Very stupid question, but since the mass of a hand is very large compared to a proton, one should be able using a solid slap to accelerate a proton to ultra-relativistic speeds. If I consider a relativistic scattering where the proton is at rest (and later treated relativistically), while the hand has Energy of the hand $E_H=M_H+P_H^2/(2M_H)\gg m_p$ one gets from energy conservation $E_H+m_p=E_p+M_H+P_H'^2/(2M_H)$ and momentum conservation $P_H=p_p+P_H'$ in 1D. Solving this using Mathematica one gets $E_p=\sqrt{m_p^2+p_p^2}\gg m_p$. Since the proton is also electrically charged it should not be able to fly through the hand. So why are we not slapping protons and then grooming the obtained beam using it for experiments?

  • $\begingroup$ I'm skeptical that $\sqrt{m_p^2+p_p^2}\gg m_p$; this won't be true unless $p_p \gg m_p$, which you haven't shown. (I'll admit I haven't done the math myself, though.) $\endgroup$ Mar 16 at 11:43
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    $\begingroup$ More generally, this result seems wrong for the following reason: suppose your hand is moving at speed $v$ before the collision. Now consider the CM reference frame, where your hand has $v_H \approx 0$. In an elastic collision in this frame, the proton will come in with some speed $v_p = -v$ and leave the collision with speed $v$. So this means that the final proton speed will be about $2v$ in the lab frame. $\endgroup$ Mar 16 at 11:45
  • $\begingroup$ Why can't you slap a pea to supersonic speed? $\endgroup$
    – John Doty
    Mar 16 at 16:17

2 Answers 2


Unfortunately, modeling the interaction of a proton and the hand as an elastic collision is very unrealistic. The interaction is highly non-elastic and non-linear, with the Bragg peak characterizing the interaction the best. This interaction is very well characterized since it is used on a daily basis in proton-beam therapy centers throughout the world.

For a low-energy proton, it will simply get absorbed by the body in a fairly short distance causing a small amount of tissue ionization. You cannot slap a low-energy proton away from your body, it will just get stuck in your body instead. And a proton with high energy will pass entirely through your hand, slowing the proton a bit, and leaving a small trail of ionization through your hand. Protons do not scatter as much as photons do, which is one reason that they are good for medical applications.

Now, beyond that, the calculation that you did was incorrect even assuming an elastic collision between the hand and the proton. In units where $c=1$ and using $E_H$ and $E_P$ for the pre-collision energy of the hand and proton respectively, $p_H$ and $p_P=0$ for the pre-collision momentum of the hand and proton respectively, and ${E'}_H$, ${E'}_P$, ${p'}_H$, and ${p'}_P$ for the post-collision quantities then conservation of energy gives $$E_H+E_P={E'}_H+{E'}_P$$ and conservation of momentum gives $$p_H={p'}_H+{p'}_P$$ and since the collision is elastic the masses are the same before and after $$ E_H^2 - p_H^2 = {E'}_H^2 - {p'}_H^2 $$$$ E_P^2 - p_P^2 = {E'}_P^2 - {p'}_P^2 $$ Then we can solve for $${v'}_P=\frac{{p'}_P}{{E'}_P}=2 \frac{(E_H+E_P) p_H}{(E_H+E_P)^2+p_H^2}$$ where in the approximation that $E_P \ll E_H$ gives us $${v'}_P=2 \frac{E_H p_H}{E_H^2+p_H^2}=2\frac{p_H/E_H}{1+p_H^2/E_H^2}=2 \frac{v_H}{1+v_H^2}$$ where in the further approximation that $v_H \ll c=1$ gives us $${v'}_P=2 v_H$$

So even assuming an elastic collision we cannot slap a proton from rest to relativistic speeds.


Consider the following:

the proton charge radius is 0.84 femtometers (a femtometer is a quadrillionth of a meter).

It is the charge radius because the dominant Feynman diagram will be with the electric field of the molecules of your hand of an area compatible (order of magnitude) with the charge area of the proton. How much of the momentum of your slap will be on that tiny area? It is quadrilliont square that will give the order of magnintude of the areas involved in you slap. You would have to have the muscles of an accelerator after all moving your hand !

A no go concept.


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