Speed of particle after decaying

[lriuui0x0]

Homework Statement

A particle $P$ of mass $M$ decays into a particle $R$ with mass $0 < m < M$ and a massless particle. Calculate the speed of $R$ in $P$'s rest frame

Relevant Equations

$m^2c^2 = E^2/c^2 - \mathbf{p}^2$

I have attempted a solution using conservation of momentum. Could people help check if this solution is correct (the result looks weird), as the problem doesn't have solution with it.

$$
\begin{aligned}
\begin{pmatrix}Mc \\ 0\end{pmatrix} &= \begin{pmatrix}E_R/c \\ \mathbf{p}_R\end{pmatrix} + \begin{pmatrix}E_\gamma/c \\ \mathbf{p}_\gamma \end{pmatrix} \\
\begin{pmatrix}Mc \\ 0\end{pmatrix} &= \begin{pmatrix}\sqrt{m^2c^2 + \mathbf{p}_R^2} \\ \mathbf{p}_R\end{pmatrix} + \begin{pmatrix}|\mathbf{p}_R| \\ -\mathbf{p}_R \end{pmatrix} \\
M^2c^2 &= m^2c^2 + 2(\sqrt{m^2c^2 + \mathbf{p}_R^2}|\mathbf{p}_R| + \mathbf{p}_R^2) \\
M^2c^2 &= m^2c^2 + 2(m\gamma v\sqrt{m^2c^2 + m^2\gamma^2v^2} + m^2\gamma^2v^2) \\
M^2c^2 &= m^2c^2 + 2(m^2\gamma cv\sqrt{1 + \gamma^2(1-\gamma^{-2})} + m^2\gamma^2v^2) \\
M^2c^2 &= m^2c^2 + 2(m^2\gamma^2cv + m^2\gamma^2v^2) \\
M^2c^2 &= m^2(c^2 + 2\gamma^2cv + \gamma^2v^2) \\
M^2c^2 &= m^2(c^2 + \frac{2cv}{1-\frac{v^2}{c^2}} + \frac{v^2}{1-\frac{v^2}{c^2}}) \\
M^2c^2 &= m^2\frac{c^4 - c^2v^2 + 2c^3v + c^2v^2}{c^2 - v^2} \\
M^2 &= m^2\frac{c^2 + 2cv}{c^2 - v^2} \\
M^2c^2 - M^2v^2 &= m^2c^2 + 2m^2cv \\
M^2v^2 + 2m^2cv + (m^2-M^2)c^2 &= 0 \\
v^2 + 2r^2cv +(r^2-1)c^2 &= 0 \\
v &= (\sqrt{r^4-r^2+1} - r^2)c \\
\end{aligned}
$$

where we have replaced $r = \frac{m}{M}$.

 

[PeroK]

I don't think that's right. I get quite a neat answer.

I would calculate the momentum and then get $v$ from that.

 

[lriuui0x0]

I don't think that's right. I get quite a neat answer.

I would calculate the momentum and then get $v$ from that.

Are you suggesting calculate the $|\mathbf{p}_R|$? May I ask what's the value of $v$ you got? Are you using the same process that I used?

 

[vela]

Homework Statement:: A particle $P$ of mass $M$ decays into a particle $R$ with mass $0 < m < M$ and a massless particle. Calculate the speed of $R$ in $P$'s rest frame
Relevant Equations:: $m^2c^2 = E^2/c^2 - \mathbf{p}^2$

I have attempted a solution using conservation of momentum. Could people help check if this solution is correct (the result looks weird), as the problem doesn't have solution with it.

$$
\begin{pmatrix}Mc \\ 0\end{pmatrix} = \begin{pmatrix}E_R/c \\ \mathbf{p}_R\end{pmatrix} + \begin{pmatrix}E_\gamma/c \\ \mathbf{p}_\gamma \end{pmatrix} \\
$$

I'd solve the problem in terms of $E_R$ and $p_R$. Don't express $E_R$ in terms of $m$ and $p_R$ yet. The algebra will be less messy that way.

Also, rewrite your equation slightly before squaring. Take advantage of the zero momentum on the lefthand side.
$$
\begin{pmatrix}Mc \\ 0\end{pmatrix} - \begin{pmatrix}E_R/c \\ \mathbf{p}_R\end{pmatrix} = \begin{pmatrix}E_\gamma/c \\ \mathbf{p}_\gamma \end{pmatrix} \\
$$

 

[PeroK]

Are you suggesting calculate the $|\mathbf{p}_R|$? May I ask what's the value of $v$ you got? Are you using the same process that I used?

Like @vela suggests.

 

[lriuui0x0]

Following the suggestion I got:$$
\begin{aligned}
\begin{pmatrix}Mc \\ 0\end{pmatrix} - \begin{pmatrix}E_R/c \\ \mathbf{p}_R\end{pmatrix} &= \begin{pmatrix}E_\gamma/c \\ \mathbf{p}_\gamma \end{pmatrix} \\

M^2c^2 + m^2c^2 - 2Mc\frac{E}{c} &= 0 \\
E &= \frac{(M^2+m^2)c^2}{2M} \\
\end{aligned}
$$

Then solve for $p^2$ we get:

$$
p^2 = \frac{(M^2+m^2)^2 c^2}{4M^2} - m^2c^2
$$

Is the above correct? Continuing on that I plan to solve:

$$
\frac{(M^2+m^2)^2 c^2}{4M^2} - m^2c^2 = m^2v^2\frac{1}{1-\frac{v^2}{c^2}}
$$

Then solve for $v$. The result doesn't look neat either, I got something like:

$$
v^2 = (1-\frac{4m^4}{M^4+2M^2m^2+m^4})c^2
$$

Is that the same as what you got?

 

[PeroK]

Following the suggestion I got:$$
\begin{aligned}
\begin{pmatrix}Mc \\ 0\end{pmatrix} - \begin{pmatrix}E_R/c \\ \mathbf{p}_R\end{pmatrix} &= \begin{pmatrix}E_\gamma/c \\ \mathbf{p}_\gamma \end{pmatrix} \\

M^2c^2 + m^2c^2 - 2Mc\frac{E}{c} &= 0 \\
E &= \frac{(M^2+m^2)c^2}{2M} \\
\end{aligned}
$$

Then solve for $p^2$ we get:

$$
p^2 = \frac{(M^2+m^2)^2 c^2}{4M^2} - m^2c^2
$$

Those are corrrect, but $p$ can be simplified. Also, note that $\frac v {c^2} = \frac p E$.

 

[vela]

Then solve for $v$. The result doesn't look neat either, I got something like:

$$
v^2 = (1-\frac{4m^4}{M^4+2M^2m^2+m^4})c^2
$$

Is that the same as what you got?

I think it's okay. Note that if you let $m=M$, you get $v=0$ as you should expect.

Note that the denominator is a perfect square. If it simplifies the way I think it should (I didn't bother working it out from your result), you should get a nice neat answer.

The way @PeroK suggested is the way I solved it. It should get you to the same answer with less algebra.

 

[PeroK]

$$
v^2 = (1-\frac{4m^4}{M^4+2M^2m^2+m^4})c^2
$$

Is that the same as what you got?

I don't think that's quite right. The $4m^4$ term in the numerator is the problem.

 

[vela]

Oh yeah, it should have been something like $4m^2M^2$.

 

[lriuui0x0]

Alright, I think I solved it. We can simplify $\mathbf{p}^2 = \frac{(M^2-m^2)^2c^2}{4M^2}$. Then:

$$
\begin{aligned}
\frac{(M^2-m^2)^2c^2}{4M^2} &= m^2\frac{1}{1-\frac{v^2}{c^2}}v^2 \\
\frac{(M^2-m^2)^2}{4M^2} &= \frac{m^2v^2}{c^2-v^2} \\
(M^2-m^2)^2c^2 - (M^2-m^2)^2v^2 &= 4M^2m^2v^2 \\
((M^2-m^2)^2+4M^2m^2)v^2 &= (M^2-m^2)^2c^2 \\
(M^2+m^2)^2v^2 &= (M^2-m^2)^2c^2 \\
v &= \frac{M^2-m^2}{M^2+m^2}c \\
\end{aligned}
$$

Is that the correct result?

 

[lriuui0x0]

Thanks for the helping!

 

[PeroK]

The way I did it, with $E, p = E_R, p_R$ and $c = 1$:

Momentum conservation gives $$p = p_{\gamma} = E_{\gamma}$$
Energy conservation gives: $$M = E + E_{\gamma} = E + p$$
This gives $$E^2 = M^2 + p^2 - 2Mp$$
Now, combining this with $E^2 = p^2 + m^2$ gives: $$p = \frac{M^2 - m^2}{2M}$$
And, we can also solve for $$E = M - p = \frac{M^2 + m^2}{2M}$$
Finally: $$v = \frac p E = \frac{M^2 - m^2}{M^2 + m^2}$$

 

[robphy]

Note that the energy-momentum diagram of this process looks like
the Doppler effect on a spacetime diagram ( a triangle with two future-pointing timelike segments from a common event, with their tips joined by a lightlike vector).
[A polygon in an energy-momentum diagram encodes the conservation of total 4-momentum.]

In analogy to $T'=kT$$ (in the Bondi calculus [my insight],
where $k$ is the Doppler factor [the k-factor]), we have $M=km$.
So, $k=\frac{M}{m}$.
Since the Doppler factor $k=\sqrt{\frac{1+v}{1-v}}$ [that is, $\exp\theta=\sqrt{\frac{1+\tanh\theta}{1-\tanh\theta}}$], we have $$v=\frac{k^2-1}{k^2+1}=\frac{\left(\frac{M}{m}\right)^2-1}{\left(\frac{M}{m}\right)^2+1}$$
in agreement with the other methods.