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
$$v_p \frac{\omega}{k} = \frac{c^2}{v}$$
$$v_g = \frac{\partial\omega}{\partial k} = \frac{\partial E}{\partial p} = \frac{\partial (\sqrt{p^2 c^2 + m^2 c^4} )}{ p }= \frac{p}{\gamma m} = v$$
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).