Capiert Posted May 11, 2017 Author Share Posted May 11, 2017 (edited) It is next to impossible to prove experimentally that anything is zero. We have shown in a lab that the mass is less that 10^-13eV/c^2 and via observations of galatic magnetic fields that the mass should be less than 10-27 eV/c^2. But you cannot prove a zero - because there are always experimental imprecisions and errors; this will lead to error bars and your figure could lurk in the error bars. Theoretically big portions of modern physics would fail - quantum chromo/electro-dynamics would fall over and no longer be re-normalisable and thus no longer give the astonishingly accurate results that it does. 10-27 eV/(c^2) is mighty small, (I must admit); but if it (=the photon mass) lies at e.g. 10-32 eV/(c^2) then we will not know till we can achieve (better than) that precision. Edited May 11, 2017 by Capiert Link to comment Share on other sites More sharing options...
Strange Posted May 11, 2017 Share Posted May 11, 2017 I cannot imagine momentum without a (physical) mass m What you can imagine is hardly relevant. That is its definition,or should I ignore that? That is one definition. For massive particles. For massless particles it has a different definition. Link to comment Share on other sites More sharing options...
Capiert Posted May 11, 2017 Author Share Posted May 11, 2017 (edited) Please proceede. Explain a bit (if you want). What you can imagine is hardly relevant.I agree, imagination & that we can understand is the hard(est )core to connecting & making sense of things. E.g. We do not write these threads for machines; they are written that people will understand, particularly the 1's asking. I find that most relevant. (Good that you pointed that out: Ridgidly relevant, (I like that).) Edited May 11, 2017 by Capiert Link to comment Share on other sites More sharing options...
Sensei Posted May 11, 2017 Share Posted May 11, 2017 (edited) "Photon has zero mass" is shortcut from "photon has zero rest-mass". Which means there is no frame-of-reference in which photon is at rest. Particles with non zero rest-mass, you can accelerate, deaccelerate etc. etc. Edited May 11, 2017 by Sensei Link to comment Share on other sites More sharing options...
frankglennjacobs@gmail.com Posted May 11, 2017 Share Posted May 11, 2017 As I understand it, the double-slit exercise proves (over and over) that light is not a particle -- and not a wave, either. But an individual photon causes a visible chemical reaction on photographic film in the double-slit exercise. And light causes the black and white vanes inside the trick light-bub thingies (photometer?) to go around. And just yesterday my mind was farther blown by the news that electrons have no size at all. (Next thing you know we will find out that the earth is NOT flat, and that space is NOT full of Aether, and that fire has nothing to do with Phlogistine.) I am a relic of a past Century, dragged kicking and screaming into these modern times! Link to comment Share on other sites More sharing options...
Strange Posted May 11, 2017 Share Posted May 11, 2017 Please proceede. Explain a bit (if you want). We know that energy is given by: [latex]e = \sqrt{m_0^2 c^4 + p^2c^2}[/latex] (http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/relmom.html#c4) When the rest mass [latex]m_0[/latex] is zero then the energy is just: [latex]e = p c[/latex] So: [latex]p = \frac e c[/latex] Or (using the relationship found by Planck: http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html#c3 ) [latex]p = \frac e c = \frac {h \nu} c = \frac h \lambda[/latex] Link to comment Share on other sites More sharing options...
Capiert Posted May 11, 2017 Author Share Posted May 11, 2017 (edited) "Photon has zero mass" is shortcut from "photon has zero rest-mass". Which means there is no frame-of-reference in which photon is at rest. Moment (=stop please).Isn't (at the speed of light, c,) light standing still wrt to the other photons? (E.g. those photons going in the same direction wrt the earth.) Isn't speed c also a reference frame? What do you mean by shortcut? Please explain (a bit). ~Particles with [non zero] rest-mass, can be accelerated & deaccelerated, etc. What is the etc? (constant speed?) Edited May 11, 2017 by Capiert Link to comment Share on other sites More sharing options...
Strange Posted May 11, 2017 Share Posted May 11, 2017 Isn't (at the speed of light, c,) light standing still wrt to the other photons? No. You can't compare the speed of photons because ... Isn't speed calso a reference frame? No. The speed of light is not a valid reference frame. Try it and you will end up dividing by zero. Link to comment Share on other sites More sharing options...
Capiert Posted May 11, 2017 Author Share Posted May 11, 2017 As I understand it, the double-slit exercise proves (over and over) that light is not a particle -- and not a wave, either.[As far as I know, the double slit experiment (results) confirms (or at least convinces me) light is a wave. Maybe a half wavelength could be interpretted as a (particlelike) impulse? But an individual photon causes a visible chemical reaction on photographic film in the double-slit exercise.? And light causes the black and white vanes inside the trick light-bub thingies (photometer?) to go around.Impulse? And just yesterday my mind was farther blown by the news that electrons have no size at all.That's wild, & blows mine too.Some of the professionals will have to comment there, I don't know that new background. (Next thing you know we will find out that the earth is NOT flat, and that space is NOT full of Aether, .. I prefer Maxwell's Ether opinion, too. and that fire has nothing to do with Phlogistine.) I am a relic of a past Century, dragged kicking and screaming into these modern times! Yes, variety is the spice of life, see that you don't suffer (integestion, heartburn) from too much spice (=new changes). Link to comment Share on other sites More sharing options...
Mordred Posted May 11, 2017 Share Posted May 11, 2017 (edited) Capiert would you please look up the definition of mass under physics. Your basic problem is your not considering what mass is defined as. Mass is resistance to inertia change. If you can't even pay attention to the very definition of mass then your never going to understand the key differences between massive and massless particles. Edited May 11, 2017 by Mordred Link to comment Share on other sites More sharing options...
Capiert Posted May 11, 2017 Author Share Posted May 11, 2017 (edited) No. You can't compare the speed of photons because ...Yes, ?, please continue. No. The speed of light is not a valid reference frame. Try it and you will end up dividing by zero.Please explain.You gave us a rooted (energy) equation, modelled perhaps? Edited May 11, 2017 by Capiert Link to comment Share on other sites More sharing options...
Mordred Posted May 11, 2017 Share Posted May 11, 2017 (edited) What Strange posted is correct he also posted relevant links to each equation Edited May 11, 2017 by Mordred Link to comment Share on other sites More sharing options...
Capiert Posted May 11, 2017 Author Share Posted May 11, 2017 What Strange posted is correct he also posted relevant links to each equation I don't doubt he is telling standard physics. Link to comment Share on other sites More sharing options...
Strange Posted May 11, 2017 Share Posted May 11, 2017 Yes, ?, please continue. Please explain. You gave us a rooted (energy) equation, modelled perhaps? If you try and use the Lorentz transform to convert between frames of reference when [latex]v = c[/latex] then [latex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - c^2/c^2}} = \frac{1}{\sqrt{1 - 1}} = \frac 1 0 [/latex] = undefined Therefore v=c is not a valid frame of reference and you cannot compare the speed of photons. Link to comment Share on other sites More sharing options...
Capiert Posted May 12, 2017 Author Share Posted May 12, 2017 (edited) Capiert would you please look up the definition of mass under physics. Your basic problem is your not considering what mass is defined as. Mass is resistance to inertia change. If you can't even pay attention to the very definition of mass then your never going to understand the key differences between massive and massless particles. In physics, mass is a property of a physical body. It is the measure of an object's resistance to acceleration (a change in its state of motion) when a net force is applied.[1] It also determines the strength of its mutual gravitational attraction to other bodies. The basic SI unit of mass is the kilogram (kg).-Wiki https://en.m.wikipedia.org/wiki/Mass Edited May 12, 2017 by Capiert Link to comment Share on other sites More sharing options...
Strange Posted May 12, 2017 Share Posted May 12, 2017 In physics, mass is a property of a physical body. It is the measure of an object's resistance to acceleration (a change in its state of motion) when a net force is applied.[1] It also determines the strength of its mutual gravitational attraction to other bodies. The basic SI unit of mass is the kilogram (kg).-Wiki And light does not accelerate and so it does not respond to force and so it has no mass. Link to comment Share on other sites More sharing options...
Mordred Posted May 12, 2017 Share Posted May 12, 2017 acceleration is inertia change Link to comment Share on other sites More sharing options...
Capiert Posted May 12, 2017 Author Share Posted May 12, 2017 (edited) If you try and use the Lorentz transform to convert between frames of reference when [latex]v = c[/latex] then [latex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}} = \frac{1}{\sqrt{1 - c^2/c^2}} = \frac{1}{\sqrt{1 - 1}} = \frac 1 0 [/latex] = undefined Therefore v=c is not a valid frame of reference and you cannot compare the speed of photons. I don't use the Lorentz transform in momentum because (it's redundant, when) I set the limit speed to c=v+v'. Similar can be done with KE (but energy gives wrong answers, m*E should be used instead). The Galilean transform was never invented with a speed limit (e.g. c) because they thought light's speed was infinite. We know better, now & can change that. Edited May 12, 2017 by Capiert Link to comment Share on other sites More sharing options...
Mordred Posted May 12, 2017 Share Posted May 12, 2017 (edited) Then your not using the correct methodology and therefore getting the wrong answers. Use the correct formulas as described by relativity and you might have a chance.. Start with listing the difference in the transformation equations of Galilean relativity and Lorentz transforms. Secondly understand that the Lorentz transforms are under constant velocity. I set the limit speed to c=v+v'. This makes absolutely no sense. All observers will always measure the velocity of a photon at c. So why would you ever add two observer velocities if both observers measure the signal at c your essentially adding c to c to equal c... obviously incorrect Edited May 12, 2017 by Mordred Link to comment Share on other sites More sharing options...
Sensei Posted May 12, 2017 Share Posted May 12, 2017 (edited) "Photon has zero mass" is shortcut from "photon has zero rest-mass". Which means there is no frame-of-reference in which photon is at rest. Moment (=stop please). Isn't (at the speed of light, c,) light standing still wrt to the other photons? (E.g. those photons going in the same direction wrt the earth.) Isn't speed c also a reference frame? What do you mean by shortcut? Please explain (a bit). "Mass" can be understood as "rest-mass", or "relativistic-mass". It depends on who is using it, and in which context. You should start from reading articles below: Rest-mass aka invariant-mass https://en.wikipedia.org/wiki/Invariant_mass Relativistic-mass https://en.wikipedia.org/wiki/Mass_in_special_relativity ~Particles with [non zero] rest-mass, can be accelerated & deaccelerated, etc. What is the etc? (constant speed?) etc. = Latin Et cetera https://en.wikipedia.org/wiki/Et_cetera Edited May 12, 2017 by Sensei Link to comment Share on other sites More sharing options...
swansont Posted May 12, 2017 Share Posted May 12, 2017 Well... In reality massless particles shouldn't exist so I still think that photons have some form of mass albeit unmeasurably small, however you might as well say they are massless. Question is is there any way to prove photons have a miniscule amount of mass, Perhaps a stream of condensed photons that impacts a silicon wafer nanometers across in a magnetic bowl (Antimatter trap)? If it moves more than a control object being put through the same experiment without the stream of photons then it would be provable through thousands of more tests that photons have miniscule amounts of mass? I personally think they do have a miniscule amount of mass as something without mass shouldn't really exist. However I could be wrong... If the mass of the photon were not identically zero, it would have measurable consequences. It seems opinions are changing. What was said in the past is not true now. (What a mess.) Is it possible we don't have enough accuracy yet? They aren't opinions. momentum mom=m*v. I cannot imagine momentum without a (physical) mass m (& speed v). That is its definition, or should I ignore that? Would you please explain that exclusion method a bit? I'm not clear yet. Momentum is only mv in the classical limit, as I have explained. It is an approximation, not the definition. Link to comment Share on other sites More sharing options...
Capiert Posted May 12, 2017 Author Share Posted May 12, 2017 (edited) What is your model, and/or what are your testable predictions based on this idea? We know, for example, that the momentum of a photon with a wavelength of 780 nm is 8.5 x 10^28 kg-m/s Don't you surely mean 8.5*(10^(-28)) kg*m/s, instead? (Negative exponent -28, instead of positive 28.) (1.56 eV=2.55*(10^(-19)) J. mom~2*E/v.) What is its mass and how fast is it moving? The photon is moving at c light's speed.The photon's mass is (the momentum divided by it's speed v=c) m=mom/v=8.5*(10^(-28)) [kg*m/s]/(2.99*(10^8)) m/s)=2.85*(10^(-36)) kg. Edited May 12, 2017 by Capiert Link to comment Share on other sites More sharing options...
swansont Posted May 12, 2017 Share Posted May 12, 2017 Don't you surely mean 8.5*(10^(-28)) kg*m/s, instead? (Negative exponent -28, instead of positive 28.) (1.56 eV=2.55*(10^(-19)) J.) The photon is moving at c light's speed. The photon's mass is (the momentum divided by it's speed v=c) m=mom/v=8.5*(10^(-28)) [kg*m/s]/(2.99*(10^8)) m/s)=2.85*(10^(-36)) kg. Yes, the "-" didn't register. 2.85*(10^(-36)) kg gets you the 1.59 eV I mentioned. But experiment says that if it's nonzero, it must be 13 orders of magnitude smaller than this. Other observations say it's 27 orders of magnitude. So your model has been shown to be wrong, in spectacular fashion. Link to comment Share on other sites More sharing options...
Capiert Posted May 12, 2017 Author Share Posted May 12, 2017 (edited) And light does not accelerate and so it does not respond to force and so it has no mass.I think your problem (there) is you don't want to admit gravity's free fall g=-9.8 m/(s^2) is an acceleration (when you say that); light falls (like any other moving object, (ruffly) without air resistance); sound does NOT. That means light accelerates. Yes, the "-" didn't register. 2.85*(10^(-36)) kg gets you the 1.59 eV I mentioned. But experiment says that if it's nonzero, it must be 13 orders of magnitude smaller than this. Other observations say it's 27 orders of magnitude. So your model has been shown to be wrong, in spectacular fashion. Please explain a bit (those experiments, so I can grasp what's happening). Spectacular sounds interesting. Edited May 12, 2017 by Capiert Link to comment Share on other sites More sharing options...
Strange Posted May 12, 2017 Share Posted May 12, 2017 I think your problem (there) is you don't want to admit gravity's free fall g=-9.8 m/(s^2) is an acceleration (when you say that); light falls (like any other moving object, (ruffly) without air resistance); Free fall is not acceleration. (Free fall means you are experiencing no force so if you had an accelerometer with you, it would read zero.) Link to comment Share on other sites More sharing options...
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