Johnny5

JC what exactly is string theory?

I have a question about something i read here. Why did they choose not to define vector division exactly? They said something about non-uniqueness, and I don't follow. To help me refresh my memory, suppose that we have two lines in an xy plane. For the sake of reality, let the plane be a real plane in real space. An equation for any line in the plane will have the following form: Ax+By+C=0 Suppose we have two lines. [math] A_1 x + B_1 y + C_1 = 0 [/math] [math] A_2 x + B_2 y + C_2 = 0 [/math] We can rewrite the equations as follows: [math] A_1 x + B_1 y = - C_1 [/math] [math] A_2 x + B_2 y = - C_2 [/math] We can rewrite this in matrix form now. [math] A \mathbf{v} = \left[ \begin{array}{cc} A_1 \ B_1 \\ A_2 \ B_2 \\ \end{array} \right] \left[ \begin{array}{c} x \\ y \\ \end{array} \right] = \left[ \begin{array}{c} -C1 \\ -C2 \\ \end{array} \right] = B [/math] Where I am using notation found here. Now the same site says, "vector division is not defined becase there is no unique solution to the matrix equation y=Ax UNLESS x is parallel to Y." So in that matrix equation they are talking about, y is a column vector, and x is a column vector right?

I have a question. You gave three properties that a multiplicative inverse of a vector A has to have... Why is property 2 there? Suppose that you remove that constraint for "multiplicative inverse" Let vector A have its tail at (a' date='b,c) and its head at (x,y,z) Now, define the dot multiplicative inverse of vector A, to be a vector with its tail at (a,b,c), and its head (x`,y`,z`) to be such that: [math'] \mathbf{A}^{-1} = \frac{1}{|\mathbf{A}|} \hat A [/math] For any vector A (except the zero vector), there is one and only one vector A-1 so that your property three is satisfied, and property 1 is also satisfied. What do you say to this? PS: To be honest with you, I am finding it a bit hard to stop thinking of the magnitude of a vector as having a length in real space.

You are coming at the problem from a purely abstract direction, that's why I worked on that problem all day yesterday. Intuitively, the answer will emerge in the solution of a physics problem, which is why I did that. Today I will look at it more mathematically. I am trying to understand what you were saying.

Tom, I have an important question. In trying to answer the question about how to define vector division, in the case of force... have you been using F=ma or F=dP/dt? Also, in the equation F=ma, how is inertial mass measured? What operational definition do you use?

You misunderstood. I was agreeing with you, that we don't start off knowing what kind of vector multiplication we are trying to find the inverse for. As to why I am analyzing bodies that can move, it's because 1. I need a problem to help me think. 2. Hopefully a problem in motion can motivate the answer. 3. I think an answer lies in this specific problem, or tiny variations of it.

No it isn't clear what kind of multiplication we are dealing with. Let me deal with a specific problem. Initially, there is an object at rest in some frame S. The center of mass is not translating in S. Now either the object is spinning or not, let us say not. And these are the initial conditions. Let it be the case that an object at rest in this frame will remain at rest in this frame, until an external agent acts upon it. Now we bring into the setup a single contact force F, we push the object say a billiard ball, with a finger. Let the applied external force F, point through the center of mass of the object. 1. If the object is on our kitchen floor, the object will begin to translate AND rotate. (It will rotate because of friction between the object and the floor) 2. If the object is in the vacuum, then the object will only translate; where we have assumed that the friction between the object and the vacuum is literally zero. Let us carry out the analysis in the vacuum. So we don't have to worry about rotation in this case, UNLESS the applied external contact force does not point right at the objects center of mass. So let us focus on the case where the contact force does not point right at the center of inertia. For simplicity, we are analyzing a spherical solid ball. Let the object have uniform density, so that the center of inertia lies at the center of the sphere. Pick an arbitrary point on the surface of the sphere, where the point of contact is. The applied force is a vector, which has some line of action. We are considering the case where the line of action does not have the exact same direction as the vector from the point of contact to the center of inertia. Let [math] \vec R [/math] denote a vector from the center of inertia, to the point of contact. The line of action of the applied force is not the same line that contains [math] \vec R [/math], so that we have two lines here, with a common point (the point of contact); so they lie in one plane. Draw the tangent to the sphere through the point of contact. Draw the normal line (line perpendicular to the tangent line, also containing the point of contact). The contact force has to come from exterior to the ball, so there are only 180 degrees spanning the direction the applied force can come from. In the case where the line of action of the contact force is the normal, there will be no rotation after contact is made, only translation. We are considering the general case. So introduce the angle of incidence. Define the angle of incidence [math] \theta [/math] to be the angle between the normal and the line of action of the applied force. So in our problem here, theta is an angle greater than zero (because we are not pushing through the center of inertia), but less than 90 (because the push is coming from outside the sphere). Now, break up the applied force vector F into two components, one which is perpendicular to the normal, and another which points right at the center of inertia (so that it is parallel to the normal). Denote the applied external force as: [math] \vec F [/math] [math] \vec F = \vec F_1 + \vec F_2 [/math] Let F1 be portion of F which acts to translate the center of inertia of the object in the CMU frame of reference. Therefore, the magnitude of F1 is: [math] F cos \theta [/math] Where F is the magnitude of the applied force. F2 is the portion of F which acts to rotate this object in the CMU frame. The magnitude of F2 is given by: [math] F sin \theta [/math] So we have: [math] \vec F = F cos \theta \hat a + F sin \theta \hat b [/math] Where a^ is a unit vector which points in the direction from the point of contact to the center of inertia, and b^ is a unit vector which points from the tail of F to the tail of vector F1. Let us focus on just the portion of F, which goes into translating the center of inertia in the CMU frame. This is vector F1. Neglecting F2 for now we have: [math] \vec F_1 = -(F cos \theta) \frac{\vec R}{|\vec R|} [/math] Now, assume that Newton's second law of motion is true, when applied to F1. Therefore, the inertia of the ball, and the resulting acceleration vector of the ball (in the CMU frame), are related by: [math] \vec F_1 = m \vec a [/math] We know the direction of a, it is going to be antiparallel to the direction of [math] \vec R [/math], and even after R vector begins to rotate in the CMU frame, the acceleration vector will still point in the direction from the point of contact (at the moment of contact) to the center of of inertia (at the moment of contact), in the CMU frame. On the other hand, the vector R will now be spinning in the CMU frame. It's tail itravels with the center of mass of the ball, and it's head is always the point of contact, even after contact is made. This vector will be spinning in the CMU frame after contact was made, but before contact was made this vector was static in the CMU frame. It is the final motion of R vector in the CMU frame which we are interested in. The tail of R is always located at the center of inertial mass of the sphere, and we can find its acceleration in the CMU frame by using: [math] \vec F_1 = m \vec a [/math] As for F2, that is the portion of the applied force F, which went into causing the sphere to start spinning in the CMU frame. So the point of contact is now orbiting the center of inertia of the sphere in the CMU frame. And since its orbiting, its direction is continuously changing in time. After the contact force is off, the orbital speed will be constant in time in the CMU frame. However, the direction will be changing, so that the point of contact will no longer be at rest, nor will it be traveling in a straight line at a constant speed in an inertial reference frame. I think (someone correct me if I am wrong)... the point of contact is going to be tracing out a wave shape, in the CMU frame. The frequency and wavelength of it are related to the acceleration given to the object by the applied external force. Now, consider things in the reference frame attached to the object, whose origin is permantly the center of mass of the object. Let the X axis of this frame be the NORMAL line, which was defined way back when we discussed the point of application of the applied force. Let the direction of increasing x coordinates in this frame, be the same direction as the acceleration of the center of mass. Keep in mind that the applied force only lasted momentarily, so that the center of inertia is no longer accelerating, it's traveling in a straight line (the normal line) at a constant speed. [math] v_f - v_i = \vec a t [/math] This formula is true in the CMU frame, and in that frame the objects center of inertia was initially at rest, hence vi=0 so that in the CMU frame the following statement is true: [math] v_f = \vec a t [/math] Where vf is the speed of the center of mass of the object in the CMU frame of reference, and a was the acceleration (in the CMU frame) given to the center of inertia of the object by the applied external contact force. That force only lasted a few moments, afterwards, the speed of the center of mass of the object is now a non-zero constant in the CMU frame. This final speed vf, will now be denoted by V. Now, let us frame switch; transfer into a frame of reference in which the speed of the center of inertia is zero, instead of V. If it is the case that this frame is also an inertial frame of reference, then we can expect Newton's laws to be true in this frame too. Let us call this new frame, frame S`. So the origin of frame S` is moving through the CMU frame with speed V, but the center of inertial mass of the sphere is permanently at rest in S` Let the X axis of S` be coincident to the Normal line, but let the point (0,0,0) be the center of mass of the sphere. Thus, the frame is rigidly attached to the sphere. Since the X axis of S` is coincident to the normal line, the tip of R vector is now tracing out a circle in S`. So this is centripetal acceleration of the point of contact (which travels with the sphere) in the reference frame S`. Centripetal acceleration is given by: [math] \vec a_c = \frac{ v^2_t}{|\vec R |} [/math] Where [math] v_t [/math] is the tangential speed or orbital speed of the contact point in S`, and is not to be confused with V, which is the speed of the center of inertia of the sphere in the CMU frame (reference frame in which the center of mass of the universe is permanently at rest, and Newton's first law is true). The details of the tangential speed vt are related to the portion of the applied force which went into causing the object to begin spinning in the CMU frame in the first place. In other words, they are directly related to F2. Now either S` is an inertial reference frame or not. Right now, the point of contact (which is at the tip of R vector) is tracing out a circle in S`. Acceleration is the time rate change of velocity. Velocity is the product of speed times direction of travel. So acceleration is zero if and only if, neither the speed, nor the direction of travel is changing. Suppose there is a particle located at the tip of R vector, it has some mass m1. The tangential speed vt of this particle is constant in time, but the direction of travel of this particle is constantly changing (the particle, located at the point of contact, is orbiting the center of mass of the sphere). Ok I think I've worked on this enough to get an idea. It seems to me, that by treating "acceleration" in Newton's second law as "acceleration of the center of inertia" is keeping things too mathematically simplistic. Certainly, the concept of "acceleration of a point" is easier to understand than "acceleration of a rigid straight line," but I seem to be interested in the motion of [math] \vec R [/math] more than just the motion of the center of inertia. So, if I had mathematics to explain the entire motion of R vector, that mathematics would include the motion of the center of inertia as well, but additionally that mathematics would also be telling me how the whole vector is moving through space, not just the tail of R vector (tail of R vector = center of mass of sphere). In other words, I want to know how the "whole vector R" moves through space, after the contact force is applied to the solid sphere. What I was originally hoping, was that somewhere in the solution of this problem, would be a motivation for a way to define vector division. So I am questioning the interpretation of acceleration in Newton's second law. Normally, when we speak of acceleration, we mean acceleration of a point mass, OR acceleration of the center of inertia. In other words, we have Newton's second law, which is this: [math] \mathbf{F} = m \mathbf{a} [/math] But that law only holds if m is independent of time. But more generally, Newtons second law is: [math] \mathbf{F} = \frac{d\mathbf{P}}{dt}[/math] So that if m is a time dependent quantity we have instead: [math] \mathbf{F} = m \mathbf{a} + \mathbf{v} \frac{dm}{dt}[/math] If I now modify the problem slightly, instead of a finger pushing the sphere, there is a tiny man standing on the sphere (and he is standing on the point of contact), his knees are bent, and he is getting ready to jump off the sphere with all his might, then the "mass of the sphere changes" because he succeeds in jumping off. So depending on how the contact force arises, we could need the more general form of Newton's second law.

Ok, so you can solve for direction, it doesn't matter that the coefficients of i^ j^ k^ were asymmetrical, since the analysis would be the same, if you just renamed the three axis, permuted x,y,z. So back to the problem of vector division. Start with [math] \mathbf{F} = m \mathbf{a} [/math] We are starting off with the case where the center of mass of an object is at rest in some reference frame S. The reference frame is 'good' in the sense that we know the laws of physics in it. It's an inertial reference frame, so therefore our formulas are true in the frame. Whether they be frame dependent formulas, or frame independent formulas is another issue entirely, but all that matters right now is that F=ma is true in S. We can view an acceleration vector, as being a magnitude times a direction. So let us write Newton's second law differently: [math] \mathbf{F} = m a \hat a [/math] Where a is the magnitude of the acceleration vector, and a^ is the direction of the acceleration vector. In the case where a body is at rest, the acceleration vector must be zero. So either a=0 or a^=0. a^ is a unit vector, so its magnitude is 1, and the previous post was meant to show that no vector can have a direction of zero. Hence, it cannot be the case that a^=0, therefore a=0. But presuming that vector division is defined, we can now divide both sides by a^, to obtain: [math] \frac{\mathbf{F}}{\hat a} = m a[/math]

Using right hand coordinate system: Right hand system Angle from z axis to vector V is theta (which varies from 0 to pi/2) Then, the angle between i^ and the projection of v onto the xy plane, is phi, which can vary from 0 to 2 pi. So let the origin of the frame be at the center of mass of the body. Any point in space could either be referenced by (x,y,z) or (R,phi,theta) Yes I remember all this. Then simply derive the relations to convert from rectangular coordinates to spherical coordinates. Offhand I don't remember how, but I could re-derive them for myself now. So we have a vector V with one endpoint at the origin, and the other at some point (x,y,z) in space, some distance R away. The direction of this vector is going to be FROM the origin TO the point (x,y,z) [math] \vec V = r \hat r [/math] Ok this is the vector using spherical coordinates, where r is the length of the vector, and r^ is direction. This is the same vector, using rectangular coordinates: [math] \vec V = x\hat i + y \hat j + z \hat k[/math] That projection has length [math] r sin \theta [/math] So now I just have to work out vector V in rectangular coordinates, and then equate to r r^, divide by r, and I should get that formula. The projection is the hypotenuse of a right triangle in the xy plane, with one side of length x, and the other side of length y. Therefore: [math] x^2 + y ^2 = r^2 sin^2 \theta [/math] and also, there is a second right triangle, with hypotenuse r, one side having the length of the projection, and the other side having length z. Therefore, in this frame, it is also true that: [math] (r sin \theta)^2 + z^2 = r^2 [/math] Well yes obviously since [math] x^2+y^2+z^2 = r^2 [/math] Oh I see. Phi is the angle such that: [math] sin \phi = \frac{y}{r sin \theta} [/math] [math] cos \phi = \frac{x}{ r sin \theta} [/math] from which we can readily see the following: [math] \vec V = r sin \theta cos \phi \hat i + r sin \theta sin \phi \hat j + z \hat j [/math] And now, all we have to realize is that [math] cos \theta = \frac{z}{r} [/math] to finally discover that: [math] \vec V = r sin \theta cos \phi \hat i + r sin \theta sin \phi \hat j + r cos \theta \hat j [/math] We can now equate this to rr^, and then divide both sides by r, to finally obtain: [math] \hat r = sin \theta cos \phi \hat i + sin \theta sin \phi \hat j + cos \theta \hat j [/math]

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.

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?

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?

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.

I'm thinking about this portion of your response first. 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-1b (a-1a)x=a-1b ex=a-1b x=a-1b

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.

Black holes aren't supposed to emit radiation. So if you put your hand close up to one, like you do a light bulb, you wont feel the heat. So you might suspect that the black hole is cold to the touch. Is it? Could it be, that past a certain mass, the black hole actually reaches absolute zero temperature? And if you are going to site the third law, are you certain it is true? Anyone care to answer? Thank you PS: I suppose, if there is internal motion, you could say that it's internal temperature isn't zero, but what about the surface? I read somewhere, that its easier to visualize internal thermal energy by focusing on the idea that the temperature inside is related to the speed of the parts in the 'gas' (Maxwellian speed distribution), but that this is too difficult to use for practical applications, and that is why the concept of entropy is used. [math] \frac{1}{T} = \frac{\partial S}{\partial U} [/math] But entropy is such a complicated variable, and I think there are problems with the mathematics which statistical mechanics is based on. Maybe you could think of this post more as, I want to understand U better.

Thanks Martin

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?

What are their physical interpretations? Also, how much will quantum gravity affect classical GR formulas?

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: And if you happen to know, how does GR deal with superconductivity... well or poorly, does it predict superconductivity is possible or impossible? Thanks

SUMMARY: If GR is correct then (matter and space are coupled) Martin, what do you think would happen if matter was UNCOUPLED from space in some local region of spacetime? Maybe I should preface this question with, "How would one go about uncoupling matter from space?" Thanks

I recently read in some post here, that according to GR, if a body of material is accelerated (really the person should have said experiences an external force) then gravitational waves are emitted. Is that accurate? Does GR imply that? If so, what mathematical part leads to this? Thank you

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?

I'm reading it now. Thanks Martin