Yes but as soon as the water ends up in the tank, it moves the tank and then the Earth. I imagine most of the energy from the KE of the water gets transferred into the KE of the Earth. Think about it. If the Earth were a ball of 10cm radius then most of the KE of the water would transfer into KE...
With regards to this setup of two point charges, I was thinking the other day (assuming both charges are the same magnitude and opposite sign...) What is the equation of motion? If we write Coulomb's Force Law then the equation seems hard to solve. Can we use energy/ work done equations to solve...
This is just common sense (well, to me anyway). Assuming all the energy gets transferred into heating the water and none of that heat gets lost, then yes, all the kinetic energy heats up the water. But this never happens in reality due to the nature of materials and their properties.
In real...
I see. Considering (a/z)<<1 or instead considering (a/z)^2<<1 leads to the same result (no information is lost w.r.t. the Binomial expansion approximation). But in both cases x = (a/z)^2
I'm not sure I get that at all. Are you sure you're not considering (a/z)^2 and therefore (a/z)^4 << 1 and similar for higher order terms? Because then the Binomial expansion formula would seem to work. I'm not used to using it with the exponent non-integer.
What I said in my last post is...
Right. At first I tried taylor series at x=0 but didn't try it at x = infinity. I guess I hadn't come across using expansion about infinity in physics yet... only complex analysis.
Maybe there is something simple I am missing but why do you automatically look for expansion about infinity...
... with constant charge density σ = Q/((pi)a^2)
The Electric field is, after some calculation, is given by E_p below:
z is the z-axis, and a is the radius of the disc.
Now for the questions at the bottom of the page, here are my thoughts:
σ is independent of a because as a->2a, Q->4Q, and...
I agree with what The Duck said. I will take this a bit further for your curiosity's sake.
Define the Parity operator P by
P|x> = |-x>.
The eigenvalues of P can be either 1 or -1 (try and prove this or tell me if you can't. Hint for proof: consider P^2 and it's eigenvalues).
These are the...
It should be noted that since an electron is a fundamental particle, it has no internal structure, and so the spin of an electron does not have any relation to it's motion internally (i.e. the electron does not spin round like the Earth spins round on it's axis).
Spin is an intrinsic property...
You should tell your lecturer the situation you're in and ask him if you'll survive (the course) if you carry on. He will know the answer better than anybody.
"Mathematical methods for physics and engineering" has a chapter dedicated to these functions. I would also recommend this book as the essential maths reference book to anyone taking a degree in a related area (maths, physics, engineering etc).
Edit: make sure you get the latest edition (if...
Most theories (/ when doing experiments we) assume that the electron, quark, etc are really point particles.
What is the evidence for this? I don't think, for example, that experiments at cern convey that they are. They could have an internal structure with a stronger force than the strong...
http://en.wikipedia.org/wiki/Almost_surely
What you're referring to is "almost never", which is a similar concept to "almost surely".
Edit: oh yeah, I forgot to mention: an event with probability 0 CAN happen. For example, in the darts scenario on the unit square (the example provided by...
Well those things aside, would just dividing by the time part of your wavefunction give your new desired wavefunction? Or do you want a single operator which does it for all wavefunctions?
I'm probably way off here, but oh well...
Doesn't this depend on if you are considering the Schrodinger picture or the Heisenberg picture?
If it is the former then H is time-independent, in the latter it is time-dependent. The time- dependence or independence of H will transfer to the time-...
1. Bear in mind that the exact point of centre of gravity of the Earth is continuously changing. Also, it depends on, for example, what matter you actually consider to be part of the Earth (how far out w.r.t. the atmosphere, etc).
2. The resultant Force on a POINT PARTICLE only will be zero...
Actually I think I was wrong before and have had a slight realization.
This is the main point.
If you act the observable A on the state |psi> then, say you get |phi>.
Using A again will not change the state, i.e. it will give you |phi>.
Now using B where [A,B] = 0 will not change...
So in the attachment, in fact (6), the formula for the propagator rectangled in red...
is the Hamiltonian ACTING on (t-t')?
is the Hamiltonian a function of (t-t')?
or should it be (this is what I think), to be more clear
U(t,t') = exp((-i/h)(t-t')H), so that when acting on a state...
Given (2) to be true, if we have another eigenstate of H with eigenvalue NOT equal to hw(n+1/2) where n is an integer, then we get a problem. Can you see what this problem is?
(so for example, say we have an eigenstate of H with eigenvalue (4 + 1/3)hw. Try and get a contradiction.)
What do you mean by this?
A wavefunction has a set of possible results if you measure one of it's properties (e.g. spin or position).
If a measurement is made, then we get our results.
If you measure the position of an electron then you will find the position.
Okay, well actually you...
andrewkirk, I read through the OP once. Before I read through it again, can i just say you have a lovely imagination. But as San k said, to "measure" a property of an electron, you must "interfere" with the electron. You can't "peek round the corner using a curved mirror" to see what the...
But what causes the motion?
We don't yet have a complete understanding of what causes motion, although lots of existing theories are extremely good at approximating motion of particles to a very good accuracy. Most theories require a "singularity" at the beginning of time, or more commonly...