Question on De Broglie

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:

granpa
are you sure it isn't v=v^2/v

Mentor
The de Broglie relations are as follows:

$$\lambda$$ = h / p

f = E / h
The first gives the de Broglie wavelength. The second relates the frequency of a photon to its energy.

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

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

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

Mentor
The relationship is not just for a photon but for any particle.
What relationship? Not E = hf.
Frequency and wavelength are related by v = f * $$\lambda$$
This is yet a third relationship, true for any wave.

Staff Emeritus
Homework Helper
Now, assuming a natural unit system with c=1

v = 1 / v

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

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.

Staff Emeritus
Gold Member
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.

lbrits
People, let's 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).