Talk:Special Relativity/Dynamics

Thanks mystery contributor (why not get a userid?). The update was good and accurate but we need to sort out the ever multiplying symbols! RobinH 20:07, 19 May 2006 (UTC)


 * I have approached this by simply redefining $$m_B$$ as $$m_B (u_B)$$ etc. This also avoids the slightly problematical reversion to rest mass and relativistic mass in the sentence where $$m_B (u_B) = m_B (0) = m_B $$. RobinH 11:02, 1 June 2006 (UTC)

We have a serious problem of chickens and eggs here! I have moved the discussion of nomenclature to the end of the section on momentum because the student will be unable to understand it until they have explored relativistic momentum. RobinH 11:25, 1 June 2006 (UTC)

Very nice job, RobinH!. I corrected a very minor error. Moriconne 06:06, 9 February 2007 (UTC)

Help...
Is there something wrong with the math? I can figure out what things meant up until $$u_{yB} = \frac{u_{yR}}{1 - u_{xR}v/c^2}$$ Then the equations I use give me results which doesn't match the statements... Can someone be more specific and tell me which formula to use... Cause I've been studying this page for ages comparing to the previous pages...Or is there really something wrong? The velocities in the y direction are related by the following equation when the observer is travelling at the same velocity as the blue ball ie: when $$u_{xB} = 0$$\,:why?

Tikai 14:55, 20 November 2007 (UTC)

Merger of Modern_Physics/Acceleration,_Force_and_Mass
I have merged this text into Special_Relativity/Relativistic_dynamics.

Error in Momentum
The momentum change by the red ball is:

$$2\mathbf{u_{yR}}m $$

The momentum change by the blue ball is:

$$2 \mathbf{u_{yB}}m $$

The situation is symmetrical so the Newtonian conservation of momentum law is demonstrated:

$$2 m\mathbf{u_{yR}} = 2m\mathbf{u_{yB}}$$

The momentum section has an error: the momentum change for the blue ball has the wrong sign and should be negative, since it receives a push downwards. The subsequent line should explain that Newton's conservation of momentum leads to equal but opposite momentum changes. Will make immediate correction. AUN4 (discuss • contribs) 14:39, 16 January 2012 (UTC)

Momentum: expression of v in terms of $$u_{xR}'$$
I think a short explanation of $$u_{xR}^' = v$$ based on the existing text is required (choice of reference frame). This is how I interpret it;

$$u_{xB}$$ can be expressed as follows (see Special Relativity/Simultaneity, time dilation and length contraction);

$$u_{xB}^' = \frac{u_{xB} - v}{1 - u_{xB}v/c^2}$$

Frame 2 has been defined such that $$u_{xB}$$ is 0, therefore $$u_{xB}^' = -v$$

Likewise, frame 1 has been defined such that $$u_{xR}^' = -u_{xB}^'$$

Therefore $$u_{xR}^' = v$$;

$$u_{xR}^' = \frac{u_{xR} - v}{1 - u_{xR}v/c^2} = v$$

Richardbrucebaxter (discuss • contribs) 12:25, 13 April 2015 (UTC)