## Classical electromagnetism and special relativity – wikipedia electricity symbols ks3

The theory of special relativity plays an important role in the modern theory of classical electromagnetism. First of all, it gives formulas for how electromagnetic objects, in particular the electric and magnetic fields, are altered under a Lorentz transformation from one inertial frame of reference to another. Secondly, it sheds light on the relationship between electricity and magnetism, showing that frame of reference determines if an observation follows electrostatic or magnetic laws. Third, it motivates a compact and convenient notation gas 0095 download for the laws of electromagnetism, namely the manifestly covariant tensor form.

Maxwell’s equations, when they were first stated in their complete form in 1865, would turn out to be compatible with 101 gas station special relativity. [1] Moreover, the apparent coincidences in which the same effect was observed due to different physical phenomena by two different observers would be shown to be not coincidental in the least by special relativity. In fact, half of Einstein’s 1905 first paper on special relativity, On the Electrodynamics of Moving Bodies, explains how to transform Maxwell’s equations.

This equation, also called the Joules-Bernoulli equation, considers two inertial frames. The primed frame is moving relative to the unprimed frame at velocity v. Fields defined in the primed frame are indicated by primes, and fields defined in the unprimed frame lack primes. The field components parallel to the electricity vocabulary words velocity v are denoted by E ∥ {\displaystyle \mathbf {E} _{\parallel }} and B ∥ {\displaystyle \mathbf {B} _{\parallel }} while the field components perpendicular to v are denoted as E ⊥ {\displaystyle \mathbf {E} _{\perp }} and B ⊥ {\displaystyle \mathbf {B} _{\perp }} . In these two frames moving at relative velocity v, the E-fields and B-fields are related by: [2] E ∥ ′ = E ∥ B ∥ ′ = B ∥ E ⊥ ′ = γ ( E ⊥ + v × B ) B ⊥ ′ = γ ( B ⊥ − 1 c 2 v × E ) {\displaystyle {\begin{aligned}\mathbf {E_{\parallel }} ‘=\mathbf {E_{\parallel }} \\\mathbf {B_{\parallel }} ‘=\mathbf {B_{\parallel }} \\\mathbf {E_{\bot }} ‘=\gamma \left(\mathbf {E} _{\bot }+\mathbf {v} \times \mathbf {B} \right)\\\mathbf {B_{\bot }} ‘=\gamma \left(\mathbf {B} _{\bot }-{\frac {1}{c^{2}}}\mathbf {v} \times \mathbf {E} \right)\end{aligned}}}

An equivalent, alternative expression is: [3] E ′ = γ ( E + v × B ) − ( γ − 1 ) ( E ⋅ v ^ ) v ^ B ′ = γ ( B − v × E c 2 ) − ( γ − 1 ) ( B ⋅ v ^ ) v ^ {\displaystyle grade 6 electricity quiz {\begin{aligned}\mathbf {E} ‘=\gamma \left(\mathbf {E} +\mathbf {v} \times \mathbf {B} \right)-\left({\gamma -1}\right)(\mathbf {E} \cdot \mathbf {\hat {v}} )\mathbf {\hat {v}} \\\mathbf {B} ‘=\gamma \left(\mathbf {B} -{\frac {\mathbf {v} \times \mathbf {E} }{c^{2}}}\right)-\left({\gamma -1}\right)(\mathbf {B} \cdot \mathbf {\hat {v}} )\mathbf {\hat {v}} \end{aligned}}}

where v ^ = v ‖ v ‖ {\displaystyle \mathbf {\hat {v}} ={\frac {\mathbf {v} }{\Vert \mathbf {v} \Vert }}} is the velocity unit vector. With previous notations, one actually has ( E ⋅ v ^ ) v ^ = E ∥ {\displaystyle (\mathbf {E} \cdot electricity symbols ks3 \mathbf {\hat {v}} )\mathbf {\hat {v}} =\mathbf {E} _{\parallel }} and ( B ⋅ v ^ ) v ^ = B ∥ {\displaystyle (\mathbf {B} \cdot \mathbf {\hat {v}} )\mathbf {\hat {v}} =\mathbf {B} _{\parallel }} .

If one of the fields is zero in one frame of reference, that doesn’t necessarily mean it is zero in all other frames of reference. This can be seen by, for instance, making the unprimed electric field zero in the transformation to the primed electric field. In this case, depending on the orientation of the magnetic field, the primed system could see an electric field, even though there is none in the unprimed system.

If S and S’ have aligned axes then: [4] u x ′ = u x + v 1 + ( v u x ) / c 2 u y ′ = u y / γ 1 + ( v u x ) / c 2 u z ′ = u z / γ 1 + ( v u x ) / c 2 {\displaystyle {\begin{aligned}u_{x}’={\frac {u_{x}+v}{1+(v\ u_{x})/c^{2}}}\\u_{y}’={\frac {u_{y}/\gamma }{1+(v\ u_{x})/c^{2}}}\\u_{z}’={\frac {u_{z}/\gamma }{1+(v\ u_{x})/c^{2}}}\end{aligned}}}

E x ′ = E x B x ′ = B x E y ′ = γ ( E y − v B z ) B y ′ = γ ( B y + v c 2 E z ) E z ′ = γ ( E z + v B y ) B z ′ = γ ( B z − v c 2 E y ) . {\displaystyle electricity rate per kwh philippines {\begin{aligned}E’_{x}=E_{x}\qquad B’_{x}=B_{x}\\E’_{y}=\gamma \left(E_{y}-vB_{z}\right)B’_{y}=\gamma \left(B_{y}+{\frac {v}{c^{2}}}E_{z}\right)\\E’_{z}=\gamma \left(E_{z}+vB_{y}\right)B’_{z}=\gamma \left(B_{z}-{\frac {v}{c^{2}}}E_{y}\right).\\\end{aligned}}}

D ′ = γ ( D + 1 c 2 v × H ) + ( 1 − γ ) ( D ⋅ v ^ ) v ^ H ′ = γ ( H − v × D ) + ( 1 − γ ) ( H ⋅ v ^ ) v ^ {\displaystyle {\begin{aligned}\mathbf {D} ‘=\gamma \left(\mathbf {D} +{\frac {1}{c^{2}}}\mathbf {v} \times \mathbf {H} \right)+(1-\gamma )(\mathbf {D} \cdot \mathbf {\hat {v}} )\mathbf {\hat {v}} \\\mathbf {H} ‘=\gamma electricity number \left(\mathbf {H} -\mathbf {v} \times \mathbf {D} \right)+(1-\gamma )(\mathbf {H} \cdot \mathbf {\hat {v}} )\mathbf {\hat {v}} \end{aligned}}}

An alternative simpler transformation of the EM field uses the electromagnetic potentials – the electric potential φ and magnetic potential A: [7] φ ′ = γ ( φ − v A ∥ ) A ∥ ′ = γ ( A ∥ − v φ c 2 ) A ⊥ ′ = A ⊥ {\displaystyle {\begin{aligned}\varphi ‘=\gamma \left(\varphi -vA_{\parallel }\right)\\A_{\parallel }’=\gamma \left(A_{\parallel }-{\frac {v\varphi }{c^{2}}}\right)\\A_{\bot }’=A_{\bot }\end{aligned}}}

where A ∥ {\displaystyle \scriptstyle A_{\parallel }} is the parallel component of A to the direction of relative velocity between frames v, and A ⊥ {\displaystyle \scriptstyle A_{\bot }} is the perpendicular component. These transparently resemble the characteristic form of other Lorentz electricity grid map uk transformations (like time-position and energy-momentum), while the transformations of E and B above are slightly more complicated. The components can be collected together as:

A ′ = A − γ φ c 2 v + ( γ − 1 ) ( A ⋅ v ^ ) v ^ φ ′ = γ ( φ − A ⋅ v ) {\displaystyle {\begin{aligned}\mathbf {A} ‘=\mathbf {A} -{\frac {\gamma \varphi }{c^{2}}}\mathbf {v} +\left(\gamma -1\right)\left(\mathbf {A} \cdot \mathbf {\hat {v}} \right)\mathbf {\hat {v}} \\\varphi ‘=\gamma \left(\varphi -\mathbf {A} \cdot \mathbf {v} \right)\end{aligned}}} The ρ and J fields [ edit ]

J ′ = J − γ ρ v + ( γ − 1 ) ( J ⋅ v ^ ) v ^ ρ ′ = γ ( ρ − J ⋅ v c 2 ) {\displaystyle {\begin{aligned}\mathbf {J} ‘=\mathbf {J} -\gamma \rho \mathbf {v} +\left(\gamma -1\right)\left(\mathbf {J} \cdot \mathbf {\hat {v}} \right)\mathbf {\hat {v}} \\\rho ‘=\gamma \left(\rho -{\frac {\mathbf {J} \cdot \mathbf {v} }{c^{2}}}\right)\end{aligned}}} Non-relativistic approximations [ edit ]

E ′ ≈ E + v × B B ′ ≈ B − 1 c 2 v × E J ′ ≈ J − ρ v ρ ′ ≈ ( ρ − 1 c 2 J ⋅ v ) {\displaystyle {\begin{aligned}\mathbf {E} ‘\approx \mathbf {E} +\mathbf electricity in water experiment {v} \times \mathbf {B} \\\mathbf {B} ‘\approx \mathbf {B} -{\frac {1}{c^{2}}}\mathbf {v} \times \mathbf {E} \\\mathbf {J} ‘\approx \mathbf {J} -\rho \mathbf {v} \\\rho ‘\approx \left(\rho -{\frac {1}{c^{2}}}\mathbf {J} \cdot \mathbf {v} \right)\end{aligned}}}