In theory one can also integrate a scalar or vector field within volume with some directional vector. This would come up if you extended the divergence theorem to a four-dimensional space.
sophiecentaur,
Ok, I see what you are saying now.
You're right that it only makes sense to pump on the face of a wave. Buf if I assume your argument for how pumping works is correct, I should be able to pump on an "ordinary" wave, since the front of non-breaking waves still move up, just not...
Quantum mechanics! It's a problem of statistical mechanics really. Individual particles don't move in a deterministic fashion, but with probabilities.
The reason there is still an interference pattern is that each particle has a probability of going on some particluar path. Since the particle...
I feel like no one read the whole thread.
sophiecentaur,
I was only making a pertubation to a simplified problem. I realize it is indeed extremely complex. But getting nowhere on an difficult problem is much worse than making a simplification.
Yes the front of a breaking wave moves upward...
I was simple making as estimation to compare how much the average kinetic energy drops if the motion goes up and down (without doing work). This was to see if it is comparable to the energy that is put in by doing the work.
Well, there is some interaction due to the edges, which is why one can obtain diffraction from a single slit. The electrons usually have low enough energy that they elastically scatter. Actually to clarify, Young's double slit experiment refers to interference, due to the two slits. It also has...
A single electon will create a single spot on a flourescence screen, hardly a diffraction pattern.
But, quantum mechanics mantains that each electron will still follow the rules of feynmann paths. Hence, it is in a superposition of all pathes and will on average create a diffraction pattern...
Yes, diffraction does occur, hence the loss in transmission. But, medical x-ray sources are not coherent, so you will not see a diffraction pattern. Secondary scattering can smear out an image, but this is a very small effect.
Yes, if the transmission is very low through some dense region...
Momentum comes from the Latin word Mótus, meaning movement or motion. Hence it is associated with velocity.
Impulse comes from the Latin impulsus meaning to push. Hence it is associated with Force.
For Vector Calculus I like J. Marsden (CalTech) & A. Tromba (UCSC). But like wolfsy said, if you are trying to conceptualize things, a book on E&M. Feynmann isn't super mathematically intensive. Griffith is a standard undergrad E&M text; some people love it some people hate it. Jackson is a grad...
Yes of course. v is the indefinate integral of v' and not over the boundary of the initial region. I should have realize this before. Thanks!
Then, is there anything to be done without knowing the exact form of f(k,x)?
Yes, it's a book. Usually for graduate level, but the section on calculus of variations is just an ellaboration of what you've probably already learned.
This method uses integrating along a parametized line, which is something you learn early on in vector calculus. I think what throws many...
Well, I'm taking u as k_x and v' as f in the standard notation
\int u v' = u v| - \int u' v
therefore I will not have a derivative of f.
Not sure what you mean by surface term. I do realize this is an integral over three variables and I should use the divergence theorem in general. But...
So, to generalize it, \int \frac{dA}{dt}B\ dt = - \int \frac{dB}{dt}A\ dt is true if one of the functions A or B vanishes at both endpoints, which is what LCKurtz showed. You just have to remember in the variational principle A and B will be derivatives.
You derivation is not quite correct. I would advise you to see Goldstein (Classical Mechanics). It has a thorough explanation of Lagrangian mechanics starting with the variational principle.
You will be making variations with respect to \varepsilon , using x(t, \varepsilon ) = x(t, 0) +...
This should be very simple, but I can find a simple example that violates my general conclusion. I clearly must be doing something wrong with my integration by parts. Any suggestions would be great.
Define a distribution such that the density;
\eta(\vec{x})=\int d\vec{k} f(\vec{x},\vec{k})...
In addition to my previous answer, the sinesoidal oscillation of a small potential energy;
{V}_{0}Cos\left[k\left(z,E_0\right)r\right], where {V}_{0}=mgz is the amplitude of oscillation and should be much smaller than the unperturbed kinetic energy, E_0; r is along the wave line and...
For simplicity, imagine the reference frame of the propagating wave in a conservative system. As a surfer slides down the face of the wave, he/she gains speed because potential energy is turned into kinetic energy. Of course, the unique nature of the surfboard in a liquid also allows for...
Ok, so then if I wanted to find the capacitance between a line charge and a cylindrical conducting shell around it, since I would have to define a potential difference, could I simply say the potential on the line is zero and the potential at the shell is the negative integral of the electric...
I know it seems a bit trivial, but what is the potential right at an infinite uniformly charged line?
Irregardless of reference point, the Potential will have a ln|s|, where s is the perpendicular distance to the line. Obviously this would result in infinity.
At the same time when I...
I am still not understanding. I completed the squares on \vec {C} \dotprod {\frac {d \vec {C}} {dt}} = 0 and get this really complex function that tells me nothing.
.5g^2[(t - \frac {3 V_0 sin \theta} {2g})^2 + (2 - \frac {9} {4} sin^2 \theta) \frac {V^2_0} {g^2}] = 0
The scalar...
Ok, so how does that help?
\sin \theta \cos \theta
is a function of maximum x-displacement.
The attached image shows the vector C that must never decrease in magnatude as the projectile moves along the arc.
Well, you got a function of theta only, but I'm not so sure this helps me. I am not trying to find an angle that will give me maximum range, but a maximum angle that will yield increasing distance from the origin.