Analyze the elastic collision of two equal-mass particles while working in a coordinate system where one of the two particles starts at rest. Assume both particles are in motion after the collision, what can be said about the angle between their trajectories? Let's analyze this in kinematics. It's an elastic collision so conservation of energy applies. We don't care about the identities of the outgoing particles. 1 + 2 → collision → 3 + 4 Without loss of generality we can assume 2 is at rest. So we have \(m_1 = m_2 = m_3 = m_4 = m \gt 0 \\ E_1 + E_2 = E_3 + E_4 \\ \vec{p}_1 + \vec{p}_2 = \vec{p}_3 + \vec{p}_4 \\ \vec{p}_2 = 0\) To work in numbers, lets decompose the vectors into components parallel and orthogonal to \(\vec{p}_1\). So we have: \(m_1 = m_2 = m_3 = m_4 = m \gt 0 \\ E_1 = E_3 + E_4 - E_2 \\ p_1 = p_{3\parallel} + \vec{p}_{4\parallel} \\ 0 = p_{3\perp} + \vec{p}_{4\perp}\) Squaring both sides, we get: \(E_1^2 = E_2^2 + E_3^2 + E_4^2 - 2E_2 E_3 - 2 E_2 E_4 + 2 E_3 E_4 \\ p_1^2 = p_{3\parallel}^2 + \vec{p}_{4\parallel}^2 + 2 p_{3\parallel} \vec{p}_{4\parallel} \\ 0 = p_{3\perp}^2 + \vec{p}_{4\perp}^2 + 2 p_{3\perp} \vec{p}_{4\perp} \) Adding the last two, we get: \(\vec{p}_1^2 = p_1^2 = p_{3\parallel}^2 + p_{3\perp}^2 + p_{4\parallel}^2 + p_{4\perp}^2 + 2 p_{3\parallel} \vec{p}_{4\parallel} + 2 p_{3\perp} \vec{p}_{4\perp} = \vec{p}_3^2 + \vec{p}_4^2 + 2 \vec{p}_3 \cdot \vec{p}_4\) or \(\vec{p}_1^2 = \vec{p}_3^2 + \vec{p}_4^2 + 2 \left| \vec{p}_3 \right| \left| \vec{p}_4 \right| \cos \theta\)
In Special Relativity, \(E^2 = \left( mc^2 \right)^2 + \left( \vec{p} c \right)^2\) So we can go about eliminating \(E_1^2\) and \(\vec{p}_1^2\) from the equations by multiplying the latter by \(-c^2\) and summing. \( E_1^2 = E_2^2 + E_3^2 + E_4^2 - 2E_2 E_3 - 2 E_2 E_4 + 2 E_3 E_4 \\ - c^2 \vec{p}_1^2 = - c^2 \vec{p}_3^2 - c^2 \vec{p}_4^2 - 2 c^2 \left| \vec{p}_3 \right| \left| \vec{p}_4 \right| \cos \theta \\ E_1^2 - c^2 \vec{p}_1^2 + 2 c^2 \left| \vec{p}_3 \right| \left| \vec{p}_4 \right| \cos \theta = E_2^2 + E_3^2 - c^2 \vec{p}_3^2 + E_4^2 - c^2 \vec{p}_4^2 - 2E_2 E_3 - 2 E_2 E_4 + 2 E_3 E_4 \\ \left( m_1 c^2 \right)^2 + 2 c^2 \left| \vec{p}_3 \right| \left| \vec{p}_4 \right| \cos \theta = \left( m_2 c^2 \right)^2 + \left( m_3 c^2 \right)^2 + \left( m_4 c^2 \right)^2 - 2m_2 c^2 E_3 - 2 m_2 c^2 E_4 + 2 E_3 E_4 \\ 2 c^2 \left| \vec{p}_3 \right| \left| \vec{p}_4 \right| \cos \theta = 2 m^2c^4 - 2m c^2 E_3 - 2 m c^2 E_4 + 2 E_3 E_4 \) This factors: \(\cos \theta = \frac{E_3 - mc^2}{ \left| \vec{p}_3 \right| c } \frac{E_4 - mc^2}{ \left| \vec{p}_4 \right| c }\) Now if \(E^2 = \left( mc^2 \right)^2 + \left( \vec{p} c \right)^2\) and \(E \vec{v} = c^2 \vec{p}\) and \(m > 0\) hold, then we have \(\vec{p} = \frac{m \vec{v}}{\sqrt{1 - \frac{\vec{v}^2}{c^2}}} , \quad E = \frac{m c^2}{\sqrt{1 - \frac{\vec{v}^2}{c^2}}}\). So \(\frac{E - mc^2}{ \left| \vec{p} \right| } = \frac{ \frac{1}{ \sqrt{1 - \frac{\vec{v}^2}{c^2} } } - 1}{ \frac{ \left| \vec{v} \right| }{c \sqrt{1 - \frac{\vec{v}^2}{c^2} } } } = \frac{c - \sqrt{c^2 - \vec{v}^2}}{\left| \vec{v} \right|} c = c \tanh \left( \frac{1}{2} \tanh^{-1} \frac{\left| \vec{v} \right|}{c} \right) \) a speed with important properties in the theory of special relativity. Thus \(\cos \theta = \frac{c - \sqrt{c^2 - \vec{v}_3^2}}{\left| \vec{v}_3 \right|} \frac{c - \sqrt{c^2 - \vec{v}_4^2}}{\left| \vec{v}_4 \right|}\) Finally in the limit of low velocities, \(\lim_{\vec{v}_3, \vec{v}_4 \to 0} \cos \theta = \frac{ \left| \vec{v}_3 \right| \left| \vec{v}_4 \right| }{4 c^2}\)
In Newtonian kinematics we don't have a concept of non-zero rest energy, so \(E_2 = 0\). Since \(E_k = \frac{1}{2} m v^2 = \frac{\vec{p}^2}{2m}\) we start with: \(m_1 = m_2 = m_3 = m_4 = m \gt 0 \\ E_1 = E_3 + E_4 - E_2 = E_3 + E_4 \\ \vec{p}_1^2 = \vec{p}_3^2 + \vec{p}_4^2 + 2 \vec{p}_3 \cdot \vec{p}_4\) Multiplying the last by \(\frac{1}{2m}\) we get \(\frac{\vec{p}_1^2}{2m} = \frac{\vec{p}_3^2}{2m} + \frac{\vec{p}_4^2}{2m} + \frac{\vec{p}_3 \cdot \vec{p}_4}{m}\) And subtracting it from the energy equation we have: \(\vec{p}_3 \cdot \vec{p}_4 = 0\). Thus in Newtonian kinematics the trajectories of the departing particles are always at right angles. This is also the low velocity limit for special relativity.
We can achieve the same results without considering explicitly energy or momentum, by working initially in the center-of-mass coordinate system and then transforming to the frame where one of the initial particles is stationary. Because we are working in the center-of-mass coordinate system, for both the incoming and outgoing trajectories, the velocities are equal and opposite. In addition, since it is elastic, all the magnitudes are the same. \( \left| \vec{v}_1 \right| = \left| \vec{v}_2 \right| = \left| \vec{v}_3 \right| = \left| \vec{v}_4 \right| = u, \quad \vec{v}_1 = - \vec{v}_2 , \quad \vec{v}_3 = - \vec{v}_4\) Working in a coordinate system, we can extract components parallel and orthogonal to \(\vec{v}_1 \). \( \begin{pmatrix} \vec{v}_{1\parallel} \\ \vec{v}_{1\perp} \end{pmatrix} = \begin{pmatrix} u \\ 0 \end{pmatrix}, \quad \begin{pmatrix} \vec{v}_{2\parallel} \\ \vec{v}_{2\perp} \end{pmatrix} = \begin{pmatrix} -u \\ 0 \end{pmatrix}, \quad \begin{pmatrix} \vec{v}_{3\parallel} \\ \vec{v}_{3\perp} \end{pmatrix} = \begin{pmatrix} u \cos \phi \\ u \sin \phi \end{pmatrix}, \quad \begin{pmatrix} \vec{v}_{4\parallel} \\ \vec{v}_{4\perp} \end{pmatrix} = \begin{pmatrix} -u \cos \phi \\ -u \sin \phi \end{pmatrix}\) The theory of transformation of velocities comes directly from the linear theory of transformation of space-time intervals for two events along the trajectory described by that velocity. \( \begin{pmatrix} \Delta t' \\ \vec{v}'_{\parallel} \Delta t' \\ \vec{v}'_{\perp} \Delta t' \end{pmatrix} = \begin{pmatrix} \Delta t' \\ \Delta \vec{x}'_{\parallel} \\ \Delta \vec{x}'_{\perp} \end{pmatrix} = { \Huge \Lambda(u) } \begin{pmatrix} \Delta t \\ \Delta \vec{x}_{\parallel} \\ \Delta \vec{x}_{\perp} \end{pmatrix} = { \Huge \Lambda(u) } \begin{pmatrix} \Delta t \\ \vec{v}_{\parallel} \Delta t \\ \vec{v}_{\perp} \Delta t \end{pmatrix} \) In Special Relativity, \( { \Huge \Lambda(u) } = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} & \frac{u}{c^2 \sqrt{1 - \frac{u^2}{c^2}}} & 0 \\ \frac{u}{\sqrt{1 - \frac{u^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{u^2}{c^2}}} & 0 \\ 0 & 0 & 1 \end{pmatrix} \) So \( \Delta t' = \frac{ \Delta t + \frac{u}{c^2} \Delta \vec{x}_{\parallel}}{\sqrt{1 - \frac{u^2}{c^2}}} = \frac{ 1 + \frac{u \vec{v}_{\parallel}}{c^2} }{\sqrt{1 - \frac{u^2}{c^2}}} \Delta t \\ \Delta \vec{x}'_{\parallel} = \frac{ \Delta \vec{x}_{\parallel} + u \Delta t}{\sqrt{1 - \frac{u^2}{c^2}}} = \frac{ \vec{v}_{\parallel} + u}{\sqrt{1 - \frac{u^2}{c^2}}} \Delta t \\ \Delta \vec{x}'_{\perp} = \Delta \vec{x}_{\perp} = \vec{v}_{\perp} \Delta t \\ \Delta \vec{v}'_{\parallel} = \frac{ \Delta \vec{x}'_{\parallel} }{\Delta t' } = \frac{ \quad \frac{ \vec{v}_{\parallel} + u}{\sqrt{1 - \frac{u^2}{c^2}}} \quad }{ \quad \frac{ 1 + \frac{u \vec{v}_{\parallel}}{c^2} }{\sqrt{1 - \frac{u^2}{c^2}}} \quad } = \frac{\vec{v}_{\parallel} + u }{1 + \frac{u \vec{v}_{\parallel}}{c^2} } \\ \Delta \vec{v}'_{\perp} = \frac{ \Delta \vec{x}'_{\perp} }{\Delta t' } =\frac{ \quad \vec{v}_{\perp} \quad }{ \quad \frac{ 1 + \frac{u \vec{v}_{\parallel}}{c^2} }{\sqrt{1 - \frac{u^2}{c^2}}} \quad } = \frac{\sqrt{1 - \frac{u^2}{c^2}}}{ 1 + \frac{u \vec{v}_{\parallel}}{c^2} } \vec{v}_{\perp} \) Thus: \( \begin{pmatrix} \vec{v}'_{1\parallel} \\ \vec{v}'_{1\perp} \end{pmatrix} = \begin{pmatrix} \frac{2 u}{1 + \frac{u^2}{c^2}} \\ 0 \end{pmatrix}, \quad \begin{pmatrix} \vec{v}'_{2\parallel} \\ \vec{v}'_{2\perp} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \quad \begin{pmatrix} \vec{v}'_{3\parallel} \\ \vec{v}'_{3\perp} \end{pmatrix} = \begin{pmatrix} \frac{( 1 + \cos \phi) u}{1 + \frac{u^2}{c^2} \cos \phi } \\ \frac{\sqrt{1 - \frac{u^2}{c^2}}}{ 1 + \frac{u^2}{c^2} \cos \phi } u \sin \phi \end{pmatrix}, \quad \begin{pmatrix} \vec{v}'_{4\parallel} \\ \vec{v}'_{4\perp} \end{pmatrix} = \begin{pmatrix} \frac{( 1 - \cos \phi) u}{1 - \frac{u^2}{c^2} \cos \phi } \\ - \frac{\sqrt{1 - \frac{u^2}{c^2}}}{ 1 - \frac{u^2}{c^2} \cos \phi } u \sin \phi\end{pmatrix}\) And: \( \vec{v}'_3^2 = \frac{( 1 + \cos \phi)^2 + (1 - \frac{u^2}{c^2})^2 \sin^2 \phi }{ \left( 1 + \frac{u^2}{c^2} \cos \phi \right)^2 } u^2 = \frac{ 2 c^4 \cos \phi +2 c^4 + u^4 \sin^2 - 2 c^2 u^2 \sin^2 \phi }{(c^2+u^2 \cos \phi )^2} u^2 \\ \vec{v}'_4^2 = \frac{( 1 - \cos \phi)^2 + (1 - \frac{u^2}{c^2})^2 \sin^2 \phi }{ \left( 1 - \frac{u^2}{c^2} \cos \phi \right)^2 } u^2 = \frac{ -2c^4 \cos \phi + 2 c^4 + u^4 \sin^2 \phi - 2c^2 u^2 \sin^2 \phi }{ \left( 1 - \frac{u^2}{c^2} \cos \phi \right)^2 } u^2 \\ \sqrt{ \vec{v}'_3^2 \vec{v}'_4^2 } = \frac{ \sqrt{ \left( 4 c^4 (c^2 - u^2)^2 + u^4 (2 c^2 - u^2)^2 \sin^2 \phi \right) \sin^2 \phi } }{ c^4 - u^4 \cos^2 \phi } u^2 \\ \vec{v}'_3 \cdot \vec{v}'_4}{u^2} = \frac{( 1 + \cos \phi)( 1 - \cos \phi) - ( 1 - \frac{u^2}{c^2} ) \sin^2 \phi }{\left( 1 + \frac{u^2}{c^2} \cos \phi \right)\left( 1 - \frac{u^2}{c^2} \cos \phi \right)} u^2 = \frac{ u^2 c^2 \sin^2 \phi }{ c^4 - u^4 \cos^2 \phi } u^2 \\ \cos \theta = \frac{ u^2 c^2 \sin^2 \phi }{ \sqrt{ \left( 4 c^4 (c^2 - u^2)^2 + u^4 (2 c^2 - u^2)^2 \sin^2 \phi \right) \sin^2 \phi } } \) The result, although parametrized in a vastly different way than above has the same properties in the limit of low velocity, for fixed choice of \(\phi\). \(\lim_{u \to 0} \cos \theta = 0\) In Galilean Relativity, \( { \Huge \Lambda(u) } = \begin{pmatrix} 1 & 0 & 0 \\ u & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \) So \( \Delta t' = \Delta t \\ \Delta \vec{x}'_{\parallel} = \Delta \vec{x}_{\parallel} + u \Delta t = ( \vec{v}_{\parallel} + u ) \Delta t \\ \Delta \vec{x}'_{\perp} = \Delta \vec{x}_{\perp} = \vec{v}_{\perp} \Delta t \\ \Delta \vec{v}'_{\parallel} = \frac{ \Delta \vec{x}'_{\parallel} }{\Delta t' } = \frac{ \quad ( \vec{v}_{\parallel} + u ) \Delta t \quad }{ \quad \Delta t \quad } = \vec{v}_{\parallel} + u \\ \Delta \vec{v}'_{\perp} = \frac{ \Delta \vec{x}'_{\perp} }{\Delta t' } =\frac{ \quad \vec{v}_{\perp} \Delta t \quad }{ \quad \Delta t \quad } = \vec{v}_{\perp} \) Thus: \( \begin{pmatrix} \vec{v}'_{1\parallel} \\ \vec{v}'_{1\perp} \end{pmatrix} = \begin{pmatrix} 2 u \\ 0 \end{pmatrix}, \quad \begin{pmatrix} \vec{v}'_{2\parallel} \\ \vec{v}'_{2\perp} \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \quad \begin{pmatrix} \vec{v}'_{3\parallel} \\ \vec{v}'_{3\perp} \end{pmatrix} = \begin{pmatrix} ( 1 + \cos \phi) u \\ u \sin \phi \end{pmatrix}, \quad \begin{pmatrix} \vec{v}'_{4\parallel} \\ \vec{v}'_{4\perp} \end{pmatrix} = \begin{pmatrix} ( 1 - \cos \phi) u \\ - u \sin \phi\end{pmatrix}\) \( \vec{v}'_3^2 = 2 u^2 ( 1 + \cos \phi ) \) \( \vec{v}'_4^2 = 2 u^2 ( 1 - \cos \phi ) \) \( \sqrt{ \vec{v}'_3^2 \vec{v}'_4^2 } = 2 u^2 \left| \sin \phi \right| \) \( \vec{v}'_3 \cdot \vec{v}'_4 = \left( 1 - \cos^2 \phi - \sin^2 \phi \right) u^2 = 0 \) \( \cos \theta = 0 \)
This doesn't look right, you should be getting: \(2 c^2 p_3 p_4 cos \theta =(m_2c^2)^2+ (m_3c^2)^2+(m_4c^2)^2-(m_1c^2)^2- 2m_2 c^2 E_3 - 2 m_2 c^2 E_4 + 2 E_3 E_4\) How did you arrive to the above? Never mind, you are dealing with the trivial case \(m_1=m_2=m_3=m_4=m\)
That is indeed an intermediate step. \( 2 c^2 \left| \vec{p}_3 \right| \left| \vec{p}_4 \right| \cos \theta = 3 (m c^2)^2 - (mc^2)^2 - 2 mc^2 E_3 - 2 mc^2 E_4 + 2 E_3 + E_4 \\ \quad = 2 m^2c^4 - 2m c^2 E_3 - 2 m c^2 E_4 + 2 E_3 E_4 = 2 ( E_3 - mc^2) (E_4 - mc^2) \\ \cos \theta = \frac{ \left( E_3 - mc^2 \right) \left(E_4 - mc^2 \right) }{ \left( \left| \vec{p}_3 \right| c \right) \left( \left| \vec{p}_4 \right| c \right)} \) Well, trivial it may be, but the goal was to get the low-velocity limit where Newtonian mechanics gives \(\cos \theta = 0 = \cos 90^{\circ}\), a result that makes one want to go out and buy an airtable. That the relativistic result gives a smaller angle between outgoing trajectories might be testable in a cloud chamber or collider. http://rspa.royalsocietypublishing.org/content/136/830/630 http://journals.aps.org/pr/abstract/10.1103/PhysRev.159.1138
....while the relativistic calculation produces trivially \(cos \theta= \frac{\gamma_3-1}{\gamma_3} \frac{\gamma_4-1}{\gamma_4} \beta_3\beta_4\). In order to see the difference, one needs to collide particles with a high \(\beta\) AFTER collision. Intuition says that this would require a very high speed BEFORE collision.
In science, you don't get credit for nitpicking the work of others; it's not sufficient to criticize, or to make sterile claims that something "isn't good enough," you have to do it better. Your repeated use of "trivial" doesn't make sense when you don't follow the algebra in post #2, use an editable written media like a forum post to correct your own misreading, and now mistranslate a quantity calculated in post #2 both in terms of energy, mass and momentum and velocity. Or even \(\cos \theta= \frac{\gamma_3-1}{\beta_3 \gamma_3} \frac{\gamma_4-1}{\beta_4 \gamma_4} \) This follows from \(\frac{E}{mc^2} = \gamma, \; \frac{p}{mc^2} = \beta \gamma\) and the calculations in post #2. But thank you for playing.
Where was I trying to take credit for your work? Yes, I misstyped, \(\cos \theta= \frac{\gamma_3-1}{\beta_3 \gamma_3} \frac{\gamma_4-1}{\beta_4 \gamma_4}=\frac{1-\sqrt{1-\beta_3^2}}{\beta_3} \frac{1-\sqrt{1-\beta_4^2}}{\beta_4} \). When \(\beta \) is very small \(cos \theta\) can be approximated by \(\frac{\beta_3 \beta_4}{4} \) , so, even in the Newtonian regime \(cos \theta \ne 0\). It is still a trivial freshman exercise, do you have a broom up your ass? Why are you so prickly?
I was simply puzzled how he got his result. Then I realized that he's dealing with the trivial case when the rest mass of the particles are identical.
It is interesting to note that you could have had a more elegant solution for the general case. Indeed, consider \(m_1=m_3\) and \(m_2=m_4\) with \(m_1 \ne m_2\). You would be getting the answer to the general case, when the rest masses of the two colliding particles aren't equal: \(\cos \theta=\frac{(E_3-m_1c^2)(E_4-m_2c^2)}{p_3p_4c^2}=\frac{(E_3-m_3c^2)(E_4-m_4c^2)}{p_3p_4c^2}=\frac{1-\sqrt{1-\beta_3^2}}{\beta_3} \frac{1-\sqrt{1-\beta_4^2}}{\beta_4}\)
I don't see that anyone accused you of taking credit, for anything! The comment was, you don't get credit for citisizing... You should also be aware that this is an open public forum and the math you characterize as, a trivial freshman exercise, is anything but trivial to many if not most of those participating or just viewing discussions on this forum. While rpenner's initial response may have seemed a little prickly, the broom it seems to me, may be up a different ass!
What is "citisizing"? Well, you could use this thread as an incentive for taking an introductory class to relativity, no?
Hmmm, got much of a chip on your shoulder? From your first post you have had quite and attitude. So far I have not seen that you have the knowledge to go with your arrogance. We shall see.
I simply asked a question about his derivation (and I figured the answer before his confirmation). I also pointed out that this is a simple exercise in introductory relativity. How is this arrogant? Looking at your posts, for a change, I see no contribution to the subject being discussed.
