swansont Posted April 6, 2005 Posted April 6, 2005 Oh it works... Let me try something.... How does simultaneity relate to whether or not I am using Euclidean space? The Pythagorean theorem does not hold in, for example, a hyperbolic geometry. I don't want you to be assuming a geometry that may not hold in a relativistic situation. I know it doesn't hold in GR. I can't recall off of the top of my head if it holds in SR.
Johnny5 Posted April 6, 2005 Posted April 6, 2005 The Pythagorean theorem does not hold in' date=' for example, a hyperbolic geometry. I don't want you to be assuming a geometry that may not hold in a relativistic situation. I know it doesn't hold in GR. I can't recall off of the top of my head if it holds in SR.[/quote'] The pythagorean theorem doesn't hold in GR because... because space can expand right? What if space cannot expand, that its just things moving away from us? Look at how I defined rigid frame. Frame: Three mutually perpendicular coordinate axes, with unit of distance chosen. A rigid frame, is a frame in which the distance between any two points obeys the Pythagorean theorem. In other words, the Pythagorean theorem is constantly true in the frame. I know what I mean really really well, I can say it this way... Given three coordinate axes, with the same unit of distance chosen on each axis... We can assign to any point at space, three numbers. No better yet do it by formula: Let (x1,y1,z1) and (x2,y2,z2) denote two arbitrary points in a frame. The three dimensional version of the Pythagorean theorem is this: [math] D = \sqrt{ (x2-x1)^2 + (y2-y1)^2+(z2-z1)^2 } [/math] Now if the truth value of that statement is constant in time, and that statement is true, then the frame is rigid, and conversely. It is plain to see that statement is not a function of time. The following statement has the same truth value as the Pythagorean statement: [math] D^2 = (x2-x1)^2 + (y2-y1)^2+(z2-z1)^2 [/math] Differentiate both sides of the equation above, with respect to time. Or just write the differentials... [math] 2D (dD) = 2[(x2-x1)(d(x2)-d(x1))+...] [/math] Dividing both sides by 2... [math] dD = [(x2-x1)(d(x2)-d(x1))+...] [/math] The left hand side is the change in the distance between the two points, over two consecutive moments in time. In a rigid frame, this is always zero. So for a rigid frame... [math] 0 = [(x2-x1)(d(x2)-d(x1))+...] [/math] So the two points are really one point iff x2=x1 y2=y1 z2=z1 The points are different if the conjunction of the three simple statements above, is false. So we can define equality of points... Let P1=(x1,y1,z1) P2=(x2,y2,z2) P1=P2 if and only if ( x1=x2 & y1=y2 & z1=z2) So two points in one frame are different (i.e. really two and not one) if and only if not (x1=x2) or not(y1=y2) or not(z1=z2). And this connects to the Pythagorean statement above, about distance between two points in one frame. Because look at the statement... [math] 0 = [(x2-x1)(d(x2)-d(x1))+...] [/math] Suppose that two points in one frame have identical y coordinates, and identical z coordinates. Since they are two, it must be the case that their x coordinates are different. It would be the case that both points lie on the x axis of the frame. The distance between them by the Pythagorean theorem is: [math] D = \sqrt {(x2-x1)^2 } [/math] We know it isn't zero, because we have assured ourselves that x1,x2 are different. So... [math] D^2 = (x2-x1)^2 [/math] Now take the derivative of each side with respect to time... [math] 2D \frac{d(D)}{dt} = 2(x2-x1)(\frac{d(x2)}{dt}- \frac{d(x1)}{dt})) [/math] Dividing both sides by two we have: [math] D \frac{d(D)}{dt} = (x2-x1)(\frac{d(x2)}{dt}- \frac{d(x1)}{dt})) [/math] Now, in the formula above, D is greater than zero for sure, and not (x2-x1=0). So either the truth value of the pythagorean theorem can vary in time or not. If it cannot, then d(D)/dt=0. If it can, then not d(D)/dt=0. I say it cannot, therefore: [math] 0 = (x2-x1)(\frac{d(x2)}{dt}- \frac{d(x1)}{dt})) [/math] And because it was stipulated that not(x1=x2) it must be the case that not (x2-x1)=0, therefore we can divide both sides of the statement above by (x2-x1), to get the following statement: [math] 0 = (\frac{d(x2)}{dt}- \frac{d(x1)}{dt})) [/math] So this is a condition for the Pythagorean theorem to always be true. And I can explain why. While there are infinitely many frames to conceive of... they all map three numbers to a unique point in space. In other words, point P might have coordinates (1,2,3) in frame S, and coordinates (5,-1,1000) in frame S`, but it is still only one point. So in other words, if the Pythagorean theorem is true in one frame, then it is true in all frames. So here is the mathematical requirement for the Pythagorean theorem to be true, over two consecutive moments in time, in at least one frame: [math] 0 = (\frac{d(x2)}{dt}- \frac{d(x1)}{dt})) [/math] And we can rewrite this as: [math] \frac{d(x2)}{dt} = \frac{d(x1)}{dt} [/math] So how are we to interpret this? Well recall that we had two points P1,P2 in one frame F1. Recall that the points had the same y coordinate, and the same z coordinate, but they had different x coordinates. The x coordinate of P1 in frame F1 was X1, and the x coordinate of P2 in frame F1 was X2. Now, d(x2) would be a differential change in the x coordinate of P2, in frame F1, and likewise for d(x1). So lets say that the point P1 has coordinates (4,0,0) in frame F1, and that the point P2 has the coordinates (6,0,0) in frame F1, at some moment in time. In other words, they have these coordinates simultaneously in frame F1. So using the Pythagorean theorem at that moment in time gives the distance between the two points as: D = 6-4 = 2 So these locattions in space are two meters apart, at that moment in time. Now, the condition for the pythagorean theorem to constantly be true in frame F1, we have already found, and it is this: If the pythagorean theorem is true at all moments in time in frame F1 then For any point P1=(x1,y1,z1) in F1, and any point P2=(x2,y2,z2) in F1: [math] \frac{d(x2)}{dt} = \frac{d(x1)}{dt} [/math] Now, either d(x2)=0 or not. d(x2) is an infinitessimal change in the x coordinate of P2, in frame F1. So if the coordinates of a point in a frame cannot change then d(x2)=0. If they can change, then they must do so in a way consistent with the constraint above. So we know that at some moment in time we have the following: P1=(4,0,0) P2=(6,0,0) So at this moment in time, the following statement is true: x2=6 Now, if d(X2) is to be equal to zero then it must be the case that at the very next moment in time, x2=6. And, if d(x2) is to be nonzero, then it must be the case that at the very next moment in time, not (x2=6). So, suppose that at the next moment in time, x2=15. Now, look for a contradiction. Now, the constraint above must be true, therefore d(x1) = d(x2). d(x2) = 15-6 = 9 = d(x1) = ? - 4 Therefore, ? = 9+4 = 13. But this would be true for all points along the x axis, not just these two. In other words, if the x coordinate of one point in space increased by 9, then the x coordinate of all points along the x axis must have increased by 9, and additionally, the y,z axes must always pass through the point (0,0,0), so that they would have moved with the point (0,0,0). So it appears that the pythagorean theorem can remain constantly true in a frame that can move. But I haven't been as rigorously logical as I am capable of. This is actually a very important question to be answered. Can the coordinates of a point P1 in frame F1, change? In other words, I have shown that if the coordinates of one point can change, then they must all change by the same amount, but this doesn't mean that the coordinates of a point in a frame can vary. It just says what happens if they can. But, in this analysis, I have ignored the location of the center of mass of the universe. I have also ignored the laws of motion. There is no great issue at stake here, other than the meaning of the term frame. All right, I have said enough about this.
swansont Posted April 6, 2005 Posted April 6, 2005 The pythagorean theorem doesn't hold in GR because... because space can expand right? Because space isn't flat - it curves.
Johnny5 Posted April 6, 2005 Posted April 6, 2005 Because space isn't flat - it curves. Ok suppose that some region of space is severely curved. And some other region is almost perfectly flat. And I have a yardstick, which is straight when its in the region of flat space. Now, I bring that yardstick to the region of space which is severely curved. What is the shape of the yardstick now?
J.C.MacSwell Posted April 6, 2005 Posted April 6, 2005 Why do you have to go and complicate things. I don't want the axes to be able to expand and contract though' date=' because there was a unit of distance chosen. I want the axes rigid. Let matter expand and contract, not the axes.[/quote'] If you find an absolute reference frame (if such a thing exists) it is unlikely to be Euclidean (or even simply Lorentzian).
Johnny5 Posted April 6, 2005 Posted April 6, 2005 If[/b'] you find an absolute reference frame (if such a thing exists) it is unlikely to be Euclidean (or even simply Lorentzian). What is an absolute reference frame?
Johnny5 Posted April 6, 2005 Posted April 6, 2005 If[/b'] you find an absolute reference frame (if such a thing exists) it is unlikely to be Euclidean (or even simply Lorentzian). What do you mean it's unlikely to be Euclidean, or Lorentzian?
J.C.MacSwell Posted April 6, 2005 Posted April 6, 2005 What do you mean it's unlikely to be Euclidean, or Lorentzian? I think that if it was obvious we would already know what it was.
swansont Posted April 6, 2005 Posted April 6, 2005 Ok suppose that some region of space is severely curved. And some other region is almost perfectly flat. And I have a yardstick' date=' which is straight when its in the region of flat space. Now, I bring that yardstick to the region of space which is severely curved. What is the shape of the yardstick now?[/quote'] Who is doing the observing? I think that if you are in the curved space, it looks flat to you locally.
Johnny5 Posted April 6, 2005 Posted April 6, 2005 Who is doing the observing? I think that if you are in the curved space, it looks flat to you locally. Oh boy, someone right next to the object, locally to it in both regions.
Johnny5 Posted April 6, 2005 Posted April 6, 2005 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 1 AU = 149' date='597,870.691 kilometers [b'] Definition:[/b] An Astronomical Unit is approximately the mean distance between the Earth and the Sun. It is a derived constant and used to indicate distances within the solar system. Its formal definition is the radius of an unperturbed circular orbit a massless body would revolve about the sun in 2*(pi)/k days (i.e., 365.2568983.... days), where k is defined as the Gaussian constant exactly equal to 0.01720209895. Since an AU is based on radius of a circular orbit, one AU is actually slightly less than the average distance between the Earth and the Sun (approximately 150 million km or 93 million miles). 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.
swansont Posted April 6, 2005 Posted April 6, 2005 It's really a wasted effort. You need to be able to move on to the next step and look at the implications of your thesis. Rigid axes that work for any observer is just another way of saying that Galilean transforms are correct, which means that the speed of light depends on your reference frame and is c+v with respect to some reference frame (the frame of the "meterstick is used to define distances throughout the entire universe"), which means you should get interference fringes in a Michelson interferometer that is moving. And you don't. Michelson and Morley measured that ~120 years ago. (We know the earth isn't in the master reference frame because of the observation of stellar aberration) You're busily trying to show mathematically that the Titanic is an unsinkable ship, while everyone else knows it's already at the bottom of the ocean.
Johnny5 Posted April 6, 2005 Posted April 6, 2005 It's really a wasted effort. You need to be able to move on to the next step and look at the implications of your thesis. Rigid axes that work for any observer is just another way of saying that Galilean transforms are correct' date=' which means that the speed of light depends on your reference frame and is c+v with respect to some reference frame (the frame of the "meterstick is used to define distances throughout the entire universe"), which means you should get interference fringes in a Michelson interferometer that is moving. And you don't. Michelson and Morley measured that ~120 years ago. (We know the earth isn't in the master reference frame because of the observation of stellar aberration) You're busily trying to show mathematically that the Titanic is an unsinkable ship, while everyone else knows it's already at the bottom of the ocean.[/quote'] I didn't know that... Rigid axes that work for any observer leads to the Galilean transformations? Can you prove that? Hmmm I did an analysis of MM about 6-7 years ago, and as I recall there was a mistake in the derivation. S and S` were used as I recall. There was some kind of mistake... I just never told anyone about though. Book was by Serway, Moses, and Moyer... That was when I was trying to understand MM by the way. I can do this...
5614 Posted April 6, 2005 Posted April 6, 2005 I believe it is something to do with if it is infinitely rigid then if you had a pole infinitely rigid and you pressed one end the other end would move immediately, it is simultaneous (you pusing one end and the other end moving).
Johnny5 Posted April 6, 2005 Posted April 6, 2005 I believe it is something to do with if it is infinitely rigid then if you had a pole infinitely rigid and you pressed one end the other end would move immediately, it is simultaneous (you pusing one end and the other end moving). 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
5614 Posted April 6, 2005 Posted April 6, 2005 How can something massless be rigid? Rigid is a physical property possesed by something with mass!
Johnny5 Posted April 6, 2005 Posted April 6, 2005 How can something massless be rigid? Rigid is a physical property possesed by something with mass! 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.
swansont Posted April 6, 2005 Posted April 6, 2005 I didn't know that... Rigid axes that work for any observer leads to the Galilean transformations? Can you prove that? Hmmm I did an analysis of MM about 6-7 years ago' date=' and as I recall there was a mistake in the derivation. S and S` were used as I recall. There was some kind of mistake... I just never told anyone about though. Book was by Serway, Moses, and Moyer... That was when I was trying to understand MM by the way. I can do this...[/quote'] If any observer has to measure the same distance, doesn't that require the spatial tranformation term to be vt? That's the Galilean tranformation.
Johnny5 Posted April 6, 2005 Posted April 6, 2005 If any observer has to measure the same distance, doesn't that require the spatial tranformation term to be vt? That's the Galilean tranformation. I can't just say yes to anything, I have to think for a moment...
swansont Posted April 6, 2005 Posted April 6, 2005 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. Is the goal to make them do what you want them to, or to reflect reality? Those are not the same thing.
J.C.MacSwell Posted April 6, 2005 Posted April 6, 2005 Definition: Let (x1' date='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. You are doing quite a job explaining Euclidean space. If you try real hard you can explain Euclidean space in a number of ways. You can come up with many more examples of Euclidean space. But it will still be Euclidean space. It will be a very good model of reality at low speeds and a very poor one when higher speeds are involved.
Johnny5 Posted April 6, 2005 Posted April 6, 2005 Is the goal to make them do what you want them to, or to reflect reality? Those are not the same thing. Do what I want them to. But since they are massless, they do what I want them to by current physical theory.
swansont Posted April 6, 2005 Posted April 6, 2005 Do what I want them to. But since they are massless' date=' they do what I want them to by current physical theory. [/quote'] Except for relativity.
Johnny5 Posted April 6, 2005 Posted April 6, 2005 If any observer has to measure the same distance, doesn't that require the spatial tranformation term to be vt? That's the Galilean tranformation. 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.
Johnny5 Posted April 7, 2005 Posted April 7, 2005 Except for relativity. But simultaneity isn't relative.
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