pzkpfw Posted April 21, 2017 Share Posted April 21, 2017 (edited) Very briefly (I don't want to distract from post #176; you should read the post by Janus, again and again). Sorry, but I ask the same question as I do not get a direct answer. No, you're not getting the answer you want. That is correct. I focus on absolutes Time dilation? Length contraction? Relativity of simultaneity? What absolutes? However, before discussing about relativity would you kindly answer the following: 1. Do you agree that with regards to B point of view, the relative speed between A and C is 1.2c? Yes or no The departure speed between A and C is 1.2 c according to B; but don't (again) confuse that with anyone actually moving at 1.2 c; B sees the distance between A and C grow at 1.2 c; but since B does not consider A as stationary, B does not consider C as moving at 1.2 c (and vice versa). 2. If so, do you agree that based on B point of view and t1 (time frame) the distance S (ac) between A to C had been increased by 1.2c * t1? Yes or no Sure. (But be sure that the speeds, time and distance are all as measured by B.) If the answers to the above are - yes, let's move to relativity: O.K. So based on relativity, the view from C isn't the same as the view from B. With regards to - Distance. And time! If we look from B point of view, the speed V (ac) between A to C is 1.2c and therefore the distance had been increased by S (ac) = 1.2c * t1. Close enough; see above. (Not "speed", "departure speed"). If we look from C point of view, the speed V (ac) between A to C is 0.88c and therefore the distance had been increased by S (ac) = 0.88c * t1. Actually not quite so easy. Who is measuring the time? Is t1 according to C the same as t1 according to B? As with your insistence that there are absolutes, you're expecting some "God's eye view" where there's one "real" truth. There's no such thing. Relativity is the opposite of "absolutivity". I had the impression that distance by definition must be absolute. All this time and you've never heard of length contraction? So, which one is the correct answer? I still don't understand how could it be that due to relativity both are correct. See post #172 by imatfaal. Day to day life doesn't prepare us to intuitively understand this stuff. People learning relativity often talk of "logic" or "common sense". Of course 1 + 1 = 2 ! Sorry, it's not that easy. Please read post #176 by Janus again. Edited April 21, 2017 by pzkpfw 1 Link to comment Share on other sites More sharing options...

David Levy Posted April 22, 2017 Author Share Posted April 22, 2017 (edited) We'll use a example that shows how, time dilation, length contraction and the relativity of simultaneity combine to the give the velocity addition result. A is an observer with a clock. B is an observer with his own clock passing A an at the rear of a rocket moving at 0.6c relative to A, by B's measurements, the rocket is 0.6 light seconds long and there is a clock at the front which is synced to his own. C is a projectile fired by B at 0.6c relative to himself towards the front of the rocket. He fires the projectile at the moment he is passing A. Thus at the start, A,B, and C are co-located. something like this. A B______| C The vertical line represents the front of the rocket. If we add the clocks it would look like this according to B at the start. A(0) B(0)____|(0) C Thanks It's fascinating example. It's quite difficult for me to trace all the time selections, so let me start with the following questions: By consider things according to B. Why B front side rocket has the same clock as it's back side although the rocket is 0.6 light sec long? Why you don't add (0) to C? So, why the starting point is not as follow: A(0) B(0)____|(0.6) C(0) With regards to factor: It is stated: "..and because A has a relative motion with respect to B, B measures A"s clock as running slow by a factor of 0.8." How did we calculate this factor? Why it is only valid between A to B? What about the factor between B to C? Edited April 22, 2017 by David Levy Link to comment Share on other sites More sharing options...

Janus Posted April 22, 2017 Share Posted April 22, 2017 (edited) Thanks It's fascinating example. It's quite difficult for me to trace all the time selections, so let me start with the following questions: By consider things according to B. Why B front side rocket has the same clock as it's back side although the rocket is 0.6 light sec long? Why you don't add (0) to C? So, why the starting point is not as follow: A(0) B(0)____|(0.6) C(0) Because, in this example, we are not as concerned with what time Anyone visually sees any other clock as reading,(which is information that, unless the clocks are co-located will be out of date), but what time the clock is reading "at the moment". I set as a starting condition that the clocks on the rocket were synchronized in the frame of the Rocket and thus that is what I showed. Besides, even if one to show what B visually sees, your diagram would be incorrect, When B reads 0, he would see the clock at the front of the rocket as reading 0.6 sec before zero. Thus given that the front of the ship is 0.6 light sec away, the light arriving from the front of the rocket left 0.6 sec ago, in which time the clock at the front has advanced 0.6 sec to read 0. I could have done all this according what A and B visually see, but this wouldn't change the outcome and would have just added an unneeded complication to the calculations. This is a common approach when dealing with Relativity examples, as it makes it clearer as to what is going on, which can get lost when you only concern yourself with what observers visually see. With regards to factor: It is stated: "..and because A has a relative motion with respect to B, B measures A"s clock as running slow by a factor of 0.8." How did we calculate this factor? By using the standard time dilation equation: [math] t = \frac{`t}{\sqrt{1 - \frac{v^2}{c^2}}}[/math] If v(the velocity of A with respect to B = 0.6c then t=`t/0.8 If `t is one second ticked off on A's clock, then is will take a time of t as measured by B, for A's clock to tick off one sec. Thus for B, it takes 1.25 sec for A's clock tick off one sec, or but another way, in one sec as measured by B, A's clock only ticks off 0.8 seconds. (I need to point out that the 0.48 sec I showed as ticking off for A as measured by B is actually incorrect, it should be 0.8 sec, The reason I didn't catch my goof until now, was because this time is not needed to work out the example) Why it is only valid between A to B? What about the factor between B to C? I never said it wasn't. In this example, I was showing that A would measure C as moving at ~0.88235c relative to himself, while B measures A as moving at 0.6c in one direction and C at 0.6 in the other. What time ticked off for C wasn't a consideration( And, as noted above, neither do you need to know how much time ticks off on A according to B) That's not to say that you can't work it into the example: According to B: Clock C, ticks at a rate of 0.8 and thus reads 0.8 sec when it reaches The front clock, which reads 1 sec. According to A: Clock C is moving at ~0.88235c Relative to A, and thus using the time dilation equation from above ticks at a rate of ~0.47059. Since A's clock ticked off 1.7 sec between the time C leaves its vicinity and reaches the front clock, 1.7*0.47059 = 0.8 sec tick off for clock C in the time it takes to reach the front clock according to A According to C: The rocket is moving at a relative speed of 0.6c, so length contraction causes C to measure the distance from B to the front clock as being 0.48 light sec, so when clock C reads zero, the front of the rocket is 0.48 light seconds away, and approaching at 0.6c. it takes 0.48/0.6 = 0.8 sec for the front clock and C to meet, so C's clock reads 0.8 sec when it reaches the front clock.( in addition, due to the relativity of simultaneity, the front clock reads 0.36 sec according to C when its own clock reads 0, and then ticks at a rate of 0.8 for 0.8 sec to accumulate an additional 0.64 sec, to read 1 sec when it and C meet up) You also could go on to work out why C would measure A as moving at ~0.88235c relative to itself, but that would be redundant (Just re-imagine that B is sitting in the middle of a 1.2 light sec long rocket with synced clocks at both ends. B is heading towards one clock and C the other. Now the same arguments applied to what velocity A measures for C can be used to show what velocity C measures for A.) Edited April 22, 2017 by Janus Link to comment Share on other sites More sharing options...

David Levy Posted April 23, 2017 Author Share Posted April 23, 2017 (edited) Perfect explanation. I'm not sure that I fully understand the time selections, but I'm sure by 100% that it is correct. It is clear to me that you are the expert. Actually all is about the same formula with different variation: By using the standard time dilation equation: We actually get at the end the same result as by using the v: Einstein's formula for composing velocities is (derived by a double application of the Lorenz transformation) [math]\frac{{v + w}}{{1 + \frac{{vw}}{{{c^2}}}}}[/math] where v and w are measured along the same line. So as the velocity of both A and C approaches light speed we have [math]\mathop {\lim }\limits_{v,w \to c} V = \frac{{v + w}}{{1 + \frac{{vw}}{{{c^2}}}}} = \frac{{c + c}}{{1 + \frac{{{c^2}}}{{{c^2}}}}} = c[/math] Check post #144. If A and B measure their relative speed as being 0.6c and B and C measure their relative speed as being 0.6c, then A and C will measure their relative speed as being: [math]\frac{0.6c+0.6c}{1+ \frac{0.6c(0.6c)}{c^2}} = 0.88235... c[/math] So, the speed is relative, the time is relative and even the distance is relative. It is clear that I need to get better understanding about this unique formula. Can you please highlight an article about it? Edited April 23, 2017 by David Levy 1 Link to comment Share on other sites More sharing options...

studiot Posted April 23, 2017 Share Posted April 23, 2017 (edited) So, the speed is relative, the time is relative and even the distance is relative. It is clear that I need to get better understanding about this unique formula. Can you please highlight an article about it? Are you looking for a derivation or some worked applications of the velocity transformations? And are you familiar with resolving velocities along the coordinate axes? Edited April 23, 2017 by studiot Link to comment Share on other sites More sharing options...

David Levy Posted April 23, 2017 Author Share Posted April 23, 2017 (edited) Please see the following article: https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations#Hyperbolic_rotation It is stated: Define this constant as δ(v)/v γ(v)= κ, where κ has the dimension of 1/v^{2}. Lorentz transformations If κ < 0, then we set {\displaystyle c=1/{\sqrt {-\kappa }}}... which becomes the invariant speed, the speed of light in vacuum. This yields κ = -1/c^{2} and thus we get special relativity with Lorentz transformation Only experiment can answer the question which of the two possibilities, κ = 0 or κ < 0, is realised in our world. The experiments measuring the speed of light, first performed by a Danish physicist Ole Rømer, show that it is finite, and the Michelson–Morley experiment showed that it is an absolute speed, and thus that κ < 0. Question: If K has the dimension of 1/v^{2 }than by definition it must be a positive no. κ < 0 means a negative no. Sorry - there is no way to convert 1/v^{2 }to a negative no. Not experiment and not any imagination!!! They have to prove how could it be that Dimension of 1/v^{2 }could represent a negative no. If K can't be a negative no by definition, than, how could it be that based on Lorentz transformations κ < 0? Without Lorentz transformations there is no room for the following formula: studiot, on 20 Apr 2017 - 7:09 PM, said: Einstein's formula for composing velocities is (derived by a double application of the Lorenz transformation) where v and w are measured along the same line. So as the velocity of both A and C approaches light speed we have So, could it be that the above formula which is based on Lorentz transformations is unrealistic? Edited April 23, 2017 by David Levy Link to comment Share on other sites More sharing options...

studiot Posted April 23, 2017 Share Posted April 23, 2017 In your post#180, you asked about detail for a formula I posted. In my post#181 I politely enquired what sort of detail. I fail to see the connection between your apparent reply, post#182 and my formula (actually it is Einstein's formula but still). Post#182 seems specifically to be asking about rotations in an article on several alternative derivations of the Lorenz transformation. My formula is not the Lorentz transformation, but is the result of applying it. Obviously those with lorentzian issues will not apply it at all. Rotations only work if the two system have a common origin, since rotation must have a centre. I would also note that hyperbolic functions involve the complex domain where k^{2} may indeed be negative. So do you want to argue with Lorenz or my formula? I only posted because I thought you wanted to have some things explained. Link to comment Share on other sites More sharing options...

David Levy Posted April 23, 2017 Author Share Posted April 23, 2017 Do appriciate your support. If I understand it correctly, the following formula is based on Lorentz transformations. However, Lorentz transformations is based on a negative K. As K must be positive, could it be that there is an error in this formula? -1 Link to comment Share on other sites More sharing options...

studiot Posted April 23, 2017 Share Posted April 23, 2017 If you wish to argue with the Lorenz formula I suggest you start your own thread, where I and others will be happy to help you. It is surely off topic of why nothing can go faster than light to argue against the mathematics upon which this assertion is based. I thought I made it quite clear that the formula for composition of velocities I showed was based on Lorenz. So if Lorenz is incorrect, then, yes, the formula would be incorrect. But consider this Thank you for the link to the Wiki article - I had not seen it before, there are so many. But look how many quite different and independent routes all lead to the same Lorenz formula. It is especially telling that some routes are purely experimentally derived whilst others are founded in the same foundations as pure maths and entirely consistent with them. Note in particular the section about the Lorenz group. Composition of a Lorenz transformation leads to another Lorenz transformation, which is why the operation can form mathematical group. I did comment on your hyperbolic question, but it will take some reading through to unravel which k you are talking about. Link to comment Share on other sites More sharing options...

Mordred Posted April 23, 2017 Share Posted April 23, 2017 (edited) Hyperbolic rotation with Lorentz calcs requires additional formulas as rotation is acceleration change. The lorentz transforms are designed for constant velocity. So naturally you will get the wrong answer. You must also account for Rapidity Edited April 23, 2017 by Mordred 1 Link to comment Share on other sites More sharing options...

studiot Posted April 23, 2017 Share Posted April 23, 2017 Thanks for that straight and simple answer, Mordred. +1 Link to comment Share on other sites More sharing options...

David Levy Posted April 24, 2017 Author Share Posted April 24, 2017 (edited) Hyperbolic rotation with Lorentz calcs requires additional formulas as rotation is acceleration change. The lorentz transforms are designed for constant velocity. So naturally you will get the wrong answer. You must also account for Rapidity Thanks However, your answer isn't fully clear to me. So let me ask the following questions: 1. Please advice if the following formula is for constant velocity (0.6c)? 2. Do you agree that this formula is based on Lorentz transformations? 3. Without Lorentz transformations, can we prove this formula based on Rapidity or some other transformations? 4. Is it correct that Lorentz transformations is based on negative K (κ < 0)? Edited April 24, 2017 by David Levy Link to comment Share on other sites More sharing options...

pzkpfw Posted April 24, 2017 Share Posted April 24, 2017 (edited) (edit: On second thought, best to step aside a while.) Edited April 24, 2017 by pzkpfw Link to comment Share on other sites More sharing options...

Mordred Posted April 24, 2017 Share Posted April 24, 2017 (edited) The answers to the above have been answered if you stop to think about it. The lorentz transformations are based on constant velocity. The minute you accelerate or change direction (acceleration) those formulas must take into account this change. The transforms with rapidity are as follows. [math]\begin{vmatrix}ct'\\x'\\\end{vmatrix}=\begin{vmatrix}cosh\phi&-sinh\phi\\-sinh\phi&cosh\phi\\\end{vmatrix}\begin{vmatrix}ct\\x\\\end{vmatrix}[/math] The formula you posted are Lorentz but only under constant velocity. The above matrix provide the changes due to rapidity. Some articles may refer to it as your hyperbolic rotation. No 4 is not based on negative k or curvature of any form. My advise to you is to step back and learn relativity through the proper stepping stones. 1) equivalence principle 2) relativity of simultaneity 3) The basic Lorentz transforms. 4) the relativistic velocity addition formulas. Master these before trying to tackle the more complex problems such as those on rapidity and hyperbolic rotation. At the moment from reading your responses this is too advanced for you till you master the above. Also get out of this habit of yours in assuming formulas are lacking or incorrect because you think you found some flaw. Relativity is a huge well tested subject. For example symmetry groups/tensors or matrixes. If you can't use the formulas as they are designed to be used. You will get the wrong answers. Not to mention drawing the wrong misconceptions and conclusions. You should focus on what everyone has been trying to teach you on how velocity addition works under relativity instead of assuming they are wrong. For example lets look at the acceleration for simplicity I will use a rocket that turns around and returns to its original start point with maximal turnaround. hyperbolic rotation (rocket acceleration due to direction change first define the four velocity. [latex]u^\mu[/latex] [latex]u^\mu=\frac{dx^\mu}{dt}=(c\frac{dt}{d\tau},\frac{dx}{d\tau},\frac{dy}{d\tau},\frac{dz}{d\tau})[/latex] this gives in the SR limit [latex]\eta u^\mu u^\nu=u^\mu u_\mu=-c^2[/latex] the four velocity has constant length. [latex]d/d\tau(u^\mu u_\mu)=0=2\dot{u}^\mu u_\mu[/latex] the acceleration four vector [latex]a^\mu=\dot{u}^\mu[/latex] [latex]\eta_{\mu\nu}a^\mu u^\nu=a^\mu u_\mu=0[/latex] so the acceleration and velocity four vectors are [latex]c \frac{dt}{d\tau}=u^0[/latex] [latex]\frac{dx^1}{d\tau}=u^1[/latex] [latex]\frac{du^0}{d\tau}=a^0[/latex] [latex]\frac{du^1}{d\tau}=a^1[/latex] both vectors has vanishing 2 and 3 components. using the equations above [latex]-(u^0)^2+(u^1)2=-c^2 ,,-u^0a^0+u^1a^1=0[/latex] in addition [latex]a^\mu a_\mu=-(a^0)^2=(a^1)^2=g^2[/latex] Recognize the pythagoras theory element here? the last equation defines constant acceleration g. with solutions [latex]a^0=\frac{g}{c}u^1,a^1=\frac{g}{c}u^0[/latex] from which [latex]\frac{da^0}{d\tau}=\frac{g}{c}\frac{du^1}{d\tau}=\frac{g}{c}a^1=\frac{g^2}{c^2}u^0[/latex] hence [latex]\frac{d^2 u^0}{d\tau^2}=\frac{g^2}{c^2}u^0[/latex] similarly [latex]\frac{d^2 u^1}{d\tau^2}=\frac{g^2}{c^2}u^1[/latex] so the solution to the last equation is [latex]u^1=Ae^{(gr/c)}=Be^{(gr/c)}[/latex] hence [latex]\frac{du^1}{d\tau}=\frac{g^2}{c^2}(Ae^{(gr/c)}-Be^{gr/c)})[/latex] with boundary conditions [latex]t=0,\tau=0,u^1=0,\frac{du^1}{d\tau}=a^1=g [/latex] we find A=_B=c/2 and [latex]u^1=c sinh(g\tau/c)[/latex] so [latex]a^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex] hence [latex]u^0=c\frac{dt}{d\tau}=c cosh(g\tau/c)[/latex] and finally [latex]x=\frac{c^2}{g}cosh(g\tau /c)[/latex], [latex] ct=\frac{c^2}{g}sinh(g\tau /c)[/latex] the space and time coordinates then fall onto the Hyperbola during rotation [latex]x^2-c^2\tau^2=\frac{c^4}{g^2}[/latex] So regardless of how fast or slow the acceleration is, the rotation itself causes changes to reference frames Now I ask you how much of that do you have a chance in understanding without first understanding the basic Lorentz formulas??? This is just one example of the complexity involved in the more advanced relativity problems Did you see me use a curvature constant k? no its not involved. Also instead of trying to learn this stuff through random google searches try a textbook. Here is a free one. http://www.lightandmatter.com/sr/ Edited April 24, 2017 by Mordred 2 Link to comment Share on other sites More sharing options...

David Levy Posted April 24, 2017 Author Share Posted April 24, 2017 (edited) Now I ask you how much of that do you have a chance in understanding without first understanding the basic Lorentz formulas??? This is just one example of the complexity involved in the more advanced relativity problems Did you see me use a curvature constant k? no its not involved. Also instead of trying to learn this stuff through random google searches try a textbook. Here is a free one. http://www.lightandmatter.com/sr/ Thanks for the explanation. It is clear that I need to improve my knowledge in relativity - and I will use your excellent advice. However, I have discovered an issue which seems to me as some sort of contradiction. Would you kindly offer a direct answers to my questions (I have only got an answer to no 4)? With regards to question n. 4. Are you sure that "No 4 is not based on negative k or curvature of any form." In the article it is stated: "If κ < 0, then we set {\displaystyle c=1/{\sqrt {-\kappa }}}... which becomes the invariant speed, the speed of light in vacuum. This yields κ = -1/c^{2} and thus we get special relativity with Lorentz transformation" So what can we understand from κ < 0 or κ = -1/c^{2?} Edited April 24, 2017 by David Levy Link to comment Share on other sites More sharing options...

pzkpfw Posted April 24, 2017 Share Posted April 24, 2017 ... Question: If K has the dimension of 1/v^{2 } than by definition it must be a positive no. ... I think you're missing what "dimension" means. It doesn't mean "value". Follow that link you quoted out of wikipedia. e.g. note that the speed of a car might be measured in km/h (or mph) regardless of whether it's going away from you or coming towards you. Link to comment Share on other sites More sharing options...

Mordred Posted April 24, 2017 Share Posted April 24, 2017 (edited) Yes I'm sure number 4 isn't based on curvature. Read your link carefully k is an inertial frame. [latex] k=(ct,x,yz) \acute{k}=(\acute{ct},\acute{x},\acute{y},\acute{z})[/latex] " consider a composition of transformations from the inertial frame K to inertial frame K′" As you can see from the quoted statement k is an assigned designation for your primed and unprimed inertial frames with the coordinates I supplied. This is the problem I am referring to. You have jumped ahead into group theory symmetry relations instead of focussing on the basic transformations. The wikipage covers literal chapters in a good GR texbook without really explaining each topic in any great detail. It is in essence telling you that it doesn't matter if your travelling toward or away from an emitter. All observers will measure the speed of light as c. (invariant) For the purpose of this thread this should be the only formulas you should focus on. https://en.m.wikipedia.org/wiki/Lorentz_transformation as well as the velocity addition formula https://en.m.wikipedia.org/wiki/Velocity-addition_formula Edited April 24, 2017 by Mordred Link to comment Share on other sites More sharing options...

studiot Posted April 24, 2017 Share Posted April 24, 2017 I would add to this a further comment to hopefully prevent a confusion often made. There is a difference between invariant and constant. Invariant means that the quantity concerned is the same in all coordinate or reference systems. So the invariant quantity,s, known as the interval and given by the pythagorean expression s^{2} = x^{2} + y^{2} +z^{2} - c^{2}t^{2} works out the same in all frames of reference. But in any given coordinate system there are many possible values of s, depending upon the points it connects. Another example is the line integral around a closed loop works out the same whether you use cartesian or spherical or cylindrical or any other coordinate system. In short an invariant is a system variable that can vary within one system but is the same when transformed to another system. Constant means it does not vary with time or position. So c is the same constant wherever and whenever you are at a point in the universe. In short a constant is the same throughout one system, but may appear different in a different system. So yes, c is both invariant and constant 1 Link to comment Share on other sites More sharing options...

koti Posted April 24, 2017 Share Posted April 24, 2017 Good to know studiot. Link to comment Share on other sites More sharing options...

swansont Posted April 24, 2017 Share Posted April 24, 2017 So, could it be that the above formula which is based on Lorentz transformations is unrealistic? ! Moderator Note This is in the relativity forum, i.e. mainstream physics. It is not the place to speculate; the default position is that well-tested physics is correct. You need to phrase questions in some other way. IOW, the issue is in your understanding, not the material. Link to comment Share on other sites More sharing options...

David Levy Posted April 25, 2017 Author Share Posted April 25, 2017 Yes I'm sure number 4 isn't based on curvature. Read your link carefully k is an inertial frame. [latex] k=(ct,x,yz) \acute{k}=(\acute{ct},\acute{x},\acute{y},\acute{z})[/latex] " consider a composition of transformations from the inertial frame K to inertial frame K′" As you can see from the quoted statement k is an assigned designation for your primed and unprimed inertial frames with the coordinates I supplied. This is the problem I am referring to. You have jumped ahead into group theory symmetry relations instead of focussing on the basic transformations. The wikipage covers literal chapters in a good GR texbook without really explaining each topic in any great detail. It is in essence telling you that it doesn't matter if your travelling toward or away from an emitter. All observers will measure the speed of light as c. (invariant) For the purpose of this thread this should be the only formulas you should focus on. https://en.m.wikipedia.org/wiki/Lorentz_transformation as well as the velocity addition formula https://en.m.wikipedia.org/wiki/Velocity-addition_formula Thanks I will read the articles as advised. However, with regards to the expansion; This formula had been developed over than 100 years ago. At that time people consider the universe as quite compact size. They didn't have any clue about the BBT or the Expansion. Therefore, I assume that they took it for granted that nothing can move faster than the speed of light. However, with all of our current knowledge, could it be that the formula is not updated for our current universe? Do we have an idea how the expansion impact this formula? Could it be that due to the expansion there are some galaxies at the far end of the Universe which are moving faster than the speed of light and therefore we can't see them? Link to comment Share on other sites More sharing options...

Mordred Posted April 25, 2017 Share Posted April 25, 2017 (edited) Expansion doesn't run counter to GR in fact the FLRW metric works beautifully with the Einstein field equations to such an extent one can choose to use either to equal accuracy. The recessive velocity exceeding c is a commonly misunderstood result of a particular formula involving extreme seperation distance to the observer. Hubbles law The greater the distance the greater the recessive velocity [latex] v_{recessive}=H_Od [/latex] This is not a true inertial based velocity but a calculation based upon mere seperation distance. Those galaxies beyond Hubble horizon with recessive velocity greater than c are not in actuality moving greater than c. It is a consequence of seperation distance not inertia which GR details. Hence their is no competition as those galaxies gain no inertia due to expansion. Now as far as GR goes our current formulas are continuosly improved when it is needed. We have taken GR to heights never imagined by Einstein but laypersons tend to get focussed on the basic equations in SR that they fail to see the later developments. I will post a 998 page texbook on GR just to illustrate this point. The article shows solutions to many of the paradoxes that plaqued GR in its earlier stages and details numerous different coordinate systems that were developed later than Einstein, Lorentz, Minkowskii etc. They supplied the stepping stones. Modern physics took those stepping stones and built a road. http://www.blau.itp.unibe.ch/newlecturesGR.pdf "Lecture Notes on General Relativity" Matthias Blau The reason you still see the formulas from Lorentz and Einstein is we proved there is no need to change them as they are incredibly accurate. If a modern day test showed them wrong. Believe me they will be replaced. However that isn't likely to happen as they are so incredibly accurate. edit I should add the later chapters cover the FLRW metric and shows some of the pertinent details of how it ties to GR and the Einstein field equations. Edited April 25, 2017 by Mordred Link to comment Share on other sites More sharing options...

David Levy Posted April 25, 2017 Author Share Posted April 25, 2017 (edited) Thanks I'm not sure that I fully understand your answer with regards to the expansion. So, please advice if the following is correct: 1. The expansion has no impact this formula. Hence, the formula is fully updated for our current universe. 2. Therefore, there is no galaxy in the whole universe which is moving away from us at a speed which is faster than the speed of light (even if it is located 40 or 100 Billion Light years away from us. Yes or no Edited April 25, 2017 by David Levy Link to comment Share on other sites More sharing options...

pzkpfw Posted April 25, 2017 Share Posted April 25, 2017 (edited) ... Yes or no It's a very sad "technique" to ask oversimplified questions and try to "trap" people into simple answers looking for a "gotcha". Current science is very clear. e.g. https://www.universetoday.com/122768/how-are-galaxies-moving-away-faster-than-light/ i.e. Nothing is moving through space faster than c, but, due to expansion of the Universe very distant objects are getting further away, faster than c. This answer has not changed since the first five times you asked it. Edited April 25, 2017 by pzkpfw 1 Link to comment Share on other sites More sharing options...

geordief Posted April 25, 2017 Share Posted April 25, 2017 (edited) . Current science is very clear. e.g. https://www.universetoday.com/122768/how-are-galaxies-moving-away-faster-than-light/ i.e. Nothing is moving through space faster than c, but, due to expansion of the Universe very distant objects are getting further away, faster than c. Is there any speed related to this expansion? I understand it is unrelated **to c but is it ,at least in theory possible to describe the rate at which this expansion is ongoing? Is there a theoretical upper limit to this expansion and does it have to be a globalized effect? Is it ,at least in theory possible that their might be localized instances of expansion ? **or is "unrelated" too strong and is there actually a formula that connects the two rates? Edited April 25, 2017 by geordief Link to comment Share on other sites More sharing options...

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