Equating E=mc^2 & E=hf: A Derivation of de Broglie's Theorem

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SUMMARY

The discussion clarifies the relationship between the equations E=mc^2 and E=hf, emphasizing that while they describe different phenomena—mass-energy equivalence and photon energy respectively—they are interconnected through de Broglie's theory of wave-particle duality. The derivation of de Broglie's wavelength equation, λ = h/p, illustrates this connection, showing that energy can be expressed in terms of mass and frequency. The concept of "relativistic mass" is deemed less useful in this context, and the discussion concludes that these equations cannot be directly equated.

PREREQUISITES
  • Understanding of mass-energy equivalence (E=mc^2)
  • Familiarity with photon energy (E=hf)
  • Knowledge of de Broglie's wavelength equation (λ = h/p)
  • Basic principles of wave-particle duality
NEXT STEPS
  • Study the derivation of de Broglie's theorem in detail
  • Explore the implications of wave-particle duality in quantum mechanics
  • Investigate the concept of relativistic mass and its applications
  • Examine the relationship between frequency, wavelength, and momentum in photons
USEFUL FOR

Students of physics, educators teaching quantum mechanics, and researchers exploring the foundations of wave-particle duality will benefit from this discussion.

repugno
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On a couple of webisites I have seen that they have equated E=mc^2 with E=hf ... i.e hf=mc^2

But how can this be done when E=hf is describing the energy of a photon and E=mc^2 isnt?

Also, can someone provide me with a derivation of de Broglie’s theorem ?

Wavelength = h/momentum
thank you
 
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repugno said:
On a couple of webisites I have seen that they have equated E=mc^2 with E=hf ... i.e hf=mc^2

I bet they're deriving the "relativistic mass" of a photon. I don't consider the concept of "relativistic mass" to be very useful in general, let alone for a photon. But some people seem to like the idea.

Of course, they might be doing something completely different, but you didn't provide any context, so it's hard to tell.

Also, can someone provide me with a derivation of de Broglie’s theorem ?

Wavelength = h/momentum

Start with the general relationship between energy, momentum and invariant mass (not "relativistic mass"):

E^2 = \sqrt {p^2 c^2 + m^2 c^4}

Set m = 0 and E = hf for a photon. Also for a photon you have good old c = f \lambda.
 


It is not entirely accurate to equate E=mc^2 with E=hf, as these equations represent different concepts. E=mc^2 is the famous mass-energy equivalence equation, which states that mass and energy are interchangeable and that a small amount of mass can be converted into a large amount of energy. On the other hand, E=hf represents the energy of a photon, where h is Planck's constant and f is the frequency of the photon. These equations cannot be directly equated because they describe different phenomena.

However, there is a connection between these two equations through the concept of mass-energy equivalence. According to de Broglie's theory, all particles, including photons, have a wave-like nature. This means that they can also be described by a wavelength, which is given by the equation λ = h/p, where p is the momentum of the particle. This equation is known as de Broglie's wavelength equation. If we substitute the value of momentum (p) in this equation with the equation p=mc, we get λ = h/mc. This is where the connection between E=mc^2 and E=hf can be seen. By rearranging this equation, we get E=hf=mc^2, which shows that the energy of a particle (represented by E) can be expressed in terms of its mass (m) and its frequency (f).

In summary, while E=mc^2 and E=hf are not equivalent equations, they are connected through the concept of mass-energy equivalence and de Broglie's theory of wave-particle duality.

As for a derivation of de Broglie's theorem, it is beyond the scope of this response. However, a simple Google search can provide you with various resources and explanations on how this equation was derived. I would also recommend consulting a textbook or seeking help from a physics tutor for a more in-depth understanding.
 

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