# Tag Info

37

If a wave $f(x,t)$ is something that satisfies the wave equation $Lf=0$ where $L$ is the differential operator $\partial_t^2-c^2\nabla^2$ then, because $L$ is linear, any linear combination $\lambda f+\mu g$ of solutions $f$ and $g$ is again a solution: $L(\lambda f + \mu g)=\lambda Lf+\mu Lg=0$. In general, there might be things that propagate (not exactly ...

27

As coconut wrote, the superposition principle comes from the linearity of the operator involved. This is the case for electromagnetic radiation in vacuum. Approximations to water waves are also linear (since it is an approximation) but probably will have small non-linear parts. Free quantum field theory is also linear, therefore you have a superposition ...

23

Impedance is usually done in Fourier analysis, not Laplace transform, but that's a quibble. The reality is that you transform the signal, not the impedance. The impedance is an operator that modifies the signal, and it's the effect that operator has on the signal in the transformed domain that we deal with. To review, the basic linear circuit elements are: ...

20

It doesn't really play a role (in a way), or at least not as far as physical results go. Whenever someone says we consider a plane wave of the form $f(x) = Ae^{i(kx-\omega t)}$, what they are really saying is something like we consider an oscillatory function of the form $f_\mathrm{re}(x) = |A|\cos(kx-\omega t +\varphi)$, but: we can represent that in ...

17

No. Despite what several answers on this thread will tell you, there are plenty of phenomena which are perfectly deserving of the term "wave" which do not satisfy the superposition principle. In technical language, the superposition principle is obeyed whenever the underlying dynamics are linear. However, there are plenty of situations that do not ...

13

Bare with me, I don't remember every little step, but I hope this derivation helps you. First remember how a wave travels through a waveguide (dielectric). $$E(x,y,z) = E^{0}(x,y)e^{-\gamma z}$$ $$H(x,y,z) = H^{0}(x,y)e^{-\gamma z}$$ Then consider Ampere's and Faraday's Laws for a source-free region. $$\triangledown \times H = j\omega\epsilon E$$ $$\... 13 Just looking at physics, if the initial problem has a solution and you rotate the plate of \pi/2 you obtain another solution with boundary conditions rotated of \pi/2. Perform the same procedure two other times and you end up with four solutions with corresponding four different boundary conditions rotated of 0, \pi/2, \pi, 3/2 \:\pi ... 13 Simply calling something a "wave" isn't enough for a superposition of solutions to satisfy the governing wave equation. When deriving wave equations linearity is achieved by requiring "small amplitude" oscillations, so in nature when large amplitudes are involved the superposition principle does not hold true in general. 12 In many cases our systems are described by linear differential equations, and these have the property that any linear combination of solutions to the differential equation is also a solution to the differential equation. This is useful because usually any arbitrary solution can be Fourier transformed to express it as a sum of plane waves. So if we can find ... 12 The tensor itself is not the model, but the the tensor is used to model (one could also say describe or quantify) the anisotropy. One example is an anisotropic electric conductor. The conductivity \sigma describes which current occurs in response to an electric field \vec j = \sigma \vec E. In isotropic materials (e.g. glass, microcrystalline metals ... 10 Since you seem happy about the internal forces let's ignore them and just set them all to zero so only consider external forces. The total momentum P is just the sum of all the individual momenta:$$ P = \sum p_i $$and we can differentiate both sides of this to get:$$ \frac{dP}{dt} = \sum \frac{dp_i}{dt} $$For any object, simple or composite, force ... 10 This is purely because of dimensional reasons. Energy has dimension \rm ML^2T^{-2}, and you can construct this dimension from different combinations of dimesionful quantities. Note that, the mass m and velocity v has dimensions \rm M and \rm LT^{-1} respectively which makes the unique combination mv^2 to have the dimension \rm ML^2T^{-2}. ... 9 When one uses complex variables in this way one never multiplies two variables because the whole system is linear: if z is the oscillating variable and you choose to represent it by a complex number, then things like z\times z don't arise in a linear equation, so you don't get the kind of contradiction you astutely and clearly pointed out above. If the ... 9 Actually, none of them satisfies the superposition completely. First, superposition requires linearity, and linearity isn't perfect in most cases. Even in the case of linear theories, the theory is only a model and it has its borders. For example, the Maxwell-equations are linear, and thus light waves superpose. If you cross two laser beams, they totally ... 9 For completeness, suppose you have any energy expression U=U(k) that you are looking at. Energy typically has to be bounded from below, otherwise thermodynamics would send the system to negative-infinite energy to spread that energy everywhere across the universe. Let's say that it attains a minimum at a special value k_0 and we define b=k-k_0, then U... 8 I) In this answer we discuss a systematic approach to linearization and stability analysis. Imagine that the physical system under consideration is described by an autonomous Lagrangian L=L(q,\dot{q}) of n generalized coordinates$$\tag{1} q~=~(q^1, \ldots, q^n)~\in~ \mathbb{R}^n.$$One of the first questions one would like to ask is, if a specific ... 8 Hints: We are evidently only supposed to solve for z-dependence (as opposed to x- and y-dependence). Note that the two variables E_z and H_z can be eliminated. In the reduced coupled ODE system of four first-order ODEs and four variables (E_x,E_y,H_x,H_y), note that the variables couple two and two together. Which pairs? Within one such pair, it ... 8 The no-cloning theorem states that it is not possible to have a quantum state |\psi\rangle evolve into two separable (non-entangled) copies described by the tensor product state |\psi\rangle|\psi\rangle. The proof boils down to the simple observation that when expressing |\psi\rangle in some basis {|0\rangle, |1\rangle, |2\rangle, ...}:$$|\psi\...

8

From a Classical perspective, this is because (as you mention) Maxwell's equations are linear. That is to say that the theory is linear (an abstract mathematical property) because it is represented by an abstract mathematical formalism which has that property. That said, perhaps you will find this more satisfying: Photons have no charge. We know that the ...

8

I would expand upon Sebastian's nice answer to point out that any orientation-sensitive quantity $f(\hat v)$ may be expanded by spherical harmonics and the symmetric rank-2 tensor can be often used to represent the first nonzero term. To understand this, start by noting that all of these spherical harmonics come with a polynomial structure factor. You may ...

7

"By the linearity of quantum mechanics" is actually a reference to the linearity of the operators used it quantum mechanics. It means that, for a linear operator $A$ (by the very definition of linearity), $$A\bigl(\alpha\lvert\Psi\rangle +\beta\lvert\Phi\rangle\bigr)=\alpha A\lvert\Psi\rangle + \beta A\lvert\Phi\rangle,$$ where $\alpha$ and $\beta$ are ...

7

As user1104 commented, you use Euler's identity: $$e^{ix} = \cos(x) + i \space \sin(x)$$ so: $$\sin(kx-\omega t) = \frac{ e^{i(kx-\omega t)} - e^{-i(kx-\omega t)}}{2i}$$ But we wouldn't normally proceed by replacing sin by this expression. Both the sin form and the exponential form are mathematically valid solutions to the wave equation, so the only ...

7

Schrodinger's equation is homogeneous -- so if $\phi_1,\phi_2,\cdots,\phi_n$ are solutions, $c_1\phi_1 + c_2\phi_2 + \cdots +c_n\phi_n$ is a solution. More importantly, if $\phi$ is a solution, $A\phi$ is a solution as well. If $A$ is the normalization constant, we see that both non-normalized and normalized versions are valid solutions of Schrodinger's ...

7

Electrostatic potentials superpose because Poisson’s equation is linear. (And it is linear because Maxwell’s equations are linear in both the fields and their sources, and the potentials and the fields have a linear relation.) Specifically, if $$-\nabla^2\varphi_1=4\pi\rho_1$$ and $$-\nabla^2\varphi_2=4\pi\rho_2$$ then $$-\nabla^2(\varphi_1+\varphi_2)=... 6 I think your qualification of "most" systems needs some clarification because really almost all of the classical universe is described by second-order, nonlinear partial differential equations. Fluids/liquids/gases and solids are described by the same set of second-order, nonlinear PDE's. Linear equations, both linear PDE's and linear ODE's, show up often ... 6 A linear system is one where, if we have two inputs x_1 and x_2, producing outputs y_1 and y_2, then the output for an input of \alpha{}x_1 + \beta{}x_2 is \alpha{}y_1 + \beta{}y_2. This is precisely the property we rely on when we apply superposition to solve a problem with inputs composed of sums of easier-to-analyze inputs. This means, ... 6 Electromagnetism is a gauge theory with the gauge group U(1), which is abelian. This means there are no terms in the EM Lagrangian where the electromagnetic field interacts with itself. L = \frac{-1}{4}F^2 + \bar{\Psi}(i\gamma^{\mu}D_{\mu} - m)\Psi, where D is the covariant derivative containing the gauge field A_{\mu}, at tree (classical) level there ... 6 To see in a very concrete case that it is not additive, consider two spherically symmetric bodies centered at the same point. If you don't like them to overlap, take a ball and a spherical shell that doesn't intersect. Then at sufficient distance we have (wikipedia) that$$R^{t}{}_{rrt} = \frac{r_s}{r^2(r_s-r)} Since $r_s$ is linear in the masses, the ...

6

Oftentimes in physics, we study phenomena that oscillate with a certain frequency, which we'll call $\omega$. These phenomena are usually modeled as the real part of a complex function $e^{i\omega t}$. These oscillations have a tendency to decay exponentially in response to damping, and so are more accurately modeled as $e^{-\beta t}e^{i\omega t}$ for some ...

6

$\boldsymbol{\S}$ A. A special case : symmetric $\Omega^{\boldsymbol{-}1}$ Let the $2\times2$ real symmetric matrices C^{\boldsymbol{-}1}\boldsymbol{=} \begin{bmatrix} \xi_1 & \xi \vphantom{\dfrac{a}{b}}\\ \xi &\xi_2 \vphantom{\dfrac{a}{b}} \end{bmatrix} \quad \text{and} \quad L^{\boldsymbol{-}1}\boldsymbol{=} \begin{bmatrix} \...

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