general relativity - Are objects in a gravitational well shortened?

Bob is in a gravitational potential well, he moves a long vertical stick up and down a distance of 1 meters. Alice observes the upper end of the stick, at upper location.

There is the phenomenon of redshift of energy, so either Alice measures a shorter motion than 1 meters, or alternatively Alice measures a weakened force.

So I'm asking: is Bob weak, or is Bob short?

( EDIT: It interests me what REALLY happens. It seems that when you fall into a black hole, you maybe become contracted: your lower part is already slowed down, while your upper part is still moving a little bit faster.

I'm certain that when you are being lowered into a gravity well, you REALLY lose energy. Therefore there REALLY is no redshift of energy when energy travels back up from a gravity well.

Now I try to make a thought experiment out of these two things: a shortened person in a gravity well, sends his reduced energy up, by poking with a stick )

EDIT 2: Question 2: Bob moves the vertical stick horizontally a 10 meters distance, what distance and what force does Alice observe at the upper end of the stick?

rocket science - Halley's Comet as a "Free Taxi"

Now that the ESA has landed a probe on a comet, namely Rosetta's Philae Probe, could we possibly land a probe on Halley's comet? Fuel seems to be a limiting factor for interstellar expeditions - could using Halley's comet as a free taxi, shutting down all equipment once landed to save fuel and energy, aid in such expeditions?

general relativity - Lagrangian for relativistic massless point particle

For relativistic massive particle, the action is $$S ~=~ -m_0 \int ds ~=~ -m_0 \int d\lambda ~\sqrt{ g_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}} ~=~ \int d\lambda \ L,$$ where $ds$ is the proper time of the particle; $\lambda$ is the parameter of the trajectory; and we used Minkowski signature $(+,-,-,-)$. So what is the action for a massless particle?

Answer

I) The equation of motion for a scalar massless relativistic point particle on a Lorentzian manifold $(M,g)$ is

$$ \tag{A} \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~\approx ~0, $$

where dot denotes differentiation wrt. the world-line parameter $\tau$ (which is not proper time). [Here the $\approx$ symbol means equality modulo eom.] Thus a possible action is

$$ \tag{B} S[x,\lambda ]~=~\int\! d\tau ~L, \qquad L~=~\lambda ~\dot{x}^2, $$

where $\lambda(\tau)$ is a Lagrange multiplier. This answer (B) may seem like just a cheap trick. Note however that it is possible by similar methods to give a general action principle that works for both massless and massive point particles in a unified manner, cf. e.g. Ref. 1 and eq. (3) in my Phys.SE here.

II) The corresponding Euler-Lagrange (EL) equations for the action (B) reads

$$ \tag{C} 0~\approx ~\frac{\delta S}{\delta\lambda}~=~\dot{x}^2, $$

$$ \tag{D} 0~\approx ~-\frac{1}{2}g^{\sigma\mu}\frac{\delta S}{\delta x^{\mu}} ~=~\frac{d(\lambda\dot{x}^{\sigma})}{d\tau} +\lambda\Gamma^{\sigma}_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}. $$

III) The action (B) is invariant under world-line reparametrization $$ \tag{E} \tau^{\prime}~=~f(\tau), \qquad d\tau^{\prime} ~=~ d\tau\frac{df}{d\tau},\qquad \dot{x}^{\mu}~=~\dot{x}^{\prime\mu}\frac{df}{d\tau},\qquad \lambda^{\prime}~=~\lambda\frac{df}{d\tau}.\qquad$$ Therefore we can choose the gauge $\lambda=1$. Then eq. (D) reduces to the familiar geodesic equation.

References:

1. J. Polchinski, String Theory Vol. 1, 1998; eq. (1.2.5).

general relativity - What techniques can be used to prove that a spacetime is not asymptotically flat?

The modern coordinate-indepenent definition of asymptotic flatness was introduced by Geroch in 1972. You can find presentations in Wald 1984 and Townsend 1997. The definition is in terms of existence of something -- a certain type of conformal compactification -- and in many cases such as the Schwarzschild metric, it's straightforward to think up a construction that proves existence. But to prove that a spacetime is not asymptotically flat, you have to prove the nonexistence of any such compactification, which seems harder.

What techniques could be used to prove that a spacetime is not asymptotically flat?

The only example of such a technique that I've been able to think of is the following. Suppose we want to prove that an FLRW spacetime is not asymptotically flat. Wald gives a theorem by Ashtekar and Hansen that says that if a spacelike submanifold contains $i^0$, it admits a metric that is nearly Euclidean, so that, e.g., the spatial Ricci tensor falls off like $O(1/r^3)$. This implies that there can't be any lower bound on the Ricci tensor, but an FLRW spacetime can have such a lower bound on a constant-time slice, since such a slice has constant spatial curvature.

Townsend, http://arxiv.org/abs/gr-qc/9707012

Wald, General Relativity

What's the longest wavelength single photon we've ever observed?

At short wavelengths, our light detectors tend to be photon counters (e.g. CCD light sensor). At longer wavelengths, we use antennas and bolometers that tend to add up energy instead of counting photons. What is the longest wavelength at which we have "single isolated photon" type detection? Or, at least low number of discrete photons counted directly?

Answer

One cannot detect single energy events when there is a lot of thermal noise. So the detector needs to be cold to detect small quanta of energy.

Fortunately, there is a nice mechanism at low temperatures: some materials go superconducting, and this opens a superconducting gap in the density of states around the Fermi level. The size of the gap $2\Delta = 3.53 k_BT_c$ according to BCS theory, which is of the order of 1 meV. Excitations across this gap can be measured electronically. Not just counted, one can even do spectroscopy.

I do not know what the current world record is. But here is a recent article measuring wavelengths up to $\lambda = 90\ \mu$m: http://ieeexplore.ieee.org/document/7151327/

lagrangian formalism - Invariance of action $Rightarrow$ covariance of field equations?

Invariance of action $\Rightarrow$ covariance of field equations? Is this statement true?

I have only seen examples of this, like the invariance of Electromagnetic action under Lorentz transformations.

How could we prove it?

The action is a scalar, $S$, so how I can't even transform it like $U^{-1}SU$...

Answer

In this answer we formally show that a (quasi)symmetry of an action implies a corresponding symmetry of its EOM$^{\dagger}$. The answer does not discuss form covariance of EOM. For further relations between symmetries of action, EOM, and solutions of EOM, see e.g. this Phys.SE post.

Let us first recall the definition of a quasi-symmetry of the action

$$\tag{1} S_V[\phi]~:=~\int_V \! \mathbb{L}, \qquad \mathbb{L}~:=~{\cal L}~d^nx.$$

It means that the action (1) changes by a boundary integral

$$\tag{2} S_{V^{\prime}}[\phi^{\prime}] +\int_{\partial V^{\prime}} \!d^{n-1}x~(\ldots) ~=~S_V[\phi]+ \int_{\partial V} \!d^{n-1}x~(\ldots) $$

under the transformation. In the following we will assume that the spacetime integration region $V$ is arbitrary.

Theorem. If a local action functional $S_V[\phi]$ has a quasi-symmetry transformation $$\tag{3} \phi^{\alpha}(x)~~\longrightarrow~~ \phi^{\prime \alpha}(x^{\prime}), \qquad x^{\mu}~~\longrightarrow~~x^{\prime \mu}, $$ then the EOM $$\tag{4} e_{\alpha}(\phi(x),\partial\phi(x),\ldots ; x)~:=~\frac{\delta S_V[\phi]}{\delta \phi^{\alpha}(x)}~\approx~0$$ must have a symmetry (wrt. the same transformation) $$\tag{5} e_{\alpha}(\phi^{\prime}(x^{\prime}),\partial^{\prime}\phi^{\prime}(x^{\prime}),\ldots ; x^{\prime})~\approx~e_{\alpha}(\phi(x),\partial\phi(x),\ldots ; x). $$

I) Formal finite proof: This works both for a discrete and a continuous quasi-symmetry.

$$ e_{\alpha}(\phi(x),\partial\phi(x),\ldots ; x) ~:=~\frac{\delta S_V[\phi]}{\delta \phi^{\alpha}(x)} ~\stackrel{(2)}{=}~\frac{\delta S_{V^{\prime}}[\phi^{\prime}]}{\delta \phi^{\alpha}(x)}~\stackrel{{\ddagger}}{\sim}~\int_{V^{\prime}}\!d^nx^{\prime}~\frac{\delta S_{V^{\prime}}[\phi^{\prime}]}{\delta \phi^{\prime\alpha}(x^{\prime})} \frac{\delta \phi^{\prime\alpha}(x^{\prime})}{\delta \phi^{\alpha}(x)}$$ $$\tag{6}~=~\int_{V^{\prime}}\!d^nx^{\prime}~e_{\alpha}(\phi^{\prime}(x^{\prime}),\partial^{\prime}\phi^{\prime}(x^{\prime}),\ldots ; x^{\prime}) \frac{\delta \phi^{\prime\alpha}(x^{\prime})}{\delta \phi^{\alpha}(x)}. $$

II) Formal infinitesimal proof: This only works for a continuous quasi-symmetry. From the infinitesimal transformation (3)

$$\tag{7} \delta \phi^{\alpha}(x)~:=~\phi^{\prime \alpha}(x^{\prime})-\phi^{\alpha}(x), \qquad \delta x^{\mu}~:=~x^{\prime \mu}-x^{\mu},$$

we define a so-called vertical transformation

$$\tag{8} \delta_0 \phi^{\alpha}(x)~:=~\phi^{\prime \alpha}(x)-\phi^{\alpha}(x)~=~\delta \phi^{\alpha}(x)-\delta x^{\mu} ~d_{\mu}\phi^{\alpha}(x),\qquad d_{\mu}~:=~\frac{d}{dx^{\mu}}, \qquad $$

which transforms of the fields $\phi^{\alpha}(x)$ without transforming the spacetime points $x^{\mu}$. The quasi-symmetry implies that the Lagrangian $n$-form $\mathbb{L}$ transforms with a total spacetime derivative

$$\tag{9} \delta \mathbb{L}~=~d_{\mu} f^{\mu}~d^nx, \qquad \delta_0 \mathbb{L}~=~d_{\mu}(f^{\mu}-{\cal L}~\delta x^{\mu})~d^nx. $$

The EOM (4) are typically of second order, so let us assume this for simplicity. (This assumption is not necessary.) Then the infinitesimal transformation of EOM (4) reads

$$ \delta e_{\alpha}(x)~=~\delta_0 e_{\alpha}(x) +\delta x^{\mu} ~\underbrace{d_{\mu} e_{\alpha}(x)}_{\approx 0}~\approx~\delta_0 e_{\alpha}(x) \qquad $$ $$~=~\frac{\partial e_{\alpha}(x)}{\partial\phi^{\beta}(x)}\delta_0\phi^{\beta}(x) +\sum_{\mu}\frac{\partial e_{\alpha}(x)}{\partial(\partial_{\mu}\phi^{\beta}(x))}d_{\mu}\delta_0\phi^{\beta}(x) +\sum_{\mu\leq \nu }\frac{\partial e_{\alpha}(x)}{\partial(\partial_{\mu}\partial_{\nu}\phi^{\beta}(x))}d_{\mu}d_{\nu}\delta_0\phi^{\beta}(x) $$ $$~\stackrel{{\ddagger}}{\sim}~ \int_V\! d^ny~ \delta_0\phi^{\beta}(y)\frac{\delta e_{\alpha}(x)}{\delta \phi^{\beta}(y)} ~=~\int_V\! d^ny~ \delta_0\phi^{\beta}(y)\frac{\delta^2 S_V[\phi]}{\delta \phi^{\beta}(y)\delta \phi^{\alpha}(x)} $$ $$~=~ \int_V\! d^ny~ \delta_0\phi^{\beta}(y)\frac{\delta^2 S_V[\phi]}{\delta \phi^{\alpha}(x)\delta\phi^{\beta}(y)} $$ $$~=~ \frac{\delta}{\delta \phi^{\alpha}(x)} \int_V\! d^ny~ \delta_0\phi^{\beta}(y)\frac{\delta S_V[\phi]}{\delta \phi^{\beta}(y)} -\int_V\! d^ny~ \frac{\delta(\delta_0\phi^{\beta}(y))}{\delta \phi^{\alpha}(x)} \frac{\delta S[\phi]}{\delta \phi^{\beta}(y)} $$ $$~\sim~ \frac{\delta(\delta_0 S_V[\phi]) }{\delta \phi^{\alpha}(x)} -\int_V\! d^ny~ \frac{\delta(\delta_0\phi^{\beta}(y))}{\delta \phi^{\alpha}(x)} e_{\beta}(y) $$ $$\tag{10} ~\approx~ \frac{\delta(\delta_0 S_V[\phi]) }{\delta \phi^{\alpha}(x)}~=~0. $$

In the very last step of eq. (10) we used that the infinitesimal variation

$$\tag{11} \delta_0 S_V[\phi]+\int_V\! d^nx~d_{\mu} \left({\cal L}~\delta x^{\mu} \right) ~=~\delta S_V[\phi]~=~\int_{\partial V} \!d^{n-1}x~(\ldots)$$

of the action is a boundary integral by assumption (2), so that its functional derivative (10) must vanish (if it exists).

--

$^{\dagger}$ Terminology and Notation: Equations of motion (EOM) means Euler-Lagrange equations (1). The words on-shell and off-shell refer to whether EOM are satisfied or not. The $\approx$ symbol means equality modulo EOM.

$^{\ddagger}$ Warning: This step is not always justified. The $\sim$ symbol indicates that we have formally integrated by part and ignored boundary contributions. Also we have assumed that the pertinent functional derivative is well-defined and exists. This caveat is the main shortcoming of the formal proof given here. The point is quite serious, e.g. in the case of a global (=$x$-independent) variation, which typically doesn't vanish on the boundary. So boundary contributions could in principle play a role.

However, instead of using functional derivatives and integrations, it is possible to prove eq. (10) $x$-locally

$$ \delta_0 e_{\alpha}(x)~=~\ldots~=~\underbrace{E_{\alpha(0)} d_{\mu}}_{=0}\left(f^{\mu}(x)-{\cal L}(x)~\delta x^{\mu}\right) - \sum_{k\geq 0} d^k\left( \underbrace{e_{\beta}(x)}_{\approx 0} \cdot P_{\alpha(k)}\delta_0\phi^{\beta}(x) \right)$$ $$\tag{12} ~\approx~ 0 $$

using only higher partial field derivatives

$$\tag{13} P_{\alpha(k)} ~:=~\frac{\partial }{\partial \phi^{\alpha(k)}}, \qquad k~\in~\mathbb{N}_0^n,$$

and higher Euler operators

$$\tag{14} E_{\alpha(k)} ~:=~\sum_{m\geq k} \begin{pmatrix} m \cr k\end{pmatrix}(-d)^m P_{\alpha(m)}, $$

that all refer to the same spacetime point $x$. This $x$-local approach circumvent the problem of un-accounted boundary contributions.

homework and exercises - Surface charge density relation with the radius of curvature at the surface of a conductor

In a text book it was given that the surface charge density of a conductor at a particular region on its surface is inversely proportional to the radius of curvature at that region. I didn't understand it. I'd like to see a proof for that can anyone please post the proof.

general relativity - Why do the Einstein field equations (EFE) involve the Ricci curvature tensor instead of Riemann curvature tensor?

I am just starting to learn general relativity. I don't understand why we use the Ricci curvature tensor. I thought the Riemann curvature tensor contains "more information" about the curvature. Why is that extra information so to speak irrelevant?

Answer

I think this question is more trivial than you think.

You should ask yourself why should the full Riemann tensor appear. I'll sketch a heuristic derivation of the field equations.

We know that with small velocities and a static field, the Poisson equation $$\Delta\phi=4\pi G\rho$$ is approximately satisfied. From special relativity we know that the mass/energy density $\rho$ must change with two Lorentz factors under a Lorentz transformation. Thus it is the time-time component of a rank two tensor $T_{\mu\nu}$. Using the equivalence principle, we promote this to a curved spacetime tensor. When we look for field equations, we demand that they be tensor equations. For one thing, this means we must have the same number of indices on both sides. We posit $$D_{\mu\nu}=\kappa T_{\mu\nu}$$ with $\kappa$ some constant. We don't know what $D_{\mu\nu}$ is, but the principle of covariant conservation fixes it to be the Einstein tensor. Note that the general form is the natural generalization of the Poisson equation.

You might propose an equation with more indices, such as $$R_{\mu\nu\rho\sigma}=\kappa'T_{\mu\nu}T_{\rho\sigma}$$ with some appropriate antisymmetrization scheme. What are the vacuum equations? They would be $$R_{\mu\nu\rho\sigma}=0$$ But this just says spacetime is flat! We know this is incorrect. Black holes are certainly vacuum solutions but are also certainly not flat spacetime solutions.

In summary, the Ricci tensor has the ability to vanish without the full Riemann tensor vanishing. The general form of the equations is determined by the Poisson equation to be a rank two equation. In my mind, these two facts are the most effective argument.

Introduction to relativity books

Possible Duplicate:
Getting started general relativity

I am an engineer who loves to read science fiction books especially when there's more science than fiction but usually I see that I lack the knowledge behind many of the relativity concepts in the novels. I've always felt curiosity in relativity so I decided that it was the time to buy an introductory book. I went to amazon and checked that there are dozens of "Introduction to general relativity" books in there. So here's my question: What book would you recommend to someone not matematician nor physicist but with (some) mathematical background as an introduction to relativity?

Answer

The answer to the question depends largely on various factors, therefore there is no point to list all books I know of. A couple books, however, stand out for an "informed" laymen. One by George Ellis and Ruth Williams, and another by Bob Geroch.

http://www.amazon.com/Flat-Curved-Space-Times-George-Ellis/dp/0198506562/ref=ntt_at_ep_dpt_1

http://www.amazon.com/General-Relativity-B-Robert-Geroch/dp/0226288641/ref=sr_1_1?s=books&ie=UTF8&qid=1316698577&sr=1-1

Both are elementary, but very lucid and didn't sacrifice sophistication of physical principles behind. I would, indeed, recommend both to any serious students of relativity.

Another good source [will be better soon] is relativity site of ComPadre, The spacetime Emporium.

http://www.compadre.org/relativity/

You may find many online sources there.

general relativity - Will an object always fall at an infinite speed in a black hole?

Most of you if not everybody will agree that the stronger the gravitational pull, the faster an object will fall. For example, on a planet with 50 times the gravity of Earth, any object will hit the ground on that planet much quicker than it would on Earth. So taking all of these into the equation, does it mean that at a blackhole, an object will fall at an infinite speed because of the infinitely strong gravitational pull of the blackhole?

Answer

does it mean that at a blackhole, an object will fall at an infinite speed because of the infinitely strong gravitational pull of the blackhole?

No.

Actually, defining exactly what you mean by the speed an object falls into a black hole is a tricky problem. In relativity you generally find that different observers observe different things. But we can work out what the various observers will see. Let's assume that the black hole is static so the geometry around it is described by the Schwarzschild metric. The task then is to calculate the orbits for objects moving in this spacetime. This is relatively simple by the standards of GR calculations, and you'll find it done in any introductory work on GR, but it's still a bit involved for non-nerds so I'll just quote the results.

If you sit a long way from the black hole and watch an object falling into it from far away then the velocity of the object will be related to distance from the black hole by:

$$ v = \left(1 - \frac{r_s}{r}\right)\sqrt{\frac{r_s}{r}}c \tag{1} $$

where $r_s$ is the Schwarzschild radius. If we graph the velocity as a function of distance from the black hole we get:

The $x$ axis shows distance in Schwarzschild radii while the $y$ axis is the speed as a fraction of the speed of light. The speed peaks at about $0.38c$ then falls as you get nearer to the event horizon and falls to zero at the horizon. This is the source of the notorious claim that nothing can fall into a black hole.

An alternative strategy might be to hover at some distance $r$ from the black hole and measure the speed at which the falling object passes you. These observers are known as shell observers. If you do this you find a completely different variation of speed with distance:

$$ v = \sqrt{\frac{r_s}{r}}c \tag{2} $$

This time the variation of the speed with distance looks like:

and this time the speed goes to $c$ as you approach the horizon. The difference between the two is because time slows down near a black hole, so if you're hovering near the event horizon velocities look faster because your time is running slower. You might be interested to note that the velocity calculated using equation (2) is equal to the Newtonian escape velocity. The event horizon is the distance where the escape velocity rises to the speed of light.

The last observer is the falling observer i.e. the one who's falling into the black hole. But here we find something even stranger. The falling observer will never observe themselves crossing an event horizon. If you're falling into a black hole you will find an apparent horizon retreats before you as you fall in and you'll never cross it. You and the horizon will meet only as you hit the singularity.

cosmology - Does an object's redshift actually decrease with time?

I am trying to determine how an object's redshift (specifically, redshift due just to the expansion of the universe) changes in time. Starting with a definition of the Hubble parameter,

$$H \equiv \frac{\dot a}{a}$$

with $a$ being the scale factor, we can write

$$\dot a = Ha~.$$

We can calculate $\dot z$ in terms of $\dot a$. Since $a=(1+z)^{-1}$,

$$\dot a = -(1+z)^{-2}\dot z~.$$

Plugging $a$ and $\dot a$ into the first or second equation I wrote here we can find

$$\dot z = - H(1+z)~.$$

This negative sign is a bit surprising to me. I would have expected that $\dot z$ would have been positive, i.e., that an object's redshift increases with time. I would have expected this from the fact alone that the universe is expanding, but perhaps I am wrong in this thought. If so, please tell me how. However, the expansion of the universe is currently accelerating, and so I would expect from this as well that $\dot z$ would be positive, since at later times things will appear to be moving away from us at a faster rate than they are now. Is there some sort of cosmological constant dependence I did not take into account in my derivation above?

My question in summary: why is there a negative sign in the equation for $\dot z$? Did I derive the expression incorrectly? Or am I wrong in thinking it should be positive?

Answer

The redshift of a source actually changes in a more complicated way: when the source entered our cosmological horizon (i.e. at the moment its light reached Earth for the first time), its redshift was $\infty$, because it was located at the edge of our observable universe. Over time, this redshift then decreases to a minimum value, but eventually the expansion of the universe causes it to increase again. In the far future, all sources will be redshifted back to $\infty$ (in the Standard $\Lambda\text{CDM}$ Model).

Let's derive the correct formula. For more details, I refer to this post: https://physics.stackexchange.com/a/63780/24142

The Hubble parameter in the $\Lambda\text{CDM}$ Model is $$ H(a) = H_0\sqrt{\Omega_{R,0}\,a^{-4} + \Omega_{M,0}\,a^{-3} + \Omega_{K,0},a^{-2} + \Omega_{\Lambda,0}}\;, $$ with $\Omega_{K,0} = 1 - \Omega_{R,0} - \Omega_{M,0} - \Omega_{\Lambda,0}$.

The observed redshift $z_\text{ob}=z(t_\text{ob})$ of a source at a time $t_\text{ob}$ is given by $$ 1 + z_\text{ob} = \frac{a_\text{ob}}{a_\text{em}}, $$ with $a_\text{ob} = a(t_\text{ob})$ the scale factor at the time of observation, and $a_\text{em} = a(t_\text{em})$ the scale factor at the time $t_\text{em}$, when the source emitted the light that was observed at $t_\text{ob}$. From this, we can write $a_\text{em}$ as a function of $z_\text{ob}$ and $a_\text{ob}$: $$ a_\text{em} = \frac{a_\text{ob}}{1 + z_\text{ob}}.\tag{1} $$ When the source moves with the Hubble flow, its co-moving distance remains constant: $$ D_\text{c}(z(t_\text{ob}),t_\text{ob}) = c\int_{a_\text{em}}^{a_\text{ob}}\frac{\text{d}a}{a^2H(a)} = \text{const}. $$ Therefore, if we treat $t_\text{ob}$ as a variable, the total derivative with respect to $t_\text{ob}$ is zero: $$ \dot{D}_\text{c} = \frac{\text{d} D_\text{c}}{\text{d} t_\text{ob}} = 0, $$ which means that, with Leibniz' integral rule, $$ \frac{\dot{a}_\text{ob}}{a_\text{ob}^2H(a_\text{ob})} = \frac{\dot{a}_\text{em}}{a_\text{em}^2H(a_\text{em})}. $$ or, with $H(a_\text{ob})= \dot{a}_\text{ob}/a_\text{ob}$, $$ \dot{a}_\text{em} = \frac{a_\text{em}^2}{a_\text{ob}}H(a_\text{em}).\tag{2} $$ We also have from eq. (1): $$ \dot{a}_\text{em} = \frac{\dot{a}_\text{ob}}{1 + z_\text{ob}} - \frac{a_\text{ob}\,\dot{z}_\text{ob}}{(1 + z_\text{ob})^2}. $$ Inserting this into eq. (2), we find $$ \dot{z}_\text{ob} = (1 + z_\text{ob})\frac{\dot{a}_\text{ob}}{a_\text{ob}} - \frac{a_\text{em}^2}{a^2_\text{ob}}(1 + z_\text{ob})^2H(a_\text{em}), $$ which simplifies to $$ \dot{z}_\text{ob} = (1 + z_\text{ob})H(a_\text{ob}) - H(a_\text{em}). $$ In particular, if we take the present day as the time of observation, we have $$ \dot{z} = (1+z)H_0 - H\!\left(\!\frac{1}{1+z}\!\right). $$ Since $H(a)$ decreases as a function of $a$, if follows that $\dot{z}_\text{ob} < 0$ if $z_\text{ob}$ is very large (and $a_\text{ob}$ is sufficiently small), and $\dot{z}_\text{ob} > 0$ if $z_\text{ob}$ is small or $a_\text{ob}$ is large.

This also means that there's a redshift at any time at which $\dot{z}_\text{ob} = 0$. Using the same values of the cosmological parameters as in my reference post, I find that this 'transition redshift' is currently $z=1.92$. In other words, the redshift of a galaxy with present-day redshift $z<1.92$ is increasing, while the redshift of a galaxy with $z>1.92$ is currently decreasing.

Also take a look at the diagram in my reference post: the dashed lines represent contours of constant $z_\text{ob}$ at a given time of observation; galaxies move vertically (dotted lines). You'll see the same thing: when a galaxy crosses the particle horizon, its redshift is $\infty$, after which it decreases, but in the (far) future it will increase again.

See also Eq. (11) in the paper Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the Universe by Davis & Lineweaver.

rotational kinematics - Elemetary Rotations -imagining differential rotations - intitutive proof of such rotations being vectors

I was reading and extract from "Fundamentals Laws of Mechanics", 1980, by I.E Irodov, § 1.2. 'Kinematics of a Solid' and came across this insane text where the author asks me to imagine a 'solid performing two elementary rotations'.

Moreover, the vector introduced ($d\boldsymbol{\phi}$) can be shown to satisfy the basic property of vectors, that is, vector addition. Indeed, imagine a solid performing two elementary rotations, $d\boldsymbol{\phi}_1$ and $d\boldsymbol{\phi}_2$, about different axes crossing at a stationary point O. The resultant displacement $d\mathbf{r}$ of an arbitrary point A of the body, whose radius vector with respect to the point O is equal to $\mathbf{r}$, can be represented as follows: $$d\mathbf{r} = d\mathbf{r}_1 + d\mathbf{r}_2 = [d\boldsymbol{\phi}_1, \mathbf{r}] + [d\boldsymbol{\phi}_2, \mathbf{r}] = [d\boldsymbol{\phi}, \mathbf{r}]$$ $where$ $$d\boldsymbol{\phi}= d\boldsymbol{\phi}_1 + d\boldsymbol{\phi}_2 \tag{1.12} $$ i.e: the two given rotations, $d\boldsymbol{\phi}_1$ and $d\boldsymbol{\phi}_2$, are equivalent to one rotation through the angle $d\boldsymbol{\phi}= d\boldsymbol{\phi}_1 + d\boldsymbol{\phi}_2$ about the exis coinciding with the vector $d\boldsymbol{\phi}$ and passing through the point O.

I need help trying to imagine/visualize two 'infinitesimal rotations'. I have managed to prove that finite rotations do not obey the law of vector addition by doing two finite rotations on an imaginary object but I am unable to do the same with infinitesimal rotations and verify that infinitesimal rotations, indeed, obey the law of vector addition.

Please explain what the author is trying to convey. I do understand that infinitesimal rotations are vectors but I find it really hard to comprehend the passage given above.

Answer

All vectors, except $\:\mathbf{r}\:$, are infinitesimals.
I wonder if the author (Irodov) makes use of this result anywhere in his textbook.

---

EDIT

The infinitesimal rotation of a vector $\:\mathbf{r}\:$ around the direction of a unit vector $\:\mathbf{n}=\left(n_{1},n_{2},n_{3}\right)\:$ by an infinitesimal angle $\:\mathrm{d}\theta\:$ may be represented as follows :

\begin{equation} \mathbf{r}' =\mathbf{r}+\mathrm{d} \mathbf{r} =\mathbf{r}+\mathrm{d}\theta\mathbf{n}\boldsymbol{\times}\mathbf{r}=\left(\mathrm{I}+\mathrm{d}\theta\mathbf{n}\boldsymbol{\times}\right)\mathbf{r} \tag{01} \end{equation}

The infinitesimal rotation matrix is explicitly expressed as

\begin{equation} \mathrm{I}+\mathrm{d}\theta\mathbf{n}\boldsymbol{\times}= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} +\mathrm{d}\theta \begin{bmatrix} 0 & -n_{3} & n_{2} \\ n_{3} & 0 & -n_{1} \\ -n_{2}& n_{1} & 0 \end{bmatrix} = \begin{bmatrix} 1 & -n_{3}\mathrm{d}\theta & n_{2}\mathrm{d}\theta \\ n_{3}\mathrm{d}\theta & 1 & -n_{1}\mathrm{d}\theta \\ -n_{2}\mathrm{d}\theta& n_{1}\mathrm{d}\theta & 1 \end{bmatrix} \tag{02} \end{equation}

Now, let the two infinitesimal rotations of the question :

\begin{align} \mathrm{I}+d\boldsymbol{\phi}_{1}\boldsymbol{\times} & =\mathrm{I}+d\phi_{1} \mathbf{n}_{1}\boldsymbol{\times} \tag{03a} \\ \mathrm{I}+d\boldsymbol{\phi}_{2}\boldsymbol{\times} & =\mathrm{I}+d\phi_{2} \mathbf{n}_{2}\boldsymbol{\times} \tag{03b} \end{align}

The composition of these rotations is : \begin{align} & \left(\mathrm{I}+d\boldsymbol{\phi}_{1}\boldsymbol{\times}\right)\left(\mathrm{I}+d\boldsymbol{\phi}_{2}\boldsymbol{\times}\right) \\ & =\mathrm{I}+d\boldsymbol{\phi}_{1}\boldsymbol{\times}+d\boldsymbol{\phi}_{2}\boldsymbol{\times}+d\boldsymbol{\phi}_{1}\boldsymbol{\times}(d\boldsymbol{\phi}_{2}\boldsymbol{\times}\\ &=\mathrm{I}+\underbrace{\left(d\boldsymbol{\phi}_{1}+d\boldsymbol{\phi}_{2}\right)}_{d\boldsymbol{\phi}}\boldsymbol{\times}+d\phi_{1} d\phi_{2}F\left(\mathbf{n}_{1},\mathbf{n}_{2}\right) \\ &=\mathrm{I}+d\boldsymbol{\phi}\boldsymbol{\times}+d\phi_{1} d\phi_{2}F\left(\mathbf{n}_{1},\mathbf{n}_{2}\right) \tag{04} \end{align}

where $\:F\left(\mathbf{n}_{1},\mathbf{n}_{2}\right)\:$ the following finite linear transformation \begin{equation} F\left(\mathbf{n}_{1},\mathbf{n}_{2}\right)\mathbf{r} \equiv \left(\mathbf{n}_{1}\boldsymbol{\cdot}\mathbf{r} \right)\mathbf{n}_{2}-\left(\mathbf{n}_{1}\boldsymbol{\cdot}\mathbf{n}_{2}\right)\mathbf{r} \tag{05} \end{equation}

In equation (04) we have the 2nd order term \begin{equation} d\phi_{1} d\phi_{2}F\left(\mathbf{n}_{1},\mathbf{n}_{2}\right) \ne d\phi_{2} d\phi_{1}F\left(\mathbf{n}_{2},\mathbf{n}_{1}\right) \tag{06} \end{equation}

So, to 2nd order the two infinitesimal rotations don't commute. But to 1rst order ($d\phi_{1} d\phi_{2} \approx 0$) since \begin{equation} d\boldsymbol{\phi} = d\boldsymbol{\phi}_{1}+d\boldsymbol{\phi}_{2} = d\boldsymbol{\phi}_{2}+d\boldsymbol{\phi}_{1} \tag{07} \end{equation}

the two rotations commute and we have a form of vector addition of infinitesimal rotations.

newtonian mechanics - How can an accelerating inclined plane prevent a block on it from sliding?

If I increase the force $F$, only the normal force $N1$ acting on $m1$ would increase which has no component along the plane, ie. along $m1gsin\theta$, so how would applying this force prevent m1 from sliding?

When I view it from the accelerated frame of the wedge,

It begins making sense. How is this possible? I'm really confused.

Answer

Diagram 1 shows the arrangement with the inclined plane stationary.
There are two forces acting on the block, its weight $mg$ and the normal reaction on the block due to the inclined plane $N_1$.
The resultant of theses two forces is $F_1 ( = mg \sin \theta)$ and this force accelerates the block down the slope.

Diagram 2 shows the situation when a force $F$ is applied to the inclined plane and there is no relative movement between the block and the inclined plane.
That is because the resultant of the weight of the block $mg$ and the now increased normal reaction $N_2$ is a horizontal force $F_2$.

If that force $F_2$ on the block produces an acceleration of the block $a$ which is the same as the acceleration of the inclined plane then the block will not move relative to the inclined plane.

When this condition is satisfied $F_2 = ma$ and $F=(m+M)a$

Note that the magnitude of $F$ controls the magnitude of $N_2$ which in turn controls the direction of $F_2$.

If $F$ is larger than in the no relative movement condition then the magnitude of $N_2$ is larger and the block accelerates up the inclined plane whilst if $F$ is smaller then the block accelerates down the slope.

Does the lagrangian contain all the information about the representations of the fields in QFT?

Given the Lagrangian density of a theory, are the representations on which the various fields transform uniquely determined?

For example, given the Lagrangian for a real scalar field $$ \mathscr{L} = \frac{1}{2} \partial_\mu \varphi \partial^\mu \varphi - \frac{1}{2} m^2 \varphi^2 \tag{1}$$ with $(+,-,-,-)$ Minkowski sign convention, is $\varphi$ somehow constrained to be a scalar, by the sole fact that it appears in this particular form in the Lagrangian?

As another example: consider the Lagrangian $$ \mathscr{L}_{1} = -\frac{1}{2} \partial_\nu A_\mu \partial^\nu A^\mu + \frac{1}{2} m^2 A_\mu A^\mu,\tag{2}$$ which can also be cast in the form $$ \mathscr{L}_{1} = \left( \frac{1}{2} \partial_\mu A^i \partial^\mu A^i - \frac{1}{2} m^2 A^i A^i \right) - \left( \frac{1}{2} \partial_\mu A^0 \partial^\mu A^0 - \frac{1}{2} m^2 A^0 A^0 \right). \tag{3}$$ I've heard$^{[1]}$ that this is the Lagrangian for four massive scalar fields and not that for a massive spin-1 field. Why is that? I understand that it produces a Klein-Gordon equation for each component of the field: $$ ( \square + m^2 ) A^\mu = 0, \tag{4}$$ but why does this prevent me from considering $A^\mu$ a spin-1 massive field?

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[1]: From Matthew D. Schwartz's Quantum Field Theory and the Standard Model, p.114:

A natural guess for the Lagrangian for a massive spin-1 field is $$ \mathcal{L} = - \frac{1}{2} \partial_\nu A_\mu \partial_\nu A_\mu + \frac{1}{2} m^2 A_\mu^2,$$ where $A_\mu^2 = A_\mu A^\mu$. Then the equations of motion are $$ ( \square + m^2) A_\mu = 0,$$ which has four propagating modes. In fact, this Lagrangian is not the Lagrangian for a amassive spin-1 field, but the Lagrangian for four massive scalar fields, $A_0, A_1, A_2$ and $A_3$. That is, we have reduced $4 = 1 \oplus 1 \oplus 1 \oplus 1$, which is not what we wanted.

quantum field theory - Physical Interpretation of Lorentz-transformed Single Particle states being linear

As in this question, let $\psi_{p,\sigma}$ be a single-particle 4-momentum eigenstate, with $\sigma$ being a discrete label of other degrees of freedom.

Weinberg discusses the effect of a homogenous Lorentz transformation $U(\Lambda, 0)$ or $U(\Lambda)$ on these states, and concludes that $U(\Lambda)\psi_{p,\sigma}$ is a linear combination of $\psi_{\Lambda p,\sigma'}$.

$$U(\Lambda)\psi_{p,\sigma} = \sum_{\sigma'} C_{\sigma'\sigma}(\Lambda, p)\psi_{\Lambda p,\sigma'}$$

Again, is there any physical information that we can extract from this? (I realize that $\psi_{\Lambda p,\sigma'}$ represent physical states after Lorentz transformation).

Answer

Yes, indeed, first of all it tells you that a particle of momentum $p$ in one frame looks like a particle of momentum $\Lambda p$ in another frame which is related to the first by Lorentz transformation $\Lambda$. That means $p$ is a Lorentz vector and hence gives you more certainty that it indeed captures momentum of the particle.

If you read a few more pages in the book, you will see that $\sigma$ would correspond to spin for a massive particle and helicity for a massless particle. So, the above equation is telling us that not only do you find a different momentum for the particle but also that you might find the particle carrying a different spin (or more concretely the probabilities to find the particle in different spin states would be changed) when you do a Lorentz transformation.

Compare this with a usual case one encounters in QM. We have a massive motionless particle as our physical system. And we now want to study how the physical state transforms under a rotation $$U(R)\psi_{p=0,\sigma} = C_{\sigma',\sigma}(R,0)\psi_{p=0,\sigma'}$$ Here $\sigma$ labels spin of the particle and $U(R)$ forms a representation of rotation group which is generated by the angular momenta $J_1, J_2, J_3$.

So, this equation is telling you that transformation of single particle states in relativistic quantum mechanics is a straightforward union of the transformations that we expect from our earlier study of relativity and quantum mechanics.

general relativity - How does Newton's 2nd law correspond to GR in the weak field limit?

I can only perform the demonstration from the much simpler $E = mc^2$.

Take as given the Einstein field equation:

$G_{\mu\nu} = 8 \pi \, T_{\mu\nu}$

... can it be proved that Newton's formulation of gravitational and mechanical force (e.g. $F = ma$) corresponds to Einstein's in the limit when masses are small and speeds are relatively slow?

electromagnetism - Electromagnetic Unruh effect?

The Unruh temperature is given by $$T=\frac{\hbar\ a}{2\pi c k_B}.$$

If we have an electron field with charge $e$ and mass $m_e$ acted on by an electromagnetic field $\vec{E}$ then very naively maybe we can write a Newton's 2nd law equation

$$\vec{E}\ e=m_e\ \vec{a}.$$

Substituting the magnitude of the acceleration $a$ into the Unruh formula we get

$$T=\frac{\hbar}{2\pi c k_B}\frac{E\ e}{m_e}.$$

If we take the electric field strength $E=1$ MV/m then the Unruh temperature is

$$T\approx 10^{-2}\ \hbox{K}.$$

Perhaps this temperature could be measured?

statistical mechanics - Relativity of temperature paradox

The imagined scenario:

Part A:

From special relativity we know that velocity is a relative physical quantity, that is, it is dependent on the frame of reference of choice. This means that kinetic energy is also relative, but this does not undermine the law of conservation of energy as long as we are consistent with our choice of frame. So far so good.

Part B:

On the other hand, from statistical mechanics, we know that the average kinetic energy of a system and its temperature are directly related by the Boltzmann constant $$ \langle E_k \rangle = \frac{1}{2}m\langle v^2 \rangle = \frac{3}{2} k_B T $$ which leads to conclude that when the notion of temperature in physics is expressed in terms of a system's kinetic energy, then it too ought to be a relative quantity, which is a bit mind-boggling, because I had always thought of temperature as absolute.

Part C:

Furthermore, we know that all objects at non-zero temperature, radiate electromagnetic energy with a wavelength given as a function of the body/object's temperature, this is the Blackbody radiation. Thus in principle, I am able to infer an object's temperature (i.e. the temperature in its own rest-frame of reference) by measuring its emitted radiation, regardless of the frame I find myself in. But this seems to violate the previously expected relativity of temperature as defined by average kinetic energy.

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Proposed resolutions:

The resolutions that I imagine to this paradox are:

• 
  

  a) Depending on the frame of reference from which I measure the emitted blackbody radiation of the object, the radiation will undergo different Doppler blue/red-shifts. Thus the relativity of the temperature in the context of blackbody radiation, is preserved due to the Doppler effect.

  
  


• 
  
  

  b) I suspect that treating temperature as nothing but an average kinetic energy does not in general hold true, and to resolve this paradox, one should work with a more general definition of temperature (which I admit I do not know how in general temperature ought to be defined, if not in terms of state of motion of a system's particles).

  
  

Case a) resolves this hypothetical paradox by including the Doppler effect, but does not contradict the relativity of temperature.

Case b) on the other hand, resolves the problem by challenging the definition that was used for temperature, which in the case that we define temperature more generally, without relating to kinetic energy, may leave temperature as an absolute quantity and not relative to a frame.

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Main question:

• But surely only one can be correct here. Which begs to ask: what was the logical mistake(s) committed in the above scenario? In case there was no mistake, which of the two proposed resolutions are correct? If none, what is then the answer here? Very curious to read your input.

Answer

Temperature is related to kinetic energy in the rest frame of the fluid/gas. In non-relatvistic kinetic theory the distribution function is $$ f(p) \sim \exp\left(-\frac{(\vec{p}-m\vec{u})^2}{2mT}\right) $$ where $\vec{u}$ is the local fluid velocity. The velocity can be found by demanding that the mean momentum in the local rest frame is zero. Then $\vec{u}$ transform as a vector under Galilean transformations, and $T$ is a scalar.

In relativistic kinetic theory $$ f(p) \sim \exp\left(-\frac{p\cdot u}{T}\right) $$ where $p$ is the four-momentum, $u$ is the four-velocity, and $T$ is the temperature scalar. The rest frame is defined by $\vec{u}=0$, and in the rest frame $f\sim \exp(-E_p/T)$, as expected.

The relativistic result is known as the Juttner distribution (Juttner, 1911), and is discussed in standard texts on relativistic kinetic theory, for example Cercignani and Kremer , equ. (2.124), and de Groot et al , equ (ch4)(25). See also (2.120) in Rezzolla and Zanotti. For an intro available online see equ. (55-58) of Romatschke's review. Neumaier notes that some (like Beccatini ) advocate defining a four-vector field $\beta_\mu=u_\mu/T$, and then define a frame dependent temperature $T'\equiv 1/\beta_0$. I fail to see the advantage of this procedure, and it is not what is done in relativistic kinetic theory, hydrodynamics, numerical GR, or AdS/CFT.

Ultimately, the most general definition of $T$ comes from local thermodynamics (fluid dynamics), not kinetic theory, because strongly correlated fluids (classical or quantum) are not described in kinetic theory. The standard form of relativistic fluid dynamics (developed by Landau, and explained in his book on fluid dynamics) also introduces a relarivistic 4-velocity $u_\mu$ (with $u^2=1$), and a scalar temperature $T$, defined by thermodynamic identities, $dP=sdT+nd\mu$. The ideal fluid stress tensor is $$ T_{\mu\nu}=({\cal E}+P) u_\mu u_\nu -Pg_{\mu\nu} $$ where ${\cal E}$ is the energy density and $P$ is the pressure. Note that for a kinetic system the parameter $u_\mu$ in the Juttner distribution is the fluid velocity, as one would expect. More generally, the fluid velocity can be defined by $u^\mu T_{\mu\nu}={\cal E}u_\nu$, which is valid even if dissipative corrections are taken into account.

Regarding the ``paradox'': Temperature is not relative, it is a scalar. The relation in B is only correct in the rest frame. The Doppler effect is of course a real physical effect. The spectrum seen by an observer moving vith relative velocity $v$ is $f\sim\exp(-p\cdot v/T)$, which exhibits a red/blue shift. The spectrum only depends on the relative velocity, as it should. Measuring the spectrum can be used to determine both the relative velocity and the temperature. However, if you look at a distant star you only measure light coming off in one direction. Then, in order to disentangle $u$ and $T$, you need either a spectral line, or information on the absolute luminosity.

optics - Reconciling refraction with particle theory and wave theory

I have searched the web for good answers to why refraction occurs when light moves from one medium to another with different density. I have limited background in physics and want to know if there is an easy way to understand this without having to go back to school for years.

Most explanations come in some variant of the "Marching line of soldiers" analogy, where a straight line of soldiers change pace when they hit an boundary. I have no problem understanding the analogy, my problem is understanding how this is even relevant to light.

• 
  

  The analogy does not work for light as a particle. The particle does not "know" of particles around it, and should follow a straight line.

  
  


• 
  

  How does a single ray of light (wave), as part of a "wave front" know what other rays are doing?

  
  

general relativity - Diffeomorphism invariance and geodesic action

I'm trying to understand the role of diffeomorphism and isometry invariance in the geodesic action in GR:

$$ S = \int_{\tau_1}^{\tau_2} \! d\tau~ g_{ab}(x(\tau)) \frac{dx^a}{d\tau} \frac{dx^a}{d\tau} $$

Now if we transform coordinates with $y = y(x)$ and apply the usual transformation laws of the metric and tangent vectors than it is clear that $$L = g_{ab}(x(\tau)) \frac{dx^a}{d\tau} \frac{dx^a}{d\tau}$$ transforms as a scalar.

I am confused because we can equally consider this coordinate change as an "active" diffeomorphism $\phi: M \to M$ and then the statement that the action transforms as a scalar is that geodesics are mapped to geodesics under an arbitrary diffeomorphism. However I expect that is not true, it should only be true for diffeomorphims such that $\phi^* g = g$ i.e. isometries.

I'd like to understand how we can properly view the transformation of the action and see that geodesics are preserved only under isometries (say one-parameter isometries generated by a vector field $\xi^a$) and not under general diffeomorphisms. I imagine it should be possible to show that this action is preserved under a one parameter group of diffeomorphisms (generated by $\xi^a$) if and only if $\xi^a$ is Killing?

In particular I'm interested to understand how I should properly apply the transformation to $S$ (either active or passive) that corresponds to an isometry? And understanding the distinction between an isometry and a diffeomorphism in both the active and passive picture - i.e. if we can view every diffeomorphism as the identity map in the passive picture then, whilst it's obviously not true, this seems to me at the moment that every diffeomorphism is an isometry - I'd like to see why that is not the case.

Answer

A general diffeomorphism does not map geodesics to geodesics. Some simple counter examples

• You can a build diffeomorphism on the Euclidean plane by imagining putting one finger on a tablecloth at point $x$ and dragging it. This map is clearly smooth, a smooth inverse is constructed by dragging your finger back. Any geodesic on the plane (a line) passing through point $x$ will certainly be mapped to some weird curve, no longer a geodesic on the plane.


• Consider the upper half plane $\mathbb E^+ \equiv \mathbb R^+ \times \mathbb R$, with the Euclidean metric. Consider the poincare half plane $\mathbb H^2$ with the hyperbolic metric. There is an obvious diffeomorphism between the two - the identity map. Under the identity map, straight lines gets mapped to ... well, themselves. So geodesics on the plane do not get mapped to geodesics in hyperbolic plane under this diffeomorphism.

In particular, the diffeomorphism invariance of the geodesic functional, which you (pretty much) correctly showed certainly doesn't imply geodesics get mapped to geodesics. So let's see what this actually implies.

Diffeomorphisms do not map geodesics to geodesics

Let $M$ be our manifold. Let $g$ be a Riemannian metric on $M$. Let $S_g$ denote the energy functional (that you wrote down) using the metric $g$. That is, let $\gamma: [0, 1] \to M$ be a smooth curve,

$$ S_g[\gamma] \equiv \int_{[0, 1]} g_{\gamma(t)}(\gamma'(\tau), \gamma'(\tau))d\tau $$

You found (by computing in local coordinates) that this is invariant under a diffeomorphism $\phi: M \to M$. This statement reads

$$ \int_{[0, 1]} g_{\gamma(t)}(\gamma'(\tau), \gamma'(\tau))d\tau=\int_{[0, 1]} g_{\phi \circ\gamma(t)}(\phi_*\gamma'(\tau), \phi_*\gamma'(\tau))d\tau $$

RHS can be rewritten in terms of the pullback metric

$$ \int_{[0, 1]} g_{\phi \circ\gamma(t)}(\phi_*\gamma'(\tau), \phi_*\gamma'(\tau))d\tau = \int_{[0, 1]} \phi^*g_{\phi \circ \gamma(t)}(\gamma'(\tau), \gamma'(\tau))d\tau $$

Comparing with our definition of $S_g$, what you have shown is $$ S_g [\gamma]= S_{\phi^*g}[\phi \circ \gamma] $$

In particular this implies, for the special case that you have a curve $\gamma$ that minimizes $S_g$ within a variational family of curves:

$$ \text{ $\gamma$ minimizes $S_g$ } \implies \text{ $\phi\circ\gamma$ minimizes $S_{\phi^*g}$} $$

Observe this does not mean $\phi \circ \gamma$ is a geodesic for the metric $g$. This says $\phi \circ \gamma$ is a geodesic for the (in general, different) metric $\phi^*g$. Therefore a general diffeomorphism does not map geodesics to geodesics.

Isometries do!

Now observe there is a special case when $\gamma$ actually does get mapped to a geodesic of $g$. Namely, when we have $$\phi^*g = g \implies S_g = S_{\phi^*g}$$

And the above implication becomes

$$ \text{ $\gamma$ minimizes $S_g$ } \implies \text{ $\phi\circ\gamma$ minimizes $S_{g}$} $$

Observe the condition $\phi^*g = g$ is exactly the statement that $\phi$ is an isometry.

quantum field theory - How does the 't Hooft renormalization scheme work?

I recently learned there is a trick called t' Hooft transformation that allows us to define a new coupling constant $g_R$ from the usual one $g$ in such a way that the beta function for $g_R$ is two-loop exact. In other words, the beta function has no terms higher than $g_R^2$.

I'm confused about how this comes about and what it implies.

We start with

$$ \beta(g) = \sum a_n g^n .$$ Then, we define $g_R \equiv g + \sum r_n g^n$ such that

$$ \beta_R(g_R) = a_1 g_R + a_2 g_R^2. $$

How does this work, i.e. why aren't there terms $\propto g_R^3,g_R^4,\ldots $ in $\beta_R(g_R)$?

I found the following statements, but wasn't really able to understand what is going on:

"The nullification of higher order coefficients of the β-function is achieved by finite renormalizations of charge." (Source)

"t Hooft' has suggested that one can exploit this freedom in the choice of $g$ even further and choose a new coupling parameter $g_R$ such that the corresponding $\beta_R(g_R) = a_1 g_R + a_2 g_R^2$ (for the above case) and thus contains only two terms in its expansion in $g_R$."(Source)

"The 't Hooft transformation is based on the observation that the first two terms in the perturbation series for $\beta$ are scheme independent. These terms are therefore invariant under a certain class of transformations on the renormalized coupling parameter $g$, a class that includes the 't Hooft transformation in which P is exactly reduced to these two terms." (Source)

PS: In addition, to my confusion how this really works, I am confused about what it implies, e.g. for the radius of convergence of the perturbation series. See this question.

newtonian mechanics - Man inside an accelerating train carriage

I'm having trouble recognizing the forces at play here.

If we have a man is standing inside a train carriage which is accelerating, and the coefficient of friction (for simplicity dynamic and static friction constants are the same) between the man and the floor of carriage isn't enough for him to stand stationary, how do we find out his resulting acceleration?

What is the force causing the resulting acceleration to be in the opposite direction? If it's the friction between the man and the moving train, isn't it the same for the opposite direction as well? What am I missing?

Answer

Ignore non-inertial frames of reference and pseudo forces - they will only confuse you.

If the man has weight $mg$ then the frictional force exerted on the man by the floor of the train is $\mu mg$, and so that man's acceleration is $\frac{\mu mg}{m}=\mu g$.

Note that this is true relative to any inertial frame of reference - acceleration is not affected by adding or subtracting a constant velocity.

The train's acceleration $a_{train}$ must be greater than $\mu g$, otherwise a frictional force less than $\mu mg$ would be sufficient to accelerate the man at the same rate as the train. So relative to the train the man is accelerating backwards at a rate $a_{train}-\mu g$.

newtonian mechanics - What do people actually mean by "rolling without slipping"?

I have never understood what's the meaning of the sentence "rolling without slipping". Let me explain.

I'll give an example. Yesterday my mechanics professor introduced some concepts of rotational dynamics. When he came to talk about spinning wheels he said something like:

"If the wheel is rolling without slipping, what's the velocity of the point at the base of the wheel?? It is... zero! Convince yourself that the velocity must be zero. Since if it wasn't zero, the wheel wouldn't be rolling without slipping. So the wheel is rolling without slipping if and only if the point at the base has velocity zero, i.e. if and only if the tangential speed equals the speed of the center of mass."

Well, what I really don't understand is this: is the "rolling without slipping" condition defined as "Point at the base has zero speed"? If not, what's the proper definition for that kind of motion?

Looking across the internet, I have found more or less the same ideas expressed in the quotation. Furthermore, if it was a definition, then it would be totally unnecessary to say "convince yourself" and improper to talk about necessary and sufficient conditions.

I'd like to point out that I'm not really confused about the mathematics behind this or with the meaning of the condition above. What puzzles me is why are those explanations always phrased as if the condition $v'=0$ (where $v'$ is the relative velocity beetween the point at base and the surface) is some necessary and sufficient condition to be "rolling without slipping". Seems to me that this is exactly the definition of "rolling without slipping" and not an "iff".

Any help is appreciated, thanks.

Answer

You can always decompose a motion like this into two parts: (1) rolling without slipping and (2) slipping without rolling.

What is slipping without rolling? It means the object moves uniformly in one direction along the surface, with no angular velocity about the object's own center of mass. For instance, a box that is pushed along the ground can easily slip without rolling.

Unfortunately, most people seem to assume that you can infer some physically important information from your own notion of what slipping is, without having to define it. I believe this is done to try to connect to intuition, but in the process, things get a lot more nebulous and ill-defined.

To me, it's easier to think about this in terms of the object's rotation--it was never obvious to me that the point in contact with the ground doesn't have velocity at the instant it touches. I prefer to think instead that an object that rolls without slipping travels 1 circumference along the ground for every for every full rotation it makes. And object that travels more than this distance (or that doesn't rotate at all) is slipping in some way.

Then, eventually, we can get to the notion that the point in contact during rolling cannot have nonzero velocity through whatever logical or physical arguments necessary.

But as is usual in physics, it's not really clear what definition should be considered "fundamental" with other results stemming from it. This emphasizes that physics is not built axiomatically.

quantum gravity - What's the recent released 750GeV particle's spin?

I was told that it has recently been confirmed to be spin-2 particle, and potentially to be graviton. I'm pretty interested in how this has been examined.

Edit: During the Moriond 2016 conference, CMS clearly stated it is a spin-2 rather than spin-0 particle (spin-1 has already been ruled out by Landau-Yang theorem [1-2]). For a review, see [3].

The yesterday report on ATLAS claims new analysis and new methods gave promising result, which is what I'm looking forward to know. Also I would like some experts to give an introduction to the physics picture how to examine the spin, and what's the theoretical model [4] means.

Edit2: on Aug 5th, ICHEP conference, the new result is gonna to be published.

Edit3: ICHEP conference released ATLAS & CMS results [5-6]. ATLAS doesn't contain newest spin-2 selection results (which could be something), while CMS gives disappointing results.

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[1] C.-N. Yang, Selection Rules for the Dematerialization of a Particle Into Two Photons, Phys. Rev. 77 (1950) 242–245. http://dx.doi.org/10.1103/PhysRev.77.242;

[2] L. D. Landau, On the angular momentum of a system of two photons, Dokl. Akad. Nauk Ser. Fiz. 60 (2) (1948) 207–209. http://dx.doi.org/10.1016/B978-0-08-010586-4.50070-5.

[3] Strumia, Alessandro. "Interpreting the 750 GeV digamma excess: a review." arXiv preprint arXiv:1605.09401 (2016).

[4] Dillon, Barry M., and Veronica Sanz. "A little KK graviton at 750 GeV." arXiv preprint arXiv:1603.09550 (2016).

[5] http://indico.cern.ch/event/432527/contributions/1072336/attachments/1321033/1981068/BL_ATLAS_HighMassDiphotons_ICHEP2016.pdf

[6] http://indico.cern.ch/event/432527/contributions/1072431/attachments/1320985/1980991/chiara_ichep.pdf

quantum mechanics - Path integral vs. measure on infinite dimensional space

Coming from a mathematical background, I'm trying to get a handle on the path integral formulation of quantum mechanics.

According to Feynman, if you want to figure out the probability amplitude for a particle moving from one point to another, you 1) figure out the contribution from every possible path it could take, then 2) "sum up" all the contributions.

Usually when you want to "sum up" an infinite number of things, you do so by putting a measure on the space of things, from which a notion of integration arises. However the function space of paths is not just infinite, it's extremely infinite.

If the path-space has a notion of dimension, it would be infinite-dimensional (eg, viewed as a submanifold of $C([0,t] , {\mathbb R}^n))$. For any reasonable notion of distance, every ball will fail to be compact. It's hard to see how one could reasonably define a measure over a space like this - Lebesgue-like measures are certainly out.

The books I've seen basically forgo defining what anything is, and instead present a method to do calculations involving "zig-zag" paths and renormalization. Apparently this gives the right answer experimentally, but it seem extremely contrived (what if you approximate the paths a different way, how do you know you will get the same answer?). Is there a more rigorous way to define Feynman path integrals in terms of function spaces and measures?

Answer

Path integral is indeed very problematic on its own. But there are ways to almost capturing it rigorously.

Wiener process

One way is to start with Abstract Wiener space that can be built out of the Hamiltonian and carries a canonical Wiener measure. This is the usual measure describing properties of the random walk. Now to arrive at path integral one has to accept the existence of "infinite-dimensional Wick rotation" and analytically continue Wiener measure to the complex plane (and every time this is done a probabilist dies somewhere).

This is the usual connection between statistical physics (which is a nice, well-defined real theory) at inverse temperature $\beta$ in (N+1,0) space-time dimensions and evolution of the quantum system in (N, 1) dimensions for time $t = -i \hbar \beta$ that is used all over the physics but almost never justified. Although in some cases it was actually possible to prove that Wightman QFT theory is indeed a Wick rotation of some Euclidean QFT (note that quantum mechanics is also a special case of QFT in (0, 1) space-time dimensions).

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Intermezzo

This is a good place to point out that while path integral is problematic in QM, whole lot of different issues enter with more space dimensions. One has to deal with operator valued distributions and there is no good way to multiply them (which is what physicist absolutely need to do). There are various axiomatic approaches to get a handle on this and they in fact do look very nice. Except that it's very hard to actually find a theory that satisfies these axioms. In particular, none of our present day theories of Standard model have been rigorously defined.

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Anyway, to make the Wick rotation a bit clearer, recall that Schrödinger equation is a kind of diffusion equation but for an introduction of complex numbers. And then just come back to the beginning and note that diffusion equation is macroscopic equation that captures the mean behavior of the random walk. (But this is not to say that path integral in any way depends on the Schrödingerian, non-relativistic physics)

Others

There were other approaches to define the path-integral rigorously. They propose some set of axioms that path-integral has to obey and continue from there. To my knowledge (but I'd like to be wrong), all of these approaches are too constraining (they don't describe most of physically interesting situations). But if you'd like I can dig up some references.

conservation laws - What does a symmetry that changes the Lagrangian by a total derivative do to the Hamiltonian $H$?

A tiny symmetry transformation may change the Lagrangian $L$ by a total time derivative of some function $f$. This is a basic fact used in the proof of Noether's theorem.

How can we see the effect of this total derivative term in the Hamiltonian framework? Is there a good example to work out? I can't think of one off the top of my head. It just seems strange to me that all this fuss about total derivatives seem to disappear in the Hamiltonian framework.

Answer

I suppose I figured out the "answer" to my very vague question, although the other answers here are also helpful. The "Hamiltonian Lagrangian" is

$$ L = p_i \dot q_i - H. $$ Say we have a conserved charge $Q$, that is $$ \{Q, H\} = 0. $$ If we make the tiny symmetry variation $$ \delta q_i = \frac{\partial Q}{\partial p_i} \hspace{1cm} \delta p_i = - \frac{\partial Q}{\partial q_i} $$ then \begin{align*} \delta L &= - \frac{\partial Q}{\partial q_i} \dot q_i - p_i \frac{d}{dt} \Big( \frac{\partial Q}{\partial p_i} \Big) + \{ H, Q\} \\ &= - \frac{\partial Q}{\partial p_i} \dot q_i - \dot p_i \frac{\partial Q}{\partial p_i} + \frac{d}{dt} \Big( p_i \frac{\partial Q}{\partial p_i} \Big) \\ &= \frac{d}{dt} \Big( p_i \frac{\partial Q}{\partial p_i} - Q\Big) \end{align*}

So we can see that $L$ necessarily changes by a total derivative. When the quantity $p_i \frac{\partial Q}{\partial p_i} - Q = 0$, the total derivative is $0$. This happens when the conserved quantity is of the form $$ Q = p_i f_i(q). $$ Note that in the above case, $$ \delta q_i = f_i(q) $$ That is, symmetry transformations which do not "mix up" the $p$'s with the $q$'s have no total derivative term in $\delta L$.

gravity - Could this fountain, under the right conditions, technically be able to lift me up?

There is a big fountain in a lake in my city.

I was talking with a friend and we were wondering whether it would be able to lift me up. I sent a few emails and obtained information about the fountain. The raw email reads in French:

Nous avons 2 pompes de 150 HP qui pousse l'eau a 195 pieds de haut et la pression par pompe à la sortie est d'environ 380 pieds de tête.

I am not knowledgable in physics in any way, but this is how I can translate it to the best of my knowledge:

We have two pumps of 150 horsepower that push water at 195 feet high and the pressure by pump at the exit is of 165 PSI

Now, could this possibly be able to lift me, a 160 pounds man? How can I find out? And under what conditions would this work?

Why is voltage described as potential energy per charge?

Voltage is often called an electromotive force since it causes a flow of charge. However, it is described in terms of Joules per Coulomb or Potential Energy per Charge.

Question: How does the potential energy associated with charge contribute to its effect on the flow of charge?

High voltage, or high electromotive force, causes high current. So this means charge with high PE will cause high current. This doesn't seem to make sense to me. Why does potential energy affect current?

I know of the PE gradient explanation, but this doesn't make sense to me. In most cases of diffusion, there is an explanation as to why it occurs: particles diffuse from high to low concentration DUE to random particle movement. Things fall from high to low gravitational PE DUE to the force of grav.

Question: Charge moves from high to low PE in a circuit, but why? What is the driving force?

quantum field theory - Renormalization is a Tool for Removing Infinities or a Tool for Obtaining Physical Results?

Quoting Wikipedia:

renormalization is any of a collection of techniques used to treat infinities arising in calculated quantities.

Is that true? to me, it seems better to define renormalization as a collection of techniques for adjusting the theory to obtain physical results. I'll explain. According to Wilson's renormalization group, a quantum field theory always inherently has a cutoff parameter, so in any case integrals should be done only up to the cutoff, so there are no infinite quantities. Yet the results are still not consistent with observation if you don't renormalize the calculations (e.g. using counterterms).

Am I correct? Is it true that the usual presentation of renormalization as a tool for removing divergences is a misinterpretation of the true purpose of it?

Answer

You're totally right. The Wikipedia definition of the renormalization is obsolete i.e. it refers to the interpretation of these techniques that was believed prior to the discovery of the Renormalization Group.

While the computational essence (and results) of the techniques hasn't changed much in some cases, their modern interpretation is very different from the old one. The process of guaranteeing that results are expressed in terms of finite numbers is known as the regularization, not renormalization, and integrating up to a finite cutoff scale only is a simple example of a regularization.

However, the renormalization is an extra step we apply later in which a number of calculated quantities is set equal to their measured (and therefore finite) values. This of course cancels the infinite (calculated) parts of these quantities (I mean parts that were infinite before the regularization) but for renormalizable theories, it cancels the infinite parts of all physically meaningful predictions, too.

However, the renormalization has to be done even in theories where no divergences arise. In that case, it still amounts to a correct (yet nontrivial) mapping between the observed parameters and the "bare" parameters of the theory.

The modern, RG-based interpretation of these issues changes many subtleties. For example, the problem with the non-renormalizable theory is no longer the impossibility to cancel the infinities. The infinities may still be regulated away by a regularization but the real problem is that we introduce an infinite number of undetermined finite parameters during the process. In other words, a non-renormalizable theory becomes unpredictive (infinite input is needed to make it predictive) for all questions near (and above?) its cutoff scale where its generic interactions (higher-order terms) become strongly coupled.

forces - Calculate the weight a simple plank can support

I'd like to build a simple desk; just a single plank of wood (or a few side-by-side) with solid supports on each end of the desk. What I'm trying to figure out is how thick a plank I want to use for the surface. Ideally, the desk will be able to support my weight (since sometimes I stand on my desk to change lights, etc), plus the weight of my computer stuff, plus some amount of safety margin.

So, supposing I have a desk surface that's eight feet wide, what equation do I need to use to figure out how thick it needs to be to support, e.g., 300lbs, placed in the middle? What properties of the wood type that I'm using do I need to determine? Is this an issue of tensile strength, or something entirely different?

This is obviously more a structural engineering problem than a physics one, but I didn't see a structuralengineering.stackexchange.com. Sorry if I posted to the wrong place.

string theory - Introduction to AdS/CFT

AdS/CFT seems like a really hot topic and I'd like to start reading about. I am looking for the best introduction at my level, i.e. I have a background in QFT, CFT and general relativity at the level of a master student in theoretical physics. What would you recommend me to start tackling the subject? I have been looking for resources and so far I have noticed:

-this synthetic introductory lectures by Horatiu Nastase: http://arxiv.org/abs/0712.0689

-the videos of lectures done by P. Vieira at Perimeter for Perimeter Scholar International students: http://www.perimeterscholars.org/341.html

-the subject starts entering the most recent textbooks on string theory. We have the Schwartz and Becker ( http://goo.gl/jh45U ) and also the Kiritsis ( http://goo.gl/ulEVw )

I probably missed a lot of resources, as the literature on the subject is already quite huge. I would really appreciate some advice on that, as I already had the frustration of losing my time on not-so-good books when started to learn something new, so if I could (to the best) avoid that this time...

lagrangian formalism - What's the point of Hamiltonian mechanics?

I've just finished a Classical Mechanics course, and looking back on it some things are not quite clear. In the first half we covered the Lagrangian formalism, which I thought was pretty cool. I specially appreciated the freedom you have when choosing coordinates, and the fact that you can basically ignore constraint forces. Of course, most simple situations you can solve using good old $F=ma$, but for more complicated stuff the whole formalism comes in pretty handy.

Then in the second half we switched to Hamiltonian mechanics, and that's where I began to lose sight of why we were doing things the way we were. I don't have any problem understanding the Hamiltonian, or Hamilton's equations, or the Hamilton-Jacobi equation, or what have you. My issue is that I don't understand why would someone bother developing all this to do the same things you did before but in a different way. In fact, in most cases you need to start with a Lagrangian and get the momenta from $p = \frac{\partial L}{\partial \dot{q}}$, and the Hamiltonian from $H = \sum \dot{q_i}p_i - L$. But if you already have the Lagrangian, why not just solve the Euler-Lagrange equations?

I guess maybe there are interesting uses of the Hamiltion formalism and we just didn't do a whole lot of examples (it was the harmonic oscillator the whole way, pretty much). I've also heard that it allows a somewhat smooth transition into quantum mechanics. We did work out a way to get Schrödinger's equation doing stuff with the action. But still something's not clicking.

My questions are the following: Why do people use the Hamiltonian formalism? Is it better for theoretical work? Are there problems that are more easily solved using Hamilton's mechanics instead of Lagrange's? What are some examples of that?

Answer

There are several reasons for using the Hamiltonian formalism:

1) Statistical physics. The standard thermal states weight pure states according to

$$\text{Prob}(\text{state}) \propto e^{-H(\text{state})/k_BT}$$

So you need to understand Hamiltonians to do stat mech in real generality.

2) Geometrical prettiness. Hamilton's equations say that flowing in time is equivalent to flowing along a vector field on phase space. This gives a nice geometrical picture for how time evolution works in such systems. People use this framework a lot in dynamical systems, where they study questions like 'is the time evolution chaotic?'.

3) Generalization to quantum physics. The basic formalism of quantum mechanics (states and observables) is an obvious generalization of the Hamiltonian formalism. It's less obvious how it's connected to the Lagrangian formalism, and way less obvious how it's connected to the Newtonian formalism.

[Edit to answer Javier's comment] This might be too brief, but the basic story goes as follows:

In Hamiltonian mechanics, observables are elements of a commutative algebra which carries a Poisson bracket $\{,\}$. The algebra of observables has a distinguished element, the Hamiltonian, which defines the time evolution via $d\mathcal{O}/dt = \{\mathcal{O},H\}$. Thermal states are simply linear functions on this algebra. (The observables are realized as functions on the phase space, and the bracket comes from the symplectic structure there. But the algebra of observables is what really matters: You can recover the phase space from the algebra of functions.)

On the other hand, in quantum physics, we have an algebra of observables which is not commutative. But it still has a bracket $\{,\} = -\frac{i}{\hbar}[,]$ (the commutator), and it still gets its time evolution from a distinguished element $H$, via $d\mathcal{O}/dt = \{\mathcal{O},H\}$. Likewise, thermal states are still linear functionals on the algebra.

Why should field operators satisfy the classical equations of motion?

To quantize a scalar field theory with the action:

$$S=\int \mathrm d^Dx\mathscr{L}(\phi,\partial_\mu\phi)=\int \mathrm dx^0L(\phi,\partial_0\phi)$$

we promote $\phi(\vec{x})$ and $\pi=\frac{\delta L}{\delta(\partial_0\phi)}=\pi(\vec{x})$ to be field operator at fixed time $x^0$ (with the canonical commutation relation $[\phi(\vec{x}),\pi(\vec{y})]=\delta^{D-1}(\vec{x}-\vec{y})$), then we use the classical equation of motion to find $\phi$ and $\pi$ at all time.

What principles lie behind this? Why can we assume that the field operators satisfy the classical equation of motion?

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By example if we have a classical theory with $S=\int \,\mathrm dtL(\phi,\partial_t\phi)$, momentum $\pi=\frac{\delta L}{\delta(\partial_t\phi)}$ hamiltonian $H=\pi\phi-L$, the classical equation of motion is:

$$\frac{\mathrm d\phi}{\mathrm dt}=\frac{\delta L}{\delta(\partial_t\phi)}$$

When we do the canonical quantization, we promote $[\phi,\pi]=\delta(\vec{x}-\vec{y})$ and the equation of motion for $\phi$ is $$\frac{\mathrm d\phi}{\mathrm dt}=i[H,\phi]$$

Why should $i[H,\phi]$ has the same form as $\frac{\delta L}{\delta(\partial_t\phi)}$? This fact comes from where?

Answer

How to see it in canonical quantization: All operators $\mathcal{O}$ in a quantum theory fulfill the Heisenberg equations of motion $$ \frac{\mathrm{d}}{\mathrm{d}t}\mathcal{O}(t) = \mathrm{i}[H,\mathcal{O}(t)]$$ where $H$ is the Hamiltonian density and which is exactly the quantum version of the classical Hamiltonian equations of motion. So the fields, as they are operators, indeed must obey the classical equations of motion.

How to see it in path integral quantization: Write $\phi'(x) = \phi(x) + \epsilon\delta(y-x)$ and observe the path integral measure is invariant under this. Expand the integrand to first order in $\epsilon$, and deduce $$ \int\left(\frac{\delta S}{\delta\phi(x)} + J(x)\right)\mathrm{e}^{\mathrm{i}S[\phi]+J\phi}\mathcal{D}\phi = 0$$ which is known as the Schwinger-Dyson equation. Setting $J=0$ gives $\delta S/\delta\phi = 0$ (inside the path integral, which is the PI version of "as an operator equation"), which is exactly the classical equation of motion.

mathematics - How $pi$ is derived from quantum mechanics

I came across this article New Derivation of Pi Links Quantum Physics and Pure Math in which they discuss about a recent discovery of deriving PI from physics. I am not a physicist or a mathematician and able to grasp only a little. Can you explain the discovery and its significance for a simpleton like me?

Please forgive me if I am wrong but $\pi$ can be derived from sphere volume also

$$V = \frac{{4\pi r^3 }}{3}$$

so why this discovery is different?

Answer

There isn't a way to derive $\pi$ because it's a fundamental constant and not something that can be derived.

However there are lots and lots of ways to approximate $\pi$. I believe that Archimedes was the first person to write down such an approximation (in 250BC), and we've been developing better and better ways of approximating $\pi$ since.

One of the approximations for $\pi$ is due to a mathematician called John Wallis in 1655 and is called the Wallis product. There's nothing special about his approximation - it's just one of the many that have been discovered. The recent paper reports that the Wallis product has turned up in some calculations about hydrogen atoms.

This sort of thing amuses both physicists and mathematicians because it's always fun when some unexpected link is made between physics and maths. However there is nothing in this that will revolutionise our understanding of physics. These sorts of links are pretty frequent and are discovered on a regular basis.

newtonian mechanics - Why is it difficult to ride a bicycle with a partially inflated or deflated tyre?

It is a common observation that riding a bicycle with an inflated tyre is easier than riding one with a deflated tyre but why is it so?

As per my knowledge in an ideal case of no deformation in tyre(when it is inflated) the torque of normal is zero(almost zero in non ideal case) about the axis of rotation whereas in the deflated case the tyre gets deformed and the normal shifts to the front of the tyre and therefore there is a torque of the normal too that our muscles need to overcome.

Is what I thought right and is there any other reason too?

Answer

There are several factors that may be taken in account, but the more important is the energy used deforming the tire.

Suppose a deflated tire. As you move forward and the tire rotates, the part of the tire that is starting to touch the ground has to be deformed (since the tire is flat). You have to use an important amount of energy for that. Note that the part of the tire that has just stopped touching the ground also has to be deformed, which recovers some energy. Nonetheless, not all energy is gained due to elastic hysteresis, so we have a net loss. That loss has to be even by the cyclist, which is why it is more difficult to ride a bike with flat tires.

Note that there are other factors that may influence this. I'm quoting Wikipedia here

Additional contributing factors include wheel diameter, speed, load on wheel, surface adhesion, sliding, and relative micro-sliding between the surfaces of contact. The losses due to hysteresis also depend strongly on the material properties of the wheel or tire and the surface.

You may want to read this article about rolling resistance. Here it suppose a solid (or perfectly inflated) tire and an elastic ground, but it is equivalent.

electricity - Potential drop in a wire having negligible resistance

What causes a potential drop in a resistor or load? Why a wire having neglible resistance have same potential across it? The given figure is of a stretched wire potentiometer. My question is why point a and A have same potential after connection through a wire of negligible resistance. Before connection there can be different potentials on poinrs a and A. So how the same potential achieved after connection. I wants it's machanism and explanation.

Answer

The potential difference, or voltage, between two points equals the work required per unit charge to move the charge between the two points. If there is negligible resistance between points a and A, the work required to move charge between the two points (potential difference) is also negligible.

Hope this helps.

fluid dynamics - Flow Rate and Orifice Diameter

I have a electrical/electronics background and have limited knowledge of Fluid Mechanics. So, i will try to be as clear as possible.

I am currently working with brake fluid in an ABS system and to effectively construct my alternative control strategy,i have following questions:

1> Flow-rate: I just need to know how to measure the flow-rate across an orifice? I know of the formula

$$Q = CA\sqrt{\frac{2\Delta P}{d}}$$

$C$ - discharge coefficient

$\Delta P$ - Change in pressure

$d$ - density of the fluid

$A$ - Area of the orifice

Now, this formula should pretty much give me the answer, but my real question is; if i want to calculate simply the flow-rate across a circular orifice of different diameters. So, $Q1$ for $D1$,$Q2$ for $D2$ ... $Qn$ for $Dn$. I am simply interested in finding the flow rate without the consideration of pressure change or velocity etc.

Just to clarify, FLOW RATE through a PIPE of varying diameters.

I will be grateful for any and every suggestion!

Answer

OK, area $A$ is just $\pi D^2/4$. The real question is: What is $C$? It depends on the shape of the orifice (and Reynolds number). There are some quick-and-dirty approximations here. It depends on the orifice geometry, like whether its edges are rounded or sharp.

quantum mechanics - The Lorentz Transformations in the Micro-World

Two particles[or micro-observers] A and B are in relative uniform motion wrt each other in the x-x’ direction. The “observer” A decides to deduce[or interpret] the Lorentz Transformations wrt to B. Accurate knowledge of the position of B makes its[that of B] momentum highly uncertain.[$\Delta x \Delta p\ge h/2\pi$]

How does the observer A go about his job? Rather how do we interpret the Lorentz Transformation in the micro-world?

renormalization - Is QCD free from all divergences?

On page 8 in http://arxiv.org/pdf/hep-th/9704139v1.pdf David Gross makes the following comment:

"This theory [QCD] has no ultraviolet divergences at all. The local (bare) coupling vanishes, and the only infinities that appear are due to the fact that one sometimes expresses observables measured at finite distances in terms of those measured at infinitely small distances. "

1) First of all, is this statement even correct?

2) Now my main question is: certainly, a naive application of the Feynman rules and regularization leads to non zero counterterms. (See chap 16 of Peskin). So what scheme could one work in such that the counterterms vanish?

electromagnetism - Refractive index of plasma dependance on temperature

How does the refractive index of plasma changes with temperature? Temperature is not high enough for new ionization. Would it be like in gas p/T dependence ?

http://farside.ph.utexas.edu/teaching/em/lectures/node100.html

thermodynamics - Confusion about Fick's first law

Consider a binary system of mass transport (A, B). Some of mass transfer books (Skelland and Welty) say that the relation $$J_A= -C D_{AB} \frac{dx_A}{dz} \tag{I}$$ is more general than $$J_A= -D_{AB} \frac{dc_A}{dz} \tag{II}$$ Where $C$ is the total concentration of system.

Now we know that $J_A+J_B=0$. Regarding this equality, I say that the first equation $(I)$ always gives $D_{AB}=D_{BA}$ since $dx_A=-dx_B$.

Second equation $(II)$ gives $D_{AB}=D_{BA}$ under the condition that $C$ is constant.

1. 
   

   So Fick says that if $C$ is constant then $D_{AB}=D_{BA}$.

   
   


2. 
   

   Skelland says that under specific operational conditions for a binary system $D_{AB}=D_{BA}$ and so Fick's relation is true if $C$ is constant.

   
   

Which statement (1) or (2) is correct? What is Bird's opinion on this matter?

Answer

If $J_A$ is the molar flux of $A$ defined with respect to the molar-averaged velocity of the mixture $A+B$ then relation (I) is the most general case as it requires no assumptions to use. Furthermore, this relation can be derived from kinetic theory.

You are correct in stating that relation (II) requires the assumption of the total concentration being constant. In gaseous system, where the pressure is relatively constant, this is a good assumption and the two relations are equivalent. Likely (although don't take my word for it), when Fick did his experiments in the distant past he formulated his law of mass transfer in systems where the assumption of total constant concentration was constant.

Note that when defined with respect to the volume-averaged mixture velocity, relation (II) is actually the correct phenomological description of the flux.

This is explained in more detail in the first chapter of Analysis of Transport Phenomena by Deen. I suggest you take a look at it.

astrophysics - Upper Mass Limit of Quark Stars

While there is no confirmation that quark stars exist, is there any theoretical limit analogous to (but different from) the Tolman–Oppenheimer–Volkoff limit for neutron stars?

In other words, what is the maximum pressure for quark matter?

Answer

The upper mass limit for a quark star depends on your assumptions and ranges between 1 and 2 solar masses (cf. this paper (arXiv link) from 2001). It seems to me that the reason for the similarity to neutron stars' mass range is that it both compact objects satisfy the TOV equation, $$ \frac{dp}{dr}=-\frac{G}{r^2}\left[\rho+\frac{p}{c^2}\right]\left[M+4\pi r^2\frac{p}{c^2}\right]\left[1-\frac{2GM}{rc^2}\right]^{-1} $$ but with different equations of state.

For the quark star, according to the aforemention paper, the pressure is defined as $$ p(\mu)=\frac{N_f\mu^4}{4\pi^2}\left[1-2\frac{\alpha_s}{\pi}-\left(G+N_f\ln\frac{\alpha_s}\pi+\left(11-\frac23N_f\right)\ln\frac{\bar{\Lambda}}{\mu}\right)\frac{\alpha_s^2}{\pi^2}\right] $$ where $G\simeq10.4-0.536N_f+N_f\ln N_f$, $\alpha_s$ the strong coupling, $N_f$ the number of flavors (often taken as 3), $\mu$ the chemical potential, and $\bar{\Lambda}$ the renormalization subtraction point (my understanding of this term is minimal, but it seems to change the size of the mass-radius relation, but not the shape).