What is the Correct Proof for the de Broglie Equation?

[Unredeemed]

I saw a proof in a book for the de broglie equation which I thought was wrong.

if l = lambda, so wavelength and then all other notation is standard they said:

E=mc^2
E=hf

therefore,

mc^2=hf

but for a photon, c=v
so:

mv^2=hf

but mv=P
so:

Pv=hf

and v=fl
so:

Pfl=hf
Pl=h

and finally:

l=h/P

Anyway, that's what they said. But I thought since a photon doesn't have a mass "mv" couldn't be its momentum. Or does it not matter?

 

[cesiumfrog]

It's correct because it uses relativistic mass, but still it doesn't prove anything about non-photons. I think de Broglie had a slightly stronger proof: relativity implies that if a particular wave packet represents a particular particle (photon or otherwise) then it will always have the same relation between wavelength and momentum, but even that doesn't require the same value of h to be shared for all particle types. Perhaps there is no theoretical a priori proof?

 

[cragar]

the mometum of a photon is P=E/c where E=hf the energy of a photon
and i thought E=mc^2 cannot be used for a photon beacuse that equation is for
particles at rest and light is never at rest.

 

[Sammy k-space]

I saw a proof in a book for the de broglie equation which I thought was wrong.

if l = lambda, so wavelength and then all other notation is standard they said:

E=mc^2
E=hf

therefore,

mc^2=hf

but for a photon, c=v
so:

mv^2=hf

but mv=P
so:

Pv=hf

and v=fl
so:

Pfl=hf
Pl=h
!
and finally:

l=h/P

Anyway, that's what they said. But I thought since a photon doesn't have a mass "mv" couldn't be its momentum. Or does it not matter?

no, for a photon v=C, so E=mCC=hn, thus mass of a photon=hn/CC (m=hn/CC) for an electron at rest E=mCC but for an electron in motion E=mCC+ mvv/2 or E=m(CC+(vv/2)) ie. the mass energy plus the kinetic energy. now take special relativity and E=hn=hn+h(nsub1-nsub2); C is just the constant of proportionality for conservation of mass/energy, think of it this way: CC=E/m or C= plus or minus the square root of E/m

 

[Sammy k-space]

E=mCC=hn or CC=E/m.
C is the constant for conservation of mass/energy and follows special relativity.
G is the constant for conservation of space/time and follows general relativity.
If one takes the mass of a photon as m=hn/CC then we have the the observed accelleration of a photon in a gravitational field and Albert's eclipse test.
Albert's photo-electric effect is just this: hn=mCC+mvv/2, if hn is greater than the fermi energy an electron is moved up.
The converse is Sammy's electro-photic effect: mCC+mvv/2=E=hn=h(nsub1+nsub2); if an electron of sufficient total energy impinges on metal then a photon is emmited; also called a light bulb.
A special case is when an anti-electron impinges on anti-metal an anti-photon is emmited; this would be an anti-flashlight or a black-hole.
Doppler is just special relativity:E=hn=h(n1-n2)=m(CC-vv/2) where the shift is the frequency change based on the velocity of the observer relative to the light source.
In Albert's eclipse test the curvature of light in a gravitational field is shown but they did not check for the frequency shift that would have been evident.
Plank's constant is the constant of proportionality between mass/energy and space/time.
Thus we have the relationship between gravity and electromagnetic events.
Next time I will address the atomic scale and do away with the misconceptions of strong and weak forces.

 

[ImAnEngineer]

the mometum of a photon is P=E/c where E=hf the energy of a photon
and i thought E=mc^2 cannot be used for a photon beacuse that equation is for
particles at rest and light is never at rest.

As I have learned from my own topic m=E/c² is valid, IF m is the relativistic mass (and not rest mass, since that is zero for a photon).

https://www.physicsforums.com/showthread.php?t=300384"

Anyway, that's what they said. But I thought since a photon doesn't have a mass "mv" couldn't be its momentum. Or does it not matter?

So photons do not have restmass, but they do have relativistic mass. Then this mass can be plugged into p=mv=mc to calculate the momentum.

 

[Sammy k-space]

the mometum of a photon is P=E/c where E=hf the energy of a photon
and i thought E=mc^2 cannot be used for a photon beacuse that equation is for
particles at rest and light is never at rest.

no, no, no! E=mCC is the energy of a particle and is equal to the mass times C squared. True this is the mass energy of a particle at rest, so for a particle in motion just add 1/2 mass times velocity squared: E total= mCC+mvv/2, but for a photon all the energy is mass in motion at the speed of light, ie. v is defined as C for a photon.
The mass of a photon is found from E=mCC=hn, thus m=hn/CC ( I use CC for C^2, it is easier to read and to type on a computer, n is nu.)

 

[Sammy k-space]

C is the constant of mass/energy; G is the constant of space/time; h is the conversion factor between the two. Plank wants quanta: how about charge? Is it 1/3 of an electron charge? Then we have quarks with 1 or 2 quanta (+ or -), electron with 3(-), proton with 3(+) and neutron with 0. So 0, 1, 2, or 3 quanta. Any thoughts on this?

 

[JustinLevy]

no, no, no! E=mCC is the energy of a particle and is equal to the mass times C squared. True this is the mass energy of a particle at rest, so for a particle in motion just add 1/2 mass times velocity squared: E total= mCC+mvv/2

No. Total energy (for a free particle) is given by:

E^2 = (mc^2)^2 + (pc)^2
This is true for all particles (massless or not).
The E = mc^2 is just a special case for a massive particle at rest.

Your equation instead tries to mix relativistic (mc^2) and non-relativistic quantities ( 1/2 mv^2 ). Please do not mix these together like that.

 

[Sammy k-space]

The mass/energy of say an electron at rest is E=mCC, now give it a shove and add the kinetic energy: E total= mCC+mvv/2 ; this is just special relativity in terms of mass; in terms of frequency it is just doppler effect. C and G are just constants, h is the convertion factor. This stuff is so simple for me, yet few can understand it!

 

[ImAnEngineer]

no, no, no! E=mCC is the energy of a particle and is equal to the mass times C squared. True this is the mass energy of a particle at rest, so for a particle in motion just add 1/2 mass times velocity squared: E total= mCC+mvv/2, but for a photon all the energy is mass in motion at the speed of light, ie. v is defined as C for a photon.
The mass of a photon is found from E=mCC=hn, thus m=hn/CC ( I use CC for C^2, it is easier to read and to type on a computer, n is nu.)

So you are saying $E=mc^2 + \frac{1}{2}mv^2$ and because for photons c = v: $E=mc^2 + \frac{1}{2}mc^2=\frac{3}{2}mc^2$. And then you say the mass of a photon can be found from $E=h\nu=mc^2$ (instead of (3/2)mc²)
That seems rather inconsistent to me, as Justin Levy pointed out as well. You aren't taking into account that $E_k_i_n=\frac{1}{2}m_0v^2$ deals with $m_0$ rather than $m_r_e_l$.

$mc^2 = m_0c^2 + \frac{1}{2}m_0v^2$ is only a good approximation at low speeds.

 

[Sammy k-space]

no, I said that e=mCC is the mass/energy, e=mvv/2 is the kinetic energy; sum of energy is E=mCC+mvv/2; also e=hn is the mass/energy of a photon, E=hn+h(n1+n2) is the mass/energy for a moving light source relative to the observer, ie. special relativity, doppler. Think of it this way: if the light source is moving towards the observer it is as if the photons have added kinetic energy, but C is constant so the energy change is expressed as an increase in frequency; if the light source is moving away from the observer then it is as if there is a decrease in kinetic energy, but again C is constant so there is a decrease in frequency. This is just doppler ( red-shift, blue-shift); this is just linear/special relativity.
For general relativity just use mass of a photon,m=hn/CC, in the gravity formula. The observed curvature of light path is an accelleration, but again C is constant along the path of light so the observer would see an increase in frequency of the photons. In the Eclipse test the curvature was observed, but I do not think they looked for the frequency shift nor could they have documented it with black and white film.
In summary, E=mCC=hn is just the conservation of mass/energy: CC=E/m=hn/m.

My first law of thermodynamics: There is one system, and this is it baby! a) there is conservation of mass/energy; b) there is conservation of space/time; c) there is conservation of angular momentum. Thus there is no need for the cat in a hat and no uncertainty.
My second law: there are two frames of reference a) this one: home of electric monopoles and magnetic dipoles b) the one that photons have: the orthogonal universe is the home of the magnetic monopoles (magnetons), gravitons, and time particles (chronotons); there may be a third frame of reference but I have not done the math yet. I can get enough dimensions to satisfy the Oiler equation. Thus no parallel universes only an orthogonal one (maybe two orthogonal universes).

You may wonder who I am; I proposed spherical carbon (bucky balls) 30 years ago and was laughed at; I did general relativity in rotational coordinates to simplefy the math but my professors could not get a handle on it. I am known as Sammy k-space because I can think in multi-dimensions. If you want to see what I see then try revolutions/sec instead of cycles/sec and photons become 3-D!

 

[Sammy k-space]

Cragar, Albert's photo-electric effect is just the opposite of Sammy's electro-photic effect; ph>metal>e- vs e->metal>ph; for a black-hole then it is: e+>anti-metal>anti-ph, the photons are moving backwards in time and thus appear to fall back into the singularity(anti-light source). Quite simple!

 

[Sammy k-space]

Here is proof of the de Broglie: Gh/CC is the constant of space/time, units are meters cubed per second, so we can warp space/time but with this constraint. This is the most significant equation since E=mCC and maybe ever. The space/time constant is proof of relativity, conservation of space/time, and m=hn/CC. Ask me how if you dare.

 

[Sammy k-space]

Yes, space/time is invariant!

 

[Sammy k-space]

So you are saying $E=mc^2 + \frac{1}{2}mv^2$ and because for photons c = v: $E=mc^2 + \frac{1}{2}mc^2=\frac{3}{2}mc^2$. And then you say the mass of a photon can be found from $E=h\nu=mc^2$ (instead of (3/2)mc²)
That seems rather inconsistent to me, as Justin Levy pointed out as well. You aren't taking into account that $E_k_i_n=\frac{1}{2}m_0v^2$ deals with $m_0$ rather than $m_r_e_l$.

$mc^2 = m_0c^2 + \frac{1}{2}m_0v^2$ is only a good approximation at low speeds.

no, v is the velocity of the observer relative to the source: I did not say 3/2, I said mCC+mvv/2=E relative for observer moving towards the source and mCC-mvv/2=E relative for observer moving away from the source. It is a vector sum.