# Is My Solution for the Speed of a Decaying Particle Correct?

• lriuui0x0
In summary, the conversation is about solving a physics problem involving conservation of momentum and calculating the speed of a particle in the rest frame of another particle. The solution is attempted using a matrix equation and the suggestion is made to rewrite the equation to make the algebra easier. The process of solving the problem involves calculating the energy and momentum in terms of the masses and then solving for the speed using the equation ##v = \frac{p}{E}##. The resulting solution is correct, and simplifying the denominator yields a neat answer.
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}##.

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

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

PeroK said:
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?

lriuui0x0 said:
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
lriuui0x0 said:
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.

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?

lriuui0x0 said:
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
lriuui0x0 said:
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
lriuui0x0 said:
$$v^2 = (1-\frac{4m^4}{M^4+2M^2m^2+m^4})c^2$$

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.

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

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
Thanks for the helping!

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
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

## 1. What is the speed of a particle after it decays?

The speed of a particle after it decays depends on the type of particle and the conditions of the decay. Some particles may have a speed close to the speed of light, while others may have a much lower speed.

## 2. Does the speed of a particle change after it decays?

Yes, the speed of a particle can change after it decays. This is because the decay process involves the release of energy, which can change the speed of the particle.

## 3. How is the speed of a particle after decaying calculated?

The speed of a particle after decaying is calculated using the laws of conservation of energy and momentum. By knowing the initial and final masses and energies of the particle, the speed can be calculated using these equations.

## 4. Is the speed of a particle after decaying always the same?

No, the speed of a particle after decaying can vary depending on the specific decay process and the conditions involved. Some particles may have a consistent speed after decaying, while others may have a range of possible speeds.

## 5. How does the speed of a particle after decaying affect its behavior?

The speed of a particle after decaying can affect its behavior in various ways. For example, a particle with a high speed may travel further and interact with other particles differently than one with a lower speed. The speed can also determine the particle's lifetime and how it interacts with other particles in the surrounding environment.

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