Proper Time and Its Relation to Velocity and Acceleration

Discussion in 'Pseudoscience Archive' started by Prof.Layman, Dec 31, 2012.

  1. Prof.Layman totally internally reflected Registered Senior Member

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    As requested by alphanumeric I will derive the proper time and show how that relates to velocity and acceleration, due to my claims that velocity is not affected by time dialation as it seems to be in the Holographic Principle that has changed how information is conserved when it encounters a black hole. I don't completely agree with the "fact" that Hawking Radiation would cause a microscopic black hole to evaporate if one was created in the LHC, but I also do not agree with the new veiwpoint that information would be stored on the event horizon of a black hole or that Hawking Radiation would be incorrect because of this. I think this could become more clear from my derivation, or that something from it could lead to disproving the Holographic Principle and information is not stored on the event horizon of a black hole.

    I will start out by deriving the proper time using algebra in a way that is different than what it would normally be derived. It is similair to the light clock example, but then it derives the proper time instead of the time dialation equation that is given in most introductory text.

    Say an object or ship is traveling at a constant speed. The distance that ship would travel would be the velocity times time, d = v t (eq. 1). This is of course because v = d/t and then you can multiply both sides of the equation by (t) in order to obtain equation 1. An observer at rest would use his own time in order to measure the ships velocity. The distance the ship travels is also relative to the distance the observer at rest measures it. So then distance and time of equation 1 is from the frame of reference of the observer at rest and is used this way in the calculation.

    Now say the ship sends a beam a light perpendicular to its direction of motion. The observer at rest will measure the distance the beam travels to be d = c t (eq. 2). He will use his own time to measure this distance but instead of velocity he replaces the speed of light as the velocity, since c = d/t. The speed of light is the distance the observer at rest or any observer would measure the distance it traveled over time.

    Now say the observer on the ship measures the speed of light of this same beam. Since he is traveling along with the ship he only measures the beam to travel straight up and down relative to himself. This turns out to be a shorter distance than the observer at rest measures it to travel. But, all observers measure the speed of light to be the same speed, so then you have to assume that the observers time is different in order for him to measure the speed of light to be the same speed even though he measures it to travel a shorter distance. So then you end up with the equation, d' = c t' (eq. 3), for the distance the beam travels relative to the observer on the ship.

    The observations from these two different frames of references can then form similar triangles and then those triangles can be translated onto each other from each coordinate system. So then you would have three sides of a triangle, (v t ) being one side (a), (c t') being the adjacent side (b), and then (c t) being the hypotenus. So then you can use Pythagorean's Theorem, a^2 + b^2 = c^2.

    ( v t )^2 + ( c t' )^2 = ( c t )^2 Then distribute the square

    v^2 t^2 + c^2 t'^2 = c^2 t^2 Then subtract both sides by (v^2 t^2)

    c^2 t'^2 = c^2 t^2 - v^2 t^2 Then factor out ( c^2 t^2) out of the right side of the equation

    c^2 t'^2 = c^2 t^2 ( 1 - v^2/c^2) Then divide both sides by ( c^2 )

    t'^2 = t^2 ( 1 - v^2/c^2 ) Then take the square root of both sides

    t' = t * sqrt ( 1 - v^2/c^2 ) Then you have it the proper time, although time cannot be negative so then you can say that the negative solution is not valid and I will call this equation 4 (eq. 4)
     
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  3. Prof.Layman totally internally reflected Registered Senior Member

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    Now that we have equation 4 (the proper time) I can now derive the proper length.

    An observer traveling with their ship would then measure distance to be d' = c t', this is because of equation 3. If that observer measures a different time and c or the speed of light is constant then they will also measure different values of distance, hence the d' in the equation. You can then substitute ( t' ) with the equation we found as the proper time.

    d' = c t' (eq. 3)
    t' = t sqrt (1 - v^2/c^2) (eq.4) now substitute ( t' ) in eq.3 with the right side of equation 4

    d' = ( c t ) sqrt ( 1 - v^2/c^2) we know that ( c t ) is equal to distance in equation 2 ( d = c t) So then you can substitute ( c t ) with d

    d' = d sqrt ( 1 - v^2/c^2) This would then be the proper length, notice how it is no different than the length contraction equation except instead I used a d for the variable. I will call this equation 5 (eq. 5) or the proper length.
     
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  5. Prof.Layman totally internally reflected Registered Senior Member

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    Now that we have the proper length and time we can now try to determine the proper velocity ( v' ). The proper velocity would be the measured velocity of another observer in motion so then, v' = d'/t' (eq.6). Then since we know what d' and t' are we can now solve for v'.

    v' = d'/t' The velocity measured by another observer.

    d' = d sqrt ( 1 - v^2/c^2 ) The proper distance we found. (eq. 5)

    t' = t sqrt ( 1 - v^2/c^2) The proper time we found (eq. 4)

    v' = [d sqrt ( 1 - v^2/c^2)] / [t sqrt ( 1 - v^2/c^2)] notice that the value [ sqrt ( 1 - v^2/c^2)] cancels out, if the velocity is not equal to the speed of light then you wouldn't have a zero in the denominator, but nothing can travel the speed of light so just cancel them!

    v' = d / t Oh look now we have the equation for the velocity of another observer and it is almost the same as the velocity an observer at rest would measure.

    v = d / t You could now substitute the velocity of the observer at rest with the value ( d / t ) in the previous equation

    v' = v Now we have discovered that the relative velocity that is measured by another observer is the same as the relative velocity someone would measure while they are at rest! I can call this equation 6 (eq. 6)

    Conclusion, the relative velocity between two objects is the same in all frames of references where there is constant motion.
     
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  7. Prof.Layman totally internally reflected Registered Senior Member

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    Now I will derive the equation for the amount of time dialation for an object in uniform acceleration. The distance an object travels in uniform acceleration can be seen as d = [t ( u + v ) ] / 2, this will be equation 7 (eq. 7). Equation 7 can then replace equation 1 in the derivation of the proper time. So then it would be (a) in the Pythagorean Theorem. Then (b) and (c) can remain the same from equations ( c t' ) and then ( c t ) respectively. The curvature of the beam of light was not detected in the M&M experiment because if it was then the light beams would have arrived at the end at different times. So then even under uniform acceleration like on the surface of the Earth, the light beams would still form a right triangle. Assuming that this type of acceleration does not cause curvature of the light beam as according to the M&M experiment and not Einstein.

    a^2 + b^2 = c^2 The Pythagroean Theorem, you can then substitute a, b, and c for the given equations of distance.

    [ [t ( u + v )] / 2 ]^2 + ( c t' )^2 = ( c t )^2 Then you can distribute the square.

    [t^2 ( u + v )^2]/4 + ( c^2 t'^2 ) = c^2 t^2 Then subtract both sides by the first value in the equation that was a^2

    c^2 t'^2 = c^2 t^2 - [[t^2 (u + v )^2]/4 Then factor out a c^2 t^2 out of the right side of the equation like we did before.

    c^2 t'^2 = c^2 t^2 [ 1 - ( u + v )^2/4c^2] Then divide both sides of the equation by c^2

    t'^2 = t^2 [ 1 - ( u + v )^2/4c^2] Then take the square root of both sides.

    t' = t sqrt [ 1 - ( u + v )^2/4c^2] Then time is not negative so then the negative answer is not valid. And we can call this equation 8.

    So now we have a time dialation equation for an object that is in uniform acceleration.
     
  8. Prof.Layman totally internally reflected Registered Senior Member

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    So then lets find the equation for the distance measured by an object under uniform acceleration. This distance could be seen as the distance the light beam travels. We know the amount of time experienced under uniform acceleration so then that value times the speed of light would be what they would measure as real distance.

    d' = c t' Equation 2

    t' = t sqrt [ 1 - ( u + v )^2/4c^2 ] Equation 8. You can now substitute the right side of equation 8 into ( t' ) in equation 2

    d' = c t sqrt [ 1 - ( u + v )^2/4c^2 ] ( c t' ) is the right side of equation 3 so then we know it is equal to ( d )

    d' = d sqrt [ 1 - ( u + v )^2/4c^2 ] So then this would be the distance measured by the object traveling in uniform acceleration that they would measure a beam of light to travel or just the proper distance. This would be equation 9.
     
  9. Prof.Layman totally internally reflected Registered Senior Member

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    So then what is the uniform acceleration of an object who's time and distance has been dialated? Velocity can be either constant motion or acceleration. So then lets try to put it into the equation v = d / t.

    v = [d sqrt [ 1 - ( u + v )^2/4c^2 ]] / [ t sqrt [ 1 - ( u + v )^2/4c^2 ]] The term sqrt [ 1 - ( u + v )^2/4c^2 ] would then cancel

    v = d / t The amount of time and distance measured under uniform acceleration would be the same velocity as the velocity of an object without considering spacetime dialation.

    Conclusion, an object under uniform acceleration would have the same relative velocity even though it was being influenced by spacetime dialation.
     
  10. Read-Only Valued Senior Member

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    Having fun just talking to yourself?
     
  11. Prof.Layman totally internally reflected Registered Senior Member

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    Kind of getting worn out but almost to the last post to start the thread. Alphanumeric wanted me to derive what force would be considering time dialation, that I was just about to get too. I told him I would have it done this evening, so I have been working on it all day. Trying to get it done before he comes back to check it out.
     
  12. Prof.Layman totally internally reflected Registered Senior Member

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    So we know that F = m a, and a = ( v - u ) / t, since the relative velocity is the same for all observers then you can replace the value of time with the proper time. So then they would measure acceleration to be a' = ( v - u ) / t' (eq. 10), so then when F = m a' you can substitute equation 10 into F = m a. And then substitute the time dialation of an object under uniform accelertion (eq. 8) for ( t' ) in equation 10.

    F = [ m ( v - u ) ] / [ t sqrt ( 1 - ( u + v )^2/4c^2 ) ] This would be the amount of force of an object in uniform acceleration with respect to the amount of time dialation that I will call equation 11.
     
    Last edited: Dec 31, 2012
  13. Aqueous Id flat Earth skeptic Valued Senior Member

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    This is confused. You are addressing special relativity (albeit in the most inelegant way). It's unclear whether you have a problem with semantics, or whether you really think you're on to something different. Different, that is, than the correct and elegant explanation of Special Relativity given in Einstein's famous 1905 paper.

    Einstein perfected what Maxwell, Michelson-Morley, Lorentz/Fitzgerald and Poincare had evolved from Newton's theory of corpuscular light. Without the physics to understand the implications of Special Relativity, which is best understood by teaching yourself a little of the history of this evolution in ideas, you are bound to miss the point and end up with faulty ideas and/or confusion about what the correct ideas really are--what they mean.

    To say velocity is not affected by time dilation is confused. The initial concept is that there is no such thing as absolute velocity. In other words, any concept of velocity is only meaningful when taken as the relative velocity between two inertial reference frames. If the meaning of inertial reference frames is not clear to you, it's due to not understanding Newtonian mechanics, also known as dynamics or kinematics. (This is late high school/first year college science. It does also require math of that level, calculus and its prerequisites, which is why science and its requisite math courses are best taken concurrently.)

    The danger of inventing one's own ideas without first understanding what has already been discovered about Nature, is that it leads to invalid, absurd or even confused conclusions. In this case, you are merely demonstrating confusion. Namely, you don't seem to take away the correct interpretation, that is, that space and time are relative, between any two observers whose inertial reference frames are moving with respect to each other.

    It makes no sense to say relative velocity is unaffected by time dilation. Relative velocity is the premise of SR, so finding the premise true after already assuming it true at the outset, is what I mean by confusion. The correct inference to draw from SR is that space and time are relative. Any other conclusion is confusion.

    It's not at all different, it's just crudely done, giving the impression that you don't understand kinematics or calculus, much less the history from Newton to Einstein, and the profound implications of his 1905 paper.

    That's just circular. The only reason you even mention proper time is because you read it somewhere. You seem to not understand that "proper time" is an idea that springs from SR, not the other way around. SR is the underlying law that describes what Nature is, what it is doing. By talking around the physics, rather than immersing yourself in it, you miss the key ideas that are so important to understanding Nature. The ideas (correct ones, not absurd or even confused ones) are the most important result of any line of inquiry.

    This is Galileo, who preceded Newton in the evolution of SR. The mass transport of his era was the ship, and he correctly demonstrated that there is no such thing as absolute velocity, having observed that a ship can seem to be moving backwards when overtaken by a faster one, blocking the view to the horizon. Then, when the faster ship clears, and the horizon reemerges, the observer will notice that he is indeed moving forward with respect to the shore. Einstein's analogy was that of a person on a moving train and another on the platform. The one on the train drops a ball, and observes that it falls vertical to the floor. The person on the platform, looking into the open car as it passes by, observes that the ball falls at a slant with respect to the Earth. Everything you do in your lengthy explanation here derives from the right triangle that Lorentz, Fitzgerald, Poincare and Einstein were aware of, the one whose vertical is the distance measured by the person on the train, whose hypotenuse is the slant measured by the person on the platform, and whose horizontal side is measured by the person on the platform as the distance traveled by the train from the time the ball drops until it hits the floor.

    The reason I said this is implausible is because it's nearly impossible to construct. Imagine trying to build an apparatus that does this, as opposed to using a stopwatch as Einstein's observer on the platform might have done. There is no true perpendicular in either case, but the slow moving ball compensates, as opposed to the fast beam of light.

    Of course you wouldn't have known for sure that light speed was constant unless you had tracked the progress from Newton to Einstein, noting the results of Michelson-Morley.

    It's not clear whether you think you are arriving at an alternative theory, because it's not. SR is the only law that governs what happens here. In case you're not aware of it, you should note that you have set your experiment up in the transverse axis ("perpendicular"). You may think that you're doing something different, since SR usually begins with the practical cases of red and blue shift, which are along the longitudinal axis. Again, this speaks to ideas, and to understanding Nature by forming the correct ones.

    From here on out you take us into a rather long winded ride using just algebra. Some of the alarms that would go off (in the heads of anyone with a science background) are that (1) a more exact explanation needs to recognize that when you write v = d / t you actually mean v = Δx / Δt, and (2), which is a simplification of v = dx/dt, which is a generalization for the Euclidean distance: d = √ (x² + y² + z²), which leads to the more general treatment done in the Lorentz transformation, and (3) which is a more compact form that covers all the bases, which is the reason it's the standard math used (matrix arithmetic) also the standard math for late high-school/early college science classes.

    That gets us to the end of your presentation, since we can use the Lorentz transformation to neatly cover all of the practical math , which is going to be in vector form (you are giving the 1-dimensional scalar form), plus we arrive at the correct and rather profound idea that you have completely missed. Let me see if I can state it very briefly for clarity:

    We start with a generalized parameter g. It may represent any of the concepts of motion you are discussing here, distance, velocity, acceleration, or even force, momentum, power, energy, etc. In the most general of these, we allow for 3 spatial dimensions, plus one of time. This yields a 4-D vector g. When we say we are going to subject g to a transformation, we mean we are going to perform an operation usually called matrix multiplication: g' = A g where A is a 4 x 4 matrix that operates as the Lorentz transform. What you want do is to pay close attention to is the form of this operator:

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    If you've ever done any graphics, geological maps, or navigation you will recognize this as the operator for projection. This is the simplest case, rotation on one axis. You can demonstrate this by holding your hand up facing your palm. At this viewing angle you receive the maximum projection of the surface area of your hand. Now rotate your wrist. Your palm still has the same surface area, but you "receive" less and less until you come to a minimum at 90°. This is what we mean by projection, and it's a fundamental idea to take away from relativity. Without this, you've lost the most basic notion of spacetime, which is, that is has the nature of a projection, one whose angle of rotation is strictly dependent on the relative velocity of the two inertial reference frames.

    If you go back and read Einstein's paper (if you can, it helps to know calculus and matrix math), and if you contemplate what he starts talking about, from the minute he mentions Lorentz, you will begin to realize that he had essentially discovered spacetime. He discovered that it's a projection. He discovered that it's subject to a variable amount of rotation, whose angle depends on the relative velocity. This is a phenomenal advance in total human awareness up to that point. His predecessors--Lorentz, Fitzgerald, Poincare and to some extent Michelson-Morley, had understood quite well the physics and the mathematical interpretation, but Einstein connected the dots by something equivalent to waking up out of a dream. Spacetime really does this. It's really a projection that curls up as observers attempt to approach the speed of light.

    Put that in your pipe and smoke it. There's alternative theory, there's a feeling of being on the brink of something new, there's confusion and error and the inability to perceive correct ideas, and then there's waking up out of a dream. All we want to do is wake up. The rest is all styrofoam.
     
  14. AlphaNumeric Fully ionized Registered Senior Member

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    Right, I've woken up a bit after NYE last night. I won't reply to everything, I'd be here till next NYE so I'll pick a few salient points.

    The most immediate and glaring problem is all of your work is based on constant motion, using equations learnt in school for Newtonian mechanics. In Newtonian mechanics you can accelerate as much as you like but in relativity there's the light speed barrier so things like \(v = u+at\) no longer apply when you consider relativistic things since if \(v\geq c\) the energy required to produce constant acceleration is infinite.

    You talk about the holographic principle and black holes but they are in general relativity while all you do is special relativity and just an application of Pythagoras. If you'd ever actually done any special relativity or non-Euclidean geometry you'd know know that the metric defines infinitesimal light elements, \(ds^{2} = g_{ab}dx^{a} dx^{b}\). The length of a curve is then defined by integrating up, \(s = \int_{\textrm{curve}}ds = \int_{L}\sqrt{g_{ab}\dot{x}^{a}\dot{x}^{b}}d\tau\). If the metric is constant and the path simple (like a line) then it simplifies a lot but that isn't so when moving in a complicated manner or in curved space-time, such as those used in the holographic principle and black hole mechanics.

    This is a mistake many people do when they first start learning non-Euclidean differential geometry, they don't get that you don't just slap down a distance function on the space, like you do for Euclidean geometry, but rather you define an infinitesimal metric and have to integrate line elements along curves. Simple rearranging the Lorentzian metric is something everyone does when first doing special relativity. It is the first step in moving beyond the Euclidean metric but your claim you're doing something new or advanced (ie close to a PhD) is laughable. I taught this stuff to 1st years when I was doing my PhD!

    As stated, without considering a black hole metric what you're doing has nothing to do with any of that.

    Nothing you've done has anything to do with those things. General relativity has local Lorentz invariance (ie looks flat in small regions) but to handle such a thing in a curved space-time requires the use of more elaborate mathematical machinery.

    This is the most trivial of things, something covered in introductory courses. Do you think no one has noticed how \(\gamma\) has close ties with Pythagoras's theorem? Relativity is all about geometry, this isn't an accident. The space-time interval \(ds^{2} = -dt^{2} + d\mathbf{x}\cdot d\mathbf{x}\) is Lorentz invariant and all you've done is give an example of this.

    The sequence of posts you're making as essentially you working through a simple homework problem, this is not new.

    Yes, a basic result in relativity. Hence the name relativ(e)ity, as what matters is relative velocity rather than some notion of absolute velocity. Likewise with the posts which follow. Expanding squares, using the SUVAT equations and applying Pythag's theorem is trivial stuff, something students will do when trying to solve such basic kinematic problems.

    It might well be the case you think this complicated or think it advanced but I cannot even begin to get across how far there is from the truth.

    I wanted you to give the work you had said you'd done, whatever it was about, to see what kind of mathematical competency you have and what you think is advanced. It would have been better for you if you'd just refused...

    The posts you've made are menial reshufflings of simple 'constant acceleration flat space' kinematical expressions. Yes, you've shown you can at least do high school algebraic manipulation but that is a long way from being able to do anything of any substance in even special relativity, never mind on to general relativity and quantum field theory in curved space-time.

    Have you ever bothered to look at any lecture notes or textbooks on say the holographic principle? Since I know you struggle to find literature yourself I'll provide this link to a review of the AdS/CFT correspondence from a decade ago. It gives a quick overview of supersymmetric gauge theory, Type II string theory and then gets into the correspondence. That is what material at PhD level and beyond looks like. And to compare here is the homework sheet for an introductory course in special relativity. Notice how the first question is considering something similar to you but does so in a more formal way? Notice how the questions only get more elaborate?

    You can do the most basic of special relativity. Whether you wish to view that as a complement (since that's beyond most people) or an insult (since its about a decade of solid work short of being worth a PhD) I don't care. If it took 6 courses in algebra to get you to that level then I question with worth of those courses. The fact of the matter is you have now shown clearly and concretely that you definitely lack even the most rudimentary quantitative grasp of anything pertaining to Hawking radiation, black holes or the holographic principle. You're welcome to your opinions on them but don't for a pico-second you have any grasp of the internals of any of them. As such hopefully your comments about them will now be tempered by this fact.
     
    Last edited: Jan 1, 2013
  15. Cheezle Hab SoSlI' Quch! Registered Senior Member

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    That is very illuminating . Especially considering that Euclidean geometry is just a special case of a vastly richer subject of non-Euclidean geometry. Which I suppose is kind of obvious. But this is a very deep subject. I am going to have to go back and think about this some more. I feel like I may be on the Threshold of a Dream. Well maybe not quite.

    Here is a video on the subject at hand. The most important thing (I think) is the relationship between the different non-Euclidean models as being projective. Different types of projections. Angle, distance, area, etc. And just how those projection are defined. I am going to have to go back and watch these videos again. Thanks again for that hint.

    [video=youtube;zHh9q_nKrbc]http://www.youtube.com/watch?v=zHh9q_nKrbc[/video]
    (50 minutes)
     
  16. AlphaNumeric Fully ionized Registered Senior Member

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    For people who are competent and comfortable doing basic calculus and vectors learning about non-Euclidean geometry by starting with the other two types of constant curvature geometries (positive curvature is spherical and negative is hyperbolic) is a good place to start.

    For example you can obtain spherical geometry by taking the definition of a sphere's embedding in flat Euclidean space and then doing what is technically known as a pull back onto the sphere's surface. For example, consider the 3d case (this generalised in an obvious way), where we embed a 2d sphere into 3d Euclidean space. The equation of a 2d sphere is \(x^{2} + y^{2} + z^{2} = R^{2}\). The sphere is 2 dimensional so we parametrise it using two coordinates which we can decide to be the polar angles via \(x = R\,\sin\theta \, \cos \phi\), \(y = R\,\sin\theta\,\sin\phi\) and \(z = \,R\cos\theta\). We know the Euclidean metric \(\delta_{ab}\), where \(ds^{2} = \sum_{a,b}\delta_{ab}dx^{a}dx^{b} = dx^{2} + dy^{2}+dz^{2}\), and we can compute the Jacobian of this transform \(J_{ia}\), which is a 2x3 matrix (pulling the 3d metric back onto the 2d surface). The metric on the surface is then \(g_{ij} = \sum_{a,b}J_{ia}J_{jb}\delta_{ab}\), which simplifies to \(g_{ab} = \sum_{a}J_{ia}J_{ja}\). If you do this you get \(ds^{2} = R^{2}(d\theta^{2} + \sin^{2}\theta d\phi^{2})\), the spherical metric!! Using similar principles you can work out area elements on the sphere too.

    It's a nice exercise and gives some insight into how to construct non-Euclidean structures with little more than a Euclidean structure and a surface Here gives 5 different ways of embedding a hyperbolic surface into Euclidean space on page 12 and then the resultant pullbacks on page 13, should anyone want to see if they can do it for themselves

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  17. Cheezle Hab SoSlI' Quch! Registered Senior Member

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    Wow, thanks. I scanned the rkenyon cannon paper and when I saw Fig. 34, I knew this was going to be a gold mine for me. I find the big picture to be the most interesting. And I love the stuff that gives you and others the enthusiasm to use multiple exclamation points. I totally understand the coolness factor. The way math and physics seem to wrap around and have all these interesting equivalences and similarities. There is enough to here to keep me busy for a long time.
     
  18. Prof.Layman totally internally reflected Registered Senior Member

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    I have always thought that to be a popular misconception. I think I have actually convinced a lot of people on the net otherwise. It has been a while since I have ran into someone that still holds on to this premis. I don't think they considered that the relative mass increase would just be relative to the observer, not the person who is doing the acceleration. If they observed a mass increase then they could observe an increase in energy as well. So say a ship was using a nuclear jet engine on the back of it, you would see the mass increase but then the mass of the particles of the engine would increase so then they would create more energy to propell the ship or the particles themselves going out the back would also be more massive so then they would create a greater force at the same rate the rest of the ships mass would increase. So then the velocity could stay the same even though it was observed to have its mass increase relative to the observer at rest.

    Different degrees of constant motion are indistinguishable from each other. If you where to accelerate from any amount of constant motion the forces felt from that acceleration would be the same. So if you went 1 m/s and then decided to go 2 m/s^2, you would feel the same force as if you where seen to be going 2m/s instead because the other object changed its velocity by 1m/s, and then if you decided to go 2m/s^2 you will feel the same force.

    The forces of acceleration will feel the same no matter what relative speed another observer is traveling. A ship couldn't do a fly by and then look out its window to squash the inhabitants of an unsuspecting civilization.

    This then lead me to thinking about galaxies on the edge of the visable universe. All of these galaxies are not almost infinite in mass. So then they are not all suppermassive black holes because their mass doesn't increase relative to other object near them, it would only increase relative to the Milky Way galaxy, so then this force of gravity would still be small since they are so far away and it wouldn't be hardly detectable.
     
  19. AlphaNumeric Fully ionized Registered Senior Member

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    No, it wouldn't. More energy wouldn't be available, it isn't like you have more Hydrogen atoms to fuse, Lorentz transforms do not alter particle numbers. Instead each atom just becomes more massive.

    You also forgot about the time dilation effect. If A and B are moving relative to one another and A is accelerating then from B's point of view A gets more and more massive so the thrust coming out of the back of the rocket increases in mass but the time dilation slows down the rate at which it is being emitted. An exhaust 10 times as massive but flowing 1/10 the speed will not provide any additional acceleration than originally but now it has to push a ship which is also 10 times as massive.

    Of course all of this is just arm waving. If the calculations are done then it is possible to show it formally. In A's frame the engines work in a clearly understood way and then you just apply the appropriate shift into B's frame (its messy because of the non-inertial nature of A's motion) and you find the behaviour of the rocket from B's point of view. It never reaches light speed, the perceived passage of time of A from B's point of view slows down more and more, never providing enough energy to get to light speed. This is why I said to keep acceleration constant, from B's point of view, you'd never an ever increasing force.

    Velocity is relative, acceleration is not.

    This is a mistake of understanding on your part. The rocket example is special relativity, moving objects in flat space-time, while the galaxies on the edge of the visible universe are getting most of their relative motion due to the expansion of space-time between us and them. They are all moving around one another in much the same way we interact with our local cluster but the space between us and them is expanding and the relative velocity over cosmological distances becomes significant. This is described in cosmology using the FRW metric, which contains a length scale parameter which defines how this distance and thus velocity behaves.
     
  20. Motor Daddy Valued Senior Member

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    5,425
    If A and B are moving relative to one another, and the closing speed is 100 MPH, how do you determine which object(s) are in motion? You do agree that each object agrees on the distance between them at every point in time, correct?

    Are you saying that the exhaust rate is proportional to the speed? Are you saying that the exhaust flows 10 times slower as the ship increases its speed and its exhaust becomes 10 times more massive??

    If the exhaust rate is proportional to the speed of the ship (slower exhaust means faster ship speed), then are you saying it's possible to know the velocity of the ship by knowing the speed of the exhaust, since the speed of the ship and the speed of the exhaust are directly proportional?

    Why is it that in the example I provided below that it's the greatest HP that accelerates at the greatest rate at every MPH and not the greatest engine torque that accelerates at the greatest rate at every MPH?????

    Engine A in second gear (1.98:1 ratio) with a 3.23:1 rear gear ratio and 26.5" diameter tire.

    2000 464 lb-ft 176 hp 24 mph 2967 RWTQ
    2500 482 lb-ft 229 hp 30 mph 3082 RWTQ
    3000 496 lb-ft 283 hp 37 mph 3172 RWTQ
    3500 511 lb-ft 340 hp 43 mph 3268 RWTQ
    4000 506 lb-ft 385 hp 49 mph 3236 RWTQ
    4500 481 lb-ft 412 hp 55 mph 3076 RWTQ
    5000 427 lb-ft 406 hp 61 mph 2730 RWTQ
    5500 363 lb-ft 380 hp 67 mph 2321 RWTQ


    Engine B in second gear (1.98:1 ratio) with a 3.73:1 rear gear ratio and 26.5" diameter tire.

    2309 410 lb-ft 180 hp 24 mph 3028 RWTQ
    2887 430 lb-ft 236 hp 30 mph 3175 RWTQ
    3464 460 lb-ft 303 hp 37 mph 3397 RWTQ
    4041 465 lb-ft 357 hp 43 mph 3434 RWTQ
    4619 465 lb-ft 408 hp 49 mph 3434 RWTQ
    5196 445 lb-ft 440 hp 55 mph 3286 RWTQ
    5773 390 lb-ft 428 hp 61 mph 2880 RWTQ
    6351 330 lb-ft 399 hp 67 mph 2437 RWTQ


    Two identical cars (except for engine and gears, same weight) are moving along at a steady 24 mph in second gear side by side. They both punch it at the same exact time. Which one will pull ahead immediately, and get to 67 mph the soonest, and cover the greatest distance in the same time period? The one with the greatest engine TORQUE, or the one with the greatest engine HP??
     
    Last edited: Jan 2, 2013
  21. Motor Daddy Valued Senior Member

    Messages:
    5,425
    Duplicate post.
     
  22. Motor Daddy Valued Senior Member

    Messages:
    5,425
    I'm having a hard time wrapping my head around your statement. Maybe you can help clear up my confusion about what you mean exactly by that statement.

    If you and I are two cars on the highway we could be each traveling at a constant velocity in the same direction down the road, one in front of the other, like in traffic. We are each doing the same speed so the distance between us is always the same. The relative velocity between us is dependent on each of our motions. Since we are traveling the same speed in the same direction there is no relative motion between us.

    If we each accelerate at the same rate at the exact same time there will be no change in distance between us.

    If you base all your measurements on the distance between us then you have no justification to claim you are accelerating. According to you, if all there is is you and me, relative velocity (closing speed) AND acceleration are relative.

    Looks like you have a bit of a contradiction on your hands.
     
    Last edited: Jan 2, 2013
  23. Prof.Layman totally internally reflected Registered Senior Member

    Messages:
    982
    Your right there wouldn't be more particles. The particles on this ship would just look to be more massive from the frame of reference of an observer at rest. I think your forgeting that E = mc^2. If the particles look to have more mass from a reference frame then they would also look like they have more energy when converted to energy. But it would only look like they had more energy from the frame of reference that is at rest, so then even though a ship was being observed by an outside observer the people on the ship would still measure the energy of their nuclear reactions to still just be goverened by E = mc^2. So then an outside observer looking into the experiments done by scientist inside the ship will not change the results the scientist on the ship would gather. The scientist will not measure their own mass to be different because of an outside observer, and the mass they measure would not be affected by multiple outside observer. There would be no need to determine what the scientist on the ship would measure because of being observed differently by outside observers. Different outside observers would all say the mass of the ship is different, but this would not change the laws of physics as seen by the scientist on the ship. So then there would be no need to corelate what they measure on the ship because outside observers didn't agree with what they measure as the mass of the ship. The laws of physics measured by the scientist on the ship would remain consistant.

    The velocity of the particles coming out of the back of the ship would remain constant. The speed of light in the equations I posted above replaces the velocity variable in the velocity equation v = d/t. So then it would be no surprise that the mathmatics that assumes the speed of light is constant would also lead to velocity being constant because the speed of light has taken the place of velocity in one of the equations. So then it is if v = c, and c being constant then makes v act like a constant. For instance you could not have a rocket ship that reached a high velocity then have the rockets perform differently because Nasa that was at rest on Earth observed what was going on inside of the rocket from base. The people in the space ship could have functioning rockets, that behaive the same way regardless if Nasa was monitoring them or not.

    Even if an object is seen to travel at 99% the speed of light, from the frame of reference on the ship it would still measure light to travel about 300,000 km/s faster than they are. So then no matter how long something accelerated at say just 1m/s they would still measure the speed of light to be the same, no matter how long they did this for. They could travel at 1 m/s^2 for one year, and then take their foot off of the accelerator to then travel at a constant speed and then still have no way to distinguish that speed from any other constant speed, and then assume they are at rest.

    If any object traveling at a constant speed can say that it is at rest then acceleration would have to be relative to this is some way. Say an object is at rest and then begins to accelerate to another degree of constant motion. It accelerates and then no longer accelerates so that it is in constant motion again. There is no way to distinguish between moving from one degree of constant motion than from another degree of constant motion. If there was you could find the absolute frame of reference. But accelerating from one constant speed is no different than accelerating from a higher constant velocity, so then there is no way to know what constant speed was actually faster or slower. If I could increase my speed to 55 mph and then travel along at 55 mph for a while and then increase to 110 mph and then feel more force from going to 110 mph from 55 mph then I would know that 0 mph was actually the slower speed.

    I was expecting you to say that relativistic effects are just local. I think that explaination is just the downfall of not being able to explain why distance galaxies do not appear to have an increase in mass. But, then I have seen some work posted on these forums recently that claim that they have detected spacetime dialation in other galaxies. I think this is a work in progress, but I think it is possible for the affects of GR to not just be local and independent of the expansion of space if it is seen to only increase in mass relative to the observer. They would be so distant that we would not notice this increase in mass because the force of gravity gets weaker with distance, and that extra force of gravity would only be felt by the other relativistic observer. I am not sure yet if galaxies traveling with the expansion of space could be left off of the hook of being affected by GR.
     

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