Energy-momentum for a point particle and 4-vectors

1effect

You seem to be missing the point. While you can surely align [tex]u[/tex] with the momentum of one particle, all the other momenta will not be necessarily aligned with [tex]u[/tex].

This is also incorrect. You can easily try it for 2 particles , take [tex]\vec{v_1}=-\vec{v_2}[/tex] and look at [tex]E_1E_2-\vec{p_1} \vec{p_2}[/tex]
You will quickly see that the expression is not invariant.

jtbell

The quantity of interest here is

[tex](m_0 c^2)^2 = E^2 - (pc)^2 = E^2 - (p_x c)^2 - (p_y c)^2 - (p_z c)^2[/tex]

A Lorentz boost along the x-direction leaves [itex]p_y[/itex] and [itex]p_z[/itex] invariant. Therefore it suffices in this case to show that [itex]E^2 - (p_x c)^2[/itex] is invariant.

yuiop

The point is that you split the momentum of each particle into x,y, and z components and then transform just the x component of each particle.

In practice you would sum all the x components and then transform that quantity. The final transformed momentum of the system is obtained from [tex] ||p '|| = \sqrt{ (\Sigma p_x ' )^2 + (\Sigma p_y ' )^2 + (\Sigma p_z ' )^2} [/tex]

1effect

yes, this would work

yuiop

When [tex]\vec{v_1}=-\vec{v_2}[/tex] the term [tex]\vec{p_1} \vec{p_2}[/tex] is zero which leaves just [tex]E^2[/tex]. It is this quantity, the energy squared of the sytem in the rest frame, that is invariant of the [tex]E_1E_2-\vec{p_1} \vec{p_2}[/tex] expression.

[tex]E_1E_2-\vec{p_1} \vec{p_2}[/tex] gives the "rest energy" of the system and not the rest mass. The only time this expression gives the rest mass is when all the particles are stationary in the rest frame.

1effect

I don't think this one is right :[tex]\vec{p_1} \vec{p_2}=-\gamma^2 m_1m_2v_1^2[/tex]

yuiop

Oops. Your right. I was talking about [tex](\Sigma E)^2-c^2||\Sigma \textbf{p}||^2[/tex]
where [tex]||\Sigma \textbf{p}||^2[/tex] goes to zero, because the momenta are summed before they are squared. [tex] [p_1 + (-p_1)]^2 = 0[/tex]

1effect

I don't think this is right either, we were taliking about a system of two particles , so [tex]\vec{p_1}+\vec{p_2}=\gamma (m_1-m_2) \vec{v_1}[/tex].

yuiop

If [itex] m_1 = m_2 [/itex] then [tex]\vec{p_1}+\vec{p_2}=\gamma (m_1-m_2) \vec{v_1} =0 [/tex]

Equal masses and opposite velocities is the simplest case and that was what I assumed we were talking about. Also remember that by definition the total momentum in the centre of mass rest frame is zero so if [itex] u_1 = -u_2 [/itex] then [itex]m_1[/itex] must equal [itex]m_2[/itex].

1effect

There was no specification of any "equal masses" anywhere. This is a system of arbitrary particles.

1effect

Yes, thank you, this would work.

yuiop

You did specify [tex]\vec{v_1}=-\vec{v_2}[/tex] so as I said in my last post that specifies equal masses by definition because the total momentum in the rest frame of the particle system is zero by definition.

yuiop

If you must do it the hard way then for two particles the invariant rest energy of the system [tex](\Sigma E)^2-||\Sigma \textbf{p}||^2[/tex] can be expressed as:


[tex] m_1+m_2+{2m_1m_2(1-u_1u_2) \over \sqrt{1-u_1^2} \sqrt{1-u_2^2}[/tex]

which reduces to

[tex] {4m_1^2 \over (1-u_1^2)[/tex]

when [itex]m_2=m_1[/itex] and [itex]v_2 = -v_1[/itex]

1effect

Why is so difficult for you to accept when you make a mistake?

yuiop

You have not pointed out where my "mistake" is.

P.S. No one else has said I have made a mistake ;)

1effect

I finally found the time to sit down and derive the momentum transformations.

[tex]\vec{p}=\gamma(v)*m_0 \vec{v}[/tex]

[tex]m_0[/tex] is the invariant mass.

In two dimensions:

[tex]p_x= \gamma(v)*m_0 v_x[/tex]

[tex]p_y= \gamma(v)*m_0 v_y[/tex]

In frame S' , moving with speed [tex]u[/tex] along the aligned x axes:

[tex]\vec{p'}=\gamma(v')*m_0 \vec{v'}[/tex]

[tex]p'_x= \gamma(v')*m_0 v'_x[/tex]

[tex]p'_y= \gamma(v')*m_0 v'_y[/tex]


where :

[tex]v'^2=v'_x^2+v'_y^2[/tex]

[tex]v'_x=\frac{v_x+u}{1+v_xu/c^2}[/tex]
[tex]v'_y=\frac{v_y \sqrt(1-u^2/c^2)}{1+v_xu/c^2}[/tex]

[tex]\gamma(v')=\gamma(v) \gamma(u) (1+v_xu/c^2)[/tex]

so:

[tex]\gamma(v')v'_x=\gamma(v) \gamma(u) (v_x+u)[/tex]

[tex]\gamma(v')v'_y=\gamma(v) v_y[/tex]

so, indeed:

[tex]p'_y=p_y[/tex]

[tex]p'_x=m_0 \gamma(v')v'_x=\gamma(v) \gamma(u) (m_0v_x+m_0u)=
\gamma(u)(p_x+\gamma(v)m_0u)=\gamma(u)(p_x+\frac {uE}{c^2})[/tex]

1effect

[tex]m_1[/tex] is different from [tex]m_2[/tex] :-)