Simply put HUP involves position and momentum, further more consider a mass of 1kg. as momentum is mass X velocity = 1X velocity = velocity for calculation purposes. now for a stationary observer the uncertainty in velocity of the observed mass is some value, if another observer is travelling at the same speed as the mass the relative velocity is zero or at least smaller than what the stationary observer would see the mass travelling at. we now have two values for the uncertainty in the velocity of the mass. they both can't be right, HUP is subject to classical Newtonian relativity?
121503
Is the uncertainty principle a technical difficulty in measurement or is it an intrinsic concept in quantum mechanics irrelevant of any measurement?  Everyone knows the thought experiment of measuring the position of an electron. One can detect electron's position by hitting it with a photon, due to Compton scattering the collision of the photon with electron will change electron's momentum. This experiment is used to explain uncertainty principle to layman, but it is over simplified, isn't it? It also gives an impression that if there was a better suited experimental method the uncertainty principle becomes irrelevant.   I personally think it is intrinsic as it arises from the non-zero commutator of position and momentum operators irrespective of the measurement process. Am I right?  EDIT: My question is similar to certain extent to  and . The answers there are nice but they focus on explaining basics of quantum mechanics more than they comment on the technical difficulty part. In the answers of question 2 there are statements like "So, it's not a knowledge limit" and "you're sort of correct when you say it's an observational limit" without further comments  To summarize, assume hypothetically we managed to find a way in the future where we can have a look at an electron without disturbing it by measurement or causing its wave function to collapse, would the uncertainty principle still hold in such a case?? Why/Why not?
74512
If a person is sitting on a chair his momentum is zero and his uncertainty in position should be infinite. But we can obviously position him at most within few chair lengths.   What am I missing? Do we have to invoke earth's motion, motion of the galaxy etc. to resolve the issue?
7035
For the sake of simplicity, let's imagine that the entire universe is empty except for a single lump of (classical) matter with mass $m$. In its center of momentum frame, it is clear that the total energy is simply $E_\text{CoM}=mc^2$. However, in a frame moving relative to it with speed $v$, we have $E_\text{moving}=\sqrt{m^2c^4+p^2c^2}&gt; mc^2$ for all nonzero $v$.  Are we to infer that the energy content of a system can be different relative to the observer? Does this not violate the first law of thermodynamics?
116428
$E = h\nu$ and $P = h\nu/c$ in vacuum. If a photon enters water, its frequency $\nu$ doesn't change. What are its energy and momentum: $h\nu$ and $h\nu/c$ ? Since part of its energy and momentum have been transferred to water, it should be less.  If water's refractive index is $n$, are the energy and momentum equal to $h\nu/n$ and $h\nu/c/n$ ?
104773
The probability amplitude for a free particle with momentum $\ p$ and energy $E$ is the complex wave function: $$\psi_{(\vec x , t)}=e^{i(\vec k\cdot \vec x -\omega t)}$$  is there any uncertainty on the free particle with a definite momentum $\vec p$!?
8542
In (non-relativistic) classical physics, if the temperature of an object is proportional to the average kinetic energy ${1 \over 2} m\overline {v^{2}}$of its particles (or molecules), then shouldn't that temperature depend on the frame of reference - since $\overline {v^{2}}$ will be different in different frames?  (I.e. In the lab frame  $K_l = {1 \over 2} m\overline {v^{2}} $, but in a frame moving with velocity $u$ relative to the lab frame, $K_u =  {1 \over 2} m \overline {(v+u)^{2}}$).
17632
There is something I don't understand about the  $H$, as it seems to clump two concepts together that I can't quite unify in my head. On the one side, we have  $$V = D H$$  which means that for a given distance $D$, there is a certain amount of new space created over time - and $H$ is simply the factor that makes this relationship work. So, for example, say we have two points 1 Mpc apart, this would mean they recede at about 70 km/s away from each other  (given our current approximation of $H$).  Now the thing I can't wrap my head around is that   $$T =\frac{1}{ H}$$  is also the age of the universe. Contrary to claims made on, say, Wikipedia this means that $H$ cannot possibly have been a constant throughout the past 13 billion years, because mathematically $1/H$ means that $H$ must be continually shrinking as the universe ages.  So if $H$ did start out as some huge value and is now shrinking over time, doesn't this mean the expansion of the universe is slowing down? Because if $H$ is shrinking, I'll get a lower value of $V$ today than I'll get tomorrow. Shouldn't the notation then be more like  $$V = DH(t)$$  So which one is it? If $1/H$ is simply the solution for $D=0$, how can we use it as the expansion-velocity-per-unit-of-distance at the same time? What's worse, how can literature say $H$ has probably been more or less constant forever and simultaneously assert that $1/H$ is the current age of the universe? What am I missing?
46553
One of the postulates of QM mechanics is that any observable is described mathematically by a hermitian linear operator.   I suppose that an observable means a quantity that can be measured. The mass of a particle is an observable because it can be measured.  Why then the mass is not described by a linear hermitian operator in QM?
79863
Is there ever a situation where the distinction between $F = m \frac{dv}{dt}$ and $F = \frac{dp}{dt}$ is important?  I can't think of a situation where one is true and not the other (assuming only conservation of momentum).  Obviously it is important to take a changing mass into account (e.g. for a rocket) when you're considering a full time evolution, i.e. $F(t) = m(t) \frac{dv}{dt}$ (or in the relativistic case, perhaps something like $F(t) = m(t) \frac{d}{dt} \left( \frac{p}{m} \right)$ with $m$ the rest-mass).  And perhaps there is a nontrivial relationship between the rate of change of mass, and the forces being exerted (again, e.g. with a rocket --- where the mass loss is tied to the propulsion).  What is not clear is that there should ever be a $F = v\frac{dm}{dt}$ term.
82475
Motivating the non-relativistic momentum $$\mathbf{p} = m\mathbf{v}$$ is quite easy: it is meant to represent the quantity of motion of the particle, and since the mass is one measure of quantity of matter it should be proportional to mass (how much thing is moving) and should be proportional to velocity (how fast and to where it is moving). Now, in Special Relativity the momentum changes. The new quantity of motion becomes $$\mathbf{p} = \dfrac{m\mathbf{v}}{\sqrt{1-\dfrac{v^2}{c^2}}}$$ Or, using $\gamma$ the Lorentz factor $$\mathbf{p} = \gamma(v) m\mathbf{v}$$ where I write $\gamma(v)$ to indicate that the velocity is that of the particle relative to the frame in which the movement is being observed. The need for this new momentum is because the old one fails to be conserved and because using the old one in Newton's second law leads to a law which is not invariant under Lorentz transformations. So the need for a new momentum is perfectly well motivated. What I would like to know is how can one motivate that the correct choice for $\mathbf{p}$ is the $\gamma(v)m\mathbf{v}$. There are some arguments using the mass: considering a colision, requiring momentum to be conserved, transform the velocity and then find how mass should transform. Although this work, it doesn't seem natural, and it is derived in one particular example. On my book there's even something that Einstein wote saying that he didn't think it was a good idea to try transforming the mass from $m$ to $M = \gamma(v)m$, that it was better to simply keep $\gamma$ on the new momentum without trying to combine it with the mass. So I would like to know: without resorting to arguments based on transformation of the mass, how can one motivate the new form of momentum that works for special relativity?
65886
$K.E=\frac{1}{2}mv^2$ , $P=mv$ thus there is a relation between them. I really can't understand why there is no decrease in momentum when kinetic energy is decreased in an inelastic collision. I am only a high school passed student(ready for college). Should I leave this question until I learn &quot;Lagrangian&quot; and &quot;Noether's Theorem&quot;? Because I literally read most of the answers about these kinds of questions and spend almost half a day on it, BUT I do not get it. Can anyone please answer this question? please help me.
117084
In Hamiltonian Mechanics, a version of Hamilton's principle is shown to evolve a system according to the same equations of motion as the Lagrangian, and therefore Newtonian formalism. In particular, letting $\delta$ indicate a variation of the path through phase space, $$ \delta \int_{t_1}^{t_2} \big(p_i \dot{q}_i - H(p_i,q_i,t)\big)dt = 0 $$ is shown to generate the same motion equations as those found by the Legendre transform of the Lagrangian equations of motion. Essentially, when we calculate the Euler-Lagrange equations for the above integrand, we find $$ \dot{p}_i + \frac{\partial H}{\partial q_i} = 0\text{, } \dot{q}_i - \frac{\partial H}{\partial p_i} = 0. $$ Now, since these are both first-order equations of motion, we require $2n$ boundary conditions to obtain a solution, where $n$ is the number of particles. This is perfectly consistent with the Lagrangian formalism which had $2n$ boundary conditions due to its $n$ distinct second-order motion equations.   An important thing to note is that since $p_i\dot{q}_i - H(p_i,q_i,t)$ contains no $\dot{p}_i$ dependence, the variation in $p_i(t)$ need not be zero at the endpoints of the path. This is not the case with $q_i(t)$ because the $\dot{q}_i$ dependance results in the appearance of boundary terms that must be set to zero in order to obtain the motion equations.  Now, as far as the boundary conditions are concerned, this makes sense. For the path to be uniquely specified given Newtonian mechanics, we require $2n$ boundary conditions. These can either be the usual Newtonian choice of initial positions and velocities/momenta, or they can be the initial and final positions. Both are perfectly acceptable mathematically speaking. If we were required, in the modified Hamilton's principle variation, to enforce that the variations in the momenta also be zero at the initial and final times, this would mean $2n$ additional boundary conditions, which would generally over constrain a Newtonian system.   However, my confusion is this: when textbooks (Goldstein in particular) consider generating functions, they require that the integrand be invariant to the addition of a total time derivative of a function $F(q_i,p_i,t)$ of the phase space coordinates. However, adding such a function will in general add some $\dot{p}_i$ dependence to the integrand, which will then add boundary terms to the equations of motion unless we require that the $p_i(t)$ variations be zero at the boundaries. This is fine as far as it goes. We can always define the variation however we like. The point is just that it spits out the right motion equations at the end of the day. But my confusion lies in how this doesn't generally over constrain the system. If we specify all the positions and momenta at both the initial and the final times, might not the required path to connect those points in phase space be non-Newtonian? Does this only matter if you're actually trying to use the stationary action principle to find the paths, rather than just using it to find the motion equations?
3483
I am taking for granted that when we say that something is conserved it is understood 'in its full integrity'.    Energy is represented by a scalar J,  and is conserved in elastic collision.  momentum is the product of a number (of Kg) by velocity which is a vector, and therefore has a direction: $p = m * v$. (In circular motion we say that speed is constant but velocity is constantly changing).   In an elastic collision if a ball A (m = 1) with v = p = +8 in the x axis (E = 32) hits a ball B at rest (M =3) , B will proceed ( p = 4 *3 = +12, E = 24)  and A will bounce in the opposite direction (p = v = -4, E = 8). We conclude that: +12 -4 = +8, momentum is conserved (and also energy: E = 8 + 24 = 32 J)  If a perfectly elastic ball is thrown against a wall energy J is conserved, $E-i = E_f$ the value of the speed is unchanged, but the direction of the vector is reversed.  I am asking: the momentum of what is conserved? if it is the momentum of the system this  is not true in the first case, if it is the momentum of the single balls this is not true in the second case.  Language of science  must be precise, I hope you will not consider this as hair-splitting, but I ask: is there a general definition of 'conserved' valid for all instances? why a broad/permissive interpretation of the principle in the case of momentum? Or, most likely, where did I go wrong?  EDIT :     This does mean that the wall contains a momentum of 2mv (for mass m   and velocity v). But note that since the mass of the wall is   incredible compared to the ball, the velocity is notably   imperceptible!   If the momentum of the wall were to be 2mv, the energy of the bouncing ball should be impercetibly less but yet less: $E - \epsilon &lt; E$ and the Energy of the ball would not be conserved. That means that a perfect elastic collision against a fixed body is not possible, in the sense that no body can have a CoR = 1. right?
38589
I'll be generous and say it might be reasonable to assume that nature would tend to minimize, or maybe even maximize, the integral over time of $T-V$.  Okay, fine.  You write down the action functional, require that it be a minimum (or maximum), and arrive at the Euler-Lagrange equations.  Great.  But now you want these Euler-Lagrange equations to not just be derivable from the Principle of Least Action, but you want it to be equivalent to the Principle of Least Action.  After thinking about it for awhile, you realize that this implies that the Principle of Least Action isn't really the Principle of Least Action at all:  it's the "Principle of Stationary Action".  Maybe this is just me, but as generous as I may be, I will not grant you that it is "natural" to assume that nature tends to choose the path that is stationary point of the action functional.  Not to mention, it isn't even obvious that there is such a path, or if there is one, that it is unique.  But the problems don't stop there.  Even if you grant the "Principle of Stationary Action" as fundamentally and universally true, you realize that not all the equations of motions that you would like to have are derivable from this if you restrict yourself to a Lagrangian of the form $T-V$.  As far as I can tell, from here it's a matter of playing around until you get a Lagrangian that produces the equations of motion you want.  From my (perhaps naive point of view), there is nothing at all particularly natural (although I will admit, it is quite useful) about the formulation of classical mechanics this way.  Of course, this wouldn't be such a big deal if these classical ideas stayed with the classical physics, but these ideas are absolutely fundamental to how we think about things as modern as quantum field theory.  Could someone please convince me that there is something natural about the choice of the Lagrangian formulation of classical mechanics (I don't mean in comparison with the Hamiltonian formulation; I mean period), and in fact, that it is so natural that we would not even dare abandon these ideas?
2714
Some people say that mass increases with speed, some people say that the mass of an object is independent of its speed.  I understand how some (though not many) things in physics are a matter of interpretation based on one's definitions. But I can't get my head around how both can be 'true' is any sense of the word.  Either mass increases or it doesn't, right?  Can't we just measure it, and find out which 'interpretation' is right? e.g.: (in some sophisticated way) heat up some particles in a box and measure their weight?  ------------------------------  UPDATE  Right, so I've got two identical containers, each with identical amounts of water, each on identical weighing scales, and each in the same g field. If one container has hotter water, will the reading on its scale be larger than the other? If the answer is yes, and g is constant, does this mean that the m in w=mg has got bigger?
52708
I am having a bit of confusion, let us say there are two observers observing two simultaneous lightning strikes with a finite distance between them. Now one observer is located such that he is in between the two lightning strikes, while the other is near to one strike and away from the other. In this scenario, the observer situated in between will perceive the lightning strikes to be simultaneous while the other will perceive them differently, since he is close to one and far from the other. Does this mean the Principle of Relativity holds even without actual motion?
781
It has been a while since I've thought about physics, however, I remember something about how if you are on a train with no windows that is going perfectly straight and is perfectly smooth, there is no experiment that you could perform to prove that you are in motion or at rest. So I have a couple of questions related to this.   Is this still held to be true? I was just wondering if anything in physics has come to light since I learned about this (about 10 years ago) that would allow an observer on the train to tell if they are moving. If this is true, why couldn't you accelerate a particle in different directions to a certain mass and measure how much energy it took to get it up to that mass. If you were moving in any direction, wouldn't you measure less energy in your direction of travel? Would having an outside observer affect the aforementioned experiment in any way?