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I'm having some awkward troubles with (relativistic) kinematics. Last problem I'm bumping my head against is the following:

a particle with mass $M$ and energy E is travelling along the z-axis. At a certain point it decays into two particles with mass $m = \frac M 3$ . Let $\theta$ be the angle between one of those particle's direction and the z-axis in the c.o.m. frame

Find:

a) $\theta$ value for which the two particles have the same velocity in the laboratory frame; evaluate the value of said velocity;

b) the angle between the two particles in the laboratory frame for a given $\theta$;

Starting to analyze the problem in the c.o.m. frame, and we have that:

conservation of momentum implies $p_1$=$-p_2$

Then, said particles will have the same energy as well, which is $e=\frac M 2$

I can evaluate $|p|$= $\sqrt{(E^2 -m^2c^2)}$

At this point I'm kinda in a dead end: I don't know how to link the conditions in the two different frames, let alone solving point B. I suppose that due to symmetry the velocities will be the same if $\theta$=$90°$, but if I were ask to write it rigorously I wouldn't be able to do it in the blink of an eye (if it's true, that is)

If I'm right then $p_1 = (e,0,p,0)$ and $p_2 = (e,0,-p,0)$ in the c.o.m. frame, and their values in the laboratory one is found after a Lorentz boost along the z-axis: $p'_1 = (e',0,p,-eβγ)$ and $p'_2 = (e',0,-p,-eβγ)$ So that's how I would "proceed", but my mind is still full of doubts and I can't write down a proper solution, anyone willing to make things a bit more clear? Thank you in advance.

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You're doing great, building up a physicist's intuition. The most basic question you can start with is always: so if this weren't a Lorentz boost, what would you do?

First off, you would know that you can break this down into (at most) a 2D problem: since $\vec p_1 = -\vec p_2$ for the particles in the COM frame, the problem is 1D in the COM frame, and then when you have a lab frame moving relative to the COM frame that is a separate direction and potentially a separate dimension. You've said that this motion is in the $z$-direction, that's fine, so let's define the $x$-direction as the direction perpendicular to $z$ such that $\vec p_1 = p_x ~\hat x + p_z ~\hat z.$ Booya, two-dimensional problem. You seem to have done this just fine!

Now after a Galilean transform, you'd have these two numbers, $(v_x, v_z)$ for the first particle and $(-v_x, -v_z)$ for the second, with some velocity $u~\hat z$ added on, leading to the magnitudes $\sqrt{v_x^2 + (v_z \pm u)^2}$, and you've been asked to make these two magintudes identical, for either $\pm$. And I think you'd agree with me that the most obvious way to do this is $v_z = 0$, and if you don't remember how to do that formally: square both sides, subtract $v_x^2$, square-root to find either $v_z + u = v_z - u$ and therefore $u=0$ or else $v_z + u = -(v_z - u)$ and therefore $v_z = 0$.

Now you should think, "I bet I want to also have $v_z = 0$ in the non-relativistic case." It comes out of a sort of symmetry, where we know that (if we forget about how $t$ gets mangled) the Lorentz transform is not going to touch $v_x$ on the x-axis but it is going to mingle $z, t$--so if we start with $z=0$ at least it will mingle those both in the same way, and we'll get the same velocity afterwards.

As for how you'd prove these, well, what's the general theorem you want to prove? Something like "If I Lorentz-boost (in the $z$-direction, by a nonzero velocity) two trajectories that go through the origin, and the particles involved both have the same speed before and after the boost, then they must have had the same $v_z$-component before and after the boost." You only need to prove the "after" part since the inverse of a Lorentz transform is a Lorentz transform; similarly that idea that the inverse of a LT is an LT leads me to this idea of "the same $v_z$ component," because we know if you start from $v_z = 0$ and boost you get the same $v_z$ component for both, so we can imagine this big family of vectors united by their parallel component under this LT.

So let's prove that. We've got these trajectories $\vec r_{1,2}(t) = \vec v_{1,2}~t$ in one frame, we boost to another frame and find out that they have the form $\vec r_{1,2}'(t) = \vec v_{1,2}'~t'$ in this other frame, and we know $|\vec v_1| = |\vec v_2|$ and $|\vec v_1'| = |\vec v_2'|.$

I think the most obvious thing to do to prove this is probably to use the fact that Lorentz transforms leave spacetime intervals unchanged. In other words consider two times $t_1, t_2$ which map to the same time $t'$ under the same Lorentz boosts of $(ct_1, \vec r_1(t_1))$ and $(c t_2, \vec r_2(t_2)):$ these both boost to $(c t', \vec r_{1,2}'(t'))$ for some $t'$. I'm going to prove that actually $t_1 = t_2.$ To see this, notice that the Lorentz transform $(t_1, \vec r_1) \mapsto (t', \vec r_1')$ must have guaranteed that $t_1^2 [c^2 - v_1^2] = (t')^2 [c^2 - (v_1')^2]$ and likewise $t_2^2 [c^2 - v_2^2] = (t')^2 [c^2 - (v_2')^2]$. Knowing that $|\vec v_1'| = |\vec v_2'|$ allows us to equate the two right-hand-sides and knowing that $|\vec v_1| = |\vec v_2| \ne c$ lets us divide out a common factor to find $t_1^2 = t_2^2$; but I was looking at forwards motion so $t_{1,2}$ are positive and hence $t_1 = t_2.$

From there it is quite straightforward to see that in the actual expression for the Lorentz transform we have $t' = \gamma_u~(t_1 + \beta_u (\hat z \cdot \vec v_1/c) t_1) = \gamma_u~(t_2 + \beta_u (\hat z \cdot \vec v_2/c) t_2),$ divide out the common $\gamma_u/c^2$ and the common $t_1=t_2$ and subtract $c^2$ from both sides to find $u ~ (\hat z \cdot \vec v_2) = u ~ (\hat z \cdot \vec v_1).$ They could only have used a common time $t_1 = t_2$ to map to a common time $t'$ under the same Lorentz boost by velocity $u~\hat z$ if either $u = 0$ or $t_1 = t_2 = 0$ or they had the same $z$-component to their velocities; anything else would have led to $t_1(t') \ne t_2(t').$

And of course that is sufficient to conclude that any other distribution of your two particles in space will fail; they have to have both $\hat z \cdot \vec v_1 = \hat z \cdot \vec v_2$ by the above theorem and yet they have to have $\hat z \cdot \vec v_1 = -\hat z \cdot \vec v_2$ by conservation of momentum, clearly these components can only be simultaneously equal and opposite is if they're zero.

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  • $\begingroup$ Thank you for you exhaustive reply! This proves $\theta$ has to be 90°, so now what's left is to evaluate the velocity: knowing that $p=mvγ$ i have $v = (\frac e'γ ,0, \frac p γ,-eβ)$ with $e'$ being $e$ after the boost $e'=γe$. I have all the infos (I got $|p|$ in the opening post) so all I have to do is to find the magnitude of $v$. $\endgroup$ – J. Doee Aug 31 '16 at 19:29
  • $\begingroup$ Now for the point B the generic p will have a z-component as well, thus $p_1 = (e,0,$p_y$,$p_z$)$ and I know the angle ( $\theta$ ) If I evaluate the boost again I'll know $p$ in the laboratory frame: once I know that I'll get the angle between the z-axis and the vectors and the sum the value to find the overall angle, I think? Does it make sense? $\endgroup$ – J. Doee Aug 31 '16 at 19:29
  • $\begingroup$ Two pieces of advice: (1) Be clear on angles; I would say $\theta$ has to be 180 degrees, but it depends whether you're measuring the angle with respect to the $z$-axis in the COM frame (which is 90 degrees) or the angle with respect to each other in the COM frame (180). Regardless, you're asked for the angle between the two, so twice the angle with $\hat z$, in the lab frame. $\endgroup$ – CR Drost Aug 31 '16 at 20:36
  • $\begingroup$ (2) I'd probably map these out into actual lines $\vec r(t)$ and Lorentz-boost those; you can treat the COM time $t$ as a parameter for parametric equations of lines for the purpose of determining the angle, so there's no huge "win" there, but the huge win comes in the second part where they ask you to work out the actual velocity in the lab frame... then it's awesome to have this expression for t in terms of t' in your back pocket! In this case it's probably just $t = t'/\gamma$ but in non-orthogonal circumstances it can be a lifesaver. $\endgroup$ – CR Drost Aug 31 '16 at 20:37

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