particle physics - What actually happens when an anti-matter projectile collides with matter?

I'm trying to understand what would really happen when large quantities (e.g., 10g) of anti-matter collide with matter. The normal response is that they'd annihilate each other and generate an expanding sphere of gamma ray photons.

However, thinking about it in more detail, what I see is that the anti-electrons annihilate first against the electrons. Let's assume the energy release in that case is not sufficient to noticeably change the momentum of the projectile. Then the nuclei penetrate the electron-annihilation plasma, and since antiprotons attract protons, their trajectory is changed. However, the nuclei are so small and so widely separated that presumably they just orbit each other as the electron clouds annihilate, and eventually enough energy is generated that the ionic plasma of nuclei and anti-nuclei just expands, with a small fraction of them actually ever combining.

In other words, we don't actually see total conversion happening -- only a small fraction of the total mass in an anti-matter/matter collision is turned into gamma rays.

Is that what actually would happen? (The ideal answer would be a video!)

Predicting stellar evolution and life cycle of a star

Is there a way to model/predict the evolution and life cycle of a star based on certain initial conditions? That is, whether it will become a red giant, brown dwarf, etc. I'm basically looking for equations defining the progress of different star-types and how varying initial conditions may affect that progress.

electromagnetism - Magnetic field strength in a solenoid

I'm pretty new to physics. I've been conducting some experiments with electromagnets. My practical results don't match up with the theory.

The magnetic field in a solenoid of length $L$ around an iron core with $N$ turns is given by: $$ B = \mu \frac{NI}{L}. $$

Assuming Ohm's law of resistance in the wire we can replace $I$ with $V/R$ to get $$ B = \mu \frac{NV}{LR}. $$

In my experiments (using batteries and copper wire around a ferrite core), doubling the number of turns in the solenoid does gives approximately (although slightly less than, as expected) double the lifting strength of my magnet.

However doubling the voltage (1.5V to 3V by connecting batteries in serial) seems to only give an increased lifting strength of about 50%.

Any ideas as to why?

Many thanks

Answer

If $B \propto I$, then by Ohm's Law $B \propto {V \over R}$

By adding a second battery you not only increase the voltage across the coil, but also the resistance of the circuit.

If the internal resistance of dry cells are large compared to solenoid wire resistance, then doubling the number of batteries will not double the magnetic flux through the solenoid since the resistance of the circuit increases accordingly. (It is not true that $no. of ~batteries \propto I$)

But if the wire resistance is small compared to battery internal resistance, then doubling the number of coils negligibly affects the resistance of the circuit, in which case you may in fact use $B\propto NI$ directly (where $I$ is constant).

Finally as you would have noticed I have not referred to magnetic force but magnetic field. Note that the magnetic force $F\propto B^2$, thereby in your case you should be expecting $F\propto N^2$, a quadrupling of holding force when number of loops are doubled. The reason why you see a linear relationship instead is due to the relative permeability of the iron core $\mu_r$ ($\mu=\mu_0\mu_r$) being variable to the magnetic fields originating in the nearby solenoid. therefore you wont see a $F\propto B^2$ until the iron core is saturated, i.e, use higher currents. Even if you used a precise voltmeter across the solenoid so you can use the first equation correctly (to avoid the assumption that current is proportional to number of batteries), the variable $\mu_r$ will still affect the results.

Refs http://en.wikipedia.org/wiki/Force_between_magnets#Force_between_two_nearby_magnetized_surfaces_of_area_A http://info.ee.surrey.ac.uk/Workshop/advice/coils/mu/index.html#mu

fluid dynamics - Bernoulli Principle and pressure function

I have read in several places now that for compressible flow the Bernoulli equation $$ \frac {v^2}{2}+ P\ + \Psi = \text{constant}$$

requires:

$$P=\int {dp\over \rho(p)} $$

But I don't get what is meant by that. Anything more, than that the density is a function of pressure? What do people want to express by this?

Answer

To derive the Bernoulli equation for inviscid fluids, the plan is to rewrite the Euler equation in such a way that we have gradients. I'll write the Euler equation with gravity here

$$\frac{\partial \vec{u}}{\partial t} + \vec{u} \cdot \vec{\nabla} \vec{u} = -\frac{1}{\rho} \vec{\nabla} p + \vec{g}.$$

Recall $g = - \vec{\nabla} \Psi$, and $\vec{u} \cdot \vec{\nabla} \vec{u} = \vec{\nabla}(\frac{1}{2} \vec{u}^2) - \vec{u} \times (\vec{\nabla} \times \vec{u})$. Now we also consider a steady flow, $\frac{\partial}{\partial t} \rightarrow 0$. Note, you can consider non-steady if you have potential flows, i.e., $\vec{\nabla} \times \vec{u} = 0 \rightarrow \vec{u} = \vec{\nabla}\phi$ (a gradient appears). Now you look at quantities along flow lines, i.e., dot the equation with $\vec{u}$, this returns

$$ \vec{u} \cdot \vec{\nabla} \left(\frac{1}{2} \vec{u}^2 + \Psi\right) = \vec{u} \cdot \left( -\frac{1}{\rho} \vec{\nabla} p\right).$$

Note that $\vec{u} \cdot (\vec{u} \times \vec{A}) = 0$, for any $\vec{A}$, including $\vec{A} = \vec{\nabla} \times \vec{u}$. Now we'd like to put that density inside that gradient so that we can have $\vec{u} \cdot \vec{\nabla}(\textrm{stuff}) = 0$, then ``stuff'' is conserved along $\vec{u}$.

Take $\rho(p,\xi)$, so a function of pressure and perhaps something else. Then let's look at how we might put $\rho$ inside the gradient but be equivalent to $\frac{1}{\rho} \vec{\nabla} p$. Consider $f = \int \frac{\textrm{d} p}{\rho(p,\xi)}$, the reason why will become apparent since we want to use the chain rule and fundamental theorem of calculus. Then

$$\vec{\nabla} f(p,\xi) = \frac{\partial f}{\partial p} \vec{\nabla} p + \frac{\partial f}{\partial \xi} \vec{\nabla} \xi = \frac{1}{\rho} \vec{\nabla} p - \int \frac{1}{\rho^2} \frac{\partial \rho}{\partial \xi} \textrm{d}p \vec{\nabla}\xi = \vec{\nabla} \int \frac{\textrm{d} p}{\rho(p,\xi)}.$$

The first step was the chain rule, and the second was the fundamental theorem of calculus to calculate $\partial_p f$. Therefore we either need that $\vec{\nabla}\xi = 0$, or $\partial_\xi \rho = 0$ so that $\vec{\nabla} f = \frac{1}{\rho} \vec{\nabla}p$. Thus either $\xi$ needs to be constant or pressure never depended on $\xi$. Therefore we see this only works for barotropic fluids, which is to equivalently say by definition, $\rho = \rho(p)$.

In conclusion

$$ \vec{u} \cdot \vec{\nabla} \left(\frac{1}{2} \vec{u}^2 + \Psi + \int \frac{\textrm{d}p}{\rho} \right) = 0.$$

A small tid-bit for anyone interested, I mentioned potential flows earlier being able to talk about time dependent flows too. I should note another nice property is that these quantities are not conserved along flow line alone, but over the entire space! The reason we considered along flow lines was the get rid of the pesky curl term, but in potential flows those drop out naturally and we never had need to consider the quantity along a flow line and the result is much more powerful. However, in a time dependent flow they need not necessarily be conserved in time (still more powerful than having to assume time independent flows).

fermions - How is Berry phase connected with chiral anomaly?

Recently I've read in one article about very strange way to describe chiral anomaly on quasiclassical level (i.e., on the level of Boltzmann equation and distribution function).

Starting from Weyl hamiltonian $$ H= \sigma \cdot \mathbf p, $$ which describes massless chiral fermions, performing unitary transformation $$ |\psi\rangle \to V|\psi\rangle, $$ where $|\psi\rangle$ is fermion state and $V$ is $2\times 2$ matrix that diagonalizes $\sigma \cdot \mathbf p$, such that $$ V\sigma \cdot \mathbf pV^{\dagger} = |\mathbf p|\sigma_{3}, $$ we obtain the expression for the matrix element $$ \langle f|e^{iH(t_{f}-t_{i})}|i\rangle \equiv \left(V_{\mathbf p_{f}}\int Dx Dp \text{exp}\left[i\int dt(\mathbf p \cdot \mathbf x - |\mathbf p|\sigma_{3}-\hat{\mathbf{a}}\cdot \dot{\mathbf p})\right]V_{\mathbf p_{i}}^{\dagger}\right)_{fi}, \quad \hat{a}_{\mathbf p} = V_{p}\nabla_{\mathbf p}V^{\dagger}_{p} $$ By choosing $+1$ helicity and neglecting off-diagonal components of $\hat{a}_{\mathbf p}$ (which is called adiabaticity approximation; it is not valid near $\mathbf p = 0$), we obtain following quasiclassical action: $$ S = \int dt (\mathbf p \cdot \dot{\mathbf x} - |\mathbf p| - \mathbf a \dot{\mathbf p}) $$ The quantity $\mathbf a$ (which is called Berry phase) plays the role of gauge field in momentum representation, with curvature $$ \mathbf b = \nabla \times \mathbf a = \frac{\mathbf p}{|\mathbf p|^{3}} $$ Effect of this Berry phase is absent when $\dot{\mathbf p} = 0$.

If we, however, turn on external EM field, it becomes to be relevant. We obtain that the invariant phase volume element is $$ dV = \frac{d^{3}\mathbf x d^{3}\mathbf p}{(2\pi)^{3}}\Omega (\mathbf p), \quad \Omega (\mathbf p) = (1 + \mathbf b \cdot \mathbf B)^{2}, $$ where $\mathbf B$ is magnetic field. This captures chiral anomaly effect, $$ \tag 1 \dot{\rho} + \nabla_{\mathbf r}(\rho \dot{\mathbf r}) + \nabla_{\mathbf p}(\rho \dot{\mathbf p}) = 2\pi \mathbf E \cdot \mathbf B \delta^{3}(\mathbf p), \quad \rho = f\Omega , $$ with $\mathbf E$ being electric field and $f$ being distribution function.

I don't understand how this berry phase leads to description of chiral anomaly. They appear due to different reasons (anomaly arises because of non-trivial jacobian of chiral transformation, while berry phase arises because of formal manipulations), anomaly has topological nature connected with difference of number of zero modes of Dirac operator, while Berry phase hasn't such origin. Finally, the rhs of $(1)$ is non-zero only when adiabaticity approximation is violated (and hence the result isn't valid).

Could someone explain me the reason due to which berry phase somehow describes effects of anomaly on quasiclassical level?

Answer

I have been collaborating with the authors of that paper. In Section 2.6 of my thesis I explained the relations between the full quantum computation of chiral anomaly, the (semi-classical) Nielsen-Ninomiya spectral flow picture and the (almost classical) Berry curvature picture to the best details of my knowledge.

Let me briefly summarize the ideas.

Nielsen and Ninomiya has provided a computation of chiral anomaly in terms of spectral flow, in the context of what is known today as Weyl semimetal. Of course, in standard texts like Peskin and Schroeder, chiral anomaly in particle physics was also taught in terms of spectral flow. The distinction is minor. We know chiral anomalies have IR and UV interpretations. The spectral flow through the zero energy (Weyl node) is an IR interpretation; but the particles "must have gone somewhere", and that is capture by the UV boundary conditions. In the Weyl semimetal, the left and right Weyl nodes become connected in the UV (deep in the valence band). On the other hand, for Weyl fermion in particle physics, we don't know what the actual UV is, of course, but we may picture an infinitely deep Dirac sea, and the UV boundary condition is just that the flow towards infinite negative energy are opposite for left and right Weyl fermions -- this assumption is needed for the conservation of the total U(1) charge. So in the two contexts the difference is only what we say about the UV, but they give essentially the same IR physics.

Once we have the picture of spectral flow, we can make easy connection to the Berry phase computation. In the Berry phase computation, the symplectic 2-form is closed except at the $p=0$ Weyl point, which means this point in the momentum space is a sink or source of Liouville measure flow. Of course, classical mechanics breaks down at this point (roughly speaking, the classical computation is good only for $\partial_x \ll p$), so what the Berry phase computation actually tells is again a boundary condition -- but now it is not a UV boundary condition, but an IR boundary condition around vicinity of the $p=0$ point (where classical mechanics fails): What we have computed is how many particles have flowed into / out of the vicinity of $p=0$. One finds the amounts are indeed opposite for left and right Weyl fermions, and the amount is in agreement with the quantum computation of chiral anomaly.

Notes added:

This agreement is expected because the (semi-)classical Berry curvature formalism can be extracted from quantum mechanics to first correction in $\hbar\partial_x$ (for instance see the paper you referred to; Section 2.4 of my thesis contains a more detailed derivation in general cases, including background electromagnetic field), which is the order of chiral anomaly $\partial_x J \sim \hbar \partial_x A \partial_x A$.

On the other hand, in gravitational field, $\partial_x J \sim \hbar (\hbar\partial_x^2 g) (\hbar\partial_x^2 g)$, there is no consistent smallness counting that makes it work. There is some counting that allows one to derive the gravitational chiral anomaly if the curvature is purely spatial, but at the sacrifice of Lorentz invariance (since this is kind of unsatisfactory, this is not written up anywhere).

A similar issue exists for non-abelian gauge field, where $\hbar F\sim \hbar\partial_x A + A^2$. For consistency of power counting, one might count $A \sim \hbar\partial_x$ (an issue that did not exist in the abelian case), hence the non-abelian chiral anomaly exhibits the same issue as the gravitational one. For instance in this paper mentioned in the other answer, they obtained the non-abelian chiral anomaly in a non-Lorentz invariant model and find the correct result; but there is no modification of the semi-classical model that can make it Lorentz invariant.

general relativity - How exactly does curved space-time describe the force of gravity?

I understand that people explain (in layman's terms at least) that the presence of mass "warps" space-time geometry, and this causes gravity. I have also of course heard the analogy of a blanket or trampoline bending under an object, which causes other objects to come together, but I always thought this was a hopelessly circular explanation because the blanket only bends because of "real" gravity pulling the object down and then pulling the other objects down the sloped blanket. In other words, to me, it seems that curved space wouldn't have any actual effect on objects unless there's already another force present.

So how is curved space-time itself actually capable of exerting a force (without some source of a fourth-dimensional force)?

I apologize for my ignorance in advance, and a purely mathematical explanation will probably go over my head, but if it's required I'll do my best to understand.

Answer

Luboš's answer is of course perfectly correct. I'll try to give you some examples why the straightest line is physically motivated (besides being mathematically exceptional as an extremal curve).

Image a 2-sphere (a surface of a ball). If an ant lives there and he just walks straight, it should be obvious that he'll come back where he came from with his trajectory being a circle. Imagine a second ant and suppose he'll start to walk from the same point as the first ant and at the same speed but into a different direction. He'll also produce circle and the two circles will cross at two points (you can imagine those circles as meridians and the crossing points as a north resp. south poles).

Now, from the ants' perspective who aren't aware that they are living in a curved space, this will seem that there is a force between them because their distance will be changing in time non-linearly (think about those meridians again). This is one of the effects of the curved space-time on movement on the particles (these are actually tidal forces). You might imagine that if the surface wasn't a sphere but instead was curved differently, the straight lines would also look different. E.g. for a trampoline you'll get ellipses (well, almost, they do not close completely, leading e.g. to the precession of the perihelion of the Mercury).

So much for the explanation of how curved space-time (discussion above was just about space; if you introduce special relativity into the picture, you'll get also new effects of mixing of space and time as usual). But how does the space-time know it should be curved in the first place? Well, it's because it obeys Einstein's equations (why does it obey these equations is a separate question though). These equations describe precisely how matter affects space-time. They are of course compatible with Newtonian gravity in low-velocity, small-mass regime, so e.g. for a Sun you'll obtain that trampoline curvature and the planets (which will also produce little dents, catching moons, for example; but forget about those for a moment because they are not that important for the movement of the planet around the Sun) will follow straight lines, moving in ellipses (again, almost ellipses).

newtonian mechanics - Why does an unhinged body rotate about its centre of mass?

If a force is applied to a body which does not act through its centre of mass, it rotates about its centre of mass and not any other point. Why?

Answer

The idea is that if there are no forces on an object, then no matter how it rotates, its center of mass must move at a constant velocity. Then in the frame of the object, the center of mass appears stationary and everything else rotates around it. In general this cannot be said of any other point in the object.

To see that the center of mass moves at a constant velocity, remember that the center of mass $\renewcommand{\r}{\mathbf{r}} \renewcommand{\rcm}{\r_\textrm{cm}} \rcm$ is defined by $\rcm = \frac{1}{M}\int \r \rho d \r$, where $\rho$ is the mass density distribution of the object and $M = \int \rho d \mathbf{r}$ is the total mass of the object. Then the center of mass velocity $\renewcommand{\a}{\mathbf{a}} \renewcommand{\acm}{\a_\textrm{cm}} \acm$ can be found by taking two time derivatives: $\acm = \frac{1}{M}\int \a \rho d \r$.

Now the integrand $\a(\r) \rho(\r)$, by newton's second law must be the total force $\renewcommand{\F}{\mathbf{F}}\F$ at the point $\r$. This force has two contributions: an external force $\renewcommand{\Fext}{\F_\mathrm{ext}}\Fext$ and an internal force $\renewcommand{\Fint}{\F_\mathrm{int}}\Fint$. Thus we have that $\a(\r) \rho(\r) = \F(\r) = \Fint(\r) + \Fext(\r)$. Plugging this back into our expression for $\acm$, we find $\acm = \frac{1}{M} \int \Fint(\r)+\Fext(\r) d \r = \frac{1}{M} \Fext + \frac{1}{M} \int \Fint(\r) d\r$, where $\Fext$ is the total external force.

Now the internal force comes from pairwise interactions with other parts of the object. So if $\Fint(\r, \r')$ is the force of the piece of the object at $\r'$ on the piece of the object at $\r$, then the total internal force at $\r$ is given by $\Fint(\r)=\int \Fint(\r,\r') d \r'$. Then the total contribution of $\Fint$ to $\acm$ can be written $\frac{1}{M} \int \Fint(\r) d\r = \frac{1}{M} \int \Fint(\r,\r') d\r' d\r$. But switching the order of integration, we find $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r = \frac{1}{M} \int \Fint(\r,\r') d\r d\r'$ by Newton's third law we have $\Fint(\r,\r')=-\Fint(\r',\r)$. Combining this with the previous equation we find $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r = -\frac{1}{M} \int \Fint(\r',\r) d\r d\r'$. But then relabelling the dummy variables on the right hand side, we find that $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r = -\frac{1}{M} \int \Fint(\r,\r') d\r' d\r$. which implies that $\frac{1}{M} \int \Fint(\r,\r') d\r' d\r =0$. Therefore $\frac{1}{M} \int \Fint(\r) d\r =0$, and so $\acm = \frac{1}{M} \Fext $. Therefore if the external force is zero, the center of mass moves with constant velocity.