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If an earthquake can destroy buildings why it cant kill us according to physics? Most earthquakes with magnitude 5.5 and higher can damage or destroy buildings. However, according to my knowledge and experience, I have never seen someone dying from an earthquake itself. Rather, they die from an associated tsunami, damaged buildings, etc.
This seems counter-intuitive, since you need much more force to destroy a building or damage it than to break the human femur or cause similar damage to other species.
| There are several differences between humans and buildings:
*
*Anchoring Suppose I were to push you backward. Just on reflex, one of your feet will move backward to catch yourself. A building, on the other hand, is anchored in place. When the ground moves, the bottom of the building has to move with it. But a human body doesn't have to move with the ground. There's only so much force that a shaking ground can apply to the human body.
*Flexibility: The human body is designed to move (anthropomorphizing evolution a bit here). Other than bones, our organs are flexible, and the bones are connected by multiple joints and that can bend without harm to the body (that what a joint is: a part of the body that's supposed to move). A building doesn't have that level of flexibility. Nowadays buildings, especially ones in earthquake-prone regions, are often designed with some flexibility, but that can't match that of a human body.
*Scaling: This is probably the largest factor. Things work differently at different scales. If you double every dimension, you multiply the cross-section by four, but you multiply the volume by eight. If a building is a hundred times as large as a human in each direction, then it's going to have a hundred times the volume per unit of the cross-section. And torque is proportional to both mass and distance, so you can get ten thousand times as much torque per area.
| {
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Is there a difference between topological defects and topological soliton? Is there a difference between topological defects and topological soliton? Or are these objects the same thing? I ask this because it very common find some papers whose the authors itself refer, for example, the domain wall's, monopoles, cosmic string, etc, as topological defects, while in others, the same objects are called of topological solitons. As the authors aren't specific about this It seems that defects and topological solitons are the same things. But will be this idea conceptually correct?
| Yes, there is. In static situations, both are strictly the same thing. But, when analyze the dynamic of these structures, we see that they are differents.
| {
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Why would entropy of a system be fixed if it can exchange energy with its environment? Entropy maximization and energy minimization are equivalent statements of the same thing, as I understand it. If the internal energy is fixed, entropy is maximized because of statiatical reasons. If the entropy is fixed, and the system can exchange energy with its environment, then the system will give energy to its environment to maximize the environments entropy (and hence total entropy).
But I haven't been able to understand why we would assume here that the entropy of a system is fixed. What physical mechanism causes this?
| I think you are misunderstanding a few things. Both the entropy maximization principle and energy minimization principle are due to statistical reasons. Suppose we have some parameter of the system $X$ (e.g. volume, particle number, etc.) that is free to vary, and the energy is fixed at $U_0$. The system will evolve to a $X_0$ such that the entropy is maximized at the value $S_0$.
Now the energy minimization principle refers to this same point $X_0, S_0, U_0$. If we change $X$ away from $X_0$ while keeping $U_0$ fixed, $S$ will decrease. So if we want to bump $S$ back up to $S_0$ we need to add energy. So any point $X$ which has the same entropy $S_0$ must have at least as much energy as $U_0$. This is all the energy minimization principle is.
This is just an alternate way of characterizing the same point $X_0, S_0, U_0$. Again, if we move away from it while keeping $S_0$ the same, we need to add energy. It is a choice we make, not some physical mechanism.
| {
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Why don't (can't) astronomers take advantage of interferometry to the extreme? Reading and watching from various sources (e.g: APOD, ESO videos, or Wiki), whenever interferometry is mentioned, it is also usually explained like this:
An astronomical interferometer consists of two or more separate telescopes that combine their signals, offering a resolution equivalent to that of a telescope of diameter equal to the largest separation between its individual elements.
So why is ESO building a humongous ELT? I mean, we can achieve even better results by using amateur telescopes separated by, say, 10km. The construction of CERN's LHC pretty long ago gave the impression that building gigantic, high-precision structures are within humans' ability. Even if the construction cost is much higher, a hypothetical 10km interferometer would have superb resolution power, dwarfing the ELT by orders of magnitude. Not to mention the upkeep cost is much lower than an ELT.
Therefore, I suppose that there must be some setbacks in interferometry that are preventing people from reaching such mentioned extremes. What are they?
| ELT is an optical telescope. The short wavelengths of light necessitate extreme precision and stability of construction.
| {
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Two-photon interference inside Mach-Zehnder interferometer Imagine there's a strong laser beam, not just an attenuated stream of single photons, entering a balanced Mach-Zehnder interferometer.
One-photon picture: Each photon interferes with itself on the second beam splitter. As a result, all photons leave out of the same output.
More realistic picture: If the intensity of the laser is strong enough, a lot of photons will enter the interferometer within the coherence time and will be therefore indistinguishable (at least by their arrival time).
How does this fact change the quantum mechanical description of the state? Shouldn’t there be a two-photon interference at the second beam splitter in this case? How do you describe the evolution of the state throughout the interferometer mathematically: two-photon interference of photons that are themselves in the same superposition of two paths?
| A MZ interferometer is very carefully setup, you can think of the path distance in 1 arm to be close to a perfect distance of n wavelengths of light and the other arm to be a n+1/2 wavelengths of light. Single photons entering the MZ interferometer will have the highest probability of transmission in the n arm and almost 0 probability in the n+1/2 arm.
Feynman's concept of the path integral and the photon wave function says the light will take the shortest path that is n multiples. If many photons are fired the result would be almost all photons in the n arm.
| {
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On proving that charge is linearly proportional to potential for a conductor In Mr. Purcell's Electricity and Magnetism, page 103, it is stated,
An isolated conductor carrying a charge $Q$ has a certain potential $\phi _{0}$, with zero potential at infinity. $Q$ is proportional to $\phi _{0}$. The constant of proportionality depends only on the size and shape of the conductor. We call this factor the capacitance of that conductor and
denote it by C. $$Q=C \phi _{0}$$
I understand that for a given charge $Q_{0}$ and its corresponding potential $\phi_{0}$ we could define a $C_{0}$ as a function of the shape and size of the conductor such that $Q _{0} =C_{0}\phi_{0}$.
When we change the charge to $Q_{1}$, the potential will become $\phi_{1}$. How can we prove that it is the same constant $C_{0}$ that will link $Q_{1}$ and $\phi _{1}$ ? In other words is charge being linearly proportional to potential an experimental result or can we prove it?
If one argues that it is the same constant because it depends only on the shape and size of the conductor, then they must also prove that this constant does satisfy $$Q=C _{0} \phi$$
for every given charge and its corresponding potential.
| we know charge and potential have linear relation , as charge is directly proportional to potential. so if you draw charge versus potential graph you'll get straight line ( notice: the value you put on graph are experimental ) . now if you compare the equation y=mx and Q=CV means C(capacitance) represent slope which comes constant . Means C doesn't depend on Q and V
| {
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Sign-Conventions for Spinor Transformations In the literature one encounters a lot of different conventions for how left-handed spinor transforms (rotation angle $\phi$, rapidity $\beta$), among them
$M_L = M_{(\frac{1}{2}, 0)} = e^{-i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} - \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$,
$M_L = M_{(\frac{1}{2}, 0)} = e^{-i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} + \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$,
$M_L = M_{(\frac{1}{2}, 0)} = e^{+i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} + \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$
Comparing various sources can be a pain, so I tried to establish a definite convention (as a reference for me), to compare against various sources, but I failed so far to do it consistently.
Specifically I tried to understand the conventions used in the following source:
Dreiner, H. K., Haber, H. E., & Martin, S. P. (2010). Two-component spinor techniques and Feynman rules for quantum field theory and supersymmetry. Physics Reports, 494(1-2), 1–196. doi:10.1016/j.physrep.2010.05.002 (http://arxiv.org/abs/0812.1594)
Is use the version with the $(-1, +1, +1, +1)$ convention that can be obtained from https://www.niu.edu/spmartin/spinors/ (whereas the arxiv version uses the (+1, -1, -1, -1) convention).
They state (page 10) that the transformation of a left-handed spinor ought to be:
$M_L = M_{(\frac{1}{2}, 0)} = e^{-i \frac{1}{2} \vec{\theta} \cdot \vec{\sigma} - \frac{1}{2} \vec{\beta} \cdot \vec{\sigma}}$
I then tried to get the corresponding lorentz-transformation by looking at the transformation of the hermitean matrix $X = \left( \begin{array}{cc} x^0+x^3 & x^1-ix^2\\ x^1+ix^2 & x^0-x^3 \end{array} \right)$
$A: X \rightarrow X' = A X A^\dagger$
To extract the lorentz-transformatin I use
${\Lambda^\mu}_\nu = -\frac{1}{2} \operatorname{tr}[\sigma^\mu A \bar{\sigma}_\nu A^\dagger]$
with $\sigma^\mu = (1, \vec{\sigma})$ and $\bar{\sigma}_\nu = (-1, -\vec{\sigma})$
This is not covered in their paper, but when I do it for
$R = e^{-i\phi\frac{1}{2}\sigma^3} = \left(\begin{array}{cc} e^{-i\frac{1}{2}\phi} & 0 \\ 0 & e^{i\frac{1}{2}\phi} \end{array}\right)$ (rotation around the z-axis)
I get the lorentz transformation
${R^\mu}_\nu = \left( \begin{array}{cc} 1& 0& 0& 0\\
0& \cos\phi& -\sin\phi & 0 \\
0& \sin\phi& \cos\phi& 0 \\
0& 0& 0& 1 \end{array} \right)$
Similar for
$B = e^{-\beta\frac{1}{2}\sigma^3} = \left(\begin{array}{cc} e^{-\frac{1}{2}\beta} & 0 \\ 0 & e^{\frac{1}{2}\beta} \end{array}\right)$ (boost in z-direction)
I get
${B^\mu}_\nu = \left( \begin{array}{cc} \cosh\beta& 0& 0& -\sinh\beta\\
0& 1& 0& 0\\
0& 0& 1& 0\\
-\sinh\beta& 0& 0& \cosh\beta \end{array} \right)$
These correspond to active rotations (counterclockwise) around the z-axis, but passive boost in the z-direction.
But this is in constrast to their statement (page 10) that corresponding infinitesimal transformations are
$\vec{x} \rightarrow \vec{x} + (\vec{\phi} \times \vec{x})$
$\vec{x} \rightarrow \vec{x} + \vec{\beta}t$
which both are active rotations and boosts.
So either:
*
*something is wrong in my calculation from $SL(2, \mathbb{C})$ to the Lorentz Group (or may be also there is a hidden convention I got wrong ?)
*or the cited paper got it wrong (unlikely, since its already in its 5th incarnation, with a list of errata [https://www.niu.edu/spmartin/spinors/])
*or I am quite generally confused about how exactely spinors should transform, and specifically how the signs are determined.
| A wrong assumption was made in the question asked.
An element $A\in SL(2,\mathbb{C})$ that induces the Lorentz transformation $X \rightarrow A X A^\dagger$ is a Spinor transformation for a right-handed Spinor (not left-handed, as claimed).
Rather, a left-handed spinor transformes with an element $A\to SL(2,\mathbb{C})$ that induces the Lorentz transformation $X \rightarrow X^{'} = \left(A^{-1}\right)^\dagger X A^{-1}$.
So everything fits together and is consistent.
| {
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Why don't we care about bound charges inside dielectric? I'm quoting Marek's answer from the question : Difference between electric field E and electric displacement field D
..materials have lots of internal charges you usually don't care about. You can get rid of them by introducing polarization P (which is the material's response to the applied E field). Then you can subtract the effect of internal charges and you'll obtain equations just for free charges
I couldn't figure out what are the exact effects he is talking about at first. After that I saw that maybe it is due to the fact that bound charges do not contribute to electric flux desnity field D outside the material (because flux line originates from bounded positive charge and terminates on bounded negative charge inside the dielectric, is it correct?). Is it one of the reasons that we are at large not interested in bound charges and have developed suitable mathematical ways to eliminate troubles caused by them? Are there any other reasons?
Thanks in advance.
| We cannot control the bound charges, but we can control the free charges. Therefore, it makes sense that we would want to refer to something that only depends on that which we can control. This is why we introduce the electric displacement
$$\mathbf D=\epsilon_0\mathbf E+\mathbf P$$
where the bound charge volume density $\rho_b$ relates to the polarization by
$$\rho_b=-\nabla\cdot\mathbf P$$
On the other hand, by construction, the electric displacement only depends on the free charge density $\rho_f$$^*$
$$\nabla\cdot\mathbf D=\rho_f$$
So you can see why this is useful. We can't control and might not even know what the bound charge is doing in our system. But we control the free charge, so it makes much more sense to use $\mathbf D$ when thinking about electric fields in matter. This is especially true if we assume that the polarization is proportional to the field for linear dielectrics.
$^*$Although if the curl of $\mathbf P$ is non-zero then we have
$$\nabla\times\mathbf D=\nabla\times\mathbf P$$
and so $\mathbf D$ is not solely determined by the free charge.
| {
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$O(p,q)$ as transformations that conserve quadratic form Let us try to define $O(p,q)$ in two different ways, which I want to show their equivalence.
*
*Define the symmetric bilinear quadratic form $[\cdot ,\cdot]$ which is given by $$[x,y]=\langle x,gy\rangle$$ where $\langle \cdot,\cdot\rangle$ is the standard inner product on $\mathbb{R}^{p+q}$, and $x,y \in \mathbb{R}^{p+q}$ and $$g={\rm diag}(1,\cdots,1,-1, \cdots, -1)$$ with $p$ number of $1$s and $q$ number of $-1$s. Then define the $$O(p,q)=\{A\in \mathbb{R}^{n\times n} | [Ax,Ay]=[x,y], \ \forall x,y \in \mathbb{R}^n \}$$ where $n=p+q$.
*Using the same quadratic form, define $$O(p,q)=\{A\in \mathbb{R}^{n} | [Ax,Ax]=[x,x], \ \forall x \in \mathbb{R}^n \}.$$
Both of them can be used to denote conserving the quadratic form. So my question is, are they equivalent or are they non-equivalent? If they are, how can we show that, and if they aren't, which one should we prefer?
| OP's two definitions of the indefinite orthogonal group$^1$ $O(p,q;\mathbb{F})$ are equivalent, which can be proven with the help of a polarization identity.
--
$^1$ Here $\mathbb{F}$ is a field with characteristic different from 2.
| {
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Some digital watches display digits in black on a white background and in the dark the backlight is reversed. How is this possible? Some digital watches display digits in black on a white background and in the dark the backlight is reversed, blue on a black background. How is this possible?
Here is an example:
https://images-na.ssl-images-amazon.com/images/I/7196xuw29OL.UL1500.jpg
Usually the backlight is black on a blue background and it's very easy to explain:
https://en.wikipedia.org/wiki/Liquid-crystal_display
https://en.wikipedia.org/wiki/Polarizer
https://www.youtube.com/watch?v=MzRCDLre1b4
But even after searching everywhere I did not find an explanation to the case mentioned
The only hypothesis I have (and which I do not really believe)
is that the electroluminescent backlight is located between the LCD and the second polarizer
and that the light it emits is polarized perpendicularly to the second polarizer
and that it is transparent when it is off
Edit:
I may have mistaken the title
In reality the background does not really change, it is the digits that go from black to light (light blue)
Besides, under an external light the effect is clearly visible
Here's a video that shows the effect I'm talking about:
around 2:36
https://www.youtube.com/watch?v=CEpXyujMHaM
By analyzing frame by frame when the backlight switch
we see that the digits change without the background is really changed
It does not seem to be a reversal of the screen when the backlight is switched on
since the bottom does not change
Sorry for the mistake
| The other way to invert an LCD so the digit segments and the background are "flipped" is to have extra LCD elements built into the background which can be switched on and off independently of the digit segments. When you want backlit segments against a black background, you de-energize the segments and energize the background. When you want backlit background and dark digit segments, you do the opposite.
| {
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Differences between charge quantity and electric charge As a senior middle school from China mainland, I am teaching physics about electric field. I work with my workmates, and we got a problem now. We cannot get an agreement. There are three viewpoints.
The first is that: electric charge is physical attribute and a physical quantity. It means electric charge is a physical quantity. The unit of electrical charge is the coulomb (symbolized C).
The second is that: electric charge is physical attribute. Charge quantity is a physical quantity. The unit of charge quantity is the coulomb (symbolized C). Electric charge has no unit.
The third is that: electric charge is physical attribute and a physical quantity. Charge quantity is a physical quantity too. The unit of electrical charge is the coulomb (symbolized C). The unit of charge quantity is the coulomb (symbolized C) too.
| I am not entirely sure of the nuances of meaning, but I hope this helps.
In English we can say "the object is charged", and we can say "the charge on an object is positive" and we can also say "the charge on the object is 3 Coulomb. I think the first and possibly the second treat charge as an attribute, and the third treats charge as a quantity, though I am not sure.
In many situations we need to specify magnitude and sign, so we say "the charge on the object is +3 Coulomb" or "the object has a positive charge of 3 Coulomb.
In Physics we rarely distinguish between an attribute and a quantity. Even when somebody says "water is wet" (an attribute) somebody else will try to find a way to measure it so wetness is not only an attribute but also a quantity.
| {
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Factor of 3 in Photon Diffusion coefficient From definition of Diffusion coefficient:
$$D = c/3(\mu_a+\mu_s),$$
where $c$ is the speed of light front, $\mu_a$ is absorption coefficient and $\mu_s$ is scattering coefficient.
I wonder where does factor of $1/3$ comes from? I assume it is coming from dimensionality, but I didn't find arguments proving that.
UPD:
In this question I refer to diffusion approximation of Radiation Transfer Equation also known as Photon Diffusion Equation.
| Here is some information about this. The factor comes from an approximation and using spherical harmonics.
| {
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Gauge fixing of Polyakov Action In the Gauge fixing of Polyakov action we do general coordinate transformation where we take the transformation stated below
$$h_{\alpha\beta} = e^{\phi(\sigma)}\eta_{\alpha\beta}.$$
But here the left side has three free parameters (one less in the 2x2 h metric as it is symmetric in the indices) while the right side only involves one parameter $\phi$ ; taking the $\eta$ metric to be constant as
$$\eta{_\alpha}{_\beta} = diag(-1,1).$$
So how can we put an equality if there are not equal free parameters on both sides? What could be the underlying reason?
| It is important to also keep track of the local symmetries present in each case. In the first case the dynamical metric $h_{\alpha\beta}$ has three gauge symmetries - two diffeomorphisms and a Weyl symmetry.
It is possible to choose the parametrisation in a particular way such that $h_{\alpha\beta} = e^{\phi(\sigma)}\eta_{\alpha\beta}$, where one drops the diffeomorphism invariance by choosing a particular parametrisation. This means that initially one has three degrees of freedom and two local symetries (diffeomorphisms), and at the end one has only a single degree of freedom and no symmetries left. It is the number of DoF - local symmetries that actually matters, as these are the physical DoF.
One can take this a step further and use Weyl symmetry to set the metric locally to $h_{\alpha\beta} = \eta_{\alpha\beta}$. This step can be even extended globally if the worldsheet has a certain euler characteristic.
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Why doesn't stochastic resonance allow us to ignore noise? I don't know how to formulate the question better. This question is inspired from here where the author states that satellite communication is sometimes lost when radio noise from the sun makes the signal impossible to detect.
My understanding of stochastic resonance is: A weak signal can be strengthened by adding noise, so that it can still he higher than the threshold needed to trigger detectors. The noise can be filtered out and the signal remains (this part I am not sure I understood correctly).
Why, then, does this help when we do it intentionally (add noise), but not when the noise comes from the environment? Does it have to with the frequency of the noise signal vs the frequency of the signal? Aren't we always talking about white noise?
| The signal is not "strengthened" by adding extra noise to it; instead the added extra noise $N_X$ and the received signal and noise $S+N$ are in different frequency bands; when the original $S+N$ is smaller than say a quantization step then adding the right amount extra noise can make that $S+N$ flip between levels and that way the quantization error is smeared out by the $N_X$ added noise and that is averaged by the post-processing filter. The environmental noise does not help here because its spectrum spreads over that of the signal. The "added noise" can also be just a sine wave of the right frequency and right amplitude.
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Magnetic Field in Perfect Electric Conductor I know that one of Maxwell's equations states that the curl of the electric field is proportional to the time derivative of the magnetic field. We know that the electric field in a perfect electric conductor is 0, so why couldn't the magnetic field just be a constant instead of 0?
Edit: There is no current everywhere in space.
| Maxwell's equations give us the divergence and the curl of the magnetic field in this case:
$$\nabla\cdot\mathbf{B}=0$$
$$\nabla\times\mathbf{B}=\mu_0\mathbf{J}$$
For a conductor with current flowing, there is obviously a nonzero, non-constant (in space) magnetic field with a finite curl. I gather that this is not the situation you asked about, since you asked specifically about differentiating between zero field and a constant field (i.e. constant in space). In this particular case, we have a conductor with no current flowing, which means:
$$\nabla\cdot\mathbf{B}=0$$
$$\nabla\times\mathbf{B}=0$$
There is a very important theorem in vector calculus (commonly called the Helmholtz theorem) that, in a particular formulation, states:
Any field that vanishes at infinity is uniquely determined by its divergence and curl.
In other words, if we find a field with zero divergence and curl everywhere that also vanishes at infinity, then we have found the only field with these properties.
Here's the important part: for any finite conductor, we expect the magnetic field to vanish at infinity. So we need a field that has zero divergence and zero curl everywhere which also vanishes at infinity. It turns out that $\mathbf{B}=0$ is exactly such a field, and by the theorem above, it is the only field configuration which can describe this situation.
The other important part: any infinite conductor is constructed as a limit of progressively larger finite conductors. Since the field is zero everywhere for all of the finite conductors, it's natural to choose the field to be zero everywhere for infinite conductors as well.
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Derivation of Conservation of Energy from Newton's Second Law Given Newtons's Second Law: $$ \frac {d}{dt} (m \boldsymbol{\dot r}) = \mathbf F $$
How is it possible to derive the conservation of energy equation with a constant mass?
That is how can you derive $ \mathbf F = - \nabla V(\mathbf r) $ where $V(\mathbf r)$ is shown to be the potential energy when the force is conservative?
Attempted Proof:
Let $KE = T = \frac {1}{2} m\boldsymbol{\dot r} \cdot \boldsymbol{\dot r} $
or
$$\frac{dT}{dt} = \frac{1}{2}m[\boldsymbol{\dot r} \cdot \frac {d \boldsymbol{\dot r}}{dt} + \boldsymbol{\dot r} \cdot \frac {d \boldsymbol{\dot r}}{dt}] = m \boldsymbol {\dot r} \cdot \frac {d \boldsymbol{\dot r}}{dt}$$
and $\nabla V = \frac {\partial V} {\partial{\mathbf r}} $
Also, a conservative force says $\frac{dE}{dt} = 0$
Newton's Second Law could also be written as: $$m\boldsymbol{\dot r} \cdot \frac{d\boldsymbol{\dot r}}{dt} = \mathbf F \cdot \boldsymbol{\dot r}$$
My question is how is $ \mathbf F = - \nabla V(\mathbf r) $ introduced to Newton's second law properly and then integrated (? maybe) to obtain the energy?
Because I can easily prove that $ \mathbf F = - \nabla V(\mathbf r) $ if $E = T + V(\mathbf r)$ but I am trying to conclude that $V(\mathbf r)$ is the potential energy, not assume it
| As stated, the answer is: you cannot. There are force fields which are not conservatives, i.e. they cannot be written as the gradient of a scalar function. In such situations energy is not conserved. Such examples are quite common, consider for example friction.
Note that, by definition, a force is said to be conservative if it can be written as the gradient of a scalar function.
| {
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Expectation value of operators with non-zero Hamiltonian commutators I'm a bit embarrassed because there seems to be an obvious thing that I'm missing, but I can't see what it is.
Consider $[\mathcal{H},A]=B$, where $A$ and $B$ are some operators and $\mathcal{H}$ is the Hamiltonian. Let's look at the expectation value of $\mathcal{H}A$, and let's use energy eigenstates.
$$\langle \psi|\mathcal{H}A|\psi\rangle=\langle \psi|B+A\mathcal{H}|\psi\rangle=\langle \psi|B|\psi\rangle+\langle \psi|A\mathcal{H}|\psi\rangle=\langle B\rangle+E\ \langle A\rangle.$$
But, can't I also write
$$\langle \psi|\mathcal{H}A|\psi\rangle=\left(\langle \psi|\mathcal{H}\right) \left(A|\psi\rangle\right)=E\ \langle \psi|A|\psi\rangle=E\ \langle A\rangle,$$
where I've made use of $\mathcal{H}| \psi \rangle=E\ |\psi\rangle \Rightarrow \langle \psi|\mathcal{H}=E\ \langle \psi|?$
I don't see what's wrong with my argument, but I feel like something is wrong because I get different answers.
| $$\left\langle B\right\rangle =\left\langle \left[H,A\right]\right\rangle=E\left\langle A\right\rangle-\left\langle A\right\rangle E=0$$
| {
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Why do things cool down? What I've heard from books and other materials is that heat is nothing but the sum of the movement of molecules. So, as you all know, one common myth breaker was "Unlike in movies, you don't get frozen right away when you get thrown into space".
But the thing that bugs me is that things in the universe eventually cool down. How is that possible when there are no other things around to which the molecules can transfer their heat?
| Everything that is not 0 Kelvin radiates electromagnetic energy. In vacuum, this is the only relevant form of heat transfer. The hotter you are, the more energy you radiate (I believe the relevant equation is given here).
The question whether you cool off or heat up in space depends on whether you absorbs more electromagnetic radiation than you give away. So, for instance, if you are orbiting Earth in plain sunlight, you may actually warm up, because there is lot of electromagnetic energy coming from the sun (much of it as visible light).
But if you are in the shadow of the Earth, the amount of energy that comes from night side of Earth and the general direction of the outer space is very low compared the the amount of energy you radiate away at 37ºC, so you cool off and eventually freeze.
| {
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Rotation of helium balloons A number of balloons are attached to a circular disk with string. Some balloons are
filled with air and some balloons are filled with helium. The disk is hung freely from
ceiling of a room and is disk is rotated about its center. Assuming that the disk
remains horizontal while rotating, describe what happens to the balloons?
(a) All the balloons move away from the center of the disk.
(b) All the balloons move towards the center of the disk.
(c) Air balloons move outwards away from the center and helium balloon move
towards the center.
(d) Air balloons move inwards towards the center and helium balloon move out-
wards the away from the center.
I am in a doubt that is the centrifugal force will dominating and all balloon goes outside or some special effect(like pressure) will affect to helium balloon to go inside(like the behaviour of helium balloon in a car at a turn)
| Since the balloons are rotating, there must be a centripetal force that is keeping them going around the circle.
This resultant force is coming from a component of the tension in the string.
The centripetal force must point inwards towards the centre of rotation, so therefore the balloon must swing outwards so that part of the tension points inwards.
I can't see any difference in the behaviour between the helium and air balloons - in the helium balloons, the vertical component of the tension counters upthrust, whereas in the air balloons, the vertical component counters weight.
I made the following diagram to aid my explanation.
*
*$T$ is the overall tension in the string.
*$F_c$ is the centripetal force (resultant).
*$F_b$ is what I am calling the "balancing force" - in the case of the air balloon, this is the opposite of the weight vector; for the helium balloon, it opposes the resultant upthrust (total thrust minus weight).
| {
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What made Born interpret $|\psi|^2$ as a probability density? What was Born reasoning when he introduced the rule that $|\psi|^2$ could be interpreted as a probability density?
| You can consider the position basis, made of delta functions $\delta(x_i) \, \forall x_i \in \mathbb{R}$ (in $1$D).
You can write your wavefunction $\psi(x)$ in terms of that basis: $$ \psi(x) = \sum_i a_i \delta(x-i) = \int \mathrm{d}x_i\, a(x_i)\,\delta(x-x_i),$$
where $a_i$ is the probability amplitude of being at a specific position $x=i$.
If you try and reverse the relation to isolate $a_i$:
$$ a(x) = \int \psi(y)\delta(x-y)\, \mathrm{d}y = \psi(x).$$
So the probability amplitude $a(x)$ in the position basis is equal to your original wavefunction $\psi(x)$.
| {
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Is it really true that valence band is completely filled at zero temperature? Is it really true that valence band is completely filled at null temperature?
Indeed, I would think that if we apply an electric field, this would give some energy to the electrons from the valence band, so would they be prevented to leave the valence band to go to the conduction band, thanks to the energy from the electric field?
I don't see why would the electrons know if the energy that they receive is due to temperature or from electric field source?
| The temperature we're talking about here isn't the temperature of the room where you do the experiment, or even the temperature of the atoms in the material. It's the temperature of the ensemble of electrons in the material.
If you had a sample of semiconductor with its electrons at 0 K, and you did something to it (apply an electric field or anything else) that promotes some electrons to the conduction band, then we'd say the electrons are no longer at 0 K.
It's possible they're not in a (quasi) thermal equilibrium state and we can't even describe them with a temperature. Or that their energy distribution is close enough to what we expect in some equilibrium state and we say they have a temperature $T$ with $T>0$.
The important thing is that temperature is a simplified description of the distribution of energy in some material or ensemble of particles. So if you change the energy distribution (by promoting some particles to a different energy level), you've changed their temperature (or possibly made it impossible to define their temperature).
| {
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Spring mass damper system: Distance from equilibrium after applying velocity to mass I have a spring fixed to a wall on one end and a mass object on the other end in its natural resting position. The question is how far does the spring stretch when a velocity $v_0$ is applied to it, assuming there is no friction.
My idea was that the spring will be stretched until the velocity $$v_0=0$$ and the kinetic energy $$k=0$$ resulting in the max $E_{pot}$.
However I can't figure out a relationship between $E_{kin}$ and $E_{pot}$ other than $E_{kin} + E_{pot} = E_{total}$ and therefore I don't know how to continue from here on.
| Initially the mass has kinetic energy $\frac 1 2 mv^2$ and the spring has potential energy $E_{pot}=0$. So $E_{total} = \frac 1 2 mv^2$.
When the spring is fully extended, the velocity of the mass is $0$, so $E_{kin}=0$, and the potential energy is a function $f$ of the spring's maximum extension $x_{max}$. So $E_{total}=f(x_{max})$.
Since there is no friction, $E_{total}$ is conserved so $\frac 1 2 mv^2 = f(x_{max})$.
Find an expression for the potential energy of the spring $f(x_{max})$ - this will involve the spring constant - and then re-arrange this equation to find $x_{max}$ as a function of $m$, $v$ and the spring constant.
| {
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Peskin's treatment of Pions as Goldstone Bosons After restoring the mass terms in the Lagrangian
\begin{align}
\mathcal{L}=\bar{u} i \not D u+\bar{d i} \not D d-m_{u} \bar{u} u-m_{d} \bar{d} d,
\end{align}
one obtains equations of motion for the quark field
\begin{align}
Q=\left(\begin{array}{l}{u} \\ {d}\end{array}\right)
\end{align}
given by
\begin{align}
i \not D Q=\mathbf{m} Q, \text{ and} \qquad -i D_{\mu} \overline{Q} \gamma^{\mu}=\overline{Q} \mathbf{m}
\end{align}
where
\begin{align}
\mathbf{m}:=\left(\begin{array}{cc}{m_{u}} & {0} \\ {0} & {m_{d}}\end{array}\right).
\end{align}
This leads to
\begin{align}
\partial_{\mu} j^{\mu 5 a}=i \overline{Q}\left\{\mathbf{m}, \tau^{a}\right\} Q,
\end{align}
where $\tau^{a}=\sigma^{a} / 2$ are the generators of SU(2). Using this equation to rewrite the matrix elements $\left\langle 0\left|j^{\mu 5 a}(x)\right| \pi^{b}(p)\right\rangle=- i p^{\mu} f_{\pi} \delta^{a b} e^{-i p \cdot x},$ where $f_\pi$ is the pion decay constant,
\begin{align}
\left\langle 0\left|\partial_{\mu} j^{\mu 5 a}(0)\right| \pi^{b}(p)\right\rangle&=- p^{2} f_{\pi} \delta^{a b}\\
%%
%%
&= \left\langle 0\left|i \overline{Q}\left\{\mathbf{m}, \tau^{a}\right\} \gamma^{5} Q\right| \pi^{b}(p)\right\rangle.
\end{align}
Peskin claims that the final line is an invariant quantity multiplied by
\begin{align}
\operatorname{tr}\left[\left\{\mathbf{m}, \tau^{a}\right\} \tau^{b}\right]=\frac{1}{2} \delta^{a b}\left(m_{u}+m_{d}\right).
\end{align}
This claim is quite non-transparent to me. I can't seem to identify where the trace might arise from, and what the invariant quantity might be.
He concludes that the quark mass terms which break the axial vector symmetry give rise to pions with masses of the form
\begin{align}
m_{\pi}^{2}=\left(m_{u}+m_{d}\right) \frac{M^{2}}{f_{\pi}},
\end{align}
where we group the remaining invariant quantity into a new parameter $M$... This is shortly after a paragraph in which he argued that an on shell-pion must be massless ($p^2=0$), as expected from Goldstone's theorem. How is this final equation consistent with this claim?
| The generator $\tau^a$ comes from the creation of the pion state. The calculation then decouples into the standard trace over all the color factors, that can be taken outside the expectation value, and an expectation value. The expectation value is an invariant for which they introduce the symbol $M$ ($M^2/2$ to be specific). They don't calculate it, just mention it can be measured
experimentally.
| {
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Is 'Curl of magnetic field in electrostatic is zero' only empirical? I was looking up on the uniqueness of the displacement current.
About the uniqueness of the displacement current this question was exactly what I was looking for, but all the answers seem to go with 'empirically, when the electric field is constant and current density is zero, the curl of the magnetic field is also zero.'
Are there any more explanation other than 'That's what we have found experimentally'? or are we solely relying on the fact the Ampere-Maxwell equation matches the experiments.
| I think that the reason is that the Maxwell equation say so, and the Maxwell equations are empirically right. I can't think of a deeper reason
| {
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Non-holonomic constraints, degree of freedom and generalized coordinates If a system has $N$ coordinates and $M$ number of holonomic constraints then number of degree of freedom $=N-M$ and generalized coordinates $=N-M$ too. But if there are $k$ non-holonomic constraints then what will be no. of degree of freedom and generalized coordinates?
| *
*Consider a classical point-mechanical system with $3N$ coordinates but only $n$ generalized coordinates $(q^1, \ldots,q^n)$, because of $3N-n$ holonomic constraints.
*Let us for simplicity assume that:
*
*All constraints are 2-sided, i.e. we do not allow 1-sided constraints (= inequalities).
*All non-holonomic constraints are semi-holonomic.
*If furthermore the system has $m$ semi-holonomic constraints, then we introduce $m$ Lagrange multipliers $(\lambda^1, \ldots,\lambda^m)$. The corresponding $n$ Lagrange equations are outlined in this Phys.SE post.
*The number of degrees of freedom (DOF) are conventionally defined as half the number of initial conditions needed for $(q^1, \ldots,q^n,\dot{q}^1, \ldots,\dot{q}^n)$, which is then $n\!-\!m/2$, which curiously could be a half-integer. (The main point is that since each semi-holonomic constraint is a 1st-order ODE, they each remove the need of 1 initial condition.)
| {
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Pressure waves for sounds Is it possible to mechanically create a pressure wave, with correct amplitude, frequency, and other noise characteristics to recreate known sounds like train horns, chime bells, gun shot, etc.?
I am interested if a vibrating body, driven by a computer with controlled forces at various places in the body, can create a specific sound that a human will recognize, even though it doesn't have any of the properties of that material.
For example, is it possible to create a train horn noise by connecting a computer to a metal plate that can create pressure waves or multiple air tubes interfering constructively? Would this noise be able to be indistinguishable to the human ear from a real train horn? Can this be done by stacking a lot of air tubes with different frequencies and amplitudes?
| A device that does this is called a synthesizer. It can be programmed to emulate many types of sounds fairly well, but complicated sounds like the human voice are harder to synthesize.
| {
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What would be the potential due to a charge $Q$ at $Q$ itself? We know that the potential due to a point charge $Q$ at any distance is given by $V=\frac{kQ}{r}$, but what would the potential be at the charge itself?
| This question goes to the centre of the problem of self interaction, which is unsolved. So it is an excellent question. Often self interaction is ignored, but in quantum field theory it is prominent. There it leads to the infamous infinities or divergences. In QFT self energy is assumed to be incorporated into mass. However even this approach is not satisfactory.
Let's take the example of the electron. Thomson argued that if the rest mass was made up entirely of self energy, which is positive as charge repels itself, then the so called classical electron radius would be about $2.8 \cdot 10^{-15}$. However experimentally we now know that the electron behaves as a point charge down to a scale of at least $10^{-22}$ m. This means that its potential does not deviate from $1/r$ down to this scale. If we then apply Thomson's approach we end up with a mass of 2 $GeV/c^2$, interesting not unlike that of a proton. This in my judgement is an important unsolved problem of physics.
| {
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Electron cloud and Quantum Physics Is it possible to detect the Electron Cloud? Also, it it possible for the Electron Cloud itself to contain any mass?
| It's always a danger to assign classical meaning to quantum objects, like electron clouds. One way to help clarify what's happening in atoms is to compare it with a another bound system: the nucleon.
Can we detect the quark-cloud in a proton? Yes, by elastic electron scattering. The angular/energy dependence cross section becomes the Fourier transformation of the charge density. This is as much a detection of the quark cloud as is Rayleigh scattering of weather radar from a rain drop a detection of the rain drop.
This has been done with electrons scattering off of bound atomic electrons:
"Collisional breakup in a quantum system of three charged particles," by T.N. Rescigno, M. Baertschy, W.A. Isaacs, and C.W. McCurdy appears in Science magazine, 24 December 1999. (https://www2.lbl.gov/Science-Articles/Archive/quantum-scattering.html)
whence the follow plot is taken:
What we see here is a prediction of what the electron cloud looks like from the paper (via angular dependence), and actual data, which qualifies (for me) as "seeing the electron cloud".
| {
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How are thermonuclear weapons possible? Given the fact that thermonuclear devices are a mix of fission through enriched uranium whose energy initiates the fusion in surrounding deuterium. Question is (how come) fission raises temperature so high that, instead of expansion of gases (possible at few hundred degree Celsius ) fusion in (,ready to expand,) gases is obtained?
| This is possible because the fission reactions occur so fast that there is essentially no time for expansion to occur on the same time scale.
In addition, the X-ray-driven implosion reaction that is initiated by the radiation output of the initial fission process also happens so quickly that it too proceeds to completion (triggering the spark plug and hence the fusion processes) before the bomb case has a chance to expand.
In the case of a fission bomb, the expansion of the bomb case and tamper cause the density of the exploding core to fall enough to snuff out the fission, but by that time, the bulk of the energy in the bomb has already been unleashed.
| {
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Why is the sky near the Earth limb warm-colored, even far from the sun? I've seen many sunset pictures with explanations. The sun is red, the areas adjacent to it is orange, then yellow... because redder lights bend less, and the air is thick at that angle. Far enough from the setting sun, the sky returns to its usual blue. OK, easy enough to understand.
So how can it explain the belt of Venus covering the whole sky, even in the east? How can pinkish light travel to our eyes from the direction opposite of the sun??
| The Belt of Venus happens before sunrise or after sunset, when the sun is below the horizon. As you know, the bluish or violet light by the sun gets scattered more than the red one, therefore when the sun is below the horizon, some blue light may be scattered and arrive to us, that's why Belts of Venus appear Violet.
Then another point is when you ask why violet light can pass through such a thick layer of atmosphere arriving to our eye. Let me try to explain it with an example: if you put a strong violet source (that only emits violet light) close to the horizon you will see it violet, because even if its light is scattered a lot, most of the light can come through. Not all blue light is scattered, it's just scattered way more than red light.
That's why you see the sun red when it is at the horizon, because the sun emits all wavelenghts (all colors), the redder ones are not scattered much, the blue ones are scattered way more and the net effect is that you see it more reddish.
| {
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Eigenstructure of the Dicke Model I am beginning a study of the Dicke model and found a very interesting publication: "The Dicke model in quantum optics: Dicke model revisited" by Barry M Garraway in Phil. Trans. R. Soc. A (2011).
I would like to understand how he developed the eigenstates in section 2 (for the 2-atom model), what each states actually means physically, and why he can say that the triplet spin state is sufficient in developing the Dicke energy levels. The latter seems an audacious claim, to me, because I'm not sure we can say that every evolution along the triplet states will lead to superradiance.
He writes n-states in terms of the spin-states. First, what is he calling the number of excitations for the 2-atom system? He has only 2 atoms, so how can he generically use |n>, |n+/-1>? Second, how did he get the coefficients for the linear combination of spin-states? Third, how does he find the eigenenergies for this new basis? This set up is very strange to me, perhaps because the Dicke model is so new to me. The biggest question is what does this new excitation number-basis say about the 2-atom model?
|
First, what is he calling the number of excitations for the 2-atom system?
There are two atoms coupled to a radiation mode, i.e. to a harmonic oscillator. The states $|n\rangle$ refer to the harmonic oscillator, the states $|S, m_s\rangle$ refer to the two atoms. (Each atom is assumed to have only two energy levels and thus behaves like a spin-1/2. Two spin-1/2 add up to total spin $S=1$ or $S=0$. Garraway here only considers the $S=1$ states.)
Second, how did he get the coefficients for the linear combination of spin-states? Third, how does he find the eigenenergies for this new basis?
He tries to find the eigenstates and eigenvalues of $a^\dagger a + S_z$.
You can easily verify that (2.1) and (2.2) are correct by checking e.g. that $(a^\dagger a + S_z) |n,+1\rangle = (n\omega_c + 2g\sqrt{n+1/2}) |n,+1\rangle$.
In order to find the coefficients, he probably made the ansatz that an eigenstate of $a^\dagger a + S_z$ will be a linear combination
$$ |\psi\rangle = \alpha |m_s=1,n-1\rangle + \beta |m_s=0,n\rangle + \gamma |m_s=-1,n+1\rangle $$
and solved the eigenvalue equation $(a^\dagger a + S_z) |\psi\rangle = \lambda |\psi\rangle$.
| {
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Why isn't this anomaly in the CMB taken more seriously? Almost every time I try to find more information about the Axis of Evil, I end up with people trying to ignore the phenomena.
The video The (Cosmological) Axis of Evil explains it.
However, it's no doubt that this discovery leads directly to earth-centered universe. Among with the many experiments that were never able to measure the movement of the earth, why isn't this discussed more seriously?
Both Einstein and Newton were able to imagine an Earth-centered universe, what about us?
If you have a good scientific argument that disproves my suggestions, please let me hear.
Also, I must ask you to forgive my English if there is anything wrong.
| The reason it is not taken too seriously, although it is still a legitimate avenue of research is that assuming we have a special place in the universe has always proved to be incorrect in the past, plus there are possibilities that remain to be explored that might explain the "anomaly". Further, it is stretching credibility to suggest that the orientation of the solar system, which is set by the pseudo-random turbulence in the giant molecular cloud that formed the Sun, could be affected by, or have any affect on, a cosmic microwave background that was formed about 8 billion years before the solar system.
The problem was reviewed by Schwarz et al. (2015). They conclude that the origin of the very weak alignment (it isn;t significant at the 5 sigma level) could be zodiacal dust or dust in the Kuiper belt. Of course the analysis teams for WMAP and Planck have tried to avoid such problems. These however seem to be plausible possibilities that cannot be wholly dismissed and therefore this seems more likely than some mystical overhaul of cosmology.
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Why is the triple point of water defined as 0.01 degrees Celsius and not 0? This was stated in my thermodynamics lecture today and I tried to ask my lecturer why it was not just defined as 0 since 0.01 seems weirdly specific. She was mentioning something about the order in which the Celsius and Kelvin scales came about but I am still uncertain about the answer.
| At very low pressures (less than 1/1000 of an atmosphere) water goes directly from a solid to a gas phase as temperature increases. At about 0.006 atmospheres, the pressure of the triple point, the pressure allows for a liquid phase to exist at intermediate temperatures between solid and gas.
The temperature at which the transition from solid to liquid happens falls very gradually as pressure increases above that of the triple point. As a result, at one atmosphere of pressure ice will melt at a slightly lower temperature than that of the triple point. The difference is about 0.01 C. So, in defining 0 C, one can pick either the temperature of the triple point, or the temperature at which ice melts at one atmosphere, but you can't pick both.
Now, the triple point is more easily controlled in a laboratory, so it is the safer place to set an exact standard. But broader considerations are that you want the standard pressure, one atmosphere, melting point to be close to zero - because that is what people are used to. So the exact standard is set at the triple point, but it involves a little fudge to make the one atmosphere point be closer to zero.
| {
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The right hand rule confusion?
I have a question regarding this problem. By using the right hand rule, I thought the answer would be A, but the answer key says it's B.
Doesn't the current come from the + side, so you wrap your fingers towards yourself(?) so that the thumb points to the left?
| The question is referring to the lines of force outside the coil. The lines of force go from the north pole to the south pole outside the magnet. The direction is the opposite inside from south to north.
Hence if you use the right hand rule correctly, you will see that the lines of force so from y to x outside the coil.
| {
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Work done by friction in a complicated path
A block of mass $M$ is taken from point $A$ to point $B$ in a complex path by a force $F$ which is always tangential to the path. We also have coefficient of friction as $K$. What will be the work done by force $F$ when it reaches point $B$ from point $A$? Given that the vertical displacement from $A$ to $B$ is $h$ and the horizontal displacement from $A$ to $B$ is $l$.
In this question, I tried solving the problem using conservation of energy, we know that the total energy will remain constant. So with that, we will have,
$$\Delta U_{gravity}+W_{friction}+W_{F} = O$$ But how do you calculate the work done by friction in this case?
Moreover, in the answer, the work done by friction is only dependent on l!!
EDIT:
1.The body is moved very slowly.
| The fact that no details on path are given implies that any path should give the same answer. Take a straight line from A to B or move horizontally and then vertically. Either way you get kmgl for the work done by friction. The work done by force F is just Fl + Fh = kmgl + mgh. (This assumes F changes and produces no noticeable acceleration.)
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Why is work equal to force times displacement? This is how I think of what work is.I am sure I am wrong somewhere because I shouldn't be coming to the conclusion that I am coming to.It would be helpful if you would point out where this conceptual misunderstanding is.
Work is just change in the energy of an object.The only way an object can gain energy is by movement.Basically if an object's velocity increases then we can say that work has been done on the object.So, work done on an object is directly proportional to the change in its velocity.Also, if the object's mass is high, then for a given change in the objects velocity, the object gains more energy than an object with lower mass, because higher the mass more will be the momentum it can transfer to other objects and therefore higher the mass of the object for a given change in velocity, more is the work done on the object.Therefore work done is directly proportional to the mass of the object.
I don't seem to find any other factor that influences the work done on an object .Hence according to me work should be equal to mass times the change in velocity.
| Let me not answer this question but only provide you with a hint. You are already close to what you are looking for.
I don't think that your approach is flawed. There are just a few gaps that you have overlooked. You are right in concluding that the work $W$ should depend on the mass $m$ and the velocity $\vec{v}$. However, you know that $W$ is a scalar. Therefore, it should depend not on $\vec{v}$ but $|\vec{v}|$ or even $|\vec{v}|^2 = v^2$. A quantity that depends on just $m$ and $v^2$ is kinetic energy. You can now start exploring if work is related to change in kinetic energy.
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How to interpret negative time in Lorentz transformation? I am somewhat confused about how to interpret negative time in Lorentz transformation. In the usual case of two reference systems S and S' where the distance X (the one that measures S) to an event, is very large with respect to the distance between S and S' (also measured by S) the time t' of S' gives a negative result.
I don't understand why if at $t = t '= 0$, S and S' were together, the event has been observed by S' before this synchronization..., so it could not be in front of S, and also, in this synchronization, S' could warn S about the event that is future for S ... which seems to me paradoxical.
Just in case my general interpretation of the Lorentz transformation is incorrect, I clarify that I assume that S and S' are in positive X, they are together at $t = t' = 0$ and after a certain time t, S observes S' at a certain distance and an event X; with X and t I can calculate that for S' the event has occurred at X' and at t '.
Then, using natural units for simplicity, if $V = 0.8$, and at $t = 7.5$, S observes an event at $X = 11$, then S' which at that time is $V * t$ away from S, observes ,applying Lorentz, $X’= 8.3333$ and $T’ = - 2.16666$
How should we interpret this result? Did S' really saw the event before synchronization?
| According to Marco's answer and comments, I tried some numbers playing with http://www.trell.org/div/minkowski.html, and now I can answer my own question.
We have two events, event $A$ in $x=x'=0$ ; $t=t'=0$ and event $B:$ $x=11$ and $t=7.5$ with $v=0.8$ ($t' = -2.1666$, $x'=8.333$)
Lorentz transformation gives you the distances regarding each observer, and what this negative time $t' = -2.1666$ means is that regarding $S'$, the event $B$ occurs before event $A$, event $B$ occurs before the synchronization... BUT, and this is the important point, S' is not going to be aware of that event until $t' = 6.1167$ ($x'=0$, $x=8.22$, $t=10.277$) so there is no way S' can warn S in the $A$ event moment of the synchronization.
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What is the time derivative of resistance? Is there a unit for $\frac{\Omega}{sec}$? I have tried looking it up, but I can’t find anything
| Resistors are usually a constant value. They may vary with temperature but you'd need to know the material the resistor is composed of. If you just search on resistors you should be able to find technical sheets for different resistors that will have information as I described. The resistors should be constant in time so $d\Omega/dt$ = 0.
| {
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Are photons locked in time, and does this explain the "delayed choice quantum eraser" experiment? I'm trying to wrap my head around the "Delayed Choice Quantum Eraser" experiment and how events in the future affect light in the past.
I'm sure I'm wrong but to me this seems to indicate that photons experience no time. Since time slows down when approaching the speed of light, a photon(which travels at the speed of light) experiences no time: the photon at the destination is the "same" photon at the source, meaning there can be no different state for the photon because inbetween source and destination time does not pass for it. The Photon would appear to be "locked" in time. So altering a photon in the future alters the photon in the past, all the way up to the source of the light.
Just for an analogy: if I was a photon and was "locked" in time, I would be like a statue not being able to change my state and if someone would draw a mustache on me, the mustache would be there yesterday and the day before that, all the way to eternity.
Now please tell me how I got it wrong :)
| The apparent retrocausality in delayed choice quantum eraser experiment is another confirmation of the theory of relativity. Because the photons travel at the maximum velocity possible, no time passes for them. So both photons hit a target, or are detected, simultaneously, as far as they are concerned. Anyone not moving at the same light speed will observe an apparent retrocausality.
| {
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The potential at a point According to my book, 'The potential at a point is said to be 1 volt when 1 joule of work is done in bringing 1 coulomb charge from infinity to that point.' But I wonder how it is possible. As the charge is being brought from infinity, the work done = force * infinity, thus, the work done would be infinity indeed. Please help me out.
| Loosely speaking, at great distance the force required gets infinitely tiny, so the two infinities largely cancel out.
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How to interpret the wave function for non point-like objects The accepted interpretation of a single-particle wave function is that it represents (among other things) the probability of finding the particle at any point. The wave function is normalised so that the probability sums to 1 over space.
In principle, how might the accepted interpretation address the hypothetical case of a non-point-like object with no component parts?
For those of you who might not see the point of the question, consider the 2D 'particle in a box' scenario, in which a point particle is constrained to lie on some portion of the X-axis by a pair of potential barriers. Now replace the point particle by a square particle the length of whose sides is greater than half the width of the box. The particle can move, but there are now values of x for which the particle will always be found, so it is no longer possible to normalise the wave function, and it is no longer possible to associate the particle unambiguously with a given value of x.
| Simply, you cannot talk about a wave function for a non point particle. A non point particle is a body, composed of many particles, and you have to write down the wave function for the coordinates of each particle composing the body. Of course, as in classical mechanics, you can make approximation, like idealizing the body as rigid. This means that you can write the wave function as a function of the center of mass coordinate, and the coordinates of the other particles are fixed once the center of mass coordinate has been chosen, but this is just an approximation.
| {
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Understanding simple LC circuits I'm trying understand the physics of simple inductor-capacitor circuits such that there is just an inductor L and a cacpacitor C and a switch.
Imagine first that the capacitor is fully charged and the switch is then closed.
I do not understand why the current increases from an initial low value as the charge difference between the plates DECREASES because this is in direct contradiction to how a capacitor discharges in isolation.
I know the solution lies in the inductor being present but I can't seem to follow the physics of cause and effect to understand it properly.
Any illumination would be appreciated.
| An inductor "resists" changes in the current through the energy required to build up the magnetic field. Much like a capacitor "resists" changes in voltage through the energy stored in the electric field between the plates.
If you connect an inductor and a resistor in series, you will get a charge/discharge curve for the current, as with a capacitor+resistor circuit, when you connect to a constant voltage/short circuit the circuit.
The slow build up of current through the inductor together with the decreasing voltage over the capacitor as more current is drawn is what gives the characteristic sinusoidal current. Incidentally the inductor's "resistance" to changes in current is also what makes the capacitor discharge "overshoot" beyond zero voltage, and thus continuing the sinusoidal current and (phase shifted voltage).
| {
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What is the difference in the way a single rope and a double rope would behave under dynamic load? Suppose I have a single rope attached to a fixed point via a load cell, which gives a number (in kN) based on the load it's experiencing. I take a weight (x) and attach it to the rope at a fixed point and raise it up to a given height and then drop it. The rope has elastic properties and has some elongation.
Now, suppose I repeat the process, but this time I use two ropes hanging in parallel (actually, it's the same rope with a mid-line knot which prevents the tension/elongation in one half being transferred to the other half).
What I need to understand is whether the number on the load cell would be different between the two scenarios, and if so, why?
| Allowing for a tidier knot than you have drawn, so the knot itself has no effect on load (i.e. the knot doesn't have any 'give' in it), then things will be as follows:
*
*While the weights are falling freely the load cells in both systems
will record zero load (for simplicity I am assuming a weightless
rope).
*When the masses reach the bottom of the rope/ropes, the doubled rope
load cell will display twice the force than the single rope load cell during the period over which the masses are being decelerated to a halt. The double force, in the folded rope system, will be displayed for half the time as the force in the single rope system. (The above assumes that the masses were dropped from the same initial high above the respective stopping heights of the two systems and that the free length of rope/double-rope is the same in both systems. If you vary the height of the drop and/or the suspension height then things will obviously be different).
*After the masses have decelerated and are hanging freely, the two
load cells will both register identical loads Mg.
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Hamiltonian capable of quantum computation Suppose we have a 1D spin chain evolving in time according to some Hamiltonian $H_t(p_0, p_1, p_2 \ldots)$, where $p_i$ are classical parameters ``set by the lab equipment". Divide time into discrete intervals of some length $\Delta t$: we are allowed to change the $p_i$ every step. In addition, assume the ability to choose whether or not to perform a projective measurement in the spin-z basis at each site at each timestep.
What is an example of a nearest-neighbour spin-chain Hamiltonian that can implement quantum computation in this way?
| Many Hamiltonians will be good. In particular, if you have control over the individual couplings, any interaction, together with a local terms in two directions, will suffice.
As an example, the Ising model with two fields,
$$
\sum p_is_z^is_z^{i+1} + \sum p_i' s_z^i + \sum p_i'' s_x^i
$$
will do: The local terms allow to implement arbitrary rotations about $X$ and $Z$, and thus any rotation, and the Ising term allows to realize a gate $\mathrm{diag}(1,i,i,1)$ (up to a phase), which is locally equivalent to a CNOT. Together, this yields a universal gate set.
| {
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Confusion about circular motion So today in class, I learnt that the angular quantities are related to the linear quantities by
$$s = r\theta \qquad \qquad v = r\omega \qquad \qquad a = r\alpha$$
where it is assumed that the object travels in a perfect circle.
My teacher then went on to derive the formula
$$a = \frac{v^2}{r}$$
where it is further assumed that the object has constant linear velocity $v$.
Then I got a bit confused. If we rearrange for $\alpha$, we get
$$\alpha = \frac ar = \frac{v^2}{r^2} = \omega^2$$
But if $\omega$ is a constant, then shouldn't we have $\alpha = 0$? What have I done wrong?
| As the other answers and comments suggest, you are mixing up different properties by accidentally giving them the same labels.
The actual formulas are:
$$a_\perp = \frac{v^2}r$$
$$a_\parallel = r\alpha$$
A parallel $_\parallel$ acceleration is along with the circular path (also called tangential).
A perpendicular $_\perp$ acceleration is sideways from the circular path (also called radial or centripetal).
When you in a circular motion have constant speed, so that $a_\perp = \frac{v^2}r$ applies, then $a_\parallel=0$. If $a_\parallel$ wasn't zero, then $v$ wouldn't be constant - and then $a_\perp = \frac{v^2}r$ wouldn't apply.
| {
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Why does water boil harder when you push a ladle to the bottom of the pot? I noticed this today while cooking. When I push a ladle to the bottom of my pot on my stove top, the boiling sound gets louder, and the bubbles rise to the top more aggressively. Can someone explain?
| I see this effect on my electric stovetop. The bottoms of my pots are no longer quite flat, and pushing downward on the pot improves the thermal contact between the pot and the heating element. The effect is present whether I push down on the bottom of the pot with a ladle, or whether I push down on the handles on the outside of the pot.
On a gas stovetop, where the heat transfer is by convection, pushing down on the pot doesn't change the thermal conductivity between the pot and the heat source and so doesn't cause any change in the rate of boiling.
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Can someone explain what is the force the ball will exert? If a ball is falling under free fall then the force exerted by the ball on the ground would be $mg$. But that's not the case in real life ball would hit with more force. But when i draw free body diagram there is only one force that is acting on it $mg$
Can someone explain what is the force the ball will exert ?
| Interesting question. Ball force affecting ground is $$ F_c = \frac{\Delta p}{t_i} = \frac{mv}{t_i} $$, because end speed is zero.
Now let's assume ball started to fall from some altitude, so it reached it's "contacting" speed $v$ according to :
$$ v = v_0 + a \space t = g \space t_f $$
Plunging this into formula above, we get :
$$ F_c = \frac{mg \space t_f}{t_i} = W \frac{t_f}{t_i} $$
So ball force affecting ground is it's weight scaled by falling time over interaction time with ground ratio.
Analogically ground will exert normal force with same magnitude on ball. That's why eggs breaks falling down :-)
Only in case when $ t_f \approx t_i $, i.e. when falling time approaches interaction time, contact force equals weight. This principle is exploited in various kinds of trampolines and Life net used by firefighters.
| {
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Would an infinitely (or very long) diffraction grating produce a diffraction pattern? Take a typical science lab diffraction grating which is producing a pattern on a screen.
Let's consider the location of say the first maximum on the right side.
Let's draw straight lines from the slits to the first maximum and think in terms of ''rays'' travelling from the slits to the maximum on the screen.
These rays all converge at the maximum and although they are close to being parallel we know that they are not perfectly parallel (because they converge).
Therefore the differences between the distances that neighboring rays have to travel will not be exactly the same.
i.e. If one ray has to travel one lambda more than its neighboring ray, then the following ray will have to travel slightly more or less than its neighboring ray.
If there are many slits very close together I can see that if the variance is small compared to the wavelength then we can ignore these differences and assume that each ray travels exactly one wavelength more/less than its neighboring ray.
But if the grating is long enough, these differences will eventually build up until they are significant enough to cause destructive interference.
My understanding is that a very long grating will have destructive interference at all locations and will not produce a pattern.
Is this true ?
What are peoples thoughts on this ?
| All secondary peaks of the pattern are going to become dimmer and dimmer until they disappear completely as the number of slits is increased. But the central peak will become more and more bright and sharp, since light from each slit interferes constructively at the center of the pattern
| {
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Why do the expressions for an object rolling down an incline not depend on the coefficient of static friction? In my physics course, we are doing an experiment rolling disks and spheres down an incline (assuming there is no slipping). In doing the derivations (assuming a moment of inertia of $\frac25mR^2$ for sphere and $\frac12mR^2$ for disk) I have derived that the final velocity should be $\sqrt{\frac{4gh}{3}}$ for the disk and its acceleration should be $\frac{2g\sin{\theta}}{3}$. I have derived similar equations for the sphere. My question relates to the coefficient of static friction.
Logically, I think that when the coefficient of static friction increases, it should increase the force that is applying torque to the rolling object and thus increase the final speed at which the object is rolling. This should result in a lower final velocity since more of the initial energy ($mgh$) is "allocated" to the rotational kinetic energy as opposed to the translational kinetic energy.
Yet, according to the derivations, the final velocity does not depend on the coefficient of static friction. Why is this? I believe my equations are correct so I know that the final velocity is entirely determined by the object's moment of inertia, and the angle of the incline, but I intuitively think that a higher coefficient of friction should change how much energy is "allocated" between translational and rotational kinetic energy.
| The coefficient of static friction predicts the maximum possible value of static friction. Static friction can have any value below that. If the required value exceeds the maximum, the the friction becomes kinetic (usually smaller).
| {
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Why is Jupiter's ring system so faint? Planetary ring systems are formed when asteroids, comets, or any other large objects pass too close to the planet and are torn apart by the planet's gravity [1]. Jupiter is often called the "punching bag" of our solar system. Because of it's massive gravity, it is impacted more often by asteroids and other objects [2].
So I would expect Jupiter to have a more prominent ring system than Saturn. Why is Jupiter's ring system not as spectacular as Saturn's?
[1] https://www.physlink.com/education/askexperts/ae188.cfm
[2] https://www.space.com/32420-jupiter-asteroid-impact-rate.html
|
So I would expect Jupiter to have a more prominent ring system than Saturn. Why is Jupiter's ring system not as spectacular as Saturn's?
The rings of Jupiter likely formed due to very different processes than that of Saturn. It is thought that Saturn's rings formed from the collapse of a large, icy body (e.g., something like half the size of Enceladus).
Jupiter's rings form from debris resulting from meteoroid impacts on the moons of Jupiter. Much of this debris is then swept up by the Jovian moons or Jupiter itself. Because the rings are made of rocky/dirty dust, they do not reflect a lot of light like Saturn's rings, which are largely made of icy rock.
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Physical Interpretation of an Elementary Inequality in Kinematics I recently came across the following question: suppose a point mass moves a distance $d_1$ for $t_1$ seconds at a velocity $v_1$, then accelerates instantaneously to a velocity $v_2$ and travels a distance $d_2$ for $t_2$ seconds at this velocity. Show that $$\frac{v_1d_1 +v_2d_2}{d_1+d_2} \geq \frac{v_1t_1+v_2t_2}{t_1 + t_2}$$
A mathematical proof is quite straightforward. What I'm wondering is is there a way to see this directly?
|
You can view it pictorally. For when the second velocity is very high, the LHS tends to the second velocity. However the RHS tends to the average of the two velocities.
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Why is the top surface of a ceiling fan more dusty than the bottom?
Top surface of a ceiling fan ( that hasn't been cleaned for a long time)
The bottom surface
The top surface seems to be way more dusty than the bottom. I reason that it doesn't have anything to do with gravity, as I've seen pretty much the same results on table fan propellors and exhaust fans, which has a horizontal axis of rotation. What else could be the reason?
| That are many factors at play here.
Let us consider the case when the fan is rotating.Since there is a suction of air towards the bottom in the top surface of the fan the dust particle which move along the air flow collides with the top surface of the fan.On the contrary the dust particles get pushed away from the bottom surface of fan.This makes the top part more dusty.
According to this idea there should not be any dust on the bottom surface at all.But this is not the case. Since the fan blade is slightly slanted some dust particles below the fan obliquely collides with it.
Now consider the case when the fan is at rest.Assuming that the composition of dust particles in the room is uniform and the motion is somewhat random,both the surfaces of the blades gather almost the same amount of dust.But amount of dust will be slightly more on the top because of gravity. This explains why the bottom surface also has a little amount of dust.
Hope this explanation is satisfactory
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Taking the speed of light into account during $n$-body simulation Currently, I compute the force between two gravitational interacting particles in a simulation with $n$ bodies according to
$$F = G\frac{m_1m_2}{r^2}.$$
Doing this, however, assumes that all bodies in the simulation interact instantly with each other meaning that I assume the speed of light to be infinity.
Let's assume I want so simulate the formation of galaxy clusters where the propagation velocity of light plays an essential role. How would one incorporate the fact that the propagation speed of light is finite into an $n$-body simulation? Any ideas?
| You could use the Einstein-Infeld-Hoffman equations, derivable from General Relativity, which have Newton’s inverse square force as their dominant term but also include first-order relativistic corrections.
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To lift an object, do we need a force equal to its weight, or greater than its weight? We have all heard people saying that to lift an object of mass $m$, you have to apply a force $F$ equal to its weight $mg$. But isn't it getting the force equal to its weight from the surface to which it is attached to (normal force). Why it is willing to change that equilibrium state by getting the same force from us as from the surface? (Consider the situation devoid of any resistance) . I think we must be applying slightly more force to it in order to move it even with constant velocity at least at the start and balancing the force of gravity afterwards.
| As has been mentioned in @Adrian's answer in the real world there is nothing rigid, neither the support where the mass m is resting on be it a table or cement bench nor the mass itself.
So there is always a certain amount of energy reserved in the support due to its deflection/settlement under the weight of the mass.
Therefore there are two forces here, let's call them,
$$ F_1 = \text {the force we exert} \\ and, \quad F_2= Kx $$
With K as spring constant of the support and x as the settlement of the support under the m*g.
So we actually need to apply just enough force (slightly less than the weight, mg) to give the spring action of the support a chance to push the object up act as a projectile moving up. And we can calculate the F1 just short of weight enough so that$$ W*H= F_1*H +\frac{1}{2}Kx^2 \\ F1= W-\frac{1}{2H}Kx^2 $$
With W as weight and H as height. And we inspect that the F1 is smaller than the wight.
And therefore we do not need any downward transient force at the top end of the trajectory.
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Does the wavelength of a particle depend on the relative motion of the particle and the observer? The de Broglie wave equation states:
$$\lambda = \frac{h}{p},$$
where $\lambda$ is the wavelength of the “particle”, $h$ is Plank's constant, and $p$ is the momentum of the particle.
Momentum is usually written $\,p=mv$, where $m$ is the mass and $v$ is the velocity of the particle. But presumably $v$ is the relative velocity between the observer and the particle.
So does this mean that the wavelength of a particle depends upon the relative motion between the particle and the observer?
Or, perhaps more accurately, when a particle is incoming to another particle, in as much as an interaction between the particles depends on their relative speed, or the energy of impact, it thus also has something to do with their relative wavelengths.
Is that a conclusion, or simply a restatement of the premise, using different words that mean the same thing?
|
So does this mean that the wavelength of a particle depends upon the relative motion between the particle and the observer?
Yes.
The second statement is more or less equivalent to the first.
Side note: The strange thing about this relationship is that it means that, e.g., wave train seen by one observer as consisting of one wavelength can be seen by another as containing some other number of wavelengths, say 3. This would be impossible if the wave was a real number whose phase was directly were observable, because you could, for example, count nodes.
| {
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Question about changing mass under conservation of momentum In class today, my professor was teaching conservation of momentum. One example she used was an open cart rolling on a frictionless track while in the rain. As the cart collects water, the mass increases; and due to conservation of momentum the velocity must decrease, in order to keep momentum constant. However, the way I understand it, since velocity is decreasing, that means it must have negative acceleration, implying net force is no longer 0, implying momentum is not constant. What am I misunderstanding here?
| The net force on the cart is not $0$. The cart is interacting with the rain in the cart, giving up some of it's momentum to the rain so that it travels the same velocity as the cart.
The net force of the cart and rain system is $0$. Any force the rain puts on the cart is directly opposed by the force of the cart on the rain inside it.
As rain enters the system, it's momentum needs to increase so that it can move with the same horizontal velocity as the cart. This comes from the momentum of the cart, therefore the cart + rain system still conserves momentum and has no net force acting on it (except for the vertical forces of rain falling; but those don't affect horizontal movement here).
| {
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Velocity not affecting heat produced by impact
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 $\text{m s}^{–1}$. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
The answer given in the book I am working from is that heat produced due to impact will be same in both the cases (when the elevator is moving with uniform velocity and when it is stationery) but I am getting different answers.
Why is the amount of heat produced the same irregardless of velocity? I don't follow the reasoning.
| I do not have the full answer, but here is what I found out so far:
If we take the elevator as a frame of reference, it is clear that everything would be identical and thus the heat produced from the impact would be the same for a stationary elevator. However, if we consider the ground as a reference frame, it would be different as you have calculated.
Although energy (kinetic in particular) is not the same when seen from different references ( https://www.physicsforums.com/threads/change-in-kinetic-energy-between-reference-frames.728195/) the energy conservation should still hold (and it should yield the same heat energy produced) (Kinetic energy with respect to different reference frames) but this is only under the condition that momentum is conserved.
So I think the problem here is that the law of conservation of momentum is not followed, since the elevator speed remains constant after the impact. Perhaps if you take your system to be the elevator and the bolt. Applying the law of conservation of momentum at the impact , the final speed would not be equal to the initial speed but rather a bit higher. I think if it is modelled like that, and both the changes in energies of the elevator and the bolt are considered, you would end up with the same value of heat generated whether the elevator is stationary or moving at const. speed.
| {
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Where do we see centrifugal acceleration? A body in circular motion always possesses centripetal acceleration which is felt by a person sitting at the center of mass. It will not be felt by a person viewing the motion from the ground frame. Then where do we feel the centrifugal acceleration? We cannot be anywhere on the body itself because at every point we will perceive the motion to be circular by considering our point as the center.
| Centrifugal force is a misnomer. The force that makes a body move on a curve is centripetal. The fact that you 'feel' a force is exactly analogous to feeling you are being forced back in your seat if you are in an accelerating car. There is no force pushing you back in your seat- the force is accelerating forwards.
| {
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Why does a chain or rope move the way it does when suspended and rotated on a vertical axis? I have always been interested in why objects like chains, ropes, etc. move the way they do when "rotated" around a vertical axis while being held only where it is suspended. It forms a shape if you will, resemblant of a "C" or a wave depending on the length. I am curious to know what laws and effects of physics are at play.
| Not a real answer but I'd like to note the model used while studying these phenomena. The shape of rotating chains is modeled by the Bessel functions. Precisely speaking, the radial displacement function $x$ is given as:
$$ r(x) = J_o ( \frac{2 \omega \sqrt{x}}{\sqrt{g}})$$
Where, $\omega_o$ is the angular frequency at which rope is spun, $g$ is gravitational constant and $x$ is the distance from the bottom of the chain
Where,
$$J_o(x) = \sum_{k=0}^{\infty} (-1)^k \frac{x^{2k}}{(k!)^2}$$
A better presentation is done by professor Ghrist
in this video
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What do we see when we observe a molecular orbital? My understanding is that when two hydrogen atoms bond (for example) the wavefunctions describing each atom combine into a single molecular wavefunction. Assuming the Copenhagen interpretation, does the wavefunction collapse into two particles whenever we look at the molecule, or is it possible for one electron to collapse while the other remains "fuzzy"? I assume this has something to do with entanglement, and electrons within a single molecule are probably all entangled, right?
| The states of the hydrogen molecule come from the molecular Hamiltonian, not a combination of the atomic states. Molecular orbital theory uses the Born-Oppenheimer approximation and the atomic orbitals to get approximate electronic states (for H2).
The states are not necessarily entangled, except in the trivial antisymmetric sense. But that probably isn't what you mean. And for identical particles, "unentangled states" are usually broadened to include the antisymmetrized product states.
Measuring one electron's position doesn't necessarily collapse the other's, but will affect it. This recent measurement of the H2 wave function may interest you.
These results are true in all interpretations.
| {
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Quantization of the Nambu bracket The most simple quantum mechanical system consists of a canonical pair of operators $\{P, Q\}$ satisfying
$$ P Q - Q P = i \hbar. $$
It is well known that there is a unique (modulo unitary maps) irreducible representation of this algebra (actually, of the Weyl algebra generated by $P, Q$), given by
$$ Q f(x) = x f(x), $$
$$ P f(x) = i \hbar f'(x). $$
The quantization of Nambu mechanics is generated by a canonical triple $\{P, Q, R\}$ and takes the following form:
$$ P Q R - P R Q - Q P R + Q R P + R P Q - R Q P = i \hbar. $$
What are the irreducible representations of this algebra? Is there a unique representation just like in the case of ordinary quantum mechanics?
| A narrow reiteration of eqn (49) in Nambu's original paper, in response to your limited
for example, an explicit realization of $P,Q,R$ would already be a very helpful answer (even with the details of how to obtain this realization missing).
You just consider the Casimir-normalized generators of so(3),
$$
P=\left(\frac{\hbar}{L^2}\right )^{1/3} {L_1}, \qquad Q=\left(\frac{\hbar}{L^2}\right )^{1/3} L_2 , \qquad R= \left(\frac{\hbar}{L^2}\right )^{1/3} L_3 ,
$$
so that the Nambu bracket you wrote reduces to
$$ P [Q, R] - Q [P, R] + R [P, Q] = i \frac{L^2}{L^2} \hbar = i\hbar. $$
So consider the irrep of your choice.
PS in response to comment. To appreciate how radically nontrivial this ternary QNB structure is, consider that a trivial $R=1\!\!1$ does not yield a vanishing QNB,
$$
[P,Q,R]= R [P,Q] \neq 0 ~~~~~!
$$
This subverts the classical solenoidal flow Nambu noticed, and makes it impossible to pair up the trivial constant $R$ with a meaningful canonical momentum.
| {
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Pendulum attached to a cart. What have I done wrong?
A bob of mass $m$ is attached to the base of a cart via a thread of length $l$. The cart is accelerating with a non-uniform acceleration with magnitude $a$. The positive x-direction is towards the right and the positive y-direction is downwards. A simple diagram of the situation is shown in the figure above. The problem which I am facing is that when I calculate the magnitude of tension $T$ in the thread using two different approaches, I get two different values. Please help me find the mistake in my solution.
First approach (Observer is stationary)
Assuming the magnitude of the displacement of the cart from its initial position to be $x_c$, the bob's $x$ and $y$ displacement from its initial position can be written as:
$x = x_c + l\sin\theta$
$y = l\cos\theta$
From Newton's laws, we will get:
$-T\sin\theta = m\ddot x = m\ddot x_c + ml(-\dot \theta^2\sin\theta + \ddot \theta\cos\theta)$
$mg - T\cos\theta = m\ddot y = ml(-\dot \theta^2\cos\theta - \ddot \theta\sin\theta)$
If we multiply the above equations by $\sin\theta$ and $\cos\theta$ respectively and add them up, we get the following equation:
$T = mg\cos\theta -ma\sin\theta + ml\dot \theta^2$
Second approach (Observer is present in the cart)
From the point of view of this observer, a pseudo force will act on the bob in the negative x-direction. From this reference frame, since there will be no displacement of the bob along the thread, from Newton's laws:
$T = mg\cos\theta - ma\sin\theta$
As you can see, we obtain two different values of $T$ using two different approaches. Please help me find out the mistake I am making.
| Your first approach is correct.
For the second approach, You have not taken into account centripetal acceleration $m\omega^2l$ which is the term you are missing in your second equation.
| {
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Poynting vector interpretation I am trying to calculate the dynamics of the power density of light in vacuum. The absorption per unit volume defined in terms of the Poynting vector is given by:
\begin{equation}
\text{Absorption} = -0.5 \nabla\cdot\vec{S}.
\end{equation}
Using $E = E_0\cos(k_yy-\omega t)\hat{z}$ and $H = (E_0/\eta)\cos(k_yy-\omega t)\hat{x}$ it follows that the Poynting vector is:
\begin{equation}
\vec{S} = \vec{E} \times \vec{H} = (E_0^2/\eta)\cos^2(k_yy-\omega t)\hat{y},
\end{equation}
leading to the absorption expression
\begin{equation}
\text{Absorption} = -(k_yE_0^2/\eta)\sin(\omega t-k_yy)\cos(\omega t -k_yy),
\end{equation}
which obviously is non-zero for vacuum. The average of the absorption, however, is zero. Is this expression correct, and if it is, what is its physical interpretation?
Thank you kindly in advance, Ian
| The divergence of the Poynting vector is a measure of the rate at which energy is arriving or departing from a given region. As a sinusoidal plane wave propagates along the $y$ axis, wave crests wash over any given point, arriving and departing. This doesn't mean that energy is being absorbed. Absorption would be occurring if the time average of the divergence were nonzero.
| {
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Equations of motion in curved spacetime I'm trying to understand how to use covariant actions to derive equations of motion. A simple example would be the free scalar field
$$
S = \int\;d^4x\; \sqrt{-g} \left( -\frac{1}{2}\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2 \right)
$$
Now from classical mechanics we know that the equations of motion pop out when we set the variation of the action to zero. How would we do that in this formulation?
The $\nabla_\mu$ s represent covariant derivatives!
What I have tried:
Using the $\partial_\mu \rightarrow\nabla_\mu $ perscription the lagrangian of the free scalar field will be $$\mathcal{L} = \left( -\frac{1}{2}\nabla_\mu\phi\nabla^\mu\phi-\frac{1}{2}m^2\phi^2 \right)$$
Hence the equations of motion will come from the modified E-L equation:
$$
\frac{\partial\mathcal{L}}{\partial\phi}=\nabla_\mu\frac{\partial\mathcal{L}}{\partial(\nabla_\mu\phi)}
$$
My question is basically how do we arrive to this equation from varying the action?
| I think it's more straightforward to vary the action with respect to $\phi$ and cancel surface terms. Also, check this: Derivation of Klein-Gordon equation in General Relativity
| {
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Why doesn't string theory predict the existence of infinitely many elementary particles? I'm a physicist, but my knowledge of string theory is extremely minimal. My naive conceptual understanding is that the vacuum is modeled as a certain topology (and geometry?) for the spacetime, and fundamental particles are explained as excitations of strings that live on this background. E.g., a certain vibrational state would be the photon, some other state would be the graviton, and so on. The actual spectrum of such vibrations is presumably something we can't calculate, because we don't know the topology of the background (i.e., which possibility it is in the string landscape).
If this is at least qualitatively correct, then why aren't there infinitely many such vibrational states, which would appear at ordinary energy scales as infinitely many elementary particles?
| For simplicity take the bosonic string theory in 26 dimensions. When you quantise the open and closed string you find excitations (states) of the string at any level $N$ with masses
\begin{equation}
M^2_\mathrm{open}=\frac{1}{\alpha '}\left(N-1\right),\qquad M^2_\mathrm{closed}=\frac{4}{\alpha '}\left(N-1\right).
\end{equation}
As $N$ takes on any non-negative value, you see that there is indeed an infinite tower of states. The tachyonic state (negative mass) at $N=0$ is a sickness of the purely bosonic string, which goes away for the superstring. At level $N=1$ you find the massless excitations (in 26 dimensions), and from the next level all states are massive. The mass is determined by the pre-factor. You can regard $\alpha '$ as the only free parameter of string theory, and it relates to the tension, $T$, of the string, which is expected to be set by the string scale, which is slightly below the Planck scale
\begin{equation}
\frac{1}{\alpha '}= 2\pi T \lesssim M_\mathrm{Pl}^2 = \left(10^{19} \, \mathrm{GeV} \right)^2 .
\end{equation}
So in conclusion, the infinite tower of massive excited states have masses at the order of the Planck scale, which means they are unobservable, and there is only a finite number of massless excitations. The same thing goes for the superstring.
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Time and distance in a photon's frame of reference Disclaimer, not a physicist.
When I look at the sky I can see the star Rho Cassiopeiae. In my frame of reference, the photon hitting my retina has traveled for 4000 years and 3.78x10^16 km. In the photon's frame of reference, there's no time elapsed and my retina and the surface of the star are at the same place (no distance traveled). Is that a correct understanding of time dilatation and space compression at relativistic speed?
| The calculation you performed in your reply to John, which related to an object moving at 0.99999999999999999999999c (or thereabouts!), has given you the right idea about the effects of relative speed when distances and elapsed times are viewed from a moving reference frame.
What you might like to think about is the fact that the effects are entirely symmetrical. Your fictitious object is at rest it its frame of reference, and sees us hurtling toward it at 0.99999999999999999999c (Apologies- I might not have the right number of 9s), so for that object it is our time which has dilated.
If you do decide to consider the reciprocity of the phenomenon, you will also need to take into account the effect known as the relativity of simultaneity. I won't try to explain that here, but it means that we and the fictitious object will have different ideas of when our respective journeys started!
| {
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How do I experimentally measure the surface area of a rock? I hope this is the right place to ask this question.
Suppose I found a small irregular shaped rock, and I wish to find the surface area of the rock experimentally. Unlike for volume, where I can simply use Archimedes principle, I cannot think of a way to find the surface area. I would prefer an accuracy to at least one hundredth of the stone size.
How can I find the surface area experimentally?
| You can put the rock in an MRI scanner and get a 3D profile of it (and therefore, the volume and surface area). If it doesn't have spins that are useful for NMR, you can dunk the rock in something that does (ie. water or mineral oil), and then image that, and the void will give you the 3D profile of the rock (which you can then use to calculate the area).
The main problem of using NMR is that if the magnetic susceptibility of your rock is very different than that of vacuum... you will get image artifacts. But there are tricks around this though.
As an example: here is a lithium dendrite inside a battery imaged using MRI.
Alternatively, you can use Xray images of your rock from many different angles and reconstruct the 3D profile of the rock using the inverse 3D Radon Transform. Having the 3D profile, you can easily calculate the area.
| {
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When the direction of a movement changes, is the object at rest at some time? The question I asked was disputed amongst XVIIe century physicists (at least before the invention of calculus).
Reference: Spinoza, Principles of Descartes' philosophy ( Part II: Descartes' Physics, Proposition XIX). Here, Spinoza, following Descartes, denies that a body, the direction of which is changing, is at rest for some instant.
https://archive.org/details/principlesdescar00spin/page/86
How is it solved by modern physics?
If the object is at rest at some instant, one cannot understand how the movement starts again ( due to the inertia principle).
If the object is not at rest at some instant, it seems necessary that there is some instant at which it goes in both directions ( for example, some moment at which a ball bouncing on the ground is both falling and going back up).
In which false assumptions does this dilemma originate according to modern physics?
| To explain this, I shall use the same example of a ball bouncing on the ground. In a perfectly ideal world, the ball will never be at rest throughout the bounce. There will be a time $t$ when the ball is going downwards. At time $t+dt$, the ball will be going upwards. This assumes that the coefficient of elasticity of the ball is exactly $1$ and the ball and ground are extremely rigid.
However this cannot happen in real life as there is no body is perfectly elastic. Also, the above case would also imply that the force applied by the ground on the ball would be ${\infty}$.
Practically speaking, the ball hits the ground and gets deformed. The velocity slowly decreases to $0$ as the kinetic energy gets used up in changing the shape. The ball will be at rest at a particular moment before bouncing up again.
| {
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What determines whether we use a vector or scalar potential? I understand that electrostatic potential is scalar because the curl of the field is zero, and this implies the electrostatic field is the gradient of the scalar potential to satisfy this. Similarly the divergence of a magnetostatic field is zero so a magnetostatic field is the curl of the vector potential.
But what actually determines when you would use which potential? Is it purely to do with these definitions in electro and magneto statics, or is it something else?
| For some vector field $\mathbf F$, you can use a scalar potential (unique up to a constant) whenever $\nabla\times\mathbf F=0$, and you can use a vector potential (unique up to the gradient of a scalar field) whenever $\nabla\cdot\mathbf F=0$. These are purely mathematical ideas, but you can see how this (very usefully) applies to the specific cases of static electric and magnetic fields as you have explained in your question.
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Why do we feel weightlessness during free fall? An object in free fall accelerates towards the Earth at a acceleration equal to $g$ (the accleration due to gravity). Now if we ignore the air resistance, why do we feel weightlessness? I could not understand the reason that we could not feel the pulling force that the Earth exerts on the object.
| It's because what you feel is distortions to your body's shape. When in free fall, when air resistance is low, every part of your body is accelerating very nearly the same amount. No difference in force between any parts of your body means your body doesn't have to exert any forces to keep its shape, so you don't feel anything.
If the gravitational field changes fast enough, a phenomenon the Earth is big enough to experience called "tidal forces" kicks in, and you very much would feel it. It is tidal forces from the black hole that cause a phenomenon called "speghettification" - when the forces between your top and bottom, or left and right, are big enough to squeeze you down into a stream, like spaghetti.
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Is there a way to construct a Hamiltonian from a set of DE? Let's say I have a set of first-order differential equations for set of position $x_i$ and conjugate momenta $p_i$, which might be complicated and time-dependent
$$ \dot{x}_i = f_i(x_j,p_j,t)$$
$$ \dot{p}_i = g_i(x_j,p_j,t)$$
and I know that these equations originate from some Hamiltonian (that is, they respect conservation of phase-space volume, for example). Is there a constructive way to find said Hamiltonian? If not, what are the minimal conditions that the set of equation have to fulfill for this to happen? I mean I know that if the equations are linear and time-independent I can in general invert them and find $H$ in a straight-forward manner but what is the general case?
| Assuming from the notation
$$ \dot{x}^i~=~f^i(x,p,t), \qquad \dot{p}_i~=~g_j(x,p,t), \tag{1}$$
that the symplectic structure is the standard canonical symplectic structure $$\omega = \sum_{i=1}^n\mathrm{d}p_i\wedge \mathrm{d}x^i,\tag{2}$$
we get that
$$\begin{align}\mathrm{d}H(x,p,t)- \frac{\partial H(x,p,t)}{\partial t}\mathrm{d}t
~=~&\sum_{i=1}^n\left(\frac{\partial H(x,p,t)}{\partial x^i}\mathrm{d}x^i+\frac{\partial H(x,p,t)}{\partial p_i}\mathrm{d}p_i\right)\cr
~=~&\sum_{i=1}^n\left(f^i(x,p,t)\mathrm{d}p_i-g_i(x,p,t)\mathrm{d}x^i\right). \end{align}\tag{3}$$
Now solve for $H(x,p,t)$, e.g.
$$ H(x,p,t)~=~H(0,0,t)+\int_0^1\!\mathrm{d}\alpha \sum_{i=1}^n\left(f^i(\alpha x,\alpha p,t)p_i-g_i(\alpha x,\alpha p,t)x^i\right).\tag{4} $$
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What is the length of the yarn in a ball of yarn? The image https://commons.wikimedia.org/wiki/File:Ball_of_yarn_10.jpg shows a typical ball of yarn. Such a spherical ball of radius $R$ has a volume $4πR^3/3$. The radius of the yarn is $r$.
How long will the yarn be on average? The length will depend on the way the ball is formed. There will be air-filled space inside the ball. Therefore yarn length $L$ is surely smaller than $V/(πr^2)$. But what is its average length for an average ball? How can one determine this average length? Equivalently: how much air is contained in an average ball of yarn?
The yarn is assumed to be unstretchable but infinitely flexible. This is a question about statistical geometry. (This is not homework; the question is about the ensemble average over all possible balls of radius $R$.) Is there some method, maybe using random walks, to estimate an expectation value for the yarn length?
P.S. A few people continue to add the "homework" tag. This is not homework level. The level is more that of a master thesis.
| I agree with you that the question is tricky. To a first order approximation you might consider the packing density of the yarn to be equivalent to the hexagonal close packing of straight lengths of tube (that's the optimal) which is about 90%. However, if the yarn is wound at random you won't achieve that optimum density- sticking with the approximation of straight tubes, it will be as if the tubes are laid across each other at odd angles rather than being packed in an orderly way into a smaller volume.
I suspect it might be that crossing two pieces of twine at right angles results in the worst use of space, while wrapping them side by side provides the best use. If you could model the worst case, then perhaps you could find an expectation value for a random winding by supposing it is midway between the two.
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Do liquids continuously boil in a pressure cooker? First, I understand that the boiling point of water is increased as the pressure is increased. At 15 PSI, the boiling point is about 250F.
My question, which seems like nobody can answer with certainty, is this: Is there a continuous boil within the pressure cooker once it reaches temp/pressure?
I understand that there needs to be a boil at some point to create the steam. But once it reaches that pressure, then what? An electric PC will turn the heating element off until the pressure drops. A user will turn the heat down on an analog PC. Does the liquid just boil constantly while the heat is off? It probably maintains temp & pressure for a lot longer in a sealed environment. In my mind, if there were a continuous boil, that steam can only build so much until it exceeds the pressure that the PC is capable of maintaining and it begins to vent. So, sure you could have a constant boil, but that means it would also be constantly venting, right?
| The way my older pressure cooker works is that, once it reaches operating temperature, it boils all the time. It has a pressure relief valve that has a weight that is placed over a conical tube sticking up in the lid. The weight has different diameter holes drilled partway through it, so that as larger a hole is placed over the conical vent tube, it will go down farther to have a larger sealing surface, so less steam pressure will lift it to release steam. This changes the relief pressure so it can be set to boil at different temperature / pressures. It is made to be used on a cook stove. If a lot of steam is coming out of the vent (rapid boil), you can reduce heat. If no steam is coming out of the vent you can increase heat, so that ideally, a little steam comes out so you know it is boiling. If it is not boiling, it may be below desired temperature.
| {
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Difference between joint eigenstate and a single eigenstate? In angular momentum, we define $m,l$ as a joint eigenstate for $J_{z}$ and $J^{2}.$ Why is there any need to introduce $2$ variables to define an eigenstate? What difference it makes if we use just a single eigenstate for $J_{z}$ and $J^{2}$. Will the expectation value of $J_{x}$ and $J_{y}$ change if we calculate them for a single eigenstate?
| Since $J_z$ commutes with $J^2$, they share a common eigenbasis (an eigenstate in this case). But that does not neccesarily mean they have the same eigenvalues (for instance, they could differ in magnitude), so you use $m$ to refer to $J_z$ eigenvalues and $l$ to $J^2$ ones.
That's why you use $m,l$ (or $m,\lambda$ depending on bibliography), but you're denoting one eigenvector anyways. It's just notation.
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Help understanding unit: Micromoles per square meter per second The context is light, illumnination, photons.
The units seem to suggest something different from the definitions I have found:
$$\frac{µmol}{m^2 s}$$
This, to me, says I have one millionth of a mole ($6.022×10^{17}$) in photons landing on a one meter square area every second.
However, the definitions I have found pretty much state something similar to the above and then add that the quantity is divided by a mole ($6.022×10^{23}$).
Frankly, not sure what that means. Is it meant to represent what fraction of a micro-mole lands on a one square meter per second? I realize mol is unit-less, which might make it confusing.
In other words, you count the photons you have in one second, divide that by Avogadro's number (or is it Avogadro's number divided by 1E6?) and that's your number.
| Well, amounts measured by given a number times a unit, so the number is the amount divided by the unit. This is, if $x=n\frac {\mu mol}{m^2s}$, then $n = x \frac{m^2s}{\mu mol}$. That's similar to how, if something is five meters long, that means that its length divided by a meter is five.
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Principle of stationary action vs Euler-Lagrange Equation I am a bit confused as to what I should use to derive the equations of motions from the lagrange equation.
Suppose I have a lagrange function:
$$L(x(t), \dot{x}(t)) = \frac{1}{2}m\dot{x}^2-\frac{1}{2}k(\sqrt{x^2+a^2}-a)$$
Method 1: Principle of least action
$$\delta L = \delta \dot{x}(m\dot{x})-\delta x \frac{kx(\sqrt{x^2+a^2}-a)}{\sqrt(x^2+a^2)}$$
$$\delta W = \int_{t_0}^{t_1} \delta L \ dt$$
After doing integration by parts, i obtain:
$$\delta W= -\int_{t_0}^{t_1} \delta x \biggl[m \ddot{x}+\frac{kx(\sqrt{x^2+a^2}-a)}{\sqrt(x^2+a^2)} \biggl] dt$$
for stationary points, $\delta W = 0$
Hence, inside the integral,
$$m\ddot{x}+\frac{kx(\sqrt{x^2+a^2}-a)}{\sqrt(x^2+a^2)} = 0$$
and this is the equation of motion.
Method 2: Euler-Lagrange Equation
Alternatively, we can consider the euler lagrange equation:
$$\frac{\partial L}{\partial x} - \frac{d}{dt}\biggl(\frac{\partial L}{\partial \dot{x}} \biggl) = 0$$
By substituting $L$ into the euler-lagrange equation, we get the same equation of motion:
$$m\ddot{x}+\frac{kx(\sqrt{x^2+a^2}-a)}{\sqrt(x^2+a^2)} = 0$$
So method 2 is a lot easier than method 1, but why do we arrive at the same answer? I have a hunch both methods are essentially calculating the same thing, but I am not sure if this hunch is right because the euler lagrange equations seems a bit too simple as compared to principle of least action. Is there something i'm missing here?
| Granted appropriate boundary conditions, the stationary action principle and the Euler-Lagrange (EL) equations are both precisely the condition that the functional/variational derivative
$$\frac{\delta S}{\delta x^j (t)} \tag{1} $$
vanishes, so they better agree!
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Why doesn't the block fall?
I came upon this question as I was going through the concepts of tension. Well according to Newton's third law- every action has an equal and opposite reaction. Here my question is that if the tension at point B balances the tension at point A then which force balances mg as it can't be balanced by the reaction force of mg which attracts the earth towards the mass as it is not in contact with the mass. Then why doesn't the mass fall?
| There are two forces acting on the mass are gravity and tension from the rope. This is seen in your free body diagram on the right side of your image. The mass does not fall (or rise) because these forces are of equal magnitude and opposite direction, thus the net force acting on the mass is $0$.
Yes, if we look at the rope the force pulling it down at $A$ is equal and opposite to the force pulling it up at $B$, but this doesn't make the force at $A$ suddenly nonexistent for the mass. There is a downward force acting on the rope at $A$, and by Newton's third law this means there is an upward force acting on the mass at $A$. This is the upward force discussed above. Note that the forces at $A$ and $B$ acting on the rope do not form a Newton's third law pair.
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Energy of a dielectric rod Let a dielectric rod kept in an electric field (initially in equilibrium) be rotated by an angle, and then released. Is some energy stored in it (the rod)? Why? What is the value of this energy? How can I come up with energy for dielectric of arbitrary shape?
| A dielectric consisting of polar molecules initially in random orientations when no electric field is applied will undergo some degree of polarization by orienting the dipole moments of the polar molecules to align them with an external electric field.
If either the rod of such a dielectric is rotated in a stationary field, or the field is rotated with the dielectric stationary, the dipoles will reorient themselves accordingly. This reorientation result in heating, sometimes referred to as dielectric heating. This is the basic principle of dielectric heating of a such a material if it is exposed to an alternating electromagnetic field, such as the electromagnetic field of a microwave oven used for cooking, or medical diathermy equipment use to heat tissue for therapeutic purposes.
Insofar as quantifying the energy losses associated with dielectric heating, you can start by Googling the subject. The Wikipedia article on "Dielectric Heating" gives one equation (though marked "citation needed").
Hope this helps.
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Isotropic moments of inertia Explicit integration can show that the moment of inertia of a Platonic solid (i.e., tetrahedron, cube, octahedron, dodecahedron, or icosahedron) of uniform density is the same around any axis passing through its center. The axis need not pass through a vertex, or midpoint of an edge, or center of a face! In tensor terms, the moment of inertia tensor is a constant times the Kronecker delta, just like for a sphere.
What is the complete class of solids with an isotropic moment of inertia, and why?
| Start with any solid whatsoever. Choose a coordinate system $x,y,z$ in which its moment of inertia tensor is diagonal. The diagonal components are
\begin{align}
I_{xx} = I_0 - \sum_n m_n x_n^2
\\
I_{yy} = I_0 - \sum_n m_n y_n^2
\\
I_{zz} = I_0 - \sum_n m_n z_n^2
\end{align}
where the $I_0$ term is the same for all three. Applying scale factors $a,b,c$ in the $x,y,z$ dimensions converts this to
\begin{align}
I_{xx}' = I_0' - a^2 \sum_n m_n x_n^2
\\
I_{yy}' = I_0' - b^2 \sum_n m_n y_n^2
\\
I_{zz}' = I_0' - c^2 \sum_n m_n z_n^2
\end{align}
where $I_0'\neq I_0$ but is still the same for all three components. Clearly we can choose $a,b,c$ (or any two of them) so that $I_{xx}'=I_{yy}'=I_{zz}'$.
This shows that any object, however irregular its shape may be, is just two ordinary scale-factors away from having a perfectly isotropic moment of inertia tensor.
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What's 'force per second'? For example, if a force of 10 N per second (10 N/s) is applied to an object, does this have a name or a definition? I'm not referring to impulse - which is Ns.
An airplane's engine thrust is simply given as a force, but this must be a force applied by the engines each second (N/s)?
Thanks!
| As per second Newton law force is defined as momentum change over time :
$$ F = \dot p $$
If force is constant, then each second it will remain the same, thus defining "force per second" doesn't make much sense.
However force can be function of time too, i.e. $F=F(t)$, then one can calculate force change over time :
$$ \frac {dF}{dt} = \ddot p $$
Then total force acted over time interval $t_2-t_1$ is :
$$ F_{~tot} = \int_{t_1}^{t_2} \ddot p ~dt $$
If force increases monotonically and linearly, then one can define force per time period, or unit force :
$$ {F}_{1s},~[N/s] = \frac {F_{tot}}{t_2-t_1} $$
Multiply that over time period elapsed and you will get force acting at that current time moment. That would be a closest interpretation of "force per second".
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What vector field property means “is the curl of another vector field?” I'm an undergraduate mathematics educator and I teach a lot of multivariable calculus. I posed this question on MSE over four years ago and I haven't gotten any definitive answers (despite 12 upvotes and a bounty posted). It could be there's no answer, but someone suggested I ask on this forum.
I know that a vector field $\mathbf{F}$ is called irrotational if $\nabla \times \mathbf{F} = \mathbf{0}$ and conservative if there exists a function $g$ such that $\nabla g = \mathbf{F}$. Under suitable smoothness conditions on the component functions (so that Clairaut's theorem holds), conservative vector fields are irrotational, and under suitable topological conditions on the domain of $\mathbf{F}$, irrotational vector fields are conservative.
Moving up one degree, $\mathbf{F}$ is called incompressible if $\nabla \cdot \mathbf{F} = 0$. If there exists a vector field $\mathbf{G}$ such that $\mathbf{F} = \nabla \times \mathbf{G}$, then (again, under suitable smoothness conditions), $\mathbf{F}$ is incompressible. And again, under suitable topological conditions (the second cohomology group of the domain must be trivial), if $\mathbf{F}$ is incompressible, there exists a vector field $\mathbf{G}$ such that $\nabla \times\mathbf{G} = \mathbf{F}$.
It seems to me there ought to be a word to describe vector fields as shorthand for “is the curl of something” or “has a vector potential.” But a google search didn't turn anything up, and my colleagues couldn't think of a word either. Maybe I'm revealing the gap in my physics background. Does anybody know of such a word?
TL;DR: gradient is to conservative as curl is to ___?
| If the domain is topologically trivial, then, as explained in the other answers, "is a curl" is the same as "incompressible," i.e., has zero divergence. So that's your answer.
In examples like the electric field of a point charge, the domain has a hole in it. This breaks the equivalence between incompressibility and is-a-curl. However, you asked this on a physics site, so you need to realize that for a physicist, features like the singularity of this electric field are unphysical idealizations. Classically, we would think of this as an idealization of the field of some charge distribution like a uniformly charged sphere. This is why physicists don't need a different name for the is-a-curl property. Space isn't a swiss cheese, and we don't have fundamental physical fields that fail to be defined at certain points.
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Slowing a beam of light In the description of the phenomenon below, apart from slowing the absorption and emission of a wave by an atom, does the wave propagation between atoms also slow down?
https://www.seas.harvard.edu/haulab/publications/pdf/Ginsberg-Garner-and-Hau-Nature-445-623-(2007).pdf
I confess without beating that my English and quantum mechanics are too weak to pick it up.
| Photons are elementary particles, massless, they always travel at speed c in vacuum, when measured locally.
Though, you are correct, light can slow down in media, but what actually slows down in media is the wavefront.
You are asking whether between atoms light travels at speed c or not.
Individual photons still travel at speed c in vacuum, inbetween the atoms of the media.
The "speed of light" quoted in the refractive index definition is not the speed of a photon in the material. Photons still travel at c; it's the overall (classical) light wave that seems to travel at a different speed. The microscopic mechanism of this is complicated, involving repeated absorption and re-emission of photons by the electrons in the material.
Acceleration of photon
A very simple answer: the photon is absorbed by the constituents of the medium, after which it's reemitted in the same direction (conservation of momentum). This process costs time, so the effective speed of light in the medium is reduced, while the speed of light between the absorption and the reemission stays equal to the speed of light in vacuum.
What really causes light/photons to appear slower in media?
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Why does foam in a rotating liquid accumulate near the centre? I first noticed this while having a coffee. When the coffee was rotating in the cup, most of its foam accumulated near the centre.
I recreated the effect with some soap and water. The accumulated foam formed a beautiful dome. You can see the dome formation in detail in this video.
Side view
Top view
I wonder what causes this pattern formation. I would expect the foam to move away from the centre due to centrifugal forces, but that's not what I see. I believe there's something else to it. So here's the question in short:
Why does foam accumulate near the centre of rotating fluids?
Also, why there is a characteristic dome shape for the accumulated foam?
| The water experiences a greater centrifugal force than the bubbles
Both the bubbles and the water experience a centrifugal force. However, since the centrifugal force is given by:
$$ F_c = m R \omega ^2 $$
You can see that a more massive object (the water) will experience a greater centrifugal force. From the perspective of the rotating frame, those forces would look like the pink arrows below:
Thus the water at the same $R$ as the bubble will flow around the bubble, shoving it closer to the center of rotation. Of course, the bubbles still rise to the surface, so they rise in a pile and cause the bubble bulge in your picture:
This is the same principle by which a centrifuge operates, but instead of throwing the heavier material to the outside of the rotating water, it throws the lighter objects towards the center of the rotating water.
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Is the relation between change in potential energy and work by internal conservative force can be used even in presence of non conservative forces? We know that work done by internal conservative forces is the negative of change in potential energy of the system stored in conservative force field. But does this logic still hold when there are non-conservative forces like friction or resistance?
Do the non-conservative forces only withdraw from the kinetic energy part and not affect the potential energy in any way? Consider for an example a system of two charges having some mass kept at a finite distance and both are free to move over a rough surface and released.
| The relation between potential energy and charge will remain. The total mechanical energy in a system is $E = KE + PE$. In a conservative system (a single particle orbiting a particle of opposite sign, not including EM radiation), that total energy remains constant. This is like a frictionless pendulum. If a system includes non conservative forces, that total energy declines as the system evolves. As this happens, $PE$ will be known as a function of the particle's position. $KE$ will then be the difference between (current) total $E$ and current $PE$. This is like a pendulum with friction. The gravitational $PE$ in a pendulum is always the same function of position ($PE = mgL(1 - \cos(\theta))$ or some equivalent formula). The $KE$ of the pendulum in a problem with friction is whatever the difference between the current total energy is and the potential energy of that known function of position.
Note that energy can also be added to a system. A rocket can be fired to increase a space craft's energy, thereby changing it to a new orbit. This is how orbital maneuvers are done. In between rocket usage, the total energy of the orbit is constant.
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How can the mechanism of electrons in an atom be explained? I am a high school student who takes both Physics and Chemistry.
Recently I learnt about the quantum mechanical point of view of looking at electrons or nuclei. I also learnt that the wave functions can be obtained by solving the Schrodinger's equation with various conditions specific to the problem (such as the particle in a box).
My shallow understanding of quantum mechanics is that we can only know the probability of an electron existing at a certain position and time, and the actual position can be determined when the 'observation' takes place.
The chemical bondings and chemical reactions are the results of electric interactions between nuclei and electrons. The Coulomb force is a function of the distance between two charges, so it is important that the exact locations of electrons should be known. But taking into consideration quantum mechanics, we don't even know where the electrons are, and we built up a subject called Chemistry, and most importantly, CHEMISTRY STILL WORKS VERY WELL.
So, what is going on?
| It's true that because of QM you can't think of the electrons in the atoms as having precise positions. It's not just position that is affected by QM but all "observable" phenomena. You are right that a theory that explains forces as a function of position is therefore likely to run into problems. But there are quantum theories of how forces like electromagnetism work, even ones that take into account special relativity, and these have been used very successfully to describe what's going on with atoms. Your understanding of QM is not wrong, it's just that it's only the start of the journey.
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How does a time varying magnetic field confined in a cylindrical region produces induced electric field even outside the cylindrical region? We know from Faraday's Law of Electromagnetic Induction that due to Time Varying Magnetic Field (TVMF), a non conservative electric field will be induced.
Now if we consider a cylindrical region in which the magnetic field is varying with time then outside the region of cylinder there is no magnetic field and no variation of it either, so how does the electric field gets induced there?
I understand that in the interiority of the cylindrical region, the electric field will be induced as there is TVMF, but how it gets induced outside it?
Isn't it true that Induced electric field should only generate at the location of time variation of magnetic field?
Isn't it strange TVMF can produce Induced Electric Field even at a distant location?
Is it a law of nature kind of thing that TVMF can produce induce electric field at a distant location?
| Faraday's Law states that a time-varying $\textbf{magnetic flux}$ through a surface induces a electric field in said surface's border.
Now, take a circumference concentric with the cylinder in where there's a TVMF. The flux - "how much the magnetic field goes through " - the surface delimited by the circumference, varies with time. In a way, if the magnetic field is changing, you can think as if the number of magnetic field lines per volume unit is changing, and thus, the amount of lines piercing the circle (inside the cylinder) is varying. Outside the cylinder there's no magnetic field and so no line pierces the surface. However, no matter how large you take the radius of the circumference, if it's concentric to the cylinder, you'll always have the magnetic field lines piercing (in the intersection of your circle and the cylinder)
Then, around your circumference, you have an induced electric field, even outside the cylinder. As the radius increases, since the flux is constant, the induced $\vec{E}\rightarrow 0$.
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What is orbit that could place satellite to be always under the Sun? I just thinking that, if we could place satellite to orbit earth in opposite direction of earth rotation, inverse of geostationary orbit. If we carefully choose a speed to sync with earth rotation, it would have that satellite stay exactly the same position in the sky as the sun
What is that altitude and are there any name of that orbit? Or is it just the same as geostationary orbit?
| I think what you're talking about are Lagrangian points:
Image Credit: Anynobody / Wikimedia Commons
They are points such that the combined gravitational effects from the Earth and the Sun give rise to areas that would have the same angular velocity as the Earth. However, out of the five lagrange points shown below: 1,2,3 are unstable
You are probably referring to the first Lagrangian point, known as $L_1$ which is roughly $0.01$ AU away from Earth.
Edit: Turns out I misunderstood what you were asking for. If you were curious if such a point can exist such that it is still "orbiting" around the Earth, the answer is no. You may think that as the orbit goes further and further out, you can eventually reduce the period to one year. However, at this point the effect from the Sun cannot be ignored and you can't realy call it "orbiting" around Earth.
We have:
$$m\left(\frac{2\pi}{T}\right)^2 r = \frac{GMm}{r^2} \to r^3 = \frac{GMT^2}{4\pi^2}$$
Plugging in $G=6.67\times 10^{-11} \mathrm{N\cdot m^2 \cdot kg^{-2}}$, $M=5.972 \times 10^{24} \text{ kg}$, and $T=31556926 \text{ s}$ gives:
$$r = 2.16 \times 10^9 \text{ m}$$
while $L_1$ is at a distance of 0.01 AU or $1.50 \times 10^9 \text{ m}$ which is slightly farther than the $L_1$ point which exists there due to the interaction between the Earth AND the Sun, so a satellite cannot orbit around Earth with a period of one year.
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How is it possible that entropy of the universe has increased since big bang but temperature of the universe has decreased? How is it possible that entropy of the universe has increased since the big bang, but the temperature of the universe has decreased?
I know that the increasing temperature of the system tends to increase in the entropy of the system thing tends to go higher entropic states.
Here System = Universe.
| Maybe this wikipedia definitions will help:
In statistical mechanics, entropy is an extensive property of a thermodynamic system. It is closely related to the number Ω of microscopic configurations (known as microstates) that are consistent with the macroscopic quantities that characterize the system (such as its volume, pressure and temperature). Entropy expresses the number $Ω$ of different configurations that a system defined by macroscopic variables could assume. Under the assumption that each microstate is equally probable, the entropy $S$ is the natural logarithm of the number of microstates, multiplied by the Boltzmann constant $k_B$. Formally (assuming equiprobable microstates),
$S=k_BlnΩ$
In an expanding universe, the number of microstates is continually growing : in the beginning ( lets skip the quantum mechanical inflation time and count entropy after that) the continuous particle exchanges in the quark gluon plasma, the continuous radiation from charged particles inevitably lead to a large increase with time of the number of microstates. There is black body radiation continuously from all masses , after all even now (which is one way of looking at starlight).
| {
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How does the Earth have enough gravity to hold on to its atmosphere? I'm looking for numbers/math that describe how earth (or any other planet) holds on to it's atmosphere.
Presumably, we would be able to determine the exact atmospheric pressure that we would expect to have on a planet of a given radius/mass/density, assuming that we start with infinite amount of atmosphere in which most of it disperses till we get into an equilibrium.
| This actually isn't too hard. The Maxwell-Boltzmann distribution provides the average velocity of gases. The most probable speed (which is only good for order of magnitude estimates) is:
$v_p=\sqrt{\frac{2kT}{m}}$
You can put in the numbers: $T$ is the temperature of the Earth's surface, about $300K$, and $m$ is the mass of whatever gas you're interested in. For example, with hydrogen, you get that the most probable velocity is about $2225 m/s$.
Compare that to Earth's escape velocity, which is calculated from Newton's laws as about $11.186 km/s$. This is much larger than the most probable velocity of hydrogen, which indicates that most hydrogen atoms would not escape. However, a substantial-enough fraction does (see the shape of the Maxwell-Boltzmann distribution). The remaining hydrogen thermalizes, which leads to more hydrogen that's moving faster than the escape velocity, etc, so that hydrogen eventually escapes the atmosphere.
You can calculate the fraction of hydrogen that's moving fast enough to escape by integrating the Maxwell-Boltzmann distribution, and the same goes for other gases.
You can also read more about the entire phenomenon at Wikipedia's article on atmospheric escape.
| {
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Does the expansion of the universe stretch distances in non-radial/angular directions? At time $t = t_0$ a galaxy is situated at a proper distance $r_0$ and has a proper diameter $D_0$. Setting $a(t_0) = 1$ the proper distance to the galaxy will then of course evolve with the expansion of the universe as:
$$r = a\chi = \frac{\chi}{1+z} = \frac{r_0}{1+z}$$
What happens to the proper diameter of the galaxy as the universe expands? Will it also scale as:
$$D = \frac{D_0}{1+z}$$
like the proper radial distance does?
And if this is true does this mean that the angle subtended by the galaxy as seen by an observer at $r = 0$ is constant as:
$$\Delta\theta = \frac{D_{em}}{r_{em}} = \frac{D_0}{r_0}$$
where 'em' designates distances at the time when the photons were emitted that are received by the observer at time $t$.
| FLRW metric can be written as,
$$ds^2 = -c^2dt^2 + a(t)^2[dr^2 + S_{\kappa}(r)^2d\Omega^2]$$
In the calculations of the angular diameter distance, we set $dt = dr = d\phi =0$ which leads to
$$ds = a(t_e)S_{\kappa}(r)d\theta$$
If the object has a diameter $D$ then we can write.
$$D = a(t_e)S_{\kappa}(r)d\theta$$
or
$$D = \frac{S_{\kappa}(r)d\theta} {1+z}$$.
I think it's not right to say the diameter of the object decreases. I think only angular size ($d\theta$) changes.
If I put a $1m$ ruler in some distance and measure its angular size, and If I move it further away, Its length will not change. However, the angular size will get smaller since its getting further away.
| {
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Constant $g$ acceleration from astronaut's frame of reference When a spaceship is experiencing a constant acceleration of $10m/s^2$, the astronauts will be moving at nearly the speed of light after about a year in the earth's reference frame. This means the spaceship's energy will start to diverge as a function of the speed $v$ so there will be a huge amount of energy necessary to increase the speed of the ship any further. This way, the speed of light can never be crossed.
All of this is clear to me, but all of this is also formulated in earth's reference frame. But from the astronaut's reference frame: the spaceship is simply accelerating at $10m/s^2$ and so the force on the spaceship is constant. Then why would we need huge amounts of energy to accelerate the spaceship?
For example, I read somewhere that the amount of energy that would be needed to accelerate a large spaceship to half the speed of light is more than 2000 times the current world annual energy consumption. How does this make sense in the astronaut's (non-inertial) frame?
| If you know how much fuel (mass) you need to travel from the earth to the next object , you can answer the question how much energy you need , remember that energy is equivalent to mass.
I found the answer in this dokument
http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html
$$\frac{m_F}{m_L}=\exp\left({\frac{a\,T}{c}}\right)-1\tag 1$$
where:
$m_F$ is the fuel mass
$m_L$ is the payload mass
$a$ is the constant acceleration
$T$ is the rocket time
$c$ is the light velocity
Example:
for:
$a=1 g$
$g\approx 1.03 [ly/yr^2]$
$c=1 [ly/yr]$
if you want to go to a distance from the earth
$d=4.3\,\, [ly]$ you need to travel $T=3.6$ years
$$d=\frac{{c}^{2}}{a} \left( \cosh \left( {\frac {aT}{c}} \right) -1 \right)
$$
thus equation (1)
for every kilogram payload ($m_L=1$) you need $m_F=10\,[kg]$ fuel . from here you can calculate the energy $E_F=m_F\,c^2$
| {
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Can the forces change with frame of reference? Consider a ball kept on man's head (mass $M$) on the Earth. Now supposing I throw the ball from height $h$ of tall building then why does he gets more hurt? Isn't the force still mg?
I would like to know what happens in ideal case (no air resistance)
and then in real case (with air resistance)
|
Now supposing I throw the ball from height ℎ of tall building then why
does he gets more hurt? Isn't the force still mg?
The impact force of the ball falling on the man's head is not the same as the weight of the ball on the persons head. This is because it takes a force to perform work in order to absorb the kinetic energy of the ball at impact. The work needed to stop the ball is given by the work-energy theorem which states that the net work done on an object equals its change in kinetic energy, or
$$Fd=\frac{mv_{f}^2}{2}-\frac{mv_{i}^2}{2}$$
Where $F$ is the average impact force on the mans head, $d$ is the stopping distance, $f$ and $i$ indicate the final and initial velocities, respectively. If the ball is brought to a stop by the head, the $v_f$ = 0, and
$$Fd=-\frac{mv_{i}^2}{2}$$
The work done is negative (meaning energy is taken away from the ball) because the force acts upwards opposite the direction of the displacement $d$. The equation assumes the displacement $d$ is small so that the change in gravitational potential energy over the stopping distance can be ignored.
I would like to know what happens in ideal case (no air resistance)
and then in real case (with air resistance)
With air resistance you have an upward force on the ball opposing the downward force of gravity. Therefore, for a given height the ball will impact the head with a lower velocity with air resistance than without. That, in turn, will reduce the average impact force on the man's head. So no air resistance is not the ideal case in terms of potential injury to the man. Air resistance reduces the impact force.
Hope this helps.
| {
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Why in mercury barometer pressure inside the glass tube at a point is same as the pressure outside the glass tube at the same height? I am quite familiar with the pascal’s law but i still think that being on the same points still they should have different pressures as above on one point is atmosphere and on the other is only the column of mercury can anyone give physical explanation for this and also if the pressure inside the torricelli vacuum is only due to mercury vapours which is quite low than why the vacuum does not get crushed under the atmospheric pressure?
|
Let understand this by the figure, if we notice that pressure at surface of mercury would be the atmospheric pressure. And same pressure at the bottom of the mercury tube and if we go a height h above.
Now the pressure in the space above the mercury is less than atmospheric pressure and equilibrium is attained when $ Mg$ (where M is the new mass of mercury which is less than initial mass in the mercury column) + P*A (where P is less than atmospheric pressure) = (Atmospheric pressure) $ A$ where A is area of cross section of tube. Or as $ρ g h$ (where h is the new height of column) = P + Atmospheric pressure
So we get. $$ \rho g h = P_\text{atm} - P $$
where $\rho$ is the density of the mercury.
| {
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Banana vs Brazilian Nuts Radiation Looking online I can find that the BED (Banana Equivalent Dose) is around 0.1uSv, but Brazilian Nuts can have up to 444Bq/kg.
I know both of those foods are safe, I just wanted to understand how those units compare to each other, how much more radioactive is a Brazilian Nut compared to a Banana.
| Answering directly to "I just wanted to understand how those units compare to each other":
You can't convert directly between Sievert (Sv) and Becquerel, since they represent different physical quantities.
Quoting the definition of Sievert from the Wikipedia:
1 Sv = 1 joule/kilogram – a biological effect.
The sievert represents the equivalent biological effect of the deposit of a joule of radiation energy in a kilogram of human tissue.
And quoting the definition of Becquerel from the Wikipedia:
One becquerel is defined as the activity of a quantity of radioactive material in which one nucleus decays per second.
So, 1 Bq = 1 s$^{-1}$.
Therefore, Sievert and Becquerel are distinct things.
| {
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Is it a typo in David Tong's derivation of spin-orbit interaction? A few lines below equation 7.8 D. Tong writes
The final fact is the Lorentz transformation of the electric field: as electron moving with velocity $\vec{v}$ in an electric field E will experience a magnetic field $\vec{B}=\frac{\gamma}{c^2}(\vec{v}\times\vec{E})$.
The note says that it was derived in another note but I couldn't find this expression.
Is this coefficient $\gamma/c^{2}$ correct? Griffiths derives this to be $-1/c^2$ and I did not find anything wrong there. See Griffiths electrodynamics, third edition, equation 12.109.
Then I looked at this book which uses Griffiths' expression in Sec. 20.5, but uses $\vec{p}=m\vec{v}$ instead to $\vec{p}=\gamma m \vec{v}$ to derive the same result. Which one is correct and why?
| $\vec{p}=\gamma m\vec{v}$ is the technically correct equation, but for non-relativistic particles where $|\vec{v}|\ll c$, the Lorentz factor becomes
\begin{equation}
\gamma=\frac{1}{\sqrt{1-v^2/c^2}}\approx 1,
\end{equation}
and so can be neglected.
For your reference, I had a quick look and I believe Eq. (6.45) of his EM notes is where this is derived.
Not sure about the negative sign in Griffiths though.
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Particle falling into a Kerr black hole Let's say that a particle starts a radial free fall towards a Kerr black hole with zero initial energy at $r\rightarrow\infty$. The initial angular momentum of the particle is zero ($p_\phi = 0)$. From the Kerr metric, $p_\phi$ and $E$ are the constants of motion for the particle geodesic. However, we know that inside the ergosphere, we have a frame dragging effect, which means that the particle will start rotating after it enters into the ergosphere. Does it mean that inside the ergosphere, the particle angular momentum $p_\phi$ is nonzero and $p_\phi$ is not a constant of motion anymore?
| No the angular momentum of the particle would remain zero (it is still a constant of motion). The angular velocity of the particle, on the other hand, would steadily increase as the particle approaches the Kerr black hole (i.e. the Newtonian relationship between angular momentum and angular velocity does not hold in GR!)
Note that the frame dragging does not start at the ergosphere. Frame dragging will start effecting the particle right from the beginning. The ergosphere is simply the region in which it becomes impossible to have a timelike curve with negative angular velocity.
| {
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Why can we assume long structural members to be in plane strain?
In plane strain, the strain in one direction is assumed to be zero. It is taught that if we have a structural member that is very long in one direction compared to the others, the strain in the long direction can be assumed to be zero. A dam is a popular example is textbooks. But why is this? I've never seen this explained very well. It is simply stated that the strain (not close to the ends) is assumed to be plane strain. Why can we assume this? What does the length of the dam have to with it?
And if this is true, why don't beams follow this same principle? If we have a long beam, we should be able to assume the normal strain along the length of the beam is zero. But according to Euler-Bernoulli beam theory, this is not the case. In fact, in E-B theory we assume to normal stress along the length of the beam is the only non-zero strain.
| I think that we can often deal with long structural members as plain strain due to friction along the length, which acts as a practical restriction to displacement.
In the case of the dam, the friction with the soil under its own huge weight avoids longitudinal displacement.
Another example is rolling of steel sheets. The roll gap, being thinner than the incoming sheet, forces a reduction of thickness. The material can flow forward much easily than sideways, due to friction with the rolls. (by the way, the process requires friction, otherwise the rolls can not "bite" the entry bar). The result is an elongated sheet with almost the same width. This is plastic deformation not elastic strain, but the reason for the outcome is the plane strain configuration.
In the case of a steel beam there is no restriction to longitudinal displacement.
| {
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Why does atmospheric pressure act on us? I have a bit of misconception about weight which I want to clarify.
The air pressure is explained as the weight of the air column above our head acting per unit area. But since air is not continuous how can the weight of all the air molecules (above our head) be acting on our head? I mean we would only feel the weight of the molecules near the surface of our head (if not then why not?) but how do we feel the weight of molecules so far away?
I have edited this question and asked this follow up question (so that it remains specific).
| The reason air pressure acts on us as humans really has nothing to do with mathematics, its science. Our bodies are more water than mass, so the pressure exerted on us by air pressure acts on our blood and other body fluids, making increases and decreases in our own pressures. Its really only common sense - this is why air pressure changes cause headaches! I should know!
| {
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Entropy of mixing Consider the case of a wall dividing a box into sections 1 and 2, each of volume $V_0$.
Let be $X$ is an ideal gases. Section 1 contains $N$ particles of $X$ and section 2 contains $N$ particles of $X$.
The entropy of mixing (removing the divider) is $0$ in the above case. I understand that an argument for this is that it is a reversible process as we can just as well place the divider back and the system will be in an identical state to its initial condition (assume indistinguishable particles).
My question is that if entropy is a measure of the number of states a particle can take on, a state being a specification of all the values of the variables needed to completely describe the particles, i.e its position and momentum.
If we remove the divider, each particle can now take up twice as many positions and so should have twice as many available states to be in, this surely increases the entropy?
I can't seem to resolve the two different ideas.
Can someone clarify on this?
| So when you mix the gases, say that the probability of any gas particle existing in the i'th state is $p_i$, so now what do you say is the probability of particles being in various states?
$\Pi_ip_i$. Now entropy is defined as the average value of $-log(p_i)$, and as $p_i$ is multiplicative, $-log(p_i)$ is additive, i.e. entropy is additive
| {
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The net charge of Earth Shouldn't the earth have an over all small positive net charge? Cosmic rays in the form of light or high energy particles can ionize atoms, when that happens the electron and the positive ion recoils and the electron gets a whole lot more velocity than the positive ion.
Meaning the electron had an easier time escaping earth, and the ones that don't will have a lower average density because of higher orbits, of this causes earth to have an over all positive charge, then the electric field grows until this effect is balanced with the electric fields potential, so that the charges are as easily ejected in the form of ions and the form of electrons. Wouldn't this mean that depending on the background ionizing radiation the earth will have a certain equilibrium positive charge?
| There are other effects at play. In particular, solar wind provides protons and electrons which can be attracted by bodies with a charge. If the earth were to develop a negative charge, more protons would be attracted to it than electrons, and the charges would neutralize.
In theory there is some equilibrium balance that is not quite perfect. This SE question suggests that the whole of the Earth and atmosphere actually has a very tiny negative charge, on the order of -1C (which is a really small charge, given how big the Earth is)
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