How can I derive the relativistic de Broglie equation?

[Boltzman Oscillation]

Homework Statement

Given wavelength = h/p, derive hc/(sqrt(K^2 + 2Kmc^2))

Relevant Equations

K = .5mv^2

I know I can get momentum in terms of kinetic energy in this way:

K = .5mv^2 => p = sqrt(2mK)

substitute into debroigles gives me:

λ = h/(sqrt(2mk)) = hc/(sqrt(2mc^2K))
which would be the nonreletivistic equation but I need the reletivistic equation. I can plug in the equation for relativistic kinetic energy K = γmc^2 - mc^2 but i do not see how that would get me closer to my desired equation. Any help would be appreciated, thanks all.

 

[Abhishek11235]

HINT:

$$E^2= (pc)^2+(m_{0})^2 c^4$$
$$p=h/\lambda$$
$$E= K + (m_{0})^2 c^4$$

Where symbols have usual meaning. Play with 1st and 2nd equation and use 3rd equation at last. The first equation is the relation connecting momentum and energy whereas you have used classical expression!

 

[kuruman]

HINT:

$$E^2= (pc)^2+(m_{0})^2 c^4$$
$$p=h/\lambda$$
$$E= K + (m_{0})^2 c^4$$

Where symbols have usual meaning. Play with 1st and 2nd equation and use 3rd equation at last. The first equation is the relation connecting momentum and energy whereas you have used classical expression!

You mean
$$E= K + m_{0} c^2$$

 

[Abhishek11235]

You mean
$$E= K + m_{0} c^2$$

Oh sorry! Use this equation.

 

[Boltzman Oscillation]

Oh sorry! Use this equation.

I see. What is mo? Rest mass? I know the following formula:

E^2 = (pc)^2 + (m^2c^4) but not the one with rest mass.

 

[Boltzman Oscillation]

Oh sorry! Use this equation.

Using those equations I get:

λ = hc/sqrt(E^2 - mc^2) where m is rest mass.
E = K + Eo and mc^2 = Eo
thus i can rewrite the denominator as sqrt(K^2 +Eo^2 +EoK - mc^2) = sqrt(K^2 +(mc^2)^2 + (mc^2)K - mc^2)
am i doing something wrong?

 

[Boltzman Oscillation]

also what program can I use to write these equations on pc?

 

[Abhishek11235]

$m_{0}$ is the rest mass. Also you have written wrong formula for $\lambda$. To correct:

$$E^2 = (hc\lambda)^2+ (m_{0})^2 c^4$$

You got mistaken at last term!

So a simple manipulation keeping third equation in mind gives your result.

also what program can I use to write these equations on pc?

See Latex guide on PF

 

[vela]

E^2 = (pc)^2 + (m^2c^4) but not the one with rest mass.

What do you think $m$ stands for here if not rest mass?

λ = hc/sqrt(E^2 - mc^2) where m is rest mass.

This can't possibly be correct because the units don't work in the denominator.

 

[Herman Trivilino]

K = .5mv^2

Not a valid (relativistic) equation.

 

[Boltzman Oscillation]

What do you think $m$ stands for here if not rest mass?This can't possibly be correct because the units don't work in the denominator.

Regular mass? Rest mass is when the particle doesn't have kinetic energy right?

 

[Boltzman Oscillation]

Not a valid (relativistic) equation.

So to derive a relativistic equation I must begin with a relativistic equation? Got it.

 

[Abhishek11235]

Regular mass? Rest mass is when the particle doesn't have kinetic energy right?

It is rest mass. It is same whether mass is moving or not. Perhaps you are considering relativistic mass which is abandoned concept.

 

[Boltzman Oscillation]

It is rest mass. It is same whether mass is moving or not. Perhaps you are considering relativistic mass which is abandoned concept.

I understand. I get the right equation using your derivation. Thank you sir!

 

[Herman Trivilino]

So to derive a relativistic equation I must begin with a relativistic equation? Got it.

To derive any equation using some theory, you must use that theory! So, to derive an expression using the theory of relativity, you must use the theory of relativity. Saying that kinetic energy equals $\frac{1}{2}mv^2$ is equivalent to saying I am going to use an approximation to that theory. Why? Because $\frac{1}{2}mv^2$ is valid only as a low-speed approximation. The fact that you learned the low-speed approximation before learning the more general theory is the reason you're using $\frac{1}{2}mv^2$ as the kinetic energy.

 

[Herman Trivilino]

Regular mass? Rest mass is when the particle doesn't have kinetic energy right?

There is no need to use any other kind of mass than the so-called rest mass. Once you fully incorporate this idea into your worldview you realize that there is no need to call it anything other than the ordinary mass, and there is no need to mark the symbol $m$ with a subscript of $o$.

This has a greater significance when considering systems that consist of a collection of particles. The rest frame of the system is a frame where the sum of the momenta of the particles in the collection is zero. The total energy of the collection, in this frame, is equivalent to the ordinary mass of the system. Make the particles in the collection move faster, in such a way that the sum of their momenta remains zero, and you increase the ordinary mass $m$. Note that we call the energy in this frame the rest energy. The existence of rest energy, and its equivalence to the ordinary mass, is one of Einstein's greatest discoveries.

Note that if your teacher is calling $m_o$ the rest mass, and $m$ the relativistic mass, they make no error in the physics. They are just using an old-fashioned terminology that was present in the textbooks they used as students, but note the overwhelming majority of college-level introductory physics textbooks no longer use that terminology. A lot of this change came about as a result of an article written by Lev Okun in the June 1989 edition of Physics Today.

 

[Boltzman Oscillation]

There is no need to use any other kind of mass than the so-called rest mass. Once you fully incorporate this idea into your worldview you realize that there is no need to call it anything other than the ordinary mass, and there is no need to mark the symbol $m$ with a subscript of $o$.

This has a greater significance when considering systems that consist of a collection of particles. The rest frame of the system is a frame where the sum of the momenta of the particles in the collection is zero. The total energy of the collection, in this frame, is equivalent to the ordinary mass of the system. Make the particles in the collection move faster, in such a way that the sum of their momenta remains zero, and you increase the ordinary mass $m$. Note that we call the energy in this frame the rest energy. The existence of rest energy, and its equivalence to the ordinary mass, is one of Einstein's greatest discoveries.

Note that if your teacher is calling $m_o$ the rest mass, and $m$ the relativistic mass, they make no error in the physics. They are just using an old-fashioned terminology that was present in the textbooks they used as students, but note the overwhelming majority of college-level introductory physics textbooks no longer use that terminology. A lot of this change came about as a result of an article written by Lev Okun in the June 1989 edition of Physics Today.

Alright thank you for clarifying that for me. That equation that relates energy and mass is:
$$E = mc^2$$
I will take a look at that article, thank you sir.

 

[Boltzman Oscillation]

To derive any equation using some theory, you must use that theory! So, to derive an expression using the theory of relativity, you must use the theory of relativity. Saying that kinetic energy equals $\frac{1}{2}mv^2$ is equivalent to saying I am going to use an approximation to that theory. Why? Because $\frac{1}{2}mv^2$ is valid only as a low-speed approximation. The fact that you learned the low-speed approximation before learning the more general theory is the reason you're using $\frac{1}{2}mv^2$ as the kinetic energy.

Yes, thank you. I only learned these equations this semester.

 

[Herman Trivilino]

That equation that relates energy and mass is:

$E=mc^2$​

If $E$ is the total energy, then the $m$ in that relation is the antiquated relativistic mass.

If $E_o$ is the rest energy and $m$ is the ordinary mass, then $E_o=mc^2$ and $E^2=(mc^2)^2+(pc)^2$.

 

[Boltzman Oscillation]

If $E$ is the total energy, then the $m$ in that relation is the antiquated relativistic mass.

If $E_o$ is the rest energy and $m$ is the ordinary mass, then $E_o=mc^2$ and $E^2=(mc^2)^2+(pc)^2$.

I see that total energy is K + Eo
Is Eo the potential energy for relativistic equations?

 

[kuruman]

I see that total energy is K + Eo
Is Eo the potential energy for relativistic equations?

No, it's the rest mass energy $m_0c^2$.

 

[Herman Trivilino]

Is Eo the potential energy for relativistic equations?

No, it's the rest energy. Which is equivalent to the mass.

 

[Abhishek11235]

I see that total energy is K + Eo
Is Eo the potential energy for relativistic equations?

$E_{0}$ is not only potential energy but total internal energy(including movement of particles,interaction,etc.)

 

[kuruman]

$E_{0}$ is not only potential energy but total internal energy(including movement of particles,interaction,etc.)

In this case we have a relativistic particle, electron, proton etc. and we want to find the relativistic equivalent of the de Broglie relation. It is a relation first derived by de Broglie in his doctoral dissertation.