swansont Posted November 19, 2012 Posted November 19, 2012 I think it's a mistake to minimize the "in terms of a few fundamental principles" part that prefaces "why the world is the way it is". Because it means you aren't asking, "Why are those the fundamental principles?" and that just reinforces StringJunky's earlier post.
michel123456 Posted November 19, 2012 Posted November 19, 2012 It has been established that a massless particle doesn't experience time. A particle of hypothetical negative mass should theoretically experience time reversal. So I understand that regular time is related to regular positive mass. IOW where you have mass you have time and you have gravitation. That should ring a bell.
swansont Posted November 19, 2012 Posted November 19, 2012 It has been established that a massless particle doesn't experience time. A particle of hypothetical negative mass should theoretically experience time reversal. So I understand that regular time is related to regular positive mass. IOW where you have mass you have time and you have gravitation. That should ring a bell. That doesn't mean that time is the source of gravitation. A massless particle still has energy, and energy is a component of the stress-energy tensor. No time, but gravity.
juanrga Posted November 19, 2012 Posted November 19, 2012 (edited) (My bold) Note: He is not addressing a fundamental article or phenomena like gravity....he is using the word world at the macro level and in the collective sense. Gravity is a part of the world and we use gravity to explain why things at the macro level do what they do. This discussion is not about the world...it is talking about a much more fundamental component of it and at that level things start to become more axiomatic ie resist deeper analysis. I think it's a mistake to minimize the "in terms of a few fundamental principles" part that prefaces "why the world is the way it is". Because it means you aren't asking, "Why are those the fundamental principles?" and that just reinforces StringJunky's earlier post. When one explains something (explanandum) in terms of something more fundamental, one take the fundamental as granted (explanan). In particular I have explained why the gravitational force is attractive using the spin of the gravitons (the force carriers). Of course, this open the door to a new question: why is the spin of gravitons just that? And again we will use something more fundamental to explain the spin of the gravitons. This is as both logic and physics work: we explain stuff using elementary stuff and next we explain elementary stuff using more elementary stuff. Regarding Weinberg, he is not referring to the world at the macroscopic level, nothing more far from reality! He is referring to fundamental aspects of the world, what he calls "principles". In fact he writes near the quote given above: I aim to present quantum field theory in a manner that will give the reader the clearest possible idea of why this theory takes the form it does, and why in this form it does such a good job of describing the real world. Emphasis in the original. Weinberg is completely right; descriptive science is just a part of science. Answering "why" is another part of science. Edited November 19, 2012 by juanrga 1
Arnaud Antoine ANDRIEU Posted November 19, 2012 Posted November 19, 2012 (edited) A particle of hypothetical negative mass should theoretically experience time reversal. Exactly. In French (again) I call this hypothetical particle "magnéto-particule". It represent the energie. So I understand that regular time is related to regular positive mass. Of course. It represent the effective "charge" moving in the time. IOW where you have mass you have time and you have gravitation. In the same way as the fusion of law. Cordialy. Edited November 19, 2012 by Arnaud Antoine ANDRIEU
michel123456 Posted November 19, 2012 Posted November 19, 2012 That doesn't mean that time is the source of gravitation. A massless particle still has energy, and energy is a component of the stress-energy tensor. No time, but gravity. that's confusing: gravity without mass. isn't that mass is the source of gravity? x I love your posts Juan. (really, no sarcasm)
D H Posted November 19, 2012 Posted November 19, 2012 (edited) that's confusing: gravity without mass. isn't that mass is the source of gravity? No. Mass-energy is the source of gravity. A bucketful of rocks gravitates, but so does a bucketful of photons. (Good luck, however, keeping the photons from being absorbed by the bucket's walls.) Edited November 19, 2012 by D H
Arnaud Antoine ANDRIEU Posted November 19, 2012 Posted November 19, 2012 Mass-energy is the source of gravity. This is true. And undeniable, I think that the realy origine provide form an electromagnétic field ; represeting the "field of energy". This electromagnétic field is a hypothetical integrated vector. It will represent the 5th type of element (invisible) and the source mechanic (mathematics). This is purely theoretical ! This "electro-field" above, can simply be the photons by example... -1
mooeypoo Posted November 19, 2012 Posted November 19, 2012 ! Moderator Note Arnaud Antoine, this thread deals with actual supported mainstream science, not personal speculations. Please keep your own interpretations of science to the speculation forum.
Arnaud Antoine ANDRIEU Posted November 19, 2012 Posted November 19, 2012 knowing that the photon dose only the visible face "in eyes" .. the rest work as well (invisible frequencies has any level we can imagine ..
D H Posted November 19, 2012 Posted November 19, 2012 I think that the realy origine provide form an electromagnétic field ; represeting the "field of energy".This electromagnétic field is a hypothetical integrated vector. It will represent the 5th type of element (invisible) and the source mechanic (mathematics). This is purely theoretical ! This is purely nonsense. We are supposed to be helping ChrisDK. Nonsense speculation is not "helping".
Arnaud Antoine ANDRIEU Posted November 19, 2012 Posted November 19, 2012 (edited) This is purely nonsense. We are supposed to be helping ChrisDK. Nonsense speculation is not "helping". oups, I'm really sorry for that. These writings are the worst I wort. of course the source of gravity (magneto-field above) works with the entropic forces. That's it. Cordialy Edited November 19, 2012 by Arnaud Antoine ANDRIEU
swansont Posted November 20, 2012 Posted November 20, 2012 When one explains something (explanandum) in terms of something more fundamental, one take the fundamental as granted (explanan). In particular I have explained why the gravitational force is attractive using the spin of the gravitons (the force carriers). Of course, this open the door to a new question: why is the spin of gravitons just that? And again we will use something more fundamental to explain the spin of the gravitons. This is as both logic and physics work: we explain stuff using elementary stuff and next we explain elementary stuff using more elementary stuff. And eventually there is no stuff that is more elementary, and "Why?" cannot be answered. 1
Art_Vandelay Posted November 20, 2012 Posted November 20, 2012 I assume the answer to this is yes, but I'm not sure why. Do electrons exert/interact with gravitational force directly? They are referred to as 'point-particles' and are not made of other particles, and are also force-carrying particles, correct? But then they also have a minute amount of mass.. so does that mean they have a higgs boson? That might be a whole other tangent, but why does an electron have mass at all, and are there any practical applications for the kinds of equations that involve it? I know I've missed something along the way
juanrga Posted November 20, 2012 Posted November 20, 2012 that's confusing: gravity without mass. isn't that mass is the source of gravity? It was explained in #4 how massless particles such as photons generate gravity and are affected by gravity. And eventually there is no stuff that is more elementary, and "Why?" cannot be answered. In the first place the logical structure is hierarchical and we can surely answer "why?" at different levels (this is why Weinberg emphasizes that theoretical physics is about answering "why?"). In the second place if you consider a linear logic, then either you consider infinite levels of "why?" or an underlying fundamental level which cannot be answered. However, if we consider nonlinear logics then it is admissible to obtain a self-contained logical structure. In fact, structures of this type are under research in meta-mathematics with the goal of giving mathematics its own mathematical foundation. The goal is to define and give a foundation for mathematics within the scope of mathematics! This involves the use of recursive theory and other advanced topics. 1
michel123456 Posted November 20, 2012 Posted November 20, 2012 (edited) It was explained in #4 how massless particles such as photons generate gravity and are affected by gravity. In the first place the logical structure is hierarchical and we can surely answer "why?" at different levels (this is why Weinberg emphasizes that theoretical physics is about answering "why?"). In the second place if you consider a linear logic, then either you consider infinite levels of "why?" or an underlying fundamental level which cannot be answered. However, if we consider nonlinear logics then it is admissible to obtain a self-contained logical structure. In fact, structures of this type are under research in meta-mathematics with the goal of giving mathematics its own mathematical foundation. The goal is to define and give a foundation for mathematics within the scope of mathematics! This involves the use of recursive theory and other advanced topics. (bolded mine) Sure you did explain, I don't understand. You wrote: The source of gravity is the stress-energy-momentum tensor whose components are energy, momentum, and stress. In the non-relativistic approximation it reduces to mass. The tensor explains why massless particles as photons are affected (and generate) gravity. (bolded mine) What is momentum? I guess not this: In classical mechanics, linear momentum or translational momentum (pl. momenta; SI unit kg m/s, or, equivalently, N s) is the product of the mass and velocity of an object. because here is mass again. You must be talking about some other momentum without mass. And I thought that photon have no rest mass (because it makes no sense) but that a bunch of traveling photons have mass. Edited November 20, 2012 by michel123456 1
swansont Posted November 20, 2012 Posted November 20, 2012 What is momentum? I guess not this: because here is mass again. You must be talking about some other momentum without mass. And I thought that photon have no rest mass (because it makes no sense) but that a bunch of traveling photons have mass. Photons are not part of classical physics, and do not conform to the classical definition. The momentum of a photon is E/c. It has no mass (by the standard definition), alone or in bunches.
juanrga Posted November 20, 2012 Posted November 20, 2012 (edited) (bolded mine) Sure you did explain, I don't understand. You wrote: (bolded mine) What is momentum? I guess not this: because here is mass again. You must be talking about some other momentum without mass. And I thought that photon have no rest mass (because it makes no sense) but that a bunch of traveling photons have mass. That paragraph and the first equation [math]\mathrm{p}=m \mathrm{v}[/math] in the wikipedia are only valid for massive particles at low velocities. For an electron moving at about [math](1/10) c[/math] its momentum is not given by the Newtonian [math]m_e \mathrm{v}[/math] but by the relativistic [math]\mathrm{p}=m_e \gamma \mathrm{v}[/math] with [math]\gamma[/math] being the time-dilation factor. Below in the section 3.2 they give the correct momentum "for massless particles such as photons". The answer is [math]p=E/c[/math]. The mass of a bunch of photons is zero and [math]P=E_T/c[/math] holds but now with [math]P[/math] being the total momentum and [math]E_T[/math] the total energy. Edited November 20, 2012 by juanrga
michel123456 Posted November 21, 2012 Posted November 21, 2012 (edited) That paragraph and the first equation [math]\mathrm{p}=m \mathrm{v}[/math] in the wikipedia are only valid for massive particles at low velocities. For an electron moving at about [math](1/10) c[/math] its momentum is not given by the Newtonian [math]m_e \mathrm{v}[/math] but by the relativistic [math]\mathrm{p}=m_e \gamma \mathrm{v}[/math] with [math]\gamma[/math] being the time-dilation factor. You will say I am stubborn but [math]m_e[/math] is mass. Below in the section 3.2 they give the correct momentum "for massless particles such as photons". The answer is [math]p=E/c[/math]. The mass of a bunch of photons is zero and [math]P=E_T/c[/math] holds but now with [math]P[/math] being the total momentum and [math]E_T[/math] the total energy. Ah, at last mass is gone. What are the units of [math]P[/math] ? N/s, or kg m/s IIRC. Edited November 21, 2012 by michel123456
D H Posted November 21, 2012 Posted November 21, 2012 What are the units of [math]P[/math] ? Don't get so hung up on units. The International System (SI) of units is predominantly a classical (pre-relativistic, pre-quantum mechanics) set of units. That photons have zero mass but non-zero momentum doesn't make a bit of sense in Newtonian physics. It makes a whole lot of sense in relativistic mechanics.
michel123456 Posted November 21, 2012 Posted November 21, 2012 (edited) Don't get so hung up on units. The International System (SI) of units is predominantly a classical (pre-relativistic, pre-quantum mechanics) set of units. That photons have zero mass but non-zero momentum doesn't make a bit of sense in Newtonian physics. It makes a whole lot of sense in relativistic mechanics. I could agree if relativistic physics had set a new unit system. But it has not. what makes "a lot of sense" to you don't make sense at all to me. As I stated above, I thought that "photons in motion" had mass and that a photon's rest mass was null (because a photon is always in motion). Is the following equation correct for a photon? [math]E= m_{rel} c^2[/math] Edited November 21, 2012 by michel123456
juanrga Posted November 21, 2012 Posted November 21, 2012 (edited) You will say I am stubborn but [math]m_e[/math] is mass. Effectively [math]m_e[/math] is electron mass. Ah, at last mass is gone. What are the units of [math]P[/math] ? N/s, or kg m/s IIRC. This is an excellent point. Precisely I clarified the meaning and usage of different 'equivalent' units in my recent article about SI derived units. Look the part starting with (bold added here) A derived unit can often be expressed in different ways by combining base units with derived units having special names. The "joule", for example, may formally be written "newton metre", or "kilogram metre squared per second squared". This, however, is an algebraic freedom to be governed by common sense physical considerations; in a given situation some forms may be more helpful than others. Consider the half dozen of examples given in the next paragraphs of the article; for example, the quantity "torque" may be thought of as the cross product of force and distance, suggesting the unit "newton metre", or it may be thought of as energy per angle, suggesting the unit "joule per radian". Regarding momentum it seems natural to use [kg m/s] for massive particles such as the electron but [J s/m] for massless particles such as the photon. The unit [N s] makes more sense when dealing with forces at the classical level (e.g., when solving Newtonian equations using the observed forces) but I would not recommend its use at quantum level (e.g. when solving the Schrödinger equation where there are no forces involved at all). Is the following equation correct for a photon? [math]E= m_{rel} c^2[/math] Formally it is correct, because it is a tautology [math]E = E[/math]; recall that relativistic mass is defined by [math]m_{rel} \equiv E/c^2[/math]. However there are a number of objections to the use of relativistic mass concept and it is deprecated in most of modern physical literature (in all of fundamental literature). Some objections to its use are given in the Am. J. Phys. paper Mass versus relativistic and rest masses. It is interesting that Einstein Never Approved of Relativistic Mass The modern equation replacing the above is [math]E= \sqrt{m^2 c^4 + p^2 c^2}[/math] For a photon [math]m=0[/math] and you obtain the well-known expression [math]E = |p| c[/math] found in the Wikipedia page that you cited above. Edited November 21, 2012 by juanrga 1
Arnaud Antoine ANDRIEU Posted November 21, 2012 Posted November 21, 2012 (edited) Consider the half dozen of examples given in the next paragraphs of the article; for example, the quantity "torque" may be thought of as the cross product of force and distance, suggesting the unit "newton metre", or it may be thought of as energy per angle, suggesting the unit "joule per radian". suche The modern equation replacing the above is [math]E= \sqrt{m^2 c^4 + p^2 c^2}[/math] [math]E= \sqrt{m^2 c^4 + p^2 c^2}[/math] or [math]E^2 = {m^2 c^4 + p^2 c^2}[/math] gives the linear momentum. I know that you know that, but for a massless particle the real ones are : (pictures below). This mean that the mass it really set to work with the only value ² above. So, for one photon we can say : 1 packet = λ (complete period of it) distance = d (m/s) E = λ d² 1 packet = "mass" or "massless" or "quantum" or "period" or "charge" ... = 1λ = 1 photon _______________________________ x = source E.utile = E - = energy E.effec. = d ¤ = photon packet = λ 1 photon : x----------¤ E = λd² 3 photons : x-------¤¤¤ E = (nλ)d² The following does work as 1λ = 4πω http://www.scienceforums.net/topic/70028-the-mass-of-photon/page__st__40__p__714470#entry714470 In conclusion this means that the energy goes faster than the speed of light. Ev=1λc² and Ev=(nλ)c² _______________________________ Edited November 22, 2012 by Arnaud Antoine ANDRIEU
juanrga Posted November 22, 2012 Posted November 22, 2012 (edited) [math]E= \sqrt{m^2 c^4 + p^2 c^2}[/math] or [math]E^2 = {m^2 c^4 + p^2 c^2}[/math] gives the linear momentum. I know that you know that, but for a massless particle the real ones are : (pictures below). As stated in my previous message the above expression also holds for massless particles. Substituting [math]m=0[/math] we obtain the well-known [math]E=|\mathbf{p}|c[/math]. The equations in your attached pictures are compatible with this. For instance using [math]E=h \nu[/math] we obtain [math]|\mathbf{p}|= E/c = h \nu/c = \hbar |\mathbf{k}|[/math]. Multiply by unit vector both sides and you obtain your second attached equation relating momentum and wave vectors. http://en.wikipedia....ical_properties Edited November 22, 2012 by juanrga 1
Arnaud Antoine ANDRIEU Posted November 22, 2012 Posted November 22, 2012 As stated in my previous message the above expression also holds for massless particles. hello juanrga. I knew you did it. This was to come to : "This mean that the "mass" it really set to work with the only value ² above". And more, I was wrong about that sentence. In fact It is the "distance" which operates ², and no "mass" in my example. Cordially.
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