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61

We don't feel any acceleration because the Earth and all of us humans on it is in free fall around the Sun. We don't feel the centripetal acceleration any more than the astronauts on the ISS feel the acceleration of the ISS towards the Earth. This happens because of the way general relativity describes motion in gravitational field. The motion of a freely ...


49

John Rennie's answer is right from the viewpoint of General Relativity -- but since the question is tagged with Newtonian mechanics, it deserves a Newtonian answer too. In the Newtonian framework, I think the best answer to "why don't we experience this force" is that we can't feel forces that apply to our body at all. What we actually experience with our ...


25

Model the tree as a point mass $m$ located some height $h$ above the ground --- that is, forget the mass of the trunk and assume all the mass of the tree is in the branches and leaves above the ground. Then the moments of inertia of the tree before and after felling are \begin{align} I_\text{tree,up} &= m \left( (R+h)\cos\theta \right)^2 \\ I_\text{...


12

According to the Equivalence Principle a free falling system cannot locally detect a gravitational field. However Earth is a large enough system such that non-local effects turn out to be appreciable. Solar tides are - although small - detectable. So in principle one can experience the Sun's gravitational field even though we are in free fall. What I claim ...


7

When we say that the electron has "spin half," we mean half of the quantum of angular momentum, $\hbar$. A good quantum mechanics text or other reference will help you derive that the Laplacian operator transforms into spherical coordinates like \begin{align} \nabla^2 &= \left(\frac\partial{\partial x}\right)^2 + \left(\frac\partial{\partial y}\right)^...


4

Even if the orbit were a perfect circle, there's some acceleration towards the sun. If there weren't acceleration then the earth would move in a straight line (instead of a circle); but it doesn't move in a straight line therefore there's acceleration. In a sense, the earth doesn't feel the acceleration because it doesn't try to resist it: if you stand on ...


4

John Rennie has answered the question in terms on general relativity, but it can also be answered with Newtonian physics. Your question is very similar to this one: Why does the moon stay with the Earth? and I can refer you to my answer there. In short, the Sun isn't only pulling on the Earth itself, it's pulling on everything on it as well, including us, ...


4

Cutting the trees and leaving them flat instead of vertical will diminish the moment of inertia of earth. The angular momentum of course will not change, but the speed of rotation will increase. However, I do not believe that the change is measurable with current instruments.


4

I know the example given sounds crazy but the physics behind it might be useful for someone learning rotational dynamics. The angular momentum of a system does not change if there are no external torques, since $$\frac{d\vec L}{dt}=\vec \tau,$$ where $\vec L$ is the total angular momentum and $\vec \tau$ is the total external torque. So if you cut the trees ...


4

Because it doesn't have total spin $s=0$ - it has total spin $s=1$, with the spin component parallel to the $z$-axis being zero. If you looked at that state in a different basis (e.g. the $x-$ or $y-$ basis) it would very clearly not have spin 0.


4

The short story is that the Hyperphysics article you link to is using classical and semiclassical heuristics to justify the numbers it present. As I'm sure you're aware, the hydrogen atom cannot be described in any rigorous detail using classical mechanics, and instead requires quantum mechanics for any appropriate treatment, particularly where the ...


4

Given an angular momentum operator with components $S_1, S_2, S_3$ and commutation relations $[S_i, S_j] = \sum_k \epsilon_{ijk}S_k$, where $\epsilon_{ijk}$ are structure constant of the $\mathfrak{su}(2)$ algebra, the Casimir operator $S^2 = S_1^2+S_2^2+S_3^2$ can be diagonalised simultaneously with any of the original components $S_j$ onto their ...


3

No you cannot assume that. The initial rotation is about the major axis, and it will continue to be so (in the absence of torque, and since you were already rotating about the major axis). Instead, since $\omega_2=\omega_3=0$, your equations for the evolution of the angular momentum don't require the moments of inertia to be the same.


2

Particles do have half-integer spin! All fermions do. It is the bosons which have to have integral spin. Spin is an intrinsic angular momentum of a particle. You can try to imagine that the particle is spinning, if this helps you sleep better, but it isn't really what is happening. You might have heard that the angular momentum is conserved because of the ...


2

I think the argument is that the proton as well as the neutron and electron are spin half particles. There is no way to add two spin half such that you get a spin half as a result. This is also a short argument why two fermion behave as a boson.


2

Displacing some mass closer to the axis of rotation reduces the moment of inertia $I$. Considering Earth as an isolated system (which is not), its angular momentum $L$ must be conserved: $$L=I\omega = \text{const}$$ Therefore if $I$ goes down, the rotation frequency $\omega$ must increase. But if we should also consider the reduction of $I$ while the tree ...


2

I would think that the whole atmosphere surrounding the earth is far heavier than the trees that were cut. The atmosphere turns with the earth and the changed position of trees would not even be noticed.


2

You might hear the story about figure skating. When a rotating person expands his/her arm, he/she can slow down rotation. Same thing can happen with earth. Assuming the tree is trillion tons and you cut it and lift it up, you can slow down the earth.


2

You can think about this in two different ways. One way is to look at the initial and final angular momentum. If you go from $L\cdot(0,1, 0)$ to $L\cdot(1,0,0)$ you need to remove the $Y$ component and add the $X$ component. If you just calculate the difference in the angular momentum, then you get $$\Delta L = L\cdot (1,-1,1)$$ which would immediately ...


2

I commented on the question but on advise, I am promoting this to an answer. A particle moving in a straight line can have non zero orbital angular momentum if the origin of the coordinate system doesn't lie on that straight line. There can also be helical modes through which light can have non zero orbital angular momentum. The first one is origin dependent,...


2

Spin operator of the total 2 electron system is tricky: the statistical requirement reduces the Hilbert space to a 3-d rather than 4-d version. Like a spin-1 system, the $S_z$, as represents in basis $|S_z=1,0,-1\rangle$ is (see http://quantummechanics.ucsd.edu/ph130a/130_notes/node247.html, for example) $$S_z=\hbar\left(\begin{matrix}1&0&0\\0&0&...


2

As a matter of fact, there is a much easier way to derive those matrices. What you are really after is the matrix representation of the $\mathfrak{su}(2)$ Lie algebra generators $J_x$, $J_y$ and $J_z$ in the irreducible $\mathrm{SU}(2)$ representation with spin $j$, given by $$ J_z|j,m\rangle = m |j,m\rangle,\quad J_\pm|j,m\rangle = C_\pm(j,m) |j,m\pm 1\...


1

Recall that the value and sight of $\mathbf{r} \times \mathbf{p}$ angular momentum depends on the point around which you chose to measure it. A free particle can (indeed, does) have angular momentum around any and all points relative which it is moving and has a non-zero impact parameters. But, frankly, that's not a very interesting statement.


1

You definitely don't need to use General Relativity to answer this question. It depends upon what you mean by "feel". If "feel" means "detectable by sophisticated instruments" then, yes, it can be "felt". But your body is not a very sophisticated detection instrument. According to what I've read elsewhere, the Earth speeds up by $1000$ $m/s$ as it moves ...


1

Sometimes a picture can help. Its not hugely rigorous (especially for a physics site) but using the vector model, pictures can be constructed to illustrate the states that two spins $\alpha$ and $\beta$ can form.


1

To formalize @knzhou's comment: The answer, in a nutshell, is no. Your basic assumption that cutting trees reduces the Earth's mass is wrong, because trees don't leave the Earth when they are cut! Even if all trees left the Earth when cut, a lot of tree cutters plant trees to replace what they cut, and each tree is such a tiny amount of mass ...


1

Assume, without the loss of generality, that the angular momentum operator has only one component, $\hat{L_z}$. It can be shown that the following eigenvalue equation holds true. $$ L_{z} |l, m \rangle = m \hbar |l, m \rangle \quad \text{such that} \quad m \in \mathbb{Z}. $$ This is to say that the eigenvalues of the orbital angular momentum operator ...


1

I want to give an answer for electrons which have both an intrinsic spin and a magnetic dipole moment. The key for the understanding why electrons in rest in relation to an external magnetic field get aligned and do not precess while moving non-paralle to the external magnetic field electrons undergo a precession is their kinetic energy in relation to the ...


1

Let $\:\mathbf{L}\:$ the vector angular momentum operator $$ \mathbf{L} =\left( L_{x},L_{y},L_{z} \right) \tag{01} $$ The eigenfunction $ | \psi_{n\ell m}\rangle\:$ is a common eigenstate of $\:H,L^{2},L_{z}$ and more precisely for our case is an eigenstate of $\:L_{z}\:$ with eigenvalue $\:m\:$, so $$ L_{z}| \psi_{n\ell m}\rangle =m\hbar|\psi_{n\ell ...


1

This turns out to have a really boring answer. We can find the two circular orbits by finding the maxima and minima of the effective potential, and we get: $$ r = \frac{L^2}{2M}\left(1 \pm \sqrt{1 - \frac{12M^2}{L^2}}\right) \tag{1} $$ where the $+$ gives the outer stable orbit and the $-$ gives the inner unstable orbit. Note that both orbits exist only ...



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