The h Bar

4:47 AM

They keep moving the world!

Q: Doesn't the assumption of equilibrium make this answer wrong as a proof of the law of the lever?

5:02 AM

@antimony This appears to the objection:

This answer was written to the question 'Proof of the Law of the Lever'. I think it is wrong because it assumes that the system is in equilibrium without logical proof. If we set the configuration as L1 F1 = L2 F2 (or in the words of the answer, m F1 = n F2), we can't prove the system is in equil...

In fact there is a good point there. Students are taught to equate the torques at the two ends of the lever, but this only applies when the lever is not accelerating. If the lever is accelerating than the torques are not equal.

Not Tfue

5:56 AM

Why do physicist disagree about time travel? Aren't we always time traveling? Like I time traveled from yesterday to today.

@NotTfue For a physicist the phrase "time travel" means a "closed time-like curve".

It's obvious you can travel in a closed loop in space because that just means you travel along some route that takes you back to the place you started.

But it takes you back to your starting point at a later time. A closed time-like curve would mean you could travel along some route and get back to the place you started at the time you started.

And this necessarily means that at some point in your journey your clock was ticking in the opposite direction to clocks of anyone watching you, or put more simply you were travelling backwards in time.

4 hours later…

10:10 AM

Hi, I'm back (I'm colourful spacetime)! Stackexchange lifted my suspension.. Hopefully I didn't miss much.

Here is my question of the day: In QFT when we are talking about the "symmetries of a system" or the "symmetries of the theory", are we referring to the symmetries of the Lagrangian or the Action? Let's say you are given a Lagrangian and someone ask you "show that the theory is invariant under SO(3) transformations", should I check that the Lagrangian or that the Action stays invariant? I found a post stating that both the Lagrangian and Action have identical symmetries, is that correct?

Q: Does the action and Lagrangian have identical symmetries and conserved quantities?

10:56 AM

From the book Introduction to Classical Mechanics With Problems and Solutions by David Morin, page 236 states: Noether's Theorem: For each symmetry of the Lagrangian, there is a conserved quantity. Whereas the Wikipedia page states: Noether's (first) theorem states that any differentiab...

lagrangian-formalism conservation-laws noethers-theorem action

11:12 AM

Yep, I read this, but I'm still not sure. Qmechanic says that a vertical symmetry of the action is not necessarily a vertical symmetry of the Lagrangian. But what about a full infinitesimal transformations? Are the symmetries then identical? What about horizontal symmetries...? And I have some trouble understanding the answer by josh; I see no equivalence between the two answers.

The Lagrangian is only defined up to a total derivative, so any given Lagrangian you take might not satisfy a symmetry but if you add a total derivative it will, so it's better to think in terms of the action rather than a Lagrangian, but if you think in terms of the Lagrangian all you need is that the end result be in the form of a total derivative, this is what they mean talking about quasi-symmetries

When you study a symmetry, you typically study $\delta S = \int d^4 x' L(x',\phi') - \int d^4 x L(x,\phi)$, vertical vs horizontal is just confusing things

If you do things in terms of vertical symmetries alone, you have to be careful, you have to figure out how things transform properly, I should have written a $\sqrt{-g}$ in those, and the vertical transform will touch those things as well etc, just keep things simple

11:47 AM

@bolbteppa Ahh I see. Usually, though, if I study the action, I'm varying it as $\delta S = \int d^4 x \delta \mathcal{L}$, where $\delta \mathcal{L} = \partial_{\mu} \mathcal{L} \delta x^{\mu} + \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + .... $. Isn't this better for calculating?

What about $x$, what if $x$ changes too

Not sure what you mean, doesn't the $\partial_{\mu} \mathcal{L} \delta x^{\mu}$ factor take that into account?

12:07 PM

Have a look around equations 3-4 here

12:19 PM

According to someone, the answer I was looking for for the general notion of going from local to global is ncatlab.org/nlab/show/Lie+integration

Right, Lie's theorems there are the big thing to know

The question is, can you get all the properties of a Lie algebra starting from a Lie group

I hope so

How about Yacobi

I don't know Mr. Yakobi

How about his brother Jacobi

12:24 PM

I hope he is doing well

@bolbteppa Are you saying that I should also vary x in $d^4 x$? Or did I not understand this correctly?

Yes, that's what those equations are accounting for in the link I sent

You apparently may indeed define the exponential map in such terms

Ok I see, thanks!

The exponential map being a... functor I guess, between the Lie algebroid to the Lie groupoid

Mapping the zero section to the unit section

"diffentiable manifolds have many units and few arrows; whereas Lie groups have many arrows and few units - actually only one."

2 hours later…