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Consider the decay of a particle $X$ to two particles $c$ and $d$ in the rest frame of $X$. Using energy and momentum (4-vector) conservation, show that the energy of particle $c$ is given by: $$E_c=\frac{\left(m_X^2+m_c^2-m_d^2\right)c^2}{2m_X}\tag{A}$$ and similary for $E_d$.


I have specific questions regarding the solution (quoted below) to the above problem which outlines an important 'recipe' for solving general problems involving particle collisions using energy-momentum four-vectors:


Let $P_X$ be the 4-momentum of particle $X$, $E_X$ its energy, and $\bf p_X$ its 3-momentum vector — and similarly for particles $c$ and $d$. From energy and momentum conservation we can write: $$P_X=P_c+P_d$$ We are not so interested in particle $d$ for now, so we isolate it on the left side: $$P_d=P_X-P_c$$ Now we square both sides and replace the 4-vector norms by the invariant masses, which is valid in all reference frames (many problems in relativistic kinematics involve these steps): $$P_d^2=P_X^2+P_c^2-2P_X \cdot P_c$$ $$m_d^2c^2=m_X^2c^2+m_c^2c^2-2P_X\cdot P_c\tag{1}$$ (You can now see why we isolated $d$: so that its information would not get caught up in the dot product...). In the frame of $X$, which corresponds to the centre-of-mass frame, in this case, $\boldsymbol{p_X} = \boldsymbol{0}$ and $\boldsymbol{p_c} = −\boldsymbol{p_d}$; we can write the two 4-vectors we need: $$P_X=\left(\frac{E_X}{c},\,\bf p_X\right)=(\color{red}{m_Xc},0,0,0)$$ $$P_c=\left(\frac{E_c}{c},\,\bf p_c\right)=(E_c/c,{p_c}^x,0,0)$$ where we defined the x-axis along the motion of $c$ and $d$. The dot product is: $$P_X \cdot P_c=m_XE_c-0=m_XE_c$$ Replacing back in the equation for $m_d$ $(1)$ this gives: $$m_d^2c^2=m_X^2c^2+m_c^2c^2-2m_XE_c$$ and, as required, $$E_c=\frac{\left(m_X^2+m_c^2-m_d^2\right)c^2}{2m_X}$$ and similary for $E_d$, by swapping $c$ and $d$: $$E_d=\frac{\left(m_X^2+m_d^2-m_c^2\right)c^2}{2m_X}$$


That is the end of the proof. I have marked in red the part for which I don't understand. Why is there a $m_Xc$ in the first element of a four-vector which (I thought) should have dimensions of energy, not momentum?

This leads me to the other question I have, it was my understanding that general four-vectors are written as $$(E,p_xc,p_yc,p_zc)$$ I thought that the elements of four vectors must all have the same dimensions and that those dimensions are energy (as above).


Update:

In the answer given by @Shrey

In the solution, they've used convention A, but you would get the same answer if you used convention B instead - it's just that all your equations would be multiplied by $c^2$ now. I suggest that you check this directly!

So I will:

$$P_X=\left(E_X,\,\boldsymbol{p_X}c\right)=(m_Xc^2,0,0,0)$$ $$P_c=\left(E_c,\,\boldsymbol{p_c}c\right)=(E_c,{p_c}^xc,0,0)$$ So $$P_X\cdot P_c=E_cm_Xc^2-0=E_cm_Xc^2$$ substituting this result in $(1)$: $$m_d^2c^2=m_X^2c^2+m_c^2c^2-2E_cm_Xc^2$$ $$\implies E_c\stackrel{\color{red}{{?}}}{=}\frac{m_X^2+m_c^2-m_d^2}{2m_X}$$

Well, this is definitely not the same answer as $(\rm{A})$. So what am I missing?

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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Commenters, please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Jul 23 '20 at 0:34
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Why is there a $m_X c$ in the first element of a four-vector which (I thought) should have dimensions of energy, not momentum?

The momentum 4-vector of a particle of mass $m$ and three-velocity $\mathbf{u}$ can be defined as:

$$P = (\gamma m c, \gamma m \mathbf{u}) = \left(\frac{E}{c}, \mathbf{p}\right) \tag{A}$$ or as

$$ P = (\gamma m c^2, \gamma m c\mathbf{u}) = (E, \mathbf{p}c) \tag{B}$$

Note that convention B is just A multiplied by $c$; all components in A have dimensions of momentum, whereas all components in B have dimensions of energy.

Also, I've entered 3-vectors into these expressions as a shorthand to represent the 3 associated components, e.g. A should really be written:

$$P = (\gamma m c, \gamma m \mathbf{u}) = (\frac{E}{c}, p_x, p_y, p_z)$$

In the solution, they've used convention A, but you would get the same answer if you used convention B instead - it's just that all your equations would be multiplied by $c^2$ now. I suggest that you check this directly!

Instead, let's do a different check to see if we recover the $E^2 = |\mathbf{p}|^2 c^2 + m^2 c^4$ relation using the invariance of the momentum four-vector norm for both cases. We're going to work out the norm squared explicitly in both cases in the rest frame of a particle of mass $m$ (so $\gamma = 1$ and $\mathbf{u} = 0$) and then equate them to a general expression in terms of $E$ and $\mathbf{p}$.

Case A: $$m^2c^2 = \frac{E^2}{c^2} - |\mathbf{p}|^2 \tag{1}$$ Case B: $$m^2 c^4 = E^2 - |\mathbf{p}|^2 c^2 \tag{2}$$

We see that equation 2 is just equation 1 multiplied by $c^2$ and that both are correct. This makes sense as convention B has just scaled the four-vector in convention A by a factor of $c$ - any equation involving the same relations between these four-vectors should then appropriately be scaled overall by some power of $c$ (which has no effect overall). The solution produces an equation involving two-fold products of these four-vectors, so using convention B will scale the equations by $c^2$.

To answer the final part of this question, you have an $m_X c$ because $E_X = m_X c^2$ in the rest frame of X.

It was my understanding that general four-vectors are written as $(E, p_x c, p_y c, p_z c)$. I thought that elements of four vectors must all have the same dimensions and that those dimensions are energy (as above).

I believe the first part of this question has now been addressed. However, I would like to add that the momentum four-vector (whichever convention you choose) is not the only four-vector that exists.

Actually, $X = (ct, x, y,z)$ is probably the most basic four-vector; this transforms as $X' = \Lambda X$ under Lorentz transformations (where X' labels the transformed components of the same four-vector in the new inertial frame and $\Lambda$ is the Lorentz transformation matrix).

Now, any 4-component object $A$ that transforms like $A' = \Lambda A$ when $X$ transforms like $X' = \Lambda X$ is a 4-vector. The elements of $A$ should all have the same dimensions, but they don't have to be energy.


$$P_X=\left(E_X,\,\boldsymbol{p_X}c\right)=(m_Xc^2,0,0,0)$$ $$P_c=\left(E_c,\,\boldsymbol{p_c}c\right)=(E_c,{p_c}^xc,0,0)$$ So $$P_X\cdot P_c=E_cm_Xc^2-0=E_cm_Xc^2 \tag{C}$$ substituting this result in (1): $$m_d^2c^2=m_X^2c^2+m_c^2c^2-2E_cm_Xc^2 \tag{D}$$ $$\implies E_c\stackrel{\color{red}{{?}}}{=}\frac{m_X^2+m_c^2-m_d^2}{2m_X}$$ Well, this is definitely not the same answer as (A). So what am I missing?

Step C is correct, but step D is incorrect: under convention B, the norm squared of the momentum four-vector is also scaled by $c^2$. This is because in the rest frame of a particle of mass $m$, $P = (mc^2, \mathbf{0})$ so the norm-squared is now $m^2 c^4$ (unlike $m^2 c^2$ under convention A).

If you use this norm, you will recover the correct answer - let's check:

$$P_d^2=P_X^2+P_c^2-2P_X \cdot P_c \tag{E}$$ is a relation between four-vectors and holds under both conventions.

Using the corrected norms and your expression for $P_X \cdot P_c$ , we find:

$$m_d^2 c^4 = m_X^2 c^4 + m_c^2 c^4 - 2 E_c m_X c^2,$$ which is just

$$m_d^2c^2=m_X^2c^2+m_c^2c^2-2 E_c m_X$$ scaled by $c^2$ as expected.

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  • $\begingroup$ Thank you very much for taking the time to write a truly excellent answer. Please see my update in the main post above. $\endgroup$
    – Electra
    Jul 23 '20 at 15:00
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    $\begingroup$ @Electra You're welcome! Your attempt was very close, but didn't take into account that the norm under convention B is different. I have updated my answer to address this. $\endgroup$
    – Shrey
    Jul 23 '20 at 15:27
  • $\begingroup$ Sorry, I just realized that I forgot to accept this as the correct answer :-) $\endgroup$
    – Electra
    Aug 1 '20 at 23:27
  • $\begingroup$ In your answer you mention "three-velocity" and "three-vectors". Is this standard terminology? When I google 'define 3-velocity' all that comes up is velocity or 4-velocity, the same comes up for "three-vectors". Could you please show me link that defines these? $\endgroup$
    – Electra
    Sep 29 '20 at 20:37
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    $\begingroup$ @Electra By "three-vector", I just mean the spatial part associated with the 4-vector (as opposed to the temporal part). This terminology is fairly standard. For example, in David Tong's special relativity notes, page 136, he refers to the spatial part of the four-momentum as the three-momentum: damtp.cam.ac.uk/user/tong/relativity/dynrel.pdf. Please also see: physics.stackexchange.com/questions/330286/… $\endgroup$
    – Shrey
    Feb 13 at 19:55
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Conventionally, the four-momentum is defined as $$P=(E/c,p_x,p_y,p_z)$$ so that all components have units of momentum. If you want to get the energy of a particle (in units of energy), you need to multiply the zero component of the four-momentum by $c$. This is one example of the central role the speed of light plays in relativity. It allows us to relate energy and momentum. Before it was realized that the speed of light was such an important fundamental quantity, there really wasn't a way to combine energy and momentum into a single concept. In classical mechanics, dimensional analysis would disallow such a relationship.

Edit: $P$ is defined to use units of momentum because this allows for a natural generalization from classical momentum. In classical mechanics, $\mathbf{p}=m\mathbf{v}$. In relativity, we write $P=mu$, where m is mass and $u$ is the four-velocity of the particle. The four-velocity combines the notion of movement through space with the notion of movement through time. We say that a particle in its rest frame travels only through time (and not through space). Its four-velocity is defined as $u=(c,0,0,0)$. The reason the speed of light enters here is because it allows us to restrict the maximum allowed velocity of the particle to be the speed of light. Any boost preserves $u^2=c^2$. Therefore, in a boosted frame, where $u=(v_t,v_x,v_y,v_z)$, we will still have $u^2\equiv v_t^2-v_x^2-v_y^2-v_z^2=c^2$, which implies that the velocity of a particle in a boosted reference frame will never be measured as being faster than light. From this definition for $u$, we get that $P^2\equiv m^2u^2=m^2c^2$ in the rest frame, which also means that $P^2=m^2c^2$ in any reference frame.

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  • $\begingroup$ Thanks for your answer. You are saying I can use either? $\endgroup$
    – Electra
    Jul 23 '20 at 0:29
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    $\begingroup$ I guess you could, but you would have to insert a factor of $\frac{1}{c}$ into any equations involving $P$ (since they are all given using the convention above). The convention to use units of momentum is nice because it allows us to define $P$ as $mu$, where $m$ is mass and $u$ is the four-velocity. This leads naturally to $P=[mc,0,0,0]$ in the rest frame because a stationary particle is travelling through time at "velocity" $c$ ($u=[c,0,0,0]$). $\endgroup$
    – Yachsut
    Jul 23 '20 at 0:33

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