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

 Quote by 1effect $$m_1$$ is different from $$m_2$$ :-)
In that case refer to post #47

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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 do not seem to accept my definition of the rest frame of a system of particles. If we have a cloud of particles all with random masses and velocities, how would you define the rest frame of that system?

If you were watching this cloud of particles and the cloud as a whole was drifting away from you (ignore any expansion) then would you still claim to be in the rest frame of the cloud of particles?

If the total momentum of the cloud is not zero by your measurements, do you accept that the cloud as a whole will drift away from you?

 Quote by kev You do not seem to accept my definition of the rest frame of a system of particles. If we have a cloud of particles all with random masses and velocities, how would you define the rest frame of that system?
Why would you care about defining a "rest frame"? The problem is asking you to derive the transformation for the momentum and energy, in any arbitrary frame because you need this for computing the norm $$E^2-...$$

 If you were watching this cloud of particles and the cloud as a whole was drifting away from you (ignore any expansion) then would you still claim to be in the rest frame of the cloud of particles? If the total momentum of the cloud is not zero by your measurements, do you accept that the cloud as a whole will drift away from you?
All I can tell you is that , your calculations for $$p'_x$$ seem incorrect.the ones for $$p'_y$$ are also incorrect but you got the desired result :-)

 Quote by DrGreg Note that the reason why it makes sense to consider $$\Sigma E$$ and $$\Sigma \textbf{p}$$ is because of conservation of energy and momentum. We want to replace our multi-particle system with a single particle which behaves like the system, as far as possible. For each particle, the four dimensional vector $$\left(E, \textbf{p}c \right)$$ is a "4-vector", which means that it obeys the Lorentz transform $$E' = \gamma_u \left(E - u p_x \right)$$ $$p_x' = \gamma_u \left(p_x - \frac{u E}{c^2} \right)$$ $$p_y' = p_y$$ $$p_z' = p_z$$
Yes, you were right all along about $$p_y' = p_y$$
$$p_z' = p_z$$, I had to do the calculations myself, it wasn't obvious.
This leads to:

$$E'^2-(p'_xc)^2=E^2-(p_xc)^2$$
and ultimately to:

$$E'^2-(\vec{p'}c)^2=E^2-(\vec{p}c)^2$$

Now I am very happy.

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 Quote by 1effect so, indeed: $$p'_y=p_y$$ $$p'_x=m_0 \gamma(v')v'_x=\gamma(v) \gamma(u) (m_0v_x+m_0u)= \gamma(u)(p_x+\gamma(v)m_0u)$$
Your final result agrees with mine and you did a better job of showing how you got there :)

Like me, you are cursed with having to prove everything to yourself from first principles.

 Quote by kev Your final result agrees with mine and you did a better job of showing you got there :)
Thank you.
Please check your formula for $$p'_x$$, it is wrong, so our results do not agree :-)

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 Quote by 1effect In frame S' , moving with speed $$u$$ along the aligned x axes: . . . . $$p'_y=p_y$$ $$p'_x=m_0 \gamma(v')v'_x=\gamma(v) \gamma(u) (m_0v_x+m_0u)= \gamma(u)(p_x+\gamma(v)m_0u)$$
 Quote by 1effect Thank you. Please check your formula for $$p'_x$$, it is wrong, so our results do not agree :-)
$$p_x ' = {m_o (v+u_x)\over \sqrt{1-u_x^2}\sqrt{1-v^2}} = { p_x\over \sqrt{1-v^2}}+{m_o v\over \sqrt{1-u_x^2}\sqrt{1-v^2}}$$ can be rewritten as

--> $$p_x ' = \gamma(u_x)\gamma(v)m_o (v+u_x) = \gamma(v)p_x+\gamma(u_x)\gamma(v)m_o v$$

--> $$p_x ' = \gamma(u_x)\gamma(v)m_o u_x+m_ov) = \gamma(v)(p_x+\gamma(u_x)m_o v)$$

You have used v for the particle velocities in the rest frame and u for velocity of frame S relative to S' while I have done the opposite. That is why the end results look different. Swap the u's and v's and they are the same.

 Quote by kev $$p_x ' = {m_o (v+u_x)\over \sqrt{1-u_x^2}\sqrt{1-v^2}} = { p_x\over \sqrt{1-v^2}}+{m_o v\over \sqrt{1-u_x^2}\sqrt{1-v^2}}$$ can be rewritten as --> $$p_x ' = \gamma(u_x)\gamma(v)m_o (v+u_x) = \gamma(v)p_x+\gamma(u_x)\gamma(v)m_o v$$ --> $$p_x ' = \gamma(u_x)\gamma(v)m_o u_x+m_ov) = \gamma(v)(p_x+\gamma(u_x)m_o v)$$ You have used v for the particle velocities in the rest frame and u for velocity of frame S relative to S' while I have done the opposite. That is why the end results look different. Swap the u's and v's and they are the same.
I don't think so. In your case :

$$p'_x$$ contains $$\gamma(u_x)$$

 $$p_y ' = {m_o u_y\sqrt{1-v^2}\over \sqrt{1-u_y^2}\sqrt{1-v^2}}$$
$$p'_y$$ contains $$\gamma(u_y)$$

This can't be right, you shouldn't have any vector components in the $$\gamma$$ expression, you should only have the vector norm.

 Quote by kev $$m_1+m_2+{2m_1m_2(1-u_1u_2) \over \sqrt{1-u_1^2} \sqrt{1-u_2^2}$$ which reduces to $${4m_1^2 \over (1-u_1^2)$$ when $m_2=m_1$ and $v_2 = -v_1$
No. The above is also wrong, check your math. The physics part is also wrong-check your dimensions.
 Blog Entries: 6 You are right, that I am wrong. You see, I can accept when I have made a mistake :P I would be interested to see how you got to: $$\gamma(v')=\gamma(v) \gamma(u) (1+v_xu/c^2)$$ from: $$v'^2=v'_x^2+v'_y^2$$ $$v'_x=\frac{v_x+u}{1+v_xu/c^2}$$ $$v'_y=\frac{v_y \sqrt{1-u^2/c^2}}{1+v_xu/c^2}$$

 Quote by kev You are right, that I am wrong. You see, I can accept when I have made a mistake :P I would be interested to see how you got to: $$\gamma(v')=\gamma(v) \gamma(u) (1+v_xu/c^2)$$ from: $$v'^2=v'_x^2+v'_y^2$$ $$v'_x=\frac{v_x+u}{1+v_xu/c^2}$$ $$v'_y=\frac{v_y \sqrt{1-u^2/c^2}}{1+v_xu/c^2}$$
Start with $$1-(\frac{v'}{c})^2$$.

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 Quote by 1effect Start with $$1-(\frac{v'}{c})^2$$.
By substituting $\sqrt{v^2-v_x^2}$ for $v_y[/tex] in $$\frac{v_y \sqrt{1-u^2/c^2}}{1+v_xu/c^2}$$ I can get a similar answer to yours except for uncertainty about the signs due to imaginary roots. I notice your answer differs from Dr Greg's transformation in sign too. Dr Greg has $$p_x ' = \gamma_u \left(p_x - \frac{u E}{c^2} \right)$$ while you appear to have $$p_x ' = \gamma_u \left(p_x + \frac{u E}{c^2} \right)$$  Quote by kev By substituting [itex]\sqrt{v^2-v_x^2}$ for $v_y[/tex] in $$\frac{v_y \sqrt{1-u^2/c^2}}{1+v_xu/c^2}$$ I can get a similar answer to yours except for uncertainty about the signs due to imaginary roots. I have no idea what you are talking about, there is no equation, so there are no roots , imaginary or otherwise. You simply need to evaluate the expression $$1-(\frac{v'}{c})^2$$ doing a few simple algebraic substitutions.  I notice your answer differs from Dr Greg's transformation in sign too. Dr Greg has $$p_x ' = \gamma_u \left(p_x - \frac{u E}{c^2} \right)$$ while you appear to have $$p_x ' = \gamma_u \left(p_x + \frac{u E}{c^2} \right)$$ This is due to: $$v'_x=\frac{v_x+u}{1+v_xu/c^2}$$ If you use instead: $$v'_x=\frac{v_x-u}{1+v_xu/c^2}$$ you will get dr.Greg's form. Recognitions: Gold Member Science Advisor  Quote by DaleSpam #29 Are you sure about that? That goes against what I understood about the conservation of the 4-momentum (of course I am not a physicist). I understood that the sum of the 4-momenta was conserved even across particle decay. IIRC, they had example problems that ran something like this: An electron at rest is anhilated by a positron moving at .6 c in the x direction. One of the two resulting photons travels only along the y axis. Find the 4-momenta of the two photons using units where c=1 and mass of an electron = 1. So the 4 momenta are as follows: electron (1, 0, 0, 0) -> mass = 1 positron (1.25, 0.75, 0, 0) -> mass = 1 system (2.25, 0.75, 0, 0) -> mass = sqrt(4.5) photon1 (e1, 0, y1, 0, 0) photon2 (e2, x2, y2, z2) system (e1+e2, x2, y1+y2, z2) So you have 6 unknowns, you get four from the conservation of 4-momentum for the system system before = system after, you get one from the norm of photon1 = 0 and one from the norm of photon2 = 0. photon1 (1, 0, 1, 0) -> mass = 0 photon2 (1.25, .75, -1, 0) -> mass = 0 system (2.25, 0.75, 0, 0) -> mass = sqrt(4.5) I think you've misunderstood my words or symbols, because your example illustrates beautifully the point I was trying to make. Yes, [itex]\sum E$ is conserved in collisions. Yes, $\sum \textbf{p}$ is conserved. Therefore $(\Sigma E)^2-c^2\|\sum \textbf{p}\|^2$ is conserved, in a closed system.

It is $\sum E^2 - c^2 \sum \|\textbf{p}\|^2$ (i.e. $c^4 \sum m^2$) that is not conserved, as your example shows: it's 2 before and 0 after.
 Recognitions: Gold Member Science Advisor After a bit of extra reading over the weekend, I should point out some extra terms and conditions. It's already been pointed out that the technique we've been using applies only to closed systems, i.e. where there's no interaction with the outside world. Note that hitting an enclosing wall counts as interaction and thus invalidates this approach unless you include the walls as part of your system of particles. It also only applies (in the form expressed so far) where each particle moves freely (inertially) between collisions. The only interaction allowed is at events (single points in spacetime); interaction over a distance (e.g. between charged particles) is forbidden. However, interaction over a distance can be included in the system by means of a potential. This means that the $\sum E$ term has to include potential energy as well as each particle's $\sqrt{\|\textbf{p}c\|^2 + m^2 c^4}$ energy. Unfortunately, that's the point where my knowledge of S.R. falls over but I believe it works. Finally, note that photons count as particles just as much as massive particles, and $E^2 = \|\textbf{p}c\|^2 + m^2 c^4$ still works for photons (with m = 0).
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