Pete

When I confirmed I was not talking about measurement, but rather the state of the particle (or thought I made that clear) you continued to disagree. If that's the confusion then we've probably been talking past each other.When I was responding to I meant that I never said it's only a property of a measurement. Can you tell me what I said that led you to this conclusion? Pete

The analogy is still valid. Just because it addresses quantized values rather than continuos ones makes no difference since what you said so far has not differentiated the two. If I explained that I'd be restating what I already did above. Its not that I really mind doing it if I believe it would be instructive to do so but I recently damaged my back and any posting I do not is at the cost of incredible amounts of agony. Sorry. I'll try to be clearer tomorrow, hoping I'll feel better. If you find that my comments in this thread are inconsistent or lead to observations that are contradictory to that predicted in QM then please specify in what way. Thanks. I never said nor implied that. In the mean time I have a question for you; Have you ever seen and/or followed the derivation of the Heisenberg-Robertson Uncertainty Relation? Analysis of the derivation will provide one with all they need to know in order to be able to interpret its meaning, application and domain of applicability. swansont - I think I might now understand what your reservations are with what I've said so far. Perhaps you believe that since individual measurements of position and momentum can be made precisely that this would lead to a classical trajectory and thus deterministic motion in QM, contrary to what one knows of quantum theory, is that the case? If so then that what I've been saying does not lead to such a conclusion. Let me elaborate. First: By simultaneous measurements of two obsevables one does not neccesarily take the measurements at exactly the same time since measurements that close to each other may be impossible in practice and it would give the wrong idea of the mearement process of incompatible observables, although not with compatible observables. What is actually meant is that one measures observable A and then, in an infinitesimally short amount of time dt later, one measures the observable B. Now let's apply this to the conjugate observables X and P. If one measures the position of a particle then the wave function collapses into an eigenstate of position. The corresponding wave function is a Dirac delta function. The Fourier transform of the Dirac delta function is a constant function. Thus when one changes from position space to momentum space one gets a uniform probability density of measuring mometum. This means that [math]\Delta P[/math] is infinite. If one then measures the momentum in an infinitesimally short amount of time later one gets a specific value of mometum. The system then collapses into an eigenfunction of mometum (another Dirac delata function). The corresponding position wave function is then a constant. Thus if one measures the position and then momentum and then position again there will be a discontinuity in the value of position. Therefore there can be no classical notion of a trajectory. If one meausures x and then p then in a second run with an identical system in which one also measures x then a subsequent postion won't neccesarily be the same value of p one got before. It would be much easier to temporarily place this QM analysis on hold and instead discuss probability theory and the meaning of uncertainty, expectation etc. first. We could then easily see what it means to measure an exact value of A when there is a non-zero uncertainty in A. The meaning will not change when we go back to quantum theory. Does anyone here who is following along have difficulty understanding the fact that situations exist where the possibility exists (in probability theory) of measuring exact values of a variable even when the uncertainty in that variable is non-zero? Pete

The term "eigen" means "proper". Seehttp://germanenglishwords.com/rlge.htm Therefore the terms proper-time and eigen-time mean the same thing. I myself only use the former. In fact I've never heard of the later being used. By the way its the ratio of total differentials not the ratio of partial derivatives that is normally found in relativity. What purpose does the ratio of partials serve?? Pete

That was an example given so as to make the physics as clear as possible. I used an observable which is quantized for clarity. The die analogy refers to quantum mechanical systems where the observable is quantized, such as spin. In such cases the uncertainty of the variable is not neccesarily zero. I think such systems are very good to think of when considering the notion of uncertainty. I'm curious. What do you think that collapsing the wave function means/refers to? I was speaking on the more general situation regarding observables in QM and the general expression for the uncertainty principle. An arbitrary quantum state can be expanded in terms of position eigenkets. The coefficients in such an expansion is the wave function evaluated at the relevant position. Once the position is measured the system falls into an eigenstate of position. This is what the term "collapse" actually means. If the quantum state is not in a superposition then it is in an eigenstate of an observable (and can therefore be repsented in another basis as a superposition of these eigenstates and therefore whether a state is in a "superposition" is really a basis dependant phenomena). If the system is in such an eigenstate then an imediate measurement of the observable will yield the cooresponding eigenvalue with certainty. If an electron is moving freely then it can be in such a superposition. If it is instead in an eigenstate of an observable of, say, position then it cannot be in an eigenstate of momentum and must therefore be in a superpostion of momentum eigenstates (when the quantum state is represented in momentum space). Let me restate that for clarity - If a system is in an eigenstate of observable A then it cannot be in an eigenstate of observable B if A and B do not commute, like x and p for example. Note - Any wave function is in a superposition of position eigenstates unless the wave function is a Dirac delta function. Pete

What does this have to do with perpetual motion? The term perpetual motion quite literally means movement that goes on forever. There are no laws in classical mechanics which prevents this. However the term usual is taken to refer to any closed system that produces more energy than it consumes. Such systems violate the second law of thermodynamics. Which meaning are you referring to? Pete

On the contrary. I know it better than you do from what you've posted so far. In the first place if one were to look that relationship I referred you to up in a physics text then it would be referred to as Coulomb's Law. The term Coulomb force is not used that often' date=' but when it [i']is[/i] used it always refers to the electric force (in the absense of a magnetic force). Its also defined in these references http://scienceworld.wolfram.com/physics/CoulombForce.html http://www.britannica.com/EBchecked/topic/140084/Coulomb-force# Are you saying that the Enclyclopedia Britannica has the wrong definition? However if you wish to prove this wrong then find an instance in a physics text/book/article and post the quote here along with the reference. You don't need to tell me what I posted and what I mean. I know it all too well. Frankly you're wasting our time and when you complain about sematics when nobody here really cares about it. All you needed to do is say "force of friction" . Regarless of what you call it you have used it incorrectly and have refused to address the proof that was given to you. Let me remind you of where this came up. In response to an erroneous assertion made by you I posted the following explaination. In response to that you responded with While I was writing this post I did a search on the internet on the term Coulomb assumption and the only thing I could find was in regards to a relationship regarding stress. I recommend that you pay more attention to your assumptions rather than criticizing what you think other people did wrong. What I explained is all about frictional forces which act when a body is sliding. Hence the term coefficient of kinetic friction. The term "kinetic" here refers to the fact that the formula pertains to sliding bodies. Something which your response indicated that you did not understand. What I explained to you is, in all instances in the physics literature, defined as the Coulomb Force. It never means Coulomb friction force or whatever. Please look these terms up in a physics text before you make any further attempts at "correcting" anybody. In any case I never heard/read the term "Coulomb friction force" before. Perhaps because no physics teacher I ever had used that term and no physics text/book/article I ever read used it. For that reason and that reason only I'm glad you mentioned it. Thanks. That is clearly wrong. There is absolutely no reason why, in principle, something can't rotate without frictional forces acting on it. Only in practice does one have to account for fruiction and then only when the frictional force are large enough to affect the results of a computation/prediction. For example; superconductors can be levitated in a magnetic field in a hard vacuum and set rotating. The frictional forces are neglegible (i.e. they are virtually absent for all practical purposes) in such cases and the rotational energy is very well defined. Rotational energy is not dependant on the presence of frictional force. There is nothing in the theory of mechanics that would imply that. Consider the rotational energy of the astronomical bodies such as moons or asteroids! There are virtually no frictional forces there either but one can't say that there is no rotational kinetic energy either. Your comments above regarding rotational energy indicate that you have a lack of understanding of basic physics so I'd like to ask you exactly how far in physics did you say you went? Did you finish the PHY 101-103 series? If so then what physics text did you use in that sequence? At least post a reference to one text that you learned/are learning from. I'd like to as the original poster what the purpose of this question was? Was the question posted to inquire about the dependance of the rider's mass on acceleration? Or was it to ask about all these frictional forces? If its the later then there is no answer until we know what frictional forces you wish to take into account, what accuracy are you considering, whether the bicycles are identical etc. I mention accuracy because if you don't have instruments which can distinguish 1 microsecond of travel time and the frictional forces turn out to effect times of less than 1 microsecond then it'd be worth knowing that. Pete

When you use the term uncertainty in that way you are using it to mean something different than it means in uncertainty in x etc. You are using it to mean something synonymous to probability. Perhaps some authors do use the term uncertainty in some instances to mean the same thing that you do, I just haven't see anyone do that. I do know that [math]\Delta x[/math] is not defined that way though. Using it to mean two different things in the same context would lead to all sorts of problems. For this reason the term probability is used for this instead. I also recall that I originally mentioned this due to someone implying that one can't precisely measure an observable quantity. People rarely, if ever, use the term "know" in quantum mechanics because it raises all sorts of philosophical problems. All that one can speak of is what is measured, its probability, is expectation and its standard deviation (i.e. uncertainty). This is equivalent to saying that when you roll a fair die and a six comes up that you really don't know its a six. That is not what quantum mechanics has to say about this. Quantum mechanics would say that the position is not determined until you measure it where at that point it becomes determined (if it wasn't already in an eigenstate of position). This phenomena is known as the collapsing of the wave function. And all this talk of uncertainty only applies to quantum states which are not eigenstates of the observable. If a state is in an eigenstate of, say, energy E, then there is no uncertainty in energy in that the value that will be measured has a 100% probability of having that value. Pete

General Relativity (GR) doesn't say that. GR states that tidal gradients and spacetime curvature are the same phenomena. I see no problem in reconciling GR and QM in this respect. By the way its quantum field theory (QFT) that speaks of gravitons. Pete

The term speed refers to a spatial displacement which occurs during a unit of time. Since there is no spatial displacements associated with what you're asking then the question makes no sense. However the rate a which time flows can certainly change with the observers choice of reference frame. Pete

Measurement. A single photon can be localized within a region arbitraryl smaller than the width of the wave as defined by the uncertainty [math]\Delta x[/math].I should mention that what I've said above applies to non-relativistic quantum mechanics. If the subject is about photons then quantum electrodynamics is the correct theory to apply and that's not something I've learned ... yet. To graphically illustrate this point I created a diagram which I'm sure plenty of you have already seen if you've studied quantum mechanics. Please see http://www.geocities.com/physics_world/qm/image_gif/qm15-im-01.gif This is a diagram which shows a screen where photons hit one at a time and their positions are recorded. Those positions are shown by the dots. Let [math]\delta x[/math] denote the error that exists in measuring the position of the photon. The physical meaning of the wave function is that the magnitude of the sqaure of it is proportional to the probability of measuring the photons position at a location x within an interval of dx. The quantity [math]\Delta x[/math] gives an idea of how the photons are not predetermined to be located at a particular place. Most of the photons are near the vertical axis and the probability of them being found further from the axis decreases with distance. The "spread" of the area where "most" of them are contained is denoted by [math]\Delta x[/math]. It is [math]\Delta x[/math] that is the uncertainty and not [math]\delta x[/math] as one might think. I was discussing this subject a while back with some people and this same subject came up. I created the following web page to explain myself on this topic. The page is at http://www.geocities.com/physics_world/qm/probability.htm It is interesting to review the example I of a single fair die. If someone asked you "What is the uncertainty in the measurement in the number shown on the die?" One might answer "Zero since when the die is rolled and I see the face I am 100% sure of what number I read." The actual answser is that there is a finite undertainty in the number of dots on the face. The example demonstrates the meaning of all this and why one has to be sure of what one is asking. The two questions Q1) What is the uncertainty in the measurement in the number shown on the die? Q2) What is the error in measurement of the number which appears on the die's face when it is rolled. The answer to Q1 is 31.4. The answer to Q2 is zero. Pete

The link that ajb gave was to the solutions for geodesics in a Schwarzchild spacetime, which is the spacetime that you are referring to. The paths that you seek are the spatial portions of the geodesics. Once those are obtained from the notes you can simply change them to Cartesian coordinates. Do you need help in solving these equations explicitly? Pete

Uncertainty is not a lack of precision, The uncertainty of an observable has a precise mathematical expression which is a function of the state vector. Do you know what that expression is and how its value is determined from measurements made in the lab? And it need not be related to a wave function. E.g. its possible that a state can exist for spin in which the uncertainty in the spin is zero, even though the value of a particular measurement of spin can only give one of two values. Can you tell me where in the Feynman lectures that is located? No. The particle property of a photon refers to the fact that a photon has no spatial dimensions. The wave packet defines a probability distribution. The smaller the wave packet the more locallized the photon is in space. The wave function is a superposition of various quantum states, each state corresponding to a photon of a particular value of energy having a specific phase. Each sinusoid corresponds to a different state where a state is defined by a freqency and phase. Only when a measurement is done will the state fall into an eigenstate corresponding to the observable measured. The wave packet will then cease to exist. Ideally the wave function becomes a Dirac delta function. Pete

If that were true then what is the distinction between them? Pete

My opinion? The definition of the term Coulomb force has rarely, if ever, be questioned. Its found in all textbooks on electrodynamics. The mathematical expression for it is given in Eq. (1) at http://scienceworld.wolfram.com/physics/CoulombForce.html Pete

Huh????? Please please post the definition of of the term uncertainty as it pertains to quantum mechanics. Pete

On the contrary, there is overwhelming evidence of what I've just explained. If you were to actually post the definition of uncertainty then what I said would be glaringly obvious. So in your the next post I request that you, or someone else, please post the definition of uncertainty. Thank you! I'm merely explaining the meaning of ther term uncertainty. Its meaning is quite often misundestood and more often than not confused with the concept of error in measurement. Did you notice what you did here? You actually agreed with me. The distribution on the photographic plate is a statistical oine. It is totally impossible to determine what the uncertainty is until a distribution pattern is established. Once that is done then it becomes quite apparent that the uncertainty in position cannot be defined for a single instance in the measurement of the location of a particle. That is not what uncertainty means. In the comment you just made you were implicitly refering to probability, not uncertainty. The uncertainty of a roll of a die cannot possiblly be measured by a single roll of the die. To determine the uncertainty one keeps rolling the die and recording the number rolled. From that data one then calculates the standard deviation of the number rolled. That will give you the uncertainty in the number rolled. It sghould be quite apparent that the uncertainty (aka standard deviation) cannot be determined by a single roll, nor would it be meaningfull to even talk about unless a large number of rolls is executed and the results recorded. Question: You do understand that the term uncertainty is merely a different name for the term standard deviation, right? To summerize - Individual measurements of observable quantities does not have an uncertainty associated with them. Only quantum states themselves have an intrinsic uncertainty associated with them. Different quantum states -> Different values of the uncertainty of a particular observable. Pete

Any decent text will refer them as Cartesian tensors rather than simply tensors and for a very good reason. But most texts will qualify it with "Cartesian" first and then drop the qualifier when it is agreed upon to be undrestood. The class of tensors which are defined in terms of generalized coordinate transformations are most often called tensors, i.e. tensor is shorthand for general tensor. It makes life easier if we agree on what the qualified is and then simply not use it. Unless a particular definition is agreed on it will be impossible to get past these semantics. You don't honestly believe that someone can succesfully argue a point by merely repeating it, do you? I mean I can keep repeating that the term "tensor" refers to a multi linear map from vectors and dual vectors to real numbers and whether it is a general tensor or a Cartesian tensor will depent on what the domain of this map is, i.e. whether it maps Cartesian vectors and 1-forms or does it map general vectors and general 1-forms. Of course all this will do is to divert our attention from the definition of tensors to the definition of vector and their duals. Can you please quote the definition of tensor the Byron and Fuller give in the text you referrenced? Thank you. If you could scan and and E-mail the scanned page then that would be even better. I'm curious as to why you didn't respond to the question I posed above. I'll repeat my question - Doesn't the fact that such well known physicists and physics texts refer to Cartesian tensors being tensors give pause and/or make you wonder about whether you should rethink your position at least a little? Let me clarify the position I have stated in more detail. Surely you're aware of the idea that there are two different but equivalent definitions of tensor in the literature. That definition refers to tensors as multilinear maps from vectors and dual vectors to real numbers. The matrix coefficients in A defined by the transformation dx' = A dx determine what kind of vector one is dealing with. The well-known change in coordinates which defines a tensor is actually what appears in the transfomation equation of tensors. I.e. to obtain the transformation of the components of a tensor all one needs is the transformation coeffcients of the basis vectors which change the old basis vectors to the new ones. This is how the elements of A come into play in the transformation law for tensors. For this reason a tensor is classified as Cartesian, Lorentzian, Affine etc. according to the domain of the vectors and 1-forms. If a map takes Lorentz vectors and Lorentz 1-forms to a real number then that tensor is called a "Lorentz tensor". For that reason a tensor is classified according to its domain. I explained all this a web page I created which is located at http://www.geocities.com/physics_world/gr_ma/tensor_via_geometric.htm Unless we agree on a particular definition then it will be impossible to get past semantics. Best wishes Pete Please note: I wanted to rewrite my last post to (1) make it more accurate and (2) more polite. Its hard to write something regarding a disagreement without comming across in way which is contrary to how it is intended (or perhaps its just me. Is there a moderator/administrator who can delete my previous attempt at this post? Thanks. Any decent text will refer them as Cartesian tensors rather than simply tensors and for a very good reason. But most texts will qualify it with "Cartesian" first and then drop the qualifier when it is agreed upon to be implicitly understood. The class of tensors which are defined in terms of generalized coordinate transformations are most often called general tensors. When it is understood that the tensor is a general tensor then qualifier general is dropped and the general tensor is then simply called a tensor. It makes life easier if we agree on what the qualified is then simply call it a tensor. From how it appears that what you are literally saying (although perhaps you are merely not implying it) is that a Cartesian tensor is not a tensor only in the sense that it is not a general tensor in that the only transformation applicable is an orthogonal transformation in E3. But since the term general tensor is sometimes meant shortened to mean tensor then it is not a tensor in that sense of the term. But to be precise that reads the same as A Cartesian tensor is not a general tensor. This has a different meaning than the statement A Cartesian tensor is not a tensor. in that the later implies something else. In such case the qualfier general should not be dropped since the resultant sentance has a different meaning than it does befoe it is dropped. In the end I believe we can both agree to the fact that the following statement is accurate A Cartesian tensor is not a general tensor. Your assertion These are not other kinds of tensors has a different meaning than A Cartesian tensor is not a general tensor where the later is the more precise statement. Qualifiers should never be dropped when their inclusion in a sentance can change its meaning and interpretation. Also the Minkowski metric denotes a special conceptualization of the concept a metric in that the concept of distance takes on a new meaning than it does in Euclidean geometry. Here is another example of when qualifiers shouldn't be dropped. Unless a particular definition and its meaning is agreed on it is impossible to get past semantics. A point cannot be succesfully argued merely by repeating it as I'm sure you know all too well. E.g. one can keep repeating that the term "tensor" refers to a multi linear map from vectors and dual vectors to real numbers. Whether it is a general tensor or not a Cartesian tensor will depend on what the domain of the map is, i.e. whether it maps Cartesian vectors and Cartesian 1-forms to real numbers or whether it map general vectors and general 1-forms to real numbers. The map itself has not changed and the map itself is the tensor. Of course all this will do is to divert our attention from the definition of tensors to the definition of vector and their duals and we are back to the same problem. I was wondering if you could do me a favor? Can you either quote the definition of tensor used in Byron and Fuller's text (or in any other text that you believe agrees with how you are interepreting all of this) or scan the relavent pages into my computer and E-mail the scanned pages? Thank you. If you could make them available to me otherwise (e.g. upload it unto a web site and I can simply download it from there myself) then that would be even better. Thank you. Let me clarify the position I have stated in more detail. Surely you're aware of the idea that there are two different but equivalent definitions of tensor in the literature, right? By this I am referring to the definition of tensor as it refers to a multilinear map from vectors and dual vectors to real numbers versus how a tensor is defined according to how its components transform from one basis to another. Let dx' = A dx define the matrix coefficients Ajk. Then the Ajk determines what kind of object/map we are discussing. The well-known change in coordinates which defines a tensor is actually what appears in the transfomation equation of tensors. I.e. to obtain the expression for the transformation of the components of a tensor all one needs is the coeffcients of the old basis vectors which are used to express the new basis vectors. This is how the elements of A come into play in the transformation law for the components of tensors. For this reason a tensor is classified as Cartesian, Lorentzian, Affine etc. according to the domain/classification of the vectors and 1-forms. If a map takes Lorentz vectors and Lorentz 1-forms to a real number then that tensor is called a "Lorentz tensor". For that reason a tensor is classified according to its domain. I explained this a web page I created which is located at http://www.geocities.com/physics_world/gr_ma/tensor_via_geometric.htm Unless we agree on a particular definition then it will be impossible to get past all this semantics. Best wishes Pete

I'm not sure what that has to do with what I posted. Can you clarify this for me? Thanks. In the meantime consider what I posted above. You wrote A photon will also have an inherent uncertainty associated with it. That is incorrect. The problem with that assertion is that you are attempting to contribute a statistical meaning to a single measurement. Doing so makes no sense at all. Recall that uncertainty refers to a standard deviation. In fact the terms uncertainty and standard deviation are actually synonymous, i.e. they both refer to exactly the same thing. In your last post you wrote A data point does not lie on a probability distribution. The probability distribution is determined by a collection of data. When the probability distribution is determined then and only then can the uncertainty can be determined. E.g. in the case of a gaussian probability density it will be about the same as the width of the gaussian curve. Note how the value of the uncertainty remains finite even if the errors in measurements were exactly zero. If one were to determine the momentum distribution then the width of that curve would be inversely proportional to the width of the first curve. This is meaning of the HUP. Pete

If you are making this statement based on the Heisenberg's Uncertainty Principle (HUP) then what you said isn't quite right. There is nothing in quantum theory that implies that you can't simultaneosly measure the position and momentum to arbitrary precision at the same time. This does not violate HUP. HUP concerns a statistical relationship between observables, i.e. it applies to a large number of measurements on an ensemble of identically prepared systems. The uncertainty is inherent in the quantum state itself and therefore as an indeterminancys which is inherent to the state itself. I disagree. Uncertainty is a statistical quantity which applies to an ensemble of identically prepared systems. A single photon cannot have an uncertainty associated with it. Uncertainty is a statistical quantity which applies only to a large number of data points. Pete

Yep. That's it! By the way. I completely screwed up bad! I made the mistake in believing that the spacetime interval does not remain invariant for an arbitrary coordinate transformation. I had a bad feeling about making that statement so I worked it out again and this time I realized that I had somehow made a mistake when I tried this month two ago. I can't understand how I made such a dumb mistake since this is a basic fact of tensor analysis! I'm glad I pressed this point because I'd otherwise still not understand in what sense a generalized coordinate transformation is used in GR. Oh well. As they say, live and learn, right? Where did you learn this from? Did you learn it from a particular text? A number of texts etc? What you are thinking of as the, presumably, "correct" definition of a tensor is really the definition of a certain type of tensor, namely what is known as a general tensor. They get their name from the type of allowable transformations they allow, i.e. generalized coordinate transformations. There are other kinds of tensors such as Lorentz tensors and of course Cartesian tensors. The former is defined in terms of Lorentz transformations while the other is defined in terms of orthogonal transformations in R3. An affine tensor is defined in terms of orthogonal transformations in Rn. Therefore it can only be said that a Cartesian tensor is not a general tensor (the term general does not refer to something which is more applicable, it only refers to the fact that the class of transformations are general). But is still a tensor, in this case a Cartesian tensor. After all it must be a type of tensor by definition. It can also be said that a Lorentz tensor is not a Cartesian tensor as that a Cartesian tensor is not a Lorentz tensor. This definition is given in most of the really good math/physics texts including Mathematical Methods in the Physical Sciences - Third Ed., by Mary L. Boas Mathematical Methods of Physics, by Mathews and Walker Classical Mechanics - Third Edition, by Goldstein, Safko and Poole Classical Electrodynamics - Third Edition by J.D. Jackson Given the fact that such well known physicists refer to Cartesian tensors a tensors make you rethink your assumption at least a little? Pete

I forgot another quantity that is referred to as a tensor and that's the susceptibility tensor. Perhaps you've heard of it? It applies to crystals and relates the electric field vector to the polarization vector. I'm not sure if this is a true tensor or not though. Pete

Before I go on let me state that, at least for the moment (until I am convinced otherwise) I firmly hold to the idea that a Cartesian tensor is a tensor in the most rigorous sense. If I'm being slow here its because I'm not used to the terminology that you are using and because you seem to be defining the term tensor in a manner which is different than every source of definition that I'm aware of. I therefore have to sort out this terminology and notation so I can determine if what you have defined here is identical to the definition of a tensor of rank (j, k) as a multilinear map of j dual vectors and k vectors to real numbers. This is the definition of the term tensor as it is normally found in the relativity and differential geometry/tensor analysis literature. For this reason I ask for your continued patience. This is where I have a problem with the definition that you are describing. You are adding in the extra criteria that the relavent coordinate transformations are all the diffeomorphisms in [math]Diff(M)[/math]. This is not part of the definition of tensor as found in texts that I'm aware of. Recall your comment above I disagree with this view. I believe the correct phrase to be used here is as follows -The Lorentz tensor is not a general tensor, but it is still a tensor, i.e. it is a Lorentz tensor. The name given to a tensor is related to the type of coordinate transformations which apply. This is a very important difference. As for the reason why the not all diffeomorphisms can be used in the set of coordinate transformations. The physical problems for which tensors were created/defined to be useful in have a restriction on the kinds of transformations which can be used. For instance, if you made the attempt to apply any sort of coordinate transformation to the components of the Minkowski metric you will find that there will be coordinate transformations which do not leave the spacetime interval invariant. I believe that this is an important restriction on what qualifies as a valid coordinate transformation. As a simple example recall that you can't use the Galilean transformation to transform the components of the Minkowski metric since this will not leave th spacetime interval invariant. It is for reasons like this that the term allowable coordinate transformation is used. In the example above the Galilean transformation is not an allowable transformation. I disagree. I know of no source which allows all diffeomorphisms to be used in coordinate transformations. I disagree with this too for the reasons stated above. You have been thinking of tensors as 'being tensors only under a given group of transformation' which is not how this works. A geometrical object is defined according to whether the components transform tensorially or not. The type of tensor is defined by the type of allowable coordinate transformations used. Thus a geometric object is not a tensor under this group of transformations etc. The correct phrase is that these geometrical objects are say, Lorentz tensors while these other geometrical objects are, say,Cartesian tensors.

Sorry but you lost me. Do you happen to know of a text which actually gives that as a definition ver batim? You're using terminology that I'm not familiar with (either I haven't learned it yet or a Iearned it and forgot it). What is "[math]Diff(M)[/math]"? What is a (passive) linear representation of [math]Diff(M)[/math]? If you could dirtect me to a text which explains this then I can take it from there. Does Wald use this terminology that you've used here? Regarding the following Can you please give me an example of a Cartesian tensor which is not a tensor? That would be strange indeed because the Cartesian tensor a tensor by definition. Pete

Yes. Please correct me if I'm wrong but that is true of all tensors, i.e. that they pertain to a certain class of transformations. Huh? I don't understand!? I'd say that all Cartesian tensors are tensors but not all tensors are Cartesian tensors. Exactly. Again, me being confused -> It seems to me that all Lorentz tensors are general tensors but not all general tensors are Lorentz tensors. I've always been confused as to what types of transformations are allowed in general tensors in general relativity though. Do you have an answer to that? Do you mean besides the examples I gave above?? Pete

For something to be considered as held together one must reject the notion that an electron is a point particle. Only then will there be two different parts which must need to held together. Since it is believed that an electron is a point particle the idea of them being held together. However there are good reasons to consider a classical electron i.e. a sphere that is charged. There is even an entire text devoted to this idea. The text is Classical Charged Particles, by Fritz Rohrlich, World Scientific Publishing Co., (2007) The name given to the force which holds the charge together is Poincare stress. The Poincare stress must be taken into account when calculating the relativistic mass of the charge since stress is a source of inertia (Feynman discusses this in The Feynman Lectures, Vol. II page 28-4 in the chapter on electomagnetic mass). Pete