The h Bar

1:03 AM

@Charlie $\partial V$ is what you write for the boundary of some other submanifold $V$, which in this case would have to be 3-dimensional

Charlie

I'm still unsure how you define that surface in such a way that you can contract it with the 2-form in the integral $f(x,y)\text dx\wedge \text dy$, unless I'm unclear on what's actually being done here

which is looking like the more likely outcome

ACuriousMind

The surface has to be some nice submanifold $\phi : N\to M$, so that you can pullback the form along $\phi$

Then, the pullback will be a top-dimensional form on $N$, and the integral of a top-dimensional form $f(x)\mathrm{d}x^1\wedge\dots\wedge\mathrm{d}x^n$ is just the integral of $f(x)$

Charlie

I've seen the word pullback used a few times now, it looked scary when I looked it up

I think this might just be something I'll have to put a pin in until I've got further into the book

JingleBells

6:26 AM

Why Does a New Car Lose Value After It's Driven off the Lot? (I googled, explanations are unclear)

I mean, what's the problem with selling a car right after I have bought it for almost the same price (taxes...) that I have just bought it at?

John Rennie

7:06 AM

@JingleBells Because the car dealer is a trusted seller and you are not.

JingleBells

7:19 AM

@JohnRennie hmm, so my car would be worth less because other people will be suspicious about why I'm selling it? yeah, I'd be too

John Rennie

If I buy from a dealer I get warranty, after sales support, free first service, etc, etc.

JingleBells

gotchya

John Rennie

If I buy from you then I get nothing but the car. Well, unless the warranty etc are transferable, but how can I be sure this is the case?

Physics Meta

11:05 AM

0

Q: This question was downvoted and the reason provided for it was not correct

I recently had asked this question The work done by Tension and Normal is not always zero It was downvoted and I was given a vague reason for the downvote. I asked the person to clarify the reason but there was no reply from that side. I asked this question to know whether the reason for downvote...

discussion

Charlie

3:28 PM

I've just read the following, in the context of electrodynamics:

> "The exterior product of a 1-form and a 2-form corresponds to the dot product. The coefficients of the resulting 3-form is equal to the dot product of the vector fields dual to the 1-form and 2-form in the Euclidean metric."

This is fine up until we talk about the "vector fields dual to the 1-form and 2-form". I understand that we get a 3-form from the exterior product of the 1-form and 2-form, and that (I guess) the hodge dual to this is a 0-form which is the correct object for the result of the dot product.

What's confusing to me is that the dot product is a map: $$\circ:V\times V\rightarrow \Bbb R,$$ but the vector "dual" to a 2-form obtained through the Euclidean metric is just a 2-blade, no? $$\omega_{\mu\nu}\delta^{\mu\alpha}\delta^{\nu\beta}=\omega^{\alpha\beta}$$

in component form at least. Unless the dual to a 2-form is somehow a vector, otherwise we don't have the correct arguments for the dot product

ACuriousMind

@Charlie In 3d, you have an equivalence between 2-blades and vectors normal to them

Charlie

oh

ACuriousMind

It's the cross product magic - instead of thinking about the blade $x\wedge y$, you can think about the vector $x\times y$

Charlie

is this why $x\times y$ is referred to as a "pseudo-vector"?

ACuriousMind

yes

Charlie