So… Well… What about linking its amplitude to the amplitude of the field for the photon. In physical theories prior to special relativity, the momentum p and energy E assigned to a body of rest mass m 0 and velocity v were given by the formulas p = m 0 v and E = E 0 + m 0 v 2 /2, where the value of the “rest energy” E 0 was undetermined. But, of course, from all that I wrote above, it’s obvious that, while these two formulas are the same from a math point of view, they represent very different things. And the probability to find it somewhere is the (absolute) square of some complex number, right? In the image above, the vertical axis should not represent some real number (and it surely should not represent a probability, i.e. In Compton's original experiment (see Fig.
You may say or think: What’s the problem here really? What’s a transient? In any case, yes, you should think of a photon as a one-cycle electromagnetic oscillation. I do not dream to understand the math any time soon. Now, you can easily see that the ‘uncertainty’ or ‘spread’ in the wavelength here (which we’ll denote by Δλ) is, quite simply, the difference between the wavelength of the ‘one-cycle wave’, which is equal to the space the whole wave packet occupies (which we’ll denote by Δx), and the wavelength of the ‘highest-frequency wave’. Heute ist man sicher und spricht auch nur noch von der Masse des Elektrons. This means you just need 80,000 times the speed, or the experimental setup size, to observe this. of normalization indeed. I reckon your confusion arises from the famous incomplete Energy-Mass relation by Einstein which says but that its effective ‘radius’ should be of the same order as the classical electron radius. Just think about it. For those who continue to be skeptical about my sanity here, I’ll quote Feynman once again: “What happens in a light source is that first one atom radiates, then another atom radiates, and so forth, and we have just seen that atoms radiate a train of waves only for about 10–8 sec; after 10–8 sec, some atom has probably taken over, then another atom takes over, and so on. OK. Could it work? Perhaps we have to stretch whatever we understand of Einstein’s (special) relativity theory, but we should be able to draw some conclusions, I feel. Indeed, the magnitude of the magnetic field vector is equal to the magnitude of the electric field vector divided by c = 3×108, so we write B = E/c.
[…] You think so? I am talking about something else here. AB and AC effects with finite photon mass 85 3.6. 0000004752 00000 n In most of the books I've come across, they just write "rest mass of photon is zero."
Using Bohr’s formulation of the Uncertainty Principle, we can see the expression I used above (Δλ = hΔp) makes sense: Δx = Δλ = h/Δp, so ΔλΔp = h. [Just to be 100% clear on terminology: a Fourier decomposition is not the same as that Fourier transform I mentioned when talking about the relation between position and momentum in the Kennard formulation of the Uncertainty Principle, although these two mathematical concepts obviously have a few things in common. Professional or amateur theories and models are welcome as long as they are backed up by actual research. So, when I write that “our concepts of amplitude and frequency of a photon are maybe not very relevant” when trying to picture a photon, and that “perhaps, it’s only energy that counts”, I actually don’t mean “maybe” or “perhaps“. [I’ll challenge you in a moment.] 0000003348 00000 n Implications of a photon mass 83 3.1. Nothing more, nothing less. Since the experiment of Weber leads to Ey/Bz = c one gets dw/dm = c2 Olympia 1993 Einstein's quanta led to wrong relativity, Olympia 1993 Einstein's quanta led to wrong relativity, That is the absorption of a photon by an electron is given by hν/m = ΔΕ/ΔΜ = c2. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. , The second reaction may be somewhat more to the point. All that I know is that there’s one thing we need to explain when considering the various possibilities: a photon has a very well-defined frequency (which defines its color in the visible light spectrum) and so our wave train should – in my humble opinion – also have that frequency. Some content on this page was disabled on June 17, 2020 as a result of a DMCA takedown notice from Michael A. Gottlieb, Rudolf Pfeiffer, and The California Institute of Technology. Indeed, note how, in the illustration above, the frequency of the component waves gradually increases (or, what amounts to the same, how the wavelength gets smaller and smaller) and how, with every wave we ‘add’ to the packet, it becomes increasingly localized. It makes the increase in kinetic energy of an object with velocity appear to be connected with some change in the internal structure of the object. The easy answer, of course, is quite straightforward: we’re not interested in the shape of a photon because we know it is not an electromagnetic wave. Therefore the math is correct but it is based on wrong hypothesis which gives fallacious results in the following process: E2 = M2c4 . You’re right. So I’ve done that, and I thought of one reason why the question, perhaps, may not make all that much sense: a photon travels at the speed of light; therefore, it has no length. That is, one observes that matter as well as light possesses both wave and corpuscular properties.
Not some amplitude associated with a path in spacetime, but a wave function giving an approximate position of the photon.
Before developing the point, I’ll raise two objections to the ‘objection’ raised above (i.e. the statement that a photon has no length). These solutions comprise pulse responses that behave as shock fronts. In other words, the probability amplitude could, perhaps, be proportional to the amplitude of E, with the proportionality factor being determined by (a) the unit in which we measure E (i.e. Remember to always back up your posts with reliable information, clear argumentation and verifiable sources. The first reaction is: “Well… I don’t know.
It has no "rest mass energy"!
;-). Hmm I thought I put "uniform gravitational field" in there, will edit. The question then is: how should we picture that photon? Also Einstein used the invalid relativistic energy due to a wrong relativistic mass M. In fact, according to our discovery of the PHOTON-MATTER INTERACTION in the Kaufmann experiment the mass M is not a fallacious relativistic mass but a variable mass due to the absorption of the photon mass.
We would say the particle had a definite momentum p if the wave number were exactly k, that is, a perfect wave which goes on with the same amplitude everywhere. . I sort of vaguely knew I had to do something with this post – because it is, effectively, one of those posts that gets many views. a photon) using the reference frame of the photon, so that we’re traveling at speed c,’ riding’ with the photon, so to say, as it’s being emitted. We also know we’re more likely to detect something with high energy than something with low energy, don’t we? So all those enigmatic statements you’ll find in serious or less serious books (i.e. I only suggested it because it has the same shape as Feynman’s representation of a particle (see below) as a ‘probability wave’ traveling through–and limited in–space. Now, that radius determines the area in which it may produce some effect, like hitting an electron, for example, or like being detected in a photon detector, which is just what this so-called radius of an atom or an electron is all about: the area which is susceptible of being hit by some particle (including a photon), or which is likely to emit some particle (including a photon).