I'm taking a course on Lagrangian and Hamiltonian Dynamics, and I would like to find a good book/resource with lots of practice questions and answers on either or both topics.
So far at my university library, I have found many books on both subjects, but not ones with good practice questions and answers. I have Schuam's outline of Lagrangian Dynamics, but didn't really find a lot of practice questions.
Any suggestions would be greatly appreciated!
Suppose we have a situation like
A ball of some mass $m$ with some velocity collides with rod hinged at point $A$. Is momentum conserved in this situation? I know that hinge will give impulsive force on rod but that is internal force when we take ball and rod together as a system so that shouldn't stop us using conservation of momentum equation
Answer
Generally speaking linear momentum cannot be conserved in this case as hinge will exert a (external) force on rod at all but one position.
Note:Angular momentum is definitely conserved about hinged point since the force applied by hinge produces zero torque here.
We may want to find that one position for which linear momentum conservation holds.We proceed as follows:
$mv_od=mv_{final}d+(M\dfrac{l^2}{3})\omega$ -----(1)
$mu=mv_{final}+M(\dfrac{l}{2}\omega)$-------------(2) [holds since force by hinge is zero for this position $d$]
Solving we get $d=\dfrac{2l}{3}$
Assuming the correctness of QM: Would the size of such a computer be smaller than the observable universe? If it were to represent all available information in the universe it seems that it's minimal necessary size would relate to a black hole of a certain mass. Would this lead us to believe this theoretical quantum computer (assuming it didn't represent itself) could be constructed smaller than the universe? Or if the information content of an object is truly expressed in relation to a surface area is the universe the object with the minimal surface required to represent the totality of it's information?
In this Math.SE question, I asked a question to which I was hoping to get a simple intuitive answer. Instead I received an otherwise perfectly correct but very mathematical one. Obviously, the words geometric and intuitive have very different meaning in that forum from what they mean to me... Can somebody help here?
There is a family of surfaces orthogonal to the vector field $\vec v \in \mathbb R^3$ iff $\vec v \cdot \operatorname{curl} \vec v = 0 $. Now the necessity part is trivial, but the proof of sufficiency I have seen in physics textbooks, e.g. Kestin: Thermodynamics, is kind of mindless integration similar to the one usually offered to prove Poincare's lemma as in Flanders, or just asserted as in Born&Wolf: Optics. Is there an intuitive and geometric interpretation of this condition that would make it obvious why its sufficiency must be true?
Answer
Edit: Now that I've read the original post more closely, I'm not sure that this answer actually addresses your question. Let me know if it doesn't and I can perhaps try again.
Remember: curl is only defined for vector fields and not for lone, individual vectors. With this in mind, curl is sometimes said to quantify the "vorticity" of a vector field (i.e. how swirly it is at a given point). Intuitively, if you imagine putting a little, tiny paddle-wheel into the vector field and think of the field as a flowing stream and the magnitudes of the nearby vectors as the strength of the current at that point, then you can tell whether the curl of the field is zero at the point described by the center of the paddle-wheel by figuring out whether or not the paddle-wheel is caused to turn by the current. Moreover, the faster the paddle-wheel turns the greater the vorticity and, thus, the greater the curl.
Now a spinning wheel lives in a plane, yes? That is to say, uniform rotation about a point is a fundamentally two-dimensional activity. Once we've observed this fact, we can easily see that the best way to define a plane is to give the components of its normal vector. Since the plane is flat and uniform out to infinity, the normal vector does not change and, more importantly, the normal vector is perpendicular to the plane at all points.
So now we see that the curl is giving us the information of a plane which we encode and compactify into a single vector. We do this again at every point in space and we get a new vector field that is the curl of the original vector field. This new vector vield has the interesting property that at any given point, the vector there is perpendicular to the vector at that point in the original field.
And the dot product of perpendicular vectors is always zero. Intuitive enough for you?
So to summarize, the curl of a vector field at a point gives you a vector that is normal to the plane of rotation of the original vector field. The magnitude of the new vector tells you how curly it is, while the direction tells you which way the original field is spinning. By the way, don't forget the right-hand rule.
In quest of confirming Higgs boson's existence, particle collisions were observed at the Large Hadron Collider. Scientists had to distinguish the observation of the background noise/fluctuation and the observation of the Higgs boson. For example, this text from Wikipedia:
"CERN reported that the Standard Model Higgs boson, if it exists, is most likely to have a mass constrained to the range 115–130 GeV. Both the CMS and ATLAS detectors have also shown intensity peaks in the 124–125 GeV range, consistent with either background noise or the observation of the Higgs boson."
What is the background noise/fluctuation? And additionally, how could a CERN scientist distinguish it from Higgs boson?
Answer
In particle physics (as with most sciences) we are rarely ever concerned with analysing single events. What we look at are distributions of the same measurement(s) made many times. From performing fits to these distributions, we can infer the number of 'signal' and 'background' events in our sample.
Signal is the process we're interested in. Background is all of the events which end up in the sample but aren't a result of the signal process. Often the largest source of background is combinatorial: i.e. random combinations of particles which happen to end up looking like the signal. Other sources include things like misidentified or misreconstructed processes.
If we're looking at the invariant mass distribution of a particular final state, the signal will often be a peak-like shape. Combinatorial backgrounds tend to follow smooth broad shapes (like exponential or simple polynomial curves). However, a statistical fluctuation in the background may result in a peak-like shape. It is important not to confuse this with signal. Statistical significance is a measure of how unlikely it is, assuming the null hypothesis, that a fluctuation will result in a peak at least as big as the one you observe. In particle physics, when you think you've found something, it's important to quote the significance of the measurement. The standard for discovery is $5\sigma$, which corresponds to a probability of 1 in 1.7 million.
It is important to note that we can never really tell exactly which particular events are signal and which are background. However, you can apply selection criteria to remove the background-like events and hopefully retain most of the signal. These can range from simple cuts on variables to training some complicated machine-learning algorithm to distinguish signal from background.
Let's take the example of searching for the Higgs boson decaying to a pair of photons ($H\to\gamma\gamma$). You start with a sample of recorded events containing two reconstructed photons and apply some selection criteria to filter out the background events which look least likely to be Higgs decays. If you plot the distribution of the invariant mass of the photon pairs, you end up with something that looks like the figure below:
We know a priori that the combinatorial background follows a nice smooth shape, so this is modelled as some sort of polynomial function. The signal will appear as a bump in the distribution, centred at the mass of the Higgs (125 GeV). This looks to be modelled as a bifurcated Gaussian of Crystal Ball function or something similar.
The fit finds a significant peak above the background at the mass of the Higgs. From this, you can say (with some quantifiable uncertainty) how many $H\to\gamma\gamma$ decays and how many background $\gamma\gamma$ pairs are in the data, but not which events are actual Higgs bosons or background events.
If two neutrons collide in 3D space and we want to determine the final velocities of both nuetrons (3 components for each neutrons), we can use the conservation of momentum equations and the conservation of energy equation. Those equations are 4 while the number of unknowns we have is 6. There are 2 free variables.
For classical perfectly spherical objects, the solution is strait forward as explained here. However, neutrons because of their quantum nature can't be treated as classical perfectly spherical objects. We can't define a precise point of collision, so the trick we use for classical objects won't work for neutrons or any quantum level particle, or will it?
If it doesn't work, does that mean we can't avoid having free two degrees of freedom in this system??
NOTE: What is meant by a collision is a momentum-exchange event
My lecture notes said that:
You cannot have two (or more) ideal voltage sources connected to the same pair of terminals.
Why is this so?
Answer
In effect you have a series circuit composed of two voltage sources of voltages $V_1$ and $V_2$.
The net voltage in the circuit is $V_1 \pm V_2$ depending on the orientation of the voltage sources.
Being ideal voltage sources there is no resistance in the circuit and so the current in the circuit would be infinity.
To only time that this is not so is if the voltage in the circuit is such that $V_1 - V_2 =0$ when the current would be zero.
This is when the positive terminals of the two voltage sources are connected together as are the negative terminals of the two voltage sources and the two voltages of the two voltage sources are the same.
This last arrangement is sometimes used with batteries so that a larger current can be provided to an external circuit.
It is not a recommended as the voltage of the two batteries are unlikely to be the same.
The batteries will also both have some source resistance.
From renormalization group equation $$ t \frac{d \bar{g}(t , g)}{dt} = \beta (\bar{g}(t , g)), \quad \bar{g}(1 , g) = g $$ (here $t$ is momentum scale factor, $g$ is initial coupling constant and $\bar{g}$ is an effective coupling constant) and $$ \tag 1 \beta (g) = ag^{2} $$ we can get $$ \bar{g}(t , g) = \frac{g}{1 - gln(t)a} . $$ For QCD $$ \tag 2 a = c\left(-11 + \frac{2}{3}n_{f} \right), $$ where $n_{f} = 6$ is the number of quark types.
How to get that for QCD we asymptotically have $(1)$ and $(2)$?
Answer
First off, equation (1) should be $\beta(g)=ag^3$. This is not an asymptotic formula. It is the approximate expression for the $\beta$ function to one loop order. Expression (2) is an exact expression at this order. The simplest method to compute this would be perform the computation in the background field gauge. Take a look a Peskin chapter 16 section 6 where this method is developed. This is a very efficient method to compute the $\beta$ function as it avoids the evaluation of numerous Feynman diagrams.
source:http://farside.ph.utexas.edu/teaching/336k/lectures/node11.html#e3.24
Consider a system of N mutually interacting point objects.
Newton's second law of motion applied to the $i$ th object yields: $$m_i \frac {d^2 \vec {r_i}}{dt^2}=\sum_{j=1,N}^{j\neq i} \vec {f_{ij}}$$ Let us now take the above equation and sum it over all objects. We obtain $$\sum_{i=1,N} m_i \frac {d^2 \vec {r_i}}{dt^2}= \sum_{i,j=1,N}^{j\neq i} \vec {f_{ij}}$$ because of newton's third law of motion , the right side of equation is equal to 0,but the question is that i can't understand how the left side of equation turn to below? $$M\frac {d^2\vec {r_{cm}}}{dt^2}=\vec 0$$ where $M=\sum_{i=1}^{N}m_i$ is the total mass. The quantity $\vec {r_{cm}}$is the vector displacement of the center of mass of the system, which is an imaginary point whose coordinates are the mass weighted averages of the coordinates of the objects that constitute the system: i.e., $$\vec {r_{cm}}=\frac{\sum_{i=1}^{N}m_i \vec {r_i}}{\sum_{i=1}^{N}m_i}$$
Answer
The position vector of the centre of mass is defined as: $$\mathbf {r_{cm}}=\frac{\sum_{i=1}^{N}m_i \mathbf {r_i}}{M}$$ i.e., $$M \mathbf {r_{cm}}=m_1 \mathbf r_1+m_2 \mathbf r_2+ m_3 \mathbf r_3+...+m_N \mathbf r_N$$ $$\Rightarrow M \frac {d^2 \mathbf {r_{cm}}}{dt^2} = m_1 \frac {d^2 \mathbf {r_{1}}}{dt^2}+ m_2 \frac {d^2 \mathbf {r_{2}}}{dt^2}+...+m_N \frac {d^2 \mathbf {r_{N}}}{dt^2}$$ (Double Differentiating both sides) Here $M=\sum_{i=1}^{N}m_i$. Since single differentiation of the position vector gives velocity $\mathbf v$ and double differentiation gives acceleration $\mathbf a$. Therefore $$\frac {d^2 \mathbf {r_{cm}}}{dt^2} = \mathbf a$$ This means that $$\sum_{i=1}^{N} m_i \frac {d^2 \mathbf {r_i}}{dt^2} = M \frac {d^2 \mathbf {r_{cm}}}{dt^2} = M \mathbf a = \mathbf {F_{net}}$$
Now if the net force acting on the object is $\mathbf 0$ then $$M \mathbf a= M\frac {d^2 \mathbf {r_{cm}}}{dt^2}=\mathbf 0$$
Note that internal forces cannot cause acceleration as they always come in action-reaction pair.
A falling object moves along a geodesic path ('straight path') in spacetime. When it comes to rest it now follows a 'curved path' through spacetime. Is the passage of time and force of gravity fundamentally affected by this difference?
To be really clear, assume one object is falling , at height x from the ground and another is hovering at height x. Do they experience the same gravitational pull? Will they age at the same rate?
I feel I have mangled the correct terminology, forgive me. Thanks!
Answer
Since from your previous questions you're obviously interested in learning how this is done I'll go into the detail of the calculation. Note that a lot of what follows can be found in existing answers, but I'll tailor this answer specifically at you.
Time dilation is calculated by calculating the proper time change, $d\tau$, using the expression:
$$ c^2d\tau^2 = -g_{ab}dx^adx^b \tag{1} $$
where $g_{ab}$ is the metric tensor. The reason we can use this to calculate the time dilation is that the proper time is an invariant i.e. all observers will agree on its value. To illustrate how we do this let's consider the simple example of an astronaut moving at velocity $v$ in flat spacetime. In this case the metric is just the Minkowski metric, and equation (1) simplifies to:
$$ c^2d\tau^2 = c^2dt^2 - dx^2 - dy^2 - dz^2 \tag{2}$$
First we do the calculation in the astronaut's rest frame. In that frame the astronaut isn't moving so $dx = dy = dz = 0$, and the proper time as observed by the astronaut is just:
$$ d\tau_{astronaut} = dt $$
Now let us here on earth calculate the proper time. We'll arrange our coordinates so the astronaut is moving along the $x$ axis, so $dy = dz = 0$. In that case equation (2) becomes:
$$ c^2d\tau^2 = c^2dt^2 - dx^2 $$
To procede we have note that if the astronaut is moving at velocity $v$ that means $dx/dt = v$, because that's what we mean by velocity. So $dx = vdt$. Put this into our equation and we get:
$$ c^2d\tau^2 = c^2dt^2 - (vdt)^2 $$
which rearranges to:
$$ d\tau_{Earth} = \sqrt{1 - \frac{v^2}{c^2}}dt $$
Because the proper time is an invariant both we and the astronaut must have calculated the same value i.e. $d\tau_{Earth} = d\tau_{astronaut}$, and if we substitute for $d\tau_{Earth}$ in the equation above we get :
$$ \frac{d\tau_{astronaut}}{dt} = \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{\gamma}$$
where $\gamma$ is the Lorentz factor. But the left hand side is just the variation of the astronaut's time with our time - in other words it's the time dilation. And the equation is just the standard expression for time dilation in special relativity that we teach to all students of SR.
The point of all this is that we can use exactly the same procedure to work out the time dilation in gravitational fields. Let's take the gravitational field of a spherically symmetric body, which is given by the Schwarzschild metric:
$$ c^2d\tau^2 = c^2\left(1-\frac{2GM}{r c^2}\right)dt^2 - \left(1-\frac{2GM}{r c^2}\right)^{-1}dr^2 - r^2 (d\theta^2 + sin^2\theta d\phi^2) \tag{3} $$
This is very similar to the equation (2) that we used in flat spacetime, except that the coefficients for $dt$ etc are now functions of distance, and we do the calculation in exactly the same way. Let's start with calculating the time dilation for a stationary astronaut at a distance $r$. Because the astronaut is stationary we have $dr = d\theta = d\phi = 0$, and equation (3) simplifies to:
$$ c^2d\tau^2 = c^2\left(1-\frac{2GM}{r c^2}\right)dt^2 $$
and this time we get:
$$ \frac{d\tau}{dt} = \sqrt{1-\frac{2GM}{r c^2}} = \sqrt{1-\frac{r_s}{r}} \tag{4} $$
where $r_s$ is the Schwarzschild radius. And that's it - calculating the time dilation for a stationary observer in a gravitational field is as simple as that. You'll find this expression in any introductory text on GR.
But the real point of your question (finally we get to it!) is what happens if our observer in the gravitational field is moving? Well, let's assume they are moving in a radial; direction at velocity $v$, so just as in the flat space case we have $dr = vdt$ and $d\theta = d\phi = 0$. We substitute this into equation (3) to get:
$$ c^2d\tau^2 = c^2\left(1-\frac{2GM}{r c^2}\right)dt^2 - \left(1-\frac{2GM}{r c^2}\right)^{-1}v^2dt^2 $$
which rearranges to:
$$ \frac{d\tau}{dt} = \sqrt{1-\frac{r_s}{r} - \frac{v^2/c^2}{1-\frac{r_s}{r}}} \tag{5} $$
And once again, it's as simple as that. If you compare this result with equation (4) you'll see the time dilation for a moving object is different because we have an extra term $\frac{v^2/c^2}{1-\frac{r_s}{r}}$ in the square root.
One last sanity check: what happens if we go an infinite distance away from the gravitating object so $r \rightarrow \infty$? Well if $r \rightarrow \infty$ then $r_s/r \rightarrow 0$ and equation (5) becomes:
$$ \frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}} = \frac{1}{\gamma} $$
which is exactly what we calculated for flat spacetime.
There are three types of neutrinos known today. When detecting them, how can we tell which type we are detecting?
Answer
Neutrino flavor is defined as agreeing with the flavor of the charged lepton participating in the interaction, so that the neutrino in the reaction $$ \nu + A \to \mu + X \,, $$ is defined to be a muon neutrino and the one in $$ \nu + n \to e + p $$ is a electron neutrino by definition.
We have no way of knowing the alleged flavor of a neutrino participating in a neutral current interaction.
As a matter of experimental fact electron and moun neutrinos (and anti-neutrinos) are easy, but tau neutrinos are much harder because demonstrating that you have a tau-lepton is hard, but both OPERA and IceCube can do that (to chose currently running experiments).
In this picture the acceleration vector $\vec{a}$ points upward when the pendulum is halfway
But according to this picture, the force acts tangentially:
Which means the acceleration should be tangential too, and never pointing upward?
So whats right?
Answer
Please note that in the picture, there are two forces acting: 1) the weight, mg, which acts vertically downward, and does not change, and 2) the tension in the string, Z, which points from the mass to the point the string connects to the ceiling, provided the string remains taut. Z varies with time periodically.
These two forces combine to give the resultant force, and it is the resultant force which occurs in the same direction as the acceleration, as seen in the gif.
The green arrows in the picture are actually just the tangential and normal components of gravity.
Edit: also, I believe the source of confusion might have lied with assuming the normal component of gravity cancels with the tension. This is not the case: you cannot use the equations of equilibrium if the system is not in equilibrium, i.e. accelerating.
I'd like some clarification regarding the roles of active and passive diffeomorphism invariance in GR between these possibly conflicting sources.
1) Wald writes, after explaining that passive diffeomorphisms are equivalent to changes in coordinates,
"This 'passive' point of view on diffeomorphisms is, philosophically, drastically different from the above 'active' viewpoint, but, in practice, these viewpoints are really equivalent since the components of the tensor $\phi*T$ at $\phi(p)$ in the coordinate system $\{y^\mu\}$ in the active viewpoint are precisely the components of $T$ at $p$ in the coordinate system $\{x'^\mu\}$ in the passive viewpoint."
(Wald, General Relativity, Appendix C.1)
2) Gaul and Rovelli write,
"General relativity is distinguished from other dynamical field theories by its invariance under active diffeomorphisms. Any theory can be made invariant under passive diffeomorphisms. Passive diffeomorphism invariance is a property of the formulation of a dynamical theory, while active diffeomorphism invariance is a property of the dynamical theory itself."
(Gaul and Rovelli, http://arxiv.org/abs/gr-qc/9910079, Section 4.1)
It seems that Wald is saying that there is no mathematical difference between the two, and that both imply the same physical consequences. It seems that Gaul and Rovelli, however, are saying that only active diffeomorphism invariance has physical meaning that may influence the dynamics of the theory. Can anyone explain?
Answer
I think the best approach is to try to understand a concrete example:
Let's look at a piece of the Euclidean plane coordinatized by $x^a=(x, y); a=1,2$ in a nice rectangular grid with Euclidean metric. Now suppose we define a transformation $$X(x,y)=x(1+\alpha y^2) $$ $$Y(x,y)=y(1+\alpha x^2) $$ $\alpha$ is just a constant, which we will take as 5/512 - for the sake of being able to draw diagrams. A point P with coordinates $(x,y)=(8,8)$ is mapped to a point P' with coordinates $(X,Y)=(13,13)$.
Here we don't think of P and P' as different points, but rather as the same point and $(13,13)$ are just the coordinates of P in the new coordinate system $X^a$. 
In the picture, the blue lines are the coordinate lines $x^a=$ const and the red lines are the coordinate lines $X^a=$ const. Metric components on our manifold $g_{ab}(x)$ get mapped to new values $$h_{ab}(X)={\frac{\partial x^c}{\partial X^a}}{\frac{\partial x^d}{\partial X^b}} g_{cd}(x) \ \ \ (1) $$ This represents the same geometric object since $$ h_{ab}(X)dX^a\otimes dX^b = g_{ab}(x)dx^a\otimes dx^b$$
One description of the active view that is sometimes used is that points are "moved around" (in some sense perhaps it's better to think just of an association between points, "moving" implies "with respect to some background"). So in our example, we'd think of the point P as having been "stretched out" to the new location P'. (These locations are with respect to the old $x$ coordinate system). 
The old (blue) $x=$ constant coordinate lines get dragged along too, into the red lines shown in the diagram. So the point P retains its old coordinate values $(8,8)$ in its new location, i.e $(X,Y)=(8,8)$. The metric is also dragged along (see for example Lusanna) according to: $$h_{ab}(X)|_{P'} \ dX^a \otimes dX^b = g_{ab}(x)|_{P}\ dx^a \otimes dx^b \ \ \ (2)$$ So the old Euclidean metric $dx^2+dy^2$ becomes $dX^2+dY^2$, i.e. still Euclidean in the new $(X,Y)$ chart - nothing has changed. So, for example, the angle between the red vectors $\frac{\partial}{\partial X}$, $\frac{\partial}{\partial Y}$ is still 90 degrees, as it was for the blue vectors $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ ! My guess is that this is what Wald means by the physical equivalence - in this example a Euclidean metric remains Euclidean.
Now, if we look at the red vectors from the point of view of the blue frame, they sure don't look orthogonal*, so from the blue point of view, it can only be a new metric in which the red vectors are orthogonal. So active diffeomorphisms can be interpreted as generating new metrics.
Now suppose we have a spacetime - a manifold with metric for which the Einstein tensor $G_{\mu\nu}$ vanishes. Applying an active diffeomorphism, we can generate the drag-along of the Einstein tensor by a rule analogous to (2). As we have discussed, if we compare the dragged along metric with the old one in the same coordinates, we see we have a spacetime with a new metric. Moreover, the new spacetime must also have vanishing Einstein tensor - by the analog of (2), the fact that it vanished in the old system means it vanishes in the new system and hence our newly created Einstein tensor vanishes too (if a tensor vanishes in one set of coordinates it vanishes in all).
In this respect, the invariance of Einstein's equations under active diffeomorphisms is special. If we take, for example, the wave equation in curved spacetime $$(g^{\mu\nu}{\nabla}_{\mu}{\nabla}_{\nu}+\xi R)\phi(x) = 0 $$ then active diffeomorphisms don't naturally take solutions to solutions - they change the metric, and the metric in this equation is part of the background, and fixed. By contrast, in Einstein's equations, the metric is what you're solving for so active diffeomorphism invariance is built in.
*Just compute the vectors $\frac{\partial}{\partial X}, \frac{\partial}{\partial Y}$ in terms of $\frac{\partial}{\partial x}$, $\frac{\partial}{\partial y}$ and test their orthogonality using the original Euclidean metric.
In high-school level textbooks* one encounters often the principles of independence of motion and that of composition (or superpositions) of motions. In this context this is used as "independence of velocities" and superposition of velocities (not of forces).
This is often illustrated by the example of the motion of a projectile, where the vertical and horizontal motions are said to be independent and that the velocities add like vectors.
Now, if $x\colon \mathbb{R} \to \mathbb{R^3}$ describes the motion of the of the considered object, it is clear that one can decompose the velocity $v = \dot x$ arbitrarily by $v = v_1 + v_2$ where $v_1$ is arbitrary and $v_2 := v - v_1$.
This leads me to my first question: Am I correct that this is purely trivial math and contains no physics at all? If so, it would not deserve to be called "principle of composition of motions" or something like that and said to be fundamental.
However it seems that one could interpret the decomposition above such that $v_1$ is the velocity of the object with reference to a frame of reference moving with $v_2$. If so, how can one see, that this goes wrong in the relativistic case?
Now suppose you have two forces $F_1$ and $F_2$ which you can switch on and off, suppose that $F_i$ alone would result in a motion $x_i$ ($i=1,2$). Newtonian Mechanics tells us the principle of superposition of forces, i.e. if you turn both forces $F_1$ and $F_2$ on, the resulting motion $x$ is the solution of the differential equation $\ddot x = \frac1m F(x, \dot x, t)$ (where m is the Mass of our object) with $F = F_1 + F_2$.
One might interpret the principle of composition of motions such that always $\dot x = \dot x_1 + \dot x_2$ holds true. This is clearly the case if $F_i$ depends linearily on $(x,\dot x)$. However I think that it doesn't need to be true for nonlinear forces. This leads me the my third question: Is there any simple mechanical experiment where such nonlinear forces occur, which shows that in this case the "principle of composition of motions" doesn't hold?
*I have found this in some (older) german textbooks, for example: Kuhn Physik IIA Mechanik, p. 107, Grimsehl Physik II p.16,17
compare also
http://sirius.ucsc.edu/demoweb/cgi-bin/?mechan-no_rot-2nd_law and Arons
On this page, the following claims are made:
You can't start a fire with moonlight no matter how big your magnifying glass is.
General rule of thumb: You can't use lenses and mirrors to make something hotter than the surface of the light source itself. In other words, you can't use sunlight to make something hotter than the surface of the Sun.
Lenses and mirrors work for free; they don't take any energy to operate. If you could use lenses and mirrors to make heat flow from the Sun to a spot on the ground that's hotter than the Sun, you'd be making heat flow from a colder place to a hotter place without expending energy. The second law of thermodynamics says you can't do that. If you could, you could make a perpetual motion machine.
(more claims which I'll omit here)
I neither believe the claim nor follow any of the reasoning.
First, I don't get the thing about the perpetual motion machine:
If this is regarding the first law of thermodynamics (conservation of energy), then it's perfectly possible to lose energy while still heating up an object, which would avoid perpetual motion, so I don't get the argument.
If this is regarding the second law of thermodynamics (increasing of entropy), then it's also invalid because the disorder in the system is still increasing.
If this is about something else, then I don't know what that is.
Second, here is a video of a guy using a mirror he can hold in his hand to light paper on fire.
Clearly the mirror itself isn't getting as hot as he's making the newspaper, and clearly the mirror is the one reflecting the sunlight.
So how can one claim that the paper fundamentally can't get hotter than the reflecting surface? Why can't the moon's surface fundamentally behave similarly (albeit with poorer reflectivity)?
Can someone explain? Is Randall confusing heat with temperature? Or maybe conduction with radiation? Or am I missing something subtle (or perhaps not so subtle)?
Can someone explain to me the origin of the exchange interaction between two electrically charged spin 1/2 fermions? Quantitative or qualitative accepted.
Answer
The wave function is antisymmetric under exchange of (all) the coordinates of each electron (we'll just call them electrons since that's shorter than "two electrically charged spin 1/2 fermions" and equivalent). We'll write the wave function as: \begin{align} \Psi(1,2) &= \psi_1(\mathbf{r}_1)\psi_2(\mathbf{r}_2)|s_1s_2\rangle -\psi_1(\mathbf{r}_2)\psi_2(\mathbf{r}_1)|s_2s_1\rangle. \end{align} (Check that this is antisymmetric under $\mathbf{r}_1 \leftrightarrow \mathbf{r}_2$ & $s_1 \leftrightarrow s_2$.)
Now let's calculate the force between these due to some two-body operator $V(\mathbf{r}_1,\mathbf{r}_2)$ (that might depend on spin). It's proportional to (I'm ignoring normalization): \begin{align} &\int d^3r_1 d^3r_2 \Psi^\dagger(1,2)V(\mathbf{r}_1,\mathbf{r}_2)\Psi(1,2)\\ &= 2\int d^3r_1 d^3r_2 \Big\{ |\psi(\mathbf{r}_1)|^2|\psi(\mathbf{r}_2)|^2 \langle s_1 s_2|V(\mathbf{r}_1,\mathbf{r}_2)|s_1 s_2\rangle\\ &- \mbox{Re}[\psi_1^*(\mathbf{r}_1)\psi_2(\mathbf{r}_1)\psi_2^*(\mathbf{r}_2)\psi_1(\mathbf{r}_2) \langle s_1 s_2|V(\mathbf{r}_1,\mathbf{r}_2)|s_2 s_1\rangle]\Big\} \end{align} The second term is the exchange term. Note that the square of the wave functions in the first term are proportional to particle densities at $\mathbf{r}_1$ and $\mathbf{r}_2$. While the second term has different wave functions, $\psi^*_1(\mathbf{r}_1)\psi_2(\mathbf{r}_1)$ at the point $\mathbf{r}_1$.
The short answer is: "The origin of the exchange interaction is the property of definite parity under coordinate exchange of the wave function." (Often the answer is given that it's the Pauli principle. But this is incomplete. Systems of bosons with internal coordinates (flavor, spin, etc.) enjoy the exchange interaction too. Prove this.) Note too that this holds for any two-body operator.
Consider a theory in $D$ spatial dimensions involving one or more scalar fields $\phi_a$, with a Lagrangian density of the form $$L= \frac{1}{2} G_{ab}(\phi) \partial_\mu \phi_a \partial^\mu \phi_b- V(\phi)$$ where the eigenvalues of $G$ are all positive definite for any value of $\phi$, and $V = 0$ at its minima. Any finite energy static solution of the field equations is a stationary point of the potential energy $$E = I_K + I_V ,$$ where $$I_K[\phi]= \frac{1}{2} \int d^Dx G_{ab}(\phi) \partial_j \phi_a \partial_j\phi_b$$ and $$I_V = \int d^Dx V(\phi)$$ are both positive. Since the solution is a stationary point among all configurations, it must, a fortiori, also be a stationary point among any subset of these configurations to which it belongs. Therefore, given a solution $\phi(x)$, consider the one-parameter family of configurations, $$f_\lambda(x)= \bar{\phi}(\lambda x)$$ that are obtained from the solution by rescaling lengths. The potential energy of these configurations is given by \begin{align} E_\lambda &= I_K(f_\lambda) + I_V(f_\lambda)\\ &=\lambda^{2-D} I_K[\bar\phi]+\lambda^{-D} I_V[\bar\phi]\tag{1} \end{align}
$\lambda = 1$ must be a stationary point of $E(\lambda)$, which implies that $$0=(D-2) I_K[\bar\phi]+D I_V[\bar\phi] \tag{2}$$
My problem is, how they got the equation(2) from (1)?
Answer
I didn't go through all of your equations. However, if you take (1), differentiate it w.r.t $\lambda$ and set $\lambda = 1$, then since $\lambda=1$ is the stationary point $E'(\lambda)|_{\lambda=1} = 0$. This is equation (2)
Suppose a wheel with radius $R$ is resting on a non-inclined surface. A torque $\tau$ is applied to the wheel center. In an attempt to prevent wheel from spinning, the ground applies a static friction force to the wheel at the contact point (parallel to the surface), then the wheel starts rolling without spinning. The same friction force also acts as a torque on the wheel around its axis. This scenario is depicted below:
I'm trying to find the magnitude of the force the ground applies to the wheel - that is, the force which drives the wheel forward (which should be the same in magnitude as the force the wheel applies to the ground at the contact point).
That's how I'm doing it:
The relation between linear acceleration $a$ and angular acceleration $\alpha$ for a pure rolling movement is given by \begin{equation} \tag{1} a = \alpha \centerdot R, \end{equation}
The relation between torque $\tau$ and angular acceleration $\alpha$ is
\begin{equation} \tau = I \centerdot \alpha \end{equation}
where $I$ is the moment of inertia of the wheel around its axis.
The relation between torque $\tau$, force $f$ and lever arm $R$ is:
$$\tau = f \centerdot R$$
Being the engine torque $\tau_e$, the friction force $f$, the counter torque due to friction force $\tau_f$ and the moment of inertia of the wheel $I$ around its axis given by $\frac{1}{2}mR^2$:
The linear acceleration of the wheel is due to the friction force only:
$$f = ma$$ $$a = \frac{f}{m}$$
This is the counter torque the ground applies on the wheel's edge (negative, because it pointing in the opposite direction of $\tau_e$):
$$\tau_f = -f \centerdot R$$
The net torque causes angular acceleration on the wheel:
$$\tau = \tau_e + \tau_f$$ $$\tau = \tau_e - f \centerdot R$$ $$\tau = I \centerdot \alpha$$ $$\alpha = \frac{\tau}{I}$$ $$\alpha = \frac{\tau_e - f \centerdot R}{I}$$ $$\alpha = \frac{\tau_e - f \centerdot R}{\frac{1}{2}mR^2}$$
Substituting $\alpha$ and $a$ in $(1)$ gives:
$$\frac{f}{m} = \frac{\tau_e - f \centerdot R}{\frac{1}{2}mR^2}R$$
Rearranging:
$$f = \frac{2}{3}\frac{\tau_e}{R}$$
And that is the force $f$ from static friction which pulls the wheel forwards without making it spin or slip, and consequently, the force the wheel applies to the road surface at the contact point.
But I found this link: http://www.asawicki.info/Mirror/Car%20Physics%20for%20Games/Car%20Physics%20for%20Games.html
And it says: "The torque on the rear axle can be converted to a force of the wheel on the road surface by dividing by the wheel radius. (Force is torque divided by distance)."
That statement doesn't match the approach I used above. If the force of the wheel on the ground was simply engine torque divided by radius (negative, since it pointing in the opposing direction):
$$f = -\frac{\tau_e}{R}$$
then the counter torque applied to the wheel would be
$$\tau_f = -f \centerdot R$$
that implies that $\tau_f = -\tau_e$.
That means the net torque would be zero, and the wheel would just slip without rotating at all.
Am I doing something wrong on my calculations?
The electron has magnetic dipole moment which points in the spin direction, which is relatively easy to understand because it mostly follows from the definition. However, why is it that the (possibly non-zero) electron electric dipole moment (EDM) also has to be collinear with the spin? i.e. why must any possible internal polarisation of the electron have to align with spin?
I have seen in many textbooks and sources which say that we can't experimentally measure potential energy but we can measure differences in potential energy.
$$\Delta U_g=-W_g$$
Choosing zero potential (reference point) at the ground.
Now if I measure change in gravitational potential energy from zero point to a point where an object thrown upwards attains zero velocity, then $U_g$ at that point would just be negative of the work done.
If potential energy at that point can be calculated then why is it said that absolute potential energy at a point can't be calculated?
Answer
Simply put, potential energy is the energy an object possesses because of its position. Position, or location, is always relative. Therefore there is no such thing as an exact or absolute position in space and consequently no exact potential energy.
Potential energy must be measured relative to something. Suppose a 1 Kg ball is suspended 1 meter above the surface of the earth. Relative to the surface of the earth it has a potential energy of 9.81 Joules. But suppose we put a 0.5 m high table underneath the ball. Relative to the surface of the table it has a potential energy of 4.9 Joules.
We haven't moved the ball, so which is the real potential energy?
So I though, and was told, that entropy is the amount of disorder in a system. Specifically the example of heat flow and it flows to maximize entropy. To me this seemed odd. This seemed more ordered to me, from two things (hot and cold) to one thing (lukewarm).
So as I have read today on wikipedia
https://en.wikipedia.org/wiki/Entropy_%28order_and_disorder%29
In recent years the long-standing use of term "disorder" to discuss entropy has met with some criticism.[20][21][22]
When considered at a microscopic level, the term disorder may quite correctly suggest an increased range of accessible possibilities; but this may result in confusion because, at the macroscopic level of everyday perception, a higher entropy state may appear more homogenous (more even, or more smoothly mixed) apparently in diametric opposition to its description as being "more disordered". Thus for example there may be dissonance at equilibrium being equated with "perfect internal disorder"; or the mixing of milk in coffee being described as a transition from an ordered state to a disordered state.
It has to be stressed, therefore, that "disorder", as used in a thermodynamic sense, relates to a full microscopic description of the system, rather than its apparent macroscopic properties. Many popular chemistry textbooks in recent editions increasingly have tended to instead present entropy through the idea of energy dispersal, which is a dominant contribution to entropy in most everyday situations.
So what this is saying basically many people mistake the macroscopic outcome of entropy in the same manner as I have above. Seemingly two dissimilar things combining and becoming one.
They say to understand it microscopically. But this still is odd to me, take for example the particles in a box, when they come to thermal equilibrium they are said to have the most entropy they can within their system (from wiki again)
Entropy – in thermodynamics, a parameter representing the state of disorder of a system at the atomic, ionic, or molecular level; the greater the disorder the higher the entropy.[7]
What is disorder in the molecular sense I guess is my question. When the particles all come to thermal equilibrium they all seem to be uniform, which I equate with ordering.
I feel disorder in the molecular sense would be fast moving particles, making a mind boggling amount of interactions with each other. Such as in a hot gas, hot gas seems super disordered, but put cold gas in thermal contact with it, let them come to thermal equilibrium, and now they are more ordered!?
Answer
Sir James Jeans has an excellent answer, without using the word "order" or "disorder". Consider a card game...¿is there anyone else on this forum besides me who still plays whist? His example is whist. I had better use poker.
You have the same probability of being dealt four aces and the king of spades as of being dealt any other precisely specified hand, e.g., the two of clubs, the three of diamonds, the five of hearts, the eight of spades, and the seven of spades. Almost worthless.
This is the microstate. A microstate is the precisely specified hand.
A macrostate is the useful description, as in the rules: two pairs, four of a kind, royal flush, flush, straight, straight flush, worthless.
You have a much lower probability of being dealt four aces than of being dealt a worthless hand.
A macrostate is the only thing about which it makes sense to say "ordered" or "disordered". The concept of "order" is undefined for a microstate. A macrostate is highly ordered or possesses a lot of information if your knowledge that you are in that macrostate implies a very high degree of specificity about what possible microstate you might be in. "Four aces", as a description, tells you a lot. "One pair", as a description, tells you much less. So the former is a state of low entropy and the latter is a state of higher entropy.
A macrostate is a probability distribution on the set of microstates. "Four aces" says that the microstate "ace, deuce, three, four, five, all spades" has zero probability. Most microstates have zero probability, they are excluded from this description. But "four aces and king of spades" has probability 1/48, so does "four aces and king of hearts", etc. etc. down to "four aces and deuce of clubs". The entropy formula then is $-k \log \frac1{48}$ where $k$ is not Boltzmann's constant. But the entropy of "one pair" is much higher: put "W" to be the number of different precisely specified hands which fall under the description "one pair". Then its entropy is $k \log W$.
Jeans makes the analogy with putting a kettle of water on the fire. The fire is hotter than the water. Energy (heat) is transferred from the fire to the water, and also from the water to the fire, by, let us assume, molecular collisions only. When we say "cold kettle on a hot fire" we are describing a macrostate. When we say "water boils" that is another macrostate. When we say "fire gets hotter and water freezes" we are also describing a possible macrostate that might result. ¿What are the entropies? They are proportional to the logarithm of the number of microstates that fall under theses three descriptions. Now, by the Maxwell distribution of energies of the molecules, there are very many high-energy molecules in the fire that come into contact with lower-energy molecules in the water and transfer energy to the water. There are very many precisely specified patterns of interaction at the individual molecular level of energy transfer from the fire to the water, so "boils" has a large entropy.
But there are some ways of freezing the water: the Maxwell distribution says that a few of the molecules in the fire are indeed less energetic than the average molecule in the water. It is possible that only (or mostly) these molecules are the ones that collide with the water and receive energy from the water. This is in strict analogy to the card game: there are very few aces, but it is possible you will get dealt all of them. But there are far fewer ways for this to happen, for the water to freeze, than for the previous process of boiling. Therefore this macrostate has less entropy than the "boils" macrostate.
This example shows that to use the definition of entropy you have to have defined the complete set of possible microstates you will consider as possible, and you have to study macrostates which are sets of these microstates. These different macrostates can be compared as to their entropies.
You cannot suddenly switch to a hockey game and compare the entropy of a full house to the entropy of "Leafs win". If you wish to make comparisons such as that, you would have to initially define an overarching system which contained both sets of microstates, define the macrostates, and even then comparisons of entropy would be merely formalistic. The laws of thermodynamics only apply when there is an interaction between all parts of the system such that any one component of the system has a chance of interchanging energy with any other part of the system. Within the time allotted. We also had to assume that each hand was equally likely, i.e., that Persi Diaconis was not dealing...some dealers know how to make some precisely specified hands less likely than others.... without these two assumptions, there is no connection between thermodynamic entropy and informational entropy, and so the Second Law of thermodynamics will not apply to informational entropy.
Better than "more ordered" would be to think "more specificity". There are fewer ways to arrange that all the slow molecules are in one body and all the fast ones in the other than there are to arrange a seemingly random mix, just as there are fewer ways to arrange that I get all the aces, than there are to arrange that each player gets one ace.
See also Does entropy really always increase (or stay the same)? where I put Sir James's exact words at length.
I was reading about interesting article here which suggests that our universe is a big computer simulation and the proof of it is a Quantum Physics.
I know quantum physics tries to provide some explanation about unpredictable behavior of universe at atomic scale but could not believe that thing has anyway relates to pixels of the universe ?
Can it be true that behavior of universe at atomic level coming from some super computer ?
The maximum obtainable angular resolution of an optical system with some given aperture is well known, but it seems to me that this isn't a real theoretical limit. The assumption is that you are going to take a picture using the system and no further processing will take place. However, given the known point spread function of the optical system, you could perform a deconvolution calculation. The final resolution of the processed image should be limited by the noise. So, what then seems to matter is the observation time (the longer you integrate the signal the better the signal/noise ratio will be).
So, what is the correct theoretical limit of the maximum resolution in terms of brightness of the two sources to be resolved, the aperture and the observation time (assuming that the only noise comes from the fluctuations in the finite numbers of photons from the sources)?
Answer
I'm going to focus on the information contained in the light field itself. This excludes from the discussion many if not all "superresolution" techniques, which directly or indirectly make use of information further to that in the imaging light field[footnote 1].
It is true that you can do a deconvolution to get somewhat below the traditional diffraction "limit" if the signal to noise ratio is very high. But there is a fundamental limit even to this, and it leads to tradeoff between distance from the object and resolution. Near field microscopy can image arbitrarily small features with light, but herein lies the catch - you need to be near to the object, and, for a given noise level, no matter how small, the imageable feature size decreases exponentially with distance from the object. The notion of what leads to the notion a hard limit arises from:
Only nonevanescent waves (corresponding to truly free photons) can convey Fourier component information to an imaging system that is arbitrarily far from the object
The phenomenon is indeed best understood through evanescent waves. If you want to encode Fourier component of a transverse feature into the light field and that component's spatial angular frequency $k_f>k$ (here $k$ is the light's wavenumber), then as the plane wave encoding this component propagates away from the object (call this the $z$ direction), its amplitude varies as $\exp(-\sqrt{k_f^2-k^2}\,z)$, i.e. the wavevector component becomes imaginary and the amplitude swiftly drops off with distance. As $z\to\infty$, only the nonevanescent waves are left, so the system transfer function looks more and more like a hard limitting lowpass filter with cutoff spatial frequency $k$ as $z$ increases. If you want to image features of characteristic length $d<\lambda$, then the loss in signal to noise ratio is:
$$\begin{array}{lcl}SNR &=& SNR_0-40\,\pi\,z\,\sqrt{\frac{1}{d^2}-\frac{1}{\lambda^2}}\,\log_{10}e\quad\text{(decibel)}\\&\approx& SNR_0-40\,\pi\,\frac{z}{d}\,\log_{10}e\quad (d\ll\lambda)\end{array}$$
where $SNR_0$ is the signal to noise ratio you would get if you held the SNOM right up on the imaged object and $z$ is the distance of the SNOM tip from the object. This is a horrifically fast dropoff. If you probe scans $1{\rm \mu m}$ from the imaged object and we wish to see $50{\rm nm}$ sized objects, the signal to noise lost by the mere $1{\rm \mu m}$ standoff is 1000 decibels (a power factor of $10^{100}$!). Practically speaking, your probe must be within a distance $d$ or less of the imaged object, where $d$ is the subwavelength feature length you wish to see; the above formula then gives an SNR dropoff of about $54{\rm dB}$ when $z=d$.
Footnotes
[1]. For example STED depletes fluorophores out of focus before taking the final light reading, thereby disabling anything more than a few tens of nanometres from the focus from registering)
I was wondering why electrons orbited protons rather than protons orbiting electrons.
My first thought was that it was due to the small amount of gravitational attraction between them that would cause the orbit to be very close to the proton (or nucleus). The only other idea that I would have is that the strong interaction between protons and neutrons have something to due with this.
I have heard that the actual answer is due to something in QM, but haven't seen the actual explanation. The only relation to QM that I can think of is that due to a proton's spin and the fact that they are fermions, the atomic orbitals should be somewhat similar. Do protons have the same types of orbitals, that are just confined by the potential of the strong force?
A related question that came up while thinking of this being due to a gravitational interaction: do orbits between protons and electrons have a noticeable rotation between each other (as the sun orbits the earth just as the earth orbits the sun), or is any contribution this has essentially nullified by the uncertainty of the location of the electron (and possibly proton as well)?
Answer
Technically the electron and proton are both orbiting the barycenter of the system, both in classical and quantum mechanics, just as in gravitational systems.
You find the same dynamics for the system if you assume the proton and electron are moving independently about the barycenter, or if you convert to a one-body problem of a single "particle" with the reduced mass $$ \mu = \frac{m_p m_e}{m_p + m_e } \approx m_e \left(1 - \frac{m_e}{m_p}\right). $$ However, the proton is nearly 2000 times more massive than the electron. If we assume that the proton is fixed and infinitely massive, and model our atom using $\mu=m_e$, we introduce errors starting in the fourth decimal place. Usually that's good enough.
This is a quote from the book A first course in general relativity by Schutz:
What we shall now do is adopt a new unit for time, the meter. One meter of time is the time it takes light to travel one meter. The speed of light in these units is $$\begin{align*}\end{align*}$$
$$\begin{align*} c &= \frac{ \text{distance light travels in any given time interval}}{\text{the given time interval}}\\ &= \frac{ \text{1m}}{\text{the time it takes light to travel one meter}}\\ &= \frac{1m}{1m} = 1\\ \end{align*}$$
So if we consistently measure time in meters, then c is not merely 1, it is also dimensionless!
Either Schutz was on crack when he wrote this, or I'm a dope (highly likely) 'cos I can't get my head around this:
The space-time interval between different events at the same location measures time, and between different events at the same time measures space. So they're two completely different physial measurents: One is a time measurement using a clock, the other a space measurement using a ruler. In which case the units of $c$ should be $ms^{-1}$
Does Schutz correctly show how $c$ can be a dimensioness constant?
We know that it takes no energy to change the direction of a vector, we know also that it takes no energy to displace a body in motion if a push is applied exactly at a right angle on its center of mass.
Considering that, if that is true, can you explain why a simple change of direction without any increase of speed is considered an acceleration?
A transparent isotropic dielectric medium moving in the negative $x'$ direction at speed $v$ in frame $S'$ is stationary in frame $S$, where it has refractive index $n$. In other words, frame $S'$ is moving at speed $v$ in the positive $x$ direction relative to frame $S$. We take the two frames' origins to coincide at $t = 0$.
Calculate the refractive index $n'$ in frame $S'$ experienced by light travelling inside the dielectric along the $y$-axis in frame $S$.
I have drawn a schematic of the setup below.
I have found two approaches to this question, both of which give me different answers. I was hoping somebody would be able to point out the flaw in one of these methods.
The 4-wavevector of a photon moving at an angle $\theta$ to the $x$-axis in the $x$-$y$ plane is
$$ K = (\omega/c, k \cos \theta, k \sin \theta, 0 )$$
By applying a Lorentz transformation to this vector we arrive at the following relations between frequency, wavenumber and angle in $S'$ and those quantities in $S$:
$$ \omega'/c = \gamma\omega/c - \gamma\beta k \cos \theta $$ $$ k' \cos \theta' = \gamma k \cos \theta - \gamma \beta \omega /c $$ $$ k' \sin \theta' = k \sin \theta$$
The refractive index in $S'$ is defined by
$$ \omega'/k' = c/n' \qquad \implies \qquad n' = ck'/\omega'$$
We set $\theta = \pi/2$, corresponding to motion in the $y$-direction. Then summing the squares of the bottom two equations to eliminate $\theta'$, square-rooting, and then dividing the result by the first equation, we find:
$$ ck'/\omega' = \frac{c}{\omega\gamma} \sqrt{k^2 + \frac{\gamma^2\beta^2 \omega^2}{c^2}}$$
If we pull a factor of $k^2$ outside we can write this thus:
$$ ck'/\omega' = n' = \frac{n}{\gamma} \sqrt{1 + \frac{\gamma^2\beta^2 }{n^2}}$$
Here I take a more 'first-principles' approach. We can identify two events in spacetime --- the point at which a particular photon enters the dielectric medium, and the point at which it leaves. Let us define the co-ordinates in frame $S$ of the first of these events to be $(0,0,0,0)$. Then in $S$ we know that the co-ordinates of the point at which the photon leaves the medium are $(ct, 0, y, 0)$, where $t$ and $y$ are related by $c/n = y/t$ --- this is just speed = distance / time. Applying a Lorentz transformation to these two points to find the co-ordinates in $S'$:
$$ \mathrm{entrance}_{S'} = (0,0,0,0) \qquad \mathrm{exit}_{S'} = (\gamma ct, -\gamma \beta ct, y, 0)$$
The total time for the photon to move through the block is hence $\gamma t$, whilst the total distance traveled is, by Pythagoras',
$$ \sqrt{\gamma^2 \beta^2 c^2 t^2 + y^2} $$
Dividing these quantities should give us $c/n'$, the speed of the photon. Hence we can write
$$n' = \frac{c \gamma t }{ \sqrt{\gamma^2 \beta^2 c^2 t^2 + y^2}} = \frac{c \gamma t }{ \sqrt{\gamma^2 \beta^2 c^2 t^2 + c^2 t^2 / n^2}}$$
So we have
$$ n' = \frac{\gamma}{\sqrt{\gamma^2 \beta^2 + 1/n^2}}$$
which is not the same as the first expression. I anticipate I've made a very silly mistake somewhere along the line here, but I can't for the life of me spot it!
Answer
The problem here isn't a simple algebra error, but rather an issue with the physics. A medium which at rest is isotropic no longer behaves as an isotropic medium when it is moving relativistically. Instead, it behaves as a nonreciprocal bianisotropic material.
In particular, the phase velocity of light at a particular frequency in a medium is no longer the same in all directions if the medium is moving. This effect can be detected even if the medium is moving at a non-relativistic speed, as in the Fizeau experiment.
Because of this, any attempt to determine one consistent number for what a refractive index turns into under a Lorentz transformation is doomed to failure, because the equivalent to the refractive index is no longer a single number in a frame in which the medium is moving. Instead, it's necessary to treat the refractive index (or really the relative permittivity, which is closely related) as a tensor.
Unfortunately, I'm unable to find a full tensor treatment of refraction in moving media online. I did find this paper on relativistic optics in moving media, but it uses Clifford algebra as an alternative to a tensor treatment, and unfortunately the paper is behind a paywall. However, the non-paywalled abstract for that paper is my source for the above claim that a medium that's isotropic at rest behaves as a nonreciprocal bianisotropic medium when moving.
It is generally believed that for the pure QCD, the classical scale invariance is broken at the quantum level (therefore anomaly rather than SSB). This breaking of scale invariance may be used to explain the quark confinement where an explicit mass scale (or mass gap for QCD) appears. Does anybody know of some references which explain or argue intuitively how this happens? Or even better, does anybody know of an argument?
Answer
The breakdown of the scale invariance in pure Yang-Mills theories takes place due to the dependence of the running coupling constant on the renormalization mass scale. This is an anomalous breakdown, since the classical theory is invariant under scale transformations. It is manifested through the formation of a nonvanishing trace to the energy momentum tensor. Hence it is given the name "trace anomaly".
The trace anomaly can be heuristically derived from the pure QCD renormalized Lagrangian:
$$\mathcal{L} = - \frac{1}{4g^2} F_{\mu \nu}^a F^{ \mu \nu}_a$$ The trace of the energy momentum tensor can be calculated as the variation of the Lagrangian by the mass scale (logarithm of the mass parameter) $$ T^{\mu}_{\mu} = \frac{\partial \mathcal{L}}{\partial \lambda} = \frac{\beta(g)}{2g^3} F_{\mu \nu}^a F^{ \mu \nu}_a$$ Where $\beta(g)= \frac{\partial g} {\partial \lambda}$ is the beta function.
Of course the above heuristic derivation does not specify the Gluon condensate upon which the trace anomaly depends.
One of the first detailed derivations of this result was given by Collins, Duncan and Joglekar
Because of my mathematical background, I've been finding it hard to relate the physics-talk I've been reading, with mathematical objects.
In (say special) relativity, we have a Lorentzian manifold, $M$. This manifold has an atlas with local coordinates.
In differential geometry, when people talk about a `change of coordinates' they mean switching from one local coordinate system in this atlas, to another. For example, one coordinate system in this atlas is a map $\phi_1: U_1 \rightarrow V$ where $V$ is an open set of $M$, and $U$ is an open set of $\mathbb{R}^4$; and if another is $\phi_2: U_2 \rightarrow W$ is another ($U_2$ and open in $\mathbb{R}^4$, and $W$ an open in $M$), then $\phi_1^{-1} |_{V\cap W}\circ \phi_2|_{\phi_2^{-1}(V\cap W)}$ is a coordinate change.
However, in physics it seems that the meaning is different. Indeed if $p \in M$ then you can have a reference frame at $p$, but you can also have a reference frame that is accelerated at $p$. I'm not sure how to interpret this mathematically! What is the mathematical analogue of having an accelerated frame of reference at a point, as opposed to having an inertial frame of reference at a point?
For context, this equation is used in the following (from Schwartz's QFT 3.44)
$$\partial_{\mu} \frac{\partial(\partial_{\beta}A_{\gamma})^2}{\partial(\partial_{\mu}A_{\nu})} = \partial_{\mu}\left[2(\partial_{\alpha}A_{\alpha})\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} g_{\beta\gamma}\right] = \partial_{\mu}[2(\partial_{\alpha}A_{\alpha})g_{\beta\mu}g_{\gamma\nu}g_{\beta\gamma}]=2\partial_{\nu}(\partial_{\alpha}A_{\alpha}).\tag{3.44}$$
But doesn't $$\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} = g_{\beta\mu}g_{\gamma\nu}$$ imply, for instance, that $$\frac{\partial(\partial_0 A_3)}{\partial(\partial_0 A_3)} = g_{00}g_{33} = -1$$ instead of $=1$? Shouldn't we instead have $\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} = \delta_{\beta\mu}\delta_{\gamma\nu}$?
Answer
Schwartz doesn't keep track of index placement. Assuming the indices on the left-hand side are placed correctly, fixing the right-hand side gives $$\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} = g_{\beta}^{\ \ \mu}g_{\gamma}^{\ \ \nu}.$$ That is, you get the metric with mixed indices, and its diagonal components are all $+1$, not a mix of $+1$ and $-1$. (You can calculate that explicitly; another way to see this is that you get to the mixed metric by raising an index on the metric by contracting with the inverse metric. In components, multiplying a matrix with its inverse gives the identity matrix.) This fixes the sign error.
You might ask why Schwartz didn't just use $\delta$ instead of $g$. That's just kind of personal preference -- the $\delta$ is usually only considered with mixed indices, while $g$ is used with indices in all positions. So Schwartz's expression generalizes to arbitrary index placement nicely.
The general moral is that it's fine to ignore index placement until you plug in actual numbers. If you don't fix them, you'll get spurious factors of $-1$.
I need to prove the commutation relation,
$$[p^2,f] = 2 \frac{\hbar}{i}\frac{\partial f}{\partial x} p - \hbar^2 \frac{\partial^2 f}{\partial x^2}$$
where $f \equiv f(\vec{r})$ and $\vec{p} = p_x \vec{i}$
I know
$$[AB,C] = A[B,C] + [A,C]B$$
Applying this, I get
$$[p^2,f] = p[p,f] + [p,f]p$$
where $p = \frac{\hbar}{i}\frac{\partial}{\partial x}$, and $[p,f] \equiv pf - fp$
using a trial function, $g(x)$, I get
$$[p^2,f] = p[p,f]g + [p,f]pg$$
$$= \frac{\hbar}{i}\frac{\partial}{\partial x}\left[\frac{\hbar}{i}\frac{\partial fg}{\partial x} - f\frac{\hbar}{i}\frac{\partial g}{\partial x}\right] + \left[\frac{\hbar}{i}\frac{\partial fg}{\partial x} - f\frac{\hbar}{i}\frac{\partial g}{\partial x}\right] \frac{\hbar}{i}\frac{\partial}{\partial x}$$
using the product rule
$$ = -\hbar^2 \frac{\partial}{\partial x}\left[g \frac{\partial f}{\partial x} + f \frac{\partial g}{\partial x} - f \frac{\partial g}{\partial x} \right] + \left[g \frac{\partial f}{\partial x} + f \frac{\partial g}{\partial x} - f \frac{\partial g}{\partial x} \right] \frac{\hbar}{i}^2 \frac{\partial}{\partial x}$$
cancelling the like terms in the brackets gives
$$= -\hbar^2 \frac{\partial}{\partial x} \left[g \frac{\partial f}{\partial x}\right] - \left[g \frac{\partial f}{\partial x}\right]\hbar^2 \frac{\partial}{\partial x}$$
using the product rule again gives
$$ = -\hbar^2 \left[\frac{\partial g}{\partial x} \frac{\partial f}{\partial x} + g \frac{\partial^2 f}{\partial x^2} \right] - \left[g \frac{\partial f}{\partial x}\right] \hbar^2 \frac{\partial}{\partial x}$$
$\frac{\partial g}{\partial x} = 0$, so
$$ = -\hbar^2 g \frac{\partial^2}{\partial x^2} - \hbar^2 g \frac{\partial f}{\partial x} \frac{\partial}{\partial x}$$
Substituting the momentum operator back in gives
$$ = -\hbar^2 g \frac{\partial^2}{\partial x^2} - \frac{\hbar}{i} g \frac{\partial f}{\partial x} p$$
The trial function, $g$, can now be dropped,
$$ = -\hbar^2 \frac{\partial^2}{\partial x^2} - \frac{\hbar}{i}\frac{\partial f}{\partial x}p$$
But this is not what I was supposed to arrive at. Where did I go wrong?
Answer
I didn't read your answer, but let's think about just computing the operator $\partial_x^2 f$. First we need to compute the operator $\partial_x f$. Now I am saying "the operator" because we are viewing $\partial_x f$ as a composition of first multiplying by $f$ and then taking the derivative. By the product rule, we know $\partial_x f = (\partial_x f) + f \partial_x$, were by $(\partial_x f)$, I really do just mean multiplication by the derivative of $f$.
Now lets try to compute $\partial_x^2 f$. It is $\partial_x [(\partial_x f) + f \partial_x] = (\partial_x^2 f) + (\partial_x f) \partial_x + (\partial_x f) \partial_x + f \partial_x^2 = (\partial_x^2 f) +2(\partial_x f) \partial_x + f \partial_x^2$.
Then $[\partial_x^2,f] = \partial_x^2f-f \partial_x^2 = (\partial_x^2 f) +2(\partial_x f) \partial_x$. If you understand this then you should get the right answer. You just need to put in the appropriate $i$'s and $\hbar$'s.
Time has many definitions per se, but the basic idea being it's "the measurement of change" so as we know, all matter looses it's ability of changing with the loss of kinetic energy. and the where it becomes zero is the absolute zero (−273.15°C). so my question would be, will time stop/freeze at Absolute zero? and does time get slow while reaching there?
Answer
You are confusing time with the flow of time.
Time is just a coordinate like the spatial coordinates, that is we label spacetime points with four coordinates $(t, x, y, z)$. Indeed, in relativity (both flavours) the time and spatial coordinates get mixed up so different observers will disagree about what is time and what is space.
But the obvious thing about time is that to us humans it appears to flow. We can stand still in space, but we have to keep moving through time at one second per second. I would guess this is what you're thinking of when you say that time is "the measurement of change". The curious thing is that the flow of time doesn't really exist in physics, and some (exactly how many I don't know) physicists believe that the flow of time is a trick of human perception and that time doesn't flow at all. Viewed in this light your question is meaningless, because you are asking about something that doesn't exist.
But even if you take the conventional view of the flow of time, motion does not stop at absolute zero. This is because quantum systems exhibit zero point energy, so their energy remains non-zero even when the temperature is absolute zero. For example helium remains fluid at absolute zero because its zero point energy is too great to allow it to crystallise. So there is no sense in which time stops as we lower the temperature to absolute zero.
I was watching a video about the swarzchild radius and it said that potential energy is $gmm/r$. This cannot be right though because potential energy goes up with distance not down. I'm assuming he meant motion due to potential energy. Also what is the intuition behind the equation $gmm/r$?
Memristor, which makes relationship between charge and flux, has not been discovered, according to Wikipedia. So I tried to implement it, here linear.
As it both stores charges and induces magnetic flux, I thought it as some chimera of capacitor and inductor, and came up with the structure below:
Though I can't tell it actually works, because of the following reasons:
Regardless of whether it is actually a memristor, I'd like to solve the system. Thanks.
Forgive me if this topic is too much in the realm of philosophy. John Baez has an interesting perspective on the relative importance of dimensionless constants, which he calls fundamental like alpha, versus dimensioned constants like $G$ or $c$ [ http://math.ucr.edu/home/baez/constants.html ]. What is the relative importance or significance of one class versus the other and is this an area that physicists have real concerns or expend significant research?
Answer
first of all, the question you are asking is very important and you may master it completely.
Dimensionful constants are those that have units - like $c, \hbar, G$, or even $k_{\rm Boltzmann}$ or $\epsilon_0$ in SI. The units - such as meter; kilogram; second; Ampere; kelvin - have been chosen partially arbitrarily. They're results of random cultural accidents in the history of mankind. A second was original chosen as 1/86,400 of a solar day, one meter as 1/40,000,000 of the average meridian, one kilogram as the mass of 1/1,000 cubic meters (liter) of water or later the mass of a randomly chosen prototype, one Ampere so that $4\pi \epsilon_0 c^2$ is a simple power of 10 in SI units, one Kelvin as 1/100 of the difference between the melting and boiling points of water.
Clearly, the circumference of the Earth, the solar day, a platinum prototype brick in a French castle, or phase transitions of water are not among the most "fundamental" features of the Universe. There are lots of other ways how the units could be chosen. Someone could choose 1.75 meters - an average man's height - to be his unit of length (some weird people in the history have even used their feet to measure distances) and he could still call it "one meter". It would be his meter. In those units, the numerical values of the speed of light would be different.
Exactly the products or ratios of powers of fundamental constants that are dimensionless are those that don't have any units, by definition, which means that they are independent of all the random cultural choices of the units. So all civilizations in the Universe - despite the absence of any interactions between them in the past - will agree about the numerical value of the proton-electron mass ratio - which is about $6\pi^5=1836.15$ (the formula is just a teaser I noticed when I was 10!) - and about the fine-structure constant, $\alpha\sim 1/137.036$, and so on.
In the Standard Model of particle physics, there are about 19 such dimensionless parameters that "really" determine the character of physics; all other constants such as $\hbar,c,G,k_{\rm Boltzmann}, \epsilon_0$ depend on the choice of units, and the number of independent units (meter, kilogram, second, Ampere, Kelvin) is actually exactly large enough that all those constants, $\hbar,c,G,k_{\rm Boltzmann},\epsilon_0$, may be set equal to one which simplifies all fundamental equations in physics where these fundamental constants appear frequently. By changing the value of $c$, one only changes social conventions (what the units mean), not the laws of physics.
The units where all these constants are numerically equal to 1 are called the Planck units or natural units, and Max Planck understood that this was the most natural choice already 100 years ago. $c=1$ is being set in any "mature" analysis that involves special relativity; $\hbar=1$ is used everywhere in "adult" quantum mechanics; $G=1$ or $8\pi G=1$ is sometimes used in the research of gravity; $k_{\rm Boltzmann}=1$ is used whenever thermal phenomena are studied microscopically, at a professional level; $4\pi\epsilon_0$ is just an annoying factor that may be set to one (and in Gaussian 19th century units, such things are actually set to one, with a different treatment of the $4\pi$ factor); instead of one mole in chemistry, physicists (researchers in a more fundamental discipline) simply count the molecules or atoms and they know that a mole is just a package of $6.022\times 10^{23}$ atoms or molecules.
The 19 (or 20?) actual dimensionless parameters of the Standard Model may be classified as the three fine-structure constants $g_1,g_2,g_3$ of the $U(1)\times SU(2)\times SU(3)$ gauge group; Higgs vacuum expectation value divided by the Planck mass (the only thing that brings a mass scale, and this mass scale only distinguishes different theories once we also take gravity into account); the Yukawa couplings with the Higgs that determine the quarks and fermion masses and their mixing. One should also consider the strong CP-angle of QCD and a few others.
Once you choose a modified Standard Model that appreciates that the neutrinos are massive and oscillate, 19 is lifted to about 30. New physics of course inflates the number. SUSY described by soft SUSY breaking has about 105 parameters in the minimal model.
The original 19 parameters of the Standard Model may be expressed in terms of more "fundamental" parameters. For example, $\alpha$ of electromagnetism is not terribly fundamental in high-energy physics because electromagnetism and weak interactions get unified at higher energies, so it's more natural to calculate $\alpha$ from $g_1,g_2$ of the $U(1)\times SU(2)$ gauge group. Also, these couplings $g_1,g_2$ and $g_3$ run - depend on the energy scale approximately logarithmically. The values such as $1/137$ for the fine-structure constant are the low-energy values, but the high-energy values are actually more fundamental because the fundamental laws of physics are those that describe very short-distance physics while long-distance (low-energy) physics is derived from that.
I mentioned that the number of dimensionless parameters increases if you add new physics such as SUSY with soft breaking. However, more complete, unifying theories - such as grand unified theories and especially string theory - also imply various relations between the previously independent constants, so they reduce the number of independent dimensionless parameters of the Universe. Grand unified theories basically set $g_1=g_2=g_3$ (with the right factor of $\sqrt{3/5}$ added to $g_1$) at their characteristic "GUT" energy scale; they may also relate certain Yukawa couplings.
String theory is perfectionist in this job. In principle, all dimensionless continuous constants may be calculated from any stabilized string vacuum - so all continuous uncertainty may be removed by string theory; one may actually prove that it is the case. There is nothing to continuously adjust in string theory. However, string theory comes with a large discrete class of stabilized vacua - which is at most countable and possibly finite but large. Still, if there are $10^{500}$ stabilized semi-realistic stringy vacua, there are only 500 digits to adjust (and then you may predict everything with any accuracy, in principle) - while the Standard Model with its 19 continuous parameters has 19 times infinity of digits to adjust according to experiments.
I know what the dual of a vector means (as a map to its field), and I am also aware of of the definition a dual of a tensor as,
$$F^{*ij} = \frac{1}{2} \epsilon^{ijkl} F_{kl}\tag{1}$$
I just don't understand how to connect this to the definition of the dual of a vector. Or are they entirely different concepts? If they are different then why use the word dual for it?
I know this is kind of mathematical or may be a stupid question, but I have a problem with understanding the need for dual of Dual Field strength Tensor in relativistic ED. I mean you could say that I am going to define another tensor using eq. (1) call it whatever we like.
Answer
The dual of a tensor you refer to is the Hodge dual, and has nothing to do with the dual of a vector. The word "dual" is used in too many different contexts, and in this case it is even used the same $*$ symbol. One usually specifies "Hodge dual", or "Hodge star operator", to avoid confusion. Both these "duals" are isomorphisms between vector spaces endowed with inner product.
The dual of a vector space $V$ is the vector space $V^*$ consisting in the linear functions (functionals) $f:V\to \mathbb R$ (or $\mathbb C$ if $V$ is a complex vector space). The dual of a vector $v\in V$ makes sense only if $V$ is endowed with an inner product $g$, and it is defined as $v^\flat\in V^*$, $v^\flat(u)=g(v,u)$. This dual is an isomorphism between the inner product vector space $(V,g_{ab})$ and its dual $(V^*,g^{ab})$.
The Hodge dual is defined on totally antisymmetric tensors from $\otimes^k V$, that is, on $\wedge V^k$. It is defined on $\wedge V\to \wedge V$, where $\wedge V=\oplus_{k=0}^n\wedge^k V$. It also requires the existence of an inner product $g$ on $V$. The inner product extends canonically to the entire $\wedge V$.
The Hodge dual is defined as follows. We construct a basis on $V$, which is orthonormal with respect to the inner product $g$, say $e_1,\ldots,e_n$. Then, for each $k$, there is a basis of $\wedge^k V$ of the form $e_{i_1}\wedge\ldots \wedge e_{ik}$. This basis is considered to be orthonormal too, and by this, $g$ defines an inner product on $\wedge^kV$.
The Hodge dual is defined first between $\wedge^kV\to\wedge^{n-k}V$, Both spaces have the same dimension, which is $C^k_n$. The canonical isomorphism between them is defined first on the elements of the basis:
$$*(e_{i_1}\wedge\ldots\wedge e_{i_k})=\epsilon e_{j_i}\wedge\ldots\wedge e_{j_{n-k}}.$$ Here, the indices $i_1,\ldots,i_k,j_1\ldots j_{n-k}$ are a permutation of the numbers between $1$ and $n$. $\epsilon$ is $+1$ if the permutation $i_1,\ldots,i_k,j_1\ldots j_{n-k}$ is even, and $-1$ otherwise. This defines uniquely the isomporphism. It extends uniquely to $\wedge V=\oplus_{k=0}^n\wedge^k V$, since this is a direct sum of vector spaces.
Please note that the vectors also admit Hodge duals, but their duals are elements of $\wedge^{n-1}V$.
Since we can consider that $\mathbb R=\wedge^0 V$, the Hodge dual of the scalar $1$ is the volume element $*1:=e_1\wedge\ldots\wedge e_n\in\wedge^n V$. In a basis, it is denoted by $\epsilon_{12\ldots n}$.
Similarly, the Hodge dual can be defined on the space of exterior forms $\wedge V^*$.
In the case of Lorentzian spacetimes of dimension $4$, the Hodge duality establishes isomorphisms between $\mathbb R$ and $\wedge^4 V$, between $V$ and $\wedge^3 V$, and between $\wedge V^2$ and itself.
An alternative way to construct the Hodge duality is for the Clifford algebra associated to $V$. In this case, there is an isomorphism (as vector spaces with inner product) between $\wedge V$ and $Cl(V)$. The Hodge dual translated to the Clifford algebra language as Clifford multiplication with the $n$-vector which corresponds to the volume element, $\gamma_1\cdot\gamma_2\cdot\ldots\cdot\gamma_n$.
Back to the confusion of terminology. For a vector $v\in V$, the dual is a covector $v^\flat\in V^*$. The Hodge dual can be obtained by constructing an orthonormal basis starting from $v$, then taking the wedge product between the other elements, and dividing by the length of $v$. The result is from $\wedge^{n-1}V$, not from $V^*$. All these three spaces are isomorphic, in a canonical way, and also with $\wedge^{n-1}V^*$, by composing the two kinds of dualities. But the two dualities refer to totally distinct isomorphisms.
In connection with Electrodynamics (in Lorentzian spacetime), the Hodge dual of the electromagnetic tensor appears in Maxwell's equations:
$$d F=0$$
and
$$d *F=*J$$
where $J$ is the current $1$-form. These two equations contain the four Maxwell equations. Here, $*F$ is the Hodge dual of the electromagnetic tensor $F$, and $*J$ of the current $1$-form $J$ (which in turn is the "dual" in the other sense of the current vector).
Added
To be more close to the question. The equation
$$*F_{ij} = \frac{1}{2} \epsilon_{ij}{}^{kl} F_{kl}$$
represents the Hodge duality between $\wedge^2V$ and itself. But
$$*F^{ij} = \frac{1}{2} \epsilon^{ijkl} F_{kl}\tag{1}$$
is a duality between $\wedge^2V$ and $\wedge^2V^*$. There is also the duality in the first sense, between $\wedge^2V$ and $\wedge^2V^*$ (extending that between $V$ and $V^*$). So, there are two distinct isomorphisms between the vector spaces with inner product $\wedge^2V$ and $\wedge^2V^*$.
I try to understand constructing of Hamiltonian mechanics with constraints. I decided to start with the simple case: free relativistic particle. I've constructed hamiltonian with constraint:
$$S=-m\int d\tau \sqrt{\dot x_{\nu}\dot x^{\nu}}.$$
Here $\phi=p_{\mu}p^{\mu}-m^2=0$ $-$ first class constraint.
Then $$H=H_{0}+\lambda \phi=\lambda \phi.$$
So, I want to show that I can obtain from this Hamiltonian the same equation of motion, as obtained from Lagrangian.
But the problem is that I'm not sure what to do with $\lambda=\lambda(q,p)$. I tried the following thing:
$$\dot x_{\mu}=\{x_{\mu},\lambda \phi\}=\{x_{\mu},\lambda p^2\}-m^2\{x_{\mu},\lambda\}=\lambda\{x_{\mu},p^2\}+p^2\{x_{\mu},\lambda\}-m^2\{x_{\mu},\lambda\}$$$$=2\lambda \eta_{\mu b} p^b+p^2\{x_{\mu},\lambda\}-m^2\{x_{\mu},\lambda\}=2\lambda \eta_{\mu b} p^b+p^2\frac{\partial \lambda}{\partial p^{\mu}}-m^2\frac{\partial \lambda}{\partial p^{\mu}},$$
$$\dot \lambda=\{\lambda, \lambda \phi \}=\{\lambda,\lambda p^2\}-m^2\{\lambda,\lambda\}=\lambda\{\lambda,p^2\}+p^2\{\lambda,p^2\}=2\lambda\eta_{ak}p^{a}\frac{\partial \lambda}{\partial x^{k}},$$
$$\dot p_{\mu}=\{p_{\mu},\lambda p^{2}-m^2\lambda \}=p^{2}\{p_{\mu},\lambda\}-m^2\{p_{\mu},\lambda\}=-p^{2}\frac{\partial \lambda}{\partial x^{\mu}}+m^2\frac{\partial \lambda}{\partial x^{\mu}}.$$
If we recall that $p^2-m^2=0$, then we get from the third equation: $\dot p=0$, and from the first: $$\dot x_{\mu}=2\lambda\eta_{ak}p^{a}.$$
So we have
$\dot x_{\mu}=2\lambda\eta_{\mu b}p^{b}.$
$\dot \lambda=2\lambda\eta_{ak}p^{a}\frac{\partial \lambda}{\partial x^{k}}.$
$\dot p=0.$
But I don't know what to do next. Can you help me?
Answer
Hints to the question (v1):
We cannot resist the temptation to generalize the background spacetime metric from the Minkowski metric $\eta_{\mu\nu}$ to a general curved spacetime metric $g_{\mu\nu}(x)$. We use the sign convention $(-,+,+,+)$.
Let us parametrize the point particle by an arbitrary world-line parameter $\tau$ (which does not have to be the proper time).
The Lagrange multiplier $\lambda=\lambda(\tau)$ (which OP mentions) depends on $\tau$, but it does not depend on the canonical variables $x^{\mu}$ and $p_{\mu}$. Similarly, $x^{\mu}$ and $p_{\mu}$ depend only on $\tau$.
The Lagrange multiplier $\lambda=\frac{e}{2}$ can be identified with an einbein$^1$ field $e$. See below where we outline a simple way to understand the appearance of the on-shell constraint $$\tag{1}p^2+m^2~\approx~0, \qquad p^2~:=~g^{\mu\nu}(x)~ p_{\mu}p_{\nu}~<~0.$$
Start with the following square root Lagrangian for a massive relativistic point particle $$\tag{2}L_0~:=~ -m\sqrt{-\dot{x}^2}, \qquad \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~<~0, $$ where dot means differentiation wrt. the world-line parameter $\tau$. Here the action is $S_0=\int \! d\tau~ L_0 $. The stationary paths includes the geodesics. More precisely, the Euler-Lagrange equations are the geodesics equations.
Introduce an einbein field $e=e(\tau)$, and Lagrangian $$\tag{3}L~:=~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2}.$$ Contrary to the square root Lagrangian (2), this Lagrangian (3) also makes sense for massless point particles, cf. this Phys.SE post.
Show that the Lagrangian momenta are $$\tag{4}p_{\mu}~=~ \frac{1}{e}g_{\mu\nu}(x)~\dot{x}^{\nu}.$$
Show that the Euler-Lagrange equations of the Lagrangian (3) are $$\tag{5} \dot{p}_{\lambda}~\approx~\frac{1}{2e}\partial_{\lambda}g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}, \qquad \dot{x}^2+(em)^2~\approx~0.$$
Show that the Lagrangian (3) reduces to the square root Lagrangian (2) when integrating out the einbein field $$\tag{6} e~>~0.$$ The inequality (6) is imposed to remove an unphysical negative branch, cf. my Phys.SE answer here.$^2$
Perform a (singular) Legendre transformation$^3$ of the Lagrangian (3), and show that the corresponding Hamiltonian becomes $$\tag{7}H~=~ \frac{e}{2}(p^2+m^2).$$ This Hamiltonian (7) is precisely of the form Lagrange multiplier times constraint (1).
Show that Hamilton's equations are precisely eqs. (4) and (5).
The arbitrariness in the choice of the world-line parameter $\tau$ leads to reparametrization symmetry$^4$ $$\tau^{\prime}~=~f(\tau), \qquad d\tau^{\prime} ~=~ d\tau\frac{df}{d\tau},\qquad \dot{x}^{\mu}~=~\dot{x}^{\prime\mu}\frac{df}{d\tau},\qquad e~=~e^{\prime}\frac{df}{d\tau},\qquad $$ $$\tag{8} p_{\mu}~=~p_{\mu}^{\prime},\qquad L~=~L^{\prime}\frac{df}{d\tau},\qquad H~=~H^{\prime}\frac{df}{d\tau}\qquad S~=~S^{\prime},$$ where $f=f(\tau)$ is a bijective function.
Thus one may choose various gauges, e.g. $e={\rm const.}$
References:
--
Footnotes:
$^1$ An einbein is a 1D version of a vielbein.
$^2$ As a consistency check of the sign (6), if we in the static gauge $$\tag{9}ix^0_M~=~x^0_E~=~\tau_E~=~i\tau_M$$ Wick rotate from Minkowski to Euclidean space, then in eq. (3), the Euclidean Lagrangian $L_E=-L_M>0$ becomes positive as it should.
$^3$ Strictly speaking, in the singular Legendre transformation, one should also introduce a momentum $$\tag{10}p_e~:=~\frac{\partial L}{\partial \dot{e}}~=~0$$ for the einbein $e$, which leads to a primary constraint, that immediately kills the momentum $p_e$ again. Note that $\frac{\partial H}{\partial e}\approx 0$ becomes one of Hamilton's equations.
$^4$ Reparametrization is a passive transformation. For a related active transformation, see this Phys.SE post.
