What are the questions surrounding the De Broglie wave equation and its proof?

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Discussion Overview

The discussion centers on the De Broglie wave equation and its proof, exploring various interpretations, mathematical formulations, and conceptual questions surrounding the relationship between energy, mass, and wavelength in the context of quantum mechanics. Participants engage with both theoretical and conceptual aspects of the equation.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • One participant questions the validity of equating the energy of a photon to mc², arguing that it implies mass destruction, which is not applicable to photons.
  • Another participant suggests that the concept of relativistic mass could be relevant to the discussion of photons and energy conservation.
  • Several participants discuss the mathematical formulation of the De Broglie wave and its relation to energy and momentum, including the use of the equation E² = m₀²c⁴ + p²c².
  • There are claims that the energy of a photon can be expressed in terms of mass or frequency, leading to discussions about the implications of mass-energy equivalence.
  • Some participants assert that the De Broglie and Einstein equations can be viewed as equivalent under certain conditions, though the specifics of this equivalence are debated.
  • One participant emphasizes the need to refer to established scientific literature rather than personal interpretations of quantum reality.

Areas of Agreement / Disagreement

Participants express differing views on the nature of mass in relation to photons, the validity of certain equations, and the interpretation of the De Broglie wave equation. There is no consensus on these issues, and multiple competing views remain throughout the discussion.

Contextual Notes

Some participants highlight the need for clarity regarding definitions of mass, energy, and acceleration, particularly in the context of photons. The discussion also reflects varying interpretations of established equations and their applicability to quantum mechanics.

Cheman
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Hi,
I recently came across a proof for the De Broglie wave equation in a book, which went as follows:
E of photon = mc2
= m*c*c
= (m*c)c
= (p)c ( ie - momentum*speed of light)
= (p)(f*lamda)
Therefore, hf = p(f*lambda)
Therefore, p = hf/ f*lambda
= h/lambda.
Therefore, Lambda (ie- wavelength) = h/ mv.

However, I'm not sure if I agree with this, so I wanted to ask a few questions. Firstly, it is true that the energy of a photon = hf, but I don't think that it should equal mc2. After all, this Einsteinian equation literally means energy produce = mass destroyed*speed of light. So, what mass has been destroyed to create the energy of the photon? Furthermore, how can we take mc to be the momentum of the photon, if this is mass destroyed, not mass of the photon?
As slightly different question, what exactly is the wavelength of a particle? If the wavelength of a wave is "the distance between two crests or adjacent points", then what is the wavelength of a particle?
Thanks is advance. :-D
 
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Really tou need the equation:

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

As a photon has zero mass:

E = pc

In your book they've defined m as relativistic mass.
 
I've had a think and could you say that, since if you accelerate and object and it gains XJoules of KE it will gain X/c2 kg of mass, that in creating the photon you have "destroyed" some mass and thus that the photon must have mass? It must do so that when it gives its energy to something else, that thing will gain energy AND mass.
 
The photon doesn't have mass as mass is usually defined as the rest mass.
 
The verificiation of the De Broglie formula (I am lead to believe) comes from considering a de Broglie wave:
$\psi ({\bf{r}},t) = Ae^{i({\bf{k}}.{\bf{r}} - \omega t)} $<br />

If you assume the relationshup E = \hbar \omega holds for material particles you then write E = \hbar \omega = \frac{{m_0 c^2 }}{{\sqrt {1 - \beta ^2 } }}, use this in the group velocity formula and obtain the required result.

However to show that E = \hbar \omega is a consitent step, you show the invariance of the quantity kr - \omega t<br /> in two inertial frames S and S'.
 
Last edited:
jcsd said:
Really tou need the equation:

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

As a photon has zero mass:

E = pc

In your book they've defined m as relativistic mass.

sorry, there is mass/energy and it is conserved. A photon has mass/energy and it may be expressed as mass or frequency. E=hn=mCC
 
Just consider that all physics equations : E = mc2 ; E = hf ; Gij = Tij; can be expressed in a single way depending on th scale. This single law is :

extensity flux = diffusivity * extensity concentration gradient.
 
Sammy k-space said:
sorry, there is mass/energy and it is conserved. A photon has mass/energy and it may be expressed as mass or frequency. E=hn=mCC

this thread is 5years old..
 
  • #10
The mass of a photon is: m=hn/CC; planks constant times the frequency divided by the speed of light squared.
Mass is mass; mass times accelleration is force but for a photon accelleration is defined as CC ie. C squared. Thus F=mCC.
 
  • #11
Sammy k-space said:
The mass of a photon is: m=hn/CC; planks constant times the frequency divided by the speed of light squared.
Mass is mass; mass times accelleration is force but for a photon accelleration is defined as CC ie. C squared. Thus F=mCC.

sammy, this is one of the most discussed things in this forum, Please read the thread that was suggested to you.

Now, WHERE and WHY should the acceleration of a photon be defined in that way? There is no reason and meaning to that, c^2 does not even have the units of acceleration.

sammy, there equation is E=m c^2, where m = \gamma m_0 and \gamma = 1/\sqrt{1-v^2/c^2}.

The equation E=m c^2 can also be written as E^2 = (pc)^2 + (m_0c^2)^2.

The energy of a photon is E = \hbar \omega

Now, try to fit the photon-energy equation to make \hbar \omega = mc^2... and you will find:

E = \hbar \omega = mc^2 = \gamma m_0c^2 now the rest mass of the photon is zero, and \gamma for the photon is infinite, since the photon moves at c. So what happened to the photon mass?

we have that m = m_0 \gamma = 0 \times \infty .. nice

BUT this is nice, since m_0 = 0:

E^2 = (pc)^2 \rightarrow
p = E/c
 
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  • #12
I agree with you, cc has not the unit of acceleration. But cc/l with "l" being wave length has the unit of acceleration. it is then easy to make de broglie and einstein equations equivallent.
 
  • #13
delplace said:
I agree with you, cc has not the unit of acceleration. But cc/l with "l" being wave length has the unit of acceleration. it is then easy to make de broglie and einstein equations equivallent.

in what sense is deBroigle and "Einstein equation" equivalent?
 
  • #14
They are the same mathematical description of the quantic reality. And you can not imagine how Sammy is close to the final solution when he writes F = m cc !
 
  • #15
delplace said:
They are the same mathematical description of the quantic reality. And you can not imagine how Sammy is close to the final solution when he writes F = m cc !

Sorry dude, but you have to refer to accepted science, published in peer review journals, not your own opinions and idea what "quantic reality" is etc.

Force is instantaneous change momentum also, so there is in the first place no need for an equation of the kind F = mc^2, since F = dp/dt will work too, using p = E/c for a photon.
 
  • #16
ok I stope. But there is no fault when you are doing true calculations directly. You do it yourself F = dp/dt... I can do for example E/V = P (Pa) = mc2/V = mc2/S.l et c2/l (m.s-2) = S/m * P
 
  • #17
This thread is from 5 years ago. There is no need to dredge up something that old, thus I am locking this.
 

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