# Question on De Broglie

1. May 24, 2008

### redtree

The wave-frequency relationship is as follows:

f = v / $$\lambda$$

Therefore:
v = $$\lambda$$ * f

The de Broglie relations are as follows:

$$\lambda$$ = h / p

f = E / h

Using some basic algebra:

v = (h / p) * (E / h)
v = E / p
v = $$\gamma$$mc$$^{}2$$ / $$\gamma$$mv
v = c$$^{}2$$ / v

Now, assuming a natural unit system with c=1

v = 1 / v

That doesn't seem to make sense. Where is my error?

Last edited: May 24, 2008
2. May 24, 2008

### granpa

are you sure it isnt v=v^2/v

3. May 24, 2008

### Staff: Mentor

The first gives the de Broglie wavelength. The second relates the frequency of a photon to its energy.

4. May 24, 2008

### redtree

Total energy is mc$$^{}2$$ not mv$$^{}2$$

5. May 24, 2008

### redtree

The relationship is not just for a photon but for any particle. Frequency and wavelength are related by v = f * $$\lambda$$

6. May 24, 2008

### Staff: Mentor

What relationship? Not E = hf.
This is yet a third relationship, true for any wave.

7. May 24, 2008

### Redbelly98

Staff Emeritus
It "makes sense" because you have chosen to have velocity be unitless. The solutions to your equation are

v = +1 or v = -1

I.e., the object moves at the speed of light.

Somewhere, you must have used a relation that holds only for objects of zero rest mass. I should be more up on this stuff than I am, and will defer to Doc Al for just where the "error" is.

8. May 24, 2008

### Gokul43201

Staff Emeritus
That's right. The error is in the equation E = hf, which is the famous Einstein relation for the energy of a photon.

More generally, the error is in writing down a string of equations without defining what any of the quantities refer to.

9. May 24, 2008

### lbrits

People, lets not get crazy and instead actually answer the fellow's question. What our friend has discovered is that the phase velocity of a deBroglie wave is indeed faster than the speed of light
$$v_p \frac{\omega}{k} = \frac{c^2}{v}$$
Instead, he should consider particles as being represented by wave packets with some spread in momentum and energy (in order to be localizable). In this case, the relevant quantity is the group velocity
$$v_g = \frac{\partial\omega}{\partial k} = \frac{\partial E}{\partial p} = \frac{\partial (\sqrt{p^2 c^2 + m^2 c^4} )}{ p }= \frac{p}{\gamma m} = v$$
I hope this clears things up.

Edit: $$E = h f$$ is true for particles of mass as well. It's a result of harmonic oscillators in general, of which quantum fields are one type (in the limit where "particle" makes sense).