## Energy-momentum for a point particle and 4-vectors

 Quote by kev Dr Greg is right that the y and z components of momentum do not change under transformation if the frame is moving in the x direction and it is correct mathematically and convenient to align one axis of the momentum components with the motion. Doing it any other way is just making life hard for yourself. $$p_x ' = {m_o (v+u_x)\over \sqrt{1-u_x^2}\sqrt{1-v^2}} = { p_x\over \sqrt{1-v^2}}+{m_o v\over \sqrt{1-u_x^2}\sqrt{1-v^2}}$$ $$p_y ' = {m_o u_y\sqrt{1-v^2}\over \sqrt{1-u_y^2}\sqrt{1-v^2}} = p_y$$ $$p_z ' = {m_o u_z\sqrt{1-v^2}\over \sqrt{1-u_z^2}\sqrt{1-v^2}} = p_z$$ Where $$u_x, u_y, u_y$$ are the velocity components of the particles in the centre of mass rest frame S.
You seem to be missing the point. While you can surely align $$u$$ with the momentum of one particle, all the other momenta will not be necessarily aligned with $$u$$.

 It seems that it is fair to say that $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$ is invariant most of the time, but care is required in its use. .
This is also incorrect. You can easily try it for 2 particles , take $$\vec{v_1}=-\vec{v_2}$$ and look at $$E_1E_2-\vec{p_1} \vec{p_2}$$
You will quickly see that the expression is not invariant.

Mentor
 Quote by 1effect You can't use the above transformation in the case of a multiparticle system because you can't arrange for $$u$$ to align with more than one momentum (if they all have different directions).In other words, you need the generic transformation , the one that changes bot $$p_y$$ and $$p_z$$.
The quantity of interest here is

$$(m_0 c^2)^2 = E^2 - (pc)^2 = E^2 - (p_x c)^2 - (p_y c)^2 - (p_z c)^2$$

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

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 Quote by 1effect You seem to be missing the point. While you can surely align $$u$$ with the momentum of one particle, all the other momenta will not be necessarily aligned with $$u$$.
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 $$||p '|| = \sqrt{ (\Sigma p_x ' )^2 + (\Sigma p_y ' )^2 + (\Sigma p_z ' )^2}$$

 Quote by kev 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.
yes, this would work

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 Quote by 1effect This is also incorrect. You can easily try it for 2 particles , take $$\vec{v_1}=-\vec{v_2}$$ and look at $$E_1E_2-\vec{p_1} \vec{p_2}$$ You will quickly see that the expression is not invariant.

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

$$E_1E_2-\vec{p_1} \vec{p_2}$$ 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.

 Quote by kev When $$\vec{v_1}=-\vec{v_2}$$ the term $$\vec{p_1} \vec{p_2}$$ is zero

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

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 Quote by 1effect I don't think this one is right :$$\vec{p_1} \vec{p_2}=-\gamma^2 m_1m_2v_1^2$$
Oops. Your right. I was talking about $$(\Sigma E)^2-c^2||\Sigma \textbf{p}||^2$$
where $$||\Sigma \textbf{p}||^2$$ goes to zero, because the momenta are summed before they are squared. $$[p_1 + (-p_1)]^2 = 0$$

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

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

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 Quote by 1effect I don't think this is right either, we were taliking about a system of two particles , so $$\vec{p_1}+\vec{p_2}=\gamma (m_1-m_2) \vec{v_1}$$.
If $m_1 = m_2$ then $$\vec{p_1}+\vec{p_2}=\gamma (m_1-m_2) \vec{v_1} =0$$

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 $u_1 = -u_2$ then $m_1$ must equal $m_2$.

 Quote by kev If $m_1 = m_2$ then $$\vec{p_1}+\vec{p_2}=\gamma (m_1-m_2) \vec{v_1} =0$$ .
There was no specification of any "equal masses" anywhere. This is a system of arbitrary particles.

 Quote by jtbell The quantity of interest here is $$(m_0 c^2)^2 = E^2 - (pc)^2 = E^2 - (p_x c)^2 - (p_y c)^2 - (p_z c)^2$$ A Lorentz boost along the x-direction leaves $p_y$ and $p_z$ invariant. Therefore it suffices in this case to show that $E^2 - (p_x c)^2$ is invariant.
Yes, thank you, this would work.

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 Quote by 1effect ... This is also incorrect. You can easily try it for 2 particles , take $$\vec{v_1}=-\vec{v_2}$$ and look at $$E_1E_2-\vec{p_1} \vec{p_2}$$ You will quickly see that the expression is not invariant.
 Quote by kev ... Also remember that by definition the total momentum in the centre of mass rest frame is zero so if $u_1 = -u_2$ then $m_1$ must equal $m_2$.
 Quote by 1effect There was no specification of any "equal masses" anywhere. This is a system of arbitrary particles.
You did specify $$\vec{v_1}=-\vec{v_2}$$ 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.

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 Quote by 1effect There was no specification of any "equal masses" anywhere. This is a system of arbitrary particles.
If you must do it the hard way then for two particles the invariant rest energy of the system $$(\Sigma E)^2-||\Sigma \textbf{p}||^2$$ can be expressed as:

$$m_1+m_2+{2m_1m_2(1-u_1u_2) \over \sqrt{1-u_1^2} \sqrt{1-u_2^2}$$

which reduces to

$${4m_1^2 \over (1-u_1^2)$$

when $m_2=m_1$ and $v_2 = -v_1$

 Quote by kev You did specify $$\vec{v_1}=-\vec{v_2}$$ 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.

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

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 Quote by kev You did specify $$\vec{v_1}=-\vec{v_2}$$ 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.
 Quote by 1effect Why is so difficult for you to accept when you make a mistake?
You have not pointed out where my "mistake" is.

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

 Quote by kev Dr Greg is right that the y and z components of momentum do not change under transformation if the frame is moving in the x direction and it is correct mathematically and convenient to align one axis of the momentum components with the motion. Doing it any other way is just making life hard for yourself. $$p_x ' = {m_o (v+u_x)\over \sqrt{1-u_x^2}\sqrt{1-v^2}} = { p_x\over \sqrt{1-v^2}}+{m_o v\over \sqrt{1-u_x^2}\sqrt{1-v^2}}$$ $$p_y ' = {m_o u_y\sqrt{1-v^2}\over \sqrt{1-u_y^2}\sqrt{1-v^2}} = p_y$$ $$p_z ' = {m_o u_z\sqrt{1-v^2}\over \sqrt{1-u_z^2}\sqrt{1-v^2}} = p_z$$
I finally found the time to sit down and derive the momentum transformations.

$$\vec{p}=\gamma(v)*m_0 \vec{v}$$

$$m_0$$ is the invariant mass.

In two dimensions:

$$p_x= \gamma(v)*m_0 v_x$$

$$p_y= \gamma(v)*m_0 v_y$$

In frame S' , moving with speed $$u$$ along the aligned x axes:

$$\vec{p'}=\gamma(v')*m_0 \vec{v'}$$

$$p'_x= \gamma(v')*m_0 v'_x$$

$$p'_y= \gamma(v')*m_0 v'_y$$

where :

$$v'^2=v'_x^2+v'_y^2$$

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

$$\gamma(v')=\gamma(v) \gamma(u) (1+v_xu/c^2)$$

so:

$$\gamma(v')v'_x=\gamma(v) \gamma(u) (v_x+u)$$

$$\gamma(v')v'_y=\gamma(v) v_y$$

so, indeed:

$$p'_y=p_y$$

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

 Quote by kev You have not pointed out where my "mistake" is.
$$m_1$$ is different from $$m_2$$ :-)