fluid dynamics - Friction loss in a vertical pipe flow

I made this problem for more understanding of pressure and pressure loss in vertical flow.

Consider the following steady system, where a fluid enters a tank and exits through a vertical pipe of length $L$ and diameter $D=2R$. Height of fluid in the tank is constant and equals to $H$. Density and viscosity of liquid are $\rho$ and $\mu$, respectively. If flow is laminar find $Q$.

Now if I write Bernoulli's equation for the free surface of tank and exiting point of pipe, then I get$$\frac{P_{atm}}{\gamma}+\frac{v_0^2}{2g}+z_0=\frac{P_{atm}}{\gamma}+\frac{v^2}{2g}+z+h_l,$$where $h_L$ is the friction loss head of exiting pipe and $v=Q/(\pi R^2)$ and $\gamma= \rho g$. We know that $v_0 \approx 0$, thus$$H+L=\frac{v^2}{2g}+h_L$$ Now we need to find another relation between $v$ and $h_L$. Can we use Darcy–Weisbach equation? I think we can not due to vertical flow! I'm interested in writing momentum balance and derive the relation between friction loss and velocity (like Hagen–Poiseuille equation), but I don't know how to treat pressure terms! Is there a pressure distribution along the exiting pipe?

Edit1: Momentum balance for the laminar flow in the pipe gives the velocity as$$v_z(r)=\frac{R^2}{4 \mu}\left(-\frac{dp}{dz}+ \rho g \right) \left(1-\frac{r^2}{R^2} \right) $$And integrating over cross section of pipe for the flow rate gives$$Q=\pi r^2 v=\int_{0}^{R} 2 \pi r v_z(r) \ dr=\frac{\pi R^4}{8 \mu} \left(-\frac{dp}{dz}+ \rho g \right)$$And finally$$-\frac{dp}{dz}+ \rho g=\frac{32 \mu v}{D^2}$$Now which one of the following is right and why?

1) $h_L=L(-dp/dz)/ \gamma=\frac{32 \mu v L}{\gamma D^2}-L$

2) $h_L=L(-dp/dz+ \rho g)/ \gamma=\frac{32 \mu v L}{\gamma D^2}$

I don't have a sense of $p$ here! Can you give a physical sense of pressure within the exiting pipe?

Edit2: The answers and discussions in this question may solve the following similar questions:

Q1, Q2, Q3.

Answer

You can use the Darcy-Weisbach equation, but you have to modify it a little for vertical flow. In vertical flow, a differential force balance on the flow gives:

$$(P(z+\Delta z)-P(z))\frac{\pi D^2}{4}+\rho g \frac{\pi D^2}{4}\Delta z=\tau_w\Delta z \pi D$$where z is the elevation above the bottom of the tube and $\tau_w$ is the shear stress at the wall. So, $$\frac{d(P+\rho gz)}{dz}=\frac{4}{D}\tau_w$$ For laminar flow,$$\tau_w=\frac{f}{4}\frac{\rho v^2}{2}$$where f is the Darcy-Weisbach friction factor. So, combining the two equations, you get: $$\frac{d(P+\rho gz)}{dz}=\frac{f}{D}\frac{\rho v^2}{2}$$ For a horizontal tube, you would have just: $$\frac{dP}{dz}=\frac{f}{D}\frac{\rho v^2}{2}$$ So, for vertical flow, you simply replace the P in the horizontal flow equation by $P+\rho gz$.

quantum mechanics - Boundary conditions of the radial Schrodinger equation

Consider the radial differential equation

$$\bigg( - \frac{d^2}{dr^2} + \frac{(\ell+\frac{d-3}{2})(\ell+\frac{d-1}{2})}{r^2} + V(r) + m^2 \bigg) \phi_\ell (r) = \lambda\ \phi_\ell (r),$$

which I've obtained by solving the Schrodinger equation in $d$ dimensions using the method of separation of variables.

Now, the boundary condition that I have been given is $$\phi_\ell (r) \sim r^{\ell+\frac{d-1}{2}},\ r \rightarrow 0.$$

However, I was expecting the boundary conditions $$\phi_\ell (0) = 0,\ \phi^{\prime}_{\ell} (0) = 1.$$

Does anybody have an idea if there is a relation of some sort between the two sets of boundary conditions in the context of the given differential operator?

Answer

I) Well, $r=0$ is the boundary of a $d$-dimensional spherical coordinate patch, i.e. an artifact of our choice of coordinate system, but $r=0$ does not correspond to a physical boundary per se, other than what we can deduce from the TISE.

The mantra is that a boundary condition (for finite $r$) should only be imposed if it is a consequence of the TISE. See also this Phys.SE post. (Of course we might have to discard solutions for other reasons, see e.g. Section III below.)

II) The regular power behavior $$\phi_{\ell}~\sim~ r^{\ell+\frac{d-1}{2}}, \qquad r~\to~ 0,\tag{1} $$ seems to be deduced under the assumption that the kinetic term dominates over the potential term $V(r)$ for $r\to 0$, i.e. by considering the free radial TISE.

III) More generally, since the radial TISE is a 2nd order ODE, there are actually two independent branches, where (1) is one of them. The other branch, which is singular, has been discarded since the wave function should be normalizable, and the kinetic energy finite. For the $d=3$ case, see e.g. this Phys.SE post and links therein.

Pauli exclusion principle in an electron beam

Do electrons in an electron beam (cathode ray) follow the Pauli exclusion principle? Or in other words, does the Pauli exclusion principle apply for the beam of electrons?

Answer

A cathode ray tube where the electrons are randomly ejected and are not constrained in space the Pauli exclusion seems irrelevant..

In structured beams though, things may be different:

The Pauli Exclusion Principle places a fundamental limit on the brightness of an electron beam. Developing a cathode which can reach this limit is useful for achieving maximum operation in current applications of electron beams, but also opens new areas of physics to be explored. When the phase space of the electron beam is filled to the maximum density, the electrons will experience a degeneracy
pressure, similar to that which keeps a neutron star from collapsing. One promising source or a quantum degenerate beam is field emission from adsorbates on carbon nanotubes. Adsorbates have been shown to provide several orders of magnitude enhancement to emission brightness, which approaches the degeneracy limit. We have developed experiments to test various adsorbates, in order to find those which bind tightest and provide the largest enhancement in brightness. Continuing work to discover better adsorbates should soon allow for the generation of a quantum degenerate electron beam .

So it seems to be an aim, to reach the degeneracy limit given by the Pauli exclusion principle.

mathematical physics - Can auxiliary fields be thought of as Lagrange multipliers?

In the BRST formalism of gauge theories, the Lautrup-Nakanishi field $B^a(x)$ appears as an auxiliary variable $$\mathcal{L}_\text{BRST}=-\frac{1}{4}F_{\mu\nu}^a F^{a\,\mu\nu}+\frac{1}{2}\xi B^a B^a + B^a\partial_\mu A^{a\,\mu}+\partial_\mu\bar\eta^a(D^\mu\eta)^a,$$ and in the superfield formalism of SUSY, the field $F(x)$ also appears as an auxiliary variable: $$\mathcal{L}_\text{SUSY}=\partial_\mu \phi\partial^\mu\phi+i\bar\psi^\dagger\bar\sigma^\mu\partial_\mu\psi+F^*F+\ldots\,.$$

It is very tempting to view $B^a(x)$ and $F$ as Lagrange multipliers since their equations of motion leads to constraints. But, these variables do not enter into the Lagrangian linearly, like a conventional Lagrange multiplier. Rather, they enter into the Lagrangian quadratically.

However, in Kugo and Ojima's paper Manifestly Covariant Canonical Formulation of the Yang-Mills Field Theores (1978), they refer to the $B^a(x)$ fields as the 'Lagrange Multiplier' fields (p.1882).

So my question is: Can these auxiliary fields be viewed as Lagrange multipliers? and in what ways do they behave differently/similar to the conventional Lagrange multipliers that enter into the function linearly?

Answer

By definition, Lagrange multipliers are only coefficients that enter the extremized quantity (action) etc. linearly – and that multiply constraints. In some exceptional cases, an auxiliary field could enter in this way. However, they typically appear in a more complicated way and bilinear terms in the auxiliary fields are a rule rather than an exception. So strictly speaking, they're not Lagrange multipliers. But they are very similar. If no derivatives of these objects appear in the action, they're also "non-dynamical" (not involving time derivatives) and the variation with respect to them implies "non-dynamical" i.e. algebraic equations of motion.

Note that in the normal treatment of extremization, we introduce Lagrange multipliers because we want to extremize the quantity given the assumption that another quantity or other quantities are kept fixed. "Kept fixed" is translated as "conservation laws" into the physics jargon. However, in physics, we rarely consider conserved quantities that are conserved because the conservation law is explicitly written down as an independent constraint. Instead, in physics we usually discover conservation laws nontrivially – the conserved quantity has to be determined by a somewhat non-trivial procedure due to Emmy Noether out of a symmetry. In almost all physical theories, conservation laws are non-trivial consequences of some other, "more elementary" equations of physics.

quantum mechanics - Understanding operator bra-ket notation

Hi I have a question that might be a bit trivial. I have just completed learning a section on the bra-ket notation. There is a statement that the following is prohibited $$\hat{A}\langle\psi|, ~|\psi\rangle |\phi\rangle,~\langle \psi| \langle\phi|, |\psi \rangle \hat{A}.$$ I understand that this can be seen when dealing with the finite-dimensional case where we can consider $\hat{A}$ as a matrix and the above then reduces to cases which are prohibited in matrix multiplication. In a set of notes I am using it proves the conservation of probability (conservation of normalization) by first stating that

$$\frac{\mathrm d}{\mathrm dt}\langle \Psi(t)| \Psi(t) \rangle = \left(\frac{\mathrm d}{\mathrm dt}\langle \Psi(t)|\right)|\Psi(t) \rangle + \langle \Psi(t)|\left(\frac{\mathrm d | \Psi(t) \rangle}{\mathrm dt}\right)$$

It seems that $\frac{\mathrm d}{\mathrm dt}$ does not satisfy the definition of an operator used in the context above so we can have that $\frac{\mathrm d}{\mathrm dt}\langle \Psi(t)|$ on the right-hand side makes sense. Would I be right in saying that this is because $\frac{\mathrm d}{\mathrm dt}$ does not satisfy the operator definition in QM which is a mapping that simply takes a ket to a ket and a bra to a bra? I know that the inner-product result used (even though I thought that this was only for finite dimensional inner product spaces) is $$\frac{\mathrm d}{\mathrm dt}\langle f,g \rangle = \langle f(t), g'(t) \rangle + \langle f'(t),g(t) \rangle.$$

Answer

Bra-ket notation is just a useful short hand for some well-defined objects in functional analysis (or linear algebra if you work in finite dimensions).

To understand what is allowed and what isn't, you would better know what those concepts are, so let's quickly recap:

• A ket $|\psi\rangle$ is just a vector in some Hilbert space $\mathcal{H}$.


• A bra $\langle \psi |$ is an object in the dual space of that Hilbert space (usually denoted $\mathcal{H}^*$), which is the space of all linear functionals on the Hilbert space. This implies that $\langle \psi |$ is a linear function, where the variable runs over all possible $|\phi\rangle\in \mathcal{H}$ and the result is a number, i.e. $\langle \psi|$ is a function $\mathcal{H}\to \mathbb{C}$.


• The dual space of a Hilbert space is itself a vector space, hence it makes sense to add bras and multiply them with scalars.


• Now, given the scalar product of the Hilbert space $\langle \cdot, \cdot \rangle$, you can easily construct such functionals: For every $|\phi\rangle\in \mathcal{H}$, the map $|\psi\rangle \mapsto \langle |\phi\rangle,|\psi\rangle\rangle$ is a linear functional. Identifying this linear functional with $\langle \phi |$ lets you write $\langle \phi |\psi \rangle$, which hints at the fact that you used the scalar product.



• According to some theorem, that's all there is to the dual space, hence the bra-ket notation captures the Hilbert space, its dual and the natural duality between the two (the inner product).


• An operator is now a linear function $A:\mathcal{H}\to \mathcal{H}$. This means that it takes vectors to vectors.


• This explains why what you write as "prohibited" is at first glance ill-defined. However, beware: some of the "prohibited" notions are actually all over other texts. For instance, $|\psi\rangle|\phi\rangle$ is often used as a shorthand for $|\psi\rangle \otimes |\phi\rangle$, but clearly, this has to be defined!

Now back to your question. The main problem is, as you say, that the differentiation is not an operator in the sense above. Why? Because it doesn't really act on your Hilbert space. $|\psi(0)\rangle\in \mathcal{H}$. Also $|\psi(t)\rangle\in \mathcal{H}$ for any fixed $t$, but the Hilbert space doesn't know anything about $t$. $|\psi(t)\rangle$ is in fact a function from some time interval $[a,b]$ to the Hilbert space $\mathcal{H}$. And it's on that parameter that the differential acts.

You can also see this with matrices: Given $\mathcal{H}=\mathbb{R}^2$ as your Hilbert space, we have for example a vector $v\in \mathcal{H}$ which is just a column vector. Also, an operator is a two-by-two matrix (that's the linear functions $\mathbb{R}^2\to\mathbb{R}^2$. Let's pick an operator and call it $A$. Then we can write:

$$ v=\begin{pmatrix}{} v_1 \\ v_2 \end{pmatrix}\quad \mathrm{and}\quad A=\begin{pmatrix}{} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} $$

But what about the $t$? Well, clearly, this means that all coordinates are dependent on $t$. You can even have operators that depend on $t$ (see for instance the Heisenberg picture):

$$ v(t)=\begin{pmatrix}{} v_1(t) \\ v_2(t) \end{pmatrix}\quad \mathrm{and}\quad A(t)=\begin{pmatrix}{} a_{11}(t) & a_{12}(t) \\ a_{21}(t) & a_{22}(t) \end{pmatrix} $$

Now if the time differentiation was an operator, you'd have to be able to write it as a two-by-two matrix. Can you? Obviously not. And the reason is that time is just a parameter. It doesn't really belong to the Hilbert space structure (it's not a dimension or anything).

From this picture, you can also see why what you wrote down is valid. How would you differentiate a scalar product of $v,w\in \mathbb{R}^2$? Well, let's write it out:

$$ \frac{\mathrm{d}}{\mathrm{d}t} (v(t)\cdot w(t)) = \frac{\mathrm{d}}{\mathrm{d}t}(v_1(t)w_1(t)+v_2(t)w_2(t))$$

Now you just have a bunch of scalar functions and you can use the usual rules of differentiation and obtain the result.

Can you make time an operator? That's an entirely different question (the answer is "not really").

newtonian mechanics - What is the function of the top point of a bouncing ball?

A ball is thrown away as parallel to x axis from M(0,h) point with speed V . After each jumping on x axis , it can reach half of previous height as shown in the figure.(Assume that no any air friction and the ball is very small.)

Question-$1$:

Find f(x) that passes from of top points ?

Question-$2$:

I have just a claim about the system but I need to prove that if we throw away the ball from a point of $f(x)$ as parallel to x axis with the speed of $V_{initial}=V\sqrt{\frac{f(x)}{h}}$, the maximum points of the ball will be on same $f(x)$. How can the claim be proved or disproved?

---

My attempt to find f(x)

$f(0)=h$

$f(x_1+x_2)=h/2$

$f(x_1+2x_2+x_3)=h/4$

$f(x_1+2x_2+2x_3+x_4)=h/8$ .

.

.

$x_1=vt$

$mgh=\frac{1}{2}mV^2$

$h=\frac{1}{2}g t^2$

After first bouncing , On first top point: N $(x_1+x_2,h/2)$:

$mgh/2=\frac{1}{2}mV_1^2$

$V_1^2=V^2/2$

$V_1=\frac{V}{\sqrt{2}}$

$h/2=\frac{1}{2}g t_2^2$

$t_2^2=t^2/2$

$t_2=\frac{t}{\sqrt{2}}$

$f(x_1+x_2)=h/2$

$f(Vt+t_2\frac{V}{\sqrt{2}})=h/2$

$f(Vt+\frac{t}{\sqrt{2}}\frac{V}{\sqrt{2}})=h/2$

$f(Vt(1+1/2))=h/2$

After second bouncing, On second top point: O $(x1+2x2+x_3,h/4)$:

$mgh/4=\frac{1}{2}mV_2^2$

$V_2^2=V^2/4$

$V_2=\frac{V}{2}$

$h/4=\frac{1}{2}g t_3^2$

$t_3^2=t^2/4$

$t_3=\frac{t}{2}$

$f(x_1+2x_2+x_3)=h/4$

$f(Vt+Vt+V_3t_3)=h/4$

$f(Vt+Vt+Vt/4)=h/4$

$f(Vt(1+1+1/4))=h/4$

On 3th top point:

$f(Vt(1+1+1/2+1/8))=h/8$

On 4th top point:

$f(Vt(1+1+1/2+1/4+1/16))=h/16$ .

.

On nth top point: $n--->\infty$

$f(3Vt)=0$

How can I find $y=f(x)$

Thanks for answers

---

EDIT: I have found $f(x)$. MrBrody is right. The function is a line.

my Proof: On nth top point:

$f(Vt.a(n))=h/2^n$

$a(0)=1$

$a(1)=1+1/2=1+1-1/2$

$a(2)=1+1+1/4=1+1+1/2-1/4$

$a(3)=1+1+1/2+1/8=1+1+1/2+1/4-1/8$

$a(4)=1+1+1/2+1/4+1/16=1+1+1/2+1/4+1/8-1/16$ .

.

$a(n)=1+1+1/2+1/4+....1/2^{n-1}-1/2^{n}$

$a(n)=1+\frac{1-(1/2)^{n})}{1-1/2}-1/2^{n}=1+2(1-(1/2)^{n})-1/2^{n}=3-\frac{3}{2^n}$

On nth top point:

$$f(Vt[3-\frac{3}{2^n}])=\frac{h}{2^n}$$

where $t=\sqrt{\frac{2h}{g}}$

$Vt[3-\frac{3}{2^n}]=x$

$3-\frac{3}{2^n}=\frac{x}{Vt}$

$1-\frac{1}{2^n}=\frac{x}{3Vt}$

$1-\frac{x}{3Vt}=\frac{1}{2^n}$

$$f(x)=\frac{h}{2^n}=h(1-\frac{x}{3Vt})=h(1-\frac{x}{3V\sqrt{\frac{2h}{g}}})$$

Now I am trying to prove or disprove my claim in second question. If I find the solution I will post it.

Answer

The key parameter here seems to be how are distributed the losses at each bounce (from what you said, there is no air friction here): how much translational and vertical speed are lost in the transition before/after the bounce. From your graphic, it seems like the hypothesis is that there is "reflection" of the ball: same angle when arriving/leaving the ground, only the norm of the speed decreases, the ratio |V_y|/|V_x| being conserved. From that, you can prove that each parabola is similar to the previous one, same ratio height/length on x axis for each one. From this you indeed get a straight line, whith its slope determined by the 2 first points, (0,h) and (x1+x1/2, h/2). Then you only need to determine x1 to determine the slope of f(x), and hence f(x) completely. I hope I didn't miss anything, as this hypothesis on the repartition of the losses on V_x and V_y at each bounce is crucial to solve the problem!

thermodynamics - EM Radiation and Heat

If there is EM radiation, will there always be a heat exchange between the wave and its surroundings? If you have an extremely energy-efficient light bulb that converts ALL the energy it receives into visible light, will it still heat up the glass bulb?

The origin of this question is a puzzle:

You are in a basement. You cannot see the above floor. You have three light switches in basement and three light bulbs in the above floor, but you do not know which switch controls which bulb. Modify the switches in any way, but you can only go to check the bulbs once. When you check the bulbs, you must immediately know which switch controls which bulb.

The solution is to turn on the switch for a long time to heat up the corresponding bulb, then turn on a second bulb. The light bulbs will be ON and HOT, ON and COLD, and OFF. This info can be used to find the corresponding switches.

This is true with incandescent light bulbs that are known for being energy inefficient, converting much of the energy it receives in to heat instead of light. If I have an incredibly energy-efficient light bulb, will this still work?

Answer

The distinction between heat and light is simply a result of the way our eyes work. Because we can't see any radiation with a wavelength longer than about 700nm we call it heat rather than light. However it's all electromagnetic radiation and it all carries energy. If you stand in front of a source emitting 100W of light it will feel just as warm as standing in front of a source emitting 100W of heat.

(Just to confuse matters a 100W light bulb doesn't emit 100W of light: it emits about 2W of light and 98W of heat.)

In this particular case you want to know how much the glass of the light bulb would heat up. This depends on how much of the EM radiation passing through the light bulb is absorbed. A quick Google found this diagram showing the absorption of light by (silica) glass:

Suppose your 100W light source emits all its energy as wavelengths longer than 4$\mu$m where the absorptance is virtually 100%. That means the glass is absorbing the whole 100W and heating up at some corresponding rate.

Between about 200nm and 2000nm the absorption is only about 10%, so if if you tweak your light bulb to emit all its energy as wavelengths in this range it would only absorb 10W. So it would still heat up but at only one tenth the speed of of the other bulb.

So you could still solve the puzzle, but you might need a sensitive thermometer. To make the puzzle impossible to solve you would need a glass that absorbs no radiation at all, and as far as I know no such glass exists.