The highest power of x in the force equation will have bound solutions if the power is odd (and the coefficient out front is negative). If the highest power of x is even, you will have the potential to have bound solutions at low energy (around the origin), but once you get to high enough energy...
Jackson suits a certain type of student very well. If you have a good, careful, thorough professor, like to read a book thoroughly (~20hours a week) and then straight up copy the methods out of the book when you are solving the exercises, you will do fine. If you are a more 'intuitive' person...
Unless you are using crazy units it is not del dot E = 4*pi*rho, it is del dot E = rho/epsilon zero. TO be honest, I am not quite sure what the question is. When you go to two dimension you can think of it as just x and y, so the z component is zero. That should pretty much answer your questions...
If there is no air resistance and you throw the ball up with a certain velocity, it stops turns around and comes back down, what will be the velocity as it passes by the spot you threw it from?
Homework Statement
Show that the anticommutator of parity and boost is zero.
Homework Equations
\{\mathcal{P},K^{i}\}=0
The Attempt at a Solution
Let the anti commutator act on a state
\{\mathcal{P},K^{i}\}\Psi(t,\vec{x})=\mathcal{P}K^{i}\Psi(t,\vec{x})+K^{i}\mathcal{P}\Psi(t,\vec{x})...
Any number that is a multiple of ten is even and divisible by 2. Then any number (base ten) denoted by ...dcba (where a is the ones place, b in the tens, etc.) will be divisible by two if a is divisible by 2 because ...+d*10^3+c*10^2+b*10^1+a represents the number and division is linear. All...
Try looking at it in this way...
Remember the following rules:
A row of a matrix can have all of its entries be multiplied by a number. The effect that this has on the determinant of this matrix is that the determinant gets multiplied by the number.
If you do this on all rows of the matrix...
Yeah, that is basically what I am doing, trying to get a good viewpoint before I start grad work. I'll try to give it a more theoretical read-through and worry about specific details later. Thanks for the help so far. Also, what other books would you recommended?
The thing that is really bothering me is the change in the type of quantity on each side of the equals sign. We start out basically by saying the D's are matrices, fine good! But then we use an identical looking relationship except that some of the D's are 'turned' into vectors. And when you...
Well, that is different! He never said in the book that we didn't want stretches or reflections. However, he does say 'Take the group elements themselves to be orthonormal basis vectors for a vector space...' but that doesn't really clue me in to what the D(g_{1}) in...
Why did he say "the dimension of a representation is the dimension of the space on which it acts"? I would say that it 'exists' in 1 dimension, but what is it acting on?
I suppose, if elements of a representation only act on other elements of the same representation, then D(a)[D(b)] would be...
Is there an errata for this book (2nd edition)? I am wondering about some of the things that are stated just in the first few pages. For instance, he says "This is Z_{3}, the cyclic group of order 3." then stuff about defining a representation and then "the dimension of a representation is the...
In my opinion, your collection of degrees is a huge hindrance. It shows at the very least a lack of the ability to realize an incorrect course, then correct it. At worst it shows that physics is yet another fleeting thought for you. You want to be a physicist but never got an undergrad physics...
If you are no good at algebra (and live in the US) go to your local community college and for a couple hundred bucks you can get a 16 week course that you can ask all of your questions in. IF you don't know calculus, go enroll in Calc 1, then Calc 2, then Calc 3 and then Differential equations...
I buy that (and that was what I was thinking intuitively). I see that the path length difference per angular displacement will be more for a thicker object and that would explain the thin bands.
I understand the idea behind the optical interference that produces colors on thin films but have never figured out the reason that the films have to by 'thin'. What is the lack of similarity that I am missing between a film and a somewhat thick sheet of glass or something that makes these...
Obviously you are opposing each other so there is no one distinct opposition, you are his opposition, he is yours... I was merely encouraging him to lay it out in detail, I don't have a horse in this race. Most of the arguments of this type (on these forums) end up with people bickering over...
Ok, but even in that setup there seems to be a spacial dependence to me. Since they move at the speed of light, and \frac{\omega}{k}=c, I think there will be a spacial dependence (as long as that state is correct).
I am not talking about subsequent measurements of the same photon, but measurements on identical photons emitted coherently from the same source but then measured at different distances.
So coherent circularly polarized photons don't have changing probabilities of linear polarizations? I find...
Don't get too discouraged Bill, there is always an opposition. Think of it as a test of your argument, not a personal attack. Now, I have not studied probability theory or local/nonlocal stuff much, so I am in no position to comment on that, but I can see that there are a lot of posts that...
I've got three questions basically.
Do photons actually have only an integer spin +/- 1, or do people really only mean the sign of its chirality? The reason I ask is that I am interested in whether the photon spin is related to its frequency. This leads to the next question.
Is the basis for...
Ben to the rescue. I am glad that this paper agrees on the decomposition to what I called 'orthogonal effects' (transformer stuff and motional stuff). They may not actually be 'orthogonal' in the sense that they have no relation to each other, but still, this is the right path I believe. I...
I like your thinking there. You have an electric field and then you have an 'induced' electric field (from the motional magnetic effects). I can't really say for sure that the equation containing vxB is 100% correct, but I think it is damn close. This moving surface business is always a problem...
Maybe I can illustrate the OP's point. The claim is that there are two ways to create an emf due to what we perceive as "flux changing". One way was termed to be "transformer induced emf" and the other was "motional emf". The OP is not arguing against both of these phenomena, he is merely saying...
Can you explain this a little more than you have. It seems that a person could clearly isolate charges and then measure their relative velocities. And when we measure stuff, at rest, they have a force on each other. We can attribute this to the E field. I'll let you comment on that if you want...
So by selecting a slope g, you are in effect picking out a line mx+b, where g=m, x(g) and -f(x(g))=b. And the same thing for the inverse but with f->F, x->g. It still seems to me that the key lies in g(x) being bijective but I think what we are saying may be equivalent, so I'll leave it at that...
Thanks for the detailed reply. I understand it well in the context of that specific definition of 'maximizing'. But the definition of the maximization seems a bit arbitrary to me, why define it that way? What if we define a new thing called the Legarden transform that is F(g)=gx(g)-f(x(g)), no...