## Electromagnetic wave equation – wikipedia, the free encyclopedia c gastronomie

In his 1865 paper titled A Dynamical Theory of the Electromagnetic Field, Maxwell utilized the correction to Ampère’s circuital law that he had made in part III of his 1861 paper On Physical Lines of Force. Gas and supply okc In Part VI of his 1864 paper titled Electromagnetic Theory of Light, [2] Maxwell combined displacement current with some of the other equations of electromagnetism and he obtained a wave equation with a speed equal to the speed of light. Gas and electric phone number He commented: The agreement of the results seems to show that light and magnetism are affections of the same substance, and that light is an electromagnetic disturbance propagated through the field according to electromagnetic laws. Electricity and magnetism equations [3] Maxwell’s derivation of the electromagnetic wave equation has been replaced in modern physics education by a much less cumbersome method involving combining the corrected version of Ampère’s circuital law with Faraday’s law of induction.

To obtain the electromagnetic wave equation in a vacuum using the modern method, we begin with the modern ‘Heaviside’ form of Maxwell’s equations. Grade 9 electricity questions In a vacuum- and charge-free space, these equations are: ∇ ⋅ E = 0 ∇ × E = − ∂ B ∂ t ∇ ⋅ B = 0 ∇ × B = μ 0 ε 0 ∂ E ∂ t {\displaystyle {\begin{aligned}\nabla \cdot \mathbf {E} &=0\\\nabla \times \mathbf {E} &=-{\frac {\partial \mathbf {B} }{\partial t}}\\\nabla \cdot \mathbf {B} &=0\\\nabla \times \mathbf {B} &=\mu _{0}\varepsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\\\end{aligned}}} These are the general Maxwell’s equations specialized to the case with charge and current both set to zero. Grade 9 electricity review Taking the curl of the curl equations gives: ∇ × ( ∇ × E ) = − ∂ ∂ t ∇ × B = − μ 0 ε 0 ∂ 2 E ∂ t 2 ∇ × ( ∇ × B ) = μ 0 ε 0 ∂ ∂ t ∇ × E = − μ 0 ε 0 ∂ 2 B ∂ t 2 {\displaystyle {\begin{aligned}\nabla \times \left(\nabla \times \mathbf {E} \right)&=-{\frac {\partial }{\partial t}}\nabla \times \mathbf {B} =-\mu _{0}\varepsilon _{0}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}\\\nabla \times \left(\nabla \times \mathbf {B} \right)&=\mu _{0}\varepsilon _{0}{\frac {\partial }{\partial t}}\nabla \times \mathbf {E} =-\mu _{0}\varepsilon _{0}{\frac {\partial ^{2}\mathbf {B} }{\partial t^{2}}}\end{aligned}}} We can use the vector identity ∇ × ( ∇ × V ) = ∇ ( ∇ ⋅ V ) − ∇ 2 V {\displaystyle \nabla \times \left(\nabla \times \mathbf {V} \right)=\nabla \left(\nabla \cdot \mathbf {V} \right)-\nabla ^{2}\mathbf {V} } where V is any vector function of space.

Gas hydrates wiki And ∇ 2 V = ∇ ⋅ ( ∇ V ) {\displaystyle \nabla ^{2}\mathbf {V} =\nabla \cdot \left(\nabla \mathbf {V} \right)} where ∇ V is a dyadic which when operated on by the divergence operator ∇ ⋅ yields a vector. Gas tax in texas Since ∇ ⋅ E = 0 ∇ ⋅ B = 0 {\displaystyle {\begin{aligned}\nabla \cdot \mathbf {E} &=0\\\nabla \cdot \mathbf {B} &=0\end{aligned}}} then the first term on the right in the identity vanishes and we obtain the wave equations: ∂ 2 E ∂ t 2 − c 0 2 ⋅ ∇ 2 E = 0 ∂ 2 B ∂ t 2 − c 0 2 ⋅ ∇ 2 B = 0 {\displaystyle {\begin{aligned}{\frac {\partial ^{2}\mathbf {E} }{\partial t^{2}}}-c_{0}^{2}\cdot \nabla ^{2}\mathbf {E} &=0\\{\frac {\partial ^{2}\mathbf {B} }{\partial t^{2}}}-c_{0}^{2}\cdot \nabla ^{2}\mathbf {B} &=0\end{aligned}}} where c 0 = 1 μ 0 ε 0 = 2.99792458 × 10 8 m/s {\displaystyle c_{0}={\frac {1}{\sqrt {\mu _{0}\varepsilon _{0}}}}=2.99792458\times 10^{8}\;{\textrm {m/s}}} The electromagnetic wave equation is modified in two ways, the derivative is replaced with the covariant derivative and a new term that depends on the curvature appears. Gas in back shoulder − A α ; β ; β + R α β A β = 0 {\displaystyle -{A^{\alpha ;\beta }}_{;\beta }+{R^{\alpha }}_{\beta }A^{\beta }=0} where R α β {\displaystyle \scriptstyle {R^{\alpha }}_{\beta }} is the Ricci curvature tensor and the semicolon indicates covariant differentiation. The generalization of the Lorenz gauge condition in curved spacetime is assumed: A μ ; μ = 0. Electricity load shedding {\displaystyle {A^{\mu }}_{;\mu }=0.} Inhomogeneous electromagnetic wave equation [ edit ] The general solution to the electromagnetic wave equation is a linear superposition of waves of the form E ( r , t ) = g ( ϕ ( r , t ) ) = g ( ω t − k ⋅ r ) {\displaystyle \mathbf {E} (\mathbf {r} ,t)=g(\phi (\mathbf {r} ,t))=g(\omega t-\mathbf {k} \cdot \mathbf {r} )} B ( r , t ) = g ( ϕ ( r , t ) ) = g ( ω t − k ⋅ r ) {\displaystyle \mathbf {B} (\mathbf {r} ,t)=g(\phi (\mathbf {r} ,t))=g(\omega t-\mathbf {k} \cdot \mathbf {r} )} for virtually any well-behaved function g of dimensionless argument φ, where ω is the angular frequency (in radians per second), and k = ( k x, k y, k z) is the wave vector (in radians per meter).

Although the function g can be and often is a monochromatic sine wave, it does not have to be sinusoidal, or even periodic. A shell gas station near me In practice, g cannot have infinite periodicity because any real electromagnetic wave must always have a finite extent in time and space.

76 gas station locations As a result, and based on the theory of Fourier decomposition, a real wave must consist of the superposition of an infinite set of sinusoidal frequencies. In addition, for a valid solution, the wave vector and the angular frequency are not independent; they must adhere to the dispersion relation: k = | k | = ω c = 2 π λ {\displaystyle k=|\mathbf {k} |={\omega \over c}={2\pi \over \lambda }} where k is the wavenumber and λ is the wavelength. Electricity usage in the us The variable c can only be used in this equation when the electromagnetic wave is in a vacuum.

C gastronomie Monochromatic, sinusoidal steady-state [ edit ] The simplest set of solutions to the wave equation result from assuming sinusoidal waveforms of a single frequency in separable form: E ( r , t ) = ℜ { E ( r ) e i ω t } {\displaystyle \mathbf {E} (\mathbf {r} ,t)=\Re \left\{\mathbf {E} (\mathbf {r} )e^{i\omega t}\right\}} where i is the imaginary unit, ω = 2 π f is the angular frequency in radians per second, f is the frequency in hertz, and e i ω t = cos ⁡ ( ω t ) + i sin ⁡ ( ω t ) {\displaystyle \scriptstyle e^{i\omega t}=\cos(\omega t)+i\sin(\omega t)} is Euler’s formula. Npower electricity bill Plane wave solutions [ edit ] Then planar traveling wave solutions of the wave equations are E ( r ) = E 0 e − i k ⋅ r {\displaystyle \mathbf {E} (\mathbf {r} )=\mathbf {E} _{0}e^{-i\mathbf {k} \cdot \mathbf {r} }} B ( r ) = B 0 e − i k ⋅ r {\displaystyle \mathbf {B} (\mathbf {r} )=\mathbf {B} _{0}e^{-i\mathbf {k} \cdot \mathbf {r} }} These solutions represent planar waves traveling in the direction of the normal vector n. Grade 9 electricity test questions If we define the z direction as the direction of n. Rahal e gas card and the x direction as the direction of E, then by Faraday’s Law the magnetic field lies in the y direction and is related to the electric field by the relation c 2 ∂ B ∂ z = ∂ E ∂ t . Electricity invented timeline {\displaystyle c^{2}{\partial B \over \partial z}={\partial E \over \partial t}.} This solution is the linearly polarized solution of the wave equations. Electricity grid australia There are also circularly polarized solutions in which the fields rotate about the normal vector.

Grade 9 electricity unit test answers Spectral decomposition [ edit ] Because of the linearity of Maxwell’s equations in a vacuum, solutions can be decomposed into a superposition of sinusoids. Gas bubble in back This is the basis for the Fourier transform method for the solution of differential equations. Electricity generation in usa The sinusoidal solution to the electromagnetic wave equation takes the form E ( r , t ) = E 0 cos ⁡ ( ω t − k ⋅ r + ϕ 0 ) {\displaystyle \mathbf {E} (\mathbf {r} ,t)=\mathbf {E} _{0}\cos(\omega t-\mathbf {k} \cdot \mathbf {r} +\phi _{0})} B ( r , t ) = B 0 cos ⁡ ( ω t − k ⋅ r + ϕ 0 ) {\displaystyle \mathbf {B} (\mathbf {r} ,t)=\mathbf {B} _{0}\cos(\omega t-\mathbf {k} \cdot \mathbf {r} +\phi _{0})} where t is time (in seconds), ω is the angular frequency (in radians per second), k = ( k x, k y, k z) is the wave vector (in radians per meter), and ϕ 0 {\displaystyle \scriptstyle \phi _{0}} is the phase angle (in radians).

The wave vector is related to the angular frequency by k = | k | = ω c = 2 π λ {\displaystyle k=|\mathbf {k} |={\omega \over c}={2\pi \over \lambda }} Assuming monochromatic fields varying in time as e − i ω t {\displaystyle e^{-i\omega t}} , if one uses Maxwell’s Equations to eliminate B, the electromagnetic wave equation reduces to the Helmholtz Equation for E: ( ∇ 2 + k 2 ) E = 0 , B = − i k ∇ × E , {\displaystyle (\nabla ^{2}+k^{2})\mathbf {E} =0,\,\mathbf {B} =-{\frac {i}{k}}\nabla \times \mathbf {E} ,} with k = ω/c as given above. Gas efficient cars under 5000 Alternatively, one can eliminate E in favor of B to obtain: ( ∇ 2 + k 2 ) B = 0 , E = − i k ∇ × B . Gas monkey monster truck {\displaystyle (\nabla ^{2}+k^{2})\mathbf {B} =0,\,\mathbf {E} =-{\frac {i}{k}}\nabla \times \mathbf {B} .} A generic electromagnetic field with frequency ω can be written as a sum of solutions to these two equations. Gas bubble in throat The three-dimensional solutions of the Helmholtz Equation can be expressed as expansions in spherical harmonics with coefficients proportional to the spherical Bessel functions.

Electricity video bill nye However, applying this expansion to each vector component of E or B will give solutions that are not generically divergence-free ( ∇ · E = ∇ · B = 0), and therefore require additional restrictions on the coefficients. The multipole expansion circumvents this difficulty by expanding not E or B, but r · E or r · B into spherical harmonics. Electricity projects in pakistan These expansions still solve the original Helmholtz equations for E and B because for a divergence-free field F, ∇ 2 ( r · F) = r · (∇ 2 F). Electric utility companies in florida The resulting expressions for a generic electromagnetic field are: E = e − i ω t ∑ l , m l ( l + 1 ) [ a E ( l , m ) E l , m ( E ) + a M ( l , m ) E l , m ( M ) ] {\displaystyle \mathbf {E} =e^{-i\omega t}\sum _{l,m}{\sqrt {l(l+1)}}\left[a_{E}(l,m)\mathbf {E} _{l,m}^{(E)}+a_{M}(l,m)\mathbf {E} _{l,m}^{(M)}\right]} B = e − i ω t ∑ l , m l ( l + 1 ) [ a E ( l , m ) B l , m ( E ) + a M ( l , m ) B l , m ( M ) ] {\displaystyle \mathbf {B} =e^{-i\omega t}\sum _{l,m}{\sqrt {l(l+1)}}\left[a_{E}(l,m)\mathbf {B} _{l,m}^{(E)}+a_{M}(l,m)\mathbf {B} _{l,m}^{(M)}\right]} , where E l , m ( E ) {\displaystyle \mathbf {E} _{l,m}^{(E)}} and B l , m ( E ) {\displaystyle \mathbf {B} _{l,m}^{(E)}} are the electric multipole fields of order (l, m), and E l , m ( M ) {\displaystyle \mathbf {E} _{l,m}^{(M)}} and B l , m ( M ) {\displaystyle \mathbf {B} _{l,m}^{(M)}} are the corresponding magnetic multipole fields, and a E( l, m) and a M( l, m) are the coefficients of the expansion.

Youtube gas pedal dance The multipole fields are given by B l , m ( E ) = l ( l + 1 ) [ B l ( 1 ) h l ( 1 ) ( k r ) + B l ( 2 ) h l ( 2 ) ( k r ) ] Φ l , m {\displaystyle \mathbf {B} _{l,m}^{(E)}={\sqrt {l(l+1)}}\left[B_{l}^{(1)}h_{l}^{(1)}(kr)+B_{l}^{(2)}h_{l}^{(2)}(kr)\right]\mathbf {\Phi } _{l,m}} E l , m ( E ) = i k ∇ × B l , m ( E ) {\displaystyle \mathbf {E} _{l,m}^{(E)}={\frac {i}{k}}\nabla \times \mathbf {B} _{l,m}^{(E)}} E l , m ( M ) = l ( l + 1 ) [ E l ( 1 ) h l ( 1 ) ( k r ) + E l ( 2 ) h l ( 2 ) ( k r ) ] Φ l , m {\displaystyle \mathbf {E} _{l,m}^{(M)}={\sqrt {l(l+1)}}\left[E_{l}^{(1)}h_{l}^{(1)}(kr)+E_{l}^{(2)}h_{l}^{(2)}(kr)\right]\mathbf {\Phi } _{l,m}} B l , m ( M ) = − i k ∇ × E l , m ( M ) {\displaystyle \mathbf {B} _{l,m}^{(M)}=-{\frac {i}{k}}\nabla \times \mathbf {E} _{l,m}^{(M)}} , where h l (1,2)(x) are the spherical Hankel functions, E l (1,2) and B l (1,2) are determined by boundary conditions, and Φ l , m = 1 l ( l + 1 ) ( r × ∇ ) Y l , m {\displaystyle \mathbf {\Phi } _{l,m}={\frac {1}{\sqrt {l(l+1)}}}(\mathbf {r} \times \nabla )Y_{l,m}} are vector spherical harmonics normalized so that ∫ Φ l , m ∗ ⋅ Φ l ′ , m ′ d Ω = δ l , l ′ δ m , m ′ . Static electricity how it works {\displaystyle \int \mathbf {\Phi } _{l,m}^{*}\cdot \mathbf {\Phi } _{l’,m’}d\Omega =\delta _{l,l’}\delta _{m,m’}.} The multipole expansion of the electromagnetic field finds application in a number of problems involving spherical symmetry, for example antennae radiation patterns, or nuclear gamma decay. Electricity receiver definition In these applications, one is often interested in the power radiated in the far-field. Electricity facts In this regions, the E and B fields asymptote to B ≈ e i ( k r − ω t ) k r ∑ l , m ( − i ) l + 1 [ a E ( l , m ) Φ l , m + a M ( l , m ) r ^ × Φ l , m ] {\displaystyle \mathbf {B} \approx {\frac {e^{i(kr-\omega t)}}{kr}}\sum _{l,m}(-i)^{l+1}\left[a_{E}(l,m)\mathbf {\Phi } _{l,m}+a_{M}(l,m)\mathbf {\hat {r}} \times \mathbf {\Phi } _{l,m}\right]} E ≈ B × r ^ . Gas oil ratio {\displaystyle \mathbf {E} \approx \mathbf {B} \times \mathbf {\hat {r}} .} The angular distribution of the time-averaged radiated power is then given by d P d Ω ≈ 1 2 k 2 | ∑ l , m ( − i ) l + 1 [ a E ( l , m ) Φ l , m × r ^ + a M ( l , m ) Φ l , m ] | 2 . Gas monkey monster truck body {\displaystyle {\frac {dP}{d\Omega }}\approx {\frac {1}{2k^{2}}}\left|\sum _{l,m}(-i)^{l+1}\left[a_{E}(l,m)\mathbf {\Phi } _{l,m}\times \mathbf {\hat {r}} +a_{M}(l,m)\mathbf {\Phi } _{l,m}\right]\right|^{2}.} Other solutions [ edit ] In spherical coordinates the solutions to the wave equation can be written as follows: E ( r , t ) = 1 r E 0 cos ⁡ ( ω t − k ⋅ r + ϕ 0 ) , {\displaystyle \mathbf {E} (\mathbf {r} ,t)={\frac {1}{r}}\mathbf {E} _{0}\cos(\omega t-k\cdot r+\phi _{0}),} E ( r , t ) = 1 r E 0 sin ⁡ ( ω t − k ⋅ r + ϕ 0 ) , {\displaystyle \mathbf {E} (\mathbf {r} ,t)={\frac {1}{r}}\mathbf {E} _{0}\sin(\omega t-k\cdot r+\phi _{0}),} and B ( r , t ) = 1 r B 0 cos ⁡ ( ω t − k ⋅ r + ϕ 0 ) , {\displaystyle \mathbf {B} (\mathbf {r} ,t)={\frac {1}{r}}\mathbf {B} _{0}\cos(\omega t-k\cdot r+\phi _{0}),} B ( r , t ) = 1 r B 0 sin ⁡ ( ω t − k ⋅ r + ϕ 0 ) . Gas efficient suv 2014 {\displaystyle \mathbf {B} (\mathbf {r} ,t)={\frac {1}{r}}\mathbf {B} _{0}\sin(\omega t-k\cdot r+\phi _{0}).} In cylindrical coordinates, the solutions to the wave equation are the ordinary Bessel function of integer order.

Electricity magnetism See also [ edit ] Theory and experiment [ edit ] Site: https://en.wikipedia.org/wiki/Electromagnetic_wave_equation