Energy-momentum for a point particle and 4-vectors

[yuiop]

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:

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

$$v'_x=\frac{v_x+u}{1+v_xu/c^2}$$

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

 

[1effect]

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 $$1-(\frac{v'}{c})^2$$.

 

[yuiop]

Start with $$1-(\frac{v'}{c})^2$$.

By substituting $\sqrt{v^2-v_x^2}$ for [itex]v_y[/tex] in $$\frac{v_y \sqrt{1-u^2/c^2}}{1+v_xu/c^2}$$ <br /> <br /> 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.<br /> <br /> Dr Greg has <br /> <br /> $$p_x ' = \gamma_u \left(p_x - \frac{u E}{c^2} \right)$$<br /> <br /> while you appear to have<br /> <br /> $$p_x ' = \gamma_u \left(p_x + \frac{u E}{c^2} \right)$$[/itex]

 

[1effect]

By substituting $\sqrt{v^2-v_x^2}$ for [itex]v_y[/tex] in $$\frac{v_y \sqrt{1-u^2/c^2}}{1+v_xu/c^2}$$ <br /> <br /> I can get a similar answer to yours except for uncertainty about the signs due to imaginary roots.[/itex]

[itex] <br /> 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.<br /> <br /> <br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I notice your answer differs from Dr Greg's transformation in sign too.<br /> <br /> Dr Greg has <br /> <br /> $$p_x ' = \gamma_u \left(p_x - \frac{u E}{c^2} \right)$$<br /> <br /> while you appear to have<br /> <br /> $$p_x ' = \gamma_u \left(p_x + \frac{u E}{c^2} \right)$$ </div> </div> </blockquote><br /> This is due to:<br /> <br /> $$v'_x=\frac{v_x+u}{1+v_xu/c^2}$$<br /> <br /> If you use instead:<br /> <br /> $$v'_x=\frac{v_x-u}{1+v_xu/c^2}$$<br /> <br /> you will get dr.Greg's form.[/itex]

 

[DrGreg]

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, $\sum E$ 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.

 

[DrGreg]

Terms and conditions apply

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).