Theory of relativity - proton entering a medium

[haruspex]

Is the entire interaction of the proton with the water somehow taken into account by just using the dispersion relation E = c_mp for the photon?

That's my understanding. I.e. it's not a 'solid' collision between the proton and water molecules, but the subtler relativistic interaction with a medium with index < 1. Not sure whether it's important that a proton carries charge - probably.

 

[sankalpmittal]

Cracked it! It also occurred to me in the middle of the night that I should just differentiate the surd equation wrt p₁ and set dp/dp₁=0. That gives p = m₀cc_m/√(c²-c_m²) = p₁. At first sight that looks useless, since it gives p = p₁, but then it dawned on me that the likely alternative to setting p₁=0 is to set p-p₁ very small. Indeed, that corresponds to the photon getting a vanishingly small amount of energy. (With that assumption in the first place, we could have expanded the surds on the basis of small δp = p-p₁ and got the result quite quickly.)
Numerically it gives 5.98.10^-19. Why so different from the given answer? Because the given answer is wrong, 3.58.10^-37 being the numerical value of p².

Same answer I get. But what I did was to write the equation in post #19 in the form, ap²+bp+c=0. Then there will be two corresponding roots of with I chose the one which seemed minimum, yet doubtful. So I considered both the roots. I differentiated both the roots w.r.t p₁, obviously putting dp/dp₁=0 and solved. Obviously I checked that if d²p/dp₁²>0 for minima. I got, I think values, out of one seems to be following the minima condition. That was the answer same as what you got. And now, the copy on which I did this all is lost.

 

[TSny]

The expression $\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)$ can be rearranged as $\sqrt{c^2p^2+m_0^2c^4} - c_mp = \sqrt{c^2p_1^2+m_0^2c^4} -c_mp_1$

Measuring $p$ in units of $m_0c$ and measuring $c_m$ in units of $c$ we have

$\sqrt{p^2+1} - c_mp = \sqrt{p_1^2+1} - c_mp_1$

Thus, defining $f(p) = \sqrt{p^2+1} - c_mp$ we have $f(p) = f(p_1)$.

A plot of $f(p)$ for water ($c_m = .767$) is attached. The intersection of a horizontal line with the graph would correspond to possible initial and final momenta of the proton, with the initial momentum being the larger momentum. The minimum of the function occurs where the initial proton momentum would have its minimum value and still be able to emit a “soft” photon. This is seen to occur near $p = 1.195$ (or, $5.98 \times 10^{-19}$ kgm/s). This, of course, agrees with haruspex’s calculation.

In a vacuum, we would have c_m = 1. The attachment shows a plot of f(p) for this case and we can see the function no longer takes on a minimum. So, as expected, the proton cannot emit a photon and conserve energy-momentum in a vacuum.

In fact, it is easy to show that the value of p that minimizes f(p) corresponds to the proton having a speed of c_m. So, only when the proton enters the water with a speed greater than the speed of light in water can the proton emit a photon.

So I guess this has has something to do with Cherenkov radiation. But I wish I understood better why simply using $E_{photon} = c_mp_{photon}$ for the photon in water automatically takes into account transfer of energy and momentum with the water. I had never seen this before.

 

[Saitama]

Thanks Haruspex and TSny for your help. Your answers are correct. The solution for this problem was posted today. The solution reaches the same answer as you. And as TSny mentioned, the solution too gives a link to Cherenkov radiation. I will quote the solution:

First simplify by considering all the particles as moving in the x-direction. The initial particle energy and momentum is a point $(p_i,E_i)$ on the curve $E_{prot}=\sqrt{c^2 p^2+m_{prot}^2c^4}$. So is the final proton energy and momentum a point $(p_f,E_f)$, with $p_f<p_i$ and $E_f<E_i$. If energy and momentum can be conserved when emitting a photon, these two points must be able to be connected by a line with slope $c_m$. The lowest value of $p_i$ where this can happen is where the slope of the $E(p)$ curve for the proton equals $c_m$. Hence we have $\frac {dE} {dp}|_{p_{min}}=c_m$ as our condition for the lowest value $p_{min}$ of the momentum for the initial proton. Algebraically solving this equation for $p_{min}$ yields the answer of $5.98 \times 10^{-19}~kg~m/s$.Finally, the pretty blue light produced by high energy particles hitting water is called Cherenkov light and can be seen here.

I don't understand the part when it says that the points connect by a line with slope c_m. From where did this come from?

Measuring $p$ in units of $m_0c$ and measuring $c_m$ in units of $c$ we have
\sqrt{p^2+1} - c_mp = \sqrt{p_1^2+1} - c_mp_1

Can you please explain this to me?

 

[TSny]

Start with
$\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)$

and divide through by $m_0c^2$. You will find that the equation simplifies to

$\sqrt{[p/(m_0c)]^2+1} - \sqrt{[p_1/(m_0c)]^2+1} = (c_m/c)[(p/(m_0c) - p_1/(m_0c)]$

So, if you measure p in units of $m_0c$ and measure $c_m$ in units of $c$, you get

$\sqrt{p^2+1} - \sqrt{p_1^2+1} = c_m(p - p_1)$

Going back to the original equation and letting $E(p) = \sqrt{c^2p^2+m_0^2c^4}$, you have $E(p) - E(p_1) = c_m(p-p_1)$. Or,
\frac{E(p) - E(p_1)}{p-p_1} = c_m
The left-hand side is the slope of the line that connects two points on the $E(p)$ vs $p$ graph corresponding to initial and final momenta ($p$ and $p_1$) of the proton. So the above equation says that the slope of the line must be equal to $c_m$ if energy and momentum are to be conserved. Thus if you draw any straight line of slope $c_m$ such that it passes through two points of the graph of $E(p)$, the two points will correspond to possible initial and final momenta of the proton such that energy and momentum will be conserved in photon emission. If you draw the line of slope $c_m$ such that it is tangent to the graph of $E(p)$, then you will have the case where the initial and final momenta of the proton are equal and the emitted photon would have zero energy. The slope of the tangent line is the derivative $E\:'(p)$. But $E\:'(p)$ can easily be shown to equal the speed of the proton. So, the minimum speed of the proton for photon emission is $c_m$.

 

[Saitama]

Start with
$\sqrt{c^2p^2+m_0^2c^4} - \sqrt{c^2p_1^2+m_0^2c^4} = c_m(p - p_1)$

and divide through by $m_0c^2$. You will find that the equation simplifies to

$\sqrt{[p/(m_0c)]^2+1} - \sqrt{[p_1/(m_0c)]^2+1} = (c_m/c)[(p/(m_0c) - p_1/(m_0c)]$

So, if you measure p in units of $m_0c$ and measure $c_m$ in units of $c$, you get

$\sqrt{p^2+1} - \sqrt{p_1^2+1} = c_m(p - p_1)$

Going back to the original equation and letting $E(p) = \sqrt{c^2p^2+m_0^2c^4}$, you have $E(p) - E(p_1) = c_m(p-p_1)$. Or,
\frac{E(p) - E(p_1)}{p-p_1} = c_m
The left-hand side is the slope of the line that connects two points on the $E(p)$ vs $p$ graph corresponding to initial and final momenta ($p$ and $p_1$) of the proton. So the above equation says that the slope of the line must be equal to $c_m$ if energy and momentum are to be conserved. Thus if you draw any straight line of slope $c_m$ such that it passes through two points of the graph of $E(p)$, the two points will correspond to possible initial and final momenta of the proton such that energy and momentum will be conserved in photon emission. If you draw the line of slope $c_m$ such that it is tangent to the graph of $E(p)$, then you will have the case where the initial and final momenta of the proton are equal and the emitted photon would have zero energy. The slope of the tangent line is the derivative $E\:'(p)$. But $E\:'(p)$ can easily be shown to equal the speed of the proton. So, the minimum speed of the proton for photon emission is $c_m$.

Great explanation. Thanks a lot TSny!