Recent content by grzz

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    B What if an exercise wasn't uniquely defined?

    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.
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    B What if an exercise wasn't uniquely defined?

    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...
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    B Real numbers and complex numbers

    Now the math is more clear in my mind. Thanks Mark44.
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    B Real numbers and complex numbers

    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.
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    B Real numbers and complex numbers

    To find √(-2)√(-3). Method 1. √(-2)√(-3) = √( (-2)(-3) ) = √(6). Method 2. √(-2)√(-3) = √( (-1)(2) )√( (-1)(3) ) = √((-1)√(2)√(-1)√(3) = i√(2)i√(3) = (i)(i)√(2)√(3) = -1√( (2)(3) ) =-√6. Why don't the two methods give the same answer? Thanks for any help.
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    I Alternative formula for the Hamiltonian

    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.
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    I Alternative formula for the Hamiltonian

    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...
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    I Alternative formula for the Hamiltonian

    I am still wondering from where did Fermi got his equation (13).
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    I Alternative formula for the Hamiltonian

    I tried the following. ## \left[ {W_a - ~eV} \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 c^2 + m^2c^4 ## ## \left[ \frac{W_a - ~eV} c \right ]^2 = \left[ p -\frac {eU} c \right ] ^2 + m^2c^2 ## ## 0 = -\frac 1 {2m}\left(\left[ \frac{W_a - ~eV} c \right ]^2 -...
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    I Alternative formula for the Hamiltonian

    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...
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    I Help required regarding deriving E-L equations for EM

    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.
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    I Help required regarding deriving E-L equations for EM

    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}) & =...
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    I Help required regarding deriving E-L equations for EM

    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}...
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    I Help required regarding deriving E-L equations for EM

    Thank you 'jedishrfu'. First I will have some practice using LaTex. Then I intend to repeat my post in a better format! Thanks again.
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    I Help required regarding deriving E-L equations for EM

    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.
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