Question regarding the general form of four-vectors in particle physics - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2022-01-25T23:14:57Z https://physics.stackexchange.com/feeds/question/567548 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/567548 0 Question regarding the general form of four-vectors in particle physics Electra https://physics.stackexchange.com/users/256509 2020-07-23T00:01:39Z 2020-09-29T23:04:53Z <blockquote> <p>Consider the decay of a particle <span class="math-container">$X$</span> to two particles <span class="math-container">$c$</span> and <span class="math-container">$d$</span> in the rest frame of <span class="math-container">$X$</span>. Using energy and momentum (4-vector) conservation, show that the energy of particle <span class="math-container">$c$</span> is given by: <span class="math-container">$$E_c=\frac{\left(m_X^2+m_c^2-m_d^2\right)c^2}{2m_X}\tag{A}$$</span> and similary for <span class="math-container">$E_d$</span>.</p> </blockquote> <hr /> <p>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:</p> <hr /> <blockquote> <p>Let <span class="math-container">$P_X$</span> be the 4-momentum of particle <span class="math-container">$X$</span>, <span class="math-container">$E_X$</span> its energy, and <span class="math-container">$\bf p_X$</span> its 3-momentum vector — and similarly for particles <span class="math-container">$c$</span> and <span class="math-container">$d$</span>. From energy and momentum conservation we can write: <span class="math-container">$$P_X=P_c+P_d$$</span> We are not so interested in particle <span class="math-container">$d$</span> for now, so we isolate it on the left side: <span class="math-container">$$P_d=P_X-P_c$$</span> 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): <span class="math-container">$$P_d^2=P_X^2+P_c^2-2P_X \cdot P_c$$</span> <span class="math-container">$$m_d^2c^2=m_X^2c^2+m_c^2c^2-2P_X\cdot P_c\tag{1}$$</span> (You can now see why we isolated <span class="math-container">$d$</span>: so that its information would not get caught up in the dot product...). In the frame of <span class="math-container">$X$</span>, which corresponds to the centre-of-mass frame, in this case, <span class="math-container">$\boldsymbol{p_X} = \boldsymbol{0}$</span> and <span class="math-container">$\boldsymbol{p_c} = −\boldsymbol{p_d}$</span>; we can write the two 4-vectors we need: <span class="math-container">$$P_X=\left(\frac{E_X}{c},\,\bf p_X\right)=(\color{red}{m_Xc},0,0,0)$$</span> <span class="math-container">$$P_c=\left(\frac{E_c}{c},\,\bf p_c\right)=(E_c/c,{p_c}^x,0,0)$$</span> where we defined the x-axis along the motion of <span class="math-container">$c$</span> and <span class="math-container">$d$</span>. The dot product is: <span class="math-container">$$P_X \cdot P_c=m_XE_c-0=m_XE_c$$</span> Replacing back in the equation for <span class="math-container">$m_d$</span> <span class="math-container">$(1)$</span> this gives: <span class="math-container">$$m_d^2c^2=m_X^2c^2+m_c^2c^2-2m_XE_c$$</span> and, as required, <span class="math-container">$$E_c=\frac{\left(m_X^2+m_c^2-m_d^2\right)c^2}{2m_X}$$</span> and similary for <span class="math-container">$E_d$</span>, by swapping <span class="math-container">$c$</span> and <span class="math-container">$d$</span>: <span class="math-container">$$E_d=\frac{\left(m_X^2+m_d^2-m_c^2\right)c^2}{2m_X}$$</span></p> </blockquote> <hr /> <p>That is the end of the proof. I have marked in red the part for which I don't understand. Why is there a <span class="math-container">$m_Xc$</span> in the first element of a four-vector which (I thought) should have dimensions of energy, not momentum?</p> <p>This leads me to the other question I have, it was my understanding that general four-vectors are written as <span class="math-container">$$(E,p_xc,p_yc,p_zc)$$</span> I thought that the elements of four vectors must all have the same dimensions and that those dimensions are energy (as above).</p> <hr /> <p><strong>Update:</strong></p> <p>In the answer given by @Shrey</p> <blockquote> <p>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 <span class="math-container">$c^2$</span> now. I suggest that you check this directly!</p> </blockquote> <p>So I will:</p> <p><span class="math-container">$$P_X=\left(E_X,\,\boldsymbol{p_X}c\right)=(m_Xc^2,0,0,0)$$</span> <span class="math-container">$$P_c=\left(E_c,\,\boldsymbol{p_c}c\right)=(E_c,{p_c}^xc,0,0)$$</span> So <span class="math-container">$$P_X\cdot P_c=E_cm_Xc^2-0=E_cm_Xc^2$$</span> substituting this result in <span class="math-container">$(1)$</span>: <span class="math-container">$$m_d^2c^2=m_X^2c^2+m_c^2c^2-2E_cm_Xc^2$$</span> <span class="math-container">$$\implies E_c\stackrel{\color{red}{{?}}}{=}\frac{m_X^2+m_c^2-m_d^2}{2m_X}$$</span></p> <p>Well, this is definitely <strong>not</strong> the same answer as <span class="math-container">$(\rm{A})$</span>. So what am I missing?</p> https://physics.stackexchange.com/questions/567548/-/567552#567552 1 Answer by Yachsut for Question regarding the general form of four-vectors in particle physics Yachsut https://physics.stackexchange.com/users/266156 2020-07-23T00:24:07Z 2020-07-23T01:38:19Z <p>Conventionally, the four-momentum is defined as <span class="math-container">$$P=(E/c,p_x,p_y,p_z)$$</span> 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 <span class="math-container">$c$</span>. 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.</p> <p><strong>Edit:</strong> <span class="math-container">$P$</span> is defined to use units of momentum because this allows for a natural generalization from classical momentum. In classical mechanics, <span class="math-container">$\mathbf{p}=m\mathbf{v}$</span>. In relativity, we write <span class="math-container">$P=mu$</span>, where m is mass and <span class="math-container">$u$</span> 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 <span class="math-container">$u=(c,0,0,0)$</span>. 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 <span class="math-container">$u^2=c^2$</span>. Therefore, in a boosted frame, where <span class="math-container">$u=(v_t,v_x,v_y,v_z)$</span>, we will still have <span class="math-container">$u^2\equiv v_t^2-v_x^2-v_y^2-v_z^2=c^2$</span>, 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 <span class="math-container">$u$</span>, we get that <span class="math-container">$P^2\equiv m^2u^2=m^2c^2$</span> in the rest frame, which also means that <span class="math-container">$P^2=m^2c^2$</span> in any reference frame.</p> https://physics.stackexchange.com/questions/567548/-/567557#567557 2 Answer by Shrey for Question regarding the general form of four-vectors in particle physics Shrey https://physics.stackexchange.com/users/193388 2020-07-23T01:12:35Z 2020-09-29T23:04:53Z <blockquote> <p>Why is there a <span class="math-container">$m_X c$</span> in the first element of a four-vector which (I thought) should have dimensions of energy, not momentum?</p> </blockquote> <p>The momentum 4-vector of a particle of mass <span class="math-container">$m$</span> and three-velocity <span class="math-container">$\mathbf{u}$</span> can be defined as:</p> <p><span class="math-container">$$P = (\gamma m c, \gamma m \mathbf{u}) = \left(\frac{E}{c}, \mathbf{p}\right) \tag{A}$$</span> or as</p> <p><span class="math-container">$$P = (\gamma m c^2, \gamma m c\mathbf{u}) = (E, \mathbf{p}c) \tag{B}$$</span></p> <p>Note that convention B is just A multiplied by <span class="math-container">$c$</span>; all components in A have dimensions of momentum, whereas all components in B have dimensions of energy.</p> <p>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> <p><span class="math-container">$$P = (\gamma m c, \gamma m \mathbf{u}) = (\frac{E}{c}, p_x, p_y, p_z)$$</span></p> <p>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 <span class="math-container">$c^2$</span> now. I suggest that you check this directly!</p> <p>Instead, let's do a different check to see if we recover the <span class="math-container">$E^2 = |\mathbf{p}|^2 c^2 + m^2 c^4$</span> 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 <span class="math-container">$m$</span> (so <span class="math-container">$\gamma = 1$</span> and <span class="math-container">$\mathbf{u} = 0$</span>) and then equate them to a general expression in terms of <span class="math-container">$E$</span> and <span class="math-container">$\mathbf{p}$</span>.</p> <p>Case A: <span class="math-container">$$m^2c^2 = \frac{E^2}{c^2} - |\mathbf{p}|^2 \tag{1}$$</span> Case B: <span class="math-container">$$m^2 c^4 = E^2 - |\mathbf{p}|^2 c^2 \tag{2}$$</span></p> <p>We see that equation 2 is just equation 1 multiplied by <span class="math-container">$c^2$</span> and that both are correct. This makes sense as convention B has just scaled the four-vector in convention A by a factor of <span class="math-container">$c$</span> - any equation involving the same relations between these four-vectors should then appropriately be scaled overall by some power of <span class="math-container">$c$</span> (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 <span class="math-container">$c^2$</span>.</p> <p>To answer the final part of this question, you have an <span class="math-container">$m_X c$</span> because <span class="math-container">$E_X = m_X c^2$</span> in the rest frame of X.</p> <blockquote> <p>It was my understanding that general four-vectors are written as <span class="math-container">$(E, p_x c, p_y c, p_z c)$</span>. I thought that elements of four vectors must all have the same dimensions and that those dimensions are energy (as above).</p> </blockquote> <p>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.</p> <p>Actually, <span class="math-container">$X = (ct, x, y,z)$</span> is probably the most basic four-vector; this transforms as <span class="math-container">$X' = \Lambda X$</span> under Lorentz transformations (where X' labels the transformed components of the same four-vector in the new inertial frame and <span class="math-container">$\Lambda$</span> is the Lorentz transformation matrix).</p> <p>Now, any 4-component object <span class="math-container">$A$</span> that transforms like <span class="math-container">$A' = \Lambda A$</span> when <span class="math-container">$X$</span> transforms like <span class="math-container">$X' = \Lambda X$</span> is a 4-vector. The elements of <span class="math-container">$A$</span> should all have the same dimensions, but they don't have to be energy.</p> <hr /> <blockquote> <p><span class="math-container">$$P_X=\left(E_X,\,\boldsymbol{p_X}c\right)=(m_Xc^2,0,0,0)$$</span> <span class="math-container">$$P_c=\left(E_c,\,\boldsymbol{p_c}c\right)=(E_c,{p_c}^xc,0,0)$$</span> So <span class="math-container">$$P_X\cdot P_c=E_cm_Xc^2-0=E_cm_Xc^2 \tag{C}$$</span> substituting this result in (1): <span class="math-container">$$m_d^2c^2=m_X^2c^2+m_c^2c^2-2E_cm_Xc^2 \tag{D}$$</span> <span class="math-container">$$\implies E_c\stackrel{\color{red}{{?}}}{=}\frac{m_X^2+m_c^2-m_d^2}{2m_X}$$</span> Well, this is definitely <strong>not</strong> the same answer as (A). So what am I missing?</p> </blockquote> <p>Step C is correct, but step D is incorrect: under convention B, the norm squared of the momentum four-vector is also scaled by <span class="math-container">$c^2$</span>. This is because in the rest frame of a particle of mass <span class="math-container">$m$</span>, <span class="math-container">$P = (mc^2, \mathbf{0})$</span> so the norm-squared is now <span class="math-container">$m^2 c^4$</span> (unlike <span class="math-container">$m^2 c^2$</span> under convention A).</p> <p>If you use this norm, you will recover the correct answer - let's check:</p> <p><span class="math-container">$$P_d^2=P_X^2+P_c^2-2P_X \cdot P_c \tag{E}$$</span> is a relation between four-vectors and holds under both conventions.</p> <p>Using the corrected norms and your expression for <span class="math-container">$P_X \cdot P_c$</span> , we find:</p> <p><span class="math-container">$$m_d^2 c^4 = m_X^2 c^4 + m_c^2 c^4 - 2 E_c m_X c^2,$$</span> which is just</p> <p><span class="math-container">$$m_d^2c^2=m_X^2c^2+m_c^2c^2-2 E_c m_X$$</span> scaled by <span class="math-container">$c^2$</span> as expected.</p>