swansont Posted November 14, 2016 Share Posted November 14, 2016 Right, distance is always positive. Time is also always positive. And gravitation always attractive. There is a "direction" in the Universe. Distance and time are scalar values, which by definition do not have a direction. Gravitational attraction is a relative direction, not an absolute one. Interesting. It sounds as if you're arguing that when we say that something is 4 high and 2 wide and we can multiply the two together to get 8 that it's simply some coincidence that the two directions are also perpendicular. I don't understand why anyone would make that argument, but there it is. It sounds incorrect to me. Orthogonality and direction are not the same thing. If you disagree, perhaps you can explain what direction the Legendre polynomial P_{2} = (3x^2-1)/2 is pointing? It is orthogonal to P_{1}=x, and P_{0}=1 (and all of the rest of them, too). Or if you'd rather use Laguerre polynomials or Bessel functions as the example, go right ahead. Link to comment Share on other sites More sharing options...

studiot Posted November 14, 2016 Share Posted November 14, 2016 (edited) Distance and time are scalar values, which by definition do not have a direction. Gravitational attraction is a relative direction, not an absolute one. Orthogonality and direction are not the same thing. If you disagree, perhaps you can explain what direction the Legendre polynomial P_{2} = (3x^2-1)/2 is pointing? It is orthogonal to P_{1}=x, and P_{0}=1 (and all of the rest of them, too). Or if you'd rather use Laguerre polynomials or Bessel functions as the example, go right ahead. To reinforce my earlier comment (post#20 ) to steveupson P0, P1 and P2 are orthogonal (perpendicular if you like) but not independent. Edited November 14, 2016 by studiot Link to comment Share on other sites More sharing options...

Butch Posted November 14, 2016 Share Posted November 14, 2016 (edited) Try this let x=time and y=space, graph this function as y = 1/x^2. Does time have a beginning? Give x limits of infinity... Does hthe universe exist 1 nanosecond ago? I am out of posts, so this will have to do. This is simply a graph of the inverse square law. Edited November 14, 2016 by Butch Link to comment Share on other sites More sharing options...

Strange Posted November 14, 2016 Share Posted November 14, 2016 Try this let x=time and y=space, graph this function as y = 1/x^2. Why? We already have a function that relates space to time, the FLRW metric. What is wrong with that one? Does time have a beginning?Give x limits of infinity... Does the universe exist 1 nanosecond ago? What is your point? As far as we know, the universe has existed for 13.8 billion years, and possibly forever. Do you have any evidence to the contrary? Link to comment Share on other sites More sharing options...

Butch Posted November 14, 2016 Share Posted November 14, 2016 (edited) I tend to lean towards forever, however physicists seem to shy away from the infinite, I have come up with 2 rules for dealing with infinity: 1. Any quantity that is a factor of infinity approaches infinity. 2. Any quantity that is not a factor of infinity approaches zero. And no I am not disagreeing with established theory at all. If you graph time v space this way and play with limits of x I think you will begin to understand the relationship of time and space. I am limited in posting until tomorrow, perhaps then I can discuss in more detail. Edited November 14, 2016 by Butch Link to comment Share on other sites More sharing options...

Strange Posted November 14, 2016 Share Posted November 14, 2016 I tend to lean towards forever, however physicists seem to shy away from the infinite Do they? It is generally accepted that the universe could be infinite in size. There are various hypotheses that lead to the universe being infinitely old. There proposals for an infinite number of universes. And so on... And no I am not disagreeing with established theory at all. Then what is the point you are trying to make? If you graph time v space this way and play with limits of x I think you will begin to understand the relationship of time and space. You need to be more explicit about what you are trying to say. What is this graph and why should it be taken seriously`/ Link to comment Share on other sites More sharing options...

studiot Posted November 14, 2016 Share Posted November 14, 2016 I tend to lean towards forever, however physicists seem to shy away from the infinite, I have come up with 2 rules for dealing with infinity: 1. Any quantity that is a factor of infinity approaches infinity. 2. Any quantity that is not a factor of infinity approaches zero. And no I am not disagreeing with established theory at all. If you graph time v space this way and play with limits of x I think you will begin to understand the relationship of time and space. I am limited in posting until tomorrow, perhaps then I can discuss in more detail. 1 and 2 are almost like finding two terms in a differential equation solution, one of which tens to infinity or dies away and the other represents the 'steady state' solution. However I must protest the slur upon physicists in your first sentence. There are an infinite number of points on the interval between 0 and 1. One of the most common processes in Physics is normalisation - or the rescaling of some quantity to lie between 0 and 1. Link to comment Share on other sites More sharing options...

Tim88 Posted November 15, 2016 Share Posted November 15, 2016 I wrote: Here we discussed a graphical presentation of equation 12. [..] If math and physics have different definitions for direction and multiplication, or different interpretations of a graph, then yes, I am definitely confused. I should perhaps have added an illustration, which I already gave in an earlier discussion. One can plot the equation F=ma in a graph, with for example F on the x axis and ma on the y axis. That gives you, as it were, a force-acceleration "world". Force is perpendicular on acceleration in that graphical presentation of the equation. That should not be confounded with the physics. As a matter of fact, the force and the acceleration do have directions, but they are in the same direction. Opposed to..."as it is"...? * Just Tim and me having our own little private chat * Exactly. It happens that by chance I agree with Einstein's remark there. "As it were": as if it were so: in a manner of speaking -http://www.merriam-webster.com Link to comment Share on other sites More sharing options...

studiot Posted November 15, 2016 Share Posted November 15, 2016 I wrote: I should perhaps have added an illustration, which I already gave in an earlier discussion. One can plot the equation F=ma in a graph, with for example F on the x axis and ma on the y axis. That gives you, as it were, a force-acceleration "world". Force is perpendicular on acceleration in that graphical presentation of the equation. That should not be confounded with the physics. As a matter of fact, the force and the acceleration do have directions, but they are in the same direction. A good idea Tim, but a better example would be the Force - Extension graph for a spring. This definitely has different physical quantities on each axis. Force - ma give you force on both axes Or I suppose you could simply drop the m. Link to comment Share on other sites More sharing options...

imatfaal Posted November 15, 2016 Share Posted November 15, 2016 A good idea Tim, but a better example would be the Force - Extension graph for a spring. This definitely has different physical quantities on each axis. Force - ma give you force on both axes Or I suppose you could simply drop the m. Yes - something in which the slope is indicative - F vs a or F vs x with m and k showing up in the slope. But it is still a nice analogy that proved the point simply and conclusively Link to comment Share on other sites More sharing options...

AbstractDreamer Posted November 16, 2016 Author Share Posted November 16, 2016 (edited) As acceleration = (distance/time)/time If acceleration has direction, then either distance has information on direction, or time has direction, or direction is emergent from a function of distance and time. If velocity has direction in the same manner as acceleration, then time cannot affect direction If velocity has direction in a different manner to acceleration, then time is some fuction of direction or direction some function of time. What properties or attributes does direction have? Edited November 16, 2016 by AbstractDreamer Link to comment Share on other sites More sharing options...

MigL Posted November 16, 2016 Share Posted November 16, 2016 Getting back to the difference between the time dimension and the spatial dimensions, and correct me if I'm wrong, but... In GR we have dS^2 = dx^2+ dy^2+ dz^2- c^2dt^2 ( sorry for the lack of LaTex) making the actual multiplier for the time variable, ic , not simply c . That means, although orthogonal to the space dimensions, time is treated mathematically as imaginary. Link to comment Share on other sites More sharing options...

swansont Posted November 16, 2016 Share Posted November 16, 2016 As acceleration = (distance/time)/time If acceleration has direction, then either distance has information on direction, or time has direction, or direction is emergent from a function of distance and time. If velocity has direction in the same manner as acceleration, then time cannot affect direction If velocity has direction in a different manner to acceleration, then time is some fuction of direction or direction some function of time. What properties or attributes does direction have? If you want the vector of acceleration you use displacement (a vector) rather than distance (a scalar). Time is a scalar. I'm not aware of any mathematical way of dividing one vector by another to give a resultant vector. Velocity and acceleration do not need to be (and generally aren't) in the same direction. A ball thrown in some parabolic arc has an acceleration pointing down, while having velocity components that can be up and in a transverse direction. Direction can be a function of time. Time is not a function of direction. Link to comment Share on other sites More sharing options...

steveupson Posted November 16, 2016 Share Posted November 16, 2016 Distance and time are scalar values, which by definition do not have a direction. Gravitational attraction is a relative direction, not an absolute one. Ah yes, there is that. Thank you swansont. Orthogonality and direction are not the same thing. If you disagree, perhaps you can explain what direction the Legendre polynomial P_{2} = (3x^2-1)/2 is pointing? It is orthogonal to P_{1}=x, and P_{0}=1 (and all of the rest of them, too). Or if you'd rather use Laguerre polynomials or Bessel functions as the example, go right ahead. Perhaps the terminology should be orientation rather than direction. I'm going to try that out and see if it makes more sense to you. Hopefully there will some facet that will distinguish the two and we will be able to ascertain precisely what that feature is. If you want the vector of acceleration you use displacement (a vector) rather than distance (a scalar). Time is a scalar. I'm not aware of any mathematical way of dividing one vector by another to give a resultant vector. Velocity and acceleration do not need to be (and generally aren't) in the same direction. A ball thrown in some parabolic arc has an acceleration pointing down, while having velocity components that can be up and in a transverse direction. In order to keep anyone from having to weed through many pages of incomprehensible arguments (mostly by myself), I'd like to include a short colloquy from another site: Me (paraphrased): There's a fairly recent paper in which they expose the need for a special treatment of direction: The Physical Origin of Torque and of the Rotational Second Law (Daniel J. Cross) Their observations seem to be correct, but the conclusion that non-rigid bodies are a requirement for manifesting torque is speculative. The math also works out if we modify our definition of direction. Benit13: "Thanks for the link. It was interesting! However, I fail to see special treatments of direction in the paper. They just use angles to denote direction. In any case, the paper is about testing the validity of the perfect rigid-body assumption under rotation. Basically, the claim made by the author is that it is impossible to rotate a perfect rigid-body that has mass because such a body would be unable to transmit force along the rigid body's length to allow a mass further down the length to accelerate (which is only possible under the so-called weak version of Newton's third law). That is, in order to get the correct equation of motion, the reactionary force from Newton's third law needs to be weighted according to its distance from the pivot, which is something that drops out naturally when you discard the perfect rigid-body assertion. Therefore, at a fundamental level, rotating objects can never be perfect rigid-bodies, they can only be assumed to be perfect rigid-bodies. I don't find that the conclusions of the paper are particularly crazy. Although most solids behave like rigid-bodies, the idea of perfect rigid-bodies only really exists in the theoretical realm. In reality, solid objects are lattices of atoms, sometimes in a regular crystalline structure, sometimes a messy mixture. All of the atoms in a rotating body will interact with each other, vibrate, wobble, translate, exchange energy, etc. Such messy systems are far removed from perfect rigid-bodies. Thankfully, the assumption is applicable (in most cases) because at the macroscopic scales we are dealing with (in most cases), we don't have to worry about deformations caused by rotational acceleration (again, in most cases!). They are negligible to the point where we could confidently say "it doesn't deform". Note that if you work under the rotational equivalent of Newton's third law, which is "if a torque is applied to an object and its rotational acceleration doesn't change, there exists an equal and opposite torque" we can assert that the force [latex]T[/latex] at [latex]m_1[/latex] gives a different torque than the force [latex]T[/latex] at [latex]m_2[/latex]. To resolve the issue we can either rescale the forces so that the torques match (which is effectively achieved in the paper with the relation [latex]T_1r_1 = T_2r_2[/latex]) or we can redefine the net torque as: [latex]\tau = \tau_1 + \tau_2[/latex] then [latex]\tau = r_1 T + r_2 (F-T)[/latex] [latex] = r_1 T + r_2 F - r_2 T[/latex] [latex] = (r_1 - r_2) T + r_2 F[/latex] [latex] = (r_1 - r_2) m_1 a_1 + r_2 (m_1a_1 + m_2 a_2)[/latex] [latex] = (r_1 - r_2) m_1 r_1 \alpha + r_2 (m_1r_1 \alpha+ m_2 r_2 \alpha) [/latex] [latex] = \alpha(m_1r_1^2 - m_1r_1r_2 + m_1r_1r_2 +m_2r_2^2)[/latex] [latex] = \alpha(m_1r_1^2 + m_2r_2^2)[/latex], which is the correct equation of motion. However, the rotational equivalent of Newton's third law is derived from the linear form. By stating the above I am actually dodging the perfect-rigid body assumption. To be consistent, we must recognise that in doing the above, we have implicitly assumed that the object is a rigid-body that is capable of transmitting force along its length." Me: "It's interesting to look at how you managed to dodge the issue. You introduced a change of direction. Could the way in which you introduced it be considered a derivative of direction? [latex] = (r_1 - r_2) m_1 r_1 \alpha + r_2 (m_1r_1 \alpha+ m_2 r_2 \alpha) [/latex] Is this even a legitimate question to be asking?" Benit13: "I didn't introduce a change in direction. In fact, I didn't even treat direction at all because I know [latex]\tau_1[/latex] and [latex]\tau_2[/latex] are pointing in the same direction, so I know I can get the right answer by using scalar operations. If I had decided to treat them as vectors, I would have to add them through vector addition which yields a [latex]\tau[/latex] that, in this case, also points in the same direction. If [latex]\tau_1[/latex] and [latex]\tau_2[/latex] had different directions, you can calculate the direction [latex\tau[/latex] would point using vector addition and then by normalising the resulting vector to give a unit vector." Me: "I hear what you're saying, but... Angular velocity is what really establishes the direction of the vector, and you've changed that." Direction can be a function of time. Time is not a function of direction. A lot more clarity can be had by restating this argument to say that direction (using a vector) is an expression of orientation, and that particular method of expressing orientation requires that a length or metric be applied first. It may be argued that the unit vector expresses direction as a value that is decoupled from length, but that isn't the case at all. The length is one, and the decoupling is simply an illusion. The argument here is that the orientation of the time scalar (and by extension, the distance or angular displacement) is already baked into the pie. Where these values appear on a number line relies on their orientation. They can be to the left or right or the inside or outside of other values. It's this orientation and how it works that is causing most of the miscommunication on my part. The fundamentals of orientation in 3D have some additional rules that don't exist in our standard methods. Incorporating these additional rules requires rethinking some things. Link to comment Share on other sites More sharing options...

swansont Posted November 16, 2016 Share Posted November 16, 2016 A lot more clarity can be had by restating this argument to say that direction (using a vector) is an expression of orientation, and that particular method of expressing orientation requires that a length or metric be applied first. It may be argued that the unit vector expresses direction as a value that is decoupled from length, but that isn't the case at all. The length is one, and the decoupling is simply an illusion. The argument here is that the orientation of the time scalar (and by extension, the distance or angular displacement) is already baked into the pie. Where these values appear on a number line relies on their orientation. They can be to the left or right or the inside or outside of other values. It's this orientation and how it works that is causing most of the miscommunication on my part. The fundamentals of orientation in 3D have some additional rules that don't exist in our standard methods. Incorporating these additional rules requires rethinking some things. The unit vector has a length of 1, and no units attached to it. It has a length of 1 no matter what you choose for your units of the magnitude. And it has a length of 1 so that the math works out; it can't modify the magnitude you've assigned. There's no inherent length to it. Scalars don't have a direction, or an orientation. It's just a number, and units. Link to comment Share on other sites More sharing options...

steveupson Posted November 16, 2016 Share Posted November 16, 2016 The unit vector has a length of 1, and no units attached to it. It has a length of 1 no matter what you choose for your units of the magnitude. And it has a length of 1 so that the math works out; it can't modify the magnitude you've assigned. There's no inherent length to it. Scalars don't have a direction, or an orientation. It's just a number, and units. The unit vector uses the exact same method to specify direction as any other vector. It's been scaled to have a magnitude of 1 but it still quantifies the direction information as a ratio between perpendicular lengths. The direction specified by an xy coordinate is simply the ratio between the x length and the y length. The same thing for xyz coordinates. In other words, the orientation of any object isn't ever specified as something independent of length. It may be independent of any coordinate system, sure, but it isn't independent of length. In 3D there exists a relationship between directions where their orientation to one another can be specified without length. Link to comment Share on other sites More sharing options...

swansont Posted November 17, 2016 Share Posted November 17, 2016 The unit vector uses the exact same method to specify direction as any other vector. It's been scaled to have a magnitude of 1 but it still quantifies the direction information as a ratio between perpendicular lengths. The direction specified by an xy coordinate is simply the ratio between the x length and the y length. The same thing for xyz coordinates. In other words, the orientation of any object isn't ever specified as something independent of length. It may be independent of any coordinate system, sure, but it isn't independent of length. In 3D there exists a relationship between directions where their orientation to one another can be specified without length. Yes, the unit vector is independent of length. The ratio of lengths is not a length. Link to comment Share on other sites More sharing options...

steveupson Posted November 17, 2016 Share Posted November 17, 2016 The ratio of lengths is not a length. There exists a relationship that allows orientation to be specified without a ratio of lengths. This can be expanded to position being specified without a ratio of lengths. Link to comment Share on other sites More sharing options...

swansont Posted November 17, 2016 Share Posted November 17, 2016 There exists a relationship that allows orientation to be specified without a ratio of lengths. This can be expanded to position being specified without a ratio of lengths. What does this have to do with time? Link to comment Share on other sites More sharing options...

steveupson Posted November 17, 2016 Share Posted November 17, 2016 What does this have to do with time? I hope that's a rhetorical question because I really have no idea. The ability to identify positions in spacetime without length must have implications about time. I just don't know what they are. I would guess the impact on time to be something geometric in substance. But that's only a guess. Link to comment Share on other sites More sharing options...

AbstractDreamer Posted November 17, 2016 Author Share Posted November 17, 2016 Given spacetime exists and is relative. Imagine a volume of space is empty. Then within that volume there is nothing for time to be relative to or between. Therefore time cannot exist within that volume. But that spacetime exists was a given. Therefore either space is more that simply volume, or any volume of space cannot be empty. Where is the fallacy? Link to comment Share on other sites More sharing options...

Strange Posted November 17, 2016 Share Posted November 17, 2016 Where is the fallacy? Here: Then within that volume there is nothing for time to be relative to or between. Therefore time cannot exist within that volume. A spacetime manifold by definition includes time. You can't wish that away with word games. Link to comment Share on other sites More sharing options...

swansont Posted November 18, 2016 Share Posted November 18, 2016 I hope that's a rhetorical question because I really have no idea. The ability to identify positions in spacetime without length must have implications about time. I just don't know what they are. I would guess the impact on time to be something geometric in substance. But that's only a guess. No, it's a serious question. And you don't seem to know why you are bringing up tangential discussion, so how about we get back to the topic at hand. Link to comment Share on other sites More sharing options...

steveupson Posted November 18, 2016 Share Posted November 18, 2016 No, it's a serious question. And you don't seem to know why you are bringing up tangential discussion, so how about we get back to the topic at hand. The OP asks about time going “forwards.” The way I interpret it, this question asks about time having a particular direction or orientation. I don’t understand how I seem to have wandered off the reservation here, in this discussion, simply because I don’t “know” the answer to your question. I doubt that anyone “knows” (can prove) the answer to your question. I do have a sound theory of how this all works, but I don’t have the algebra skills to show how the math works. Therefore I can’t “know” one way or the other. That doesn’t make the issue tangential at all, afaict. The likely scenario is that every particle (or position in spacetime, for that matter) has a future and all of these futures are connected by a coherent entity. This coherent entity is direction (or we can use “orientation” if that nomenclature is less offensive.) There is a scalar value for direction. I don't think that this fact in any way invalidates the traditional way of representing direction as a vector. Link to comment Share on other sites More sharing options...

AbstractDreamer Posted November 18, 2016 Author Share Posted November 18, 2016 Does super-symmetry include time? Does anti-time flow "backwards"? How can time not have direction in this regard? Link to comment Share on other sites More sharing options...

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