Energy-momentum for a point particle and 4-vectors

yuiop

In that case refer to post #47

yuiop

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?

1effect

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 [tex]E^2-...[/tex]

All I can tell you is that , your calculations for [tex]p'_x[/tex] seem incorrect.the ones for [tex]p'_y[/tex] are also incorrect but you got the desired result :-)

1effect

Yes, you were right all along about [tex] p_y' = p_y [/tex]
[tex] p_z' = p_z [/tex], I had to do the calculations myself, it wasn't obvious.
This leads to:

[tex]E'^2-(p'_xc)^2=E^2-(p_xc)^2[/tex]
and ultimately to:

[tex]E'^2-(\vec{p'}c)^2=E^2-(\vec{p}c)^2[/tex]


Now I am very happy.

yuiop

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.

1effect

Thank you.
Please check your formula for [tex]p'_x[/tex], it is wrong, so our results do not agree :-)

yuiop

[tex]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}} [/tex] can be rewritten as

--> [tex]p_x ' = \gamma(u_x)\gamma(v)m_o (v+u_x) = \gamma(v)p_x+\gamma(u_x)\gamma(v)m_o v [/tex]

--> [tex]p_x ' = \gamma(u_x)\gamma(v)m_o u_x+m_ov) = \gamma(v)(p_x+\gamma(u_x)m_o v) [/tex]

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.

1effect

I don't think so. In your case :

[tex]p'_x[/tex] contains [tex]\gamma(u_x)[/tex]

[tex]p'_y[/tex] contains [tex]\gamma(u_y)[/tex]

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

1effect

No. The above is also wrong, check your math. The physics part is also wrong-check your dimensions.

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:

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

from:

[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]

1effect

Start with [tex]1-(\frac{v'}{c})^2[/tex].

yuiop

By substituting [itex]\sqrt{v^2-v_x^2}[/itex] for [itex]v_y[/tex] in [tex]\frac{v_y \sqrt{1-u^2/c^2}}{1+v_xu/c^2}[/tex]

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

[tex] p_x ' = \gamma_u \left(p_x - \frac{u E}{c^2} \right) [/tex]

while you appear to have

[tex] p_x ' = \gamma_u \left(p_x + \frac{u E}{c^2} \right) [/tex]

1effect

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 [tex]1-(\frac{v'}{c})^2[/tex] doing a few simple algebraic substitutions.



This is due to:

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

If you use instead:

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

you will get dr.Greg's form.

DrGreg

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[/itex] is conserved in collisions. Yes, [itex]\sum \textbf{p}[/itex] is conserved. Therefore [itex](\Sigma E)^2-c^2\|\sum \textbf{p}\|^2[/itex] is conserved, in a closed system.

It is [itex]\sum E^2 - c^2 \sum \|\textbf{p}\|^2[/itex] (i.e. [itex]c^4 \sum m^2[/itex]) that is not conserved, as your example shows: it's 2 before and 0 after.

DrGreg

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 [itex]\sum E[/itex] term has to include potential energy as well as each particle's [itex]\sqrt{\|\textbf{p}c\|^2 + m^2 c^4}[/itex] 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 [itex]E^2 = \|\textbf{p}c\|^2 + m^2 c^4[/itex] still works for photons (with m = 0).