# De broglie and light

1. Oct 27, 2007

### ankitpandey

i have a doubt.... can someone please tell me where i am going wrong?
by Einsteins calculations, rest mass of photon or something moving with v=c has zero rest mass, else its mass at motion tends to infinity, which is not possible. NOW substituting this in de broglies wavelenght eqn,
lambda=h/mv
=h/mc (for photon moving with light speed)
=h/0*c (since rest mass of the photon is zero)
=infinity
but for no photon it is possible that wavelenght is infinite. please tell my mistake.

2. Oct 27, 2007

### meopemuk

The formula $p = mv$ is from non-relativistic physics. It cannot be used for photons. Three formulas are useful for photons. One formula connects momentum p and energy E

$$E = pc$$

(this is, actually, a limiting case of the most general formula $$E = \sqrt{m^2c^4 + p^2c^2}$$ when $m = 0$)

Another formula connects photon's energy with the light frequency

$$E = h \nu$$

The wavelength can be obtained by the third equation

$$\lambda = \frac{c}{\nu} = \frac{h c}{E} = \frac{h}{p}$$

Eugene.

3. Oct 27, 2007

### Staff: Mentor

[added... aha, I now see that Eugene was typing at the same time that I was!]

The classical formula for momentum, p = mv, doesn't work in relativity, at least not when m is the "rest mass" as you're using it.

The general relationship between energy, momentum and rest mass in relativity is $E^2 = (pc)^2 + (mc^2)^2$. For a photon, m = 0 so this becomes E = pc or p = E/c. So, starting with de Broglie's formula:

$$\lambda = \frac{h}{p} = \frac{hc}{E}$$

Now use Planck's equation E = hf:

$$\lambda = \frac{hc}{hf} = \frac{c}{f}$$

which is just the familiar wave equation $c = f \lambda$.

Last edited: Oct 27, 2007
4. Oct 27, 2007

### Phred101.2

I believe Mr. deBroglie was the first to propose this very fact, btw.

5. Oct 27, 2007

### rbj

sure it can. you use relativistic mass.

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

where $m_0$ is the rest mass.

and that can derived from:

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$E = m c^2$$

and

$$p = m v$$

$$E^2 = m_0^2 c^4 + p^2 c^2$$

solve for momentum, and define the inertial mass to be whatever the momentum of the body is (from the POV of some other inertial frame of reference) divided by the velocity (from the POV of the same inertial frame of reference), and you will get a quantity of dimension mass. that mass, when multiplied by $c^2$, is the total energy (rest energy plus kinetic energy) of the body from the POV of that same inertial frame of reference. and that mass fits the relativistic mass formula above. dunno why the concept of relativistic mass is so deprecated here.

Last edited: Oct 27, 2007
6. Oct 27, 2007

### meopemuk

There are two equivalent approaches to relativistic kinematics. These approaches differ by the way they define mass. In your approach mass is defined as a velocity-dependent quantity. In my approach, the rest mass (which is independent on the velocity of the body or velocity of the observer) is used everywhere. Formulas can be written both ways, but physical meaning doesn't change.

Eugene.

7. Oct 27, 2007

### rbj

i dunno if the two approaches are equivalent pedagogically.

$$m = \frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

$$E = m c^2$$

$$p = m v$$

and get:

$$E^2 = (m_0 c^2)^2 + (p c)^2$$ .

but i do not know how to get the bottom equation independently (although i realize that this is likely done in a formal SR class). but, even if you do come up with it independently, then you define inertial mass as whatever scaler you need to multiply the velocity vector with to get the momentum vector (from the POV of the inertial frame of reference that this body with rest mass $m_0$, is flying by). that's what the definition is, allowing for the possibility for it to be a function of velocity, but velocity dependent is not in the definition, but is a consequence of the definition (momentum divided by velocity) and further derivation.

8. Oct 27, 2007

### meopemuk

Such an independent derivation can be done within Wigner's theory of unitary representations of the Poincare group. (This theory is usually formulated within relativistic quantum mechanics, but it is also possible to do a purely classical derivation). This derivation involves the following major steps:

1. The Hilbert space of any isolated system carries an unitary representation of the Poincare group.

2. Representations corresponding to elementary particles are irreducible.

3. Generators of time and space translations in this representation ($H$ and $\mathbf{P}$, respectively) are identified with Hermitian operators of energy and momentum.

4. It can be shown that operator $$M = c^{-2}\sqrt{H^2 - \mathbf{P}^2c^2}$$ commutes with all generators of the Poincare group (it is one of the two Casimir invariants), so it corresponds to an invariant property, which is naturally identified with the rest mass of the system.

5. It then follows that $$H = \sqrt{M^2c^4 + \mathbf{P}^2c^2}$$

Eugene.

9. Nov 2, 2007

### ankitpandey

new q

k i thank everyone. but i have another doubt. energy of de broglie waves of matter much slower than light will be hc/lambda or hv/lambda? or no rule has been discovered so far

10. Nov 4, 2007

### learningphysics

Use the second de broglie relation:

$$E = hf$$

$$\gamma mc^2 = hf$$

$$f = \gamma mc^2/h$$

The first de broglie relation:

$$\lambda = \frac{h}{\gamma mv}$$

multiplying the two (f by lambda), we get

phase velocity = $$\frac{c^2}{v}$$

where v is the velocity of the particle of matter.

this is a "phase velocity"... it doesn't violate special relativity.

11. Nov 4, 2007

### country boy

You are correct, and your explanation in fact shows a way to derive the bottom equation "independently." Start with ds^2 = (cdt)^2 - dx^2 and divide through by dt^2 to get velocities, (ds/dt)^2 = c^2 - v^2. Now multiply by a quantity m that has the property mds/dt = invariant. m is the relativistic mass and mds/dt is the rest mass * c. This then gives (E/c)^2 = (mo*c)^2 + (mv)^2, as required. Maybe not rigorous, but the essence is there.

12. Nov 7, 2007

### ankitpandey

thanks learninphysics n countryboy.... so then i guess it must be correct to
say MC^2=HC^2/LAMBDA*V where V is velocity of particle of matter right?thanks
a lot

13. Nov 7, 2007

### country boy

Yes. This also comes easily from the expression for the de Broglie group velocity v:

v = c^2/(lambda*frequency)

frequency = c^2/(lambda*v)

mc^2 = E = h*frequency = hc^2/(lambda*v)

14. Nov 10, 2007

### ankitpandey

from that we can derieve lambda=h/mv -de broglie eqn.
so then i guess this is a safe conclusion-
lambda=h/mv is true with m as rest mass
in case of all non-zero-rest-mass particles
and also true with m as mass at motion in case of
zero-rest-mass particles like photons. i want to
confirm this by receiving a reply from you.

Last edited: Nov 10, 2007
15. Nov 10, 2007

### country boy

m is the total relativistic mass. This applies to both zero and non-zero rest mass particles.