/sci/ - Science & Math

(Still looking. I've got two solutions already, but none of them is straightforward enough to be posted here. Also, screw Peskin.)

>>4094076
are they in textbooks? you can post the book name and chapter and ill see if i can find it.

bump

erm, to me it looks like they want you to expand

<span class="math">U\left(y,x\right) \to e^{i \alpha \left(y\right)}U \left(y,x\right)e^{-i \alpha \left(x\right)}[/spoiler]

in x the way they do with U only, and then you compare first order coefficients, i.e. the terms with \epsilon. From U only 1 survives multiplied with the first order derivative of the expansion of \alpha(x).

bumping for Jo's reply

For a good book on Gauge field theories, I recommend Pokorsky (or Pokorski?).

Regarding your question. There is some part missing, I presume. As far as I remember, you read off the transformation behavior of the gauge field by plugging in the transformation into your Lagrangian - and then simply by demanding that your L stays invariant, A' follows rather intuitively.

Oh hey cool anon. I totally agree with you
>>4094101
and then i saw this thread and was like d'oh

I will write my work down starting now.

i got it! it was basically>>4094097 but i was going at it from the wrong angle the whole time. ill post how i did it for anyone whos interested.

thanks to everyone who responded.

Just transform your gauge field with a <span class="math">SU(3)xSO(3)[/spoiler] group as your coefficient.

Before I start interpreting the QM2 lecture notes again: the solution's in there, start reading at page 66, "local transformations". The lecture notes assume the non-abelian case, i.e. <span class="math">[A,A]\neq0[/spoiler], but your result is obtained as a special case in equation (3.52).
Hope it helps,
http://www.physik.uni-wuerzburg.de/fileadmin/11030200/Ohl/Lehre/Quantenfeldtheorie_II/2011/qft2.pdf

>>4094106
That's usually one clean way to do it. You don't have or want to use lagrangians to introduce connection in a fibre bundle. Therefore the Peski Schröder way taken here is conceptially fruitful, I'd say. However, by calculating in terms of A_{mu} we are of course using components (as physicists do), which, I agree, isn't as nice looking as a pertubation of the lagrangian or action.

Ralated to the non-mechanical view, and especially to the guy how posted here
>>4094110
I could recommend this blog post by Terence Tao

http://terrytao.wordpress.com/2008/09/27/what-is-a-gauge/

The beginning is easy, at least.

Other than that I'd say research some Cartan Geometry <3

>>4094122
>not naming your fields

>>4094106
That's usually one clean way to do it. But you don't have or want to use lagrangians to introduce connection in a fibre bundle. Therefore the Peskin Schröder way taken here is conceptially fruitful, I'd say. By calculating in terms of A_{mu} we are of course using components (as physicists do), which, I agree, isn't as nice looking as a pertubation of the lagrangian or action.

Raleted to the non-mechanical view, and especially to the guy how posted here
>>4094110
I could recommend this blog post by Terence Tao

http://terrytao.wordpress.com/2008/09/27/what-is-a-gauge/

The beginning is easy, at least.

Other than that I'd say research some Cartan Geometry <3

>>4094122
>not naming your fields

<div class="math"> U\left(x+\epsilon n,x\right) \to e^{i \alpha \left(x+\epsilon n \right)}U \left(x+\epsilon n,x\right)e^{-i \alpha \left(x\right)} </div>
<div class="math"> 1-ie \epsilon n^\mu A_\mu (x) \to e^{i \alpha \left(x+\epsilon n \right)}\left[ 1-ie \epsilon n^\mu A_\mu (x)\right]e^{-i \alpha \left(x\right)} </div>
expanding the first exp around x:
<div class="math"> e^{i \alpha \left(x+\epsilon n \right)} = e^{i \alpha \left(x\right)} +i \alpha'(x) e^{i \alpha \left(x\right)} </div>
inserting we have:
<div class="math"> e^{i \alpha \left(x+\epsilon n \right)}\left[ 1-ie \epsilon n^\mu A_\mu (x)\right]e^{-i \alpha \left(x\right)} = \left[ e^{i \alpha \left(x\right)} +i \alpha'(x) e^{i \alpha \left(x\right)} \right]\left[ 1-ie \epsilon n^\mu A_\mu (x)\right]e^{-i \alpha \left(x\right)} = \left[ 1 +i \alpha'(x) \right]\left[ 1-ie \epsilon n^\mu A_\mu (x)\right] = 1+i \alpha'(x) \epsilon n -i e \epsilon n^\mu A_\mu</div>
thus
<div class="math"> 1-ie \epsilon n^\mu A_\mu (x) \to 1+i \alpha'(x) \epsilon n -i e \epsilon n^\mu A_\mu</div>
thus
<div class="math"> A_\mu (x) \to - \frac{\alpha'(x)}{e} + A_\mu</div>

i condensed some of the notation such as the derivatives ans skipped some steps, but this it the general outline. actually, im tired and could have subconsciously cheated in this derivation to get the right answer.

So my math is wrong somewhere obviously, but maybe someone can correct my mistake. Assuming that that big term that i have does become d sub u a(x) (which i dont know what d sub u represents, its a partial derivative of something i suppose holding u constant, but idunno what) this should be pretty close.

Also pardon my shitty mspaint math

oh ok its done. oops again

>>4094126
thanks, ill take a look at that. it seems to handle it alot more formally than my book.

and also thanks to the mod for not banning this thread.

>>4094159
look up the Einstein summation convention. you sum over the index so
<span class="math"> a^i b_i =a^0b_0+a^1b_1+a^2b_2+a^3b_3[/spoiler]

>>4094188
you forgot at least one minus sign there pal
I usually write three minus signs

>>4094198
no i didnt, you are amusing im working in a 3+1 time-space, but im working in a 4+0 one. so there!

>>4094188
I dont know anything about physics, really, I just know some math (though even my algebra could use a good bit of work).
What part of this is the einstein summation notation used for?

So is Josef like the official /sci/ advanced math homework-doer?

Why not just gb2 mathoverflow where you at least get points for it?

>>4094479

because those points are probably just a pointless as posting here.