Actually, something that doesn't make sense to me...why is it that my first attempt at the problem (see original post) gives you the same answer even though we're treating the flux as though it were crossing some spherical region and not a plane?
My latest thoughts:
F = \int{I cos^2\theta dA/\rho^2}
dA = 2\pi \rho d\rho
F = 2\pi I \int{cos^2\theta d\rho /\rho} = 2\pi I \int{cos^3\theta d\rho /z} = 2\pi I \int{cos\theta sin\theta d\theta} = -I\pi cos^2\theta
Or, when evaluated from 0 to pi/2:
F = I\pi
And I think this is where you...
Ok, great. So then I think up to here I'm good:
F = \int{I cos^{2}\theta dA / \rho^2}
One thing that just occurred to me from my last post is that dA is the infinitesimal area through which the flux is being calculated, in our case, the plane at z=Z. I was treating it as an infinitesimal area...
It seems like we disagree right off the bat with the equation for flux? Isn't this how it's defined in radiative processes? Where d\Omega = dA/r^2. dA is an "infinitesimal amount of surface area that is located a distance r from the source and oriented perpendicular to the position vector r."...
\rho^2 = r^2 + z^2
2\rho d\rho = 2rdr
since z is constant.
I think I understand most of your reasoning, but I'm still not sure how to go about the problem. A disk as viewed from some point P that is off axis will appear as an ellipse -- ok, that much makes sense. So that means the projected...
Hm, ok. Let's see...
F = \int{I \cos{\theta} d\Omega}
d\Omega = dAcos{\theta}/\rho^2
F = I \int{cos{\theta} dA cos{\theta}/\rho^2}
cos{\theta} = z/\rho
F = I \int{(z/\rho)^2 dA/\rho^2} = I \int{(z/\rho)^2 2\pi r dr/\rho^2} = 2\pi I z^2 \int{\rho d\rho / \rho^4} = -\pi I z^2/\rho^2
Not sure...
Ok...so is what I did above in calculating the flux not the same? The cos term arrises from the projection into the line of sight direction. I am incredibly confused.
Ok, fair enough. That being said, I don't think I understand the drawing. I'm not sure what that length (perpendicular from X1 to QX2 is?
Thanks for the help by the way!
Ok, I think you're essentially asking me to do this (page 32), right? I don't think I fully grasped the concept of solid angle. It's the differential of the projected area (projected onto the direction of the observer) divided by the radius of a sphere squared. So, above, I was regurgitating the...
Ah, you mean expressing the solid angle in cylindrical coordinates and in turn the flux? So d\Omega ={z d\phi d\rho}/\rho^{2}?
The problem also says that by assuming Z>>r, you can use the small angle approximation, so in the expression for flux can I let cos\theta = 1, and then F is entirely in...
Homework Statement
Suppose you have a disk of radius r at x=y=z=0 with its normal pointing up along the z-axis. The disk radiates with specific intensity I(\theta) from its upper surface. Imagine the observer plane is at z=Z, where Z is much greater than r. Let I(\theta) = I = constant...
Thanks for the reply.
You're right, I missed the square roots on z-basis expression. Thanks!
I haven't checked whether \Psi is really the same. I know that the inner product of \Psi with itself should give me 1 (if indeed it's the same). But wouldn't I need to somehow go between x and z to do...
Homework Statement
It is known that there is a 36% probability of obtaining S_z = \hbar/2 and therefore a 64% chance of obtaining S_z = -\hbar/2 if a measurement of S_z is carried out on a spin 1/2 particle. In addition, it is known that the probability of finding the particle with S_x =...
Hello all,
I'm trying to compile a list of schools (mainly in the US) that offer Ph.D. programs with research in radio astronomy.
Here's what I've come up with so far:
Caltech
Cornell
Harvard
New Mexico Tech
Ohio State University
UC Berkley
UC Los Angeles
UC Santa Cruz
University of Arizona...
Hm, I was able to work through the problem, correcting for the period, and it looks like I'm now only off by a factor of 2. A friend of mine is also having the same problem. Not sure if we can chalk it up textbook error, or just something we're not seeing...
Ah nevermind, feeling rather silly...
Yep, I just got that. Makes sense now. Thank you so much!
Although, can I ask, how would one know (without knowing the answer) that the period is in fact 2l and not l, like I wrongly assumed it was, just from reading the problem/looking at the image. Ah man, feeling rather stupid.
http://en.wikipedia.org/wiki/Fourier_series
Everything I've looked up suggests the 2 should be there. I'm also pretty sure I've been using the 2 in all my Fourier coefficients calculations up until now. :( I thought the two arises from the fact that ω=2pi/T.
I'm sorry. I guess I'll run...
I'm almost certain the 2pi factor belongs there. At least it certainly does in the equations for the coefficients of a Fourier series, where the 2pi emerges from ω. Unless for some reason those equations are modified in this case? I know the book tends to leave off the 2pi, but that's when...
So actually, my An includes the prefactor of 2h/\pi^{2}. There's on the other hand, is independent of the constants out front. That's why I was thinking that maybe the two answers are equivalent, in that they factored those pre-factors out, but then the fact that our sin terms are different is...
I used y= 4hx/l for 0<x<l/4
2h-4hx/l for l/4<x<l/2
0 for l/2<x<l
So this basically gave me two equations for An, where the first is equal to
An = 2/l (from 0 to l/4)∫(4hx/l)*sin(2\pinx/l)dx
and the second, for l/4<x<l/2
An = 2/l ∫(2h-4hx/l)*sin(2\pinx/l)dx
Do those look ok...
Homework Statement
A string of length l has a zero initial velocity and a displacement y_{0}(x) as shown. (This initial displacement might be caused by stopping the string at the center and plucking half of it). Find the displacement as a function of x and t.
See the following link for...
Hm, ok, I'm not sure I follow here. If I computed the dot product of two cross products, isn't that wrong, in that I should first compute the cross product of (BxC) x (CxA), before taking the dot product that I did in the step before?
Thanks for the response!
Ok, well, I know that (AxB)_{i} = ε_{ijk}A_{j}B_{k}. And I can say that, (BxC)_{i} = ε_{ilm}B_{l}C_{m}.
If I write that all as one term,
ε_{ijk}A_{j}B_{k}ε_{ilm}B_{l}C_{m}
then that equals,
(δ_{jl}δ_{km} - δ_{jm}δ_{kl})A_{j}B_{k}B_{l}C_{m}
and I know...
Homework Statement
Hi all,
Here's the problem:
Prove, in tensor notation, that the triple scalar product of (A x B), (B x C), and (C x A), is equal to the square of the triple scalar product of A, B, and C.
Homework Equations
The Attempt at a Solution
I started by looking at the triple...
Whoops, that was a bad mistake. Thanks for catching that.
Though I'm still not really sure what to do. My potential can then be written in terms of x and y components, right?
For instance,
U_x = 1/2 * k * rx^2, but I don't really see what r ought to be. I understand that there should be some...
Homework Statement
A puck with mass m sits on a horizontal, frictionless table attached to four identical springs (constant k and unstreched length l_0). The initial lengths of the spring a are not equal to the unstretched lengths. Find the potential for small displacements x,y and show that...
Hi all,
My classmates and I have been at this problem for some time, and it doesn't look like we're getting anywhere. We'd really love any help in the right direction!
Homework Statement
A lawn sprinkler is made from a spherical cap (max angle = 45\)
with a large number of...
Homework Statement
Find the center of mass of a solid of density \delta = 1 enclosed by the spherical coordinate surface \rho = 1-cos\phi.
Homework Equations
The Attempt at a Solution
I'm a bit confused about how to start here, mainly because the surface is defined by spherical...
Ah ok. How's this?
$$\int_{0}^{\pi/2}\int_{0}^{1}\int_{0}^{r} r(6+4rsinθ) dzdrdθ$$
Thanks again for all the help by the way. Definitely starting to feel a bit better about these conversions.
Yeah, you're absolutely right. I realized that right after I submitted that last post.
Here are my new integrals.
Cartesian Coordinates:
$$\int_{0}^{1}\int_{0}^{\sqrt{1-y^{2}}}\int_{0}^{\sqrt{x^{2}+y^{2}}} 6+4y dzdxdy$$
Cylindrical Coordinates:
I'm not sure about this one, particularly...
Hm, I am thoroughly confused now, haha. I thought the region was the inside of the cone?
What I'm picturing: If I drew a cylinder and placed a cone directly inside it, so that the top of the cone meets the top of the cylinder, the region I'm looking at is inside the cone.
Should your theta limits be from 0 to $$\theta/2$$ since it's only the first octant?
So this is what my integral in cartesian coordinates looks like:
$$\int_{0}^{1}\int_{0}^{\sqrt{1-y^{2}}}\int_{\sqrt{x^{2}+y^{2}}}^{1} 6+4y dzdxdy$$
I guess I don't quite understand ##\rho##, in that case. I...
Hehe fair enough. I get 2pi^2.
Looks like a nice and neat enough answer -- hopefully that's what you got!
I went through the problem a second time, though, and now I'm finding myself confused about one of the trig substitutions I made. Once I've expanded (1-cos(2ϕ)/2)^2, I'm left with a few...
Homework Statement
Set up triple integrals for the integral of f(x,y,z)=6+4y over the region in the first octant that is bounded by the cone z=(x^2+y^2), the cylinder x^2+y^2=1 and the coordinate planes in rectangular, cylindrical, and spherical coordinates.
Homework Equations...
Ok, so when I expand (1-cos(2\phi)/2)^2, I get 1/4 - (cos(2\phi))/2 + (cos^2(2\phi))/4
Replacing the last term with a half-angle formula, I have
1/4 - (cos(2\phi))/2 +1/8 + (cos(2\phi))/8
At that point, I take the integral of that with respect to \phi.
I wind up with 1/4*\phi -...
Ok, so as far as my limits of integration go, those are ok?
I also get (1/3)\rho^3, and evaluating at 2sin\phi I get 8/3∫∫sin^4(\phi) d\phid\theta. I'm not sure where I ought to be getting 16 from?
As for ∫∫sin^4(\phi) d\phid\theta, I think I'm simply screwing up some basic calculus. I...
Homework Statement
Find the volume enclosed by the spherical coordinate surface ρ = 2sin∅
Homework Equations
dV = ∫∫∫(ρ^2)sin∅dρd∅dθ
The Attempt at a Solution
(Sorry about my notation!)
Alright, here's what I've done so far...
Since the region is a torus, centered...