E=hf for massive particles, but which Energy exactly?

In summary, the E=hf equation for mass particle (f for frequency) and the Lorentz Invariant (E^2 -p^2c^2=m^2c^4) both refer to total energy, including rest energy. The de Broglie frequency and wavelength for a massive particle at rest are not 0, but rather positive and infinite, respectively. This is not an issue as the absolute level of energy can be chosen arbitrarily without changing any physics. For photons, as their de Broglie frequency and wavelength are related through lambda times omega equals c, as lambda approaches infinity, omega must approach 0. However, this is not a problem for massive particles as they do not have this relationship and mass
  • #1
AlonZ
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which energy is it in E=hf and in lorentz invariant?
Hi there, I'm a bit confused about the E=hf equation for mass particle(f for frequency), and Lorentz Invariant (E^2 -p^2c^2=m^2c^4).
The question is, which energy is it? Total Energy- Kinetic plus Rest, or only kinetic energy.
Now, if it's total energy, then you get that a particle at rest (Ek=0) has frequency which can't be right by De-Broli's equation p=h/lambda.
And if it's only kinetic energy then I get the following: lambda=h/p=hv/mv^2--> 2Ek=mv^2=hv/lambda=hf (v for speed, f for freq).
so both ways don't add up for me, i have to be wrong somewhere, will be glad for your help.
Thanks a lot.
 
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  • #2
In relativistic physics it's always convenient to work with Minkowski scalars, vectors, and tensors. That's why for a free particle you define the energy as you wrote as ##E=\sqrt{m^2 c^4+\vec{p}^2 c^2}##, i.e., including rest energy. Correspondingly for the corresponding energy-momentum eigenstates of free particles you have the equation
$$-\hbar^2 \Box \psi(x)=m^2 c^2 \psi(x),$$
where ##\psi(x)## is any kind of relativistic free wave function (Klein-Gordon, Dirac, etc.).

You can of course always redefine the absolute level of energy by an arbitrary constant, i.e., you can use $$E_{\text{kin}}=\sqrt{m^2 c^4+\vec{p}^2 c^2}-mc^2.$$
This then refers to new wave functions only changed by a phase factor ##\exp(-\mathrm{i} m c^2 t/\hbar)## relative to the covariant wave functions defined above. As in classical (relativistic or Newtonian) physics the absolute level of energy is not observable, only energy differences.
 
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  • #3
AlonZ said:
The question is, which energy is it?
It is total energy.

AlonZ said:
Now, if it's total energy, then you get that a particle at rest (Ek=0) has frequency which can't be right by De-Broli's equation p=h/lambda.
You have a misunderstanding of de Broglie’s relations. See: https://en.m.wikipedia.org/wiki/Matter_wave

The de Broglie frequency is given by ##E=\hbar \omega## and the de Broglie wavelength is given by ##\vec p=\hbar \vec k## where ##\lambda=2\pi/k##. So for a massive particle at rest ##p=0## and ##E>0## so ##\lambda = \infty## and ##\omega>0##. I am not sure why you think that is an issue.
 
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  • #4
Dale said:
It is total energy.

You have a misunderstanding of de Broglie’s relations. See: https://en.m.wikipedia.org/wiki/Matter_wave

The de Broglie frequency is given by ##E=\hbar \omega## and the de Broglie wavelength is given by ##\vec p=\hbar \vec k## where ##\lambda=2\pi/k##. So for a massive particle at rest ##p=0## and ##E>0## so ##\lambda = \infty## and ##\omega>0##. I am not sure why you think that is an issue.
What's troubling me is how does a particle at rest has frequency different than 0, as you said w>0.
I'm attaching a file so you can see exactly where my problem is.
if lambda goes to infinity, doesn't frequency has to go to zero? As it does not in the attached photo.
Thanks a lot for the informative replies.
 

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The plane-wave solution of the Klein-Gordon equation is
$$u_{\vec{p}}(t,\vec{x})=\frac{1}{\sqrt{(2 \pi)^3 2 E_{\vec{p}}}} \exp(-\mathrm{i} E_{\vec{p}} t+\mathrm{i} \vec{p} \cdot \vec{x}),$$
where I use natural units with ##\hbar=c=1##. The normalization is chosen in the relativistic covariant convention in QFT. The energy is given by the "on-shell condition",
$$E_{\vec{p}}=\sqrt{\vec{p}^2+m^2},$$
i.e., such that ##(p^{\mu})=(E,\vec{p})## transforms as a four-vector. For ##\vec{p}=0## you get ##E_0=m##. There's no problem with that. You can choose the absolute level of the energy as you like without changing any physics, and here it's convenient to choose it in this way such that ##(p^{\mu})## is a four-vector.
 
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  • #6
AlonZ said:
if lambda goes to infinity, doesn't frequency has to go to zero?
Not in general, no. For a photon ##\lambda \omega =c## so for them it is true that as ##\lambda\rightarrow\infty## we must have ##\omega\rightarrow 0##. But massive particles don’t have such a relationship so it isn’t a problem for massive particles. And massless particles can never be at rest so it isn’t a problem for photons either.
 
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Related to E=hf for massive particles, but which Energy exactly?

1. What is the meaning of "E=hf" for massive particles?

According to the theory of quantum mechanics, the equation E=hf represents the energy of a photon (light particle) where E is the energy, h is Planck's constant, and f is the frequency of the photon's wave. This equation can also be applied to massive particles, such as electrons, to calculate their energy levels.

2. How does the equation E=hf apply to massive particles?

The equation E=hf can be applied to massive particles by considering them as having both wave-like and particle-like properties. The frequency, f, in the equation represents the frequency of the particle's wave function, which describes its probability of being in a certain state. The energy, E, is related to the frequency through Planck's constant, h.

3. Can the equation E=hf be used for all types of particles?

Yes, the equation E=hf can be used for all types of particles, including massive particles. However, it is most commonly used for particles with wave-like properties, such as photons and electrons.

4. How is the energy of a massive particle calculated using E=hf?

The energy of a massive particle can be calculated by multiplying its frequency, f, by Planck's constant, h. This will give the particle's energy in units of joules (J).

5. Is the equation E=hf for massive particles exact, or are there other factors to consider?

The equation E=hf is an approximation for massive particles, as it does not take into account other factors such as the particle's momentum and mass. For more accurate calculations, other equations, such as the relativistic energy-momentum equation, must be used.

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