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

by 1effect
Tags: 4vectors, energymomentum, particle, point
P: 321
 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.

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

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

Now I am very happy.
P: 3,967
 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.
P: 321
 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 :-)
P: 3,967
 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.
P: 321
 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.
P: 321
 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.
 P: 3,967 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: $$v'^2=v'_x^2+v'_y^2$$ $$v'_x=\frac{v_x+u}{1+v_xu/c^2}$$ $$v'_y=\frac{v_y \sqrt{1-u^2/c^2}}{1+v_xu/c^2}$$
P: 321
 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 $$1-(\frac{v'}{c})^2$$.
P: 3,967
 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 $$p_x ' = \gamma_u \left(p_x + \frac{u E}{c^2} \right)$$ P: 321  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. Sci Advisor PF Gold P: 1,847  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.
 Sci Advisor PF Gold P: 1,847 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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