Mordred Posted June 12, 2023 Share Posted June 12, 2023 4 minutes ago, Logicandreason said: The onus is on me to prove that the Math in Einstein's paper is wrong, which I have done many time already. 100 percent correct. That math includes the c-v and c+v relation which you won't grasp how it's being used Link to comment Share on other sites More sharing options...

Logicandreason Posted June 12, 2023 Author Share Posted June 12, 2023 (edited) I proved that Swansont is mathematically wrong, and his claims are based on the approved narrative as taught by experts on Einstein's theories. So now what? Can anyone show that my criticism of Swansont has a fault? 5 minutes ago, Mordred said: 100 percent correct. That math includes the c-v and c+v relation which you won't grasp how it's being used Explain then how its being used. c+v or c-v can only have one meaning. Show me an alternative meaning. And then indicate exactly where in the confines of the Rod experiment Einstein clarifies which of the alternative meanings of c+v he has used. Edited June 12, 2023 by Logicandreason Link to comment Share on other sites More sharing options...

Mordred Posted June 12, 2023 Share Posted June 12, 2023 (edited) You want this again I already did so but whatever. lets go through it again. Lets simplify it however use just one emitter and send the signal bouncing between two mirrors mounted on a moving rod. Forget all about observers we or relativity of simultaneity. mirror A back of the train mirror B front of the train. on a static train entire length of train to truly simplify the math lets say its the time it would take light 1 second to travel on a non moving train. Static velocity=0. now lets say the train travels at a velocity 0.5 c set the train moving from A to C. now send a quick signal pulse from B to A. time for signal to arrive at A from B is 1 second on a non moving train. However the train is moving while the pulse is in flight. It is moving at 0.5 c so the pulse will hit mirror A at 0.5 seconds and not 1 second because mirror A towards the signal while the signal was moving toward mirror A. (C-V) in math speak. 1-0.5=0.5 or more accurately \((ct_{sl}-v)\)where the subscript " sl " denotes the interval length it would have taken on a stationary train CLASSICAL VECTOR ADDITION....Forgot to add we already established the train length being the equivalent of d=1 light second so I simplified Edited June 12, 2023 by Mordred Link to comment Share on other sites More sharing options...

swansont Posted June 12, 2023 Share Posted June 12, 2023 1 hour ago, Logicandreason said: Just as I suspected. You surely have no qualifications as a Mathematician, probably failed high school math by the look of your attempt at the logic of algebra. Just so you know, it’s the insults that will be why you get banned. 1 hour ago, Logicandreason said: The GLARING ERROR in your really silly algebra attempt, is here: d = ct, so r + vt = ct Its so bizarre that a grown up, in a healthy state of mind, would make the error that is obvious here. equating ct with r + vt is the error. ct is a variable, not a constant, because the elapsed time t, is a different value in the two instances that you mention in the one breath. t is a variable. c is a constant. I never claimed that ct was constant. v is a variable, too. 1 hour ago, Logicandreason said: So the correct reference to ct must show that in this set of equations that there has to be another time period, a t' for example. t is the time it takes for the light to get from one end of the rod to the other. There is only one value for this time. 1 hour ago, Logicandreason said: so if you insist on using ct rather than the more convenient v, c, d, and l, then you must have a ct and also a different distance ct'. Because first you said that ct was the distance covered by light in time t which was equal to the rod length. I did not say it was equal to the rod length. Quote But then you also say they its the distance of the rod length as well as the distance the rod moved. Yes, I did say that. Quote The TIME required for light to span the Rod is t, so if now the light has to span that rod PLUS some extra distance, then it will take longer, therefore we need a t'. So there are TWO time variables in you method not one. But you only try to use one, thus there is the reason your algebra is nonsense The rod is moving. In the time it has moved a distance r, the rod will have moved, so light will have not reached the end. I asked you to derive this expression, but you did not. All you’ve done is insult people. ! Moderator Note So, one chance: you derive the expression for the time it takes for the light to go from A to B, in terms of the known terms of r and v, as well as the speed of light. (i.e. the terms in Einstein’s equation.) for the co-propagating case, as I did. 35 minutes ago, Logicandreason said: The onus is on me to prove that the Math in Einstein's paper is wrong You haven’t done so until you give the correct expression in terms of the same parameters. Link to comment Share on other sites More sharing options...

Logicandreason Posted June 12, 2023 Author Share Posted June 12, 2023 1 hour ago, Mordred said: You want this again I already did so but whatever. lets go through it again. Lets simplify it however use just one emitter and send the signal bouncing between two mirrors mounted on a moving rod. Forget all about observers we or relativity of simultaneity. mirror A back of the train mirror B front of the train. on a static train entire length of train to truly simplify the math lets say its the time it would take light 1 second to travel on a non moving train. Static velocity=0. now lets say the train travels at a velocity 0.5 c set the train moving from A to C. now send a quick signal pulse from B to A. time for signal to arrive at A from B is 1 second on a non moving train. However the train is moving while the pulse is in flight. It is moving at 0.5 c so the pulse will hit mirror A at 0.5 seconds and not 1 second because mirror A towards the signal while the signal was moving toward mirror A. (C-V) in math speak. 1-0.5=0.5 or more accurately (ctsl−v) where the subscript " sl " denotes the interval length it would have taken on a stationary train CLASSICAL VECTOR ADDITION....Forgot to add we already established the train length being the equivalent of d=1 light second so I simplified Great. you just confirmed what classical physics says, that we have to use c+ v and c- v. But then you muck it up with (ctsl−v) because while c + or - v equates to a velocity, but ct or ctsl are distances. And you cant deduct a velocity from a distance. 1 hour ago, swansont said: Just so you know, it’s the insults that will be why you get banned. t is a variable. c is a constant. I never claimed that ct was constant. v is a variable, too. t is the time it takes for the light to get from one end of the rod to the other. There is only one value for this time. The rod is moving. In the time it has moved a distance r, the rod will have moved, so light will have not reached the end. I asked you to derive this expression, but you did not. All you’ve done is insult people. ! Moderator Note So, one chance: you derive the expression for the time it takes for the light to go from A to B, in terms of the known terms of r and v, as well as the speed of light. (i.e. the terms in Einstein’s equation.) for the co-propagating case, as I did. You haven’t done so until you give the correct expression in terms of the same parameters. I did not insult anyone. I merely suggested that is SEEMS to me that, because on your use of Algebra, you can't have attended a school. This is not an insult, its just me posturing a likely cause for some algebra that I considered is invalid. 1 hour ago, swansont said: t is the time it takes for the light to get from one end of the rod to the other. There is only one value for this time. Here is your error AGAIN. Consider this: " t is the time it takes for the light to get from one end of the rod to the other. There is only one value for this time." BUT there are TWO time periods in the Rod Experiment, if you insist in replacing (for no practical reason) the simple term r (for rod length), with the unnecessary term ct. So the first t is how long it took light to span the rod length. Unless you believe that the light can also span that same rod length plus another distance, in the same time? This is like saying light takes 1.5 seconds to go to the moon from earth, AND it also takes 1.5 second to go to the Sun. Because DISTANCE = ct. So if you insist on using ct to replace rod length as well as the distance the rod moved, then you MUST, MUST have two different variables called t and t' and they MUST have different values. Anyway, ct is not appearing in the equations Einstein gave which he claimed was pure classical Physics. Its obviously not classical physics and its not even a correct derivation. What's this about ONE CHANCE? I have given you many opportunities to show where my claims are wrong, and you have tried, but so far failed. How many chances do I have to allow you? But Ive already given the correct equation, fixing Einstein's error, and also proven that there will be no possible way that two observers can end up with different values for that Rod. here is the correct equation yet again, Let t = one second. R is length of Rod measured as 10 units Let c = 10 units per second velocity ct is the length of the rod, thus ct = 10 units of distance THUS we confirm that t is one second exactly vt is the distance travelled by the rod NEXT BUT vt is the distance travelled by the rod and ct is the Rod length. Now because we have TWO different distances in the equation, one is Rod length, and the other is distance the rod moved "ct can not be both the length of the rod as well as the distance the rod moved + the Rod length. Because there are two totally different distances here. ct doesn't cover both of them when t remains the same value. Then as you insist on using the constant "c" and a time period to measure distances, then we MUST have two different periods of Time to consider. so to reflect this, we MUST have t and t prime Light cannot take 1 second to span the Rod's length PLUS the distance the rod has moved in that same time period it took to span just the Rod alone. So we MUST have another variable for TIME. call it t' or t Prime or Z if you like. WHERE IS THE MISSING Time variable in your Algebra? I will give YOU ONE MORE CHANCE to prove that your algebra is sound. As t = distance travelled/ velocity THEN we can only derive Link to comment Share on other sites More sharing options...

Logicandreason Posted June 12, 2023 Author Share Posted June 12, 2023 My previous comment accidently included a bit that I forgot to remove at the very end... Please ignore the lines 1 hour ago, Logicandreason said: As t = distance travelled/ velocity THEN we can only derive Which appear as the last lines in the comment. It was part of my draft. Just ignore it. Seems I cant go back and edit the comment. Link to comment Share on other sites More sharing options...

swansont Posted June 12, 2023 Share Posted June 12, 2023 9 hours ago, Logicandreason said: Here is your error AGAIN. Consider this: " t is the time it takes for the light to get from one end of the rod to the other. There is only one value for this time." BUT there are TWO time periods in the Rod Experiment, What are the two times? We are considering the case where the light and rod are traveling in the same direction. 9 hours ago, Logicandreason said: if you insist in replacing (for no practical reason) the simple term r (for rod length), with the unnecessary term ct. I have not replaced r with ct. I replaced d with ct. d is the distance the light travels as it gets to the end of the rod. This should be blatantly obvious. It’s simple algebra. 9 hours ago, Logicandreason said: So the first t is how long it took light to span the rod length. That time does not appear in my derivation, as it’s unnecessary. If you want to use it in yours, go ahead. 9 hours ago, Logicandreason said: Unless you believe that the light can also span that same rod length plus another distance, in the same time? This is like saying light takes 1.5 seconds to go to the moon from earth, AND it also takes 1.5 second to go to the Sun. I never claimed this. You’re misreading the derivation. 9 hours ago, Logicandreason said: Because DISTANCE = ct. So if you insist on using ct to replace rod length as well as the distance the rod moved, then you MUST, MUST have two different variables called t and t' and they MUST have different values. I’m not doing this, so this is moot. ct is the distance the light travels. vt is the distance the rod moves. (v is the speed of the rod) Since the light has to go the length of the rod plus the distance the rod moves, it travels a total distance of r + vt But we know the total distance is ct Thus we know that ct = r + vt 9 hours ago, Logicandreason said: Anyway, ct is not appearing in the equations Einstein gave which he claimed was pure classical Physics. No. But it’s used in derivation of the equation. He rearranged the terms. But if you multiplied both sides by c-v, you would end up with a ct term, as anyone who can do basic algebra can see. Can you do basic algebra? 9 hours ago, Logicandreason said: What's this about ONE CHANCE? I have given you many opportunities to show where my claims are wrong, and you have tried, but so far failed. How many chances do I have to allow you? It’s about enforcing the rules of the forums, which I do, and you do not. 9 hours ago, Logicandreason said: But Ive already given the correct equation, fixing Einstein's error, and also proven that there will be no possible way that two observers can end up with different values for that Rod. here is the correct equation yet again, Let t = one second. R is length of Rod measured as 10 units Let c = 10 units per second velocity ct is the length of the rod, thus ct = 10 units of distance THUS we confirm that t is one second exactly vt is the distance travelled by the rod That’s not an equation. 9 hours ago, Logicandreason said: NEXT BUT vt is the distance travelled by the rod and ct is the Rod length. ct is not the rod length. You berated me for allegedly saying this (I didn’t) and now you are claiming it. Shouldn’t you be yelling about how you can’t think your way out of a paper bag, or something? 9 hours ago, Logicandreason said: Now because we have TWO different distances in the equation, one is Rod length, and the other is distance the rod moved "ct can not be both the length of the rod as well as the distance the rod moved + the Rod length. Because there are two totally different distances here. ct doesn't cover both of them when t remains the same value. Then as you insist on using the constant "c" and a time period to measure distances, then we MUST have two different periods of Time to consider. so to reflect this, we MUST have t and t prime Light cannot take 1 second to span the Rod's length PLUS the distance the rod has moved in that same time period it took to span just the Rod alone. So we MUST have another variable for TIME. call it t' or t Prime or Z if you like. WHERE IS THE MISSING Time variable in your Algebra? You are doing the problem in a different fashion. There’s often more than one way to do it. But what I asked for was the equation for the time (not numbers) using the same terms Einstein used, for the case of the light moving in the same direction as the rod. Here’s the thing, though: I don’t think you are conversant in algebra, and I think your reading comprehension is poor. I think you haven’t done what I ask because you don’t have the ability to do it. Prove me wrong. If you don’t derive the equation, I will lock the thread. 9 hours ago, Logicandreason said: I will give YOU ONE MORE CHANCE to prove that your algebra is sound. You don’t get to dictate this. 12 hours ago, Logicandreason said: The GLARING ERROR in your really silly algebra attempt, is here: d = ct, so r + vt = ct Its so bizarre that a grown up, in a healthy state of mind, would make the error that is obvious here. equating ct with r + vt is the error. ct is a variable, not a constant, because the elapsed time t, is a different value in the two instances that you mention in the one breath. There is only one instance: the light travels from A to B There is a different time for light traveling from B to A, but I have not derived that. You might notice that is a different equation Link to comment Share on other sites More sharing options...

Mordred Posted June 12, 2023 Share Posted June 12, 2023 10 hours ago, Logicandreason said: Great. you just confirmed what classical physics says, that we have to use c+ v and c- v. But then you muck it up with (ctsl−v) because while c + or - v equates to a velocity, but ct or ctsl are distances. And you cant deduct a velocity from a distance. You already know the velocity of Light. You enter it where the c is then multiply that by time to get the interval length. So where is the issue in using that value for a vector ? If you have a velocity and you are graphing that velocity as a vector you require the length of the vector. Knowing the distance travelled by an object in a specified time frame and direction of travel is precisely how you graph a velocity as a vector. Do I assume you have never drawn velocity as a vector ? Link to comment Share on other sites More sharing options...

Logicandreason Posted June 12, 2023 Author Share Posted June 12, 2023 23 minutes ago, swansont said: What are the two times? We are considering the case where the light and rod are traveling in the same direction. I have not replaced r with ct. I replaced d with ct. d is the distance the light travels as it gets to the end of the rod. This should be blatantly obvious. It’s simple algebra. That time does not appear in my derivation, as it’s unnecessary. If you want to use it in yours, go ahead. I never claimed this. You’re misreading the derivation. I’m not doing this, so this is moot. ct is the distance the light travels. vt is the distance the rod moves. (v is the speed of the rod) Since the light has to go the length of the rod plus the distance the rod moves, it travels a total distance of r + vt But we know the total distance is ct Thus we know that ct = r + vt No. But it’s used in derivation of the equation. He rearranged the terms. But if you multiplied both sides by c-v, you would end up with a ct term, as anyone who can do basic algebra can see. Can you do basic algebra? It’s about enforcing the rules of the forums, which I do, and you do not. That’s not an equation. ct is not the rod length. You berated me for allegedly saying this (I didn’t) and now you are claiming it. Shouldn’t you be yelling about how you can’t think your way out of a paper bag, or something? You are doing the problem in a different fashion. There’s often more than one way to do it. But what I asked for was the equation for the time (not numbers) using the same terms Einstein used, for the case of the light moving in the same direction as the rod. Here’s the thing, though: I don’t think you are conversant in algebra, and I think your reading comprehension is poor. I think you haven’t done what I ask because you don’t have the ability to do it. Prove me wrong. If you don’t derive the equation, I will lock the thread. You don’t get to dictate this. There is only one instance: the light travels from A to B There is a different time for light traveling from B to A, but I have not derived that. You might notice that is a different equation Its late here, I'm off to bed, so ill address this tomorrow. But briefly, for tonight's effort, as you insist in measuring distances using the speed of light and time rather than a simple ruler, then lets go all out and measure ALL distance using light and time, agreed? This will simplify things a lot. Here is the meaning of the necessary variables for distances when we use light speed and time to determine distance: There are 3 different distances. The "total distance" (rod + how far it moved) as you have identified is ct. The Length of the rod has to be ct^{A.} The distance the Rod moves has to be something else, ct^{B} So clearly ct is equal to the sum of ct^{A + }ct^{B} ct= ct^{A + }ct^{B} So as these three distances are not equal to each other, then there has to be 3 different time periods here. Seems to be pretty clear that we can not have only ONE variable for t in your equations, because the only difference between your equation and this one is that we have measured the rods distance moved by means of light velocity instead of its velocity of the rod. Of course we can measure anything moving or stationary using light and elapsed time. In my equation we need two measures to calculate the value of ct. But the equation I supplied can now be reduced to simply t = tA + tB ( Total length of rod and rods motion = rod length plus rods distance moved.) But if I had the time to substitute real values to replace your variables it would be revealed (proven by demonstration) that your equations are unbalanced. Hence there is an error in your equations. 26 minutes ago, Mordred said: You already know the velocity of Light. You enter it where the c is then multiply that by time to get the interval length. So where is the issue in using that value for a vector ? If you have a velocity and you are graphing that velocity as a vector you require the length of the vector. Knowing the distance travelled by an object in a specified time frame and direction of travel is precisely how you graph a velocity as a vector. Do I assume you have never drawn velocity as a vector ? Ok, ct can be the "interval length" meaning its the total distance of the rod and the distance that the rod moved. But its a distance. You can not deduct a measure of speed, from a measured distance. V is a speed, ct is a distance. You wrote the function (ct_{sl} - v) Show me how you can deduct 29 mph from 69 miles. Link to comment Share on other sites More sharing options...

Mordred Posted June 12, 2023 Share Posted June 12, 2023 Wow ok obviously you have never used vectors before. What determines the length of a vector with velocity as the vector ? Link to comment Share on other sites More sharing options...

swansont Posted June 12, 2023 Share Posted June 12, 2023 47 minutes ago, Logicandreason said: Its late here, I'm off to bed, so ill address this tomorrow. But briefly, for tonight's effort, as you insist in measuring distances using the speed of light and time rather than a simple ruler, then lets go all out and measure ALL distance using light and time, agreed? This will simplify things a lot. We don’t have a simple ruler, we have equations. 47 minutes ago, Logicandreason said: Here is the meaning of the necessary variables for distances when we use light speed and time to determine distance: There are 3 different distances. The "total distance" (rod + how far it moved) as you have identified is ct. The Length of the rod has to be ct^{A.} which is r, so this is unnecessary 47 minutes ago, Logicandreason said: The distance the Rod moves has to be something else, ct^{B} This is vt, so also unnecessary 47 minutes ago, Logicandreason said: So clearly ct is equal to the sum of ct^{A + }ct^{B} ct= ct^{A + }ct^{B} So as these three distances are not equal to each other, then there has to be 3 different time periods here. Seems to be pretty clear that we can not have only ONE variable for t in your equations, because the only difference between your equation and this one is that we have measured the rods distance moved by means of light velocity instead of its velocity of the rod. Of course we can measure anything moving or stationary using light and elapsed time. In my equation we need two measures to calculate the value of ct. The goal is to get to an equation that only has the total time, r, c and v in it, so you need to eliminate these other times from your derivation. 47 minutes ago, Logicandreason said: But the equation I supplied can now be reduced to simply t = tA + tB ( Total length of rod and rods motion = rod length plus rods distance moved.) But if I had the time to substitute real values to replace your variables it would be revealed (proven by demonstration) that your equations are unbalanced. Hence there is an error in your equations. I will wait for you to show this 47 minutes ago, Logicandreason said: Ok, ct can be the "interval length" meaning its the total distance of the rod and the distance that the rod moved. But its a distance. You can not deduct a measure of speed, from a measured distance. Good thing I didn’t do this, then 47 minutes ago, Logicandreason said: V is a speed, ct is a distance. You wrote the function (ct_{sl} - v) Show me how you can deduct 29 mph from 69 miles. Show me where I did this. Link to comment Share on other sites More sharing options...

Mordred Posted June 12, 2023 Share Posted June 12, 2023 3 hours ago, swansont said: Show me where I did this. Lol he keeps mixing us up. But he doesn't seem to understand how it applies to a vector so I'm seeing if he can get the length of a vector when he is given velocity (v) and the duration of the velocity. (@Logicandreason hint hint) reminder the sl subscript is just an identifier. I posted what the identifier means previous. Link to comment Share on other sites More sharing options...

Logicandreason Posted June 12, 2023 Author Share Posted June 12, 2023 6 hours ago, Mordred said: Lol he keeps mixing us up. But he doesn't seem to understand how it applies to a vector so I'm seeing if he can get the length of a vector when he is given velocity (v) and the duration of the velocity. (@Logicandreason hint hint) reminder the sl subscript is just an identifier. I posted what the identifier means previous. Swansont and Mordred. Mordred refereed to ct_{sl } Mordred wrote: Forget all about observers we or relativity of simultaneity. mirror A back of the train mirror B front of the train. on a static train entire length of train to truly simplify the math lets say its the time it would take light 1 second to travel on a non moving train. Static velocity=0. now lets say the train travels at a velocity 0.5 c set the train moving from A to C. now send a quick signal pulse from B to A. time for signal to arrive at A from B is 1 second on a non moving train. However the train is moving while the pulse is in flight. It is moving at 0.5 c so the pulse will hit mirror A at 0.5 seconds and not 1 second because mirror A towards the signal while the signal was moving toward mirror A. (C-V) in math speak. 1-0.5=0.5 or more accurately (ctsl−v)where the subscript " sl " denotes the interval length it would have taken on a stationary train So, Swansont, do you agree with Mordred or disagree? And Mordred, please clarify exactly what you mean by the term ctsl.. If we have ct (a distance), then if c=10 velocity units, and t = one time unit then ct = 10 units of distance. then what do we get for ct_{s}_{l.} ? Just to make things unambiguous we need to pin this down. And Mordred is the person who needs to explain how he can deduct a velocity from a distance. He wrote, "or more accurately (ctsl−v)" 10 hours ago, swansont said: We don’t have a simple ruler, we have equations. which is r, so this is unnecessary This is vt, so also unnecessary The goal is to get to an equation that only has the total time, r, c and v in it, so you need to eliminate these other times from your derivation. I will wait for you to show this Good thing I didn’t do this, then Show me where I did this. Mordred seems to he presenting a different argument than you, and I did get all these versions mixed up between you two. I will try to address Mordred first, and reply to you after we have clarified what exactly he is claiming. Link to comment Share on other sites More sharing options...

swansont Posted June 12, 2023 Share Posted June 12, 2023 10 minutes ago, Logicandreason said: So, Swansont, do you agree with Mordred or disagree? I haven’t been paying attention to Mordred’s line of discussion. I’m more interested in your contortions to deny the algebra I presented. If you can’t come up with the equation I requested -it’s pretty simple - you really have no business arguing that my derivation is wrong (and it isn’t; but you have to not fabricate statements that you attribute to me) Link to comment Share on other sites More sharing options...

Logicandreason Posted June 13, 2023 Author Share Posted June 13, 2023 10 hours ago, swansont said: The goal is to get to an equation that only has the total time, r, c and v in it, so you need to eliminate these other times from your derivation. To Mordred and Swansont, Agreed. I will try to isolate your arguments from Mordred's in my thoughts, the mix-up was unintentional. Anyway, I now have to admit that I am mathematically wrong with my claim that Einstein's equation t = rod length / c-v is incorrect. It is correct. I was wrong. The classical position is that there will be two different time periods one for the forward trip of light and a different time elapsed for the return trip of light is correct as Einstein says. But despite my error about the need to have the rod length AND the distance moved as the numerator, I still believe there is an issue with SR theory. Which I'm sure can be revealed. Its error of logic is caused by Einstein's insistence that just because the elapsed times for forward and return are different, he claims that somehow now the observers clocks must therefore now be out of sync. I believe this can't be shown, and Einstein never showed that this will be the result. He claims it is, but I cant see where he shows it. Can you point out how he comes to the conclusion that the clocks now must be showing different times on their faces? As far as I can see, all the clocks just measured different things, a forward elapsed time and a return elapsed time period. But measuring elapsed time periods doesn't equate the actual Time of day changing. I was trying to show this when I falsely believed I found that error in the equation. I was wrong on that. But none the less, I still would like to know how or where in the paper, Einstein explains how measuring durations of different events, can alter the Time on the faces of clocks. He makes that claim at the end of Section 2, immediately after that equation that I was having trouble with. But its not clear how or why he could make that conclusion. Everyone both moving any stationary had synchronized clocks, then they took two different recordings of elapsed times, so how does that cause the display time to get unsynchronized? He doesn't explain it at all. He just states it, then moves on, assuming that he has shown that there is a problem with classical physics. The observer on the station, is recording the forward trip, (c-v) and the person on the Train is measuring the return trip (c+v) BUT in fact, the person on the station can just as easily measure BOTH the forward and return trips as c-v and then c+ v, AND the person on the train can also measure the forward and return trips of the light. And according to the classical Math that you have just corrected me on, they all will get the same elapsed time periods, so they will both agree. Einstein thinks the will not agree As well as they clock hand will now move to unsynchronized locations of the faces. Show me in the text, where this is explained, as I can't see it. Which was what led me on the quest to understand the theory in the first place. Remember Einstein is still at this point explaining that there is a problem in classical Physics, by revealing that problem using classical physics. There is no such thing at this stage as Special Relativity. Please explain where he spells out how two different measures of two different speeds can mean that this is some problem. Its exactly what we expected, not a flaw. Where is the issue? P.S. As you are aware, I'm not good at math, but I do think, and know when something appears bizarre. . I'm self taught in Math so will make errors. But I will tell you that My son is now 18 and in college and is at the top of his school in Math and most other subjects, A+ scores. He's living with his mum, my Ex wife, who happens to be a resident Professor of Mathematics in a Large University in China, having also worked in a University in Australia and got her Ph.D. while studying in Edinburgh, Scotland. She has recently retired. So I do have access to advice from both my son and my ex, however as I prefer to learn on my own, try to understand a subject fully by applying my own thought processes, I have not bothered them. If I get really stuck, I will ask them. But you can see that I've now realised that my math was wrong. Took a while. However the issue remains that the verbal, textural conclusions of Einstein's about there being a synchronicity issue on those clocks is not revealed by his Math, nor explained in the text. If it is, show me where. Section one and Two are mostly text, with little math. It seems that he just made a statement that the clocks are unsynchronised but never explained why he came to that conclusion. It was all concluded very quickly. Immediately following the equation that I was labouring over, he simply states, "Observers moving with the moving rod would thus find that the two clocks were not synchronous, while observers in the stationary system would declare the clocks to be synchronous" But this is not the conclusion that a classical physics can come to. He would just say the observers measured two different events and got two different durations. And all the clocks tick on in perfect synchronization. The clocks were only recently synchronized, so measuring the duration of two different events makes them get unsynchronized? 1 Link to comment Share on other sites More sharing options...

swansont Posted June 13, 2023 Share Posted June 13, 2023 21 minutes ago, Logicandreason said: Its error of logic is caused by Einstein's insistence that just because the elapsed times for forward and return are different, he claims that somehow now the observers clocks must therefore now be out of sync. It’s more than this; it’s that each of the times in the moving frames are different than in the rest frame, and more importantly, the round-trip times will be different in the two frames. 21 minutes ago, Logicandreason said: I believe this can't be shown, and Einstein never showed that this will be the result. He claims it is, but I cant see where he shows it. Can you point out how he comes to the conclusion that the clocks now must be showing different times on their faces? Einstein showed that the time for a tick is different in the rest frame (t_{rest} = 2r/c if a tick happens once per round trip) than the ticks in the moving frame because the round-trip time will differ. He didn’t explicitly show this because it’s pretty obvious. The people one would expect to read his paper (physicists) would know this. Link to comment Share on other sites More sharing options...

Logicandreason Posted June 13, 2023 Author Share Posted June 13, 2023 9 minutes ago, swansont said: It’s more than this; it’s that each of the times in the moving frames are different than in the rest frame, and more importantly, the round-trip times will be different in the two frames. Einstein showed that the time for a tick is different in the rest frame (t_{rest} = 2r/c if a tick happens once per round trip) than the ticks in the moving frame because the round-trip time will differ. He didn’t explicitly show this because it’s pretty obvious. The people one would expect to read his paper (physicists) would know this. Referencing your first paragraph: Well of course they are going to be different, they are measuring different events. Not a problem for classical physics. About your second Paragraph. A tick is a second. So one second in the rest frame, is supposed to equal a different number of seconds in a moving frame? How so? Only if the distances covered by light are identical can you come to this conclusion. (Meaning that there is an error in classical Physics) In reality, the distances assessed by the stationary observer is not referring to the same distance that the moving Observer is interested in. Because you already said that as far as the Stationary observer is concerned, the rod is in motion, but as far as the moving Observer is concerned he believes that the Rod moved nowhere, so he never had that as a variable in his equations. He is only measuring how long it takes light to span the Rod length. Whereas the stationary observer is accounting for Rod length as well as Rod change in location, relative to the Light. One observers account of distance traversed by light, is different that the other observers account of how far light traversed, simply because they each were using different goal posts. Different distances covered by light at a constant speed must give different answers, and that's what Classical Physics repots, It is not an error. Link to comment Share on other sites More sharing options...

Mordred Posted June 13, 2023 Share Posted June 13, 2023 (edited) Ok it takes a person of strong character to admit being in error and further more wanting to better understand to avoid future errors. So +1 on that. Now we can move onto Relativity of Simultaneity. As Swansont mentioned above its not explicitly shown in the paper. However you can trust me on one detail, its a topic that went through decades of contested debate in numerous papers and methodologies. Time was until then considered absolute. However studies started to show that this wasn't accurate. It wasn't even Einstein that first noticed this. Poincare also made note of it prior to SR. Anyways without going into the history per se. (lol we have numerous forum members far more familiar with the history than I ). It might be best to examine what relativity of simultaneity entails. For this we do not need any eather based theory though Lorentz himself was a supporter of eather. Its an unnecessary complication. Are you in agreement to stick with the moving rods for examination to ensure you understand relativity of simultaneity ? (keep in mind this experiment itself won't show time dilation ) as both observers are moving at the same velocity. Its used in the paper more of a reminder of what classical physics (Galilean relativity would show) Edited June 13, 2023 by Mordred Link to comment Share on other sites More sharing options...

Logicandreason Posted June 13, 2023 Author Share Posted June 13, 2023 7 minutes ago, Mordred said: Ok it takes a person of strong character to admit being in error and further more wanting to better understand to avoid future errors. So +1 on that. Now we can move onto Relativity of Simultaneity. As Swansont mentioned above its not explicitly shown in the paper. However you can trust me on one detail, its a topic that went through decades of contested debate in numerous papers and methodologies. Time was until then considered absolute. However studies started to show that this wasn't accurate. It wasn't even Einstein that first noticed this. Poincare also made note of it prior to SR. Anyways without going into the history per se. (lol we have numerous forum members far more familiar with the history than I ). It might be best to examine what relativity of simultaneity entails. For this we do not need any eather based theory though Lorentz himself was a supporter of eather. Its an unnecessary complication. Are you in agreement to stick with the moving rods for examination to ensure you understand relativity of simultaneity ? I am in agreement, However I will accept that I am indeed of strong character, but that comes with a price, I'm also very stubborn, so I don't give in easily. So to win me over, you will need patience and a really really great logical argument. I doubt that a math equation solution that is devoid of a logical explanation will be sufficient to win the day. As I've recently said, I STILL believe that there is an issue in SR somewhere, but it was not in that Equation. I was barking up the wrong tree, but I see more trees out there. Link to comment Share on other sites More sharing options...

Mordred Posted June 13, 2023 Share Posted June 13, 2023 (edited) Oh I will try to keep the math as light as possible, Hence I'm also going to keep this in slow stages to make sure each stage is understood. lets start with a stationary rod. We both know light takes time to travel from A to B. So lets set a clock at each end. Now there is only one distance from each clock where an observer will read the same time on both clocks. Can you tell me where it is ? (classical physics only needed here). Yes this stage seems redundant but its important to understand relativity of simultaneity itself. The obvious answer is any point where the distance between the observer and both clocks are equal. It could be at the midpoint between each clock L/2 or it can be any distance where the length between observer and clock A =the length between observer and clock B. We just finished the related equations for when the train is in motion. That details how the trains motion will alter the previous relation. Now we have to account for the trains motion as well as c. Thankfully we just went through that. What this shows is that there are coordinates where an observer will agree on the time where each event occurs at the same time. This is extremely useful. So I'm going to stop here and make sure your good up this point. Edited June 13, 2023 by Mordred Link to comment Share on other sites More sharing options...

Logicandreason Posted June 13, 2023 Author Share Posted June 13, 2023 18 minutes ago, Mordred said: Oh I will try to keep the math as light as possible, Hence I'm also going to keep this in slow stages to make sure each stage is understood. lets start with a stationary rod. We both know light takes time to travel from A to B. So lets set a clock at each end. Now there is only one distance from each clock where an observer will read the same time on both clocks. Can you tell me where it is ? (classical physics only needed here). Yes this stage seems redundant but its important to understand relativity of simultaneity itself. Yep, the middle of the Rod. Now I have a clarification question: Is that radiated light from each clock face, that reaches the observer, relative to the clock faces? Seems not, as Einstein and classical Physics agree, "light speed is not influenced by the motion (or Not) of the source." 1 Link to comment Share on other sites More sharing options...

Mordred Posted June 13, 2023 Share Posted June 13, 2023 (edited) 22 minutes ago, Logicandreason said: Yep, the middle of the Rod. Now I have a clarification question: Is that radiated light from each clock face, that reaches the observer, relative to the clock faces? Seems not, as Einstein and classical Physics agree, "light speed is not influenced by the motion (or Not) of the source." We cross talked while I was editing your correct in the section you quoted the rest of my above post contains an edit where I added details on the moving train. When the train moves Right down the middle is no longer true and we must now account for the train motion. All events (a term that includes observer-emitter-coordinates-signals ie light etc) are always relative to the observer. Yes an observer will measure the velocity of light at c regardless of the velocity of the emitter. This is an invariant quantity. It never varies regardless of observer Nor how fast the emitter is travelling. If your good up to hear I suggest to better understand the math of SR we cover principle of equivalence next. If you have further questions on this stage ask away. As we can see though there is only limited positions where the two clock are simultaneous and where those point lie will vary. Edited June 13, 2023 by Mordred Link to comment Share on other sites More sharing options...

Logicandreason Posted June 13, 2023 Author Share Posted June 13, 2023 3 minutes ago, Mordred said: We cross talked while I was editing your correct in the section you quoted the rest of my above post contains an edit where I added details on the moving train. When the train moves Right down the middle is no longer true and we must now account for the train motion. All events (a term that includes observer-emitter-coordinates-signals ie light etc) are always relative to the observer. Yes an observer will measure the velocity of light at c regardless of the velocity of the emitter. This is an invariant quantity. It never varies regardless of observer Nor how fast the emitter is travelling. If your good up to hear I suggest to better understand the math of SR we cover principle of equivalence next. Nope, you lost me with this claim: Quote: "Yes an observer will measure the velocity of light at c regardless of the velocity of the emitter. This is an invariant quantity. It never varies regardless of observer Nor how fast the emitter is travelling." So I agree with this statement up to where you said, " It never varies regardless of observer". But I cant agree with that part. I also agree with the remainder of the statement, "Nor how fast the emitter is travelling." Can you clarify, Are you referring to the fact that light' condition of motion (specifically its velocity) is unvarying, or are you meaning that any MEASURE of that velocity is going to be the same, "irrespective of the observer"? Because a second ago you said that QUOTE: "All events.... ie light.... are (is) always relative to the observer." So I'm confused, is measurement of of Light speed "relative to the observer" as in your original statement, or is it "not relative to the observer" as in your second statement.? Link to comment Share on other sites More sharing options...

iNow Posted June 13, 2023 Share Posted June 13, 2023 Light speed is invariant, constant relative to EVERYTHING. It is absolute, regardless of the characteristics of the observer. Rejecting that is like rejecting the sunrise in the morning. It’s true whether or not you accept it. Link to comment Share on other sites More sharing options...

Mordred Posted June 13, 2023 Share Posted June 13, 2023 An observer can himself be at near c and still measure the velocity of a light signal as being c even if the ship with the emitter is also going near c. Regardless of the velocity of emitter or observer any measurement of the velocity will equal c. (this is the part where the length contraction and time dilation kicks in of the Lorentz transformations. ) This is also the point where the deviations occur from the classical physics. LOL is typically also the hardest to get people to accept. Hence all the research and studies, this required a huge burden of proof. So far its tested as true to extremely high precision Link to comment Share on other sites More sharing options...

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