cosmology - Isn't Big Bang theory a violation of law of conservation of linear momentum?

What Big Bang theory assumed in the formation of universe?

Answer:-At its simplest, it says the universe as we know it started with a small singularity, then inflated over the next 13.8 billion years to the cosmos that we know today.

Now This nothing must be there outside this singularity as everything is contained in universe itself.So let's take the singularity as our system.Then,$$\vec{F_{ext}}=0$$

It means linear momentum of the system must be conserved.So,the centre of mass must not move even after 13.8 billion years as it was initially at rest.I am saying that it was initially at rest because it would never have acquired a velocity because nothing pushed it from outside because every force is an internal force for the universe.Initially,$$\vec{v_{com}}=0$$ $$\vec{p_{com}}=0$$

The velocity of centre of mass of universe must be zero for all time.Now after 13.8 billion years,the velocity of centre of mass by its definition must be:-

$$\vec{v_{com}}=\frac{\sum_{i=1}^{\infty}m_{i}\vec{v_{i}}}{\sum_{i=1}^{\infty}m_{i}}$$

$$\vec{v_{com}}=\frac{\sum_{i=1}^{\infty}\vec{p_{i}}}{\sum_{i=1}^{\infty}m_{i}}$$

$$\vec{v_{com}}=\frac{0}{unmeasurable}$$

It is an undetermined form in mathematics because dividing zero by a unmeasurable number is meaningless.So linear velocity of center of mass is not zero which is a complete violation of law of conservation of linear momentum.With due respect,it means our assumption of universe as a singularity may be wrong.

My question is,isn't Big Bang theory a violation of law of conservation of linear momentum?

EDIT

I made a major mistake in the above problem so I have corrected it now. The mistake was I quoted that $\sum_{i=1}^{\infty}m_{i}$ tends to be infinity which is a limiting process and it can't be a limiting process as universe is still expanding so it's mass must be undefined as we can't calculate the mass of this universe due its expansion.It is meaningless to say something about the mass of universe.

Answer

You’ve made two mistakes.

First, you can’t use Newtonian physics to understand the entire universe. If the universe had a center of mass (which it doesn’t), what would it be moving, or not moving, relative to? In mainstream cosmology, you have to use Riemannian geometry and General Relativity to understand the dynamics of the universe.

Second, zero divided by infinity isn’t undetermined. It’s well-determined and is just zero.

classical mechanics - Why does friction cause a car to turn?

I've had a lot of difficulty conceptually understanding the physics of how a car turns on an unbanked curve, so I'm hoping you could help me out. When a car is moving in uniform circular motion, we know that $|\vec{a}| = \frac{v^2}{r}$, and the direction of acceleration is towards the radius of the circle about which the car is moving. Drawing a free body diagram for the car shows that there are only three forces acting on it: gravity $(\vec{F_g})$, the normal force $(\vec{F_n})$, and friction $(\vec{F_f})$. Since gravity and the normal force negate each other, the car isn't accelerating in the $y$ direction. Because it is in uniform circular motion we know it is accelerating in the $x$ direction, and summing up the forces in this direction yields $$\vec{F_{net x}}=m\vec{a}=\vec{F_f}$$ which implies that the centripetal acceleration is due to the frictional force.

What I am having difficulty understanding is why this intuitively makes sense. I've read some other people's answers on this question but I haven't found anything satisfactory. In particular, many people talk about how wheels "are pushing the pavement to the left or right", and this causes the pavement to exert a force on the car wheels by Newton's third law, but this hasn't made sense to me.

Another way of putting this might be that I don't understand why friction should be directed inwards towards the center of the circle about which one is turning. I would expect that, since the wheels have been turned, that friction would be directed in the opposite direction of where the car is moving to prevent the car from continuing to move forward and skidding on the road.

I hope this makes sense, thanks.

Answer

I had fun trying to make this as intuitive as possible. I hope I've succeeded without doing the physics of the situation much injustice.

When a car is driving straight ahead, the plane in which the wheels are rotating is aligned with the direction of movement. Another way of saying this is that the rotation axis is perpendicular to the momentum vector $\vec{p}=m\vec{v}$ of the car. So the friction merely makes it harder for the car to move, which is part of the reason why you need to put your foot on the gas pedal to maintain a constant speed. At the same time, the friction is what allows you to maintain that constant speed because the rotating tires sort of grab onto the ground, which is the intuitive picture of friction. The tires grab the ground and pull/push it backwards beneath themselves, as you would do when dragging yourself over the floor (if it had handles to grab onto). Those grabbing and pulling/pushing forces are what keeps you going.

Things change when the wheels are turned. The plane in which they are rotating now is at an angle with the direction of motion. Alternatively but equivalently, we could say the rotation axis now makes an angle with the momentum vector of the car. To see how friction then makes the car turn, think again in terms of the wheels grabbing onto the ground. The fact that they now make an angle with the direction of motion, means the force the tires are exerting is also at an angle with the direction of motion - or equivalently, the momentum vector.

Now, a force is a change in momentum$^1$ and so (because the wheels are part of the rigid body that is a car) this force will change the direction of the car's momentum vector until it is aligned with the exerted force. Imagine dragging yourself forward on a straight line of handles on the floor and then suddenly grabbing hold of a handle slightly to one side instead of the one straight ahead. You'll steer yourself away from the original direction in which you were headed.

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$^1$ Mathematically: $$\vec{F}=\frac{d\vec{p}}{dt}$$

Proof that $(1/2,1/2)$ Lorentz group representation is a 4-vector

Taken from Quantum Field Theory in a Nutshell by Zee, problem II.3.1:

Show by explicit computation that $(\frac{1}{2},\frac{1}{2})$ is indeed the Lorentz vector.

This has been asked here:

How do I construct the $SU(2)$ representation of the Lorentz Group using $SU(2)\times SU(2)\sim SO(3,1)$ ?

but I can't really digest the formality of this answer with only a little knowledge of groups and representations.

By playing around with the Lorentz group generators it is possible to find the basis $J_{\pm i}$ that separately have the Lie algebra of $SU(2)$, and thus can be separately given spin representations.

My approach has been to write $$J_{+i}=\frac{1}{2}(J_{i}+iK_{i})=\frac{1}{2}\sigma_{i}$$ $$J_{-i}=\frac{1}{2}(J_{i}-iK_{i})=\frac{1}{2}\sigma_{i}$$ which implies that $$J_{i}=\sigma_{i}$$ $$K_{i}=0$$ However I don't really get where to go next.

solid state physics - When I stretch a rubber band, it breaks. When I hold the broken ends together, why doesn't it join again?

The question is simple. When we join the two broken surfaces, what is it that keeps the surfaces from connecting with each other, while earlier they were attached to each other? Also, would the two sides join again if I hold them together for an infinite amount of time?

electromagnetism - Second law of thermodynamics and a bunch of magnets

Say I put a bunch of powerful square magnets on a nearly frictionless table in a disordered fashion. The second law of thermodynamics states that the system shall spontaneously get more disordered, yet the magnets will attract each other and form a chain (typically), thereby increasing the order of the system-and, seemingly, decreasing its entropy.

It would seem to me that the system is closed and the lattice is indeed the equilibrium state. Therefore, I suspect that by attracting each other, the magnets increase their own entropy by a larger amount than the decrease in entropy caused by the lattice formation. Is it true? If yes, what are the thermodynamics of magnets responsible for this? Is there a microscopic explanation?

Thanks!

Answer

The magnets will indeed attract each other. This attraction will put them in motion, and they will head towards each other, converting electromagnetic energy into kinetic energy. Then they will collide, and loose their kinetic energies in the collision, finally coming to rest in a more ordered, low-energy state.

In terms of energy, the outcome of the experiment is that you have converted electromagnetic energy into heat: the heat released in the collisions. This conversion creates far more entropy than the entropy lost by arranging the magnets in a more ordered fashion.

mass - Why are scientists involved in the Avogadro Project using silicon-28 atoms instead of carbon-12?

My question is, why use silicon-28 atoms to calculate the kilogram when you already have carbon-12 atoms defining the constant?

Does the Avogadro Project intend to define the constant by replacing the idea of carbon-12 and putting silicon-28 in its place?

Answer

The idea is to create a sphere of about 1 kilogram and then both weight it and count the number of atoms in it. This is only possible by using crystalline matter, by taking advantage of the regular arrangement of the atoms.

Diamond would be indeed a perfect candidate but machining diamond is a hell of a lot more difficult than machining a crystal of silicon, because of the huge difference in hardness, a problem which pales in comparison with the sheer impossibility of making a diamond mono-crystal weighing one kilogram! The world record is about 20 grams. The difficulty is the need to apply a pressure of the order of 100,000 atmospheres, which is only possible in too small a volume for the target weight of 1 kilogram, which would be a cube with 6.5 cm sides. We could imagine settling for a bunch of smaller diamonds of course but this would introduce an extra source of uncertainty. Since it is possible to make a monocrystal of silicon weighing one kilogram, by using refinements of the growth methods developed and refined by the electronic industry, it would not make sense to consider diamond.

Why not graphite instead? Unlike diamond, it is possible to carefully make big enough mono-crystals. Unfortunately, graphite is made of a regular arrangement of carbon atoms strongly bonded in sheets, and those sheets do then stack up and keep together because of weaker forces between them: in particular they can easily shift with respect to each other. As a result, this makes graphite much less suitable for the described precision experiment where pinpointing the atomic arrangement is key.

With graphite and diamond, we have exhausted the crystalline phases of carbon. Thus exit carbon!

particle physics - Is the "protophobic fifth force" observed in nuclear decays a new fundamental force?

There is a recent paper in PRL, also on the ArXiv that claims the discovery of a fifth force, mediated by a 17 MeV boson, which explains an anomaly in nuclear decays.

Is this a real effect? Could the claimed particle be some sort of composite particle such as a pion?

lagrangian formalism - How can Noether's Theorem be used to prove that the probability density satisfies a continuity equation?

How can I use Noether's Theorem to show that the probability density $\rho (x)=|\psi(x)|^2$ for a wave function $\psi(x)$ satisfies the continuity equation $\frac{\partial \rho}{\partial t}+\nabla \cdot\vec{j}=0$, where $\vec{j}$ is the probability current defined in quantum mechanics?

I have solved this problem before by other means but I don't think I understand Noether's Theorem well enough to apply it in this case. Any help would be greatly appreciated.

Answer

First note that Schrödinger's equation can be understood to come from an action. The Lagrangian is $$L = \int~\mathrm d^3x \,\,\psi^†(x) \left(i \frac{\partial}{\partial t} - \frac{\nabla^2}{2m}\right)\psi(x) - \psi^†(x)\psi(x)V(x)$$

The Euler-Lagrange equation for $\psi^†(x)$ is exactly the Schrödinger equation. Since the dynamics of $\psi(x)$ are determined by Lagrangian mechanics in this way, Noether's theorem applies without any caveats.^^

In particular, this Schrödinger Lagrangian has a $U(1)$ symmetry corresponding to $\psi(x) \mapsto e^{i\alpha}\psi(x)$. The corresponding conserved charge current density is $$\rho = j^0 = \frac{\partial L}{\partial \dot{\psi}}\delta \psi = \psi^†\psi(x)$$ $$\vec{j}^i = \frac{\partial L}{\partial_i\psi}\delta \psi+\frac{\partial L}{\partial_i\psi^†}\delta \psi^†=\frac{i}{2m}\left((\partial^i\psi^†)\psi-\psi^†\partial^i\psi\right),$$ which is the well-known probability current density.

^^ In non-relativistic quantum mechanics the wavefunction $\psi(x)$ is a "classical" variable in that it is simply a function from space and time to $\mathbb{C}$. Noether's theorem works exactly the same for it as in classical mechanics. In quantum field theory the relevant objects $\psi(x)$ become quantum operators and the usual arguments have to be modified somewhat.

newtonian mechanics - Why is force a vector?

"We have focused our discussion on one-dimensional motion. It is natural to assume that for three-dimensional motion, force, like acceleration, behaves like a vector."- (Introduction to Mechanics) Kleppner and Kolenkow

We learn it very early in the course of our study that Force is vector; But, if I were the physicist defining the the Newton's second law (experimentally) and analysing the result F=ma, how would I determine whether Force is vector or scalar(especially in 3-D).

Actually, when I read the aforementioned sentences from the book, I wanted to know why do the authors expect it to be natural for us to think that in 3-D "Force" behaves like a vector. I know a(acceleration ) is vector and mass a scalar and scalar times vector gives a new vector but is there another explanation for this?

Answer

Uhm ... you start with an object at rest and notice that if you push on it in different directions it moves in different directions? Then notice that you can arrange more than two (three for planar geometries and four for full 3D geometries) non-colinear forces to cancel each other out (hopefully you did a force-table exercise in your class and have done this yourself).

The demonstration on an object already in motion is slightly less obvious but you can take the ideas here and generalize them.

In a sense this is so obvious that it's hard to answer because almost anything you do with forces makes use of their vector nature.

homework and exercises - Geometric optics question (from 2001 GRE)

In the diagram above, two lenses with focal lengths $f_1 = 20$ cm and $f_2 = 10$ cm are placed $40$ cm and $70$ cm from an object $O$, respectively. Where is the final image formed by the two-lens system, in relation to the second lens?

(a) 5 cm to its right

(b) 13.3 cm to its right

(c) infinitely far to its right

(d) 13.3 cm to its left

(e) 100 cm to its left

I can use the thin lens equation $$\frac{1}{s} + \frac{1}{s'} = \frac{1}{f}$$ (where $s$ is the object distance, $s'$ is the image distance, and $f$ is the focal length, up to sign, which is what makes this business confusing) to determine that in the absence of a second lens there would be an image $40$ cm to the right of the first lens.

According to this online solution we must take the image of the first lens to be the object of the second lens and use the above equation again to get choice (a).

How is that allowed? Doesn't the second lens totally interfere with/ obstruct the formation of the first image, making this a complicated problem?

Answer

It works because the thin lens equation works for virtual images (i.e. $s'<0$) and virtual objects (i.e. $s<0$). To see why that is true, lets derive the thin lens equation with the following two assumptions:

1. A light ray passing through the center of a lens does not bend.


2. a light ray entering normal to the lens bends and passes through the focal point.

The rays originate from a point on the object, pass through the lens, then converge at a point that generates an image, like shown for a convex lens:

The similar triangles with angle $\alpha$ have legs that are in proportion: $$ \frac{y_o + y_i}{s + s'} = \frac {y_o}{s} $$
And the similar triangles with angle $\beta$ have legs that are in proportion: $$ \frac{y_o + y_i}{s'} = \frac{y_o}{f} $$
Dividing these two equations and rearranging gives the thin lens equation: $$ { 1 \over s } + {1 \over s'} = {1 \over f} $$
Note that $s,s',y_o,f$ are all positive while $y_i$ is negative in this example. What would happen if we made $s$ negative? That would amount to placing $s$ on the same side as $s'$:

The object at $s$ is called virtual because it is located where the light rays would converge to if the lens wasn't there. The above rules for (1) and (2) rays still apply however, so we can generate a real image. And so long as you treat $s$ as negative (i.e. let $s - s' = -(s+s')$ ) you will get the thin lens equation by analyzing the two triangles.

newtonian mechanics - Violation of energy conservation during collisions of a particle with different sections of a rod

Imagine a homogenous rod with total mass $M$ and length $l$ floating in free space without any force or constraint acting on it. Then, think about two possible scenarios. In the first, a particle with lineal momentum $mv$ impacts right at the center of the rod. In the second, the same particle with same lineal momentum impacts right at one of the ends of the rod.

First case

In the first case there will be no rotation, only translation. We apply conservation of momentum (where I think there are two cases, one with the particle having final lineal momentum $v_f$ and the other with $-v_f$, but i will omit the first one).

$$mv = MV - mv_f \quad \quad V= \frac{m}{M}(v+v_f)$$

Then the final energy is:

$$E_f = \frac{1}{2} MV^2 + \frac{1}{2} mv_f^2 = \frac{m^2}{2M} (v+v_f)^2 + \frac{1}{2} mv_f^2$$

Second case

In the second case there will only be rotation (or so I think, but I'm getting really confused so that's why I've come up with this scenarios to try to prove it) without translation. Again we apply conservation of momentum:

$$mv = MV - mv_f \quad \quad V= \frac{m}{M}(v+v_f)$$

Let me explain what $V$ represents in this case. Since the rod rotates all the points in it don't have the same velocity, but, since the velocity increases linearly with the radius and the rod is homogenous, if the velocity of the ends is $2V$ (which I compell it to be this way) then the average speed of thte whole rod is $V$. Then we proceed (the cross product of $L$ doesn't matter):

$$I \omega = L = pr = MV l/4 = \frac{ml(v+v_f)}{4}$$

So the final energy of the whole system is (having $I=\frac{1}{12}Ml^2$):

$$E_f= \frac{1}{2} \frac{(I\omega)^2}{I} + \frac{1}{2} mv_f^2 = \frac{3m^2}{8M} (v+v_f)^2 + \frac{1}{2} mv_f^2$$

So comparing the two cases, there's a difference in the final energy, specifically a difference of $\frac{m^2}{8M} (v+v_f)^2$, which doesn't make sense.

Conclusions

This difference can mean various things, and I don't know which of this ones is, or even if it implies something else I haven't thought about. But before, there's something I don't understand and then there would be no need of other explanations.

As I've said the average velocity of the rod is $V$ so why should it have different speed when rotating than if the whole rod is advancing? In the end, the product mass times velocity is the same. So that's one thing I don't understand.

If that's not the case and the kinetic energy is different for some reason here's the explanations I've thought about for this difference in the energies:

1. In the second case there's also kinetic energy in the form of translation of the rod, and not only rotation, specifically with a translational kinetic energy of $\frac{m^2}{8M} (v+v_f)^2$, but why this specific quantity?


2. Actually, apart from the rotation, the rod has the same translational kinetic energy as in the first case, and $v_f$ in the second case is what causes the difference, being much smaller, so in this case the rod would have much more kinetic energy in the end when having the particle impact at one of the ends than having it impact at the very center.


3. There's no translation, which would agree with my hypothesis, and $v_f$ would be bigger in the second case than in the first.

Answer

In both cases if you apply the conservation of linear momentum then the relationship between the velocity of the centre of mass of the rod $V$ and the final velocity of the particle $(-) v_{\rm f}$ is $V = \frac mM (v+v_{\rm f})$.

However the final velocities of the particle $v_{\rm f1}$ and $v_{\rm f2}$ and the rod $V_1$ and $V_2$ in the two cases are not necessarily the same.

As there are no external torques acting on the rod and the particle, the conservation of angular momentum must be satisfied which when applied about the centre of mass in the second case gives $mv\frac L2= mv_{\rm f2} \frac L2 + I_{\rm c}\omega$ where $I_{\rm c}$ is the moment of inertia of the rod about its centre of mass and $\omega$ is its angular speed.

The final kinetic energy for the second case is $\frac 12 MV_2^2+ \frac 12 m v_{f2}^2 + \frac 12 I_{\rm c}\omega^2$ whereas in the first case it was $\frac 12 MV_1^2+ \frac 12 m v_{f1}^2$.

If you want kinetic energy to be conserved during the collision then the final translational velocities of the rod and the particle are less in the second case as compared with the first case.

electromagnetism - When to use method of images in Electrostatics?

I am a bit confused about when to use the method of images in E&M? For example, in Griffith's Electrodynamics Example 3.2, the problem reads:

A point charge $q$ is situated a distance $a$ from the center of a grounded conducting sphere of radius $R$. Find the potential outside the sphere."

But in Example 2.7, which says,

Find the potential of a uniformly charged spherical shell of radius $R$

or Problem 3.1, which says,

Find the average potential over a spherical surface of radius $R$ due to a point $q$ located inside

the method of image is not used to solve the problem. In Example 2.7, this equation for the potential is used instead: $$ V\left(\mathbf r\right)=\frac{1}{4\pi\epsilon_0}\int\frac{\sigma}{r}da' $$

Is it because the object is "conducting" that we use method of images? And if so, why? Could someone help me clear up my misunderstanding?

Answer

For problem solving, the method of images comes in handy in the following cases: conducting sphere, conducting cylinder, conducting ellipsoid, and conducting plane. Another example is two regions of dielectrics, with different $\epsilon$ (permittivity). And that is it. This general rule should save you some time.

So, the first problem you mentioned can be solved with images, the second and third cannot (there is no conductor).

A caveat about using the method of images, which I have seen frequently confuse students is as follows. In the method of images, there is always two distinct regions of space. For example, inside and outside the conductor, left and right of the conducting plane, or in dielectric region 1 and dielectric region 2. The imperative point is that, when you want to find the potential in one region, the image charges are not allowed to be in that region; they should be situated in the other region. For example, if there is a conducting sphere, not grounded, and there is a point charge inside of it (not at the center), and you are asked to find the potential for every point inside the sphere, then you should situate the image charge outside the sphere.

I suggest not confusing yourself too much with the method of images, and remembering that only the above-mentioned geometries have neat closed-form image solutions.

particle physics - What is the relationship between string net theory and string / M-theory?

I've just learned from this one of Prof. Wen's answers that there exists a theory called string net theory. Since I've never heard about this before it picks my curiosity, so I`d like to ask some questions:

How is string net theory related to the "usual" string /-M-theory framework? I mean if there is a relationship between them ...

Are the strings in string net theory and in string / M-theory the same?

What are the differences in the goals one wants to achieve or phenomena in nature one can describe with those two theories?

Answer

Goals one wants to achieve with those two theories are similar.

We know that superstring theory is a potential theory of everything. One may want to ask what is the difference between the string-net-liquid approach and the superstring approach? Our understanding of the superstring theory has been evolving. According to an early understanding of the superstring theory, all the elementary particles correspond to small segments of superstrings. Different vibration modes of a small superstring result in different types of elementary particles. This point of view is very different from that of the string-net liquid. According to the string-net picture, everything comes from simple qubits that form the space. No qubits no space. The "1" qubits form string-nets. The strings can be as long as the size of universe, which fill the whole space. Light (photons) correspond to the collective motion of the large string-nets and an electron corresponds to a single end of string. (See a picture of string-net "vaccum". See also a talk) A modern understanding of the superstring theory is still under development. According to Witten, one of the most important questions in superstring theory is to understand what is superstring. So at this time, it is impossible to compare the modern understanding of the superstring theory with the string-net theory. In particular it not clear if the superstring theory can be viewed as a local bosonic system (ie a qubit system). The string-net theory is fundamentally a local bosonic system (ie a qubit system).

So, if superstring theory is a qubit model (or a quantum spin model in condensed matter physics), then superstring theory and the string-net theory is the same, since the string-net theory is a qubit model (or a quantum spin model in condensed matter physics).

spacetime - Is the Planck length the smallest length that exists in the universe or is it the smallest length that can be observed?

I have heard both that Planck length is the smallest length that there is in the universe (whatever this means) and that it is the smallest thing that can be observed because if we wanted to observe something smaller, it would require so much energy that would create a black hole (or our physics break down). So what is it, if there is a difference at all.

Answer

Short answer: nobody knows, but the Planck length is more numerology than physics at this point

Long answer: Suppose you are a theoretical physicist. Your work doesn't involve units, just math--you never use the fact that $c = 3 \times 10^8 m/s$, but you probably have $c$ pop up in a few different places. Since you never work with actual physical measurements, you decide to work in units with $c = 1$, and then you figure when you get to the end of the equations you'll multiply by/divide by $c$ until you get the right units. So you're doing relativity, you write $E = m$, and when you find that the speed of an object is .5 you realize it must be $.5 c$, etc. You realize that $c$ is in some sense a "natural scale" for lengths, times, speeds, etc. Fast forward, and you start noticing there are a few constants like this that give natural scales for the universe. For instance, $\hbar$ tends to characterize when quantum effects start mattering--often people say that the classical limit is the limit where $\hbar \to 0$, although it can be more subtle than that.

So, anyway, you start figuring out how to construct fundamental units this way. The speed of light gives a speed scale, but how can you get a length scale? Turns out you need to squash it together with a few other fundamental constants, and you get: $$ \ell_p = \sqrt{ \frac{\hbar G}{c^3}} $$ I encourage you to work it out; it has units of length. So that's cool! Maybe it means something important? It's REALLY small, after all--$\approx 10^{-35} m$. Maybe it's the smallest thing there is!

But let's calm down a second. What if I did this for mass, to find the "Planck mass"? I get: $$ m_p = \sqrt{\frac{\hbar c}{G}} \approx 21 \mu g $$

Ok, well, micrograms ain't huge, but to a particle physicist they're enormous. But this is hardly any sort of fundamental limit to anything. It isn't the world's smallest mass. Wikipedia claims that if a charged object had a mass this large, it would collapse--but charged point particles don't have even close to this mass, so that's kind of irrelevant.

It's not that these things are pointless--they do make math easier in a lot of cases, and they tell you how to work in these arbitrary theorists' units. But right now, there isn't a good reason in experiment or in most modern theory to believe that it means very much besides providing a scale.

electricity - Electrical Resistance

1)I have read that the cause for resistance is charges colliding with atoms. But what exactly are the charges colliding with. Surely charges cannot be straight up hitting the nucleus/electrons.

2) Also if the cause of resistance is purely based on the random( as it seems to me) incident of collision. How is it that we have well defined formulas like V=IR and R=rhol/A. And most importantly Heat energy=VIt. How can we be so sure that all of the loss in PE of charges between 2 points gets converted to heat. How can we say that there will be no gain in KE at all? Thanks

Answer

1) [...] But what exactly are the charges colliding with. Surely charges cannot be straight up hitting the nucleus/electrons.

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  It is easy and intuitive to think of both charges and atoms as spherical balls. The charge-balls collide with the atom-balls.

  
  


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  If you wish to get closer to reality but still keep an intuitive visualization, then think of a charge (electrons e.g.) as a tiny "spot" in the middle of an electric field "bubble". Likewise with atoms (their electron shells give them a similar negative electric field "bubble"). When a charge gets close to an atom, their fields meet and interact. The charge field might therefore be repelled / pushed. In this way the "collision" is nothing more than a field interaction. The field is like a "pillow" and the collision is like the two pillows bumping together, prevents the charge from ever coming into contact with the atom.

  
  


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  If you want an even closer-to-reality picture, then the intuitive visualization becomes tough. The electron size is so tiny that quantum probability becomes relevant. Now we can't think of an electron as a spherical ball, but rather as a "probability cloud". This "cloud" carries the charge - this "cloud" is the electron. The cloud is the electron's "soul" and the electron has no "body". The electron is not at a particular point inside the cloud, but there is a certain probability for it to be at each point constituting this cloud. Same goes for each particle of the atom - the electrons in so-called shells should now rather be thought of as several "merged" or "squeezed" "probability clouds", with various shapes depending out the configuration. The attraction/repulsion between this atom and the incoming charge is still the factor that causes deflection of the charge.

  
  

All these are more and more detailed models or ways to think. Common for them all is that temperature is defined as the Ångström-scale motion and vibration of atoms.

Anytime a charge interacts with an atom according to any of the above models, that atom is pushed/pulled a bit and gains some kinetic energy. It isn't ripped free from the crystal/molecular structure, so it just vibrates in its place. (Were it ripped free, then that corresponds to the material breaking apart/melting.)

This increased vibrational motion corresponds to a temperature increase, eventually leading to a heat loss to the surroundings. Simultaneously, the charge looses some kinetic energy corresponding to a loss in potential (stored or carries electrical energy).

So, no, there is very rarely any electrons colliding with a nucleus.

2) Also if the cause of resistance is purely based on the random( as it seems to me) incident of collision. How is it that we have well defined formulas like V=IR and R=rhol/A. And most importantly Heat energy=VIt. How can we be so sure that all of the loss in PE of charges between 2 points gets converted to heat. How can we say that there will be no gain in KE at all? Thanks

Yes, it is random. But random with a "tendency".

We can think of the collisions as being a mix straight-on or sliding off or impacts at different angles, all causing more or less transfer of energy. But there is an average. If you look at 10 electrons, you can find the average collision. If you look at 100 electrons, the average collision is more accurate. If you look at billions and billions of electrons, you are pretty sure that the average is quite accurate. A good representation of the overall collisions that happen.

This is why we can establish relationships such as Ohm's law. The individual electron may not behave strictly according to this law, but the overall average electron will. And therefore the resulting current, consisting of billions and billions of electrons, will.

Accessible resources for learning Bohmian mechanics? (Undergrad)

I am an undergraudate physics and math major. For context, I've taken senior level quantum 1 as well as real and complex analysis. I'd be really interested in focusing on understanding the mathematical formulation of Bohemian mechanics for a research project in my quantum mechanics 2 class but I can't seem to find any good introductory resources on the topic.

Any help would be appreciated!

renormalization - Would Color Confinement apply in higher dimensions?

As I understand it color confinement comes from the fact that as the distance between two color charges increases the color potential energy increases, instead of decreasing, and the energy needed to pull two quarks apart is the same as the energy needed to create two new quarks. The way the color potential energy between two color charges is related to the fact that gluons themselves have color charge. For the electric potential energy between two electric charges the relationship between distance and potential energy depends on the number of dimensions, and for the spacetime curvature around a massive body the relationship between the distance from the massive body and the spacetime curvature depends on the number of dimensions, but I'm not sure if the relationship between distance and color potential energy depends on the number of dimensions?

Would color confinement apply in $n+1$ dimensions, in which $n>3$, or could color charges be free particles in higher dimensions?

quantum field theory - Solving the Klein-Gordon equation via Fourier transform

I have been writing a personal set of notes on QFT and I'm currently writing up a section on solving the Klein-Gordon (K-G) equation. I many texts that I've read, the author starts by expressing the ansatz for solution $\phi$ in terms of a spatial Fourier transform for fixed times $t$, $$\phi(\mathbf{x},t)=\int_{-\infty}^{\infty}\frac{d^{3}k}{(2\pi)^{3}}\widetilde{\phi}(\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{x}}$$ they then proceed to solve the K-G equation by noting that $$(\Box + m^{2})\phi = (\partial_{t}^{2}-\nabla^{2}+m^{2})\phi=\int_{-\infty}^{\infty}\frac{d^{3}k}{(2\pi)^{3}}e^{i\mathbf{k}\cdot\mathbf{x}}\left[\partial_{t}^{2}+k^{2}+m^{2}\right]\widetilde{\phi}(\mathbf{k},t)=0$$ But this confuses me, as aren't we considering $\phi$ at fixed instants in time $t$? Is the above just formal manipulation, considering the time derivative $\partial_{t}^{2}\widetilde{\phi}$ to be evaluated at a fixed instant $t$ (i.e.$\partial_{t}^{2}\widetilde{\phi}\big\vert_{t}$)?

The steps after this I understand, as $e^{i\mathbf{k}\cdot\mathbf{x}}$ is non-zero we require that in order for the integral to be zero $\forall\;\mathbf{k}$, that $$\left[\partial_{t}^{2}+k^{2}+m^{2}\right]\widetilde{\phi}(\mathbf{k},t)=0$$ and this gives us the general solution $$\widetilde{\phi}(\mathbf{k},t)=a(\mathbf{k})e^{-iEt}+b(\mathbf{k})e^{iEt}$$ where $E\equiv k_{0}=\sqrt{k^{2}+m^{2}}$.

It is really that first part that is causing me problems.

Answer

If at every time $t$, $\phi(\mathbf{x},t)$ is a nice enough function that it has a Fourier transform, then $$\phi(\mathbf{x},t)=\int_{-\infty}^{\infty}\frac{d^{3}k}{(2\pi)^{3}}\widetilde{\phi}(\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{x}},$$ where $\widetilde{\phi}(\mathbf{k},t)$ is just the Fourier coefficient at that time $t.$

But now you ask that the whole function $\phi(\mathbf{x},t)$ (at every time) be a solution to the Klein-Gordon equation. Which means the function at different times need to have the right derivatives. If at every time there is a Fourier transform, then there are Fourier coefficients at every time. So if the wave evolves a certain way in time, then the Fourier coefficients need have values at different times need to be just right to let he wave have the right temporal derivatives.

OK, so the Klein-Gordon equation is second order so we can find the initial function $\phi(\mathbf{x},t=0)$ and its Fourier transform, call it $\theta(\mathbf{k})$ and we can take the initial temporal derivative $\partial_t \phi(\mathbf{x},t)\big\vert_{t=0}$ and it's Fourier transform, call it $\omega(\mathbf{k}).$ Then we know the initial Fourier coefficients and we know their derivatives and the second derivative is enforced to make it satisfy the Klein-Gordon equation, so

$\widetilde{\phi}(\mathbf{k},t)=\theta(\mathbf{k})\cos(E_kt) + \frac{1}{E_k}\omega(\mathbf{k})\sin(E_kt).$

Why? Because it has the right initial values and the right initial temporal derivative and it satisfies the Klein-Gordon equation when $E_k=\sqrt{m^2+k^2}.$

So at each time we have Fourier coefficients of our wave. They are designed so that the functions satisfies the Klein-Gordon equation when the Fourier coefficients evolve in time according to a second order equation.

I may have misunderstood your question. But the idea is that from the initial wave and the initial temporal derivative of the wave you get enough initial conditions to know the initial Fourier coefficients and the initial time rate of change of the Fourier coefficients which is all the freedom you have. The rest is determined that the Fourier coefficients have to evolve temporally a certain way so the wave evolves a certain way temporally.

edit to respond to comments

At each time you get a Fourier transform. And you then ask yourself how those Fourier coefficients depend (in time) on each other. In order for the wave to evolve in time by a second order equation, the Fourier transform needs to evolve in time by a second order equation. But the transform gets to do a pointwise evolution that is simpler and that's why we do it.

When you take the Fourier transform at each time, you get a Fourier transform at each time, so you get a $\widetilde{\phi}(\mathbf{k},t)$ that is itself a function of time. So it has a partial derivative $\partial_t \widetilde{\phi}(\mathbf{k},t)\neq 0.$

edit to further respond to comments

Let's say we want to solve $$\left(\partial_t^2-\nabla^2+m^2\right)\phi=.0$$ We start by noting that $\left(A_ke^{i\mathbf{k}\cdot\mathbf{x}}+B_ke^{-i\mathbf{k}\cdot\mathbf{x}}\right)\cos(t\sqrt{m^2+k^2})$ and $\frac{\left(C_ke^{i\mathbf{k}\cdot\mathbf{x}}+D_ke^{-i\mathbf{k}\cdot\mathbf{x}}\right)}{\sqrt{m^2+k^2}}\sin(t\sqrt{m^2+k^2})$ are both solutions for any $\mathbf{k},$ any $A_k, B_k$ and any $C_k, D_k$.

Great.

And the first solution has zero time rate of change at $t=0$ and the second has zero value at $t=0.$ So if our initial wave was $\phi(\mathbf{x},t=0)=3\cos(3x)$ and had an initial time rate of change equal to $\partial_t\phi(\mathbf{x},t)\big\vert_{t=0}=4\cos(2y)$ then we know exactly what the solution to $\left(\partial_t^2-\nabla^2+m^2\right)\phi=0$ is: $$\phi(\mathbf{x},t)=3\cos(3x)\cos(t\sqrt{m^2+3^2})+\frac{4}{\sqrt{m^2+2^2}}\cos(2y)\sin(t\sqrt{m^2+2^2}).$$

And any finite linear combination of $e^{\pm i\mathbf k \cdot \mathbf x}$ for the initial value $\phi(\mathbf{x},t=0)$ is as easily manageable by having a finite linear combination of solutions like $e^{\pm i\mathbf{k}\cdot\mathbf{x}}\cos(t\sqrt{m^2+k^2}).$ And similarly if the initial time rate of change $\partial_t\phi(\mathbf{x},t)\big\vert_{t=0}$ is a finite linear combination of $e^{\pm i\mathbf k \cdot \mathbf x}$ then we add terms that are a finite linear combination of $\frac{e^{\pm i\mathbf{k}\cdot\mathbf{x}}}{\sqrt{m^2+k^2}}\sin(t\sqrt{m^2+k^2}).$

All we are doing is taking solutions and adding them up in combinations that give us the right initial values and right initial time rate of changes. And it is super easy if the initial conditions happen to be a finite linear combination of sines and cosines.

But wait. What if instead of being a finite linear combination of terms like $e^{\pm i\mathbf k \cdot \mathbf x}$ the initial condition was just a function that has a Fourier Transform? Then you can try the same thing. Write your initial value $\phi(\mathbf{x},t=0)$ and your initial time rate of change $\partial_t\phi(\mathbf{x},t)\big\vert_{t=0}$ as inverse Fourier transforms of sines and cosine spatially. Then replace a spatial $\sin(\mathbf k \cdot \mathbf x)$ with the function $\sin(\mathbf k \cdot \mathbf x)\cos(t\sqrt{m^2+k^2})$ and replace the spatial function $\cos(\mathbf k \cdot \mathbf x)$ with the function $\cos(\mathbf k \cdot \mathbf x)\cos(t\sqrt{m^2+k^2}).$ Why? Because each of those satisfies the Klein-Gordon equation. And so for $t=0$ the inverse Fourier transform of those will be the initial value of the wave $\phi(\mathbf{x},t=0).$ So you are taking a $t=0$ spatial Fourier transform of $\phi(\mathbf{x},t=0)$ then replacing every spatial Fourier component $e^{\pm i \mathbf k \cdot \mathbf x}$ with $e^{\pm i \mathbf k \cdot \mathbf x}\cos(t\sqrt{m^2+k^2}).$ This solves the Klein-Gordon equation, has the right initial values and has $\partial_t\phi(\mathbf{x},t)\big\vert_{t=0}=0.$

Next, take your initial time rate of change $\partial_t\phi(\mathbf{x},t)\big\vert_{t=0}$ as an inverse Fourier transforms of sines and cosine spatially. Then replace a spatial $\sin(\mathbf k \cdot \mathbf x)$ with the function $\frac{\sin(\mathbf k \cdot \mathbf x)}{\sqrt{m^2+k^2}}\sin(t\sqrt{m^2+k^2})$ and replace the spatial function $\cos(\mathbf k \cdot \mathbf x)$ with the function $\frac{\cos(\mathbf k \cdot \mathbf x)}{\sqrt{m^2+k^2}}\sin(t\sqrt{m^2+k^2}).$ Why? Because each of those satisfies the Klein-Gordon equation. And yet for $t=0$ if you take the inverse spatial Fourier transform you get a function that has value zero at $t=0$ and has an initial time rate of change that equals $\partial_t\phi(\mathbf{x},t)\big\vert_{t=0}$ of the wave $\phi(\mathbf{x},t=0).$

Why have two solutions? Because this second one also solves the Klein-Gordon equation and has an initial value of zero but has an initial time rate of change that equals $\partial_t\phi(\mathbf{x},t)\big\vert_{t=0}.$ And the first one solves the Klein-Gordon equation, has an initial value that equals $\phi(\mathbf{x},t=0)$ and has a zero initial time rate of change.

So if you add those two solutions together you get a function that (1) solves the Klein-Gordon equation (2) has the right initial value and (3) has the right initial time rate of change. That's what you wanted all along.

If you understand that when the initial wave was $\phi(\mathbf{x},t=0)=3\cos(3x)$ and had an initial time rate of change equal to $\partial_t\phi(\mathbf{x},t)\big\vert_{t=0}=4\cos(2y)$ then he solution was $\left(\partial_t^2-\nabla^2+m^2\right)\phi=0$ is: $$\phi(\mathbf{x},t)=3\cos(3x)\cos(t\sqrt{m^2+3^2})+\frac{4}{\sqrt{m^2+2^2}}\cos(2y)\sin(t\sqrt{m^2+2^2}).$$ If you understand that, then everything else is the same idea applied to finite and "infinite" linear combinations of $e^{\pm i \mathbf k \cdot \mathbf x}$ for the initial conditions.

general relativity - Is "now" or "the present moment" properly defined in GR?

My question is about the extent to which "now" is defined in GR.

In Minkowski spacetime, it's possible to define a "now" for an inertial observer by finding a spacelike 3-plane such that, in the observer's frame, all 4-vectors in the 3-plane have zero time component (or something like that - apologies, my geometry is a bit rusty - anyway, a 3-plane in which all vectors are orthogonal to the tangent vector of the observer's world line). This plane can be defined globally, so that my "now" is a slice through the whole of Minkowski space.

My question is whether it's possible to define such a thing in general relativity. So, for instance, can I meaningfully speak about what the Andromeda Galaxy is doing "right now"? Or is the "present moment" something that can only be locally defined? I remember seeing something in a Roger Penrose book about this, but I can't find the reference (if anyone knows it, please let me know!)

cosmology - Why did dark matter interact with the Standard model particles in the early Universe but not now?

In the thermal dark matter (DM) scenario, it is the interaction of the DM particles with the Standard Model (SM) particles that kept the DM particles in kinetic equilibrium with the thermal plasma.

Why is it that today the DM particles do not have appreciable non-gravitational interaction with the SM particles like they had in the early Universe? Do I misunderstand something?

Answer

Dark matter have clearly negligible cross-section with baryonic matter at current low temperatures, otherwise we would have detected it by now in experimental searches. But that cross-section can be higher at higher energies. In the early universe, there was enough pressure and temperature for this to happen often enough

wavefunction - Shape of Hydrogen atom in excited state with nonzeros angular momentum

Solution of Schrödinger's equation for an Hydrogen atom is well known: $$\Psi_{n,l,m} (r, \theta, \phi) = N e^{\frac{-r}{n r_1}} R_n^l (r) P_l^m(cos \theta) e^{im\phi} \,.$$

If we interested with state with nonzeros angular momentum (say, $l=1$), the wave functions are: \begin{align} \Psi_{2,1,0} &= \frac{1}{4\sqrt{2\pi}a_0^{3/2}}\frac{r}{a_0}e^{-\frac{r}{2a_0}}\cos\theta\\ \Psi_{2,1,\pm1} &= \frac{1}{8\sqrt{\pi}a_0^{3/2}}\frac{r}{a_0}e^{-\frac{r}{2a_0}}\sin\theta e^{\pm i\phi} \end{align}

This wave functions descrie the state with certain projection of angular momentum to some Cartesian axes, i.e. it is an eigenfunctions functions of an operator $\hat L_z$. So, if we switch on the magnetic (or electric) field, the atom will be in some state of three possible, and the shape of its cloud will have some form that will not be spherically symmetric (for example, look here here).

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But if there is no such direction, then what will be the shape of the hydrogen atom?

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In chemistry, not the wave functions themselves are used, but their linear combinations, which are real:

\begin{align} p_z = \Psi_{2,1,0} &= \frac{1}{4\sqrt{2\pi}a_0^{3/2}}\frac{r}{a_0}e^{-\frac{r}{2a_0}}\cos\theta\\ p_x = \frac{1}{\sqrt{2}} (\Psi_{2,1, + 1} + \Psi_{2,1, - 1} ) &= \frac{1}{8\sqrt{2\pi}a_0^{3/2}}\frac{r}{a_0}e^{-\frac{r}{2a_0}}\sin\theta \cos\phi\\ p_y = \frac{i}{\sqrt{2}} (\Psi_{2,1, + 1} - \Psi_{2,1, - 1} ) &= \frac{1}{8\sqrt{2\pi}a_0^{3/2}}\frac{r}{a_0}e^{-\frac{r}{2a_0}}\sin\theta \sin\phi\\ \end{align}

These functions are no longer the eigenfunctions of the operator $\hat L_z$ and describe states with an indefinite projection of the angular momentum. But what exactly do they describe? If there is no dedicated direction, then all three functions $p_x, p_y, p_z$ must be equiprobable, that is, the state of the electron must be a linear combination of these three functions, and the shape of the cloud must therefore be spherically symmetric. How, then, can the directional valence be explained to form a chemical bond?

Answer

Just to be clear, even without a magnetic field, the eigenstates of the hydrogen atom are simply not all spherically symmetric. The space of eigenstates of a given energy is spherically symmetric. Spherical symmetry means that if we transform an eigenstate by rotating it, it will give another eigenstate, not that each eigenstate is spherically symmetric.

Now, if we take a hydrogen atom at a particular energy, and then let it sit in a spherically symmetric environment with which it somehow interacts weakly for a long time, it should end up in a spherically symmetric superposition of eigenstates of the energy we started with, i.e. it should end up looking like a sphere. If we then measure the angular momentum we are equally likely to find it pointing in any direction.

The wavefunctions you mentioned as used in chemistry -- let's focus on the two ''new'' ones $p_x$ and $p_y$ -- describe energy eigenstates with total angular momentum $1$ and moreover angular momentum $0$ in the $x$ and $y$ direction respectively (you missed a factor of $i$ in the definition of $p_y$ btw). The simplest way to see this is to consider the space of states of total angular momentum $1$ as a three dimensional vector space (this makes sense as long as you only look at the angular momentum operators). In this vector space, the angular momentum operators act like matrices of the form $$L_x \sim \begin{pmatrix} 0 & 0 & 0 \\0 & 0 & i \\ 0 & -i & 0 \end{pmatrix},\quad L_y \sim \begin{pmatrix} 0 & 0 & -i \\0 & 0 & 0 \\ i & 0 & 0 \end{pmatrix},\quad L_z \sim \begin{pmatrix} 0 & i & 0 \\-i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ and the $\Psi_{2,1,x}$ are represented by $$ \Psi_{2,1,0} \sim \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}, \quad \Psi_{2,1,1} \sim \begin{pmatrix} 1 \\ -i \\ 0 \end{pmatrix}, \quad \Psi_{2,1,-1} \sim \begin{pmatrix} 1 \\ i \\ 0 \end{pmatrix}.$$ Now if you take the linear combinations of $\Psi_{2,1,\pm1}$ that you write, you can easily see that they are in the null space of $L_x$ and $L_y$ respectively.

As soon as you put more than one hydrogen atom in your universe, there can be a preferred direction (as there definitely is in molecules), and this will put your hydrogen atom in a particular not-spherically-symmetric eigenstate, which allows for the typical picture of a valence bond.

particle physics - In general what will holding an anti-hydrogen atom for more than a 1/10th of second allow scientists to discover?

In general what will holding an anti-hydrogen atom for more than a 1/10th of second allow scientists to discover? Specifically, given that they can hold one for <1/10th of a second, what would they discover that have not previously been able to determine. Or if not known, what have they been able to discover to date?

Answer

The ultimate goal is to be able to do precision spectroscopy of antihydrogen, to make sure that the energy states are the same as in ordinary matter. If there are differences between the energy levels of ordinary hydrogen and antihydrogen, that would violate "CP" symmetry, which says that if you change the sign of all the charges in some system, and invert the parity, every interaction should be the same.

We know that CP violation occurs in nature-- it's been observed in kaon decay, among other things-- and it's related to the observed asymmetry between matter and antimatter in the universe. The known sources of CP violation are not enough to explain the matter-antimatter imbalance in the universe, though, so there have to be other forms of it out there that have yet to be discovered.

From what we known about the interactions of matter and antimatter, any differences in the antihydrogen states would have to be very small, but laser spectroscopy can be used to do measurements of astonishing precision-- there are single-ion atomic clocks that are good to a few parts in $10^{18}$. Having the target atoms trapped for only a tenth of a second complicates matters, but a group at Argonne National Lab did spectroscopic measurements of the charge radius of unstable helium isotopes that don't last very much longer than that, so it's a good step toward the goal of doing spectroscopy.

Another thing that people talk about testing with anti-atoms is the behavior of gravity. Again, you need to have trapped neutral atoms for this, because electrostatic forces are thirty-some orders of magnitude stronger than gravity. That will also require extreme precision, and many more atoms than have been trapped to date, but the recent experiments are a good start, and the remaining issues are mostly technical, not fundamental.

thermodynamics - Connection between entropy and energy

An isolated system $A$ has entropy $S_a>0$.

Next, the isolation of $A$ is temporarily violated, and it has entropy reduced $$S_b ~=~ S_a - S,\space\space\space S\leq S_a.$$

Is it true to say: the process of lowering entropy of a system requires work and energy?

I am not sure if energy of the system must be changed when entropy is reduced. However, energy certainly is required – changing entropy is work and uses energy?

quantum mechanics - Prove that angular momentum commute with the hamiltonian of a central force

I'm trying to prove that $$[\hat{L}_i,\hat{H}]=0$$ for $\hat{H}$ the hamiltonian of a central force $$\hat{H}=\frac{\hat{p}^2}{2m}-\frac{\alpha}{r}.$$

I'm getting this:

$$[\hat{L}_i,\hat{H}]=[\hat{L}_i,\frac{\hat{p}^2}{2m}-\frac{\alpha}{r}] = [\hat{L}_i , \frac{\hat{p}^2}{2m}] - [\hat{L}_i , \frac{\alpha}{r}].$$

Already prove that the first one is zero (it's a known result too), but the second one, I dont know what to do with it, there is no way for me. I was doing something like this:

$$[\hat{L}_i , \frac{\alpha}{r}] = [\epsilon_{ijk} r_j p_k , \frac{\alpha}{r}] = \epsilon_{ijk}r_j [p_k ,\frac{\alpha}{r}] + \epsilon_{ijk} [r_j , \frac{\alpha}{r}]p_k.$$

Obviously, the last term is zero, but the other one, I do something that has no sense to be zero. Have any idea for this? I'm I doing wrong? All of this is to prove that Laplace-Runge-Lenz operator commute with hamiltonian of a central force.

nuclear physics - how to measure activation energy of fission?

i heard that fission activation energy of (235)U is less than of neutron separation energy of (236)U so this must the reason that (235)U is fission able

$$E_s+(236)U\to (235)U+n$$ in this interaction $E_s$ is neutron separation energy ( energy required to separate neutron )

1. 
   

   how to measure activation energy of fission?

   
   


2. 
   
   

   is there any relation or equation?

   
   

Answer

Here is a tutorial on fission .

The concept of "activation energy" is not utilized in the study of fission. There are energy levels that can be modeled with nuclear potential well models.Mass is turned into energy in fission.

What happens is that there is a high cross section for U235 to capture a thermal neutron and become a U236 in an excited state. There are more than one energy levels where this could happen.

The configuration is unstable enough that instead of falling down with gamma rays to a ground state of U236, which is fairly stable, the system (U235+neutron) breaks up into Kr92 Ba141 and 3 neutrons which can with suitable engineering be thermalized to create a chain reaction in a fission reactor.

thermodynamics - Is there a phase transition between a gas and plasma?

Does a phase transition occur as a gas is heated to create a plasma? If so, is this a first or second order phase transition?

Also, does the presence of a phase transition depend on the pressure or composition? It seems to me that in the dilute limit (i.e. low pressures), no phase transition should occur because the fraction of atoms that are ionized will follow a Boltzmann distribution, which is a smooth function of temperature. However, the presence of phase transitions in Debye-Hückel theory seems to suggest that a gas-plasma transition could occur at higher pressures.

newtonian mechanics - Internal potential energy and relative distance of the particle

Today, I read a line in Goldstein Classical mechanics and got confused about one line.

To satisfy the strong law of action and reaction, $V_{ij}$ can be a function only of the distance between the particles: $V_{ij} = V_{ij}(|{\bf r}_i-{\bf r}_j|)$.

What confuses me is that I can't see the logic between these two statements. Obviously, I understand strong law of action and reaction and Internal energy. But why the strong law of action and reaction leads to internal energy only depending on relative distances?

I prefer to receive mathematical proof (not thorough, but provide a direction so that I can know where I'm going); yet, intuitive illustration is also welcome.

Answer

The quote is taken from just above eq. (1.32) in Ref. 1:

[...] If the internal forces are also conservative, then the mutual forces between the $i$th and $j$th particles, ${\bf F}_{ij}$ and ${\bf F}_{ji}$, can be obtained from a potential function $V_{ij}$. To satisfy the strong law of action and reaction, $V_{ij}$ can be a function only of the distance between the particles: $$V_{ij} ~=~ V_{ij}(|{\bf r}_i-{\bf r}_j|). \tag{1.32}$$

The structure of internal forces among $N$ point particles can be quite rich in general, see e.g. this Phys.SE post. However the first sentence in the quote makes it clear that Ref. 1 is additionally assuming:

1. 
   

   that the internal forces on one particle is a sum of forces from the other particles. Thus it is enough to study the internal force ${\bf F}_{ij}$ from the $i$th particle on the $j$th particle.

   
   


2. 
   
   

   that ${\bf F}_{ij}({\bf r}_i,{\bf r}_j)$ only depends on the two positions ${\bf r}_i$ and ${\bf r}_j$ of the $i$th and $j$th particles, respectively.

   
   


3. 
   

   that ${\bf F}_{ij}({\bf r}_i,{\bf r}_j)$ is a conservative force, meaning that there exists a potential $V_{ij} = V_{ij}({\bf r}_i,{\bf r}_j)$ such that $$ {\bf F}_{ij}~=~-{\bf \nabla}_{j} V_{ij}. $$

   
   


4. 
   

   that the potential $$V_{ij}({\bf r}_i,{\bf r}_j)~=~V_{ji}({\bf r}_j,{\bf r}_i)$$ is symmetric.

   
   

The weak form of Newton's 3rd law then implies that $$ {\bf 0}~=-{\bf F}_{ij}-{\bf F}_{ji}~=~ {\bf \nabla}_{j} V_{ij}+{\bf \nabla}_{i} V_{ji}~=~ ({\bf \nabla}_{i}+ {\bf \nabla}_{j}) V_{ij},$$ which in turn implies that the potential $V_{ij} = V_{ij}({\bf r}_{ij})$ only depends on the difference ${\bf r}_{ij}:={\bf r}_j-{\bf r}_i$ in positions.

Finally the strong form of Newton's 3rd law implies that $${\bf r}_{ij}~\parallel~ {\bf F}_{ij}~=~-{\bf \nabla}_{j} V_{ij}({\bf r}_{ij}),$$ which in turn implies that the potential $V_{ij} = V_{ij}(|{\bf r}_{ij}|)$ only depends on the distance $|{\bf r}_{ij}|$. [The last point follows from the fact that a scalar function $V:\mathbb{R}^3\to \mathbb{R}$, with the property that the gradient $${\bf \nabla} V({\bf r})~\parallel~ {\bf r}$$ is parallel to the position vector ${\bf r}$, can only depend on the length $|{\bf r}|$. Can you see why? Hint: Decompose the gradient in spherical coordinates.]

References:

1. Herbert Goldstein, Classical Mechanics, Chapter 1.

electromagnetism - How electromagnetic fields travel through vacuum?

How electromagnetic fields travel through vacuum is it right to say that they propagate or travel? do they alternately form each other by charged particles?

Answer

from a, b and c:

An electromagnetic field is propagating by changing the field that's generated by electrically charged particles pass through the air and the space that is devoid of particles of space.

An electromagnetic wave propagates, not an electromagnetic field. An EM wave is a propagating disturbance in the existing electromagnetic field.

Consider the field of an isolated charged point particle - the field extends over all space.

Should the particle abruptly accelerate for some short time, the disturbance in the electromagnetic field of the particle will propagate 'to infinity' at the speed $c$.

From the classic textbook "Gravitation":

For an animated visualization, try the Java applet here.

newtonian mechanics - What happens when a force is not applied on the centre of the mass?

The answer to this question

force applied not on the center of mass

states that for a force applied not directly on the center of mass will have the same linear acceleration as one applied on the center of mass. If this is true, then consider the following:

A disc with center of mass on the dot. Firstly, does that mean that force a will result in purely downward motion, as if a could be translated towards the center of mass (of course with some torque as well), or will it be at an angle (if so, what angle?). Secondly, does force a balance with force c (no net linear force) or does force a balance with force d (for linear) or do both pairs balance? I'm supposing that a and c will generate no torque and a and d will, but I am only really interested in linear motion for now.

Answer

Force a alone will produce both downward motion (as if the force were applied at the center of mass) and rotational motion, because it causes torque. But the linear acceleration is the same, regardeless of where you apply the force. For traslation, both c and d balance a, so the disk will not translate in either case, but it will rotate with d because there will be net torque.

cosmology - Why isn't dark matter just ordinary matter?

There's more gravitational force in our galaxy (and others) than can be explained by counting stars made of ordinary matter. So why not lots of dark planetary systems (i.e., without stars) made of ordinary matter? Why must we assume some undiscovered and unexplained form of matter?

Answer

There is a very precise reason why dark planets made of 'ordinary matter' (baryons - particles made up of 3 quarks) cannot be the dark matter. It turns out that the amount of baryons can be measured in two different ways in cosmology:

• By measuring present-day abundances of some light elements (esp deuterium) which are very sensitive to the baryon amount.


• By measuring the distribution of the hot and cold spots in the Cosmic Microwave background (CMB), radiation left over from the early universe that we observe today.

These two methods agree spectacularly, and both indicate that baryons are 5% of the total stuff (energy/matter) in the universe. Meanwhile, various measures of gravitational clustering (gravitational lensing, rotation of stars around galaxies, etc etc) all indicate that total matter comprises 25% of the total. (The remaining 75% is in the infamous dark energy which is irrelevant for this particular question).

Since 5% is much less than 25%, and since the errors on both of these measurements are rather small, we infer that most of the matter, about 4/5 ths (that is, 20% out of 25%) is 'dark' and NOT made up of baryons.

atoms - Why the properties of a substance like color changes when we go into the nano scale?

I have heard that the physical as well as chemical properties of a substance changes when we go to the nano scale. For example, Gold is considered an inert material in that it doesn’t corrode or tarnish. Normally, gold would be a silly material to use as a catalyst for chemical reactions because it doesn’t do much. However, break gold down to nano size (approximately 5 nano meters) and it can act as a catalyst that can do things like oxidizing carbon monoxide.
Why the same material behaves differently at different sizes?

Atoms in an element are identical. Then why they behave differently if we go to a scale containing a few atoms from a scale containing billions of atoms?

Answer

Atoms in an element are identical. Then why they behave differently if we go to a scale containing a few atoms from a scale containing billions of atoms?

It is all due to orbitals, atomic and molecular orbitals, which are the probability loci of where the electrons may be found in space.

The five d orbitals in ψ(x, y, z)2 form, with a combination diagram showing how they fit together to fill space around an atomic nucleus.

The spatial structure allows for an aggregate of atoms to be able to bind with each other because where the probability for electrons is low, the field from the positive nucleus will not be cancelled and will be high, thus there is attraction and repulsion , and minimum quantum mechanical levels can be attained, depending on how the atoms "fit" in the sense of LEGO brick building, except quantum mechanically.

Because this is at the basic quantum mechanical frame, molecular orbitals, i.e. probability loci will be created for the given combination, with different energy levels, different LEGO bricks . For example:

Suitably aligned f atomic orbitals overlap to form phi molecular orbital (a phi bond)

Again the new spatial geometry will give new possibilities for molecules fitting with each other .

In bulk matter, where ~10^23 molecules exist in a mole, thermodynamic equations of heat exchanges treat atoms and molecules statistically very successfully. The great variety of forms of how atoms( as of gold) and molecules can bind with each other in space , is homogenized and the statistically most probable form will dominate a bulk sample of matter.

With the current technologies, specific three dimensional shapes have been observed and artificially formed, taking advantage of the great variety of possible bindings , leading to the great field of nanotechnology.

Why the properties of a substance like color changes when we go into the nano scale?

Color as frequency is dependent on transitions between quantum mechanical energy levels. If one has a substance which predominantly has a specific nanocluster solid geometry, it will give its specific energy levels when interacting with light, different from another permutation of solid geometry for the same molecules/atoms.

One has to realize that at the nano scale the quantum mechanical framework dominates and it is probability amplitudes and energy levels that characterize the clusters, and differentiate their properties at the cluster level.In a sense it is one huge new molecule: even though the same number of atoms , different spatial orientations/fits generate a new substance, with different behaviors that have to be studied.

Relation of conformal symmetry and traceless energy momentum tensor

In usual string theory, or conformal field theory textbook, they states traceless energy momentum tensor $T_{a}^{\phantom{a}a}=0$ implies (Here energy momentum tensor is usual one which is symmetric and follows conservation law) conformal theory. (i.e, see page 3 )

I wonder how they are related to each other.

I found similar question Why does Weyl invariance imply a traceless energy-momentum tensor? and get some idea about weyl invariance.

and get some another useful information from Conformal transformation/ Weyl scaling are they two different things? Confused! which dictates that conformal transformation and weyl transformation is totally different things .

molecular dynamics - What is water pressure, really?

It's easy for me to imagine gas pressure on a molecular level. More pressure means that more molecule bounce against the thing you are measuring, or the molecules hit it with more speed or the molecules are heavier.

But I can not imagine how water pressure works on a molecular level. Water is not compressible, so there can not be more water molecules per volume to increase pressure. But if the velocity increases, this should also increase the water temperature. However the bottom of a pool of water is not hotter than the surface. So I just don't understand it.

Answer

The fundamental difference between a liquid and an ideal gas is that in the gas, the molecules fly around free and only occasionally and briefly interact with each other, whereas in a liquid, the molecules are close enough together that they are always within the electromagnetic fields of their neighbors. If two molecules in a liquid move apart a bit, there is a weak electromagnetic attractive force between them. If they move a bit closer together, there arises a repulsive force between them that grows rapidly as the separation decreases. So it doesn't take much compression (but some) for the pressure to increase markedly. Though things at the molecular scale are governed by quantum mechanics, it is fair to imagine a little free-body diagram for a molecule of water which is closest to the "thing you are measuring". There would be a force on the molecule from the "thing", but also forces, on the whole in the opposite direction, from the neighboring molecules.

newtonian mechanics - Two pendulums, same pendulum length, same bob mass, but one bob's full of water

Suppose, a ideal pendulum which has a pendulum lenghth $L$ and a bob of mass $m$, another one whose bob has same mass and same effective length. But the second one's bob is hollow and and the hollow is full of water.

Will the periodic time $T$ of the two pendulums be same?

Answer

If we can neglect friction with air, the formula for the period of a pendulum will be:

$T= 2\pi \sqrt {\frac {l}{g}}$

If $g$ is costant, as in this case, the period of the pendulum will only depend on the lenght of the string.

Because the center of mass of both bobs lays in the middle, the effective lenght (lenght of the string + distance of CM from the top of the bob) will be the same, ergo the two pendula will have the same period.

It is also interesting to see how the period of the pendulum changes when we ad different amounts of water in the bob.

1. At first, with no water, the center of mass of the bob lays in the middle (distance of CM from the top of the bob= Radius). The period is T.

2.When we star adding water, the CM of the system will go below the original one (distance of CM from the top of the bob= Radius +y ).The new period of the bob will be $T_f>T$. The period will rise as long as we keep adding water below the original CM of the bob. When the water fills in half the bob, the period will reach a maximum. As we add more water, the period will start to decrease again.

3. Eventually,when we fill the bob (completly) with water, the CM of the system will lay in the same positon of the beginning. At this point, the effective lenght will be the same and so will be the period.

magnetic monopoles - A question from Professor Anthony Zee's book: "Quantum Field Theory in a Nutshell"

When Professor Anthony Zee is discussing the topological monopole in page 309 of his famous book "Quantum Field Theory in a Nutshell", Princeton University Press (2010), he writes:

But I am still not clear how the mass of the topological monopole, which is related to the mass of intermediate vector boson of the weak interaction, explains that why the monopole has not yet been discovered. Could anyone help me to explain this? Thanks very much!

Answer

The $W$ boson has a mass of about 80 GeV, 137$M_W$ would make a mass of about 11 TeV which is way above the mass of the highest-energy particles recently detected, such as the Higgs Boson which weighs about 125 GeV.

nuclear physics - Which nucleus is the most resilient against gamma-induced fission?

To state the title question perhaps more precisely:

What is the largest photon energy $E_{\gamma}$ and the corresponding mass number $A$ and atomic number $Z$ of a suitable nucleus ${}^A_ZX$ (presumably in a ground state) such that the hypothetical reaction

$$ {}^A_ZX + \gamma \rightarrow {}^{(A - a)}_{(Z - z)}Y + \text{whatever remains (with combined charge} +z \text{)}$$

is "kinematically" forbidden for any values $1 \le a < A$ and $Z \ge z \ge Z + a - A$,

while conforming to the standard model?

Edit
Changed the question title (removed the parenthetical qualification "whether otherwise stable or not"): for any unstable nucleus the stated question and condition is not meaningful and not relevant.

nuclear physics - Why is caesium-137 more stable than caesium-134?

Caesium-133 is stable. Caesium-134 and caesium-137 are radioactive isotopes of caesium with half-lives of 2.065 years and 30.17 years respectively.

Why does caesium-137 have a longer half-life if it contains three more neutrons than caesium-134 and four more neutrons than stable caesium?

It would seem to me that caesium-134 would have the longer half-life given that it only contains one more neutron than stable caesium.

Answer

As noted in the comments, all of the various Cs isotopes I'll mention decay by emitting a beta, converting the Cs isotope to a Ba isotope. Now, while details of nuclear decays are not necessarily touched on unless you are in a nuclear physics course, they are at least somewhat analogous to electron or photon decays. What I mean by that is that you are, at a hand-waving level, looking at an initial state (the Cs), final states (Ba in various possible energy levels), and any applicable quantum numbers you would like to try and conserve (like nuclear spin).

So, lets take a tour of the isotopes, relying mainly on data from nuclear data sheets. Start with Cs-134 (you probably did not know there was a journal called Nuclear Data Sheets). Going down to page 69, one finds that the Cs-134 nucleus has a spin of 4. It can decay to any one of 6 possible Ba-134 nuclear energy levels (the ground state and 5 excited states). The majority of the decays go through an excited state, which also has a nuclear spin of 4. The half life is 2 years.

Cs-135 is listed with a nuclear spin of 7/2. There is only one available Ba-135 level to decay to, and it has a nuclear spin of 3/2. The half life of this decay is 2.3 million years. Only one state to go to, and a spin mismatch to slow it down.

Cs-137 has a nuclear spin of 7/2. It can decay in to 3 different Ba-137 levels, the ground state and two excited states. The majority go through an excited state with spin 11/2. The other two states have spin 1/2 or 3/2 (the ground state). So, a few more states to decay to, but some pretty large spin mismatches on several of them. The half life is 30 years.

Cs-139 has a nuclear spin of 7/2. It can decay in to one of 60 (!) different Ba levels, with most decays being to the ground state which has a spin of 7/2. The half life is 9 minutes.

Taken all together, what do we see?

More available levels to decay to increases the chance of decaying. Closer spin values between the parent and daughter nuclei increases the chance of decaying. To go much deeper requires diving deeper in to nuclear physics.

statistical mechanics - Moments of a Distribution via Laplace Transforms and Wick Rotations

On a mathematical level, the statistical mechanical partition function is just a Laplace transform of the microcanonical probability distribution, i.e. it's moment generating function. Understanding the mathematical and physical motivation and meaning behind this is something many have unfortunately been unable to help me with, hence I ask here:

What is the intuitive meaning of the n'th moment of a function or probability distribution?

• I understand that "Broadly speaking a moment can be considered how a sample diverges from the mean value of a signal - the first moment is actually the mean, the second is the variance etc... ", but using this interpretation I can't make sense of what the 5'th moment of $f(x) = x^2$ is, why it's useful, what it says about anything, and how in the world such an interpretation falls out of the mathematical derivation.

Why would one ever say to oneself that they should decompose a function into it's constituent moments via a moment generating function?

• A Fourier series/transform of a function of, say, time is just an equivalent way to characterize the same function in terms of it's amplitude, phase and frequency instead of time (as a sine wave is uniquely characterized by these quantities, and to a physicist a Fourier series is more or less just a big sine wave where these numbers vary at each point). A Geometric interpretation of every term in the expansion of the Fourier series of a function can be given. Intuitively, just looking at, say, light or sound there is an immediate physical motivation for using a Fourier transform because it decomposes them into their constituent frequencies, which exist physically. A m.g.f. is a bilateral Laplace transform, and it decomposes a function/distribution into it's moments. Thus the Laplace transform should have an almost similar story behind using it, one that should explain the reason for wanting the n'th moment of $x^2$. (An answer saying that it uniquely characterizes "nice" functions/distributions is to miss the point of my question).

Why is it that the Fourier series of an exponentially damped distribution (the moment generating function or Bilateral Laplace Transform) decomposes a function into it's moments?

• I understand the mathematical derivation, but not the motivation for doing any of this, or why it should work, i.e. why wouldn't the regular Fourier transform just be enough? There has to be meaning behind it, so taking the fact we're using a Fourier series/transform literally, the question arises as to why it is that the constituent waves of an exponentially damped version of a function/distribution allow for an interpretation in terms of moments? (Note "exponentially damped" means using a minus sign in the m.g.f.)

Why does a Wick rotation of the Moment generating function give the Characteristic function?

• The characteristic function is the Fourier transform of a probability distribution, but it is also a wick rotation of the Fourier transform of an exponentially damped distribution. I understand we are analytically continuing the argument to a new axis, but is there any meaning behind this, does it say anything about the relationship between the Laplace transform's moments and the composite waves that a Fourier series decomposes a function into, and does it help explain why there is a mathematical relationship between the Feynman Path integral and the statistical mechanical Partition function?

In other words, regarding the partition function: What are we doing, why are we doing it, why does a particular method for doing it work & does it explain an interesting connection between two other things? Thanks

quantum mechanics - Getting the density of states for photons

I know that the density of states $g(\epsilon)d\epsilon$ is the number of states in the energy range $[\epsilon, \epsilon + d\epsilon]$.

I considered a system of non-interacting free photons in 3 dimensions.

To calculate the density of states we just need:

1) The energy of our system. In this case we are dealing with the energy of photons, so:

$$\epsilon = \hbar \omega $$

2) The number of states with energy $\le \epsilon$ (let's call it $N(\epsilon))$. As 3 dimensions are considered, the number of states is enclosed in a sphere:

$$N(\epsilon) = \frac{4}{3} \pi R^3$$

Now I need to have the radius R in function of the energy so as to be able to compute the density of states:

$$g(\epsilon) = \frac{dN}{d\epsilon}$$

As I considered a free photon, its momentum is:

$$p = \hbar k$$

$$p = \frac{h}{L} (n_x + n_y + n_z)$$

Then, the energy can be rewritten as:

$$\epsilon (n_x, n_y, n_z) = \frac{p \omega}{k} = \frac{hc}{L}(n_x + n_y + n_z)$$

This expression should be correct (I checked dimensions), but it is not function of the radius.

But I suspect that $(n_x + n_y + n_z)$ has something to do with a geometric figure, which encloses the number of states...

Now I work out another example, completely independent of the one above.

COMPARISON WITH ANOTHER CASE TO SEE WHAT IS GOING ON

Actually I tested if $(n_x + n_y + n_z)$ had something to do with the radius of a sphere (I assumed 3D) using the free particle's case (without relativistic effects):

$$\epsilon = \frac{p^2}{2m}$$

For the energy, I got:

$$\epsilon (n_x, n_y, n_z) = \frac{h^2}{2mL^2}(n_x^2 + n_y^2 + n_z^2)$$

At this point, I had the following thought:

$$(n_x^2 + n_y^2 + n_z^2) = R^2$$

And I gave it a shot... and ended up getting the correct density of states for the free particle (I also ignored spin as it is just about multiplying the final answer by a number).

So I would say that $(n_x + n_y + n_z)$ has something to do with the radius.

To sum up, I have the following troubles:

1') How can I get the radius from this equation $\epsilon (n_x, n_y, n_z) = \frac{p \omega}{k} = \frac{hc}{L}(n_x + n_y + n_z)$? Technically I cannot do $(n_x + n_y + n_z) = R$. Actually, I tried it but did not get it (at the end of my post I have written the density of states I should get).

2') In the $\epsilon = \frac{p^2}{2m}$ case I got:

$$\epsilon (n_x, n_y, n_z) = \frac{h^2}{2mL^2}(n_x^2 + n_y^2 + n_z^2) = \frac{h^2 R^2}{2mL^2}$$

But this expression does not have dimensions of energy! I mean, if I regard $(n_x, n_y, n_z)$ as axis of my phase space I have no problems with dimensions but once I regard $(n_x^2 + n_y^2 + n_z^2)= R^2$ dimensions do not match in the equation for the above energy... but I get the right result! what is going on here?

3) Why the dimension of the density of states is time in photons and in the free particle case (assuming nonrelativistic quantum mechanics) is $\frac{T^2}{ML^2}$?

I know that the density of states of free photons (assuming Plank's energy; see Griffiths, page 244; EQ 5.112; second edition):

$$g(\omega) = \frac{V \omega^2}{\pi c^3}$$

Any help is appreciated.

Answer

The problem with your approach is that you ignore $\boldsymbol{k}$ being a vector. In fact

$$\boldsymbol{p}=\hbar \boldsymbol{k}=\frac{h}{L}\boldsymbol{n}$$

where $\boldsymbol{n}=\left(n_{x},n_{y},n_{z}\right)$. This gives you

$$R=\left|\boldsymbol{n}\right|=\frac{L}{h}\left|\boldsymbol{p}\right|=\frac{L}{hc}\varepsilon=\frac{L}{2\pi c}\omega$$

so (the factor of $2$ accounts for the two polarizations of the photon)

$$N\left(\omega\right)=2\cdot\frac{4}{3}\pi R^{3}=\frac{8\pi L^{3}}{3 \left(2\pi\right)^{3} c^3}\omega^{3}=\frac{V}{3 \pi^{2} c^3}\omega^{3}$$

$$g\left(\omega\right)=\frac{\partial N\left(\omega\right)}{\partial\omega}=\frac{V}{\pi^{2} c^3}\omega^{2}$$

cosmology - Is the observable region of the universe within the event horizon of a super-massive black hole?

Observations:

1. 
   

   I have read that for a free-falling observer within the event horizon of a black hole that all lines of sight will end at the singularity which is black.

   
   


2. 
   
   

   I also look up and see that the sky is mostly black.

   
   


3. 
   

   I also know that by measuring the red-shifts of the galaxies that they are all accelelerating towards that blackness. (rather than accelerating away from a big-bang event, which makes less intuitive sense)

   
   


4. 
   

   The red-shifts are greatest for those galaxies closest to the singularity (i.e. furthest away from us).

   
   

I therefore conclude from these consistent observations that all the matter in at least the finite observable region of the universe in which we are located, is accelerating due to gravity towards the singularity of a super-massive black hole.

In the spirit of good science I will happy accept being wrong, but I need to understand why I am wrong.

Predictions from the theory:

The light from the universe outside the event horizon of the black hole should be visible as a small disc or point light source in the direction facing away from the singularity. Of course, if this happens to be inconveniently located on the far side of the milky way's core from us, then we will be unable to observe it.

Further Speculations:

It may be that in regions of the universe far ouside the influence of this black hole (e.g. >50 billon light years away) that the sky is white hot due to Olber's paradox.

The black hole may have been formed from a large quantity of anti-matter, which might explain the local perponderance of ordinary matter.

The microwave background radiation may be a result of Hawking radiation at the event horizon (which we are viewing fromn the inside), or, depending upon the non-euclidian geometry, we may be observing the result of matter being destroyed at the singularity.

Answer

This is a case of an unwisely chosen simile taken waaaay too far. This idea, that the entire universe could be inside the event horizon of not a supermassive, but rather a superduperultrahypermegastupendouslymassive black hole, is usually introduced in introductory classes about general relativity. The instructor in this case is trying to make clear that, contrary to a fairly popular misconception, the event horizon of a black hole is locally flat. That is, there are no CGI-fireworks, nor any kind of hard "surface", nor anything else particularly special, in the immediate vicinity of the event horizon. The only special thing that happens is a long distance effect, like noticing that every direction now points off in the distance towards the singularity.

The simile is also used to point out that, at the event horizon, even second-order, nearly local effects (that is, curvature of spacetime, or tidal effects in other words) become less pronounced the more massive the black hole is. (As an aside, this also explains why Hawking radiation is more intense for smaller black holes) So... as the simile suggests, if the black hole were massive enough, we might not even be able to detect it.

The key, though is massive enough. First of all, the whole beauty of the Einstein curvature tensor (the left side of Einstein's equation) is that it is Lorentz invariant, so it can be calculated in any reference frame, including one that is hypothetically based inside an event horizon.

The curvature can still be unambiguously calculated, so when you suggest that it may be only an optical illusion, you are also suggesting that all the scientists who do that type of large-scale curvature calculation (not me personally) are totally incompetent. Just so you know. I would suggest not mentioning that at any conferences on cosmology. One of the enduring mysteries of modern cosmology is that the large-scale curvature of the Universe seems to be open (like the 3-space-plus-one-time dimensional analog of a saddle or Pringle potato chip) and not flat (like Euclidean geometry) or closed (like a sphere). The last is what we would calculate if the visible Universe were inside a black hole.

So, for the visible Universe to be inside an event horizon, the Cosmic Acceleration we have seen thus far would have to actually just be one small, contrarian region inside an even larger event horizon of globally closed curvature. Just to make the event horizon radius 13.7 giga-lightyears (a bare minimum starting point that excludes all manner of things that make the real situation many orders of magnitude worse*), you would need over 8E52 kg of mass in the singularity. This would require over 5E79 protons, where I have heard that the entire visible Universe only has about 10^80 particles, total, and I think I heard that there are about 10^18 photons for every proton, or maybe even all other particles. Somebody can look that up if they want to, but it's definitely a big number. The upshot is that there would have to be an amount of mass, all crammed into one singularity, that would render the total mass of every single thing we can see a barely detectable rounding error. Monkeying with all those dark matter and even dark energy theories is less of a leap than that.

Your prediction doesn't actually predict anything, since you account for either its presence or its absence.

Speculation 1: Olber's Paradox is already solved for accepted theories of cosmology, so pointing out that your theory can also resolve it is nice but doesn't score any points.

Speculation 2: Are you suggesting that the singularity is where all the antimatter to match the Universe's matter went? Remember the singularity dwarfs the visible Universe. That still doesn't explain the asymmetry, it only pushes the question back by one logical step: Why did the antimatter go into the big singularity and not the matter?

Speculation 3: Hawking radiation for the big singularity's event horizon lends whole new meaning to the term negligible. See my previous aside. Also, we can't observe matter being destroyed at the singularity. That's information flowing the wrong way. Also, that negates the previous assertion that the sky is black because it's towards the singularity.

*Like cosmic expansion, just how small our contrarian region is, compared to the whole event horizon, and probably some other, subtler things.