Hot answers tagged classical-physics
12
This was something that confused me for awhile as well until I found this great set of notes: homepages.physik.uni-muenchen.de/~helling/classical_fields.pdf
Let me just briefly summarize what's in there.
The free Klein-Gordon field satisfies the field equation
$(\partial_{\mu} \partial^{\mu} +m^2) \phi(x) = 0$
the most general solution to this equation ...
9
As it is, differential forms don't tell you the whole story--strictly speaking, differential forms only deals with covectors and wedge products of covectors and then uses the hammer of the Hodge star to be able to clumsily do inner products. To me, it is too far removed from the vector calculus you may already know.
Instead, I strongly urge you to look ...
7
I would really recommend the book by Frankel, The Geometry of Physics. He deals with all the fundamental concepts of topology and differential geometry, but gives clear and detailed applications to classical mechanics, electromagnetism, GR and QM. He is not too formal, but develops really a lot of useful tools using differential forms.
Another book, which ...
6
There is a very easy way to see this and it is through an $\hbar$ series. This claim can be traced back to Sydney Coleman and states that in the ultraviolet one is doing an expansion with $\hbar$ going to zero. A previous answer cited these lectures on classical fields but I would like to start from the generating functional of the scalar field theory and ...
5
Yes, the invariance of the action follows from special relativity – and special relativity is right (not only) because it is experimentally verified. All the equations of motion may be derived from the condition $\delta S = 0$, the action is stationary (which usually means it has the minimum value on the allowed trajectory/history among all ...
3
This may not be exactly what you are looking for, but I am going to recommend two specific texts.
Misner, Thorne, and Wheeler, Gravitation, Chapters 4, 9, and end of 14
Solidly in the realm of physics but they have a lot of tidbits of interpretation in there.
Choquet-Bruhat and DeWitt-Morette, Analysis, Manifolds, and Physics, Chapter IV.C
I mean, this ...
3
This paper (http://www-stat.stanford.edu/~cgates/PERSI/papers/dyn_coin_07.pdf) shows that the probability distribution of getting a head, if I toss with the head side up is given by:
$p(ψ, φ) =\frac{1}{2}+\frac{1}{\pi}
\sin^{-1}
(\cot(φ) \cot(ψ))$ if $(\cot φ)(\cot ψ) ≤ 1$,
=1 if $\cot(φ) \cot(ψ) ≥ 1$
where $\phi, \psi$ are the Euler angles.
3
No, it is not possible, and the argument is simple--- there is no dimensional parameter with unit of length, so if there were a stable equilibrium at one radius, there would be many such equilibria obtained by rescaling the original solution to a one-parameter family of solutions.
In fact, it is easier to see that the stable solution is for the electron to ...
3
The particles that communicate the Weak interaction, i.e W Bosons and Z bosons are massive. So unlike Electromagnetism which is communicated by massless particles(Photons), the weak interaction has a very short range.
For Massive particles the Potential of interaction falls as
$V(x) = -K \frac{1}{r} e^{-m r} $
The range of this force is approximately ...
3
Yes, of course, classical strings are just the $\hbar\to 0$ limit of "ordinary strings" and they do interact although the rate goes to zero in the limit, too. The local picture of the interactions used for "quantum strings" should be interpreted literally and it does allow strings to interact even if they're "classical":
On the picture, you see the "type ...
2
1) Usually, but not always, the word classical in physics refers to the limit $\hbar\to 0$.
2) Perturbative string interactions of open and closed strings are governed by the open and the closed string coupling constants, respectively, which are independent of Planck constant $\hbar$.
2
They're two different limits in which two different constants are sent to zero and the resulting limiting theory has different names.
However, both of them are limits for dimensionful constants and the analogy is perfect.
The properly derived $\hbar\to 0$ limit of a quantum mechanical theory is a classical theory – its classical limit – in the very same ...
2
Yes, when people talked about the thermodynamic limit, they are always refer to the limit of large number of particle, or $N \to \infty$.
The role of $k_B$ is to link the microscope quantities and macroscopic thermodynamic quantities. In Newtonian Mechanics, we have already defined the momentum and kinetic energy very clear. On the other hand, Thermodynamic ...
1
Your premise violates Newton's first law of motion:
If there is no net force on an object, then its velocity is constant. The object is either at rest (if its velocity is equal to zero), or it moves with constant speed in a single direction.
For an object (a body) to be accelerating there must be an external force applied. One of the reasons for ...
1
New answer
What you've done here is just dimensional analysis. But you've gone a little too far. In particular, just because two things have the same dimensions doesn't mean that they are equal. If you want to expand $F/A$ a little more, you can choose your favorite from
\begin{equation}
F = \frac{d p}{dt} = \frac{d}{dt}(m\, v) = m\, \frac{d}{dt} v = m\, ...
1
When looking at phase in a sine wave, for example when you are interested in wave interactions such as comb filtering, values are between 0deg and 360deg (you can normalize between 180 and -180). You are looking for the distance between peaks.
When looking at phase on a real source, such as a mono recording playing on two speakers, phase can easily exceed ...
1
The kinetic energy the ball has when it hits the block is $mgh$. We have the relation:
$$
\frac{mv^2}{2}=mgh
$$
This can be rewritten:
$$
m^2v^2=2m^2gh
$$
Which means the downward momentum of the ball is:
$$
p=mv=m\sqrt{2gh}
$$
The block will excert a force on the ball to cancel this momemntum and give it a momentum of $e\cdot p$ in the opposite ...
1
1st: you must take care that friction is a impulsive force and momentum never remains conserved when $f$ changes.
2nd:the normal reaction will become $mg+I/\Delta t$ where $I$ is the impulse imparted on the ball ie:$mv(1+e);v=\sqrt{2gh}\ and\ \Delta t .$is the small time of impact.So, $$N=m(g+v(1+e)/\Delta t)$$
As,the momentum and energy both are not ...
1
One way for finding the field of a magnet is to model it (as a polarized material inside volume $V$ ) with magnetic dipoles , as lots of dipoles near each other , and then sum the produced fields of all dipoles at the desired point.
To find the field of a dipole, You can model it as two (to date, fictitious) magnetic monopoles and use coulomb force law ...
1
There are classical systems without trajectories with the particles 'going through' all possible classical paths. Check for instance Poincaré resonances and the limits of trajectory dynamics.
The concept of trajectory is an approximation both in quantum and classical mechanics (check above ref.); we recover trajectories when the states are localized $\sigma ...
1
I will give it a shot. Spoiler: I did this in the body frame so that the moment of inertia is time independent, before you get excited...
Starting with Euler's equations:
$$
I_i\dot{\Omega}_i+(I_j - I_k)\Omega_j \Omega_k = 0
$$
and taking cyclic permutations of $i,j,k$ to get the three of them; and in the absence of torques (I ignore air friction). It's a ...
1
The solution is not as simple as you wrote, it is a sum over discrete $\lambda_n$.
To get the right solution, you first introduce a shifted $u$: $u'=u-T_0$. For $u'$ you will get $B=0$ and an equation for finding the discrete spectrum of $\lambda$. Then you make a superposition with different $A_n$ and make it obey the initial condition. This permits to ...
1
OP wrote(v1):
What would happen if I considered a system in which the potential is [velocity-dependent] $U(q, \dot q)$?
Well, if OP already knows that the generalized force$^1$
$$\tag{1} Q_j~=~\frac {d}{dt} \frac {\partial U}{\partial \dot q^j}- \frac {\partial U}{\partial q^j}$$
is given in terms of a velocity-dependent potential $U=U(q, \dot q, ...
1
This is nonsense--- the "quantization of energy" you are referring to in classical theories is not a quantization at all, it is equipartition. It is only true that independent degrees of freedom have an average energy which is roughly quantized in classical mechanics.
You can't make a classical equilibrium between a field and an atom, because the field ...
1
Equilibrium will be reached when the net torque on the armature is zero. Since, as we will see below, it will be impossible to have the net torque vanish over an extended interval of time, we'll look for the situation when the torque averaged over time vanishes.
The Lorentz force law tells us that for a wire of length vector $\vec{l}$ carrying current $I$ ...
1
The classical analogue of quantum $\Phi^4$ theory is classical $\Phi^4$ theory, with the same action. There are no particles, but there is still scattering of waves! The correspondence between tree-level QFT and classical fields is on the level of fields only. (Particles make their appearance in classical field theory only in the limit where geometric optics ...
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