Energy-momentum for a point particle and 4-vectors

[jtbell]

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.

 

[yuiop]

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 $$||p '|| = \sqrt{ (\Sigma p_x ' )^2 + (\Sigma p_y ' )^2 + (\Sigma p_z ' )^2}$$

 

[1effect]

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

 

[yuiop]

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 system 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.

 

[1effect]

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 :$$\vec{p_1} \vec{p_2}=-\gamma^2 m_1m_2v_1^2$$

 

[yuiop]

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. $$[p_1 + (-p_1)]^2 = 0$$

 

[1effect]

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 $$\vec{p_1}+\vec{p_2}=\gamma (m_1-m_2) \vec{v_1}$$.

 

[yuiop]

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 $m_2$.

 

[1effect]

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.

 

[1effect]

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.

 

[yuiop]

...
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.

...
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 $m_2$.

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.

 

[yuiop]

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 $v_2 = -v_1$

 

[1effect]

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?

 

[yuiop]

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?

You have not pointed out where my "mistake" is.

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

 

[1effect]

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:

$$p'_y=p_y$$

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

 

[1effect]

You have not pointed out where my "mistake" is.

$$m_1$$ is different from $$m_2$$ :-)

 

[yuiop]

$$m_1$$ is different from $$m_2$$ :-)

In that case refer to post #47

 

[yuiop]

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?

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]

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 :-)

 

[1effect]

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.

 

[yuiop]

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.

 

[1effect]

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 :-)

 

[yuiop]

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

$$p'_y=p_y$$

$$p'_x=m_0 \gamma(v')v'_x=\gamma(v) \gamma(u) (m_0v_x+m_0u)= \gamma(u)(p_x+\gamma(v)m_0u)$$

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.

 

[1effect]

$$p_x ' = {m_o (v+u_x)\over \sqrt{1-u_x^2}\sqrt{1-v^2}} = { p_x\over \sqrt{1-v^2}}+{m_o v\over \sqrt{1-u_x^2}\sqrt{1-v^2}}$$ 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.

 

[1effect]

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

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.

 

[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 $$\frac{v_y \sqrt{1-u^2/c^2}}{1+v_xu/c^2}$$

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

$$p_x ' = \gamma_u \left(p_x + \frac{u E}{c^2} \right)$$

 

[1effect]

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

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.

 

[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).