Recent content by Frostman

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    Evaluate the outgoing radiation from an optical fiber on a surface

    The geometric configuration that I am adopting is the following, I hope you understand. The optical fiber is positioned relative to the bottom surface at a height ##a## and an angle ##\alpha## with respect to the y-axis in the yz-plane with x = 0. ##b## is the distance between the origin and...
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    Charge invariance with Heaviside's function

    In the first step you indicated all the primate terms, except ##\eta_\beta## because I must be able to reduce to the definition of charge as a three-dimensional integral of ##J'^0##. I then observe that in reality the other primate quantities are Lorentz invariants. When I exchange the...
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    Charge invariance with Heaviside's function

    $$d^4x'=\bigg|\frac{\partial x'}{\partial x}\bigg|d^4x$$ Where ##\bigg|\frac{\partial x'}{\partial x}\bigg|## is the determinant of jacobian of Lorentz-transformation, for the Lorentz proper transformation ##det\Lambda^\mu{}_\nu=1##. The demonstration I am following is that of Weinberg chapter...
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    Charge invariance with Heaviside's function

    Perfect, one last step and I believe I am there. Given this definition of charge Q we have that $$Q'=\int d^4x J^\alpha(x)\partial_\alpha \theta(\eta_\beta'x^\beta)$$ For ##J_\alpha\partial_\alpha## I understood because the product of a covariant and contravariant four-vector is a relativistic...
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    Charge invariance with Heaviside's function

    Yes, I corrected, I forgot a zero. Excuse me, but this is my first time working with the Heaviside and Dirac delta function together. $$\partial_0\theta(x^0)=\delta(x^0)$$ Is it correct? $$\int d^3x \int dx^0J^0(x^0, \vec x) \delta(x^0)$$ Or can I suppose that ##\delta(x^0) =...
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    Charge invariance with Heaviside's function

    Can I proceed like this to reduce it to the 3D relationship? $$Q=\int dx^0d^3x J^0(x)\partial_0\theta(x^0)=\int d^3x J^0(x)dx^0\partial_0\theta(x^0)=\int d^3x J^0(x)$$
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    Charge invariance with Heaviside's function

    The only derivative ##\partial_\alpha## that can act is only this one ##\partial_0##. So it remains $$Q=\int d^4x J^0(x)\partial_0\theta(x^0)$$
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    Charge invariance with Heaviside's function

    The only term that remains is ##x^0## because ##\eta_\beta## is defined as ##\eta_\beta=(1,0,0,0)## So ##\theta(\eta_\beta x^\beta)=\theta(\eta_0x^0)=\theta(x^0)##
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    Charge invariance with Heaviside's function

    I hope I understand the definition of the Heaviside function well For ##\beta=0##, ##\theta(\eta_\beta x^\beta)=1## For ##\beta\neq0##, ##\theta(\eta_\beta x^\beta)=0## Right?
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    Charge invariance with Heaviside's function

    I followed a demonstration in one of my electromagnetism books, but it is not clear to me. My problem is at the starting point. The book begins by considering the office defined in the following way: $$Q=\int d^4xJ^\alpha(x)\partial_\alpha\theta(\eta_\beta x^\beta)$$ where...
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    Conservation of charge with Dirac delta

    So that factor ##3## really isn't there, because there's ##\sum_n## in front of everything, right?
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    Conservation of charge with Dirac delta

    Hello, I was reviewing a part related to electromagnetism in which the charge and current densities are defined by the Dirac delta: ##\rho(\underline{x}, t)=\sum_n e_n \delta^3(\underline{x} - \underline{x}_n(t))## ##\underline{J}(\underline{x}, t)=\sum_n e_n \delta^3(\underline{x} -...
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    De-excitation of a moving atom with photon emission

    I tried, but the more I go I get lost in the accounts. I'm not even rewriting what I wrote here on the forum because it seems to me just a waste of time, but to show that I tried to do the math I am attaching the spreadsheets. As this is an exam to be done along with 2 exercises in 2 hours...
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    De-excitation of a moving atom with photon emission

    The information I have are the following: ##p^\mu=(E, p, 0, 0)## ##p'^\mu=(E', p'\cos\beta, -p'\sin\beta,0)## ##k^\mu=\tilde{E}(1, \cos\alpha, \sin\alpha, 0)## Where: ##E=\sqrt{M^2+p^2}## ##E'=\sqrt{m^2+p'^2}## Using the conservation of the four-momentum ##p^\mu=p'^\mu+k^\mu##...
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    Magnetic field generated by a current in a wire - special relativity

    First I wrote in ##S'##, by using Gauss theorem $$ \int_{\Sigma} \underline E' \cdot \hat n d\Sigma = \frac Q {\varepsilon_0} \rightarrow E'(r)2\pi rH=\frac{\lambda'H}{\varepsilon_0} $$ $$ \underline E'(\underline r)=\frac{\lambda'}{2\pi\varepsilon_0r}\hat r $$ Its components are...
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