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Fundamental question unable to find answer

  • Thread starter cjf
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cjf
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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.
 

Answers and Replies

Dick
Science Advisor
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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.
 
cjf
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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
Science Advisor
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Ok. So this is not a question. It's an advert for your own theory of relativity.
 
cjf
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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?
 

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