# The relativistic de Broglie equation

• I
• mjda
In summary, the 4-wave vector is a vector that is invariant with respect to the inner product of its components. It is derived from the 4-momentum vector by considering the scalar product of the two vectors. The de Broglie equation can be derived from the inner product of the 4-wave vector with itself, and the 4-momentum vector.
mjda
I'm having trouble understanding how the 4-wave vector is derived, and also how it is then used alongside the 4-momentum vector to formulate the relativistic de Broglie equation.

The inner product of the 4-momentum vector with itself, is an invariant quantity. If we define the 4-momentum vector, P, as:

P = (p''' , iE/c) ---- where p''' is just the 3 dimensional momentum vector.

P.P = p2 - (E/c)2 <-- this is invariant

The 4-wave vector, N, is defined as:

N = f (c/w , 1) <-- how does one derive this?

From what I have read, you should be able to derive the de Broglie equation by considering the inner product of P and N, with P.P?

The result being:

cP = hN

Has anyone seen this notation before and able to show this result is true? I can't find it anywhere in this form!

First of all, you have to realize that the de Broglie relation is an empirical relation. It was historically convenient to consider particles having a wavelength proportional to its momentum. For the same reason, you should not expect to ”prove” the relativistic version of the relation.

What you can do is to look at plane wave solutions to the Klein-Gordon or Dirac equations. For a plane wave, you can easily derive the 4-frequency as the gradient of the phase function and relate it to how the solution behaves under space-time translations, which are generated by the momentum and energy operators.

mjda said:
From what I have read, you should be able to derive the de Broglie equation by considering the inner product of P and N, with P.P?

Mister T said:

It is in a question from a past paper I found. Maybe my wording is a bit misleading, but the exact phrasing is:

By considering the scalar product of P both with itself, and with N, deduce the deBroglie equation.

mjda said:
It is in a question from a past paper I found.

What paper? Please give a reference.

Relativistic relation

$$E^2-p^2c^2=m^2c^4$$

replacing E and p by QM operators

$$[-\frac{\partial^2}{c^2\partial t^2}+\mathbf{\nabla}^2-\frac{m^2c^2}{\hbar^2}]\psi=0$$

$$(\mathbf{\square}^2+\frac{m^2c^2}{\hbar^2})\psi=0$$

Are you looking for something like this?

sweet springs said:
Relativistic relation

$$E^2-p^2c^2=m^2c^4$$

replacing E and p by QM operators

$$[-\frac{\partial^2}{c^2\partial t^2}+\mathbf{\nabla}^2-\frac{m^2c^2}{\hbar^2}]\psi=0$$

$$(\mathbf{\square}^2+\frac{m^2c^2}{\hbar^2})\psi=0$$

Are you looking for something like this?
That is pretty sexy looking. Would you mind telling me what the square symbol is called so I can look it up? Wild guess is it has something to do with a wave since there is a second order partial derivative there.
Edit- Nevermind, I found it. It is indeed related to waves, specifically the four dimensional, Minkowski space variety. https://en.m.wikipedia.org/wiki/D'Alembert_operator

sweet springs
The ##\square## is a second order differential operator, the d'Alembert operator ##\square = g^{\mu\nu}\partial_\mu \partial_\nu##. Note that there should not be a square on the ##\square##, i.e., the equation should contain ##\square##, not ##\square^2##.

sweet springs and Sorcerer
Orodruin said:
The ##\square## is a second order differential operator, the d'Alembert operator ##\square = g^{\mu\nu}\partial_\mu \partial_\nu##. Note that there should not be a square on the ##\square##, i.e., the equation should contain ##\square##, not ##\square^2##.
I really need to take the time to learn Einstein summation convention. But four dimensional partial differential equations and matrices are hard enough without bringing weird dot product rules and tensor stuff into it, let alone short hand for it. ;) *sigh* So much work to be done...

But thanks for the insight! Slowly but surely I’ll learn, until I can get back into school.

sweet springs said:
Relativistic relation

$$E^2-p^2c^2=m^2c^4$$

replacing E and p by QM operators

$$[-\frac{\partial^2}{c^2\partial t^2}+\mathbf{\nabla}^2-\frac{m^2c^2}{\hbar^2}]\psi=0$$

$$(\mathbf{\square}^2+\frac{m^2c^2}{\hbar^2})\psi=0$$

Are you looking for something like this?

I am trying to find it in the form of:

cP = hN

where P and N are both the 4-Momentum, and 4-Wave vectors.

It can also be written in the form:

cm( u , ic ) = hf( cn / w , 1 )

c - speed of light
m - mass
u - 3-D velocity vector
i - imaginary component
f - frequency
h - Planck's const.
w - ang. freq.
n - unit vector

I suppose you would say
4-momentum
$$mc^2u^\mu=mc^2(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}},\frac{v_x/c}{\sqrt{1-\frac{v^2}{c^2}}},\frac{v_y/c}{\sqrt{1-\frac{v^2}{c^2}}},\frac{v_z/c}{\sqrt{1-\frac{v^2}{c^2}}})$$
equals 4-wave vector
$$c\hbar k^\mu =c\hbar (\omega/c,k_x,k_y,k_z)$$
in de Broglie relation.

Last edited:

## 1. What is the relativistic de Broglie equation?

The relativistic de Broglie equation is a mathematical equation that relates the momentum of a particle to its wavelength, taking into account the effects of special relativity. It is often written as p = h/λ, where p is the momentum, h is Planck's constant, and λ is the wavelength.

## 2. Who developed the relativistic de Broglie equation?

The relativistic de Broglie equation was developed by French physicist Louis de Broglie in 1924. He proposed that particles, such as electrons, could exhibit both wave-like and particle-like behavior, and that their momentum could be described by their wavelength.

## 3. What does the relativistic de Broglie equation tell us about particles?

The relativistic de Broglie equation tells us that particles can have a wavelength associated with them, similar to how light can behave as both a particle and a wave. This concept is known as wave-particle duality and is a fundamental principle in quantum mechanics.

## 4. How is the relativistic de Broglie equation different from the classical de Broglie equation?

The classical de Broglie equation, also known as the non-relativistic de Broglie equation, only applies to particles with non-relativistic speeds, meaning speeds much slower than the speed of light. The relativistic de Broglie equation takes into account the effects of special relativity and can be applied to particles with any speed.

## 5. What are some applications of the relativistic de Broglie equation?

The relativistic de Broglie equation has many applications in modern physics, including in quantum mechanics and particle physics. It has been used to explain the behavior of particles in various experiments, such as the double-slit experiment, and is also a key component in the development of technologies such as electron microscopes and particle accelerators.

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