Johnny5

Yes please. What is the problem at hand? I know this field equation here: [math] G_\mu_\nu = \frac{8 \pi G}{c^4} T_\mu_\nu [/math] The quantity on the LHS is called the Einstein tensor, and the quantity on the RHS is a scalar quantity whose units are that of force (Kg m/s^2), times the Maxwell stress-energy tensor.

Done. Therefore, in SR: [math] e_1 = \hat i [/math] [math] e_2 = \hat j [/math] [math] e_3 = \hat k [/math] and i suppose that: [math] e_0 = [/math] direction of time? Is there a way to use the notation to express the fact that the time coordinate t of a frame changes unidirectionally?

Right. So if your vector is entirely in the XY plane of some reference frame, then the vector will only have two components. So that the indice i is going to go from 1 to 2. So suppose that the symbol [math] \vec V [/math] was used to denote the vector. Let it be the case that: [math] \vec V = x \hat i + y \hat j [/math] So as a time saving technique, we are going to write that using the summation convention as: [math] \vec V = V^i e_i [/math] Where we have used ei to represent the i'th basis vector. So if we let i range from 1 to 3 we get this: [math] \vec V = V^i e_i = v1 \hat i +v2 \hat j + v3 \hat k = v1 \hat e_1+v2\hat e_2+v3\hat e_3 [/math] Just trying to work out some equivalences right now. Now, change to SR theory..., which means go from zero... [math] \vec V = V^i e_i = v0\hat e_0 + v1 \hat e_1+v2\hat e_2+v3\hat e_3 [/math]

Ok I got you. When you see the same index raised, and lowered, you don't have to write the giant Sigma symbol, you are to know that you are summing over the index. In other words, there is a semantical equivalence of exactly what you wrote, namely... [math] \sum_{i} A_iB^i = A_iB^i [/math] In other words, when Einstein summation convention is used, the meaning of the left hand side of the expression above is equivalent to the meaning of the RHS. A shorthand notation. So that is for when you see the same index.

Metric has something to do with distance. I am not familiar with the Einstein summation convention, however I do remember reading that he thought it was his greatest contribution to physics. No, I don't have a perfect definition of "tensor" yet. Let me go get one. Here is mathworld's definition of tensor. It will take me a few seconds to make sense out of that. I will learn what a tensor is as fast as possible. Here is the first sentence: An nth-rank tensor in m-dimensional space is a mathematical object that has n indices and mn components and obeys certain transformation rules. Let me fix m at 3. An nth-rank tensor in 3-dimensional space is a mathematical object that has n indices and 3n components and obeys certain transformation rules. Suppose that n=0. Then a zero rank tensor in three dimensional space is a mathematical object that has 0 indices and 1 component, and obeys certain transformation rules. This mathematical object is a scalar. Suppose that n=1. Then a first rank tensor in three dimensional space is a mathematical object that has one indice and 3^1=3 components, and obeys certain transformation rules. This mathematical object is a vector. Suppose that n=2. Then a second rank tensor in a three dimensional space is a mathematical object that has two indices, and 9 components, and obeys certain transformation rules. Here is what it says about notation... The notation for a tensor is similar to that of a matrix (i.e., [math] (A_i_j) [/math], except that a tensor may have an arbitrary number of indices. The article makes it clear that whether an indice is a superscript or subscript matters in the symbol's interpretation. contravariant... upper indice covariant... lower indice The author of the article then says something very important... While the distinction between covariant and contravariant indices must be made for general tensors, the two are equivalent for tensors in three-dimensional Euclidean space, and such tensors are known as Cartesian tensors. Tensor notation Consider our vector u from earlier. [math] \vec u = u1 \hat i + u2 \hat j + u3 \hat k [/math] In the mathworld article at Wolfram, we are told the following: An nth-rank tensor in m-dimensional space is a mathematical object that has n indices and components and obeys certain transformation rules. Fixing the dimension of the space at 3, leads to: An nth-rank tensor in 3-dimensional space is a mathematical object that has n indices and 3^m components and obeys certain transformation rules. Now, keeping in mind that a vector is a first rank tensor, it follows that: An first-rank tensor in 3-dimensional space is a mathematical object that has 1 indice and 3 components and obeys certain transformation rules. So if we use lower indices, the notation for our vector u is: [math] \vec u = u_1 \hat i + u_2 \hat j + u_3 \hat k [/math] instead. The author then says this Tensor notation can provide a very concise way of writing vector and more general identities. For example, in tensor notation, the dot product is simply written [math] \vec v \bullet \vec u = u_i v^i [/math] So far the author has been consistent. A vector in three space will have one indice (apparently either upper or lower, the choice is yours), and it will have three components. The logic of tensor notation seems quite convoluted. Actually, what I can do is this... [math] \vec u = \sum_{i=1}^{i=n} u_i [/math] Now I just need to get the basis vectors into the notation somehow. Here is the notation for an arbitrary point in three dimensional space, in a frame of reference with an x axis, y axis, and z axis P = (x,y,z) Note that this is a three tuple, and what that means is this... that means that order matters. In other words... not( (2,-4,3) = (3,2,-4) )

The addition takes awhile, but I made an improvement on the binomial expansion formula.

The integral is absolutely trivial. What you need is the series expansion of: [math] (1+U)^\alpha [/math] Where U is a function of x, that is U=U(x). I can help you later. In your problem, alpha = 1/2, and U(x) = x^3. Your final answer will be a series, and if you can put it into a program to rapidly do the addition for you. If you want an answer written in terms of elementary functions, then you will either need to use a different approach, or figure out a way to transform the final series which you got after integrating, back into elementary functions. This isn't always possible though. Regards

Here's what I know so far... I know what the equation looks like. And don't worry about me getting offended, it wont happen. I know the future, well a very tiny portion of it anyways. Kind regards Actually, I know a tiny bit about the stress energy tensor. I do know that it is present in what everyone is calling the field equations. Also, that isn't the formula that Martin, or Tom Mattson showed me (theirs had three terms), and I know there is one that involves the ricci tensor. I don't know what the stress tensor represents though. Let me see if i remember what Martin showed me... [math] G_{\mu \nu} = \frac{8 \pi G}{c^4} T_{\mu \nu} [/math] G/c^4 is supposed to have units of inverse force I believe, and... G mu nu is called the umm The Einstein tensor. And T mu nu is called the stress-energy tensor, but as I said I don't know what that is supposed to symbolize. I also know that G is the Newtonian gravitational constant, and c is the speed 299792458 meters per second. I also know that the Einstein tensor can be written in terms of the ricci tensor and the ricci scalar, but I don't remember what they look like, Tom showed me. And I also remember that Martin was eager to mix hbar in somehow. He gave me a series of problems to solve. And I do know a fair amount of electrodynamics, and I remember seeing comments about the Maxwell stress-energy tensor, when I first took the class, but we skimmed over that part. And I also remember that Martin said that sometimes the 8 pi is present in things, and sometimes not. And I know what a positive definite matrix is. It has a positive trace, the trace being equivalent to the sum of the components down the main diagonal. That is: [math] Tr(A) = \Sigma A_{ii} [/math] That's about all I know about it off the top of my head.

Well its my money, and I don't mind losing. But you never know. Actually I already know, but whatever, my proof skills need work.

I am going to try again to prove this. The goal is to construct the simplest proof possible for it. It has to be made rigorously logical first though. For reference, here are the Galilean transformations: [math] x^\prime = x - vt [/math] [math] y^\prime = y [/math] [math] z^\prime = z [/math] [math] t^\prime = t [/math] Swansont mentions the spatial transformation term vt. Let us consider two arbitrary moments in time t2,t1, with t1 before t2, in frame S. The Galilean transfomations require that the following statements be true at moment in time t1, in frame S: [math] x1^\prime = x1 - vt1 [/math] [math] y1^\prime = y1 [/math] [math] z1^\prime = z1 [/math] [math] t1^\prime = t1 [/math] In the statements above, (x1,y1,z1) denotes the coordinates of an arbitrary point in frame S, at moment in time t1, and (x1`,y1`,z1`) denotes the coordinates of that very same point, but in frame S`, which is moving relative to S. Now, consider things at moment in time t2. Using the same logic as before, the following statements are simultaneously true in both frames (S,S`), at moment in time t2: [math] x2^\prime = x2 - vt2 [/math] [math] y2^\prime = y2 [/math] [math] z2^\prime = z2 [/math] [math] t2^\prime = t2 [/math] In the statements above, (x2,y2,z2) denotes the coordinates of an arbitrary point in frame S, at moment in time t2, and (x2`,y2`,z2`) denotes the coordinates of that very same point, but in frame S`, which is moving relative to S. Now, in frame S, the location where the axes of S intersect is the point (0,0,0), and this statement is always true in frame S. In S`, the location where the axes of S` intersect is the point (0`,0`,0`), and this statement is always true in frame S`. So consider the location of the origin of S in frame S`, at moment in time t1: By the Galilean transformations the following statements are true in both frames: [math] x1^\prime = 0 - vt1=-vt1 [/math] [math] y1^\prime = 0 [/math] [math] z1^\prime = 0 [/math] [math] t1^\prime = t1 [/math] So, the coordinates of the origin of frame S in frame S` are: (-vt1,0,0) Now consider the location of the origin of S` in frame S, at moment in time t1: By the Galilean transformations the following statements are true in both frames: [math] 0^\prime = x - vt1=-vt1 [/math] [math] 0^\prime = y [/math] [math] 0^\prime = z [/math] [math] t1^\prime = t1 [/math] [math] vt1+ 0^\prime = x [/math] So, the coordinates of the origin of frame S` in frame S are: (vt1,0,0) Now, at moment in time t2, we have the following statements being true in both frames: [math] x1^\prime = 0 - vt2=-vt2 [/math] [math] y1^\prime = 0 [/math] [math] z1^\prime = 0 [/math] [math] t1^\prime = t2 [/math] So, the coordinates of the origin of frame S in frame S` are: (-vt2,0,0) Now consider the location of the origin of S` in frame S, at moment in time t1: By the Galilean transformations the following statements are true in both frames: [math] 0^\prime = x - vt2=-vt2 [/math] [math] 0^\prime = y [/math] [math] 0^\prime = z [/math] [math] t2^\prime = t2 [/math] [math] vt2+ 0^\prime = x [/math] So, the coordinates of the origin of frame S` in frame S are: (vt2,0,0) Now, focus upon the location of the origin of frame S` in frame S, at both moments in time. At t1, the coordinates of the origin of frame S` in frame S were computed to be: (vt1,0,0) At t2, the coordinates of the origin of frame S` in frame S were computed to be: (vt2,0,0) Using the finite discrete difference calculus, the instantaneous change in any quantity Q, is found by subtracting the earlier value from the latter, over two adjacent moments in time. So if Q initially has value Q1, and then one moment in time later it has value Q2, then the change in Q is defined to be Q2-Q1. That is, we have the following definition: Definition: [math] \Delta Q \equiv Q2-Q1 [/math] So in frame S, we can compute the following differences: The change in the z coordinate of the origin of frame S` in frame S, is by definition: [math] \Delta Z \equiv 0 - 0 = 0 [/math] The change in the y coordinate of the origin of frame S` in frame S, is by definition: [math] \Delta Y \equiv 0 - 0 = 0 [/math] The change in the x coordinate of the origin of frame S` in frame S, is by definition: [math] \Delta X \equiv vt2 - vt1 = v(t2-t1) [/math] Now, focus upon the location of the origin of frame S in frame S`, at both moments in time. At t1, the coordinates of the origin of frame S in frame S` were computed to be: (-vt1,0,0) At t2, the coordinates of the origin of frame S in frame S` were computed to be: (-vt2,0,0) Notice, that we could have also said that: At t1`, the coordinates of the origin of frame S in frame S` were computed to be: (-vt1`,0,0) At t2`, the coordinates of the origin of frame S in frame S` were computed to be: (-vt2`,0,0) And the reason that we can also say this, is because for any moment in time t, t=t`, since we have assumed that the Galilean transformations are true. Now, in frame S`, we can compute the following differences: The change in the z` coordinate of the origin of frame S in frame S`, is by definition: [math] \Delta Z^\prime \equiv 0 - 0 = 0 [/math] The change in the y` coordinate of the origin of frame S in frame S`, is by definition: [math] \Delta Y^\prime \equiv 0 - 0 = 0 [/math] The change in the x` coordinate of the origin of frame S in frame S`, is by definition: [math] \Delta X^\prime \equiv vt2^\prime - vt1^\prime = v(t2^\prime-t1^\prime) =v(t2-t1) [/math] Since moment in time t2, is different from moment in time t1, we can divide both sides of the equation above by (t2-t1) or (t2`-t1`), to obtain an expression for v. Incidentally, t1 has to be a number, and t2 has to be a number, therefore, we shall assume that there is a clock/clocks at rest in frame S, which map natural numbers onto moments in time. In the case where the amount of time has to be determined by different clocks at rest in a frame, as long as the clocks are initially in sync, they remain synchronous, since they tick at the same rate. So we have: [math] \frac{\Delta X^\prime}{t2-t1} = \frac{\Delta X^\prime}{t2^\prime -t1^\prime} = v [/math] Similarly we find that: [math] \frac{\Delta X}{t2-t1} = \frac{\Delta X}{t2^\prime -t1^\prime} = v [/math] Now, using the transitive property of equality, and the few relations just given, we have: [math] v = \frac{\Delta X}{(t2^\prime -t1^\prime)} = \frac{\Delta X^\prime}{(t2-t1)} [/math] To requote swansont So we are to focus on the spatial transformation term vt. Multiplying both sides of the equation above by (t2-t1) we obtain: [math] v (t2-t1) = \frac{(t2-t1) }{(t2^\prime -t1^\prime)} \Delta X = \Delta X^\prime [/math] Now, denoting (t2-t1) by Dt we have: [math] v \Delta t = \frac{(t2-t1) }{(t2^\prime -t1^\prime)} \Delta X = \Delta X^\prime [/math] The LHS is not quite the spatial transformation term, but it is close. But the thing that I would point out now, is that to even define "relative speed v" you need two moments in time, not one. Now might be a good time to close the scope of any assumptions. What I am trying to do right now, is determine whether or not what swansont said is true. What was asked was this, if different observers must measure the same distance, mustn't it be the case that the Galilean transformations are true? Symbolically, that means this: ? [math] \Delta X = \Delta X^\prime [/math] iff GT are true. I am going to have a look at that in two parts: Part 1: If [math] \Delta X = \Delta X^\prime [/math] then GT are true. Part 2: If GT are true then [math] \Delta X = \Delta X^\prime [/math] From one of the relations above, and the assumption that t=t`, we can see that it must be the case that delta X = delta X`. So all we have proven thus far, is the following: If the Galilean transformations are true then observers with relative speed v must measure the same amount of change in the distance between them. The argument above did not prove the converse, which is part 1 above, namely: Converse: If two arbitrary frames S,S` with relative speed v must measure the same amount of change in the distance between them, then the Galilean transformations are true. Before trying to tackle the converse, let us be clear as to what is meant by relative speed v. In some frame S, something moves, lets say along the X axis. The distance it moves as measured by rulers at rest in S, is to be denoted by: [math] \Delta X [/math] And the time coordinate of this frame is to be denoted by t. So, for something that moves in this frame, the amount of time of travel, as measured by clocks at rest in frame S, is going to be denoted by: [math] \Delta t [/math] So, for two consecutive moments in time t2,t1, with t1 before 2 in this frame, we have: [math] \Delta t = t2 - t1 [/math] The relative speed is then defined to be the ratio of the distance traveled in the frame, divided by the time of travel in the frame, that is: [math] v \equiv \frac{\Delta X}{\Delta t} [/math] Now, the definition of relative speed must be the same in both frames. So in frame S`, the origin of S moved a distance which is denoted by: [math] \Delta X^\prime [/math] And that distance is to be measured by rulers at rest in S`. And the time coordinate of frame S` is denoted by t, and the amount of time of travel of the origin of S through S` as measured by clocks at rest in S` is denoted by: [math] \Delta t^\prime [/math] And using the same definition of relative speed v that an observer at rest in S must use, we have: [math] v^\prime \equiv \frac{\Delta X^\prime}{\Delta t^\prime} [/math] And the relative speeds must be equal, that is v=v` therefore: [math] v \equiv \frac{\Delta X}{\Delta t} = \frac{\Delta X^\prime}{\Delta t^\prime}[/math] And the preceding equation is true, regardless of whether or not the Galilean transformations are true. From the equation above, we see that if the numerators are equivalent, then the denominators are equivalent, and if the denominators are equivalent, then the numerators are equivalent. That is, measurments of distances are equal if and only if measurements of "amounts of time" are equal. If [math] \Delta X = \Delta X^\prime [/math] then GR are true. Let the relative motion be along coincident X axes of two arbitrary frames S,S`. Denote the relative speed by v. Assumption 1: [math] \Delta X = \Delta X^\prime [/math] The LHS of the equation above, is the change in the X coordinate of the origin of S` in S. Let X1 denote the initial X coordinate of frame S` in frame S, at moment in time t1, and let X2 denote the X coordinate of frame S` in frame S, at the very next moment in time, t2. The RHS of the equation above, is the change in the X coordinate of the origin of S in S`. Let X1` denote the initial X coordinate of frame S in frame S`, at moment in time t1`, and let X2` denote the X coordinate of frame S in frame S`, at the very next moment in time, t2`. Therefore: [math]\Delta X = x2-x1 [/math] [math]\Delta X^\prime = x2^\prime - x1^\prime [/math] Now, the speed of the origin of S` in S is by definition: [math] v = \frac{\Delta X}{\Delta t} = \frac{x2-x1}{\Delta t} =\frac{x2-x1}{t2-t1} [/math] And the speed of the origin of S in S` is by definition: [math] v^\prime = \frac{\Delta X^\prime}{\Delta t^\prime} = \frac{x2^\prime-x1^\prime}{\Delta t^\prime} = \frac{x2^\prime-x1^\prime}{t2^\prime-t1^\prime} [/math] And the relative speeds must be equal, that is v=v`. Therefore: [math] \frac{x2-x1}{t2-t1} = \frac{x2^\prime-x1^\prime}{t2^\prime-t1^\prime} [/math] So, the assumption that both observers must measure the same distance travelled, that means that the numerators must be equal. Therefore, the denominators are equal. Therefore, under the assumption that the distance traveled is the same in both frames, it necessarily follows that: [math] t2-t1 = t2^\prime - t1^\prime [/math] Now, let the two clocks be synchronized at moment in time t1=t1`, and let them read the number 0. Therefore, [math] t2-0 = t2= t2^\prime - 0^\prime = t2^\prime [/math] Therefore, for any moment in time, which comes after t1=t1`, the clocks will still be synchronous. This is one of the four Galilean relationships. There are three left to verify. Now, the relative motion was stipulated to be along the common X axis of both frames, therefore the following statements were stipulated to be true: [math] y^\prime = y [/math] [math] z^\prime = z [/math] And the relation [math] x2-x1 =x2^\prime - x1^\prime [/math] was assumed to be true. So the only Galilean relation which is left to verify is the following one: [math] x^\prime = x - vt [/math] Once the previous relation is verified, we will have proven the following: Theorem: If [math] \Delta X = \Delta X^\prime [/math] then the Galilean transformations are true, in the case where S,S` are in uniform relative (non-rotational) motion, with speed v. Which is what I think swansont asked me about. So, the answer is yes to his question. That is, if two observers in uniform relative motion with constant speed v must agree upon distance measurements, then they must also agree upon time measurments, which is equivalent to saying that the Galilean transformations are true. Since the relative speed computations must be equal we have: [math] v = \frac{x2-x1}{t2-t1} = \frac{x2^\prime -x1^\prime}{t2^\prime-t1^\prime} [/math] Under the assumption that delta X = Delta X` we can reverse the numerators above, as follows, so that the following statements are true under the single assumption: [math] v = \frac{x2^\prime -x1^\prime}{t2-t1} = \frac{x2-x1}{t2^\prime-t1^\prime} [/math] Let the clocks by synchronized at t1=t1`, and let them both read zero. Therfore we have: [math] v = \frac{x2^\prime -x1^\prime}{t2} = \frac{x2-x1}{t2^\prime} [/math] And we have already shown that under the assumption that delta x = delta x`, that we must have t=t` for any moment in time t. Therefore: [math] v = \frac{x2^\prime -x1^\prime}{t2} = \frac{x2-x1}{t2} [/math] We have not caused division by zero error. Multiplying by t2 we have: [math] vt2 = x2^\prime -x1^\prime= x2-x1 [/math] Therefore, if delta X = delta X` then [math] x1^\prime= x2^\prime -vt2[/math] This argument is way too long, and it involved meandering, but I just figured out how to finish it off. We are trying to prove this: [math] x^\prime = x - vt [/math] Now, at moment in time t1=t1`=0, the equation above leads to: [math] x1^\prime = x1 [/math] And at moment moment in time t2, the equation above leads to: [math] x2^\prime = x2 - vt2 [/math] So here are our initial conditions for S,S`: t1=t1`=0 And this For any point X1 on the X axis of S, and any point X1` on the X axis of S`: x1=x1`. Now here is the only assumption to be made during the argument: x2-x1=x2`-x1` And we have already shown that: If x2-x1=x2`-x1` then for any moment in time t2>t1 it must be true that t2=t2`. All that remains, is to use the information properly, to formulate the first of the four Galilean relations, and then clean up the argument, so that the final argument is in its simplest form. It should minimize the number of mental steps necessary to reach the conclusion. We start off with this: Theorem: Initial conditions: t1=t1`=0 And this For any point X1 on the X axis of S, and any point X1` on the X axis of S`: x1=x1`. (What these initial conditions do, is stipulate that at moment in time t1=t1`, the frames S,S` are equivalent. In other words, their axes not only overlap, but the positive x,y,z axes of each also correspond. If " moment in time t>t1: (y`=y and z`=z and Dx = Dx` then " moment in time t>t1: [math] x^\prime = x - vt [/math] [math] y^\prime = y [/math] [math] z^\prime = z [/math] [math] t^\prime = t [/math] We can now use first order logic to solve the problem. Solution. We have the following initial condition: For any point x on the x axis of S, and any point x` on the x axis of S` x`=x Now, suppose the following: x2-x1=x2`-x1` It was previously shown that: If x2-x1=x2`-x1` then " moment in time t>t1: t`=t Therefore: t`=t Therefore vt`=vt Therefore: x2-x1 - vt =x2`-x1` -vt` The initial conditions are such that when t1=t1`= 0 x1=x1` but at the very next moment in time, it is no longer the case that x1=x`, because the frames are in relative motion with speed v. At all moments in time y`=y, z`=z, and under the single assumption, we also have t`=t, all that remains to be shown, is that for any moment in time after t1=0, we have x`=x-vt as the coordinate transformation from one frame to another. At moment in time t2, the relation above gives: x2-x1 - vt2 =x2`-x1` -vt2` Now, at moment in time t1=t1`=0, we have from the initial conditions x1=x1`. Now, at moment in time t2, a point on the x axis of frame S` has moved along the x axis of frame S, and the distance traveled is equal to the relative speed times the time of travel. The time of travel is given by t2-t1=t2-0=t2 So that the distance travelled in frame S, by a point (x`,y`,z`) that is fixed in S` is given by vt2. So consider the fixed point (4,0,0) in frame S`. Initially, its coordinates in frame S were the same, namely (4,0,0). Now, at the very next moment in time, it moved a positive distance vt2, in frame S. Suppose the distance traveled in this time was three units. Then this fixed point in S`, no longer has an x coordinate of 4, its x coordinate is now displaced from that by 3 units. Therefore, either its new location in S is 7, or its new location in S is 1. So now, we need one more initial condition. We need to know the direction of motion of a fixed point in S`, in frame S. Thus, we need to know not just the relative speed, but the relative velocity of the frames. Suppose a fixed point in S` moves in the i^ direction of frame S. Then at moment in time t2, the point (4,0,0) fixed in S`, has had its x coordinate in frame S increased by amount vt2=3. Therefore its new coordinate in frame S is 7. And therefore, in frame S`, a fixed point in S moved with the following velocity: [math] \vec v^\prime = -v \hat i^\prime = -\frac{\Delta x^\prime}{\Delta t^\prime} \hat i^\prime [/math] And of course i^ = i^` therefore we can just write: [math] \vec v^\prime = -v \hat i = -\frac{\Delta x^\prime}{\Delta t^\prime} \hat i[/math] And therefore, in frame S, a fixed point in S` moved with the following velocity: [math] \vec v= v \hat i = \frac{\Delta x}{\Delta t} \hat i [/math] So, we have supposed that a fixed point in S` moves in the i^direction of frame S, and that a fixed point in S moves in the -i^` direction of frame S`. So here is the idea behind the Galilean transformation. Consider a fixed point in reference frame S, say (5,0,0). At the moment in time t1=t1`=0, the initial conditions are such that the frames overlap perfectly, in other words S=S` when t=t1=t2=0. So, at that moment in time the coordinate of that point in S` is also (5,0,0) Now, we know the relative velocity, I just wrote it. So as the time variable t increases, the point (5,0,0) which is fixed (at rest) in frame S, has moved in frame S`. And it has moved in the -i^` direction of frame S`. So the x` coordinate of (5,0,0) is no longer 5, it is something less than 5, because it moved in the -i^` direction. The distance it has moved in frame S` is given by vt`, but since t=t` in the Galilean transformations, we can just as well say that the distance it has moved in frame S` is given by vt. So that its new position in frame S` one moment in time in the future is given by: 5 - vt Its original position in frame S` was 5. So over the first two consecutive moments in time the following statement is true: x` = 5-vt We can write this a little more explicitely as follows: [math] x^\prime = 5 - \frac{(x2-x1)t}{t2-t1} [/math] Now, lets look at the first few instantiations... when the variable t is at t1=0, we have: [math] x^\prime = 5 - \frac{(x2-x1)t1}{t2-t1} = 5 - 0 [/math] This is correct, because initially the frames are equivalent, that is they are really only one frame, and the fixed point (5,0,0) in frame S, has the following coordinates in frame S` (5,0,0), which is what the statement above says. Now, when the variable t is instantiated by t2, we have: [math] x^\prime = 5 - \frac{(x2-x1)t2}{t2-t1} [/math] And since t1 is stipulated to be zero we have: [math] x^\prime = 5 - \frac{(x2-x1)t2}{t2} [/math] And let us stipulate that the relative velocity doesn't change. We can now instantiate the variable t with t3, and obtain: [math] x^\prime = 5 - \frac{(x2-x1)t3}{t2} [/math] Thus, we can write x` as a function of t, as follows: [math] x^\prime(t) = 5 - vt [/math] Where we understand that the relative velocity is constant in time. And the statement above would be true for any fixed point on the x axis of frame S, not just the point (5,0,0). So for any point on the x axis of frame S (x,0,0) the following statement is true: [math] x^\prime(t) = x - vt [/math] Under the single assumption made. Therefore we are done. We have proven the following fact: If delta x=delta x` then the Galilean transformations are true. Specifically: [math] x^\prime(t) = x - vt [/math] and y`=y z`=z and t`=t

But simultaneity isn't relative.

This has been on my mind for two hours now. Your statement needs to be converted to a statement using first order logic. The word observer needs to be replaced. If any observation frame... Let there be two points in frame F1. P1=(x1,y1,z1) P2=(x2,y2,z2) Axiom I: Points in a frame cannot move in the frame. By axiom 1, the distance between the two points is constant in time. There is no change in that distance, over any two consecutive moments in time. It now must be the case that the Pythagorean theorem is constantly true in the frame. "P1 in F1 "P2 in F1: [math] D(P1,P2) = \sqrt{(x2-x1)^2+(y2-y1)^2+(z2-z1)^2} [/math] and [math] \frac{d(D)}{dt} = 0 [/math] Definition: The distance between two arbitrary points is one unit long if and only if: [math] D(P1,P2) = \sqrt{(x2-x1)^2+(y2-y1)^2+(z2-z1)^2} = 1 [/math] Let us have been given a solid object, whose length in time is almost always equal to one unit of frame length. At least to a point beyond sensory perception to judge otherwise. A real ruler. Now, this ruler can move within the frame. So the coordinates of its center of mass, can be plotted as a function of discrete moments in time. It can move any way that real rulers move, and only in a way that a real ruler can move. It can rotate, translate, expand, and contract. Anything that a real ruler can do, this given ruler can do. So now we have to define the measurement process. Let there be two points in the frame, which are one unit of frame length apart. That is the distance between them. In other words, D=1. Also, the length of this ruler is to within a good approximation, also equal to one. now, we want to move the ruler up to the points, and measure the distance between them. If the ruler starts off at rest between the two points, then the measurement process is over. The ruler will be at rest in the frame, and it will have it's proper length for sure, independently of the theory of SR, and hence the measurement will agree with the Pythagorean theorem, to within a small tolerance, which is beyond our ability to percieve, or measure. Now, we have to consider the case where the ruler is not aligned with the two points that we wish to measure the distance between. As long as we can translate the ruler, and rotate is as necessary, we can get it next to the points, the distance between we wish to measure. And even if its length is a function of speed in a frame, as long as we can bring it to rest in the frame, it will again have its proper length, as in the previous case, and correctly measure the distance to be 1, within allowable tolerance of real ruler contraction, and expansion, due to temperature. Axiom II: Any ruler initially moving in a frame, can be brought to rest in the frame. The goal is to prove this: If any observer must measure the distance between the two points as D, doesn't that imply that the Galilean transformations are true statements. There needs to be a perfect logical link between everything. So what are the Galilean transformations? They are: [math] x^\prime = x - vt [/math] [math] y^\prime = y [/math] [math] z^\prime = z [/math] [math] t^\prime = t [/math] These transformations are for two frames in rectilinear motion. The relative speed is v. Consider things at the moment in time when t=0 here is what the transformations say at that moment in time... [math] x^\prime = x [/math] [math] y^\prime = y [/math] [math] z^\prime = z [/math] [math] t^\prime = t=0 [/math] The four statements above, are actually saying that the frames overlap, at this moment in time. Let us call the unprimed frame S, and the the primed frame S`. Consider the point (1,1,1) in frame S. At the moment in time when t=t`=0, we have for this point: x=1 y=1 z=1 Now, we ask for the coordinates of this point in frame S`. The Galilean transformations above, stipulate that when t=0 the following statements are true: [math] x^\prime = 1 [/math] [math] y^\prime = 1 [/math] [math] z^\prime = 1 [/math] So right now, given the coordinates of any point in S, the coordinates of the exact same point in S`, are identical to S. This means chiefly that the axes of both frames coincide. But more than that, it means that the basis vectors point in the same direction. So that at this moment in time the two frames are really one frame, in other words if t=0 then S=S`. Now consider things at the very next moment in time. Increment the time coordinate of frame S by one unit. Thus, at the next moment in time we have t=1. Here are what the Galilean transformations tell us are true at this moment in time: [math] x^\prime = x - v [/math] [math] y^\prime = y [/math] [math] z^\prime = z [/math] [math] t^\prime = t =1[/math] Now, in order to define instantaneous relative speed v, you need two consecutive moments in time. The speed of the origin of S` through S, must be equivalent to the speed of the origin of S through S`, since speed is relative. Now here is what we can infer from the Galilean transformations... Consider the point which is still at (1,1,1) in frame S. Therefore, the coordinates of this point in S are given by x=1,y=1,z=1. According to the Galilean transformations, the coordinate of this point in frame S` are: [math] x^\prime = 1 - v [/math] [math] y^\prime = 1 [/math] [math] z^\prime = 1 [/math] [math] t^\prime = t =1[/math] If v=0, then the frames are not in relative motion, and they are still one frame. Let it be stipulated that the relative speed is nonzero. Since speed is a strictly positive quantity, it now follows that the speed is greater than zero, i.e. v>0. For the sake of clarity, let the frame length of S be one meter. Since the frames are stipulated to overlap at t=0, the frame length of S` is also one meter. Let the unit of time in frame S be the second. Now, for the sake of clarity, let the relative speed v, be equal to one meter per second. Therefore, the Galilean transformations predict that at the very next moment in time we have: [math] x^\prime = 1 - 1 = 0 [/math] [math] y^\prime = 1 [/math] [math] z^\prime = 1 [/math] [math] t^\prime = t =1[/math] So how did the frames just move in relation to one another? Well... The point (1,1,1) in S has coordinates (0,1,1) in S`. So what does that mean? That means this... The point on the positive x axis of frame S which is permanently located one meter away from the origin of frame S, is now located at the origin of frame S`, but also... There are an infinite sequence of true statements. For example, consider the point on the positive x axis of frame S which is permanently located two meters away from the origin of frame S. In frame S its coordinates are (2,0,0). Using the Galilean transformations, its coordinates in S` are (1,0,0). And now, consider the point on the positive x axis of frame S which is permanently located three meters away from the origin of frame s`. Its coordinates in S are (3,0,0). Using the Galilean transformations, its coordinates in S` are: (2,0,0) So here is what happened. Suppose that at the previous moment in time, the x axes of both frames pointed from left to right as illustrated below: ---------------------------------------------------> To an observer at rest in observation frame S, the origin of S` is currently located at (1,0,0). This took one second, as measured by a clock at rest in S, and it also took one second as measured by a clock at rest in S` (since t=t`). The speed of the origin of S` through S is defined as the distance traveled in S (as measured by a ruler at rest in S), divided by the time of travel (as measured by a clock at rest in S). The distance traveled over the two consecutive moments in time was one meter, the amount of time this took was one second (the second is really composed of an enormous number of moments in time, but that isn't relevant to the argument). So the speed of the origin of S` through S has a computation that looks like this: [math] v = \frac{x_f-x_i}{t1-t0} = \frac{1-0}{1-0} = 1 [/math] Which we stipulated to be true. In the computation above, xi was the initial x coordinate of the origin of frame S` in frame S (which was at xi=0), and xf was the location of the origin of S` in frame S at the very next moment in time. So focus on the ZY plane of frame S`. Every point in that plane just moved one meter along the x axis of frame S.

Do what I want them to. But since they are massless, they do what I want them to by current physical theory.

I can't just say yes to anything, I have to think for a moment...

I was trying to say that mathematically before. Well if you define it right, you can make the axes do what I want them to.

yeah ok, that fits in there somewhere. Unless you stipulate that the axes are massless. These axes aren't real solid objects. They allow you track the motion of real things though. In other words... I am someplace in the universe, and I have an origin of a frame where I am. And I can rotate my frame, so that the positive x axis passes through the center of inertia of our sun. So when I rotated my frame, to align it in this manner, points on the x axis 1 trillion light years away from my current location, rapidly moved... lol But the axes arent real. So you twist your frame, and make the center of mass of our sun lie on the x axis... it is then 8 minutes away, at the speed of light And we can check this, to get to meters... 8(60)(299792458) = 143900379840 143900379840 = 1.4 x 1011 meters

Definition: Let (x1,y1,z1), (x2,y2,z2) denote the coordinates of a point in some frame. Let D be defined as follows: [math] D = \sqrt{(x2-x1)^2+(y2-y1)^2+(z2-z1)^2} [/math] The axes are rigid if and only if d(D) = 0 Ok, I'm not going to let JC or Swansont confuse me. Here is the deal. I have an actual meterstick. This one single meterstick is used to define distances throughout the entire universe. So if I want to know the distance from my current location to the center of this universe, what I want to know is how many metersticks I would have to place in a striaght line, to reach from here to there. If I want to know the distance from here to the sun, again, I want to know how many metersticks I need. If I want to know the distance from the sun to some other star, again, I want to know how many metersticks I need. So the Pythagorean theorem is irrelevent. Again, if I want to know the distance from my current location to the sun, I want to know how many meter sticks i need, to rightnow simultaneously place in a straight line pointing right at the sun. In other words I want to know how many metersticks I can fit in between my current location, and the sun. So for example, here is the current approximate distance from the earth to the sun: Astronomical Unit So there it is. 150 million kilometers And since 1000 meters = 1 kilometer The approximate distance from the earth to the sun is [math] 1.5 \times 10^8 Km \frac{1000 m}{Km} = 1.5 \times 10^{11} m [/math] So 150 billion metersticks would be needed. And this gives the mind a sense of the distance. So this is why we need rigid axes. Real rulers must be substitutable for the axes. But not rulers that can expand or contract though. When we were interested in the distance from here to the sun, we used rigid rulers. And the length of a real ruler doesn't vary much over large amounts of time. And any change in its length is certainly imperceptible. So this is the empirical thought behind rigid axes. In order to say this logically, it suffices to say that the distance between any two points on an axis is constant in time. So given any frame, we are given three coordinate axes. The distance between any two points on an axis is found by subtracting the lesser coordinate from the greater. So the distance between the point 4 on the x axis, and the point 7 on the x axis is simply 7-4=3. And this distance never changes. If this distance could change, then the axis wouldn't be rigid. The key thing here, is temporal constant. This is the concept to be developed.

I said that several days ago, but I went back and re-read it and I think I meant this... You don't have to be serious about physics. No one has to. But, for those who are serious about the question, it falls into the domain of physics.

I've read about Emmy Noether before. What is exactly is a continuous symmetry transformation? It didn't say clearly there. I'm still looking it over though.