For some information on tides, here's a website:
http://www.lhup.edu/~dsimanek/scenario/tides.htm
For some mathematical detail (just algebra):
http://mb-soft.com/public/tides.html
You only need gravity to explain the tides.
Omega is a constant, equal to the angular velocity of the earth...
I'm pretty confused by the post - I think you might have a lot of misconceptions.
First, tidal forces have nothing to do with centripetal acceleration. Tidal forces are due to gravity.
Second, if you're standing still on the earth, unless you're at the poles, there is a centripetal force on...
You can prove it with some clever thermodynamical arguments, but I always find those to be dissatisfying.
Here's my attempt at explaining it. Imagine an object that is painted white. That means if you shine light on it, the light will reflect off it. Now imagine all the atoms inside the object...
Apologies for digging up this old thread, but I have some questions on some of the responses:
But why doesn't the metallic screen inside the microwave window get burnt, as does aluminum foil when you put it in the microwave?
What you say makes a lot of sense. However, what about the...
Well it's been awhile since I looked at thermodynamics, but I thought a Debye solid didn't require periodic boundary conditions, just vanishing at the endpoints.
So you have a solid, and the frequencies it can vibrate at correspond to wavelengths that are 2L/n for positive integers n and...
I'm not sure why you would need to decompose it. Assuming a Biot-Savart field, the flux through a circle whose center goes through the current carrying wire is zero, since the magnetic fields circle around the wire, so that they are always tangential to planes perpendicular to the direction of...
I'm not sure I know the answer at the level you want it. This is kind of like the example you see in textbooks of charging a capacitor. If you form a loop around one of the wires leading up to the capacitor, and calculate the line integral of the magnetic field, then you get an answer if you use...
Kinetic energy would not be a scalar because of Galilean boosts. If you're standing still, a tree has very little kinetic energy. If you're in a moving car, the tree is moving very fast. However, the kinetic energy is a scalar in the sense that is remains invariant under rotations. So it doesn't...
I think that's an interesting historical question. The equations for electricity and magnetism, for example, are not the same when you make a Gallileo boost of the form y=x+vt. People tried to save it by introducing an ether fluid, and so when you boost you also have to boost the ether fluid...
You have to make some assumptions with Gallilean boosts. For example, if you have a force that depends on velocity, say a drag force instead of a spring force:
-kx'=mx''
The transformation y=x+vt (where v is velocity, t is time) results in the equation:
-k(y'-v)=my''
which is not -ky'=my''...
That's just a consequence of using vectors. If you have a vector equation, such as Newton's law, then rotations and translations result in the same equation.
If in addition you make other assumptions, such as homogeneous forces, you can stretch and dilate. For example -kx=mx'', you can make...
Ampere's law is:
B*2 \pi R=\frac{1}{c^2}\frac{d}{dt} \int \frac{q}{4 \pi \epsilon_0 r^2} \hat{e_r}\cdot d\vec{A}
where if you imagine the charge moving along a line, and r is the point you want to calculate the field, then R is the perpendicular distance from r to the line. The surface of...
Because technically when your object is at the focus of a lens, the magnification is infinite, but the image is infinitely far away, so what's important is the angular magnification or what's in the parlance, magnifying power, which is the ratio of the angular size of the object with the lens...
Yes, if it's biconvex or biconcave. Say light starts at the left and travels rightward where it goes through the lens, and then leaves the lens. Say the light ray has an initial angle that's upwards. When the light hits the convex surface on the left, then it is bent down a little bit. If it...
This is something I never really understood. Isn't the point of a lens to have a small focal point? Because that's what you want to do, to bend light?
So it seems to achieve the lowest focal point, you would choose a biconvex or biconcave lens. What's the point of the rest?
I took coherence to mean for a given circumstance, not for all circumstances. So sunlight for example is regarded as spatially incoherent, but technically if the area that it's focusing on is less than 10-3 mm^2, then it's coherent.
I'm not sure how something can have a short coherence time and...
Re: Waves
Yeah, power radiated happens to be proportional to amplitude of current squared, just like impedance.
Quick question. The load at the end of a transmission line is transformed to a new impedance, and it is this impedance that the transmitter sees. Is the impedance that the...
Your eyes are a thin lens, with focal length 1.85 centimeters. So another way to approach this problem is with the thin lens equation, that gives you the magnification of an object as a function of distance.
Essentially, m=xi/xo, but 1/xi+1/xo=1/1.85, so m=1.85/(xo-1.85). But the object...
Re: Waves
If you have an antenna that is three quarter-wavelength long, then what would it behave like? I would like to say short circuit, but the fact that energy is radiating into space suggests that it can't be a short circuit, that the resistance has to be at least the radiation...
Re: Waves
I'm a little confused about this. Isn't the impedance of an antenna infinity, since an antenna is an open circuit? Therefore an antenna always has full reflection of the transmitter wave, and you get standing waves both in the antenna and in the line leading up to the antenna. Or...
I think your first integral is zero, even though there is a discontinuity at the boundary. The only possible way that your first integral could be non-zero is if the discontinuity jumped to infinity.
As for your second integral equaling your third integral, that's correct.
I'm probably being nitpicky here, but optics textbook probably ought to have
\nabla \cdot D =0
since you will have charges on boundaries, although they're not "free" ( i.e., they're not put there by hand).
Inertia is mass. Mass is scalar, the same no matter how fast observer.
Momentum is vector. Momentum depends on velocity of observer.
I find that interesting, but I'm not sure I follow. How do particles transform through their gauge group as they propagate in time? Don't mass matrices...
I believe the correct terminology is helicoid.
A Riemann surface is something else entirely. I'm not a mathematician, but as I understand it, a Riemann surface is just a way to visualize different branches of a function and how they are related to each other.
If friction steals away all the energy of the particle before it reaches at most an equilibrium position of x_e=\frac{mg(sin \theta +\mu cos\theta)}{k} , where xe is the distance from the starting position of the particle and k is the spring constant, then there is no oscillation. The...
I meant voltage. For a capacitor, Q=CV, where C (the capacitance) is real. So whatever phase the voltage has, the charge Q has the same phase. A Hertzian dipole (i.e. a very tiny antenna) is pretty much an AC generator connected to a capacitor. If you want you can put two spheres at the end of...
Your equation (11.9) says that one of the quantities must be imaginary. So it's either Q or I.
A Hertzian dipole behaves like a capacitor, where the current leads the voltage. Now for a capacatitor Q follows V. So you expect the current to lead the charge. Since a hertzian dipole behaves like...
You derived this:
i(t)=d[q(t)]/dt=ℜe[Qd(ejωt)/dt]=ℜe[Qjωejωt]=−ωQsinωt
which is incorrect.
Q is imaginary, so you can't just take it out of the Re[]. What you do is plug in Q=+- I/(jw), and then take the real part. Then you should get something like: I(t)=Icos(wt)