Particle Disintegration and Momentum Conservation in Relativity

[etotheipi]

I'm making a mistake somewhere, I hoped someone could point it out? Say a particle of mass $2m$, initially at rest, disintegrates into two particles of mass $m$ that move at $v_{1x} = v$ and $v_{2x} = -v$. Then we switch to a frame moving at $-U$ along the $x$ axis, so that$$v'_{1x} = \frac{U+v}{1+Uv}$$ $$v'_{2x} = \frac{U-v}{1-Uv}$$The total momentum in S' is$$\begin{align*}p_x' = \frac{\frac{U+v}{1+Uv}}{\sqrt{1-\left( \frac{U+v}{1+Uv} \right)^2}}m + \frac{\frac{U-v}{1-Uv}}{\sqrt{1-\left( \frac{U-v}{1-Uv} \right)^2}}m &= \frac{2mU}{\sqrt{(1-v^2) -U^2(1-v^2)}} \\ &= \frac{2mU}{\sqrt{(1-v^2)(1-U^2)}} \\ &= \frac{1}{\sqrt{1-v^2}} [2m \gamma_U U]\end{align*}$$Why does this not equal $2m \gamma_U U$, but only approaches it in the limit $v \rightarrow 0$? The initial momentum in S' would surely be $2m \gamma_U U$, wouldn't it - surely the 3-momentum would be conserved? Please let me know if I've done something dumb. Thanks!

 

[PeroK]

What happens if you impose conservation of energy?

Perhaps is energy is not conserved in one frame, then more generally energy and momentum are not conserved in other frames?

In what way is this related to length contraction and time dilation?

 

[Ibix]

To save you a bit of algebra, if you have a particle of momentum $p=\gamma mv$ in the $x$ direction and energy $E=\gamma m$ (note that I'm using $c=1$) then its four momentum is $P=(E,p,0,0)$. The Lorentz transforms can be written as a matrix:$$\Lambda=\left(\begin{array}{cccc}
\gamma_U&-\gamma_UU&0&0\\
-\gamma_UU&\gamma_U&0&0\\
0&0&1&0\\
0&0&0&1\end{array}\right)$$You can then calculate the four momentum in the transformed frame as $\Lambda P^T$ using usual matrix multiplication rules. Obviously the y and z components are trivial in this case.

You can easily confirm that you get the same result as you did above, and it saves mucking around with the velocity transforms.

 

[etotheipi]

What happens if you impose conservation of energy?

In S, problems, because...$$2m = 2\gamma_v m \implies \gamma_v = 1$$i.e. it's only valid if $v_{1x} = v_{2x} = 0$?

Perhaps is energy is not conserved in one frame, then more generally energy and momentum are not conserved in other frames?

So if energy is not conserved in the CM frame, it won't be conserved in other frames?

In what way is this related to length contraction and time dilation?

This is not obvious to me at the moment.

 

[Ibix]

So if energy is not conserved in the CM frame, it won't be conserved in other frames?

Hint: where does nuclear energy come from?

 

[etotheipi]

Hint: where does nuclear energy come from?

So we require that the sum of the final masses are slightly less than the original mass? That is to say that the rest energy of the system (invariant mass) is conserved, but the masses not so much?

 

[PeroK]

In S, problems, because...$$2m = 2\gamma_v m \implies \gamma_v = 1$$i.e. it's only valid if $v_{1x} = v_{2x} = 0$?So if energy is not conserved in the CM frame, it won't be conserved in other frames?

To impose conservation of energy you need the final rest masses to be less than $m$.

If energy is not conserved in the CoM frame then it won't be conserved in any other frame. But, energy and momentum are unified (like space and time) into energy-momentum. What is "purely" energy in one frame is a combination of energy and momentum in another frame.

 

[Ibix]

So we require that the sum of the final masses are slightly less than the original mass? That is to say that the rest energy of the system (invariant mass) is conserved, but the masses not so much?

Yes. The four momentum of the original mass was (2M,0,0,0). You can write down the four momentum of the daughter particles (say they have mass M'), you can add four vectors just like any other, and the total four momentum is conserved. Then transform to your moving frame and you should get consistent answers.

 

[etotheipi]

Edit, said something silly. I'll plug the numbers in first and check it works . Thanks!

 

[PeroK]

Okay, that makes sense. I'll try plugging the numbers in and see if it works.

So generally, it is the 4-mentum (street slang) that is conserved, but not any of its components? That seems right, but I did see on Wikipedia and it said that the three space momentum $\mathbf{p}$ is conserved, as well as the total energy $E^0$, i.e. the components of the 4-mentum.

Am I misinterpreting that?

All four components of energy-momentum are conserved in a closed system. You do not have a closed system here as extra energy came from somewhere external.

The loss of conservation of energy in the CoM frame, however, leads to the loss of conservation of energy and conservation of three-momentum in other frames. The analogy with spacetime is apt. What can be described purely in terms of the time (energy) component in one frame is described in terms of both time and space (energy and momentum) components in another frame.

 

[PeroK]

PS If you look at the transformation laws for energy-momentum this becomes clear immediately.

 

[DrGreg]

@ etotheipi , if each particle after disintegration has mass $m$, the only thing wrong with your original post is the assumption that the particle before disintegration has mass $2m$. You can use conservation of energy (with $U=0$) to work out what the original mass should be, and then I think you'll find the rest of your calculation is correct.

 

[etotheipi]

All four components of energy-momentum are conserved in a closed system. You do not have a closed system here as extra energy came from somewhere external.

The loss of conservation of energy in the CoM frame, however, leads to the loss of conservation of energy and conservation of three-momentum in other frames. The analogy with spacetime is apt. What can be described purely in terms of the time (energy) component in one frame is described in terms of both time and space (energy and momentum) components in another frame.

Thanks, that's a nice explanation. Yeah I dropped a negative sign from the matrix $\Lambda$ that @Ibix wrote down in #3 and spent 5 minutes panicking about what I'd done wrong () but it does work when I re-insert that lost minus sign back in. So if in doubt, the 4-mentum is always conserved (for a closed system). Thanks for the help everyone!

 

[robphy]

One can draw an energy-momentum diagram of the particle-disintegration process.
Note that this diagram is essentially the spacetime-diagram of the clock-effect/twin-paradox.

From my https://www.physicsforums.com/insights/spacetime-diagrams-light-clocks/

Reinterpret as follows:
Let $\overrightarrow {OZ}$ be the initial particle's 4-momentum. (In the diagram, this has magnitude 10 [unit-mass diamonds].)
Let the identical-mass daughter-particles have 4-momenta $\overrightarrow {OQ}$ and $\overrightarrow {QZ}$ (say, with opposite velocities $\pm(3/5)c$ in the initial-particle frame).
so that
$\overrightarrow {OZ}=\overrightarrow {OQ}+\overrightarrow {QZ}$ (conservation of total 4-momentum).
So, in special relativity, this implies that
$\left|\overrightarrow {OZ}\right| \geq \left|\overrightarrow {OQ}\right|+\left|\overrightarrow {QZ}\right|$ (reverse-triangle inequality)
from the Law of [Hyperbolic] Cosines [in simpler notation]:
$OZ^2=(OQ^2+QZ^2+2\ OQ\ QZ\ \cosh\theta_{OQ,QZ}) \ \geq \ ( OQ^2+QZ^2+2\ OQ\ QZ) = (|OQ|+|QZ|)^2$

From the diagram, the daughter-masses are 4 and 4 (which satisfies $10 > 4 + 4$).
You can check that
$10^2= 4^2+ 4^2+ 2(4)(4)\cosh( {\rm arctanh}(3/5)-{\rm arctanh}(-3/5))$
( https://www.wolframalpha.com/input/?i=4^2++4^2++2(4)(4)cosh(+arctanh(3/5)+-+arctanh(-3/5))
)