[1] The F_\mu\nu term does not interact with \phi. He added it b/c A_\mu "...contributes by itself..." to the Lagrangian.
[2] What is the motivation to say a gauge field "...contributes by itself..." to the Lagrangian? Do all gauge fields contribute by themselves to the Lagrangian? Are there...
That term is necessary to include the E- and B-fields in the quantized version of E&M.
Here is a thread which discusses this term: https://www.physicsforums.com/archive/index.php/t-37318.html
The terms which contain the fields themselves, such as \phi, A_\mu and \phi^* are interaction...
After deriving the Lagrangian for the electromagnetic field using only gauge invariance of the action, the result is: (i.e. to say \delta L \equiv 0.)
L \equiv (\partial_{\mu} + ieA_{\mu})\phi(\partial^{\mu} - ieA^{\mu})\phi^* - m^2\phi^2.
Et. viola'. Done. Finished. Complete. Let's go home...
Well then, what does this mean?
BTW, I moved on knowing how Ryder makes statements that are clear to him (and strong students) that leaves me in the dark. You know my previous post about the mystery with the \xi? It turns out that the way I described it in the end - as being utterly...
I did verify all that.
But let me ask you this: do I need all that stuff with \xi? The reason is that I just went over it again and found that I can deduce the exact same results skipping all that voodoo by just starting with the Pauli matrices, which are a basis, and r is a vector and then...
From your three posts I have made progress. I still have one question remaining. But first let me summarize to see if I have it correct:
1) If one performs a rotation in R3 then one is also simultaneously performing a rotation in the space for SU(2).
2) This is b/c 'h', a 2x2 matrix in SU(2)...
I have been struggling with this for a long time. I gave up to review GR and came back to Ryder. I started in Chapter 2 and the material was easier and more intuitive on this second pass. But the same topic, "SU(2) and the rotation group" has trapped me yet again. I am bogged down specifically...
That minus sign is common to both t+ and t-. The only thing that changes is that the results are switched. Thanks for that. But I still think that something is weird about the "square both sides of the inequality" step.
Thanks for all your help.
-joe
Ok, here goes...
t_{+} = \frac{v_{oy}}{g} + \frac{\sqrt{v_{oy}^2 + 19.6\Delta y}}{g} > 0
Canceling the common denominator, g, and bringing the voy term to the other side then I get:
\sqrt{v_{oy}^2 + 19.6\Delta y} > -v_{oy}
Squaring both sides I get:
v_{oy}^2 +...
Using the quadratic solution for the time (link is to the above formula):
https://www.physicsforums.com/latex_images/28/2874555-0.png
I set t+ > 0 and t_ > 0 and look for the necessary condition to ensure this.
For t+ > 0, yf > yi, and
for t_ > 0, yi > yf.
But this was my...
If the initial and final positions are fixed, then solving for the time of flight should yield only one positive time.
yf = yi + voy*t - 4.9*t2
The solution of this is the quadratic formula:
t = \frac{-v_{oy} \pm \sqrt{v_{oy}^{2} + 19.6*\Delta y}}{-g}
There are two answers: t+ and...
I explained to my class that in projectile motion one always chooses the positive time answer with the negative time answer being dropped. Then a student asked if two positive time answers were possible. I just looked at it assuming that t+/- are both positive and got the result that for t+ to...
Yes! the (uw-wu) is anti-symmetric in a,b and the e_a,b is symmetric in a,b. That's a good one to remember.
Thanks for all your kind help and patience with me. May God richly bless you, in Jesus' name, amen.
-joe
[1] By "Levi-Civita connection" do you mean:
a) the "connection coefficients" aka
b) the affine connection aka
c) the Christoffel symbols??[2] In a coordinate basis, the commutator of the partials is zero. The second term is not a commutator of the partials - but of the components of the...