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.
As the secant AE is moved downwards, the exterior angle remains equal to the same interior angle, with the result that as the secant becomes a tangent, the cyclic quadrilateral disappears and the exterior angle becomes equal to the angle in the alternate segment. pdf is attached.
It is...
Thanks 'fresh_42'.
That's the idea that crossed my mind.
But then then when one talks about 'limits' one may say that limits are approached as far as required ... but not attained.
I am very sorry.
I meant comparing the graphs of y = x - 1 and y = (x - 1)2.
Both equations have the solution, i.e. x = 1 but that of the tangent is x = 1 (twice).
I am asking if there is some deeper meaning for that 'twice'.
The graph of y = x - 1 CUTS the x-axis at x = 1 while the graph of y = x2- 1 TOUCHES the x-axis at x = 1.
The point at which the tangent touches the curve is shown mathematically by having two solutions of x, i.e. x = 1 (twice).
Is there some deeper meaning to these two identical solutions for x?
Thanks so much to you all.
It seems that the first method is the better.
As for the post of DarkoDornel, I am happy to say that I do not have to worry about "they always ask why that, and why not this" because I am a retired high school teacher who does not intend to work towards any thesis.
When discussing Lorentz transformations some books use
xμ/ = Λμ/αxα
while some other books use
x'μ= Λμαxα.
Perhaps someone can help me understand which is the better notation.
Thanks.
A simpler notation may help.
One can call the final velociies u and w and substitute for the masses right away.
Then conservation of momentum and conservation of KE will give two equations in u and w which can be solved.
You imagined correctly.
I think that as for (h) and (i) you do not have to substitute any numbers. Just consider KEf = ( M1/(M1 + M2) )KEi.
Then ignore the smaller mass when compared to the bigger one in (h) and in (i) and make the obvious deductions as regards what happens to the KEf and KEi.
For part (h),
imagine a very large heavy truck moving and colliding with a very small car. Do you think the initial KE of the truck will decrease much?