# Fundamental question unable to find answer

To improve my understanding of physics I was unable find the answer to this
question on the web after a lot of searching so I'll try here.

let v= net velocity of the x direction and the y direction

You take a particle at rest with mass m and you accelerate it to 80% of c in the x direction it appears to have the mass = m/(1-v^2/c^2}^1.5 when
resisting acceleration in the x direction.

Now you accelerate the particle moving at 80% of c in the x direction with an electric field which now accelerates the particle in the y direction with no force acting in x direction,
my question is does the particle appear to have the mass=m/(1-v^2/c^2)^1.5 or mass=m/(1-v^2/c^2)^.5 when resisting acceleration in the y direction.

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Dick
Homework Helper
It has the same inertial mass in all directions. Mass isn't a vector and doesn't have different components in different directions. And where did you get the funny exponent (1.5)? 0.5 is the correct one.

where the 1.5 comes from

If you accelerate a particle in the x direction only you will find that the
particle appears to have a mass = m/(1-v^2/c^2)^1.5 when resisting acceleration in the x direction. This is verified by E=mC^2 which is
easy to do,you take a fractional change in speed from which you calculate
the energy change and therefore compute the force*distance required to make that energy change, the particle does not appear to have the
resitive mass=m/(1-v^2/c^2)^.5 when making a speed change.

Dick
Homework Helper
Ok. So this is not a question. It's an advert for your own theory of relativity.

this relationship is well known in linear acceleration, what I don't understand
if you accelerate in the y direction and decelerate in the x direction if you used a magnetic field in place of an electric field the resistance to
acceleration,or mass appears to be in the y direction, in a magnetic field, is mass=m/(1-V^2/c^2)^.5 if in the case of a electric field if mass appears to be mass=m/(1-v^2/c^2)^.5 in acceleration in the y direction, then e=mc^2
would not be valid in x y acceleration. For e=mc^2 to be valid in the case
of the electric field, the resistance to acceleration would have to be mass=m/(1-V^2/c^2)^1.5 in both the x and y direction.Why the difference in what
mass appears to be in the magnetic and electric field in acceleration in the y direction?