time, then for part of that time the velocity is no long purely in the x-direction, so the angle between the force and the motion is no longer 90 degrees, so the sine term is no longer zero and the work is also non-zero, and the KE rises.

You should compare this to the case of uniform circular motion where the force is always perpendicular to the motion and the KE is constant.

user8718165

@dmckee yes sir. I got the confusion while thinking of it.

dmckee

YEah. Sometimes just framing the question carefully to ask someone else is enough to work out the solution for yourself.

Programmers call that "rubber duck debugging". The notion being that you should explain the problem to the rubber duck (or other totem) that you keep around for the purpose.

user8718165

@dmckee yeah sir... I'm having a new idea just now. If we continue applying the force, it will stop being perpendicular. Is it correct sir?

@dmckee Thanks for telling this. Got it :-)

dmckee

Yep. The path of the object's motion bends in the direction of the applied force, and then the force is providing a change of speed as well as change of direction.

user8718165

@dmckee I got it. Thank you very much sir for replying to me. Whenever I talk, nobody replies. That's why I was a bit afraid at the beginning. Thank you so much for help.

dmckee

04:09

This week isn't a great time for chatting.

Many regulars will have interruptions from travel and unaccustomed social events.

user8718165

@dmckee Yeah. (belated) Merry Christmas sir, any a very happy new year and holidays 😊🎄

@dmckee Yeah. Got it.

dmckee

Personally I have been way down on chat time since I was able to bring my family out to the location of my new job this summer.

@user8718165 Indeed, and to you, too.

user8718165

@dmckee Wow. That's really awesome. Travelling :-)

@dmckee Thanks sir.

Secret

05:43

@SirCumference that is an unfortunate fact that I used d for the distance lol

I should have used s

Sir Cumference

06:21

@Secret Or $\dot{d}$ ;)

actually on second thought the dot looks pretty ugly

skillpatrol

08:14

Notation for differentiation

In differential calculus, there is no single uniform notation for differentiation. Instead, several different notations for the derivative of a function or variable have been proposed by different mathematicians. The usefulness of each notation varies with the context, and it is sometimes advantageous to use more than one notation in a given context. The most common notations for differentiation (and its opposite operation, the antidifferentiation or indefinite integration) are listed below. == Leibniz's notation == The original notation employed by Gottfried Leibniz is used throughout mathematics...

Novalium Company

08:24

In ML, one of the applications of autoencoders is dimensionality reduction. My question is, when I input a 6x6 image (36 input neurons), the output will still be 36 neurons. How is dimensionality reduced here? I understand that the compressed output image is much more simplified than the original and that the whole autoencoder can reduce noise but how are the dimensions reduced?

Slereah

10:15

@NovaliumCompany It depends on the autoencoder you're using I suppose?

IIRC sometimes it's the number of channels that is reduced

ie 3 channels for RGB pixels to 1 channel

Secret

11:06

After reading process philosophy, it now make sense to me why the action is just the time integral of the lagrangian

The lagrangian is the tangent bundle of the configuration space, meaning that for any paths in confuguration space, the lagrangian captures all its possible infinitesimal changes

Slereah

I mean it's just a functional of the configuration space

Therefore any configuration has an associated value

That's why it's so easy to make zero-dimensional field theory

Secret

Right

Slereah

since it's just a linear functional from $\{ p \}$ to $\mathbb{R}$

Which is isomorphic to $\mathbb{R}$

Secret

11:28

Thus in the special case where the configuration space is just 1D space, classical with no potentials (I.e. free particle), the tangent bundle is just all its tangent vectors at each position. (And then I am missing something crucial here), and hence the lagrangian here is just the kinetic energy

(I don't just get the kinetic energy right away with the tangent vectors. I think I need to make use of the fact that it is a functional and act on the configuration space somehow to recover the expected expression of p^2/2m)

Slereah

You need a whole bunch of assumptions to deduce that the proper Lagrangian is $\dot{x}^2$

Secret

Hmm... seems I cannot just combine the ontology and the mathematical formulation together directly. I better testify classical mechanics again to ensure I am clear of all the assumptions

But basically, the original claim I want to deduce is that in process philosophy, the most basic unit are ongoing processes instead of substances. There they have a notion of action which is a process over a extended period of time. This reminds me of action in physics somewhat and I suspect it can be given an interpretation as the ongoing infinitesimal changes in configuration space over a time interval

and hence it can justify ontologically why action is the time integral of lagrangian and why there needs to be stationary action to recover the equation of motion

(Because if the action is not stationary, then the configuration space has not settle down and the dynamics of the system will not be determined to allow experimental observation)

But I guess this is not a very good argument if it cannot recover all the assumptions used in the lagrangian formalism

Typo: testify should read "reread"

Slereah

The action is a time integral because the configuration space is a function of time

That's basically what it is

When the configuration space is a function of spacetime, the lagrangian is an integral over spacetime

And when the configuration space isn't a function at all, the action is not an integral

In the $0$-dimensional case you simply have $A = P[x]$, for $P$ some polynomial of the field values

The simplest such polynomial that has a critical point is $mx^2$, for which the Euler-Lagrange equation is just $x = 0$

Or $mx^2 + bx + c$ really, in which case $x = m/b$

Secret

11:58

I see

Pulasthi Udugamasooriya

Can somebody please answer my question

1

Q: Tipping while moving forward?

(This is a follow-up to this question I asked earlier, but here let us consider a more general situation where the center of gravity is at a height $H$ above the base, and the force is applied at a height $h$.) I understand that if the force applied is greater than $\dfrac{mgr}{h}$, i.e., if t...

Pulasthi Udugamasooriya

12:16

...

Aaron Stevens

14:16

@PulasthiUdugamasooriya From the chat guidelines Don't advertise your recent questions. If you just posted something on the main site, give it some time, and don't tell people to go there and look at it...Those who can answer are already watching the queue on the main site!

Physics Meta

0

Q: Creating a community wiki post for special relativity questions about comparing events in different reference frames

As I am sure many are aware, we get a fair amount of questions about special relativity that are essentially variances of Alice sees this event happen at this time over here, and Bob sees this event happen at this time over there, so then how can Alice see this if Bob sees this? Of course q...

discussion community-wiki

Yuvraj Singh...