Relativity, force, acceleration, momentum

[DonnieD]

let q=relativistic factor
then F=q*m0*a (where m0 is rest mass and a=acceleration)

if E=m*c^2 =>dE=dm*c^2 (i'm not sure about this)
and dL=F*dx (is the infinitesimal work completed by an external strenght F on the system)

F*dx=dm*c^2 =>

q*m0*a*dx=dm*c^2 => dm/m0=(q*a/c^2)*dx

m'(x)=q*m0*a(x)/c^2 ==> F=m'(x)*c^2 .

a=m'(x)*(c^2/(q*m0))

is right?

 

[m_haseli]

No,I suggest you to study "Introduction to special relativity"
Robert Resnick.
I can provide you the detailed computations,if you want.

 

[DonnieD]

someone can explain easy, where's the mistake?

 

[Hootenanny]

someone can explain easy, where's the mistake?

Well, this is wrong for a start

let q=relativistic factor
then F=q*m0*a (where m0 is rest mass and a=acceleration)

I suggest you read http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html" , specifically the addendum at the end, to begin with.

 

[lightarrow]

someone can explain easy, where's the mistake?

F is defined as dp/dt and p = qm0v. But v is contained in the factor q so when you derive p you actually obtain:

F = qm0a + q^3(m0/c^2)(v°a)v

where ° means scalar product.

 

[Hootenanny]

Just a light correction on lightarrow's post (and to make things a little easier to read);

$$\vec{F} = \gamma m_{0}\vec{a} + \gamma^{3}m_{0}\frac{\vec{v}\cdot\vec{a}}{c^2}\vec{v}$$

Edit: Ahh lightarrow's on the ball today