ydoaPs Posted October 16, 2012 Share Posted October 16, 2012 Only slightly, and only in some situations. Read the bit after the last comma. No, it's completely different as it does not provide the same answer. Give us an equation. Link to comment Share on other sites More sharing options...
A-wal Posted October 16, 2012 Author Share Posted October 16, 2012 (edited) I'm not going to start copying and pasting equations from special relativity. Ah, I can't keep up. I edited my last post. Edited October 16, 2012 by A-wal Link to comment Share on other sites More sharing options...
ydoaPs Posted October 16, 2012 Share Posted October 16, 2012 Read the bit after the last comma. That's precisely why the equations are different. Your version is wrong. Link to comment Share on other sites More sharing options...
A-wal Posted October 16, 2012 Author Share Posted October 16, 2012 That's precisely why the equations are different. Your version is wrong. No it isn't, GR is wrong! That isn't why the equations in GR are different. According to GR free-fall is inertial. That's simply not true! Link to comment Share on other sites More sharing options...
uncool Posted October 16, 2012 Share Posted October 16, 2012 Torus, that's a donut shape isn't it? Informally, yes. It can also be considered as the product of two circles. I wasn't very clear there. What I meant is that any valid coordinate system for describing space-time should be applicable to the whole of space-time. If it stops at an event horizon that that coordinate system predicts that space-time stops at an event horizon. And as I said, the torus experiences the same problem. Consider a circle. It cannot be covered in a 1-1 way by the real line. However, if you consider a set of 2 charts, they do cover the entire circle and fully characterize it as a manifold. Neither chart applies to the whole of space-time, but both are valid and necessary. =Uncool- Link to comment Share on other sites More sharing options...
mississippichem Posted October 16, 2012 Share Posted October 16, 2012 You can't just take the bits that you've bolded and ignore the rest. It's even in the same sentence. Fair enough I'm not being in the slightest bit "handwavy". You've completely ignored what I'm saying. Using mass to accelerate is the EXACT EQUIVALENT of using energy to accelerate, and all the maths is the same, accept that energy pushes and mass pulls. Let's see the math then. Do you know how to use LaTeX? If not there is a tutorial available here on site. Did that really make a difference? I'm not going to start copying and pasting equations from special relativity. [/explicit snip for honesty] Don't copy and paste. Do the derivations. Use equations and the relevant prose. This is standard stuff. You made the claim, you construct an original derivation showing the equivalence of your theory with SR. It's very simple really. 1 Link to comment Share on other sites More sharing options...
ydoaPs Posted October 16, 2012 Share Posted October 16, 2012 No it isn't, GR is wrong! That isn't why the equations in GR are different. According to GR free-fall is inertial. That's simply not true! Then do the math instead of making bombastic claims without any support. Put up or shut up. Your version CLEARLY requires very different math. Link to comment Share on other sites More sharing options...
A-wal Posted October 16, 2012 Author Share Posted October 16, 2012 (edited) And as I said, the torus experiences the same problem. Consider a circle. It cannot be covered in a 1-1 way by the real line. However, if you consider a set of 2 charts, they do cover the entire circle and fully characterize it as a manifold. Neither chart applies to the whole of space-time, but both are valid and necessary. Yes, I understand that, but space-time is not shaped like a donut. Don't listen to Homer Simpson, he's wrong. You simply don't need multiple manifolds to describe space-time. Let's see the math then. Do you know how to use LaTeX? If not there is a tutorial available here on site. Did that really make a difference? Don't copy and paste. Do the derivations. Use equations and the relevant prose. This is standard stuff. You made the claim, you construct an original derivation showing the equivalence of your theory with SR. It's very simple really. Did what really make a difference? I don't have to construct an original derivation to make a valid claim! That's even more very simple really. Then do the math instead of making bombastic claims without any support. Put up or shut up. Your version CLEARLY requires very different math. No it doesn't! Acceleration is acceleration. Why would it CLEARLY require very different maths? My whole point is that they're equivalent! Edited October 16, 2012 by A-wal Link to comment Share on other sites More sharing options...
mississippichem Posted October 16, 2012 Share Posted October 16, 2012 Yes, I understand that, but space-time is not shaped like a donut. Don't listen to Homer Simpson, he's wrong. You simply don't need multiple manifolds to describe space-time. How about a derivation for that as well. Still waiting. 1 Link to comment Share on other sites More sharing options...
uncool Posted October 16, 2012 Share Posted October 16, 2012 (edited) Yes, I understand that, but space-time is not shaped like a donut. Don't listen to Homer Simpson, he's wrong. You simply don't need multiple manifolds to describe space-time. First, you're misusing terminology. A manifold is the entire object; a chart or coordinate system is what you should be referring to. This is one reason why people insist that you learn the math - so that you can understand the terminology used in the first place. Second, this is an assertion with no support. Not just that, but it's an assertion that is not verifiable and makes no specific predictions. Special relativity can take place on a 4-torus. Not on a 4-sphere, but yes on a torus. This is because a torus can be considered a quotient of 4-Euclidean space, which is the traditional place where relativity takes place. =Uncool- Edited October 16, 2012 by uncool 1 Link to comment Share on other sites More sharing options...
A-wal Posted October 16, 2012 Author Share Posted October 16, 2012 How about a derivation for that as well. Still waiting. Don't be impatent, I needed a smoke. Do you really expect me to learn the maths, learn LaTeX and then post it in less than four minutes? Do you think you're being fair? I've never done that kind of maths in my life. That doesn't invalidate my claims, no matter how much you try to claim that it does! First, you're misusing terminology. A manifold is the entire object; a chart or coordinate system is what you should be referring to. This is one reason why people insist that you learn the math - so that you can understand the terminology used in the first place. Sorry, I'm trying to use the correct terminology but I'm not as used to this as you are. It doesn't mean that I don't understand it. Second, this is an assertion with no support. Not just that, but it's an assertion that is not verifiable and makes no specific predictions. 1. The rope parodox. 2. The equivelence in the horizons when accelerating. 3. The strength of gravity relative to electro-magnatism. I've just mentioned all three in the last couple of hours. Have you not been reading? Special relativity can take place on a 4-torus. Not on a 4-sphere, but yes on a torus. This is because a torus can be considered a quotient of 4-Euclidean space, which is the traditional place where relativity takes place. Can be does not equate to does! Special relativity takes place in four dimensional space-time, not a donut! Link to comment Share on other sites More sharing options...
mississippichem Posted October 16, 2012 Share Posted October 16, 2012 Don't be impatent, I needed a smoke. Do you really expect me to learn the maths, learn LaTeX and then post it in less than four minutes? Do you think you're being fair? I've never done that kind of maths in my life. That doesn't invalidate my claims, no matter how much you try to claim that it does! My apologies. I just figured it would be a trivial task for someone who so confidently asserts that "the math is exactly the same". So now it comes out that you don't know the math is exactly the same, by your own admission. I can't see how you can know that it is the same as SR when you just admitted to not know the math of SR . But that's just me being frustrated by intellectual dishonesty and is of little consequence to the validity of your assertion (though it sure doesn't inspire confidence). Please, by all means, take all the time you need. I'm not going anywhere. Let me know if you need any help with the LaTeX. Link to comment Share on other sites More sharing options...
swansont Posted October 16, 2012 Share Posted October 16, 2012 Using mass to accelerate is the EXACT EQUIVALENT of using energy to accelerate, and all the maths is the same. Then how can there be any test that shows a difference? You claimed there was. 1 Link to comment Share on other sites More sharing options...
uncool Posted October 16, 2012 Share Posted October 16, 2012 Sorry, I'm trying to use the correct terminology but I'm not as used to this as you are. It doesn't mean that I don't understand it. It's certainly a strong indicator that you don't. It's not only that you're "not as used to this as [i am]"; the fact that you mixed these up shows that you're not used to dealing with manifolds at all. This isn't by itself a problem; what makes it a problem is that you're trying to make claims that are based in your lack of experience. 1. The rope parodox. 2. The equivelence in the horizons when accelerating. 3. The strength of gravity relative to electro-magnatism. I've just mentioned all three in the last couple of hours. Have you not been reading? You've misunderstood the antecedent again. The antecedent to "this" is clearly is "space-time is not shaped like a donut". Can be does not equate to does! Special relativity takes place in four dimensional space-time, not a donut! This is false. Special relativity can take place on "a donut". =Uncool- Link to comment Share on other sites More sharing options...
ydoaPs Posted October 16, 2012 Share Posted October 16, 2012 No it doesn't! Acceleration is acceleration. Why would it CLEARLY require very different maths? My whole point is that they're equivalent! Because vectors pointing in opposite directions have opposite signs. You said one of them pulls while the other pushes. It's bleeding obvious. Consider a circle. It cannot be covered in a 1-1 way by the real line. While not a 1-to-1 (1-to-1 functions are boring, anyway), a circle is a function if you do the relation in polar co-ordinates. Link to comment Share on other sites More sharing options...
A-wal Posted October 16, 2012 Author Share Posted October 16, 2012 The England match got cancelled. We should get a 3-0 win for that! My apologies. I just figured it would be a trivial task for someone who so confidently asserts that "the math is exactly the same". So now it comes out that you don't know the math is exactly the same, by your own admission. I can't see how you can know that it is the same as SR when you just admitted to not know the math of SR . But that's just me being frustrated by intellectual dishonesty and is of little consequence to the validity of your assertion (though it sure doesn't inspire confidence). Please, by all means, take all the time you need. I'm not going anywhere. Let me know if you need any help with the LaTeX. I haven't once been dishonest! I said "Using mass to accelerate is the EXACT EQUIVALENT of using energy TO ACCELERATE, and all the maths is the same" because acceleration using mass is mathematically no different to acceleration using energy, accept that energy is stronger (how much by is a prediction) and that it curves outwards instead of inwards, creating the same two horizons, one approaching from in front of an accelerating object as they start to catch up to their own light at the exact same progressively slower rate as an object approaching the event horizon of a black hole, and one approaching from behind an accelerating object as their own light starts to catch up to them at the exact same progressively slower rate (the Rindler horizon) that information from over certain distance away will never reach the accelerating object as long as they keep on free-falling at at least the same rate are equivalent (is a better prediction), as I've said over and bloody over. That's how the maths is the same. Pretty isn't it? It makes a butterfly shape. Then how can there be any test that shows a difference? You claimed there was. (: What? I said that there was way to test the difference between GR, and free-fall being inertial! You're referring to the fact that there's no way to distinguish acceleration from mass and acceleration from energy, and free-fall from inertia. You've misunderstood the antecedent! (; It's certainly a strong indicator that you don't. It's not only that you're "not as used to this as [i am]"; the fact that you mixed these up shows that you're not used to dealing with manifolds at all. This isn't by itself a problem; what makes it a problem is that you're trying to make claims that are based in your lack of experience. Not really. It just indicates that I'm really up on the terminology. In fact I'm not convinced I even made a mistake. I think you might have misunderstood the antecedent. I can easily clarify what I meant. You can cover the entire manifold (all of space-time, that's what I meant by manifold) using a single coordinate system. The Schwarzschild coordinates do exactly that. You've misunderstood the antecedent again. The antecedent to "this" is clearly is "space-time is not shaped like a donut". Yes and the next sentence shows just how serious I was, "Don't listen to Homer Simpson, he's wrong."! The next sentence explains what I was saying, "You simply don't need multiple manifolds to describe space-time." accept that I accidently said manifolds instead of coordinate system. That was just a stupid mistake because I was replying to multiple posts at once as they were coming in, you don't need to crucify me for it. This is false. WHAT? WTF "Can be does not equate to does! Special relativity takes place in four dimensional space-time, not a donut!" Is false? How? What part? If it's can be does not equate to does then how exactly does can be equate to does? If it's SR takes place in four dimensional space-time, not a donut then please provide evidence, or better yet, equations that you've copied and pasted. Special relativity can take place on "a donut". I never claimed it couldn't! In fact I think a donut is an inside out sphere, so they should be the same. Oh no, a negative sphere is saddle shaped isn't it. Because vectors pointing in opposite directions have opposite signs. You said one of them pulls while the other pushes. It's bleeding obvious. I'm no mathematition but that hardly seems different. Will you please all stop deliberately taking and quoting everything I'm saying out of context! It's really annoying. And you accuse me of being dishonest! Link to comment Share on other sites More sharing options...
ydoaPs Posted October 16, 2012 Share Posted October 16, 2012 Will you please all stop deliberately taking and quoting everything I'm saying out of context! It's really annoying. And you accuse me of being dishonest! All of my quotes have been entirely in context. And you don't see how big of a difference a sign makes in equations? You really must not know much math. a-b is not the same as a+b. axb is not the same as ax(-b). a/b is not the same as a/(-b), etc. Link to comment Share on other sites More sharing options...
uncool Posted October 16, 2012 Share Posted October 16, 2012 Not really. It just indicates that I'm really up on the terminology. In fact I'm not convinced I even made a mistake. I think you might have misunderstood the antecedent. I can easily clarify what I meant. You can cover the entire manifold (all of space-time, that's what I meant by manifold) using a single coordinate system. The Schwarzschild coordinates do exactly that. The Schwarzschild coordinates cover an example of a possible spacetime. That doesn't mean it's possible to do it on any spacetime. Yes and the next sentence shows just how serious I was, "Don't listen to Homer Simpson, he's wrong."! And yet you apparently thought it serious enough as a response to the point that most manifolds do need multiple charts to be covered that you didn't address it otherwise. Any manifold that isn't homeomorphic to Euclidean space needs multiple charts. The next sentence explains what I was saying, "You simply don't need multiple manifolds to describe space-time." accept that I accidently said manifolds instead of coordinate system. That was just a stupid mistake because I was replying to multiple posts at once as they were coming in, you don't need to crucify me for it. I'm not "crucifying" you at all. I'm pointing out that it's an error that someone comfortable with the material would not have made. There's no problem with not being comfortable with the material. The point is that given your lack of comfort with the material, it's more likely that your disagreements come from misunderstandings than that your disagreements are actual problems with the theory. WHAT? WTF "Can be does not equate to does! Special relativity takes place in four dimensional space-time, not a donut!" Is false? I apologize; I misread what you said. I never claimed it couldn't! In fact I think a donut is an inside out sphere, so they should be the same. Oh no, a negative sphere is saddle shaped isn't it. Here, again, is where you are showing mathematical ignorance. What do you mean by "an inside out sphere"? As a side note: a sphere can be turned inside out smoothly. =Uncool- Link to comment Share on other sites More sharing options...
Bignose Posted October 16, 2012 Share Posted October 16, 2012 Can I please have an example of an experiment that matched the predictions of general relativity, but at the same time didn't match the predictions of treating gravitation acceleration due to mass as the equivalent of acceleration caused by energy? This should be an excellent start... http://relativity.livingreviews.org/Articles/lrr-2001-4/ It is a journal article solely on the topic of discussing the experiments that support GR. 1 Link to comment Share on other sites More sharing options...
swansont Posted October 17, 2012 Share Posted October 17, 2012 (: What? I said that there was way to test the difference between GR, and free-fall being inertial! You're referring to the fact that there's no way to distinguish acceleration from mass and acceleration from energy, and free-fall from inertia. You've misunderstood the antecedent! (; I asked you how "treating gravitation acceleration due to mass as the equivalent of acceleration caused by energy" differed from GR, and you gave an answer But OK, there's no difference, according to you. (I don't see how, though) How do you do a test to differentiate one from the other? Link to comment Share on other sites More sharing options...
mississippichem Posted October 17, 2012 Share Posted October 17, 2012 (edited) I haven't once been dishonest! I said "Using mass to accelerate is the EXACT EQUIVALENT of using energy TO ACCELERATE, and all the maths is the same" because acceleration using mass is mathematically no different to acceleration using energy, accept that energy is stronger (how much by is a prediction) and that it curves outwards instead of inwards, creating the same two horizons, one approaching from in front of an accelerating object as they start to catch up to their own light at the exact same progressively slower rate as an object approaching the event horizon of a black hole, and one approaching from behind an accelerating object as their own light starts to catch up to them at the exact same progressively slower rate (the Rindler horizon) that information from over certain distance away will never reach the accelerating object as long as they keep on free-falling at at least the same rate are equivalent (is a better prediction), as I've said over and bloody over. That's how the maths is the same. Pretty isn't it? It makes a butterfly shape. Response to me quoted in its entirety and bolded by me You said that the math (in your proposal) is exactly the same as in SR: I'm not being in the slightest bit "handwavy". You've completely ignored what I'm saying. Using mass to accelerate is the EXACT EQUIVALENT of using energy to accelerate, and all the maths is the same, accept that energy pushes and mass pulls. Then you said: Don't be impatent, I needed a smoke. Do you really expect me to learn the maths, learn LaTeX and then post it in less than four minutes? Do you think you're being fair? I've never done that kind of maths in my life. That doesn't invalidate my claims, no matter how much you try to claim that it does! You see? You made a claim that you had knowledge you didn't have. You've never that kind of math in your life, yet you claim to know it well enough to know that the math in SR is no different from the math in your theory! What part of that is not dishonest? Edited October 17, 2012 by mississippichem 2 Link to comment Share on other sites More sharing options...
ydoaPs Posted October 19, 2012 Share Posted October 19, 2012 Still missing from this thread: derivation of the "proper" equations so that we can compare answers to the data and see which is more accurate Link to comment Share on other sites More sharing options...
ydoaPs Posted October 24, 2012 Share Posted October 24, 2012 Any update? Link to comment Share on other sites More sharing options...
A-wal Posted December 4, 2012 Author Share Posted December 4, 2012 All of my quotes have been entirely in context.Really? That's almost the exact opposite of true. Mathematics is a language just like any other except it's better suited for the task. You not understanding it doesn't mean it doesn't carry as much understanding as English or French. In fact, it carries more. You tell me in English how electric fields and magnetic fields relate. I guarantee the Maxwell Equations carry more understanding than your English. If you can't do the math, you don't understand what's going on. Then do the math instead of making bombastic claims without any support. Put up or shut up. Your version CLEARLY requires very different math. Because vectors pointing in opposite directions have opposite signs. You said one of them pulls while the other pushes. It's bleeding obvious. While not a 1-to-1 (1-to-1 functions are boring, anyway), a circle is a function if you do the relation in polar co-ordinates. Please explain how exactly each of those quotes was taken out of context. And you don't see how big of a difference a sign makes in equations? You really must not know much math. a-b is not the same as a+b. axb is not the same as ax(-b). a/b is not the same as a/(-b), etc. Now this is what I was talking about. Thats a proper example of taking something out of context. The Schwarzschild coordinates cover an example of a possible spacetime. That doesn't mean it's possible to do it on any spacetime. And yet you apparently thought it serious enough as a response to the point that most manifolds do need multiple charts to be covered that you didn't address it otherwise. Any manifold that isn't homeomorphic to Euclidean space needs multiple charts. Its possible to completely cover any genuine space-time using a single coordinate system. I'm not "crucifying" you at all. I'm pointing out that it's an error that someone comfortable with the material would not have made. There's no problem with not being comfortable with the material. The point is that given your lack of comfort with the material, it's more likely that your disagreements come from misunderstandings than that your disagreements are actual problems with the theory. You might want to check the relativity thread. I wrote thread instead of forum by mistake. I just wrote the wrong word, I do that sometimes, but I didn't misunderstand its meaning and it doesn't mean I'm uncomfortable using a website. Here, again, is where you are showing mathematical ignorance. What do you mean by "an inside out sphere"?A black holes event and Rindler horizon are an inside out version of the Rindler horizon and the speed of light when using energy to accelerate rather than mass. I asked you how "treating gravitation acceleration due to mass as the equivalent of acceleration caused by energy" differed from GR, and you gave an answerI don't think any of that has anything to do with whether or not free-fall is inertial. But OK, there's no difference, according to you. (I don't see how, though) What are you having trouble with? I'll try to clarify. How do you do a test to differentiate one from the other?GR claims that acceleration and free-fall are distinct because using energy to accelerate isn't inertial according to GR but it doesnt even suggest why they should be. What tests does it do to differentiate them from each other and show the differences? Response to me quoted in its entirety and bolded by me You said that the math (in your proposal) is exactly the same as in SR: Then you said: You see? You made a claim that you had knowledge you didn't have. You've never that kind of math in your life, yet you claim to know it well enough to know that the math in SR is no different from the math in your theory! What part of that is not dishonest? All of it. Using mass to accelerate is the equivalent of using energy to accelerate rather than being inertial. The Rindler horizon works in exactly the same way when the acceleration is caused by mass and so does the speed of light (the event horizon). If an object is being pulled towards a very powerful magnet then its velocity would increase relative to a non-magnetic object in exactly the same way as the velocity of an object accelerating towards a black hole would increase relative to an object maintaining a constant distance from the black hole (using energy to accelerate in the opposite direction to perfectly balance the acceleration from the mass of the black hole and the energy from their engines). In both cases there is a point behind the accelerating object (the Rindler horizon from SR in the case of energy) where even the speed of light wouldn't be enough to ever catch up to the accelerating object as long as they carry on accelerating at at least the same rate, and in both cases it approaches from behind them at the exact same progressively slower rate as their acceleration increases. The same thing happens in front of an accelerating object. If an accelerating object is shining a light in front of it then the beam will be moving away from them slower than the speed of light, and at a constant rate if they keep their acceleration constant. The lights speed relative to the accelerating object decreases as their acceleration increases and increases as their acceleration decreases, approaching the speed of light as their acceleration approaches zero. As they increase their acceleration they gain on their own light at a progressively slower rate, making it impossible to ever catch up to their own light, whether it's using energy to accelerate or using mass (approaching an event horizon). This also exactly describes the relationship of an inertial objects velocity relative to another object. The same amount of acceleration increases the relative velocity between them at a progressively slower rate as the velocity between them increases, so acceleration can be accurately described as velocity relative to energy. A black hole in GR has an expanding horizon but there's absolutely no way for a an expanding event horizon to have any kind of affect on any object before the event horizon reaches the object because the event horizon is moving at the speed of light, so it can't possibly expand to reach any object because if it could then anything the event horizon touches would have to be instantly accelerated to the speed of light relative to objects further away because no information from the black hole can reach them before the event horizon does because they're both moving towards objects at the speed of light locally and slower from a distance as an inverse square so that it would be moving outwards at a quarter of it's current speed if the observing object were to double its distance from the black hole. Gravity is supposed to be time reversible but in GR it isn't. Gravity is still an attractive force when the arrow of time is reversed but that doesn't apply to black holes in GR. Instead you get an impossible object known as a white hole that has no way to form and in fact describes exactly half of a true black hole. The whole thing in space and time is a four dimensional sphere. It expands outwards and then back again, all at the speed of light, so the first thing that anything can see is a full size black hole that's shrinking at the speed of light. The event horizon moves inwards slower from the perspective of more distant objects as it gets more length expanded and time, er, extended from their perspective. Once the event horizon has expanded to its maximum size, determined by its mass, it then contracts inwards at the speed of light locally and again, slower from a distance as an inverse square. A white hole has a contracting event horizon, so its a white hole, not black by the time any object can become aware of it. Once its reached it maximum size it becomes a white hole and then it starts contracting, but its influence carries on expanding outwards at the speed of light and accelerates any object it comes into contact with towards it. This makes the whole thing a perfect four dimensional sphere. The event horizon still cant be reached once it starts contracting because it contracts faster as its approached and would be moving at the speed of light if any object were able to reach it. This isnt at all surprising because a black/white hole isnt a real object as we normally think of one, its just acceleration. Thats why theres an equivalent of a Rindler horizon behind any free-falling object marking the point where not even something moving towards them at the speed of light would ever be able to catch up to them as long as they keep accelerating at at least the same rate. The acceleration required to move away from a singularity increases as an inverse square of the distance between the falling object and the singularity because that's the rate that gravitational acceleration increases. If an event horizon were reachable it would mean the acceleration required to move away would suddenly jump to infinity. The event horizon would be the point where an object would be accelerated to the speed of light relative to all other objects if it were reachable, and the singularity would be the point where an object would be accelerated to infinite velocity, or zero velocity because it would have nowhere else to go, but singularities are a single point in space and time and so dont have any length at all in any dimension, including time. Singularities exist for a single instant but their volume in space and time appears to be larger from the perspective of more distant objects. They take up more space and time the further away you observe them from as a square of the distance, so their volume in space-time changes less dramatically over the same distance the further away they're being observed from because length contraction and time dilation increase more dramatically over the same distance as an observers distance decreases because the gravitational acceleration increases more dramatically over the same distance the closer the observer. The event horizon marks the point that an object would be accelerated to the speed of light (if it were reachable) relative to the black hole and any other object not heading towards the black hole so it also marks the closest point any object could have got towards the singularity at that time. Curved space-time can be used to describe acceleration due to energy just as easily as it can describe acceleration due to mass. The only differences are that gravity is much weaker than the other forces because of the energy/mass equivalence (E=mc^2) and mass curves inwards towards itself and energy curves outwards away from itself, but theres a limit to how much space-time can be curved. It has to be a smooth curve, so there cant be any sharp angles and it cant curve past ninety degrees because the dimensions are at right angles to each other and it would mean objects travelling back in time relative to other objects, which is what would happen if objects were able to reach the speed of light relative to other objects, either by accelerating using energy or by accelerating using mass by reaching the event horizon of a black hole. You can look at it as objects in flat space-time being pulled and pushed by mass and energy or you can look at it as objects following straight paths through space-time that's curved inwards by mass and outwards by energy, it makes absolutely no difference. When an object follows a curved path in two spatial dimensions it feels a force pushing it in the same two spatial dimensions in the opposite direction. The exact same thing happens when one of the dimensions is time. The object gets pushed backwards in the opposite direction in the two dimensions that the curved path is moving through, and so is being pushed backwards in time as well as space and therefore ages slower than an inertial object. You can see this for yourself by drawing a simple diagram. Just draw a circle with a vertical and horizontal line going all the way through so they cross in the centre. This isnt an ordinary diagram or I wouldnt be using it. Its using two dimensions at right angle to each other to represent exactly that. We can use one of those directions to represent time because theres no real difference between space and time. The vertical line represents time and the horizontal line represents one of the three spatial dimensions. We only need one spatial dimension because you can always draw a one dimensional straight line between any two objects. Objects always move through space-time with a combined velocity of the speed of light. Now use a pencil. An inertial objects path through space-time is represented by a vertical line because theyre travelling through time at the speed of light and not travelling through space at all from their own perspective. Other inertial objects paths through space-time are represented by lines running parallel to the vertical line. Their distance in space is represented by its distance across the horizontal spatial line and its relative velocity is represented by its distance across the vertical temporal line. Other inertial objects moving at different relative velocities are moving through time at the same speed as each other but are at different points along the vertical line and so are looking at each other at a skewed angle and see each other as shortened in time. They would see them as shortened in the spatial dimension that theyre moving through relative to them if they werent looking at a one dimensional straight line, but you cant have everything. You could use a sphere I suppose but let's not. This shortening of length from each others perspective represents Doppler shift. Infinitely far away in the vertical direction representing time on the circle would represent an object moving at the speed of light and one that's vertically running exactly parallel would represent an inertial object at rest relative to them. An accelerating objects path is represented by an angled line, like the hand of a clock. 12:00 is inertial and 3:00 is infinite acceleration because the dimensions are at right angles to each other. An accelerating objects path is a diagonal line moving across both dimensions of the circle and their speed through the horizontal spatial dimension means that their speed through the vertical temporal dimension is slower than an inertial object to keep their speed through space-time constant at the speed of light. An object that's increasing or decreasing its rate of acceleration creates a curved line. It takes more energy to increase its angle of acceleration the harder it accelerates approaching infinity as it approaches 3:00, which represents an event horizon as well as the speed of light because they're the same thing. This works just as well with gravity. If an object were able to cross an event horizon and move relatively faster than light then it would be moving down the circle instead of up which would mean it would be moving backwards in time relative to other objects. Acceleration and curvature are the same thing. They fact that most galaxies are red shifted and the further away they are the more red shifted they are shows that the universe is curved like the surface of the Earth, but with an extra dimension. Objects don't just vanish as they get further away, they get red shifted. They get more red shifted the further away they are because the observer is looking over a greater amount of curved space-time. If an object were to travel in a straight line in any direction then it would eventually end up back at its starting point in the same way as on Earth, and this also applies to time. This creates a kind of pseudo singularity if you look back in time which prevents information from reaching all the way around the universe to the same point in time and creating a paradox, it wouldnt look like a singularity if you were there. The universe is also a four dimensional sphere. If you turn a sphere inside out you still have a sphere, as the video posted earlier shows. The difference is that we are on the inside of a four dimensional that we can't reach the edge of and a black hole is inside out, which puts us on the outside of a four dimensional sphere that we can't reach the edge of. I think that's what you call checkmate. What I'm describing here is completely mathematical and it's the same in both situations because they're equivalent. It's just that I'm using words to describe it instead of numbers and symbols. It helps to keep things in context anyway. You should now be able to clearly see that the maths of acceleration is the same in both cases and it's so simple. Now please stop calling me dishonest and stop taking everything that I say out of context. I'm not the one being dishonest here. -1 Link to comment Share on other sites More sharing options...
mississippichem Posted December 4, 2012 Share Posted December 4, 2012 I guess you never learned the math or the LaTeX. That post is so full of misconceptions that I really don't know where to start. Have you ever even read a book on SR? I'm not insulting you here intentionally, just trying to point out that you clearly don't understand, even heuristically, the way in which SR is formulated. I imagine any further discussion will not be fruitful until you stop hand waving and pony up the math. 1 Link to comment Share on other sites More sharing options...
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