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I'm having trouble with an exercise from Zee's book:

We observe an experimentalist moving by with 4-velocity $u^\mu$ and a particle zipping by with 4-momentum $p^\mu$. Show that the magnitude of the particle's 3-momentum as seen by the experimentalist is given by $$ \vert \vec{p}\vert = \sqrt{(p\cdot u)²+(p\cdot p)²}$$

I've used the formula $ w = \dfrac{u+v}{1-uv}$ and found $ p_{exp} = mw = \dfrac{mu + p}{1-uv}$, but its magnitude doesn't look like what I'm looking for.

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Take a step back. Don't jump right to calculating relative velocities. Instead, use vectors and the nice properties of vectors. The nice property of vectors that you're looking for is invariance of the dot product: no matter which frame you evaluate the dot product in, the answer will always be the same.

Now, try to think of how you would define the three-momentum $\vec{p}$, as seen by the experimentalist. It's just the part of the four-momentum $p$ that is not along the experimentalist's time direction. But the experimentalist's time direction is just $u$. I don't know about Zee, but I use $u \cdot u = -1$, so I would say that \begin{equation} \vec{p} = p + (p \cdot u) u. \end{equation} If you use $u \cdot u = +1$, just flip that plus to a minus. In any case, you can check that you've made the right choice by checking that the component of $\vec{p}$ along $u$ is zero: \begin{equation} \vec{p}\cdot u = p\cdot u + (p \cdot u) u\cdot u = p\cdot u - p\cdot u = 0. \end{equation} So now what do you get for $\vec{p} \cdot \vec{p}$? Take the square-root of that, and you've got your answer!

Now, hopefully you're convinced that this is a better approach to solving the problem. But you might still wonder what was wrong with your approach.

First of all, you should remember that the relativistic momentum isn't $m \vec{v}$; it's $m\, u$. That four-velocity brings in a factor of $\gamma$. But in your formulas, you'd have two different gammas because you have two different speeds — one for the speed of the particle relative to the experimentalist $\gamma_e$, and one for the speed of the particle relative to the frame in which you're doing the calculation $\gamma_c$. So you can see things starting to get ugly already.

Second, the problem (at least as you've presented it) didn't specify that the speeds were colinear. Nor did it specify that they were in opposite directions, rather than the same direction. But you assumed both of those things when you used the velocity-addition formula.

A good rule of thumb is that, if at all possible, you should try to use vectors (and preferably four-vectors) to do problems in physics. It may be possible to skip vectors, but it will always be possible to use them because they're more general than lots of the specialized results that you may remember — like the velocity-addition formula. It might also take a little more work to go with vectors, but it's good practice for when you have no choice.

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  • $\begingroup$ Wait, so $p$ and $u$ in the answer are 4-vectors? $\endgroup$
    – user174295
    Commented Dec 15, 2017 at 16:54
  • $\begingroup$ Yes, just as they are in $ \lvert \vec{p} \rvert = \sqrt{(p\cdot u)^2+(p\cdot p)^2}$. $\endgroup$
    – Mike
    Commented Dec 15, 2017 at 18:00
  • $\begingroup$ Oh, I thought they were supposed to be 3-vectors, since $\vec{p}$ is a 3-vector. But I'm still confused... you also said that $u$ is the experimentalist's time direction, why is it a 4-vector? $\endgroup$
    – user174295
    Commented Dec 15, 2017 at 18:30
  • $\begingroup$ I haven't read Zee, but I assume he discusses the four velocity. Usually, it's defined as the tangent vector to the object's worldline. For an object at rest, the components of four velocity are (1, 0, 0, 0). In particular, those are the components of $u$ in the experimentalist's frame. So if you want the part of a vector that the experimentalist would consider purely spatial, you just remove the part along $u$. If you still don't understand, you should probably google for more information — like this page. $\endgroup$
    – Mike
    Commented Dec 15, 2017 at 19:40
  • $\begingroup$ As for why $p$ and $u$ are four-vectors: In relativity (both special and general), four-vectors are the only things that make sense unless you've picked out a particular frame. In this particular case, you are picking out the experimentalist's frame. But usually, physics doesn't care what frame you're using, so you want to discuss things in terms of four-vectors. Thus, you should always assume things are four-vectors unless specifically told otherwise. An arrow over $\vec{p}$ is a typical way to specifically tell otherwise. But $p \cdot p$ should be interpreted as $p^\mu\, p_\mu$. $\endgroup$
    – Mike
    Commented Dec 15, 2017 at 19:48

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