Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 Why is it you chose not to accept physics (such as SR + GR) which most physicists in the world happily accept? And when nearly all physicists who are working in the field of physics prove them correct on a daily basis? This is a good question. Well for one, my mind seems to have no tolerance for space bending, and from what I can tell, GR says that space can bend. Also, I have found an error in the equivalence principle, using logic (and electrodynamics). So that explains this. As for SR, the same thing goes. The barn and ladder paradox... They can continue to use whatever makes them happy, that's inconsequential to me. There are so many things that physicists of today do not know, that I am sure to spot one or two of them in my time. Kind regards Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 No, you've changed the conditions of the experiment. The two are moving at some speed relative to each other. No acceleration. They start off in the same frame, with no relative speed. In order to have the relative speed v between them be greater than zero, one of them must be subjected to an external force, in order so that the relative speed become greater than zero. And so if we start with them at rest in some frame, then once the external force acts upon one of the twins, that twin will have acceleration a=dv/dt in that frame. Let that applied force last for time delta T. Then turn off the engines, and then he is coasting at constant relative speed v to the other twin. But beforehand, he must accelerate at least for a little while. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 And yet you've assumed X and Y are the same for both observers. It's X before Y and Y' before X' No. And I can show you what I did explicitely. I still have my paper. It is interesting that you primed things, without knowing the event which I analysed. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 4, 2005 Share Posted April 4, 2005 Then you must show it to be wrong' date=' not assume it to be wrong, and come up with a contradiction. The contradiction only means that one part is false - and you cannot assume that your assumption isn't it. You also have to be willing to follow your assumptions to all of their conclusions. You seem to disappear when that happens, only to pop up in a new thread. "Alas, to wear the mantle of Galileo it is not enough that you be persecuted by an unkind establishment; you must also be right." (Robert Park, of the American Physical Society)[/quote'] In the proof I made one assumption which was that the Lorentz contraction formula is true in any inertial reference frame. From that assumption I was able to arrive at the following statement, which was supposed to be true, under the single assumption: V = 0 and not (V=0) where V was the uniform relative speed of two rulers with the same proper length. There was only one assumption to be negated, Lorentz contraction. Regards. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 5, 2005 Share Posted April 5, 2005 Originally Posted by swansont If you assume absolute reference frames, you find that relativity doesn't hold. That's your contradiction. One of those two is false, but you haven't shown your assumption to be true. Originally Posted by Johnny5 What assumption? Read it again, please... Ok, I have read it again, so now I have to ask you, "What is an absolute reference frame?" Link to comment Share on other sites More sharing options...

swansont Posted April 5, 2005 Share Posted April 5, 2005 They start off in the same frame, with no relative speed. In order to have the relative speed v between them be greater than zero, one of them must be subjected to an external force, in order so that the relative speed become greater than zero. And so if we start with them at rest in some frame, then once the external force acts upon one of the twins, that twin will have acceleration a=dv/dt in that frame. Let that applied force last for time delta T. Then turn off the engines, and then he is coasting at constant relative speed v to the other twin. But beforehand, he must accelerate at least for a little while. No, you are discussing a different example. Once you accelerate you lose the symmetry. But I already pointed this out in post #4, and stated the conditions I was discussing: They both can say they are stationary and the other one is moving. Both see the other's clock slow down. And that's valid - until one of them accelerates. The time dilation equation applies to bot reference frames when there is a relative speed between them. It's special relativity, and you have to limit discussion to the conditions under which SR applies. add: janus pointed this out, too: If the twin accelerates then S' is not an inertial reference frame and remain permanently attached to the twin. If it is an inertial refernce frame it cannot remained attached to twin. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 5, 2005 Share Posted April 5, 2005 No' date=' you are discussing a different example. Once you accelerate you lose the symmetry. But I already pointed this out in post #4, and stated the conditions I was discussing: [i']They both can say they are stationary and the other one is moving. Both see the other's clock slow down. And that's valid - until one of them accelerates.[/i] The time dilation equation applies to bot reference frames when there is a relative speed between them. It's special relativity, and you have to limit discussion to the conditions under which SR applies. Well I was discussing the "twin paradox," as that was the original poster's question. You are saying things wrong. You say, and I quote, "Once you accelerate you lose the symmetry." What you should say is this, "Once one of the ships is sujected to an external force, symmetry is lost". Force and acceleration are not the same thing. This is why I think superacceleration is possible, rather than impossible. They can both say they are stationary, and the other is moving. I didn't say they could not do that. But they cannot each say that the other person's clock is slowing down. For suppose they can... John,Jim Then in the one frame, the statement that Jim turned 40 when John turned 31 will be true, and in the other frame, the statement that, "John turned 40 when Jim turned 31" will be true. This is a temporal paradox. And it is caused by assuming that the time dilation formula is true in both frames. Link to comment Share on other sites More sharing options...

swansont Posted April 5, 2005 Share Posted April 5, 2005 In the proof I made one assumption which was that the Lorentz contraction formula is true in any inertial reference frame. From that assumption I was able to arrive at the following statement' date=' which was supposed to be true, under the single assumption: V = 0 and not (V=0) where V was the uniform relative speed of two rulers with the same proper length. There was only one assumption to be negated, Lorentz contraction. Regards.[/quote'] But you used a non-inertial reference frame when you added an acceleration, which invalidates your assumption. No valid conclusion can be reached. And I can't find any other mention of V=0 in this thread. You made an analysis with X before Y and Y before X and not (X=Y), and in an inertial system where SR applies, you will have X before Y and Y' before X', because the two observers will measure different things. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 5, 2005 Share Posted April 5, 2005 But you used a non-inertial reference frame when you added an acceleration, which invalidates your assumption. No valid conclusion can be reached. And I can't find any other mention of V=0 in this thread. You made an analysis with X before Y and Y before X and not (X=Y), and in an inertial system where SR applies, you will have X before Y and Y' before X', because the two observers will measure different things. But this example problem isn't how I arrived at the contradiction. I used barn/ladder. What I did is not in this thread, nor any other. But it essentially is barn/ladder. I just happen to have good logical form. Actually, there was some new physics in what I did, but I saw it as intuitive. I really didn't have to introduce it, but it simplified the explanation. I broke the whole event analysed, into two sub events, whose union was the whole. Link to comment Share on other sites More sharing options...

swansont Posted April 5, 2005 Share Posted April 5, 2005 But this example problem isn't how I arrived at the contradiction. I used barn/ladder. What I did is not in this thread, nor any other. But it essentially is barn/ladder. I just happen to have good logical form. There is no question that you made a mistake, and your "logical form" is not as good as you think it is. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 5, 2005 Share Posted April 5, 2005 There is no question that you made a mistake, and your "logical form" is not as good as you think it is. Do you have the fortitude to follow my argument? Link to comment Share on other sites More sharing options...

swansont Posted April 5, 2005 Share Posted April 5, 2005 Do you have the fortitude to follow my argument? If it's as unnecessarily verbose as some of your other posts, probably not. Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted April 5, 2005 Share Posted April 5, 2005 Then in the one frame, the statement that Jim turned 40 when John turned 31 will be true, and in the other frame, the statement that, "John turned 40 when Jim turned 31" will be true. This is a temporal paradox. And it is caused by assuming that the time dilation formula is true in both frames[/b']. Johnny, this is how these frames work by definition. If you somehow come up with absolute frames that work differently that would be great. I won't say it is impossible. But, the way these frames are defined, that is the way they work. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 5, 2005 Share Posted April 5, 2005 If it's as unnecessarily verbose as some of your other posts, probably not. It isn't long at all. I wrote two papers actually, the second a revised edition of the first. The second one included some material that wasn't in the first one. The proof was 2 pages, and the steps were algebraic. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 5, 2005 Share Posted April 5, 2005 Johnny' date=' this is how [b']these frames [/b] work by definition. If you somehow come up with absolute frames that work differently that would be great. I won't say it is impossible. But, the way these frames are defined, that is the way they work. First swansont, and now you. What is an absolute frame? Link to comment Share on other sites More sharing options...

5614 Posted April 5, 2005 Share Posted April 5, 2005 Well for one' date=' my mind seems to have no tolerance for space bending, and from what I can tell, GR says that space can bend. Also, I have found an error in the equivalence principle, using logic (and electrodynamics). So that explains this. As for SR, the same thing goes. The barn and ladder paradox...[/quote'] So as soon as it is something you cannot easily imagine (space bending, barn/ladder etc) you don't like it? Have you ever considered that maybe all physicists have been presented with the same problems and rather than saying "I can't see a 40ft ladder in a 20ft barn therefore it can't be right" they've learnt the mathematics and the theories and now understand it. As dave said, it's all theories, maybe you dislike the fact that physics is all theories and quote dave: "you can't really prove anything to be 100% right in Physics" maybe the uncertainty (in that we cannot be 100% certain) is what you dislike? And as I said, I do like what you are doing (learning physics) just not the fine details! I'm not trying to attack you, when Phi threatened you with suspension a while back I PMed him saying not to because you bring a lot into this forum, some of it may be wrong and some is controversial but it's still good.... I'm just wondering why you who wants to learn physics can't accept physics like the majority of people do? Link to comment Share on other sites More sharing options...

5614 Posted April 5, 2005 Share Posted April 5, 2005 First swansont, and now you. What is an absolute frame[/i']? An absolute frame of reference is one which is stationary relative to all movement. On the earth we use the ground as an absolute frame of reference. All velocities are measured relative to Earth, which is considerd to be stationary. No such reference frame exists for the universe in general. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 5, 2005 Share Posted April 5, 2005 An absolute frame of reference is one which is stationary relative to all movement. On the earth we use the ground as an absolute frame of reference. All velocities are measured relative to Earth' date=' which is considerd to be stationary. [/color'] No such reference frame exists for the universe in general. I have a few questions about this... The definition of "absolute frame of reference" is still unclear to me. Any frame of reference has it's stationariness, or non-stationariness only in relation to another frame. I would suggest using first order logic to define what you mean. Let the domain of discourse be frames. In fact, you can use multiple domains of discourse, and I could still follow. Let A denote the origin of one frame. Let B denote the origin of another frame. Let v denote the relative speed of A to B. All points in frame A are rest relative to the point A. All points in frame B, are at rest relative to the point B. But the point B can be moving through coordinates of frame A, and vice versa, with some relative speed v. You cannot say frame B is absolutely moving. You cannot say frame A is absolutely moving. You cannot say frame B is absolutely at rest. You cannot say frame A is absolutely at rest. Suppose that the relative speed v is equal to zero. There is still the possibility that frame B is spinning, in frame A. To take that possibility away, there need to be three non-collinear points in B which at rest in frame A. Then points in B are neither rotating or translating in frame A. Let that be the case. Then we can say frame B is at rest in frame A. We can also say that frame A is at rest in frame B. Now, suppose that v is nonzero. v is speed which cannot be negative, therefore in the case where v is nonzero, v is positive. So therefore, the point B is in relative motion to the point A. So that A is moving relative to B, and B is moving relative to A. In other words relative speed is a relation. Now, you say, "An absolute frame of reference is one which is stationary with respect to all movement." In order for me to understand you, you need to formulate the definition using first order logic, as I have already said. Then it will make sense to me. Kind regards Link to comment Share on other sites More sharing options...

5614 Posted April 5, 2005 Share Posted April 5, 2005 This is why you agree with SR... you agree that motion is relative. SR says an absolute frame of reference cannot exist for the exact reason you said: "Any frame of reference has it's stationariness, or non-stationariness only in relation to another frame" I should have made it clearer in my last post... absolute frames of reference are not possible unless you disgard even the basics of SR. At the same time sometimes it is useful to adopt one. e.g. measuring the speed of vechiles (cars etc) travelling down a road on earth. It is illogocial to say I was travelling down a road at 1000mph and the person overtaking me was doing -5mph (possible due to different frames of reference) so we take the road (earth) as a stationary frame, we use it as an absolute frame of reference in that it is still and we are moving relative to it. Obviously no such thing really exists as an absolute frame of reference. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 5, 2005 Share Posted April 5, 2005 This is why you agree with SR... you agree that motion is relative. SR says an absolute frame of reference cannot exist for the exact reason you said: "Any frame of reference has it's stationariness' date=' or non-stationariness only in relation to another frame" I should have made it clearer in my last post... absolute frames of reference are not possible unless you disgard even the basics of SR. At the same time sometimes it is useful to adopt one. e.g. measuring the speed of vechiles (cars etc) travelling down a road on earth. It is illogocial to say I was travelling down a road at 1000mph and the person overtaking me was doing -5mph (possible due to different frames of reference) so we take the road (earth) as a stationary frame, we use it as an absolute frame of reference in that it is still and we are moving relative to it. Obviously no such thing really exists as an absolute frame of reference.[/quote'] I don't follow, this is why I asked you to use first order logic. It sounds harder than it is, and I know you don't fully follow me either. Firstly, you need to be clear what a frame is. Sometimes it is defined as synonymous with a coordinate system, other times its a set of coordinate systems. It is easiest to consider a frame to be a set of three mutually orthogonal infinite straight lines, with coordinates on them, so that to any point in space, there correspond three numbers, which locate that point in space (or center of inertia of object) in relation to the origin of the frame. These three numbers make sense only in this frame, and no other. And so there is a unit of distance assigned as well. The meter works. So now the question is, how do you discuss frames using first order logic? I kind of already know the answer. Let F1 denote a frame. Let F2 denote a frame. We now have to define what is meant by relative motion of the frames. Two questions arise, that of translation, and that of rotation. Let R denote the position vector of the origin of frame F2 in frame F1. Thus, the R vector points from the origin of frame 1 towards the origin of frame 2. Let the coordinates of the origin of frame 2 in frame one be denoted by (x,y,z) The coordinates of the origin of frame F1 in F1 are (0,0,0). The vector from the origin of F1 to the origin of F2 is defined as follows: [math] \vec R = (x-0) \hat i + (y-0) \hat j + (z-0) \hat k = x \hat i + y \hat j + z \hat k [/math] Now, the velocity of the origin of F2 in F1 is defined as follows: [math] \vec V \equiv \frac{d \vec R}{dt} [/math] So we have: [math] \frac{d \vec R}{dt} = \frac{dx}{dt} \hat i + \frac{dy}{dt} \hat j + \frac{dz}{dt} \hat k [/math] So we have: [math] \vec V = \frac{dx}{dt} \hat i + \frac{dy}{dt} \hat j + \frac{dz}{dt} \hat k = v_x \hat i + v_y \hat j + v_z \hat k [/math] Now, in the case where the origin of F2 is at rest in frame F1, the speed of F2 in F1, is equal to zero. In that case, we would say that F2 is at rest in F1. Of course the coordinates of F2 could be spinning in F1, but for my purpose we can neglect that. If it bothers you, then define non-spinning as well, by saying that there are at least three non-collinear points in F2, which are at rest in F1. Meaning that the speed of at least three points in F2 is zero, in F1. Then all the coordinates of F2 are neither spinning, nor translating in F1. Now, the magnitude of the velocity vector defined above, is the speed of the origin of F2 in F1. The magnitude of the velocity vector is defined by: [math] |\vec V| \equiv \sqrt {\vec V \bullet \vec V} = \sqrt{v_x^2+v_y^2+v_z^2}[/math] The quantity above will have units of meters per second, where second is the unit of time, as measured by a clock at rest in frame F1. Actually, because of the way I have done things, it is the instantaneous speed of the origin of F2 in F1. That means that the amount of time which has passed, for the origin of F2 to move relative to F1 is infinitessimally small. Now, notice this: [math] \sqrt {\vec V \bullet \vec V} = \sqrt{(\frac{dx}{dt}) ^2+(\frac{dy}{dt})^2+(\frac{dz}{dt})^2} = \frac{\sqrt{(dx) ^2+(dy)^2+(dz)^2} }{dt} [/math] So at some moment in time in frame F1, the location of the origin of F2 is (x,y,z), and then an infinitessimal amount of time dt later, the origin of the frame is now (x2,y2,z2), so that the infinitessimal change in the x coordinate of the origin of F2 is given by: [math] dx = x_2-x [/math] And likewise: [math] dy = y_2-y [/math] [math] dx = z_2-z [/math] So that we may write this for the speed of the origin of F2 in F1: [math] |\vec V|= \frac{\sqrt{(x2-x) ^2+(y2-y)^2+(z2-z)^2} }{dt} [/math] You should recognize the numerator as the distance traveled by the origin of frame F2 over two consecutive moments in time, in frame F1. Now, because motion is relative, we must get the same value if we define the speed of the origin of F1 in frame F2. The origin of F2 in F2 is (0,0,0)` Let an arbitrary point in F2 have coorinates (x`,y`,z`). Now, we can define the position vector of frame F1 in F2 as we defined the position vector of frame F2 in F1. So, the position vector of F1 in F2 is the vector from the origin of F2 to the origin of F1. Now, at the moment in time when the origin of F2 had coordinates (x,y,z) in frame F1, let the coordinates of the origin of F1 in F2 have been (x`,y`,z`). At that moment in time, the position vector of the origin of F1 in F2 was: [math] \vec R^\prime = x^\prime \hat i + y^\prime \hat j + z^\prime \hat k [/math] The magnitude of the vector above, is the original distance D between the two origins. The magnitude of the vector above is given by: [math] |\vec R^\prime| = \sqrt{(x^\prime)^2+ (y^\prime)^2 + (z^\prime)^2} [/math] Likewise the magnitude of the original R is given by: [math] |\vec R| = \sqrt{x^2+y^2+z^2} [/math] The quantity above, is the distance D between the origins, as measured by a ruler at rest in F1. So that we have: [math] D = \sqrt{x^2+y^2+z^2} [/math] And [math] |\vec R^\prime| [/math] is the distance between the origins as measured by a ruler at rest in F2. These distances must be equal, because if not simultaneity is relative. Therefore: [math] \sqrt{(x^\prime)^2+ (y^\prime)^2 + (z^\prime)^2} = \sqrt{x^2+y^2+z^2} [/math] Now, at the very next moment in time, the origin of F1 has moved in F2. Let the new coordinates of the origin of F1 in F2 be given by: (x2`,y2`,z2`) Thus, the distance the origin of F1 moved in F2 is given by: [math] \sqrt{(x_2^\prime - x^\prime)^2 + (y_2^\prime - y^\prime)^2 + (z_2^\prime - z^\prime)^2 } [/math] Now, let the time of travel as measured by a clock at rest in F2, be given by dt`. Thus, the speed of the origin of F1 in F2 is given by: [math] |\vec V^\prime|= \frac{\sqrt{(x_2^\prime - x^\prime)^2 + (y_2^\prime - y^\prime)^2 + (z_2^\prime - z^\prime)^2 } }{dt^\prime} [/math] Since speed is relative we have to have this: [math] |\vec V|=|\vec V^\prime}| [/math] Ok so, if the numerators are equal then the denominators are equal, and if the denominators are equal then the numerators are equal. Therefore, the numerators are equal if and only if the denominators are equal. In the remainder of the post, I will just represent the relative speed by v. Now, it was others who used the term "absolute frame of reference" not me. All I ask is that if you define it, use first order logic in some manner. For example... Let F1 denote an inertial reference frame. That means that Sir Isaac Newton's three laws of motion are true in frame F1. One could reasonably ask what are the conditions for F2 to also be an inertial reference frame. The answer is of course, that Newton's three laws of motion must be true in F2. Now, one of Newton's three laws, is a restatement of Galileo's law of inertia. Newton's First Law of Motion (The law of inertia): An object at rest will remain at rest, or in motion in a straight line at a constant speed, unless acted upon by an external force. So let the origin of F2 be a particle. Since F1 was stipulated to be an inertial reference frame, then if there are no forces acting upon this particle, then the particle should either remain at rest in F1 (if it was previously at rest in F1) or continue moving in a straight line at a constant speed (if it was previously moving in a straight line at a constant speed), unless acted upon by an external force. So I have permanantly fixed the origin of F2 to be located where the particle is. Wherever the particle goes, it is located at the origin of F2. Now, if the particle is subjected to an external force, then F2 cannot be inertial. So let us stipulate that the particle located at the origin of F2 is a free particle, meaning that there are no external forces acting upon it. Now, either the particle is initially at rest, or the two frames are in relative motion, with nonzero relative speed v. Case I: Particle initially at rest in frame F1. Since there are no external forces acting upon the particle, the particle should remain at rest, over two consecutive moments in time. That means that the numerator in the following formula must be zero... [math] |\vec V|= \frac{\sqrt{(x2-x) ^2+(y2-y)^2+(z2-z)^2} }{dt} [/math] Now, the only way that numerator can be zero, is if the location of the particle in F1 didn't change over two consecutive moments in time. That condition means this: [math] x2 = x [/math] [math] y2 = y [/math] [math] z2 = z [/math] But the denominator cannot be zero. And the relative speed is v, therefore: [math] v = |\vec V|=|\vec V^\prime}| = 0 [/math] Therefore, we must also have this: [math] 0 = |\vec V^\prime|= \frac{\sqrt{(x_2^\prime - x^\prime)^2 + (y_2^\prime - y^\prime)^2 + (z_2^\prime - z^\prime)^2 } }{dt^\prime} [/math] The only way for the quantity above to equal zero, is for the numerator to be equal to zero. The conditions for that are: [math] x_2^\prime = x^\prime [/math] [math] y_2^\prime = y^\prime[/math] [math] z_2^\prime = z^\prime[/math] So this handles the case where the origin of F2 is at rest in F1, but I have neglected whether or not the axes of F2 are spinning in F1. Suppose that the axes of F2 are spinning in frame F1. Is this sufficient to make F2 a non-inertial reference frame? Well suppose that there is an eight ball in frame F1, and that the center of inertia of frame F1 is located at the center of inertia of the eight ball, and let the eight ball be free from all external forces. Since we have stipulated that frame F1 is an inertial reference frame, the center of mass of the eight ball will obey Galileo's/Newton's law of inertia. That means that either the center of inertia of the eightball will remain at rest if initially at rest, or continue to move in a straight line at a constant speed. Let the eight ball initially be at rest in frame F1. Therefore, until such time as an external force acts upon the eight ball, the center of inertia of the eight ball will remain at rest in frame F1. But what about whether or not the eight ball is spinning? Let any point on the surface of the eightball also be at rest in reference frame F1. Let the position vector from the origin of F1 to the origin of F2 actually pass through the center of the number (8) which is on the eightball. Now, if the axes of frame F2 are spinning in frame F1, then the eightball will appear to be orbiting the origin of reference frame F2. But it was stipulated that there are no external forces acting upon the eightball. So, for example, if the axes of F2 are spinning at constant angular velocity in frame F1, the eightball will appear to have a centripetal acceleration of: [math] \vec a^\prime = \frac {\vec V^\prime \bullet \vec V^\prime}{|\vec R^\prime|} \hat r^\prime [/math] In the formula above, V` is the velocity of the center of inertia of the eightball, as defined using the coordinates of reference frame F2. The symbol R` is the position vector of the origin of frame F1, as defined using the coordinates of frame F2. Since the eightball was stipulated to be currently at rest in frame F1, you could also say that R` is the position vector of the eightball, as defined using the coordinates of frame 2. But the point is, that if the eightball is orbiting the origin of frame F2 in the way described, then frame F2 cannot possibly be an inertial reference frame, since it was stipulated that there are no external forces acting upon the eightball in it's own rest frame (F1). So... In frame F2, there is an object upon which no external forces are acting, which is neither at rest nor having its center of inertia moving in a straight line at a constant speed. Which is sufficient reason for F2 to be non-inertial. Also, Newton's third law is untrue in frame F2, since there is an object which is accelerating in absence of an external force. Acceleration is change in velocity, and velocity is speed times direction, and even though the speed is constant, the direction of motion is changing. And this is happening without an external force. So this also leads to the conclusion that F2 is non-inertial. So, if the axes of F2 are spinning in frame F1, at a constant angular speed of w then F2 is a non-inertial reference frame. And if the angular velocity is changing, again Newton's first law is untrue. Therefore, part of the condition for F2 to also be an inertial frame, is for its axes to be at rest in frame F1. Actually, the correct thing to say is that this is a necessary and sufficient condition, for F2 to also be an inertial reference frame. So this was the case where v=0. Case II: Particle not initially at rest in frame F1. Since the particle at the origin of F2 does not have any external forces acting upon it, in this case the origin must be moving in a straight line at a constant speed through F1. Since F1 was stipulated to be inertial. So in order for F2 to also be inertial, it must be the case that the center of inertia of the eightball is also moving in a straight line at a constant speed in reference frame F2. And this would not be the case if the axes of F2 are spinning in frame F1. So in both cases, we have the same necessary and sufficient conditions for F2 to be inertial. So now we can use first order logic to say this. I just want to say it in the simplest way possible. As for what an absolute frame of reference is, I still have no idea what that is. Ok here it is... Suppose that reference frame F1 is an inertial reference frame. Another reference frame F2, will also be an inertial reference frame if Either all points in reference frame F2 are at rest in reference frame F1 OR any point in reference frame F2 is moving in straight line at a constant speed through F1. So let v denote the velocity of an arbitrary point in reference frame F2. if |v|=0 for any arbitrary point in F2 then F2 is inertial. Now consider the case where |v| isn't zero. Then in order for F2 to be inertial, we need the point to be moving in a straight line at a constant speed. That means its direction of motion through F1 must be constant, and its speed through F1 must be constant. That means the product of its speed and its direction must be constant. The product of speed and direction of motion is velocity. So... Here is one way to say what I was trying to say using first order logic: Suppose that F1 is an inertial reference frame. Let v denote the velocity of an arbitrary point in F2, through frame F1. If dv/dt = 0 then F2 is inertial, otherwise F2 is non-inertial. For any inertial reference frame F1: and any reference frame F2: If for any point (x`,y`,z`) in F2, dv/dt = 0 then F2 is inertial. Where v is the velocity of the point through F1. As you can see, I have used two domains of discourse, a set of frames, and a set of points. Now, here is the definition of the velocity of a point in F2, through F1: [math] \frac{d \vec R}{dt} = \frac{dx}{dt} \hat i + \frac{dy}{dt} \hat j + \frac{dz}{dt} \hat k [/math] So in frame F1 (which is stipulated to be inertial) the condition for frame F2 to be inertial is this: For any point at rest in frame F2, with coordinates (x`,y`,z`) in F2 and coordinates (x,y,z) in F1... [math] \vec a = \frac{d \vec v}{dt} = \frac{dx}{dt} \hat i + \frac{dy}{dt} \hat j + \frac{dz}{dt} \hat k = 0 [/math] In other words, no point in F2 can be accelerating in F1. Another way to say this is... No point in F2 can have its velocity (as defined using the coordinates of F1) change. Now, the velocity of a point (x`,y`,z`) of F2 through frame F1, is equal to its speed through frame F1, times it's direction of travel through frame F1. Its instantaneous speed through frame F1 is easy enough to define. Its just the magnitude of its change in its position in F1, divided by the time this took, as measured by a clock at rest in F1. So here is what we can do. We need to focus on equations of lines in three dimensional space. In a plane the answer is easy... Ax+By+C=0 or y = mx +b or y - y1 = m(x-x1) So, let (x`,y`,z`) denote an arbitrary point in reference frame F2. Let F1 be an inertial reference frame. Let the position of (x`,y`,z`) in frame F1 at some moment in time be given by (x1,y1,z1). Then, at the very next moment in time, let its position in frame F1 be given by: (x2,y2,z2) Now, both (x1,y1,z1) and (x2,y2,z2) are at rest in frame F1. They are fixed positions in that frame. Therefore, there is a unique distance from one to the other, which can be measured by a ruler at rest in frame F1. But also the Pythagorean theorem is true, so we can use that to express the distance between the two points as: [math] D = \sqrt{ (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2} [/math] But these are infinitessimals, so we can also write: [math] D = \sqrt{ (dx)^2 + (dy)^2 + (dz)^2} [/math] Now, we also need to define the direction of motion as well. Let R1 denote the initial position vector of the point (x`,y`,z`), and let R2 denote the position vector at the very next moment in time. Thus, we can write a formula for the infinitessimal displacement vector. In the case where the points (x`,y`,z`) velocity v through F1 is nonzero, we have a vector triangle. [math] \vec R_1 + \vec {dR} = \vec R_2 [/math] Therefore: [math] \vec {dR} = \vec R_2 - \vec R_1 [/math] Now, we already know what the magnitude of this vector is, from the Pythagorean theorem... [math] D = \sqrt{ (x2-x1)^2 + (y2-y1)^2 + (z2-z1)^2} [/math] or this [math] D = \sqrt{ (dx)^2 + (dy)^2 + (dz)^2} [/math] So D is the magnitude of the infinitessimal displacement vector dR. Now, as with any vector, dR is equal to its magnitude times its direction. So, if we divide dR by D, we are left with its direction over two consecutive moments in time. So here is the initial direction of the infinitessimal displacement vector: [math] dir (\vec dR) = \frac{\vec dR}{D} [/math] To recapitulate: We are given an arbitrary inertial reference frame F1, and we wish to express the conditions under which another reference frame F2 is also inertial. The conclusion reached, is that F2 will also be inertial, provided that any point (x`,y`,z`) which is at rest in F2, isn't accelerating in F1. This means the same thing as saying that any point (x`,y`,z`) in F2, cannot have a velocity vector (defined using the coordinates of F1) which is changing. This means two things. The speed of (x`,y`,z`) in F1 must be constant in time, and the direction of motion of (x`,y`,z`) in F1 must also be constant in time. Now, we can differentiate both sides of the direction of dr with respect to time, and then set that equal to zero to say that the direction of motion is constant. [math] \frac{d}{dt} dir (\vec dR) = \frac{d}{dt} \frac{\vec dR}{D} [/math] Notice that the derivative is defined using the "time coordinate" of frame F1, not the "time coordinate" of frame F2. So we can use quotient rule. d/dx (f/g) = (f`g-g`f)/g^{2} [math] \frac{d}{dt} \frac{ \vec dR}{D} = \frac{1}{D^2}[D \frac{d(\vec dR)}{dt} + \vec dR \frac{d(D)}{dt}][/math] So the condition for the direction of motion not to change is: [math] \frac{1}{D^2}[D \frac{d(\vec dR)}{dt} + \vec dR \frac{d(D)}{dt}] = 0 [/math] Or equivalently: [math] [D \frac{d(\vec dR)}{dt} + \vec dR \frac{d(D)}{dt}] = 0 [/math] Now D is a pure number I believe, with units of meters. So I think d(D) =0 , but I need a moment to think about it. What I now think I should have done, is focus on three consecutive moments in time, instead of two, but I am not wanting to go back and re-do everything. I know what I should have done. I should have specified the future position of an arbitrary point (x`,y`,z`) of F2 in frame F1, and left it as a variable, while fixing the initial velocity of it. Also, maybe I should be using the difference calculus, instead of the differential calculus. I think I am right about d(D) =0. D is just a number, the initial distance travelled by (x`,y`,z`) in frame F1. So that means the condition for the direction of motion to be constant in time is: [math] D \frac{d(\vec dR)}{dt} = 0 [/math] Dividing both sides by D we have: [math] \frac{d(\vec dR)}{dt} = 0 [/math] The meaning of the formula above is unclear to me. [math] \vec {dR} = \vec R_2 - \vec R_1 [/math] Lets focus on three consecutive moments in time, instead of two. Denote the position vector of (x`,y`,z`) at the third of these three moments in time be denoted by: [math] \vec R_3 [/math] There is now a simple way to say that the velocity of (x`,y`,z`) was constant over these three moments: [math] \vec R_3 - \vec R_2 = \vec R_2 - \vec R_1 [/math] And so on. So at the fourth moment in time we have to have... [math] \vec R_4 - \vec R_3 = \vec R_3 - \vec R_2 [/math] Now, looking back we have: [math] \vec R_3 = 2 \vec R_2 - \vec R_1 [/math] [math] \vec R_4 = 2\vec R_3 - \vec R_2 [/math] [math] \vec R_4 = 2(2 \vec R_2 - \vec R_1 ) - \vec R_2 [/math] [math] \vec R_4 = 4\vec R_2 - 2\vec R_1 - \vec R_2 [/math] [math] \vec R_4 = 3\vec R_2 - 2\vec R_1 [/math] And if the velocity is constant over 5 consecutive moments in time, we have to have: [math] \vec R_5 - \vec R_4 = \vec R_2 - \vec R_1 [/math] [math] \vec R_5 = \vec R_4 + \vec R_2 - \vec R_1 [/math] [math] \vec R_5 = 3\vec R_2 - 2\vec R_1 + \vec R_2 - \vec R_1 [/math] [math] \vec R_5 = 4\vec R_2 - 3\vec R_1 [/math] So there is a recursive pattern here. [math] \vec R_4 = 3\vec R_2 - 2\vec R_1 [/math] [math] \vec R_5 = 4\vec R_2 - 3\vec R_1 [/math] [math] \vec R_6 = 5\vec R_2 - 4\vec R_1 [/math] And so on. The main conclusion has already been reached anyway. Which is that for F2 to also be inertial, any point (x`,y`,z`) at rest in F2, must have a constant velocity in F1. Link to comment Share on other sites More sharing options...

swansont Posted April 5, 2005 Share Posted April 5, 2005 It isn't long at all. I wrote two papers actually, the second a revised edition of the first. The second one included some material that wasn't in the first one. The proof was 2 pages, and the steps were algebraic. Then post it somewhere and let people pick it apart. Peer review and all that. Link to comment Share on other sites More sharing options...

Johnny5 Posted April 5, 2005 Share Posted April 5, 2005 Then post it somewhere and let people pick it apart. Peer review and all that. I guess I could do that, but where would I post it? Link to comment Share on other sites More sharing options...

J.C.MacSwell Posted April 5, 2005 Share Posted April 5, 2005 First swansont, and now you. What is an absolute frame[/i']? You will be telling us at your Nobel prize acceptance speech! (I do not know but I'm sure it's not Euclidean) Link to comment Share on other sites More sharing options...

5614 Posted April 5, 2005 Share Posted April 5, 2005 Think this'll provide some reading material! http://www.pinkmonkey.com/studyguides/subjects/physics/chap35/p3535401.asp Note the index on the right... should be an interesting read, I've read some not all of the site. Link to comment Share on other sites More sharing options...

swansont Posted April 5, 2005 Share Posted April 5, 2005 These distances must be equal' date=' because if not simultaneity is relative. [/quote'] So, you're assuming absolute simultaneity... Link to comment Share on other sites More sharing options...

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