Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result in accumulation of liquid inside the pipe which is not possible. Does streamline terminates or it changes its direction at the open end of pitot tube?

homework and exercises - Moment of inertia of a hollow sphere wrt the centre?

I've been trying to compute the moment of inertia of a uniform hollow sphere (thin walled) wrt the centre, but I'm not quite sure what was wrong with my initial attempt (I've come to the correct answer now with a different method). Ok, here was my first method:

Consider a uniform hollow sphere of radius $R$ and mass $M$. On the hollow sphere, consider a concentric ring of radius $r$ and thickness $\text{d}x$. The mass of the ring is therefore $\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi r\cdot\text{d}x$. Now, use $r^2 = R^2 - x^2:$ $$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \left(R^2 - x^2 \right)^{1/2}\text{d}x$$ and the moment of inertia of a ring wrt the centre is $I = MR^2$, therefore: $$\text{d}I = \text{d}m\cdot r^2 = \frac{M}{4\pi R^2}\cdot 2\pi\left(R^2 - x^2\right)^{3/2}\text{d}x $$ Integrating to get the total moment of inertia: $$I = \int_{-R}^{R} \frac{M}{4\pi R^2} \cdot 2\pi\cdot \left(R^2 - x^2\right)^{3/2}\ \text{d}x = \frac{3MR^2 \pi}{16}$$

which obviously isn't correct as the real moment of inertia wrt the centre is $\frac{2MR^2}{3}$.

What was wrong with this method? Was it how I constructed the element? Any help would be appreciated, thanks very much.

Answer

The mass of the ring is wrong. The ring ends up at an angle, so its total width is not $dx$ but $\frac{dx}{sin\theta}$

You made what I believe was a typo when you wrote

$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \left(R^2 - x^2 \right)\text{d}x$$

because based on what you wrote further down, you intended to write

$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \sqrt{\left(R^2 - x^2 \right)}\text{d}x$$

This problem is much better done in polar coordinates - instead of $x$, use $\theta$. But the above is the basic reason why you went wrong.

In essence, $sin\theta=\frac{r}{R}$ so you could write

$$\text{d}m = \frac{M}{4\pi R^2}\cdot 2\pi \frac{r}{sin\theta} \ \text{d}x \\ = \frac{M}{4\pi R^2}\cdot 2\pi \frac{r}{\frac{r}{R}} \ \text{d}x\\ = \frac{M}{4\pi R^2}\cdot 2\pi R \ \text{d}x\\ = \frac{M}{2 R} \ \text{d}x$$

Now we can substitute this into the integral:

$$I = \int_{-R}^{R} \frac{M}{2 R} \cdot \left(R^2 - x^2\right)\ \text{d}x \\ = \frac{M}{2R}\left[{2R^3-\frac23 R^3}\right]\\ = \frac23 M R^2$$

newtonian mechanics - Under what conditions does the relation $vec{L} =I vec{omega}$ holds good?

if $\vec L=I\vec ω$ holds good in all the cases then directions of angular momentum and angular velocity must be parallel always which is not true in some of the situations. So under what conditions does the relation $\vec L=I\vec ω$ holds good ?

electromagnetism - How will SR EM Lagrangian change if we find a magnetic charge?

When we introduce electromagnetic field in Special Relativity, we add a term of $$-\frac e c A_idx^i$$ into Lagrangian. When we then derive equations of motion, we get the magnetic field that is defined as $$\vec H=\nabla\times\vec A.$$

If we now take divergence of both sides of this definition, we automatically get

$$\nabla\cdot\vec H=0,$$

which is equivalent to inexistence of magnetic charges.

But suppose we've found a magnetic charge. What will change in our Lagrangian or in definition of electric and magnetic fields in this case to make $\nabla\cdot\vec H=\sigma$?

In this Phys.SE answer it's asserted that magnetic field would get an additional term "gradient of a scalar potential". Is this "a" scalar potential instead "the" $A^0$ potential?

Answer

In the absence of magnetic monopoles, Maxwell's equations are

$$ \begin{align} \text d F &= 0 ,\\ \text d{\star F} &= J_e , \end{align} $$

where $J$ is the 4-current 3-form due to electric charges (assuming a metric with signature $(-,+,+,+)$). For cohomological reasons, from the first equation one can asserts that there exists a 1-form $A$ such that $F = \text d A$, and $A$ is the interpreted as the 4-potential $(\phi,\mathbf A)$ (up to the musical isomorphism between tangent and cotangent bundle to Minkowski spacetime). In the presence of magnetic monopoles (or charge, to even symmetrise terminology) the above equations would become

$$ \begin{align} \text d F &= J_m ,\\ \text d{\star F} &= J_e , \end{align} $$

where $J_m$ is the 4-current for magnetic charges. Therefore in this extended theory of electrodynamics both the Faraday tensor $F$ and its Hodge dual $\star F$ (sometimes also denoted by $G$) figure in constitutive equations.

Since $F$ is no longer a closed form, its expression must be modified by the introduction of a non-exact part, say $C$, so that

$$F = \text d A + C.$$

Since the equations are symmetric in $F$ and $\star F$ we can postulate there exist 1-forms $B$ and $D$ such that

$$\star F = \text d B + D,$$

and assume that $C$ depends on $B$, while $D$ depends on $A$. But since $\star\star = -1$ in special relativity, we conclude that

$$F = \text dA - \star\text dB,$$

which can be related to the Helmholtz decomposition into polar and axial part for twice differentiable vector fields.

The Lorentz force for a particle with electric charge $q_e$ and magnetic charge $q_m$ would be $$K = \iota_u(q_e F + q_m G),$$ where $u$ is the particle's 4-velocity vector and $\iota$ denotes the interior product. The extra term can then be reproduced with a Lagrangian containing the extra term $B_\mu u^\mu$.

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To make contact with the usual vector notation, observe that the Faraday tensor has the covariant matrix representation $$F = \begin{bmatrix}0&-~\mathbf E^T\\\mathbf E&\star\mathbf H\end{bmatrix}$$ where $\star\mathbf H$ is the Hodge dual of the magnetic field $\mathbf H$, and can be thought as the linear map $(\star\mathbf H)\mathbf v = \mathbf v\times\mathbf H$ for any $\mathbf v\in\mathbb R^3$. Skew-symmetric tensors as the one above are then represented by a polar vector $\mathbf E$ and an axial vector $\mathbf H$, and can be denoted as $F=(\mathbf E,\mathbf H)$. Having defined this notation, the action of the Hodge dual is then $\star(\mathbf E,\mathbf H) = (\mathbf H,-\mathbf E)$ (up to a sign which I can't be bothered remembering). The exterior derivative of the 4-current $A$ is a tensor of the form above, and it turns out that $$\text dA = \left(\nabla A^0+\frac{\partial\mathbf A}{\partial t},\nabla\times\mathbf A\right),$$ where the first component is the polar part and the second one is the axial part. Hence with no magnetic charges we recover the electric and magnetic fields. Now for the extra potential $B=(B^0,\mathbf B)$ we have, using the rule for the Hodge dual discussed a few lines above, $$\star\text dB = \left(\nabla\times\mathbf B, - \nabla B^0 - \frac{\partial\mathbf B}{\partial t}\right)$$ Remark Here $\mathbf B$ is an extra vector potential, not to be confused with the magnetic induction.

Reconstructing the Faraday tensor according to the prescription $F=\text dA - \star\text dB$ given above we then have, in terms of polar and axial parts $$F = \left(\nabla A^0 + \frac{\partial\mathbf A}{\partial t} - \nabla\times\mathbf B, \nabla\times\mathbf A + \nabla B^0+\frac{\partial\mathbf B}{\partial t}\right),$$ whence $$\mathbf E = \nabla A^0 + \frac{\partial\mathbf A}{\partial t} - \nabla\times\mathbf B$$ and $$\mathbf H = \nabla\times\mathbf A + \nabla B^0 + \frac{\partial\mathbf B}{\partial t}.$$

cosmology - How can a quasar be 29 billion light-years away from Earth if Big Bang happened only 13.8 billion years ago?

I was reading through the Wikipedia article on Quasars and came across the fact that the most distant Quasar is 29 Billion Light years. This is what the article exactly says

The highest redshift quasar known (as of June 2011[update]) is ULAS-J1120+0641, with a redshift of 7.085, which corresponds to a proper distance of approximately 29 billion light-years from Earth.

Now I come to understand that the Big Bang singularity is believed to be around 13.8 Billion years ago.

So how is this possible? Does the presence of such a quasar negate the Big Bang Theory?

I'm not a student of Physics and was reading this out of (whimsical) curiosity. Is there something I'm missing here or the "proper distance" mentioned in the fact is a concept that explains this?

Edit: My Bad! Here's how..

A simple google search led me to this article which says the farthest quasar found is 12.9 billion LYs and not 29 billion.
So in the end we have just proven that wikipedia needs more moderation.

quantum field theory - What does it mean to say that "the fundamental forces of nature were unified"?

It is said that immediately after the Big Bang, the fundamental forces of nature were unified. It is also said that later they decoupled, becoming separate forces.

Indeed, if we look at the list of states of matter on Wikipedia we see:

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  Weakly symmetric matter: for up to $10^{−12}$ seconds after the Big Bang the strong, weak and electromagnetic forces were unified.

  
  



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  Strongly symmetric matter: for up to $10^{−36}$ seconds after the Big Bang, the energy density of the universe was so high that the four forces of nature — strong, weak, electromagnetic, and gravitational — are thought to have been unified into one single force. As the universe expanded, the temperature and density dropped and the gravitational force separated, a process called symmetry breaking.

  
  

Not only is it said that the forces were once unified, but this is also somehow related to the states of matter.

I want to understand all of this better. What does it truly mean, from a more rigorous standpoint, to say that the forces were unified and later decoupled? How this relate to the states of matter anyway?

Answer

When we say that the forces were unified, we mean that the interaction was described by a single gauge group. For example, in the original grand unified theory, this group was $SU(5)$, which spontaneously broke down to $SU(3) \times SU(2) \times U(1)$ as the universe cooled. These three components yield the strong, weak, and electromagnetic forces respectively.

I'll try to give a math-free explanation of what this means. To do so I'll have to do a decent amount of cheating.

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First, consider the usual strong force. Roughly speaking, the "strong charge" of a quark is a set of three numbers, the red, green, and blue color charges. However, we don't consider the strong force three separate forces because these charges are related by the gauge group: a red quark can absorb a blue anti-red gauge boson and become blue. In the case of the strong force, we call those bosons gluons, and there are 8 of them.

At regular temperatures, the strong force is separate from the electromagnetic force, whose charge is a single number, the electric charge, and whose gauge boson is the photon. There is no gauge boson that converts between color charge and electric charge; the two forces are independent, rather than unified.

When we say all the forces were unified, we mean that all of the Standard Model forces were described by a common set of charges, which are intermixed by 24 gauge bosons. These gauge bosons are all identical in the same way that the 8 gluons are identical. In particular, you can't point at some subset of the 24 and say "these are the gluons", or "this one is the photon". They were all completely interchangeable.

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As the universe cooled, spontaneous symmetry breaking occurred. To understand this, consider slowly cooling a lump of iron to below the Curie temperature. As this temperature is passed, the iron spontaneously magnetizes; since the magnetization picks out a specific direction, rotational symmetry is broken.

In the early universe, the same process occurred, though the magnetization field is replaced with an analogue of the Higgs field. This split apart the $SU(5)$ gauge group into the composite gauge group we have today.

The process of spontaneous symmetry breaking is closely analogous to phase transitions, like the magnetization of iron or the freezing of water, which is why we talk about 'strongly/weakly unified' matter as separate states of matter. Like the iron, which state we are in is determined by the temperature of the universe. However, a exact theoretical description of this process requires thermal quantum field theory.

newtonian mechanics - How is a running man able to Accelerate himself?

Suppose a man is running and he gradually speeds himself up . For this he applies a force on the ground backward and the ground pushes him forward . This is probably due to to friction between the shoes of the man and the ground . The friction acting in this case is Kinetic friction which has a constant magnitude. Since the magnitude of frictional force is constant, how is the man able to accelerate himself ? Who is providing this force?

Answer

Don't think kinetic friction, just because some part of him is moving. Is it also kinetic friction if I stand still but swing my arms? Many particles in him or elsewhere might or might not move. They are irrelevant.

Only the particles in contact with the ground are relevant.

And they are not moving. His foot is not moving during the step. It is stationary and not sliding while touching. There is static friction here.

And static friction can vary easily.