How special relativity leads to Lorentz Invariance of the Maxwell Equations. Differential form will suffice.

Deborah Proctor
2022-10-12
Answered

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Kenley Rasmussen

Answered 2022-10-13
Author has **13** answers

Due to the constraint $\mathrm{\nabla}\cdot B=0$, there exists a vector potential $A$ such that $B=\mathrm{\nabla}\times A$ and ${E}_{j}={\mathrm{\partial}}_{0}{A}_{j}-{\mathrm{\partial}}_{j}{A}_{0}$. In other words, $E$ and $B$ assemble into a "field strength tensor" ${F}_{\mu \nu}={\mathrm{\partial}}_{\mu}{A}_{\nu}-{\mathrm{\partial}}_{\nu}{A}_{\mu}=d{A}_{\mu \nu}$. This is the correct object to reason about when thinking relativistically. It's just a $2$-form, so its transformation rules are simple. We can write Maxwell's equations as

$d\ast F=0$

here $\ast $ is the Hodge star. This is clearly invariant under isometries (Lorentz transformations).

$d\ast F=0$

here $\ast $ is the Hodge star. This is clearly invariant under isometries (Lorentz transformations).

Lance Liu

Answered 2022-10-14
Author has **2** answers

Another way to see it is when deriving the EM wave equation from Maxwell equations. The Lorentz Invariance means that the amplitude should be symmetric under translations of space and time and rotations. In the wave eq. you have only 2nd derivatives (time and space), hence symmetry under Lorentz is preserved.

asked 2022-05-17

if $u$ and ${u}^{\prime}$ are a velocity referred to two inertial frames with relative velocity $v$ confined to the $x$ axis, then the quantities $l$, $m$, $n$ defined by

$(l,m,n)=\frac{1}{|u|}({u}_{x},{u}_{y},{u}_{z})$

and

$({l}^{\prime},{m}^{\prime},{n}^{\prime})=\frac{1}{|{u}^{\prime}|}({u}_{x}^{\prime},{u}_{y}^{\prime},{u}_{z}^{\prime})$

are related by

$({l}^{\prime},{m}^{\prime},{n}^{\prime})=\frac{1}{D}(l-\frac{v}{u},m{\gamma}^{-1},n{\gamma}^{-1})$

and that this can be considered a relativistic aberration formula. The author gives the following definition for $D$, copied verbatim.

$D=\frac{{u}^{\prime}}{u}(1-\frac{{u}_{x}v}{{c}^{2}})={[1-2l\frac{v}{u}+\frac{{v}^{2}}{{u}^{2}}-(1-{l}^{2})\frac{{v}^{2}}{{c}^{2}}]}^{\frac{1}{2}}$

Why is that better than the second expression?

Also, in case it's not clear, $\gamma =1/\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$ and $|u|=|({u}_{x},{u}_{y},{u}_{z})|=\sqrt{{u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}}$

$(l,m,n)=\frac{1}{|u|}({u}_{x},{u}_{y},{u}_{z})$

and

$({l}^{\prime},{m}^{\prime},{n}^{\prime})=\frac{1}{|{u}^{\prime}|}({u}_{x}^{\prime},{u}_{y}^{\prime},{u}_{z}^{\prime})$

are related by

$({l}^{\prime},{m}^{\prime},{n}^{\prime})=\frac{1}{D}(l-\frac{v}{u},m{\gamma}^{-1},n{\gamma}^{-1})$

and that this can be considered a relativistic aberration formula. The author gives the following definition for $D$, copied verbatim.

$D=\frac{{u}^{\prime}}{u}(1-\frac{{u}_{x}v}{{c}^{2}})={[1-2l\frac{v}{u}+\frac{{v}^{2}}{{u}^{2}}-(1-{l}^{2})\frac{{v}^{2}}{{c}^{2}}]}^{\frac{1}{2}}$

Why is that better than the second expression?

Also, in case it's not clear, $\gamma =1/\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}$ and $|u|=|({u}_{x},{u}_{y},{u}_{z})|=\sqrt{{u}_{x}^{2}+{u}_{y}^{2}+{u}_{z}^{2}}$

asked 2022-08-10

$\rho ={\varphi}^{\ast}\varphi \phantom{\rule{thinmathspace}{0ex}}$

$\rho ={\varphi}^{\ast}\varphi \phantom{\rule{thinmathspace}{0ex}}$

......

$J=-\frac{i\hslash}{2m}({\varphi}^{\ast}\mathrm{\nabla}\varphi -\varphi \mathrm{\nabla}{\varphi}^{\ast})$

with the conservation of probability current and density following from the Schrödinger equation:

$\mathrm{\nabla}\cdot J+\frac{\mathrm{\partial}\rho}{\mathrm{\partial}t}=0.$

The fact that the density is positive definite and convected according to this continuity equation, implies that we may integrate the density over a certain domain and set the total to 1, and this condition will be maintained by the conservation law. A proper relativistic theory with a probability density current must also share this feature. Now, if we wish to maintain the notion of a convected density, then we must generalize the Schrödinger expression of the density and current so that the space and time derivatives again enter symmetrically in relation to the scalar wave function. We are allowed to keep the Schrödinger expression for the current, but must replace by probability density by the symmetrically formed expression

$\rho =\frac{i\hslash}{2m}({\psi}^{\ast}{\mathrm{\partial}}_{t}\psi -\psi {\mathrm{\partial}}_{t}{\psi}^{\ast}).$

which now becomes the 4th component of a space-time vector, and the entire 4-current density has the relativistically covariant expression

${J}^{\mu}=\frac{i\hslash}{2m}({\psi}^{\ast}{\mathrm{\partial}}^{\mu}\psi -\psi {\mathrm{\partial}}^{\mu}{\psi}^{\ast})$

1. What exactly are ${\mathrm{\partial}}_{t}$ and ${\mathrm{\partial}}^{\mu}$?

2. Are they tensors?

3. If they are, how are they defined?

$\rho ={\varphi}^{\ast}\varphi \phantom{\rule{thinmathspace}{0ex}}$

......

$J=-\frac{i\hslash}{2m}({\varphi}^{\ast}\mathrm{\nabla}\varphi -\varphi \mathrm{\nabla}{\varphi}^{\ast})$

with the conservation of probability current and density following from the Schrödinger equation:

$\mathrm{\nabla}\cdot J+\frac{\mathrm{\partial}\rho}{\mathrm{\partial}t}=0.$

The fact that the density is positive definite and convected according to this continuity equation, implies that we may integrate the density over a certain domain and set the total to 1, and this condition will be maintained by the conservation law. A proper relativistic theory with a probability density current must also share this feature. Now, if we wish to maintain the notion of a convected density, then we must generalize the Schrödinger expression of the density and current so that the space and time derivatives again enter symmetrically in relation to the scalar wave function. We are allowed to keep the Schrödinger expression for the current, but must replace by probability density by the symmetrically formed expression

$\rho =\frac{i\hslash}{2m}({\psi}^{\ast}{\mathrm{\partial}}_{t}\psi -\psi {\mathrm{\partial}}_{t}{\psi}^{\ast}).$

which now becomes the 4th component of a space-time vector, and the entire 4-current density has the relativistically covariant expression

${J}^{\mu}=\frac{i\hslash}{2m}({\psi}^{\ast}{\mathrm{\partial}}^{\mu}\psi -\psi {\mathrm{\partial}}^{\mu}{\psi}^{\ast})$

1. What exactly are ${\mathrm{\partial}}_{t}$ and ${\mathrm{\partial}}^{\mu}$?

2. Are they tensors?

3. If they are, how are they defined?

asked 2022-07-13

It is experimentally known that the equation of motion for a charge $e$ moving in a static electric field $\mathbf{E}$ is given by:

$\frac{\mathrm{d}}{\mathrm{d}t}(\gamma m\mathbf{v})=e\mathbf{E}$

Is it possible to show this using just Newton's laws of motion for the proper frame of $e$, symmetry arguments, the Lorentz transformations and other additional principles?

$\frac{\mathrm{d}}{\mathrm{d}t}(\gamma m\mathbf{v})=e\mathbf{E}$

Is it possible to show this using just Newton's laws of motion for the proper frame of $e$, symmetry arguments, the Lorentz transformations and other additional principles?

asked 2022-08-12

Is it allowed to have the zeroth-component of a four-velocity be negative?

This is referring to $V^0$ for a curved space metric with signature $-+++$.

This is referring to $V^0$ for a curved space metric with signature $-+++$.

asked 2022-08-25

A proton accelerated with electric field gives off E.M. radiation and therefore should lose mass. Larmor's formula gives us a value for the power emitted (varies as acceleration squared). However, as the proton picks up speed, it also gains mass. Now, say I set up an immense electric field which provides an immense acceleration to the proton. In the initial moments of motion, even though its acceleration is extremely high, its velocity is low. In those moments, does the proton lose mass faster than it gains?

asked 2022-08-18

How to write ${F}^{\mu \nu}{F}_{\mu \nu}$ in terms of ${F}_{\mu \nu}{F}^{\mu \nu}$. How to do it?

asked 2022-09-16

$E=(t{)}^{0}(m{)}^{1}(c{)}^{2}$. Here, $m$ = mass of the body. $c$ = velocity of light. Is $t$ the time?