JohnLesser Posted May 3, 2017 Author Share Posted May 3, 2017 Impossible if two people get the same measurement. Then both measurements are the same. But 2 people , get different measurements, already proven . Link to comment Share on other sites More sharing options...
Strange Posted May 3, 2017 Share Posted May 3, 2017 But the background does not contract, only the virtual coordinates? What? If you have time dilation then you also have length contraction. Because the two effects always go hand in hand. Because the speed of light is defined to be constant. Therefore it is impossible to have different observers measure different values for the speed of light. Link to comment Share on other sites More sharing options...
JohnLesser Posted May 3, 2017 Author Share Posted May 3, 2017 You might try listening as I have a master degree in physics. Some of those relativity tests I have personally performed in a lab. Have I once said your test are wrong? NO I have said things are misinterpreted. Link to comment Share on other sites More sharing options...
Strange Posted May 3, 2017 Share Posted May 3, 2017 So they measure the same speed but think it is different. Got it. yes exactly That doesn't make any sense. If they measure the same speed, then it is the same speed. How would you feel if you were driving down the road at 40 MPH and the police used a radar gun to measure your speed as 40 MPH but then prosecute you for driving at 120 MPH (because they think the speed is different from what they measure). The whole point about measurements is that they are supposed to be objective. If you are going to say that, even if they measure the same speed it is not actually the same, then you are no longer doing science. Link to comment Share on other sites More sharing options...
JohnLesser Posted May 3, 2017 Author Share Posted May 3, 2017 What? If you have time dilation then you also have length contraction. Because the two effects always go hand in hand. Because the speed of light is defined to be constant. Therefore it is impossible to have different observers measure different values for the speed of light. What are you contracting? the length of space? the length of light? If you contract the length then you can't have a time dilation anyway, you are just saying it travels a shorter length in less time than it would take travel a longer length. It is impossible to measure the same value of the speed of light if using two different ''rates'' of time . That doesn't make any sense. If they measure the same speed, then it is the same speed. How would you feel if you were driving down the road at 40 MPH and the police used a radar gun to measure your speed as 40 MPH but then prosecute you for driving at 120 MPH (because they think the speed is different from what they measure). The whole point about measurements is that they are supposed to be objective. If you are going to say that, even if they measure the same speed it is not actually the same, then you are no longer doing science. Sorry it was ambiguous of me, they both can conclude that the speed of light is constant, but neither can agree on a speed. example: Alan times light to travel dx takes 1 second John times light to travel dx takes 2 seconds Link to comment Share on other sites More sharing options...
Strange Posted May 3, 2017 Share Posted May 3, 2017 What are you contracting? The distance travelled (dx, in your notation). Link to comment Share on other sites More sharing options...
Mordred Posted May 3, 2017 Share Posted May 3, 2017 What are you contracting? the length of space? the length of light? If you contract the length then you can't have a time dilation anyway, you are just saying it travels a shorter length in less time than it would take travel a longer length. It is impossible to measure the same value of the speed of light if using two different ''rates'' of time . https://en.m.wikipedia.org/wiki/Length_contraction https://en.m.wikipedia.org/wiki/Time_dilation Both these occur at the same time. Study relativity before trying to say its wrong. Your just making yourself look the fool by ignoring this detail. Link to comment Share on other sites More sharing options...
JohnLesser Posted May 3, 2017 Author Share Posted May 3, 2017 The distance travelled (dx, in your notation). So you are contracting a length of space , therefore for that to be objective reality you would have to prove space was made of something to contract. There is no evidence that suggests space is made of anything. Contracting of the ''background'' doe's not happen. https://en.m.wikipedia.org/wiki/Length_contraction https://en.m.wikipedia.org/wiki/Time_dilation Both these occur at the same time. Study relativity before trying to say its wrong. Your just making yourself look the fool by ignoring this detail. Again ,what are you contracting? you are contracting subjective virtual lines. Space itself has no solidity to contract. Link to comment Share on other sites More sharing options...
Strange Posted May 3, 2017 Share Posted May 3, 2017 (edited) So you are contracting a length of space , therefore for that to be objective reality you would have to prove space was made of something to contract. There is no evidence that suggests space is made of anything. Contracting of the ''background'' doe's not happen. So you are saying that time is made of something? Why do you accept time dilation and reject length contraction when they are the same thing? Edited May 3, 2017 by Strange Link to comment Share on other sites More sharing options...
JohnLesser Posted May 3, 2017 Author Share Posted May 3, 2017 How would you feel if you were driving down the road at 40 MPH Alan measured 40 mph d/t John measures 80 mph d/t Sorry officer, you are relatively wrong. Link to comment Share on other sites More sharing options...
Strange Posted May 3, 2017 Share Posted May 3, 2017 Alan measured 40 mph d/t John measures 80 mph d/t Sorry officer, you are relatively wrong. OK. So the analogy went completely over your head. Link to comment Share on other sites More sharing options...
Mordred Posted May 3, 2017 Share Posted May 3, 2017 (edited) Here is the full transforms. Lorentz transformation. First two postulates. 1) the results of movement in different frames must be identical 2) light travels by a constant speed c in a vacuum in all frames. Consider 2 linear axes x (moving with constant velocity and [latex]\acute{x}[/latex] (at rest) with x moving in constant velocity v in the positive [latex]\acute{x}[/latex] direction. Time increments measured as a coordinate as dt and [latex]d\acute{t}[/latex] using two identical clocks. Neither [latex]dt,d\acute{t}[/latex] or [latex]dx,d\acute{x}[/latex] are invariant. They do not obey postulate 1. A linear transformation between primed and unprimed coordinates above in space time ds between two events is ( this below doesnt have curvature as SR assumes Euclidean) [latex]ds^2=c^2t^2=c^2dt-dx^2=c^2\acute{t}^2-d\acute{x}^2[/latex] Invoking speed of light postulate 2. [latex]d\acute{x}=\gamma(dx-vdt), cd\acute{t}=\gamma cdt-\frac{dx}{c}[/latex] Where[latex] \gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}[/latex] Time dilation dt=proper time ds=line element since[latex] d\acute{t}^2=dt^2[/latex] is invariant. an observer at rest records consecutive clock ticks seperated by space time interval dt=d\acute{t} she receives clock ticks from the x direction separated by the time interval dt and the space interval dx=vdt. [latex]dt=d\acute{t}^2=\sqrt{dt^2-\frac{dx^2}{c^2}}=\sqrt{1-(\frac{v}{c})^2}dt[/latex] so the two inertial coordinate systems are related by the lorentz transformation [latex]dt=\frac{d\acute{t}}{\sqrt{1-(\frac{v}{c})^2}}=\gamma d\acute{t}[/latex] Here is relativity of simultaneaty coordinate transformation in Lorentz. [latex]\acute{t}=\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}[/latex] [latex]\acute{x}=\frac{x-vt}{\sqrt{1-v^2/c^2}}[/latex] [latex]\acute{y}=y[/latex] [latex]\acute{z}=z[/latex] Both these occur simultaneous. [latex]\acute{t}=\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}[/latex] [latex]\acute{x}=\frac{x-vt}{\sqrt{1-v^2/c^2}}[/latex] So you coordinate length contracts as well as your time coordinate changes. Edited May 3, 2017 by Mordred Link to comment Share on other sites More sharing options...
JohnLesser Posted May 3, 2017 Author Share Posted May 3, 2017 So you are saying that time is made of something? Why do you accept time dilation and reject length contraction when they are the same thing? They are not the same thing, I neither accept the present interpretation of time dilation. Why do people feel the need to try to force me to accept ''your way or the high way? '' Here is the full transforms. Lorentz transformation. First two postulates. 1) the results of movement in different frames must be identical 2) light travels by a constant speed c in a vacuum in all frames. Consider 2 linear axes x (moving with constant velocity and \acute{x} (at rest) with x moving in constant velocity v in the positive \acute{x} direction. Time increments measured as a coordinate as dt and d\acute{t} using two identical clocks. Neither dt,d\acute{t} or dx,d\acute{x} are invariant. They do not obey postulate 1. A linear transformation between primed and unprimed coordinates above in space time ds between two events is ( this below doesnt have curvature as SR assumes Euclidean) [latex]ds^2=c^2t^2=c^2dt-dx^2=c^2\acute{t}^2-d\acute{x}^2[/latex] Invoking speed of light postulate 2. [latex]d\acute{x}=\gamma(dx-vdt), cd\acute{t}=\gamma cdt-\frac{dx}{c}[/latex] Where[latex] \gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}[/latex] Time dilation dt=proper time ds=line element since[latex] d\acute{t}^2=dt^2[/latex] is invariant. an observer at rest records consecutive clock ticks seperated by space time interval dt=d\acute{t} she receives clock ticks from the x direction separated by the time interval dt and the space interval dx=vdt. [latex]dt=d\acute{t}^2=\sqrt{dt^2-\frac{dx^2}{c^2}}=\sqrt{1-(\frac{v}{c})^2}dt[/latex] so the two inertial coordinate systems are related by the lorentz transformation [latex]dt=\frac{d\acute{t}}{\sqrt{1-(\frac{v}{c})^2}}=\gamma d\acute{t}[/latex] Here is relativity of simultaneaty coordinate transformation in Lorentz. [latex]\acute{t}=\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}[/latex] [latex]\acute{x}=\frac{x-vt}{\sqrt{1-v^2/c^2}}[/latex] [latex]\acute{y}=y[/latex] [latex]\acute{z}=z[/latex] Both these occur simultaneous. [latex]\acute{t}=\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}[/latex] [latex]\acute{x}=\frac{x-vt}{\sqrt{1-v^2/c^2}}[/latex] So you coordinate length contracts as well as your time coordinate changes. I appreciate you know your stuff, so the coordinates contract, not the space? Link to comment Share on other sites More sharing options...
DrP Posted May 3, 2017 Share Posted May 3, 2017 but the co-ordinates define the space right? Link to comment Share on other sites More sharing options...
Delta1212 Posted May 3, 2017 Share Posted May 3, 2017 They are not the same thing, I neither accept the present interpretation of time dilation. Why do people feel the need to try to force me to accept ''your way or the high way? '' I appreciate you know your stuff, so the coordinates contract, not the space? Because it's a science forum and your way isn't science? 1 Link to comment Share on other sites More sharing options...
Mordred Posted May 3, 2017 Share Posted May 3, 2017 (edited) In the formulas its your coordinate length. However we have evidence this applies as observations show length contraction in redshift and temperature. Time dilation cannot account for the energy level changes involved in the above two. A change in density via the length contraction however does. If you think about light bending around a planet. Time dilation can't cause this without length contraction as well. Edited May 3, 2017 by Mordred 1 Link to comment Share on other sites More sharing options...
JohnLesser Posted May 3, 2017 Author Share Posted May 3, 2017 but the co-ordinates define the space right? No, coordinates define virtual points and vectors, the lines do not actually exist. Space itself, the background to the 2d interpretation , does not contract. like the background behind these words is relative ''stationary''. Link to comment Share on other sites More sharing options...
Delta1212 Posted May 3, 2017 Share Posted May 3, 2017 No, coordinates define virtual points and vectors, the lines do not actually exist. Space itself, the background to the 2d interpretation , does not contract. like the background behind these words is relative ''stationary''. What evidence is there for such a static background? Link to comment Share on other sites More sharing options...
JohnLesser Posted May 3, 2017 Author Share Posted May 3, 2017 In the formulas its your coordinate length. However we have evidence this applies as observations show length contraction in redshift and temperature. Time dilation cannot account for the energy level changes involved in the above two. A change in density via the length contraction however does. I am not saying what you have is wrong, in the conclusion of my theory I fix it all, a simple fix in changing the definition of time dilation to a timing dilation, fixes the problem sort of. Timing dilation then can be assumed to be correct , but this still does not solve the speed of light problem. Link to comment Share on other sites More sharing options...
Mordred Posted May 3, 2017 Share Posted May 3, 2017 (edited) Incorrect I have provided 3 phenomena where space (real space ) is being affected. 1)temperature 2) light path curvature 3) redshift Show that you can solve those three observer dependant measurements with just time dilation. I can honestly tell you, it won't work you need length contraction as well Edited May 3, 2017 by Mordred 2 Link to comment Share on other sites More sharing options...
JohnLesser Posted May 3, 2017 Author Share Posted May 3, 2017 What evidence is there for such a static background? The inflating balloon in a box, the space passes through the balloons rubber solid surface and the surface passes through the space, the air in the box is displaced but the space is not. The space in the box ends up inside the balloon, not by magic , the balloons skin offers no permeability or permitivity resistance to space. It is as if space has 0 ''viscosity'' . Incorrect I have provided 3 phenomena where space (real space ) is being affected. 1)temperature 2) light path curvature 3) redshift Show that you can solve those three observer dependant measurements with just time dilation. I can honestly tell you, it won't work you need length contraction as well You are discussing apples when I am discussing pairs. Please try to think about what I am saying instead of thinking about what you already know. Link to comment Share on other sites More sharing options...
Mordred Posted May 3, 2017 Share Posted May 3, 2017 (edited) No I am discussing observational proof that you are incorrect. Simply ignoring evidence because it doesn't suit your theory won't cut it. Solve those three problems without length contraction and using just time dilation and you may have something. As it is you have your declarations with no evidence to back up your assertions. Edited May 3, 2017 by Mordred 1 Link to comment Share on other sites More sharing options...
JohnLesser Posted May 3, 2017 Author Share Posted May 3, 2017 The inflating balloon in a box, the space passes through the balloons rubber solid surface and the surface passes through the space, the air in the box is displaced but the space is not. The space in the box ends up inside the balloon, not by magic , the balloons skin offers no permeability or permitivity resistance to space. It is as if space has 0 ''viscosity'' . You are discussing apples when I am discussing pairs. Please try to think about what I am saying instead of thinking about what you already know. I have already said you are correct by your interpretation of time, however the notion is your interpretation is incorrect. You quite clearly are not considering why it is incorrect which I have explained several times already. Do you agree or disagree in the following statement? If two observers measure two different frequencies of time and use these independent measurements to measure the speed of a car travelling distance x, t1/dx will be different to t2/dx Yes or no Link to comment Share on other sites More sharing options...
Strange Posted May 3, 2017 Share Posted May 3, 2017 If two observers measure two different frequencies of time and use these independent measurements to measure the speed of a car travelling distance x, t1/dx will be different to t2/dx Yes or no Not if dx changes as well. (Which, if there is time dilation, it will.) This wilful ignorance and refusal to engage with the evidence is probably the reason your previous posts got shut down. 1 Link to comment Share on other sites More sharing options...
Mordred Posted May 3, 2017 Share Posted May 3, 2017 (edited) your dx is changing what part of that don't you understand. You have dx and [latex] d\acute {x}[/latex] Look over the transforms I posted and read the two wiki links. This is all contained in every textbook published on relativity. Edited May 3, 2017 by Mordred 1 Link to comment Share on other sites More sharing options...
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