In the text, it reads that the momentum of a particle will change if it is moving at speed close to light speed. In the general case, the wavelength is given as $$ \lambda = \frac{h}{p} $$ and $$p = \frac{mv}{\sqrt{1-v^2/c^2}}$$ when $v \to c$, $p\to\infty$, so is it say that the wavelength is ZERO? I don't understand why the wavelength will change to zero if it is moving at speed very close to light speed?
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
Lorentz contraction! The measured de Broglie wavelength in the direction of propagation vanishes because that's what special relativity says happens. The wavelength has to go as $h/p$ as you wrote, so why does it surprise you that when $p$ gets large the wavelength gets small?
-
$\begingroup$ Thanks for your reply. I am thinking two factors. If the particle is photon, so the wavelength gives the color of the light. So when the speed close to speed of light, the color will change? Physically, why does it happen? Also, in a book, it state that when the speed extremely close to speed of light, we have pc = K.E. + m_0c^2 so it is not infinite, in a word, $\lambda = \frac{hc}{pc}$ not approaching zero. Why the contradiction? $\endgroup$ Apr 7, 2013 at 5:59
-
$\begingroup$ Photons always move at the speed of light - they have no rest frame. Their momentum $p$ is exactly equal to their energy, so $p = \hbar k/c$. I'm not sure where the contradiction is coming from. As $p$ gets large, the kinetic energy gets large as well. $\endgroup$– webbApr 7, 2013 at 6:15
-
$\begingroup$ Thanks webb. My confusion is like this. In the initial equation I gave $p = mv/sqrt(1-v^2/c^2)$, so when $v\to c$, $p\to\infty$. But the equation equation given in the book for extreme-high-speed case gives $pc=K.E.+m_0c^2$, so here $p$ actually will not be infinity because K.E. is speed is at maximum c not infinity. So which one should I use for momentum when it extremely close to light speed? If I use the first one, wavelength will be 0. If I use the second one, the wavelength is finite! $\endgroup$ Apr 7, 2013 at 16:26
$\begingroup$
$\endgroup$
2
Yes. The energy-momentum equation $E^2=p^2c^2+m^2c^4$ says that a massive object's mass (relativistic mass), momentum and energy approaches $\infty$ as a particle is accelerated towards $c$. There's nothing obvious about the fact that it requires infinite energy to accelerate it to $c$. This strictly means that you can't accelerate the object to speed-of-light...
Substituting Planck's wave equation $h\nu$ in $E=pc$ for photons, we get $$\lambda=\frac{hc}{pc}$$
Or in case of particles with rest energy where we can substitute for total energy, $$\lambda=\frac{h}{\sqrt{E^2-{(m_0c^2)}^2}}=\frac{h}{\sqrt{p^2-2mE_0}}$$
This equation says that matter waves are observable only for matter (massive) particles which always travel at $v<c$. In other words, there is no matter wave for such objects.
answered Apr 7, 2013 at 6:48
-
$\begingroup$ Thanks. I know the non-relativistic wavelength is $\lambda = h/p$. I still confuse if what's the main difference if I go extreme high speed or at high speed but not that close to speed of light? In hyperphysics, I found the equation for two cases: for the extreme high speed, $pc = KE + m_0c^2$ and for the ordinary relativistic case, we have the approximation $pc=\sqrt{2KE*m_0c^2}$. So if you plug that into $\lambda = hc/pc$, kinetic energy is always there so the mass will appear. But in your formula, there is no mass? I wonder if the hyperphysics equation is right or not? $\endgroup$ Apr 7, 2013 at 16:21
-
$\begingroup$ @user1285419: You can go any velocity below $c$ as you're massive. The usage of $h/p$ or the relativistic correction is based on your velocity. And, Yes - It's totally right. In fact, I got confused (I've corrected now). But, the formula is still valid. Anyways, there's always mass in the denominator. So, no massive objects..! ;-) $\endgroup$ Apr 7, 2013 at 16:53