I am inclined to be in favour with the reply of "etotheipi', though I totally missed the perfectly logical reply of 'Vanadium 50'!
Anyway, thanks for all contributions.
A math exam question asked for the ratio of two wheels given the required information about the number of revolutions made along a certain distance by the two wheels of a penny-farthing bike.
Some students gave the ratio of the radii while others gave the ratio of the areas.
What should an...
I thank all those who posted their replies.
But am I correct to argue as follows?
Using only real numbers,
√( (-2)(-3) ) = √6 is correct but √(-2)√(-3) does not exist as a real number.
Hence Method 1 and Method 2 in my first post cannot be compared.
Thank you both, 'nrqed' and 'kimbyd'!
Yes, my mistake was assuming that ##W_a## includes the rest mass energy.
I derived it again and it turned out as the one given by Fermi, as 'nrqed' said.
Thanks 'Kimbyd' for your interest.
I am using the usual
##E^2 = p^2 c^2 + {m_o}^2c^4,##
and replacing E by ##W_a##, assuming ##W_a## stands for the total energy of the particle.
I am also putting in the scalar potential V and vector potential U since the charged particle is in an...
In his article 'Quantum theory of radiation', Reviews of Modern Physics, Jan 1932, volume 4, Fermi gives the relativistic hamiltonian function ##W_a## for a point charge by equation (13),
## 0 = - \frac 1 {2m} \left( \left[ mc +\frac {W_a - ~eV} c \right ]^2 - \left[ p -\frac {eU} c...
I think that it all depends on with whom I share my work.
Perhaps if I share it with a beginner like me, it is good to show all steps.
Thank you for your help, including your last hint.
Thanks 'Orodruin".
I am not so fluent with tensors and hence,
$$
\partial F_{\alpha\beta}F^{\alpha\beta} = 2F_{\alpha\beta} \partial F^{\alpha\beta} = 2F^{\alpha\beta} \partial F_{\alpha\beta},
$$
is not so obvious to me and I have to work it out.
##\begin {align}
∂(F_{αβ}F^{αβ}) & =...
I am following the suggestion by 'jedishrfu' and repeating my original post but removing the pdf attachment and trying LaTex instead.
The langragian for an electromagnetic field with sources is,
$$\mathcal{L}= - \frac {1}{4} F_{αβ}F^{αβ} + j^μA_μ.$$
Hence,
##\mathcal{L} = - \frac {1}{4}...
I am a retired High School teacher trying to use tensors in getting the Euler-Lagrange equations from the em lagrangian density.
I attached a document in my post since I am not fluent in writing LaTex.
Can anyone, please check my work.
Thanks.