I have a feeling that a general physics book, such as Serway's Physics for Scientists and Engineers with Modern Physics would be helpful for you. And no need to go with the latest edition. You can find older editions (used) for next to nothing.
You seem to be thinking that in the frame where he particle is at rest, since it is at rest and not accelerating, it cannot emit radiation. But that is only true for particles at rest in inertial frames.
That's not correct. ##x## is here the coordinate (in position representation; more generically, it would have been better to use here ##\hat{x}##, the position operator). It has no time dependence. For a quantum mechanics oscillator, its position is in its state, that is, in its wave...
A partial derivative is a derivative with respect to a variable where other variables are held constant. If ##x=t^2##, then ##x## isn't held constant.
Note that this has nothing to do with quantum mechanics. It is calculus. For a given ##f(x,y)##, ##\partial f / \partial y## can have only one...
You are considering a diatomic ideal gas, so the internal energy is
$$
U = \frac{5}{2} PV
$$
This equation works if the pressure changes linearly, which is not true for adiabatic compression. You have to calculate
$$
W = - \int_{V_i}^{V_f} P \, dV
$$
In order to save oneself the trouble of...
##x## appears on both sides of the equality, so this is incomplete.
Consider first the latter: ##x## is obtained by solving
$$
x^2 + x + 2 = (z-x)^2
$$
then ##\sqrt{x^2 + x + 2}## is obtained by substituting the result for ##x## in ##z-x##.
The generic equation for the total number of fermions is
$$
N = \int_0^\infty f(\varepsilon) D(\varepsilon) d\varepsilon
$$
where
$$
f(\varepsilon) = \frac{1}{e^{\beta(\varepsilon + \mu)} + 1}
$$
is the Fermi-Dirac distribution and ##D(\varepsilon)## is the density of states. The degeneracy...
You can't leave n undefined. Mathematica can't find the rule for arbitrary n by itself. You have to calculate the result for explicit values of n, and then find by yourself the general rule.
Pedantic point: the ##\hat{S}_\xi## operators are not 2x2 matrices, but they can be represented as 2x2 matrices (for a spin-1/2 system). But this representation is not unique, as it will depend on the choice of basis.
As for ##\hat{S}^2##, the thing is that ##\hat{\mathbf{S}}## it is a vector...
I recommend also the books by Walter Greiner:
https://link.springer.com/book/10.1007/978-3-662-00707-5
https://link.springer.com/book/10.1007/978-1-4612-0827-3
I have just stumbled upon the following article, which may be of interest:
How to realize a universal quantum gate with trapped ions
https://arxiv.org/abs/quant-ph/0312162
The mass of the object doesn't disappear. The gas molecules retain the momentum of the original ice. This water vapor will push against the air in the craft, which will push against the walls, and so on.
Yes it does. Why is it counter-intuitive to you? Imagine a block of ice in the middle...
Gates are not devices, they are manipulations of the qubits. Depending on the actual qubits, they can be for example implemented by laser pulses or by varying electric and magnetic fields.
The most recent news articles I could find are from September 2022, and it appears that the baby was still on life support at that time.
Here are some links:
https://www.standard.co.uk/news/uk/baby-breathe-declared-dead-high-court-case-guys-st-thomas-hospital-london-b1021291.html...
There are no Kelvin "degrees". The kelvin is a unit, like the newton or the joule, and there is no reason to affix it the term degree.
Second, temperatures in kelvin can be negative, and then correspond to a system that is "hotter" than any system with a positive temperature, see...
The phrasing of the answer lead to a misunderstanding. It is never four straight lines, it is either four circles or three circles and a straight line. If A, B, and C are collinear, then you can draw a straight line as you did, and three circles concentric with the circles passing through ABD...
If you want to understand more fundamentally where the MB distribution comes from, starting from the barometric equation makes no sense, since it itself comes from a more fundamental basis. (The ideal gas law, which is the other thing used in the video, can also be derived from more basic...
I assume that the latter is correct (plus sign, not minus, in front of ##2N##).
Considering that the answer is given with two significant figures and that ##0.4## only has one, this looks like a typo (missing 1).
Yes. My point was that this is not how things are usually done (you don't consider individual protons).
I should also have added that I have never seen a definition of a dipole moment that took into account the mass of the charges.
My guess is that using a point dipole is the only way to get an analytical solution. It is an interesting toy problem, but I don't think anybody uses such a model for actual molecules.
Air travels far and wide (think about the short time it took for radioactive material go from Chernobyl to the farthest reaches of western Europe). Mixing is also quite good.
You are right that water can be localized, but overall, because of evaporation, you end up in the same situation as air.
Atmospheric pressure is 101325 Pa. The room will contain about 1300 moles of gas. Even with these numbers, I get that 1000 ppm of CO2 will be reached in less than 1 hour. But in anything but a sealed room, air exchange with the outside will be important.
In the PhysicsForums way of doing things, I will try and guide you to the solution.
How many molecules/moles is that?
Setting a temperature, you can use the ideal gas law to convert that to a number of molecules/moles. You can then compare with the number above (we can assume that all that...
I don't know which equation you should start from since I don't know what is to be taken for granted in your course. My starting point would have been to start from entropy as a function of the partition function.
From the integration by parts (integral of ##T^{-2} dT##).
Edit: To be clear, I moved the factor in front of the integral, but it appears in ##1/2T## in the image you posted.