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}##.

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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?

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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
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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?

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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
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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.

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

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.

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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?

PeroK
lriuui0x0
Thanks for the helping!

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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: $$v = \frac p E = \frac{M^2 - m^2}{M^2 + m^2}$$

TSny
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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.

Last edited:
PeroK