Classical Physics is coming back, RELOADED!!!

1100f,

Oh yes, how I understand!

Tell me which are the sources of the fields for those solutions you mention. (If there are such many solutions because from what I know the mathematical recognized solution is that of plane waves only!).

Each electric or magnetic field MUST have a source, there is NO field if there is NO source.

Now tell me about the predicted sources for each case of solution you mention!

 

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Because an EM wave is NOT a particle, there is no particle to be destroyed. The energy in an uncertrain(not unknown) point in a wave function is transfered to an electron in the atom of the recieving matter, if the energy gab is smaller than E=hv from the EM wave.

 

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First, it isnot a photon that is created in the photoelectric effect, but it is absorbed.
And yes, the total angular momentum of the system before the photoelectric effect (photon + metal in which the photon will be absorbed) and after the photon is absorbed is conserved.

 

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1100f,

What about the photons created for example in an ordinary light lamp?
"Photoelectric emission" of course, where photons are created!

How?

 

imaplanck,

So there are no photons for you. What about those photons ("quantums") Einstein introduced by his famous (Nobel prize) photoelectric effect?
Means nothing for you?

 

I found it in two different online references for the equation. See equation 8.5 here and equation 60 here. But you are correct, most sites use the classical expression just as most sites use F=ma rather than the four-vector equivalent.

martillo, please read more carefully in the future. I specifically and explicitly mentioned that I was using units where c=1 and that I would not be writing the factors of c, but that they were there wherever necessary to make the units work out. So, E=mc²=mc=m. I wrote it that way simply to make clear that the conservation of the relativistic four-momentum encompassed the classical conservation of energy, mass, and momentum.

No |v|=c for all objects, massive and massless. Don't forget that time is part of spacetime and that when you sit at rest you are still traveling through time at c. As you increase your spatial velocity your time dilates so you still travel at c through spacetime.

The wave four-vector is in units of spatial frequency (as is its norm) so think of it as the wavenumber in spacetime. So the norm of the wave four-vector is a frame-invariant measure of the frequency through spacetime rather than the traditional frame-variant measures of frequency through time or wavelength through space.

Why should that be strange? Wikipedia is a popular reference with little or no editorial control, it is not a scientific reference. I use it often, but you must use it understanding what it is and what it is not. Also, as I mentioned above most sites and most people use the classical expressions of any given law rather than the relativistic formulation, but the common usage does not mean that the general relativistic laws do not exist nor does the common usage mean that the classical expressions are "general laws of physics" in the sense of the first postulate.

-Dale

 

You see this is what so much contridicts any theory that a photon is a particle....Where does the particle go to?.....How was it created?
The answer is no particle is created or anihelated because there is no particle, only a particle like behaviour of a wave.

 

If one says a photon is a particle then there is indeed no photons. If you say a photon is the particle behaviour of a wave-particle as Einstein then photons do exist.

 

imaplanck,

Well we are analyzing here the "particle" behaviour of your "wave-particle duality", this means the photons which have a spin...

 

Dalespam,
How do you obtain λ from the four-vector formulation?
Which is the predicted λ with that four-vector formulation in the referential at rest with the particle/object being analyzed?

 

k=(ω,k) and |k|=1/λ.

So the spacelike part of the four-vector formulation gives you the traditional deBroglie relationship as follows:

p=hk
(E,p)=h(ω,k)
p=hk
|p|=h|k|
p=h/λ

-Dale

 

Dalespam,
I appreciate your four-vector formulation developments you gave.
I think its fine for me to study them. It seems the formulation is right and according with the relativistic invariance the formula is the same in any referential.
There is only one thing I don't understand:
Why, if I calculate λ starting from a referential where the object/particle has velocity v and making the Lorentz transforms for it (as a lenght = distance) to the object/particle "rest" frame, I get a different result?

I know about the mathematics in change of coordinates and I find essential for the formulas to be invariant in the change that both results must be the same: the calculation of a variable directly from the formula and its calculation by the coordinate transform.

I don't know what is exactly wrong but something is not well.

What do you think about?

 

If you know, in an atom you have discrete energy levels, but each ebergy level is degenerate, there are several states which have the same energy level (in fact, these ldegenerate levels have not exactly the same energy, because of the magnetic interaction between the electrons and the magnetic field of the nucleus). not all transitions are allowed, the transitions that are allowed (when I say allowed I mean that the theory allows these transitions and experiments show that indeed these are the only transitions that occur) are the transitions that leave unchanged the total angular momentum. The total angular momentum of the atom before the transition is the same as the total angular momentum of the photon and the atom after the transition.
Furthermore, quantum theory will not only give the allowed transitions but also gives their relative intensity (google clebsch gordan coefficient)

 

I am not 100% sure I understand your objection here, but it seems like you are describing a very typical mistake. Specifically it looks as though you are not doing the full Lorentz transform, but are only doing the length contraction part of the transform. This is obviously incorrect and often causes errors, which is why I am constantly suggesting on this forum that people use the full Lorentz transform instead of just bits and pieces of it.

For a more concrete example, let's consider some transformations on the four-velocity and for brevity let's consider a single spatial dimension. As I mentioned above |v|=c in all cases, so in the rest frame v=(1,0). There exists a frame where the particle's velocity is .6c in the positive direction. In this frame the four-velocity is v'=(γ,γv')=(1.25,0.75). Here, if we just look at the spatial component we see that we have v=0 and v'=.6 which is obviously not explained simply by applying the length contraction formula (0 contracts to 0, not .6). Instead, if we generate the full Lorentz transform matrix we see that it is A=((1.25,.75),(.75,1.25)) so that v'=A.v and we see that the timelike component in the rest frame is the source of the "extra" spacelike component in the moving frame. Everything transforms properly when we do the full Lorentz transform instead of just doing a third of it.

Similarly, here we have a momentum four vector. In the particle's rest frame there is still the timelike component, which corresponds to the objects mass. When Lorentz-transforming from the rest frame to some other frame this component rotates into the spacelike part and causes the discrepancy you were noticing by trying to do only part of the transform.

I hope that example helps illustrate the importance of doing the whole transform.

-Dale

 

Dalespam,
I Find in the four-vector formulation the same problem I have with the classic formula as described in Section 1.1-B!

As I said in my previous post the problem is the following:
Is essential (necessary condition) for any formula to be invariant in the Lorentz change of coordinates that both results must be the same: the calculation of a variable directly from the formula and its calculation by the coordinate transform.

Now if we consider a change of referential from that where the object has velocity v to its "rest" referential and calculate λ with both procedures we have:
1)directly from the formula: λ' = infinite since at rest v=0
2)Applying Lorentz Transform (matricial for four-vector) to λ as a lenght (distance in the x direction) : λ' = λ/s
where s= root(1-v2/c2)

This mean that the four-vector formulation is also not invariant since one of the variables present a not consistent calculation in the change of coordinates (Lorentz Transformation)!

It is exactly the same problem as with the classical formulation.

I don't understand, you say the law in its four-vector formulation is invariant while my calculations show it isn't!

 

You are making an error in your calculation. Please show your work and I should be able to find it.

-Dale

 

why do you think that λ should change according to the formula: λ' = λ/s?

 

Let consider:
R: referential where the object moves at velocity v
R’: referential where the object is at rest.

Lorentz transformation for four-vector formulation is:

|ct’| | γ -βγ 0 0 | |ct |
|x’ | |-βγ γ 0 0 | |x |
|y’ | | 0 0 1 0 | |y |
|z’ | | 0 0 0 1 | |z |

Where γ = 1/s = root(1-v2/c2), and β = v/c2

λ as a lenght in the x direction at t=0 is the vector:
|0 |
|λ |
|0 |
|0 |

multiplying it by Lorentz matrix we get λ’ as:
|-βγλ|
| γλ |
|0 |
|0 |

λ’ is the second component of the vector and λ’ = γλ = λ/s


The value calculated applying the four-vector formula.as you stated is:

k=(ω,k) and |k|=1/λ.

”So the spacelike part of the four-vector formulation gives you the traditional deBroglie relationship as follows:”

p=hk
(E,p)=h(ω,k)
p=hk
|p|=h|k|
p=h/λ

p=0 implies λ=infinite

The two values are different and the formula is not invariant.

 

martillo,

I think this makes you the first anti-relativityist I've seen actually use the Lorentz transform

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I suggest to you, in order to find your error, to try to see the meaning of the formula p = h/λ yo try to find how this goes without using relativity.

Suppose you have a free particle that has a mass m and moves with velocity v. It has a wavelength λ
Do you think that in another reference frame it will have the same wavelength?

(I remind you, for the moment, do not use relativity)


Edit - Added:

Or in other words, do you think that the wavelength should be treated as some kind of solid ruler (which has a constant length for non relativistic physics) or as the distance between crests (at some instant of time) of a wave whose frequency and velocity are given in the original frame and might be different in another frame?
And the same, in the relativistic case, do you think that the wave length should be treated as the length of some ruler which gets contracted or should it be treated as the distance between two crests (at some instant of time) of some wave whose velocity and frequency are given in the original reference frame?