Relativity, force, acceleration, momentum

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Discussion Overview

The discussion revolves around the concepts of relativity, force, acceleration, and momentum, particularly in the context of relativistic physics. Participants explore the relationships between these concepts and challenge each other's formulations and interpretations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes a formulation of force in relativistic terms as F=q*m0*a, where q is the relativistic factor, m0 is the rest mass, and a is acceleration.
  • Another participant suggests that the relationship between energy and mass, E=m*c^2, leads to an expression dE=dm*c^2, questioning its validity.
  • A participant challenges the initial formulation, indicating that it is incorrect and suggests reading a specific resource on special relativity for clarification.
  • Another participant argues that force should be defined as dp/dt, where p is momentum, leading to a more complex expression involving relativistic factors.
  • A later reply provides a corrected expression for force in terms of relativistic mass and acceleration, indicating a more nuanced understanding of the relationships involved.

Areas of Agreement / Disagreement

Participants express disagreement regarding the initial formulations and interpretations of relativistic force and momentum. Multiple competing views remain, with no consensus reached on the correct approach or formulation.

Contextual Notes

Some limitations include potential misunderstandings of the definitions of relativistic mass and force, as well as unresolved mathematical steps in the derivations presented by participants.

DonnieD
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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?
 
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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:
someone can explain easy, where's the mistake?
 
DonnieD said:
someone can explain easy, where's the mistake?
Well, this is wrong for a start
DonnieD said:
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.
 
Last edited by a moderator:
DonnieD said:
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.
 
Last edited:
Just a light correction on lightarrow's post (and to make things a little easier to read);

[tex]\vec{F} = \gamma m_{0}\vec{a} + \gamma^{3}m_{0}\frac{\vec{v}\cdot\vec{a}}{c^2}\vec{v}[/tex]

Edit: Ahh lightarrow's on the ball today :wink:
 

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