# Matter waves and Energy

1. Mar 23, 2006

### arunbg

At school, I am taught that the "derivation" of de Brogle wavelength of a matter wave is as follows.

$$E=mc^2, E= h\nu$$

So
$$mc^2=h\nu$$

Then setting c=v by analogy derive the exp. for de Broglie wavelength.

Here's the problem, I can't understand in what context the two energies are equated.

I only know that the first eqn is mass energy equ. and the second eqn. is energy of a discrete photon.So obviously m must be the mass of photon
right? Now is this mass the rest mass of photon ? I don't even know much about SR other than the basic postulates.

Also what is the exact nature of matter waves?
Do they transport energy or are they probability functions?
Is it possible to approximate how much of a system is wave like and and how much of it is particle like ?

I don't think it is really right to deal with these topics before SR as they do in my syllabus. What do you think?

Thanks in advance for the replies.Any links will also be appreciated.

Arun

2. Mar 23, 2006

### jack47

you're right to question this 'derivation'... because its stupid!
E doesn't equal mc^2 for a photon, so imo you cant really equate the first two formulae you've quoted

3. Mar 23, 2006

### ZapperZ

Staff Emeritus
I agree. It is a very dubious "derivation".

Take a look here, for example.

http://hyperphysics.phy-astr.gsu.edu/Hbase/debrog.html

You will notice that the de Broglie wavelength is associated with the MOMENTUM of the particle, and not with the rest mass.

Zz.

4. Mar 23, 2006

### arunbg

To further the "derivation" what is done is that c->v

$$mv^2=h\frac{v}{\lambda}$$

so $$\lambda=\frac{h}{mv}=\frac{h}{p}$$

And it is not specified what mass is m .
Students in my school (my final exams got over today, phew!) simply mug up the formula and apply to problems.