I haven't understood what you say here. Are you talking of a mass increase of a particle with speed or what?
Yes. I'll restate it to make it clearer. In what follows what I mean by the use of the term mass I mean inertial mass in the Newtonian case and relativistic mass in the relativistic case.
In Newtonian mechanics mass does not increase with speed.
In relativistic mechanics mass does increase with speed.
Simple, yes?
And what are you saying about a mass spectrometer?
That's one way to measure the relativistic mass of a charged particle.
That you could measure a particle's mass increase with speed using that device? If you are saying this, it's incorrect, and not because it would be too small, but because you can't.
That's incorrect. In fact Wikipedia states that in 1901 Walter Kaufmann used a mass spectrometer to measure the relativistic mass increase of electrons. See http://en.wikipedia.org/wiki/History_of_mass_spectrometry
See also the article in American Journal of Physics about this particular point at
http://gabrielse.physics.harvard.edu/gabrielse/papers/1995/RelativisticMassAJP.pdf
However a mass spectrometer is in essence a cyclotron which can measure a mass increase due to relativistic mass.
The physics principles are the same as they are in a cyclotron. The particle is charged and that makes it move on a circle in a magnetic field. The radius of the circle it moves in. The physics for a cyclotron and a mass spectrometer is in the following webpage in my website: http://home.comcast.net/~peter.m.brown/sr/cyclotron.htm
If you are saying this, it's incorrect, and not because it would be too small, but because you can't.
Please provide a proof of this. A derivation whose results show what you assert would be fine. Thanks.
In order to analyze relativistic dynamic experiments, what we need is:
E
^{2} = (c*
p)
^{2} + (m*c
^{2})
^{2} (1)
v = c
^{2}*
p/E (2)
(
p and
v are momentum and velocity vectors, " " means modulus of the vector).
From these two equations we can deduce other relativistic equations, for example:
E = mc
^{2}*γ
p = M*
v*γ
Where γ = 1/sqrt{1  (
v/c)
^{2}}.
We also need the law:
F = d
p/dt (3)
which is the definition of force in relativistic dynamics.
Let’s analyze the motion of a charged particle in a magnetic field.
We know that (Lorentz force):
F = q*
v x
B (4)
Where “x” means vectorial product.
So
F is always orthogonal to
v, and orthogonal to
p too, according to eq. (2). It follows that the modulus of
p is constant (and according to eq. (1) even E is constant):
d
p
^{2}/dt = 2
p.(d
p/dt) = 2
p.
F = 0.
(the dot between two vectorial factors means scalar product).
Defining the angular speed as: ω = 
v/r, according to kinematics we have:
d
p/dt = ω*
p
(This equation is valid in galileian as well as in relativistic kinematics).
For the sake of simplicity, let’s assume
v and
p both orthogonal to the magnetic field
B.
Then the trajectory is a circle, traveled of uniform motion and which lays in a plane orthogonal to B.
So, using equations (2), (3), (4) and the expression of the (modulus of) Lorentz force on a charged particle:

F = q*
v*
B
we easily deduce :
ω*
p = q*
v*
B (5)

p/r = q*
B →
p = q*B*r (6)
and it’s this last equation that we can use for a mass spectrometer, measuring r and knowing 
B, to compute 
p and so to find the mass m according to eq. (1), knowing the total energy E. The last can be found using also:
E = E
_{k} + m*c
^{2} (7)
where the kinetic energy E
_{k} is set by the potential difference V applied to the charge q: E
_{k} = q*V.
Substituting in eq. (7) and then in eq. (1):
(q*V + m*c
^{2})
^{2} = (c*
p)
^{2} + (m*c2)
^{2} from which you can find the invariant mass m of the charged particle.
No need of relativistic mass.
N.B. (Most of what written up is not mine, it comes from another source, but it's rather simple to understand.
The source is an italian thread started from prof. Elio Fabri:
https://groups.google.com/d/msg/free.it.scienza.fisica/UyfGXka6rgQ/MiEiUTVexIsJ).
But if you intended that the mass spectrometer allows you to find the momentum 
p and so the total energy
E and you identify the last with relativistic mass (multiplied c
^{2}), then we again come back to what I have said before several times in various threads, even answering to you, that is that relativistic mass is nothing else than another name for total energy E.
Edit: coloured in blue eq. (6)  01/12/2014

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