What is the De Broglie-Wave Problem in Special Relativity?

[fluidistic]

Homework Statement

Consider a particle whose rest mass is $$m_0$$. By analogy of $$E=h \nu$$ for the electromagnetic field, de Broglie assumed that there existed some kind of intrinsic oscillatory motion with frequency $$\nu _0$$ associated to the particle at rest, where $$h \nu _0=m_0 c^2$$.
Assuming that the particle is moving with a velocity v with respect to an inertial frame of reference:
1)Show that for an observer in the fixed inertial reference frame the oscillatory motion of the particle is described by a progressive wave whose phase velocity is $$\frac{c^2}{v}$$.
2)Deduce the relation $$\lambda =\frac{h}{p}$$.
3)Show that the total energy of the particle satisfies $$E=h \nu$$ in any intertial reference frame, where $$\nu=\gamma \n_0$$ and $$\gamma$$ is Lorentz factor.

Homework Equations

Not sure.

The Attempt at a Solution

For 1) I should maybe find something of the form $$A \cos (bx+ct)$$. But I really don't see how to even start. I'd like to solve 1) first and then proceed further.
I'd love a tip just to get me started... thank you very much.

 

[sgd37]

a de Broglie matter wave is of the form

$$e^{-i\vec{p} \cdot \vec{x}}$$

where

$$\vec{p} = (E, p_x , p_y , p_z )$$ and $$\vec{x} = (t , x ,y ,z)$$

so that you can see for one dimension it is a wave of the form $$e^{-i(\omega t - k x)}$$

the phase velocity is defined as $$v_p = \frac {\omega}{k}$$ now use special relativity relations for the rest

 

[fluidistic]

Thanks for helping! I appreciate your time and help.

a de Broglie matter wave is of the form

$$e^{-i\vec{p} \cdot \vec{x}}$$

Nice to know, I never seen this before.

where

$$\vec{p} = (E, p_x , p_y , p_z )$$ and $$\vec{x} = (t , x ,y ,z)$$

So what are $$\vec p$$ and $$\vec x$$? They seem like the momentum vector and the position vector but extended with energy and time? I never seen that either before. I'd like to know how do you call them.

so that you can see for one dimension it is a wave of the form $$e^{-i(\omega t - k x)}$$

I try to follow you on this but doing the dot product and considering only 1 dimension I get $$\vec p \cdot \vec x=(Et,xp_x)$$. With the data of the problem I could simplify it to $$\vec p \cdot \vec x=(m_0c^2t,x\gamma m_0 v)$$, unfortunately nothing looking like $$\omega t-kx$$.
Seems like I need further assistance.

the phase velocity is defined as $$v_p = \frac {\omega}{k}$$ now use special relativity relations for the rest

Perfect.

 

[sgd37]

These are vectors commonly encountered in relativistic physics called the four momentum and the four dimensional space-time vector. Note that i have missed out constants of c so that the units of E and t should have the dimensions of momentum and space respectively.

The dot product of two vectors is defined as $$\vec{a} \cdot \vec{b} = (a_1 , a_2) \begin{pmatrix} b_1 \\ b_2 \end{pmatrix} = a_1 b_1 + a_2 b_2$$

in relativistic physics the dot product is defined differently. The difference being that the spatial parts have a minus in front of them

 

[fluidistic]

Thanks a lot! My bad I'm so rusty how could I forget that a dot product between 2 vectors gives a number and not a vector... ouch. :shy:

Ok so I reach $$\vec p \cdot \vec x =m_0 c^2t-x \gamma m_0 v$$. Setting $$\omega =m_0 c^2$$ and $$k=\gamma m_0 v$$, I get $$v_p=\frac{c^2}{\gamma v}$$ instead of $$\frac{c^2}{v}$$.
Does this mean I should have considered the classical momentum $$p_x=m_0 v$$ instead of the relativistic one $$p_x=\gamma m_0 v$$? I don't think so, thus I don't know what I did wrong.

 

[sgd37]

you're missing a factor of gamma in the energy term SR energy is given by $$E = \gamma m c^2$$ remember the particle is moving the energy relation you used is only true for a particle at rest

 

[fluidistic]

Ok good so that solves part 1).
I've been playing with equations for part 2) and I can't reach the answer.
I must deduce that $$\lambda =\frac{h}{p}$$. Therefore that $$h=\lambda p$$.

On one hand I have that $$E=\gamma h \nu _0=m_0c^2 \gamma \Rightarrow h=\frac{m_0 c^2}{\nu _0}$$.
On the other hand I have that $$\lambda p = \frac{2\pi}{k} \cdot \gamma m_0 v=2\pi$$ which of course does not match the value of h. (Edit: Hmm now that I think, it might match the value of h but how to show this?)
I have made the use of the relation $$k=\frac{2\pi}{\lambda}$$. I can see no flaw in what I did, yet I do not get the result. Where did I go wrong?

 

[sgd37]

you just derived the relation you need

$$\frac{E}{p} = v_p$$

knowing that $$E = h \nu$$ and $$v_p = \nu \lambda$$ you can derive the wavelength momentum relation

 

[fluidistic]

Thank you, I solved part 2).
For part 3 I made a type in the latex formula, I forgot a n. I must show that $$E=h \gamma \nu_0$$; something I've been assuming till here. I feel like turning in circles. What should I assume, what to start with?

 

[sgd37]

you haven't assumed that at all the only thing you have assumed is that $$E = h \nu$$ for arbitrary frequency now using the given assumption for a stationary particle $$E = h \nu_0 = mc^2$$ all you have to do is relate the energy of a stationary particle to that of a moving one

 

[fluidistic]

you haven't assumed that at all the only thing you have assumed is that $$E = h \nu$$ for arbitrary frequency now using the given assumption for a stationary particle $$E = h \nu_0 = mc^2$$ all you have to do is relate the energy of a stationary particle to that of a moving one

Hmm I'm all confused. To me when you/I write $$\nu$$, I understand it as $$\gamma \nu _0$$. The same apply for $$m=\gamma m_0$$.
Seems like I shouldn't have thought this way?!
Just to be sure, when you write $$E = h \nu_0 = mc^2$$, do you mean $$E = m_0 \gamma c^2$$?

 

[sgd37]

yeah that is what i mean. They shouldn't really teach that special relativity stuff you don't see that gamma ever again after the first year