Pete

To provide an answer you need to define the term natural reaction force since I too have never heard of this term. Pete

Your analogy is quite flawed. There is no uncertainty intrinsic to Newtonian mechanics whereas in quantum mechanics there most definitely is. The very meaning of the wave function as the square being a probability density demands uncertainty. It can't be taken out no matter how its formulated unless the meaning of the wave function is changed, i.e. if the postulates of quantum mechanics were different. Pete

That is not true for all physicists. In fact if one were to look in the textual literature they'd see that m is often used to mean relativistic mass. And it would be unwise to forget about it because no other definition works in all generality, especially not the "mass = proper mass" definition. That only works for particles and closed systems. In fact its used in over 50% of the textual literature (more like 70% in fact). Gary Oas did a study on this. His results are at http://arxiv.org/abs/physics/0504110 A bibliography of his work is at http://arxiv.org/abs/physics/0504111 Pete

Not at all. The very definition of uncertainty is statistical by nature. Is this what you think the definition of uncertainty is? If so then its wrong. There is a very precise mathematical formula for the uncertainty of a physical observable. It is dependant on the particular initial state that the system is in. A different initial state thus gives a different uncertainty for the observable. No. This is all wrong. Even throwing a die has a non-zero undertainty associated with the outcome. I'm not sure what you're asking in this question but let me say this - If you were to consider a state whose position wave function which has a width of delta x and and whose momentum wave function has a width of delta p then its the product of those widths that satisfies the uncertainty principle. Pete

I don't know people here that well so I'll be unable to determine who is doing things like that. Even then I'll fail to understand the reason! Pete

As far as uncertainty goes, no. That is incorrect. Just to be sure we are on the same wavelength please post the definition of the term uncertainty as you understand the term. Thanks. But I don't understand what you mean when you say that that one can't ever be free from having to deal with superposion. Any quantum state for a free particle can be in an eigensate corresponding to one physical observer but be in a superposision of the eigenstates of a complementary observable. No. I think that you have the wrong idea what uncertainty is. That's not far from the truth. Pete

To be precise; a photon has zero proper mass and a well-defined, non-zero inertial mass (as well as a non-zero active gravitational mass and non-zero passive gravitational mass) which I refer to simply as mass. I don't understand your reason for saying this. What exactly do you mean when you say photons do no exert Newtonian Gravity? If you mean that photons don't exert a gravittional forces on other particles then that is clearly wrong. Of course I'm assuming that a photon can be treated as a localized EM field that propagates through space with speed c. You do understand that a beam of light generates a gravittional field, don't you, Pete

The title of the section in which this is found is entititled Relativist mass. Pete

I agree. It is true that the majority of people believe that relativistic mass is defined as m = E/c2. That is unfortunate. When momentum is defined as p = mv where m is a proportionality factor the m is called the mass of the particle. When it is defined as such people often refer to it as relativistic mass. The term is well defined in the relativity literatare as the m in p = mv. For example; if you have the text Special Relativity, by A.P. French then take a look at the footnote on page 16 which reads If you have The Feynman Lectures on Physics - Vol. I then turn to page 16-6 which reads Similarly, from The Theory of Relativity, by C. Moller, page 65 Those are just three examples. It can be traced back to Richard Tolman who seems to be the first physicist to employ this concept. Herman Weyl defined mass in this way. In what sense? Absolutely. A similar notion is employed if one were to defined inertial mass as the ratio m = F/a. One takes the limit a -> 0. The answer to this question is identical to the question What is mass good for? An important fact in relativity is that stress/pressure contributes to inertia, i.e. the momentum of a body is a function of both energy and stress. This fact would not be reflected if one were to define inertial mass (aka relativistic mass) as the ratio E/c2. This definition holds in all conceivable cases. Using this definition it becomes obvious that pressure is a source of gravity. That's how one obtains the expression rho = (u + 3p) for the active gravitational mass of a body. Strangely enough the inertial mass density is rho = (u + p) I haven't figured out why there is a difference yet. One of the puzzles I am seeking to find a solution to. Does anybody know why there is a difference? It is important to keep in mind that m = p/v is the definition of relativistic mass whereas m = \gamma m0 is an equality. Once mass is defined by m = p/v the equality can be derived by applying the principle of conservation of momentum to two particles which undergo an elastic collision. It can then be shown that, under certain circumstances, the relativistic mass is related to total inertial energy by E = mc2. If the system under consideration is a particle then that relation will hold. If the body is an isolated system then it will hold in that case too. However if a body is under stress then E = mc2 will not hold. However the mass is still related to the momentum as m = p/v. It will always hold in fact because its a definition and not an equality. I imagine that what I said above regarding the fact that E = mc2 is not always true will come as a surprise to most people. For that reason I think it'd be a good idea if I quote a special relativity text. From Introduction to Special Relativity, by Wolfgang Rindler, page 150 That this is true is readily seen by examining the components of the stress-energy-momentum tensor. In a frame of reference which is moving relative to a body the momentum density will be a function of the x-component of stress. This is also the case in the new relativity FAQ regarding the relativistic mass of the photon which is at http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/photon_mass.html Pete

The gravitational field of the sphere is the same as that of a particle. However the force on one sphere due to another is not the same as that of two point particles. This is because the sphere is an extended body and as such the tidal effects of one sphere will have an effect on the other sphere. Pete

I disagree. In the first place Einstein never discussed a photon in a box. Einstein's derivation centered on the emission of radiation. Although the radiation can be thought of as consisting of a stream of photons it is not neccesary to obtain this radiation from an atomic transition. The radiation could have been produced by accelerating charges. There is an approximation that Einstein used. I.e. he treated the box as if were a rigid body. However the non-rigidity of the box can be ignored. If that were true the the pinciple of conservation of momentum would be violated. All that is required is that energy be conserved. There is no requirement which says that all the energy must be carried off by the photon. Some of the energy can go into the kinetic energy of the radiating atom. I.e. the change in energy equals the sum of the photon energy plus the atoms's kinetic energy. Einstein's photo-in-a-box thought experiment has been analyzed by several physicists in the past. There is an article in the American Journal of Physics which explains the experiment from several points of view, including the radiating atom idea that you're talking about. The article is Inertia of energy and the liberated photon by Adel F. Antippa. Am. J. Phys. Vol. 44, No. 9, September 1976 Section 7 is labeled The Radiating Atom Pete

I don't know about whether its not used to a great extent or not (because I'm unsure as to the precise meaning of that phrase). That's difficult to say. I myself have researched this subject and found that its not possible to define mass to refer to what is otherwise called rest mass (which I myself refer to as proper mass for various reasons). The mass = relativistic mass is completely consistent, logical and works in all possible circumstances. I've explained my reasons in a paper I wrote on the subject. For those who would like to take a look at it then you can find it online at - http://arxiv.org/abs/0709.0687 There have been many articles written on this topic. One is online at http://arxiv.org/abs/physics/0504110. The article is called On the Abuse and Use of Relativistic Mass, by Gary Oas, last revised 21 Oct 2005. In this article the author makes a very relevant observation, i.e. The results of the authors research shows that 75% of the literature in the survey employs the concept of relativistic mass. Scientific journals were left out of the survey so as to keep the survey of a manageable size. I'm not sure what is meant by "generally be assumed." Each instance must be evaluated individually. One can't merely assume that it means rest mass. I'm sure you didn't mean that though. I believe that you meant that if one were to take a survey then you meant that in most cases the term would mean rest mass. However since most of the literature in the survey employs relativistic mass, and since authors who do so simply call it mass, then your assumption is generally not true. Especially since a great deal of the relativity literature was published several decades ago. Some of the most important relativity texts of all use the term "mass" to mean relativistic mass and they didn’t even mention the term "relativistic mass" at all. Case in point; Gravitation by Misner, Thorne and Wheeler. I examined that text for such usage and came across two important examples. The first had to do with the symmetry of the stress-energy-momentum tensor. The second was in their statement of the source of gravity. In those two, very important, cases they used the term mass unqualified to mean relativistic mass. I myself did a short survey and have recorded the results in this web page http://www.geocities.com/physics_world/ref/relativistic_mass.htm Pete

The relativistic mass of an object is defined as the ratio of the magnitude momentum of that object to the object's speed. Inertial energy (defined as the sum of kinetic energy and rest energy) is defined otherwise. For that reason relativistic mass it not another name for energy. In fact energy includes potential energy of position which is not included in relativistic mass. Relativistic mass is not even proportional to inertial energy in general. I.e. if one were to calculate the general expression for mass density and compare it to the general expression for inertial energy density they'd see that they are quite different. Pete

No. It is possible to measure both the position and momentum of a particle at the same time with arbitrary precision (but not with exactness). Uncertainty is related to an inherent property of the state itself. A different state will have different values of, say, the uncertainty in position. Uncertainty is determine solely by what state the system is in. It is a statistical quantity. An individual measurement has nothing to do with uncertainty. I.e. you can't reduce the uncertainty in position by using more precise instruments. The only way to reduces uncertainty is to utilize a different quantum state, one for which the position is more localized. Consider a spin-1/2 system which is in a superposition of the spin-up and spin-down states. Even though the spin itself can be measured with exact results (there are only two possible values that can be measured) the uncertainty in the spin is not zero. Pete

Photons have both active and passive gravitational mass. That means that they generate gravitational fields and are effected by gravitational fields. Mass is the source of gravity. The stress-energy-momentum (SRM) tensor provides a complete description of gravity. It is not enough to associate the components of the SEM tensor as sources because that is insufficient. Pressure is also a source of gravity. The active gravitational mass density rho of a relativistic fluid is a function of the energy density u and its pressure p. i.e. (let c = 1) rho = u + 3p No. Photons moving parallel to each other generate a gravitational field which doesn't exert a gravitational force on photons moving in the same direction (the gravitational force is velocity dependant). If they are moving in opposite directions then there will be a force. Yes. Note that the only thing particle physicists work with are particles. As such it makes no difference what you call mass. But in general it does make a difference and physicists who work with relativity consider more than just particles and as such proper mass (aka rest mass) is not even meaningfull in the most general circumstances. Pete

That is correct. That is true. A vague question but I'll say no. The energy of a spaceship cannot come from itself. If it did then the energy of the spaceship would remain constant and its speed would never change. The way a spaceship accelerates is by ejecting matter. The momentum given to the ejected matter is balanced by the increased momentum of the ship. Hopefully the above explanation will help. What exactly is the source of your objection? Time dilation is the phenomena in which two clocks of identical construction tick at different rates. In order for there to be gravitational time dilation there must be two different clocks being compared. When the two clocks are at two different gravitational potentials then the clocks will run at different rates. Remember that we have to refer to two clocks to speak about time dilation. If we are speaking about a distant clock which is outside a gravitational potential Call this clock S) then a clock on the Earth's surface will run slower than a clock on the surface of Mars as compared to the clock S. Yes. To be precise - A uniform gravitational field is equivalent to a uniformly accelerating frame of reference. Or, similarly, the gravitational force is an inertial force. Yes, in the sense described above. No. No. In fact the twin paradox doesn't even need acceleration for it to occur. All that is required is non-symmetry of worldlines. If we concern ourselves only with the time readings on clocks then we can observe the readings on two clocks that are synchronized as they pass each other. E.g. let a clock be whizzing by the Earth in the direction of the destination and let the clock be set to zero when the clock passes Earth. When it passes by the destination let there be a second clock passing by the destination in the direction of Earth and let the clocks be synchronized as they pass. When the second clock passes by the Earth it will have a smaller reading on it than the clock on the Earth. Proper mass is invariant and relativistic mass is not. What "mass" means when unqualified can always be determined by the context in which it is used. The mass of a body is a function of its speed. The larger its value the greater the momentum of the body, i.e. mass defines momentum. When the momentum is larger then it takes a larger force to change it by the same amount that the same force changes it when the speed is less. The value of the mass is a function of velocity and becomes infinite as v -> c. Explained in another way - mass is a body's measure of its resistance to changes in speed. The greater the speed the greater the mass and thus the harder it is to accelerate. As the speed approaches the speed of light the energy required getting it to that speed approaches infinity. Time dilation and length contraction are geometric concepts and not dynamic ones and as such there is no relationship between them and energy. Only when one starts to bring energy into it can relationships between them be discussed. Pete

Conservation means does not change with time while invariance means does not change upon change in coordinates. Energy is conserved but not invariant. It seems as if you're thinking of an absolute rest frame in which kinetic energy is to be evaluated. That is not the case. There are no absolute frames of rest. What is this proof that you're referring to? Pete

A good understanding of vector calculus is required. For a very good understanding of EM one should learn special relativity and tensor analysis. Pete

Newton's law of gravitation F = Gm1m2/r2 is the force between two point particles. The force between two objects is found by using the principle of superposition and thus integrating over the source body and the body which the source is acting on. Pete

The value of the magnetic field between the poles increases. A full description would require a picture though. Pete

The Jahrbuch article was published in 1907, not 1909. Pete

Howdy My name's Pete. I'm into relativity and quantum mechanics. Nice to be here.

I don't understand what that means. Yep. That's the idea of Mach's Principle. How do you think the gravitational field is defined in Einstein's general relativity? Are you familiar with the principle of equivlence? It states that a uniformly accelerating frame of reference is equivalent to a uniformly accelerating frame of reference. Therefore if you're in such a field and drop a ball you can't tell whether there is a gravitational field present or if you're in an accelerating frame of reference with respect to an inertial frame. The same idea extends to arbitrary inertial frames and to non-uniform gravitational fields. Pmb

According to Einstein's General Theory of Relativity (GR)' date=' yes. In Newtonian physics, yes. In Einstein's GR, no. In GR there are no special frames of reference as their are in Newtonian physics. As such all frames of reference are legitimate and as such one needs to explain why a particle is experiencing an inertial force in any given frame. For a rotating frame the cause is that you're rotating with respect to the "distant stars" and as such the matter in the universe which is rotating around you gives rise to a gravitational field. This was demonstrated by Einstein quite early on in GR. E.g. suppose you're inside a hollow massive spherically symmetric shell and the shell is very massive. If you now started to rotate the shell the matter inside would start to rotate as well (something that doesn't happen in Newtonian physics). This is called "frame dragging". If an observer inside decided to stay in your (the outside distant observer's) frame, whereby I mean that he choose a frame in which he is not rotating with respect to you, then to him there will be a measureable gravitational field. This is identical to the situation of a person choosing a frame which is rotating with respect to an inertial frame. No. Newton would agree with you. Einstein would disagree with you. Einstein touched on this point in an essay he wrote in the February 17, 1921 issue of Nature More on this can be found in The Meaning of Relativity, A. Einstein, pp 100-102 Pete

No. The term cannot apply in this manner according to the current concepts of space and time and the meaning of "move". Pmb