Recent content by Frostman

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...

$$d^4x'=\bigg|\frac{\partial x'}{\partial x}\bigg|d^4x$$ 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...

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...

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) =...

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

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

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)##

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?

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...

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

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} -...

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...

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##...

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...

Okay, it's an hyperbola. Furthermore, the more the acceleration value is greater, the more the trend is that of: ##x(t)=|t|## and this one is the trend of the light signal. The two trends, as I have drawn them, never meet.