# Relativity, force, acceleration, momentum

1. Jan 12, 2007

### DonnieD

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

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?

2. Jan 13, 2007

### m_haseli

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

Last edited by a moderator: Jan 15, 2007
3. Jan 15, 2007

### DonnieD

someone can explain easy, where's the mistake?

4. Jan 15, 2007

### Hootenanny

Staff Emeritus
Well, this is wrong for a start
I suggest you read this tutorial, specifically the addendum at the end, to begin with.

5. Jan 15, 2007

### lightarrow

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.

Last edited: Jan 15, 2007
6. Jan 15, 2007

### Hootenanny

Staff Emeritus
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