Re: New version of a relativity FAQ

On Jun 21, 8:21 pm, "Pmb" <peter.m.br...@xxxxxxxxxxxxx> wrote:


Ok, I had a look at your solution : http://www.geocities.com/physics_world/sr/cyclotron.htm

No wonder you were sa cagey in sharing it, there are quite a few
mistakes. I hope that you will take the following as constructive
criticism and that you'll correct the page. Here it goes:

1.It is quite clear that you can't solve the problem by using
relativistic mass, the "m" in your equations is rest mass. The reason
is quite simple, starting from eq.1 you should be writing:

d(\gamma*m*v)/dt=q(E+vxB)

2. You keep avoiding writing the ODE above, instead you go through all
kinds of girations and you arrive to...

3. ....the following commedy of errors:


"From the diagram it is easy to see that a positive charge q moves in
a clockwise motion around the circle. w is known as the cyclotron
frequency. To find an explicit expression for w we start with the
expression for the magnitude of the Lorentz force which is F = qvB.
Since the motion is transverse to the acceleration we also have F = ma
= mRw2 = qvB = qRwB or w = mv/qB = p/qB. "

4. You are now mixing relativity with Newtonian mechanics (F=ma),
exactly like Juanshito did earlier in this thread. This is a "no-no".
You simply cannot have any "F=ma" in a relativistic page, if you want
to be taken seriously.

5. This is the first time that mass, "m" appears in your solution (it
should have appeared from the very beginning, in eq.1) but you were
too focused on avoiding it. If you had it early on, in the ODE
d(\gamma*m*v)/dt=q(E+vxB), you would have avoided this embarassing
error.


6. Now, as a result of the error at point 4, you have a completely
wrong formula for the "cyclotron frequency". Contrary to what you
derived, w is NOT equal to mv/qB = p/qB.
Nowhere close.

7. Now, you have spent quite a lot of effort writing a lot of formulas
but you are missing the most basic thing (aside from botching the
formula for w). The particle moves in a circle, so what is the value
for its radius? You come close to calculating the radius from eq.13
but not quite.


Pete,

You are not a bad guy but you have a bee under your bonnet when it
comes to relativistic mass. If you were doing the computations
honestly you would have arrived to the correct results. In order to do
that , you need to use rest mass, ok? I know that this pains you but
this is the truth.

Now, I will also transfer to Dave the solution for the case when you
have BOTH a non-null B AND a non-null E. Don't even dream of trying to
solve this problem by using relativistic mass, you can't. On the other
hand, if you start with d(\gamma*m*v)/dt=q(E+vxB) and you pay
attention to the math, you might get the correct result. This is a
tougher problem, so I suggest that you get cracking. Oh, and you will
need to use the rest mass m, ok :-)

.

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