pljames Posted February 28, 2005 Share Posted February 28, 2005 I love the word and all it pertains to. But does the word relavitist have anything to do with relativity? I keep hearing it. I agree with the word tho need more information about the meaning. Paul/pljames Link to comment Share on other sites More sharing options...

Tom Mattson Posted February 28, 2005 Share Posted February 28, 2005 It depends on the context. "Relativist" could pertain to one who works with Einstein's relativity. But it could also refer to the philosophical school(s) of thought that fall under the banner of Relativism. So at best I can give you a very firm "I don't know." Link to comment Share on other sites More sharing options...

Guest iceman12387 Posted March 2, 2005 Share Posted March 2, 2005 I have a question is it possible to create a space distortion usion a lot of energy and magnets Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 7, 2005 Share Posted March 7, 2005 In General Relativity, any energy density affects the curvature of spacetime. So the answer is 'yes'. Link to comment Share on other sites More sharing options...

Johnny5 Posted March 8, 2005 Share Posted March 8, 2005 In General Relativity, any energy density affects the curvature of spacetime. So the answer is 'yes'. Would it be wrong to say that general relativity states that the ratio of the stress energy tensor to the curvature tensor is a constant? How do you interpret the numerator (stress energy tensor) and denominator (curvature), in some local region of space of cubical volume V? Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 8, 2005 Share Posted March 8, 2005 Would it be wrong to say that general relativity states that the ratio of the stress energy tensor to the curvature tensor is a constant? Yes' date=' it would be wrong. GR states that: [math'] R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi GT_{\mu\nu} [/math] where R_{uv} is the Ricci tensor, R is the Ricci scalar, g_{uv} is the metric tensor, and T_{uv} is the stress-energy tensor. If we redefine the left side as follows: [math] G_{\mu\nu}\equiv R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} [/math] where G_{uv} is called the Einstein tensor, then we have the GR field equations in the more compact form: [math] G_{\mu\nu}=8\pi GT_{\mu\nu} [/math] On the other hand, we have the Riemann curvature tensor which is: [math] R^{\sigma}{}_{\mu\alpha\beta}=\partial_{\alpha} \Gamma ^{\sigma}_{\mu \beta}-\partial_{\beta} \Gamma^{\sigma}_{\mu \alpha}+ \Gamma^{\sigma}_{\lambda \alpha} \Gamma^{\lambda}_{\mu \beta}- \Gamma^{\sigma}_{\beta \lambda} \Gamma^{\lambda}_{\mu \alpha} [/math] This tensor does not explicitly appear in the GR field equations, although the Ricci tensor is a contraction of it. It should be readily seen that, if anything, GR states that the ratio of the Einstein tensor to the stress-energy tensor is a constant. But I don't think that quotients of tensors are referred to as ratios. How do you interpret the numerator (stress energy tensor) and denominator (curvature), in some local region of space of cubical volume V? Does this question change based on my answer to your first question? Link to comment Share on other sites More sharing options...

Johnny5 Posted March 8, 2005 Share Posted March 8, 2005 It should be readily seen that' date=' if anything, GR states that the ratio of the Einstein tensor to the stress-energy tensor is a constant. But I don't think that quotients of tensors are referred to as ratios. [/quote'] What are quotients of tensors referred to as? Well aren't scalars zero rank tensors, so you can divide one tensor by another, at least in the case of zero rank tensors (division by zero error avoided of course). And vectors are first rank tensors, and if two vectors have the same direction, I don't see any problem with dividing one by the other, and so here is another case where one tensor can be divided by another. And if I gave it enough thought, I could probably figure out a useful way to define vector division, in the case where the vectors do not have the same direction, although not too long ago someone told me it's impossible. I just don't have a reason to work it out... or is the answer to this a tensor of rank 2?????????? It seem's to me that you can do it. Consider this [math] \vec F = m \vec a [/math] In the case of pure translation (contact force aimed right at the center of inertial mass), F,a have the same direction, and you have this: [math] m = \frac {\vec F}{\vec a} [/math] The direction of F, cancels with the direction of a, you are left with a scalar on the RHS, and the LHS is inertial mass, which is a scalar. I see no problem here. I don't really have any use for a higher rank tensor yet. I seem to be able to make my way around Euclidean three space pretty well, without tensors of rank greater than 1. The only reason I asked was something that Martin wrote in another thread. G_{ab} = (1/F) T_{ab} LHS is curvature' date=' RHS is energy density (actually the stress energy tensor whose unit of measure is energy density) divided by a universal constant force F = c[sup']4[/sup]/(8piG) I like simplicity, this formula seems like only three things. curvature =LHS = constant times stress energy tensor. But now you are telling me that the LHS is the Einstein tensor, which is actually the Ricci tensor minus the Ricci scalar times the metric tensor. Did I get all this right? Also, in that thread, Martin used the word 'couple' and said that according to GR, matter and space couple. Is there any way to decouple them allowed by GR? Also, is this site reputable: Dicussion Of Couple In particular, note where he says the following: 1. The centre of inertia is the same point as the centre of gravity. This similarity does not give us the right to unify inertia with gravitation as Einstein did. 2. Physicists who still believe in the validity of Einstein's relativity theories still do not understand what mass or inertia really is. And if you happen to know, how does GR deal with superconductivity... well or poorly, does it predict superconductivity is possible or impossible? Thanks Link to comment Share on other sites More sharing options...

Johnny5 Posted March 8, 2005 Share Posted March 8, 2005 Yes' date=' it would be wrong. GR states that: [math'] R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=8\pi GT_{\mu\nu} [/math] where R_{uv} is the Ricci tensor, R is the Ricci scalar, g_{uv} is the metric tensor, and T_{uv} is the stress-energy tensor. What are their physical interpretations? Also, how much will quantum gravity affect classical GR formulas? Link to comment Share on other sites More sharing options...

Johnny5 Posted March 8, 2005 Share Posted March 8, 2005 How do you interpret the numerator (stress energy tensor) and denominator (curvature)' date=' in some local region of space of cubical volume V? [/quote'] Does this question change based on my answer to your first question? Yes it does, you can ignore this question for now, because you have made the whole thing intensely complex. Right now I am more concerned with dividing one vector by another vector, in general. For example, suppose that an externally applied force is applied to a steel ring, with uniform density, so that the center of inertia of the ring is at the center of the ring. Someone pushes the ring, but let the externally applied force not be aimed right at the center of inertia. This is the case where F and a don't have the same direction yes? Link to comment Share on other sites More sharing options...

swansont Posted March 8, 2005 Share Posted March 8, 2005 you have made the whole thing intensely complex. I think the subject is inherently complex. Don't shoot the messenger. Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 8, 2005 Share Posted March 8, 2005 Yes it does, you can ignore this question for now, because you have made the whole thing intensely complex. Funny, I was thinking that you made it intensely oversimple. I'll get to the rest later. Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 8, 2005 Share Posted March 8, 2005 What are quotients of tensors referred to as? As far as I know' date=' they aren't dealt with at all. Well aren't scalars zero rank tensors, so you can divide one tensor by another, at least in the case of zero rank tensors (division by zero error avoided of course). Yes, but that's just division on the reals. It's considerably more complicated with tensors of higher rank, as will become apparent when I answer the next part... And vectors are first rank tensors, and if two vectors have the same direction, I don't see any problem with dividing one by the other, and so here is another case where one tensor can be divided by another. And if they aren't pointing in the same direction, what then? Even if they are pointing in the same direction, there are problems. Take the scalar equation ax=b. To solve for x, we divide both sides by a so that x=b/a. As long as a isn't zero, there's no problem here. But what is division anyway? Division by a is really multiplication by the multiplicative inverse of a, which is a^{-1}. Letting e stand for the identity element (e=1 in R) and rewriting our solution in these terms, it becomes: ax=b a^{-1}ax=a^{-1}b ex=a^{-1}b x=a^{-1}b Now one important property of multiplication on the reals is closure. That is, for any elements {a,b} in R, the product ab is also in R. So a^{-1}b is a real number, as expected. Now let's try this with vectors, as we attempt to solve for x. We'll have to assume that a "multiplicative inverse" a^{-1} exists for a. I put the word multiplicative in quotes, because it is not clear what kind of multiplication we are dealing with. For now, I'll denote it by the pound sign (#). ax=b a^{-1}#ax=a^{-1}#b Now we have 2 ways to multiply vectors: The dot product and the cross product. Let's see if either one could possibly stand for #. Dot product: a^{-1}^{.}ax=a^{-1}^{.}b Now we hit our first roadblock. What exactly is a^{-1} with respect to the dot product? A multiplicative inverse is supposed to have 3 properties: 1. It has to return an identity element when multiplied by a. 2. It has to be in the same set as a and the identity. 3. It has to be unique. There is no vector a^{-1} that satisfies all these properties. To prove this, assume that an identity element e exists such that a^{-1}a=e. You could try to define a^{-1} as a vector which when put in a dot product with a, yields the scalar "1". But there are two problems with this. First, there are an infinite number of such vectors, which violates #3. And second, vectors aren't closed under the dot product, which violates #2. And of course, if we look at the other case, namely that in which an identity element e satisfying a^{-1}a=e does not exist, then we have contradicted condition 1 above. Thus, we cannot define vector division from the dot product. Moving on... Cross product: Now let # stand for X. ax=b a^{-1}Xax=a^{-1}Xb We immediately run into 2 problems here. First, being obvious that a^{-1}Xa is a vector, I can tentitively call that product e, the identity vector. Let us first assume that not all of the components of e are zero. Our first problem comes from the antisymmetry of the cross product. That is, aXb=-bXa. Now look at the action of taking the cross product of a and its inverse from each side: a^{-1}Xa=e aXa^{-1}=-e The problem here is that multiplication of a by its inverse should yield the identity element no matter which side you multiply from. But that's not the case here, unless we define 2 identity elements for this multiplication. But in that case, there can't be any sense in which this could be called "multiplication" in the usual sense, because in the usual sense there can only be one multiplicative identity. Now consider the second case: the components of e are all zero. That solves the problem above, but the problem now is that our multiplicative identity is the zero vector, which will lead to the division by zero problem. Thus, we cannot define vector division from the cross product. And if I gave it enough thought, I could probably figure out a useful way to define vector division, in the case where the vectors do not have the same direction, How? although not too long ago someone told me it's impossible. I just don't have a reason to work it out... or is the answer to this a tensor of rank 2?????????? No, rank 2 tensors won't get you out of the problems detailed above. The closest thing to "vector division" that I have ever encountered is "phasor division", which necessitates the use of complex functions. It seem's to me that you can do it. Consider this [math] \vec F = m \vec a [/math] In the case of pure translation (contact force aimed right at the center of inertial mass), F,a have the same direction, and you have this: [math] m = \frac {\vec F}{\vec a} [/math] The direction of F, cancels with the direction of a, you are left with a scalar on the RHS, and the LHS is inertial mass, which is a scalar. I see no problem here. What makes you think that the direction of F can cancel with anything? You're taking an operation on the reals, and applying it to something that is not even a mathematical object. At least, it isn't a mathematical object until you define its properties. Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 8, 2005 Share Posted March 8, 2005 What are their physical interpretations? Roughly speaking: The right side describes the curvature of spacetime' date=' and the left side describes the matter and energy distribution. In the language of PDEs, the right side is the source term for the left side. Also, how much will quantum gravity affect classical GR formulas? That depends on what approach you look at. If you look at string theories, then that is a generalization of QFT, which will contain GR as a special case. If you look at loop quantum gravity, then that is a direct quantization of the GR theory. Among other things, general covariance and background independence will be retained from GR. But of course GR is not a quantum theory, so that aspect will require a major overhaul. Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 9, 2005 Share Posted March 9, 2005 The only reason I asked was something that Martin wrote in another thread. G_{ab} = (1/F) T_{ab} LHS is curvature' date=' RHS is energy density (actually the stress energy tensor whose unit of measure is energy density) divided by a universal constant force F = c[sup']4[/sup]/(8piG) I like simplicity, this formula seems like only three things. curvature =LHS = constant times stress energy tensor. But now you are telling me that the LHS is the Einstein tensor, which is actually the Ricci tensor minus the Ricci scalar times the metric tensor. Did I get all this right? Yes, that's right. The curvature tensor is actually a rank-4 tensor. By contracting 2 indices together, you get the Ricci tensor, which appears in the field equations of GR. Also, in that thread, Martin used the word 'couple' and said that according to GR, matter and space couple. Is there any way to decouple them allowed by GR? In principle, yes: In the low-speed, low-energy density limit. If the energy (I'm lumping mass in with energy) density goes to zero, you recover SR from GR. Space and time are still coupled, but now at least the spacetime is flat. Taking it one more step: If the speed scale (that is, a scale which is indicative of typical speeds of objects) goes to zero (or equivalently, as c goes to infinity), you recover Galilean relativity from SR. So, in a universe in which nothing moves, and nothing exists, Galilean relativity is true, and space and time are completely decoupled. In other words, there is no way that space and time can be decoupled in the real universe. Also, is this site reputable: Dicussion Of Couple Absolutely not. It's pure crackpottery. That much is obvious from the abstract, found here: http://www.hyperinfo.ca/LivingAtom/ The solar system is intelligent? The atom is intelligent? There are contradictions in relativity? "Living atom theory" can logically explain telepathy? The bozos who wrote that website not only have no real understanding of physics, but they also have serious misunderstandings of cognitive science if they really believe all this crap. And if you happen to know, how does GR deal with superconductivity... well or poorly, does it predict superconductivity is possible or impossible? I don't know of any general relativistic treatment of superconductivity, but I do know that superconductivity is understandable in terms of quantum statistical mechanics, and I know that GR is perfectly compatible with QM (that is, their axioms are all consistent). The only problem I can see is that GR might contradict the quantum field theoretic (as opposed to quantum mechanical) description of superconductivity. But we already know that GR and QFT don't get along, so this isn't all that Earth shattering. Link to comment Share on other sites More sharing options...

Johnny5 Posted March 9, 2005 Share Posted March 9, 2005 Thank you Tom, I just briefly read all of your replies, I'm going to take a lot of time, and read through your two proofs until I understand them. After I do understand them, I will ask you another question. I will also inspect all your other answers, and if I have any further questions about those, I will ask them. Thank you for taking so much time to answer my questions. I will take the time to read your answers. Link to comment Share on other sites More sharing options...

Johnny5 Posted March 9, 2005 Share Posted March 9, 2005 I'm thinking about this portion of your response first. Even if they are pointing in the same direction' date=' there are problems. Take the scalar equation ax=b. To solve for x, we divide both sides by a so that x=b/a. As long as a isn't zero, there's no problem here. But what is division anyway? Division by a is really multiplication by the multiplicative inverse of a, which is a[sup']-1[/sup]. Letting e stand for the identity element (e=1 in R) and rewriting our solution in these terms, it becomes: ax=b a^{-1}ax=a^{-1}b ex=a^{-1}b x=a^{-1}b Now one important property of multiplication on the reals is closure. That is, for any elements {a,b} in R, the product ab is also in R. So a^{-1}b is a real number, as expected. Yes, for scalars we do have closure under addition, and closure under multiplication. That is: Closure under +,* If X is an element of the real number system, and Y is an element of the real number system then A+B is an element of the real number system and A*B is an element of the real number system. But in the example [math] m = \frac{\vec F}{\vec a} [/math] vector division isn't necessarily closed under division. Here we have division of one vector F, by another vector a, and the answer is a scalar. Is there any reason we have to force closure properties on vector division? The strange thing is, lately I have read in a few places that m is a vector. I still don't understand how anyone concluded that inertial mass is a vector. To even say that, would make inertial mass a direction dependent quantity. I am not insisting that it isn't, I just don't see any reason to stipulate that it is yet. But, supposing for a second that inertial mass m must be a vector, then that means that my example (which came from F=ma) doesn't do what it is designed to, so if you have any comment on whether or not inertial mass is a vector, now would be the time to make it. PS: You know you used the associative property in the LHS of step three above, don't you? (So you have vector multiplication/division being associative already, just so you know.) (ax)=b a^{-1}(ax)=a^{-1}b (a^{-1}a)x=a^{-1}b ex=a^{-1}b x=a^{-1}b Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 9, 2005 Share Posted March 9, 2005 Is there any reason we have to force closure properties on vector division? It's not that we're forcing closure on vector division' date=' it's that there's no way to define vector division in terms of multiplication such that vectors are closed under multiplication [i']and[/i] the necessary properties of multiplicative inverses and identities are satisfied. If you conclude that m=F/a from F=ma, then there has to be a multiplicative inverse a^{-1} such that: F=ma Fa^{-1}=maa^{-1} Fa^{-1}=me Fa^{-1}=m and then define Fa^{-1}=F/a, it turns out that you cannot do this with either the dot product or the cross product, as I've shown. So what multiplication rule do you propose we use to define vector division? But, supposing for a second that inertial mass m must be a vector, then that means that my example (which came from F=ma) doesn't do what it is designed to, so if you have any comment on whether or not inertial mass is a vector, now would be the time to make it. The "m" in F=ma cannot be a vector. If it were, then neither F nor a could be vectors. Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 9, 2005 Share Posted March 9, 2005 PS: You know you used the associative property in the LHS of step three above' date=' don't you? (So you have vector multiplication/division being associative already, just so you know.) [/quote'] I did not use the associative property when "#" still stood for an arbitrary multiplication. I used it when discharging my hypothetical supposition: Vector division can be defined from the dot product or the cross product. I used associativity there because I know that it applies with respect to those two operations. Then I showed by contradiction that the negative of the supposition is actually the case: Vector division cannot be defined from the dot product and it cannot be defined from the cross product. Link to comment Share on other sites More sharing options...

Johnny5 Posted March 9, 2005 Share Posted March 9, 2005 So what multiplication rule do you propose we use to define vector division? Nothing is going to just jump out at me right away' date=' but let me ask you this. Vector is a very general term. There are different kinds of vectors. I am reading a good book on it now (Author: Louis Brand). At the very beginning, he discusses line vectors, which is the kind of vector an application force is. In general, a 'vector' is completely defined by it's magnitude and direction, so that two vectors are equal if and only if they have the same magnitude and direction. But in the case of line vectors, they are equal if and only if they have the same magnitude and direction, and line of action (if I am remembering Brand correctly) So my question is, what kind of 'vector' are you thinking about in F=ma? Regards PS: I would think specifically, that in the case of a contact force, the vector is defined by the location of its endpoints in real 3D space, so that the only vector it is equivalent to is itself. It may have the same magnitude and direction as other vectors elsewhere, but in the analysis of how the whole body is going to move after the contact force is applied, what centrally matters is [i']where [/i] that contact force is applied in relation to the center of mass of the body. So line vectors are what we want, because in the analysis in the rest frame of the body which receives the applied force, we are going to keep the point of contact fixed, but vary the angle at which the force is applied. Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 9, 2005 Share Posted March 9, 2005 Nothing is going to just jump out at me right away' date=' but let me ask you this. Vector is a very general term. There are different kinds of vectors. I am reading a good book on it now (Author: Louis Brand). At the very beginning, he discusses line vectors, which is the kind of vector an application force is.[/quote'] This is a good point. There is a sense in which the real numbers are vectors (that is, R^{1} is a vector space). To tighten things up, I am specifically talking about vectors in R^{n} for n>1. In general, a 'vector' is completely defined by it's magnitude and direction, so that two vectors are equal if and only if they have the same magnitude and direction. No, in general a vector doesn't need a direction. You're talking about vectors in R^{n}. All that is required for an object v to be a 'vector' is that it be a member of some set that satisfies Definition I.1 on page 80 of the following textbook: ftp://joshua.smcvt.edu/pub/hefferon/book/book.pdf But indeed, it would be best to stick with R^{n}. But in the case of line vectors, they are equal if and only if they have the same magnitude and direction, and line of action (if I am remembering Brand correctly) So my question is, what kind of 'vector' are you thinking about in F=ma? I regard the vectors in that equation as elements of the vector space R^{3}. edit: So line vectors are what we want, because in the analysis in the rest frame of the body which receives the applied force, we are going to keep the point of contact fixed, but vary the angle at which the force is applied. Even if you use line vectors, you are going to have the same problems satisfying the requirements for multiplicative identities and inverses. Link to comment Share on other sites More sharing options...

Johnny5 Posted March 9, 2005 Share Posted March 9, 2005 I regard the vectors in that equation as elements of the vector space R^{3}. Me too. Let me ask you something, do you already know the answer to this 'vector division' question, or are you really as clueless as me right now? Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 9, 2005 Share Posted March 9, 2005 All I know is that you can't define vector division in terms of cross products or dot products. I'm still working on the more general problem. Link to comment Share on other sites More sharing options...

Johnny5 Posted March 9, 2005 Share Posted March 9, 2005 A multiplicative inverse is supposed to have 3 properties: 1. It has to return an identity element when multiplied by a. 2. It has to be in the same set as a and the identity. 3. It has to be unique. F=m a Initially let some body be at rest in some reference frame S. All this means is that the center of mass of the body is at rest in frame S. But from this' date=' it follows that the acceleration of the body is zero in frame S. [b']a [/b] = zero vector = 0 Let us stipulate that the body isn't spinning either, thus any point (x,y,z) on the surface of the body is also at rest in frame S. Let S be an inertial reference frame, so that the body remains at rest in S until an external force acts upon it. So there are no external forces currently acting on it so F=0 too, hence regardless of m, the following statement is true in S: F = m a Now I am just going to repeat your argument: Solve for m. First, let us presume that there is a multiplicative inverse a^{ -1 } Here there is a snag, because we would be dividing by the zero vector, which is tantamount to the division by zero error of ordinary algebra. Let me think about this for a moment. A vector has magnitude and direction. Let there be a multiplicative process between these two different quantities, so that they couple. [math] \vec V = (magnitude) \bullet (direction) [/math] where [math] \bullet [/math] is being used to show that we have joined them in some mathematical sense. So zero vector. I want to say that a vector is equal to the zero vector if either its magnitude is the number zero, or its direction is the "direction zero". Of course this doesn't make sense because "zero direction" doesn't make sense. But if it did make sense we would have an argument like this: F = m a In the case where a is the zero vector, either the magnitude of a is zero, or the direction of a is zero, or both. If the magnitude is zero, we are prevented from multiplying each side by the multiplicative inverse of a because we will have divided by the scalar zero. But if the reason a is equal to the zero vector is because its direction is the zero direction, then the magnitude of a could be nonzero, and we could divide both sides by the magnitude of a. All right I have a question. Take a body at rest, and not spinning. Its center of inertia is at rest. So we say that the object isn't accelerating in the frame. Absolutely this must mean the following: [math] \vec a = \vec 0 [/math] But, in general a vector has magnitude and direction. So we could have: [math] \vec 0 = (magnitude) \bullet (direction) [/math] Which could be zero if the magnitude is zero, but what if... So, here is my question... Is there any such idea out there of the zero direction? Link to comment Share on other sites More sharing options...

Tom Mattson Posted March 9, 2005 Share Posted March 9, 2005 Is there any such idea out there of the zero direction? If V=(magnitude)^{.}(direction)' date=' then I can already tell you what (direction) is. In R[sup']3[/sup] we have: [math] \mathbf {V}=| \mathbf {V} | cos( \phi ) sin( \theta ) \mathbf {i} + | \mathbf {V} | sin( \phi ) sin( \theta ) \mathbf {j} + | \mathbf {V} | cos (\theta) \mathbf {k} [/math] Factoring out |V| we have: [math] \mathbf {V}=| \mathbf {V} | (cos( \phi ) sin( \theta ) \mathbf {i} + sin( \phi ) sin( \theta ) \mathbf {j} + cos (\theta) \mathbf {k} ) [/math] Since the |V| on the right hand side is (magnitude), it follows that the other quantity is what you call (direction). As one might expect, it is a unit vector in the direction of V. Furthermore, it is not possible for all 3 components to simultaneously be zero, for any choices of theta or phi. Link to comment Share on other sites More sharing options...

Johnny5 Posted March 9, 2005 Share Posted March 9, 2005 If V=(magnitude)^{.}(direction)' date=' then I can already tell you what (direction) is. In R[sup']3[/sup] we have: [math] \mathbf {V}=| \mathbf {V} | cos( \phi ) sin( \theta ) \mathbf {i} + | \mathbf {V} | sin( \phi ) sin( \theta ) \mathbf {j} + | \mathbf {V} | cos (\theta) \mathbf {k} [/math] Factoring out |V| we have: [math] \mathbf {V}=| \mathbf {V} | (cos( \phi ) sin( \theta ) \mathbf {i} + sin( \phi ) sin( \theta ) \mathbf {j} + cos (\theta) \mathbf {k} ) [/math] Since the |V| on the right hand side is (magnitude), it follows that the other quantity is what you call (direction). As one might expect, it is a unit vector in the direction of V. Furthermore, it is not possible for all 3 components to simultaneously be zero, for any choices of theta or phi. There is something to this... how did you derive this formula (something is asymmetric about the coefficients of the three unit vectors): direction = [math] (cos( \phi ) sin( \theta ) \mathbf {i} + sin( \phi ) sin( \theta ) \mathbf {j} + cos (\theta) \mathbf {k} ) [/math] How did you arrive at the coefficients of the three unit vectors? PS: This formula leads to the conclusion that a vector is equivalent to the zero vector if and only if its magnitude is zero. Link to comment Share on other sites More sharing options...

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