jv1 Posted May 8 Posted May 8 Observing pedestrian crossing the street - thought experiment. The experiment is split in three systems: System 1 There is two pedestrians at position A. Pedestrian 1 is moving at speed of v1=1m/s and goes to position B and return backwards to position A. The distance d1=2m time t1=2sec speed v1 is constant. This is all observed by non moving pedestrian 2. System 2 The pedestrian 1 is not moving. The road is moving at speed of v2=1 m/s The pedestrian 1 moves from point A to point E. the distance d2=2 m Time t2=2s The speed v2 =1 m/s is constant This is observed by pedestrian 2. The third system is system 1 and system 2 taking place simultaneously. The trajectory of movement of pedestrian 1 has zig zag pattern - this is observed by pedestrian 2. The total length dt=2.84m The total time travelled is tt=2.84sec the total observed speed is vt=dt/tt=1 m/s The speed of pedestrian 1 is 1 m/s - constant - in all three systems . The observed distance was increased because system 1 and system 2 are applied simultaneously. It will take longer time for pedestrians 1 ,traveling at constant speed to move on Bigger distance of observed trajectory 2.84 m. There is difference in observed time and length but the observed speed is constant. Opinions? -1
Halc Posted May 9 Posted May 9 This whole topic seems to be related to Galilean relativity and Newtonian physics. There is no relativity theory going on in any of this. I am assuming that points A through F are painted points marked on the sides of the path/road. If you mean otherwise, it isn't indicated. 11 hours ago, jv1 said: System 2 The pedestrian 1 is not moving. The road is moving at speed of v2=1 m/s The pedestrian 1 moves from point A to point E. ... This is observed by pedestrian 2. How does ped1 move from A to E if he's not moving? This is a contradiction. Given the pic, the road is moving to the left, and spot E painted on the road moves left to the pedestrian over 2 seconds. Anything other than that and the ped is not stationary. I take it ped2 is always with ped1 in this scenario? If not, then ped1 would be moving, and you say he isn't, as observed by ped2. 11 hours ago, jv1 said: The third system The trajectory of movement of pedestrian 1 has zig zag pattern - this is observed by pedestrian 2. The total length dt=2.84m The total time travelled is tt=2.84sec the total observed speed is vt=dt/tt=1 m/s So ped2 stays at point A, the road is stationary, and ped1 takes this diagonal path to D, then E. The distance is under 2.83, but close enough. Speed is as you say. If ped2 is not staying at A, then none of the above works. Ped2 is your reference frame. You have an orange line going from A to E but that's apparently not ped2, it's just a line. 11 hours ago, jv1 said: The speed of pedestrian 1 is 1 m/s - constant - in all three systems . No it isn't. You have ped1 stationary in system 2, but you also contradict that. I am presuming that we're using the frame of the road for system 3. The speed values make no sense otherwise. Thus, since ped2 defines the frame, ped2 stays at location A of the road. 11 hours ago, jv1 said: It will take longer time for pedestrians 1 ,traveling at constant speed to move on Bigger distance of observed trajectory 2.84 m. There is difference in observed time and length but the observed speed is constant. Yes, it takes 2.83 seconds to go 2.83 meters at 1 m/sec. This is news? It isn't a difference in observed time, it is a difference in path taken. ped1 goes straight across the meter path, or he takes a diagonal, a longer path. All observers will measure 2.83 seconds in this 3rd scenario. Changing observers (frames) will change the measured speed and distance, but it won't change the 2.83 seconds. Not at these speeds anyway. A car might go by and measure ped1 moving at 100 m/sec relative to the car. That guy will still see it take him 2.83 seconds to go from A to E.
jv1 Posted May 9 Author Posted May 9 Hi Halc may I first explain situation 2 The pedestrian 1 is positioned on the road. Pedestrian 2 is positioned beside the road. The pedestrian 1 is not moving at all. The road is moving to the right at speed v2=1m/s. And road is transporting pedestrian 1 from A to E(marks on the non moving part beside the moving road). In all three systems we can observe(measure ) ditance and time . And from these two values we calculate the speed . Pedestrian 2 has coordinates (t,x,yz)=(0,0,0,0) in all systems. in system one pedestrain 2 observes movement of pedestrian 1. Only pedestrian one is moving. pedestrian one has coordinates A(0,0,0,0) B(1,1,0,0) d1=1 m in on x t1=1 sec on x v1=d1/t1=1 m/s System 2 Pedestrian 1(was stationary in the road- the road moves- pedestrian 2 is observing movement of pedestrian 1) A(0,0,0,0) B((1,0,1,0) E(2,0,2,0) d2=2m t2=2s v2=1m/s Both systems moving at the same time . Pedestrian 2 is observing movement of pedestrian 1 pedestrian 1 has coordinates A(0,0,0,0) D(1,1,1,0) E(2,0,2,0) From these value we calculate observed distance to be d3=1.402x2=2.84m t3=1.42x2 =2.84m v3=1 m/s And these are all Galilean transformations. And the time will be the same 2.83 for any observer . The distance is only thing changing. (Speed is d/t). Now let’s try to say that speed v=1 m/s is maximum speed in universe . And let’s try to do the same experiment for speed v2=1/3x10e8=0.3x10e-8m What would be distance and time increase ? And one more time we have to movements Observed - movement of pedestrian on axis x (vertical up and down) and movement of road (from left to right) Transporting pedestrian on y axis . Now let’s compare this thought experiment with baseball thrown up and down by the driver in the car traveling at 100km/s. Base ball moves up/down car is perpendicular transpiring the ball and the driver. Or train with light clock on it. Light dot is moving up and down -vertically and train moves light clock horizontally. Is there any similarities? tx
Halc Posted May 9 Posted May 9 (edited) 1 hour ago, jv1 said: may I first explain situation 2 The pedestrian 1 is positioned on the road. Pedestrian 2 is positioned beside the road. The pedestrian 1 is not moving at all. The road is moving to the right at speed v2=1m/s. OK. And all this is 'as observed by' ped2, which means ped2 is by definition stationary, and is with ped1 the whole time. The road moves to the right (as the arrow in the picture shows). Both pedestrians are at location A the whole time, since you said they were stationary. Locations A-F are coordinate locations relative to ped2, our definer of the frame. The locations are not painted on the road as I had earlier guessed. 1 hour ago, jv1 said: And road is transporting pedestrian 1 from A to E(marks on the non moving part beside the moving road). No it isn't. You said he was stationary. If the road is transporting him anywhere, he isn't stationary. You're being inconsistent. If you want him to ride with the moving road, then don't say he's stationary. 1 hour ago, jv1 said: Pedestrian 2 has coordinates (t,x,yz)=(0,0,0,0) in all systems. As I said, ped2 seems to be there to define the frame. He cannot remain at (0,0,0,0) since after 1 second he will be at (1,0,0,0). But he starts at that event in all systems. I am presuming location A is at x=y=0. Apparently you've assigned the x to vertical (across the road, to B) and y horizontal (along with it), towards C,E and such. 1 hour ago, jv1 said: System 2 Pedestrian 1(was stationary in the road- the road moves- pedestrian 2 is observing movement of pedestrian 1) This is a contradiction. Ped1 is stationary relative to ped2 or he's moving. You have to pick one or the other. Scenario 3: 1 hour ago, jv1 said: pedestrian 1 has coordinates A(0,0,0,0) D(1,1,1,0) E(2,0,2,0) OK. He's moving faster than 1 m/sec then because he's getting to D (1.41 away) in only one second. Your OP description said he was moving at 1/sec, not 1.4 1 hour ago, jv1 said: From these value we calculate observed distance to be d3=1.402x2=2.84m t3=1.42x2 =2.84m That last line is wrong. You show time 2 above in the E coordinate, not 2.84. 2.84 is not computed from the above values as you assert. Maybe you should fix all these problems before moving on to the silly parts. 1 hour ago, jv1 said: Now let’s try to say that speed v=1 m/s is maximum speed in universe . And let’s try to do the same experiment for speed v2=1/3x10e8=0.3x10e-8m You can't. You said the max speed was 1 (harsh cops, I tell ya), and then immediately suggest going faster than that. Edited May 9 by Halc
jv1 Posted May 10 Author Posted May 10 Hi again I was trying to say that : Two systems (objects) are observed at the same time by pedestrian 2. Object one : pedestrian 1 moving up and down on road. Object two: road moving in direction to the right (pedestrian 1 is located on the road and he is not moving relative to the road) The road is transporting pedestrian 1 to the right direction and away from pedestrian2 So pedestrian 2 observes in case 1 (system 1) only movement of pedestrian 1 (up / down) In system 2 pedestrian 2 observes movement of road- and the pedestrian 1 is stationary only to demonstrate that pedestrian 2 is not observing road - he is only observing pedestrian 1 . He is not aware that road is moving. system 3 Both road moving to the right away from pedestrian 2 for 2 seconds and pedestrian 1 moving on the road (first second away in vertical direction and back for another second) The pedestrian 2 is not aware that road is moving. These two separate movements are cratering chevron like trajectory. The length of trajectory is 1.42m- this is hypothenuses of triangle with sides of 1 m each. d total =2.84m The time is is the same triangle and t total =1.42 s x2=2.84m These are scalar quantities and we can do these calculations. The observed speed is vtotal=dt/tt=1 m/s Triangle with vertical speed v1 and horizontal v2 can not be cretaed because These are vectors of two separate - not related systems (or phenomena ) the pedestrian 2 is not aware of this . He just sees the movement of pedestrian 1. If pedestrian 1 has a jet pack and his thrusters are pushing him at the same time at horizontal and vertical direction with speed v1=v2 =1 m/s than we would be able to say that speed is v3=1.42m/s This is the most important thing. If the speed v=1 m/s is the maximum speed than the speed v1=v/3x10e8 =0.3x10e-8m is the speed of road . so vertical speed of pedestrian 1 is huge compared to the movement of road . This movement is insignificant.The pedestrian one is 0.7 m wide. If the road speed is 0.1m and up - all the way to 1 m/s we will see effects of speed on observed distance and time . This was supposed to make you think speed of light vertical ,up 2 down movement ,of light dot in light clock and movement of train at speed of 1 m/s away from observer beside the train tracks in original tile dilation experiment . I hope that those other things look a bit less silly now. Again the biggest confusion is that trajectory of pedestrian 1 , light dot in the train,baseball in the car, astronaut in space shuttle , observed by not moving onserver (or pedestrian 2 ) Is created by two movements of two different objects. So again for time dilation (Lorentz transformation) to apply to all of these objects , these objects need “Jett pack “ to push only the observed object at vertical and horizontal speed vectors v1 and v2 That is so easy to overlook and so hard to see after 120 years. Cheers -1
jv1 Posted May 10 Author Posted May 10 Here is very rough drawing of pedestrian 1 with jet pack . I hope it is visible that with changing thrust nozzle angle we change the direction of speed v=1 m/s There is no simultaneous movement of road and pedestrian. Pedestrian 2 observes only the motion of pedestrian 1 / Both vertical and horizontal components of v have value of 0.7 m/s
Halc Posted May 10 Posted May 10 (edited) 12 hours ago, jv1 said: I was trying to say that I understand what you seem to be trying to say. I am trying to help you use consistent terminology to do so. You've been saying ped1 is stationary and immediately saying he has a nonzero speed. That's a contradiction, and it has to be fixed. Motion is relative, and ped2 is your designated reference. If motion is specified relative to anything other than ped2, you need to explicitly say this, else the statement is not even wrong. It renders further discussion impossible. Your scenario seems to take place at an airport with moving sidewalks. Ped2 is standing outside of one of them, and ped1 is standing on the sidewalk. Sometimes the sidewalk is on (cases 2 & 3) and sometimes off (case 1). The points in space (A-F) are all locations relative to Ped2, not relative to anything else. 12 hours ago, jv1 said: two: road moving in direction to the right (pedestrian 1 is located on the road and he is not moving relative to the road) Oh, that's much better. You added 'relative to the road', which is the alternate frame reference that was missing before. Now it's not a contradiction. 12 hours ago, jv1 said: 3 Both road moving to the right away from pedestrian 2 for 2 seconds and pedestrian 1 moving on the road (first second away in vertical direction and back for another second) The length of trajectory is 1.42m- this is hypothenuses of triangle with sides of 1 m each. d total =2.84m The time is is the same triangle and t total =1.42 s x2=2.84m This part is still contradictory. You said it takes 2 seconds to do the two diagonals, one second apiece. Now you assert a different time of 2.8 s. That's a contradiction. It takes 2 seconds to go 2.8 meters at 1.4 m/sec which is the speed ped1 is moving relative to ped2. You also get units wrong. You multiply 1.4 seconds by two, which should get 2.8 seconds, but you call it 2.8 meters. I suspect the unit thing is just a typo. 12 hours ago, jv1 said: The observed speed is vtotal=dt/tt=1 m/s This is still wrong. Ped1 moved from A to D in one second, and that distance is 1.4 m, so his speed relative to Ped2 is 1.4 m./sec, not 1. You did not show your calculation of from where this 1 m/s comes. 12 hours ago, jv1 said: the pedestrian 2 is not aware of this . It really doesn't matter if he's aware of how ped1 is accomplishing his motion. Point is, he sees him 1.4m away after 1 second, and his speed is thus 1.4 m/sec, regardless of what the road is doing. You seem to think it does matter, and that a ped3 doing the same thing with a jetpack, but always being at the same location as ped1, is somehow moving at a different speed relative to Ped2. This is wrong. If you insist otherwise, well then perhaps this topic needs to be moved to strange claims section. 12 hours ago, jv1 said: If the speed v=1 m/s is the maximum speed than the speed v1=v/3x10e8 =0.3x10e-8m is the speed of road so vertical speed of pedestrian 1 is huge compared to the movement of road . This movement is insignificant. The pedestrian one is 0.7 m wide. If the road speed is 0.1m and up - all the way to 1 m/s we will see effects of speed on observed distance and time . This is really hard to parse, but apparently you've imposed a speed limit of 1, and the road moves incredibly slow. I think this is an attempt to an analogy of using natural units for speed rather than m/sec, but given your inability to describe simple Newtonian motion above, I think I won't go there and take your description as worded. "The pedestrian one is 0.7m wide" : What does that mean? He is now on a somewhat narrower super-slow moving sidewalk? Legal, but why? The road speed cannot be 0.1m since meters is not a unit of speed. You perhaps mean 0.1m/sec, so it moves perceptibly fast now. I think a typical airport sidewalk goes about 2 m/sec, but we're in the slow and safe airport today. Why the speed limit? Relative to what? Ped2? What if ped2 decides to walk left? Can nobody put a box on the sidewalk now? What prevents it? 4 hours ago, jv1 said: I hope it is visible that with changing thrust nozzle angle we change the direction of speed v=1 m/s How very non Newtonian. The abrupt change of velocity at point D is straight downward (south), so the jet pack should be pointing south, not to the southeast as drawn. Better to just put a wall at D and have him bounce off it. No jet pack needed at all then. He can just be an inertial object in freefall the whole time. Edited May 10 by Halc
jv1 Posted May 11 Author Posted May 11 Halc ,thank you so much for your time and good points. You are correct airport moving walkway is inspiration. I tried to explain this : The pedestrian 1 moving up down on non moving road is system 1 . pedestrian 1 is moving relative to pedestrian 2( non moving observer - reference point) Speed v1=1m/s has only vertical component - horizontal component of speed is zero. system 2 Is movement of road relative to pedestrian 2. The road moves horizontally at speed v2=1 m/s I mentioned non moving pedestrian 1 - relative to road. Pedestrian 1 movement (together with road)observed by pedestrian 2 is 2 m horizontally. The same pedestrian 1 we can put in non moving position at any position between A,B . The observed horizontal distance would be 2 m That is movement of road - relative to pedestrian 2 Conclusion : There is two simultaneous movements in the same reference frame (pedestrian 2) vertical movement of pedestrian 1 and horizontal movement of road Again ,both observed by pedestrian 2 at the same time . For system 3 I used jet pack and nozzle to emphasize the importance of nozzle angle Alfa. When two speeds vertical and horizontal Acting at the same time they create observed trajectory to be seen by pedestrian 2 as a diagonal . Diagonal which is 1.42 m long. This diagonal angle rotation is equal to angle Alfa of nozzle 45 degrees . The tan angle alfa =dv/dh dv=d1 =vertical distance of 1m dh=d2=horizontal distance 1 m The road moving horizontally and the pedestrian moving vertically on the road(picture at the top of thread ) at the same time will create the same trajectory as a trajectory cretaed by nozzle on the jet pack moved CW (picture at the bottom of the thread ). When we now the length of trajectory (1.42m) this is only one half of total distance ,this half goes up at 45 degrees - the other half is 1.42 going down at 45 degrees . So total length is 2.84 m . This is length observed by pedestrian 2 - observing only pedestrian 1. Now the most important part speed vector v3 on the chevron trajectory of 2.84 m is v3=1 m/s. This v3 speed vector is created by vertical component v3 v=0.7 m(originates from vertical v1 - system 1) and horizontal component v3h=0.7 m(originates from v2 - system 2). The total length d3/v3=t3=1.42 Times 2 is total time t3=2.84s. And now the vertical component v3 v =0.7m is the observed vertical speed of pedestrian 1 (by pedestrian 2) It will take exactly d1 /v3v =t3v=1.42 sec for pedestrian 1 to travel distance d1 =1 m. This is slowing down of the clock. Here is explanation: If instead of speed v1=1 m/s as a fastest speed in universe we use c-speed of ligh=v1 And instead of pedestrian 1 we use light clock 1 with L=3x10e8 m(distance between mirrors). Instead of road let’s use train. Theoreticaly(it is impossible ) If train moves at speed v2=c The total time t3 will be t3=2.84s - the same as for pedestrian 1 . The slowest tick will be 0.7 s Not 0 - like we think - at speed of light (hypothetically) . in my humble opinion , time dilation does not happen . actually overlooked increase in observed trajectory length (chevron system 3) d3 Is the reason for observed time increase . The original tile dilation experiment used light clocks with L=1m Distance between mirrors L. Light travels that distance in one second 3x10e8 times. It is very hard to observe that chevron d3 trajectory length d 3 only increased to d3=1.00000001 m from original d1=1 m But it is calculated - as a time dilation . I would like to say again - what we call time dilation is mislabeled increase in observed chevron trajectory length. -1
Halc Posted May 11 Posted May 11 (edited) 11 hours ago, jv1 said: speed vector v3 on the chevron trajectory of 2.84 m is v3=1 m/s. This v3 speed vector is created by vertical component v3 v=0.7 m(originates from vertical v1 - system 1) and horizontal component v3h=0.7 m Speed is not a vector. You seem to be saying that moving 2.8 meters in 2 seconds is 1 m/sec. That's simple arithmetic that you're getting wrong. You also are still labeling a speed in meters where you say v3h=0.7m Do I have to point this out every post? The vertical component of each velocity vector is 1 and -1 per second respectively, not 0.7. Ditto with the h component, which is the speed of the road which you said is 1, regardless of whether ped2 knows about it or not. I tire or repeating this. You seem bent on just asserting the same wrong numbers over and over no matter how many times the mistakes are identified. You seem to be doing this deliberately, perhaps in order to argue some mathematically inconsistent point that you think has some unfathomable bearing on relativity. It doesn't. Relativity doesn't support your numbers at all, nor does it give different numbers for this scenario than does Newtonian physics. Edited May 11 by Halc
jv1 Posted May 11 Author Posted May 11 To explain more closely my thought train I will use two light clocks : First : Light clock 1 with distance between magnets L=3x10e8m The vertical speed v1 (of system one- pedestrian 1)is v1=c The light clock mirrors (road)are moving at horizontal-speed of v2=0.5c This is observed by pedestrian2(jv). The angle of chevron trajectory created(observed) by these two speed systems is: tan Alfa=0.5c/c=0.5 Angle Alfa =26.5 degrees The second : Light clock with L=3.6cm distance between mirrors. The speed v1=c The speed v2=0.5c The angle Alfa is the same Alfa=26.5 degrees. The “small “ light clock vertical distance d1=3.6cm the time t1=0.036/3x10e8=0.012x10e-8m For the same time the hirizontal distance d2=0.5x3x10e8x0.012x10e-8 d2=0.018m(1.8cm) The chevron trajectory distance will be d3^2=12.96+3.24 d3=4.02cm The time difference delta t Delta t=d3/c=0.0402/3x10e8=0.0134x10e-8s Why is this relevant? I am going to try to explain how the cesium nuclear clock is working - with a twist. The frequency of radiation of ceasing inside the nuclear clock is f=9.1922633177GHz the wave length l=3.26cm the speed of this microwave range photon is v1=c Now let’s put two polished copper plates at distance L=3.6 cm Between these two plates When caesium atom passes between them, it will emit photon and this photon will be “trapped “ and continue to move vertically up and down between copper plates. What do we got? Ceasium version of light clock. Now let’s put this ceasium light clock on the space ship. The space ship speed is v2=0.5 c (horizontal speed) If you have a look at the small light clock2 from above ,what do we get? If jv is on the space ship with the cesium light clock- the dial will show the time 1 sec. But the signal - microwave photon traveling between copper plates will travel longer trajectory and the time on dial will be wrong. To compensate for this - we have to readjust trajectory we are traveling on (GPS satellites come to mind) Conclusion: the time is imaginary phenomena. The light will travel distance 3x10e8 m For period of time we call 1 second . If there is change on the measuring instrument - clock- it is because the distance light signal,or microwave signal, or laser signal is changed. Tx
TheVat Posted May 11 Posted May 11 On 5/8/2024 at 6:08 AM, jv1 said: Opinions? The pedestrians actions still don't answer the eternal question - why did the chicken cross the road? 1
jv1 Posted May 12 Author Posted May 12 Hi halc and Thevat I was hoping to explain that movement of pedestrian 1 is just reference frequency f=1 . This vertical movement is not related at all to horizontal movement of road . What it does it gives us reference distance travelled in time period. In pedestrian 1 case , this frequency f=1 d1 =1 m is travelled in t1=1sec. f=1/t If we do not know the horizontal speed v2=? We can observe - measure the distances Between points ABCEF… These are real distances - nothing relative about them. From triangles made of distances AB,CD…. we can calculate the speed v2. This is the point of having reference frequency(pedestrian 1). Now if we substitute pedestrian 1 for light clock - the reference frequency is not f=1 the reference frequency couod be f=1 for light clock L=3x10e8m All the way to f=3x10e8 for light clock with L=1 m And we calculate the v2 of object in real world - using reference points ABCD…. or (t xyz) from the observer (reference frame). I was hoping somebody will see this ,without me pointing this out. The light clock ,same as pedestrian 1 has nothing to do with horizontal speed v2 - the speed of objects in real world. We measure distance in light years . The biggest light clock with L=3x10e8m is actually reference distance of One light second the light travels 3x10e8 in one second . Inside the distance of 3x10e8m from the light clock we can use this reference distance - one light second. For the wast universe we use light years . Could you please spare few more minutes and let me explain caesium atomic clock A bit more. Atomic clock in the oven heats caesium And creates beam of atoms traveling Through the resonator. Inside resonator the atoms are excited and spectated . Than excited atoms hit the metal wire. They release photon - micro wave photon with wave length l=3.6 cm The distance this photon travels from metal wire to counter is equivalent to d1 in light clock. When photon reaches the optical optical counter it does not deflect- it goes through it. This is very modern version of light clock. And as I said above - it he distance d1 increases to distance d3 when horizontal speed is applied to light clock This is the real increase in length- nothing relative or imaginary about this In atomic clock processor unit which counts the number of 3.6 cm photons is pre calibrated to work at v2=0m/s(horizotal speed). Any increase in horizontal speed v2 from 0 to c will literally increase the distance of photon 3.6cm(the wave length). So the display on the atomic clock will show WRONG time . Now there is only one thing left to do let’s revisit Hafele- Keting experiment. Or even better ,can you guys try to tell me what really happened on those flights? May be just may be we can answer why chicken cross the road - for no reason . The same way why was increase in distance travelled by photons in light clock called TIME DILATION- for no reason
jv1 Posted May 12 Author Posted May 12 I was thinking how do I explain this in a more simple way. Here is another try: Case 1 pedestrian 2 will observe two movements occurring separately. The pedestrian 1 goes from A to B,observed - measured distance is 1 m. The measured time is 1 sec. pedestrian 1 stops. Than the road moves to the right - transporting (without his good will) pedestrian 1 from point B to point D Distance is 1 m ,time is 1 sec. the pedestrian 1 moved from A to E in total time tt=4sec pedestrian 1 was traveling for dt=4 m the average speed Vt=dt/tt=1 m/s Observed measured and confirmed - speed v is Constant in all distances . Case two: Both systems - vertical movement of pedestrian 1 and horizontal movement of road (transporting pedestrian 1 - without his own will) horizontally- this is observed by pedestrian 2. The total distance travelled is dt=2.84m(calculated by Pitagora rules for distance . tt=2.84sec total speed is Vt=dt/tt=1 m/s The speed vectors horizontal and vertical are working non related to each other . We can not apply pitagora rules for velocity But , without a will of pedestrian 1 - is moving on trajectory d3=1.42x2 And time spent is 1.42x2 the total speed is vt=1 m/s calculated from observed (measured ) length of trajectories. From reference time / distance vertical movement of pedestrian1 we can calculate any speed v2 (horizontal speed ) form v2 =0 m/s to 1 m/s based on angle alfa=v2/v1 when instead of pedestrian 1 we use light clock , or modern version of light clock (caesium atomic clock) as a reference we can calculate any speed from 0 to c observed by pedestrian 2 anywhere in universe . I hope this clears things a bit better .
exchemist Posted May 12 Posted May 12 2 hours ago, jv1 said: Hi halc and Thevat I was hoping to explain that movement of pedestrian 1 is just reference frequency f=1 . This vertical movement is not related at all to horizontal movement of road . What it does it gives us reference distance travelled in time period. In pedestrian 1 case , this frequency f=1 d1 =1 m is travelled in t1=1sec. f=1/t If we do not know the horizontal speed v2=? We can observe - measure the distances Between points ABCEF… These are real distances - nothing relative about them. From triangles made of distances AB,CD…. we can calculate the speed v2. This is the point of having reference frequency(pedestrian 1). Now if we substitute pedestrian 1 for light clock - the reference frequency is not f=1 the reference frequency couod be f=1 for light clock L=3x10e8m All the way to f=3x10e8 for light clock with L=1 m And we calculate the v2 of object in real world - using reference points ABCD…. or (t xyz) from the observer (reference frame). I was hoping somebody will see this ,without me pointing this out. The light clock ,same as pedestrian 1 has nothing to do with horizontal speed v2 - the speed of objects in real world. We measure distance in light years . The biggest light clock with L=3x10e8m is actually reference distance of One light second the light travels 3x10e8 in one second . Inside the distance of 3x10e8m from the light clock we can use this reference distance - one light second. For the wast universe we use light years . Could you please spare few more minutes and let me explain caesium atomic clock A bit more. Atomic clock in the oven heats caesium And creates beam of atoms traveling Through the resonator. Inside resonator the atoms are excited and spectated . Than excited atoms hit the metal wire. They release photon - micro wave photon with wave length l=3.6 cm The distance this photon travels from metal wire to counter is equivalent to d1 in light clock. When photon reaches the optical optical counter it does not deflect- it goes through it. This is very modern version of light clock. And as I said above - it he distance d1 increases to distance d3 when horizontal speed is applied to light clock This is the real increase in length- nothing relative or imaginary about this In atomic clock processor unit which counts the number of 3.6 cm photons is pre calibrated to work at v2=0m/s(horizotal speed). Any increase in horizontal speed v2 from 0 to c will literally increase the distance of photon 3.6cm(the wave length). So the display on the atomic clock will show WRONG time . Now there is only one thing left to do let’s revisit Hafele- Keting experiment. Or even better ,can you guys try to tell me what really happened on those flights? May be just may be we can answer why chicken cross the road - for no reason . The same way why was increase in distance travelled by photons in light clock called TIME DILATION- for no reason Hafele-Keating or Dunning-Kruger? 1
Phi for All Posted May 12 Posted May 12 On 5/10/2024 at 7:33 PM, jv1 said: I tried to explain this : On 5/11/2024 at 7:40 AM, jv1 said: To explain more closely my thought train 3 hours ago, jv1 said: I was hoping to explain 1 hour ago, jv1 said: I was thinking how do I explain this in a more simple way. ! Moderator Note This doesn't seem to be working at all. You keep trying to explain something that others have found flaw with. You need to address the flaws mentioned instead of trying to say it in a different way.
jv1 Posted May 13 Author Posted May 13 Dunning - Kruger,very interesting. According to the special theory of relativity introduced by Albert Einstein, it is impossible to say in an absolutesense that two distinct events occur at the same time if those events are separated in space. If one reference frame assigns precisely the same time to two events that are at different points in space, a reference frame that is moving relative to the first will generally assign different times to the two events (the only exception being when motion is exactly perpendicular to the line connecting the locations of both events). THE ONLY EXCEPTION BEING WHEN MOTION IS EXACTLY PERPENDICULAR TO THE LINE CONNECTING LOCATION OF BOTH EVENTS. Here verybsimple question from guy who does not know much about theory of relativity - The event 1 - pedestrian 1 motion is perpendicular to event 2 motion of the road? Or is the motion of light dot (in light clock) event 1 Perpendicular to motion of train- event 2 ? I would like to say thanks to science forums for giving me opportunity to present,explain and defend this very simple Overlook. i would like to thank moderators and all of the people who spend time and effort to point out the flows. I was lucky not to know theory of relativity on the level you guys do- because I would not be able to see this very simple overlook. Cheers Motion of space shuttle is perpendicular to the motion of light dot in the light clock relative to the line of observer
jv1 Posted May 15 Author Posted May 15 The speed of light is constant and it is never at rest. Here is one more thought experiment . The distance 3x10e8m travelled in 1 second could be used as a reference distance to confirm is the the object of system 2 in relative motion to reference point A. As you can see sped of light c is constant . Time t1 for both systems is 1 second. The system one is light clock with distance between mirrors L=3x10e8m And system two can be train,space shuttle, road The light clock is used as a range/speed finder device So I would call this thought experiment light clock as a range /speed finding device . If we apply The Dunning-Kruger effect DUNNINO-KOUSER occurs when a person's lack of knowledge and skill in a certain area causes them to overestimate their own competence. By contrast, this effect also drives those who excel in a given area to think the task is simple for everyone, leading them to underestimate their abilities. May be we can call this thought experiment Dunning -Kruger’s effect of speed of light
Janus Posted May 15 Posted May 15 The light clock experiment with 2 light clocks( red and blue) starting at the same point: The expanding circles highlight how the the light pulses travel at c. The same experiment, but now, we are "riding along" the blue light clock. From the inertial frame where the red clock is stationary, the blue clock ticks slower, and the from the inertial frame where the blue clock is stationary, the red clock ticks slower. Keep in mind this is the same scenario, just viewed from different inertial frames of reference. 3
jv1 Posted May 15 Author Posted May 15 Hi Janus These gifs are awesome. There is no easy way to say this. The movement - observed trajectory/ distance of light dot depends on two speeds. Speed of light c and speeds v1 - horizontal speed of red mirrors and v2=horizontal speed of blue mirrors . speed c is distance d=3x10e8m which light moves in 1 sec. v1 is distance d1 red mirror moves in 1 sec v2 is distance d2 blue mirrors moves in 1 sec. Now again these are two separate systems Movement of light dot inside the clock has constant distance and time of 1 sec. the speed of light c is constant . The distance between mirrors L=3x10e8m is the way to go to see that . The distance d1 and distance d2 are accomplished by something transporting the mirrors - train ,road ,space shuttle…… The mirrors can not move on their own. The overlook is that we have two Newtonian events happening at the same time . The interaction between these two systems is by changing distances between observer 1 And blue clock light dot. The light dot has one distance dc=3x10e8m And distance of d2 =1 m/s(for example) If these events are not simultaneous- the total time to travel distance dc+d2 =d t=3x10e8 +1 m the total time will be tt=2 sec I mean light dot travels 3x10e8 m for 1 second - stops than perpendicular to it blue mirrors transport the dot 1 m to the right in 1 sec. The total speed vt=dt/tt=300000001/2=1500000000.5m/s If these two distances are travelled at the same time the observed trajectory will be Hypotenuse of triangle created by dc and d2 Total dt^2=dc^2+d2^2 The total time tt=1 .42 s tan Alfa =d2/dc for time of 1 sec d=v we can see that tan Alfa =v2/c Than total vt on hypotenuse will be vt=c x tan alfa I used Newtonian physics to solve the theory of relativity problem. How ? Over look is that two systems observed are Newtonian events happening at the same time . And increase in distance in observed trajectory - was misunderstood as a time dilation. -2
jv1 Posted May 16 Author Posted May 16 Here is the closest situation of two Newtonian events happening at the same time . The fast flying aircraft can be seen as a light dot of second (observed light clock) speed is max.(we can see it as c) The slow flying aircraft is seen as a mirrors of second observed light clock. his speed is v2 This situation is the second half of chevron trajectory of light dot. The shortest distance betwen aircrafts (red line ) is L=distance between mirrors of both clocks. We can see that these two aircrafts are traveling at their own trajectories- not related to each other. When fast aircraft travels on red trajectory it travels at max speed (c)and it takes time t1. d1 =cxt1 When fast aircraft travels in angled trajectory it will move at max speed (c) but the trajectory is longer . The time t2>t1 the distance d2=cxt2 we observe the fast aircraft- we do not observe the slow aircraft at all. the slow aircraft always travels at speed v2 and it will be at colossi on point x in time t3 And distance d3=v2xt3 From these totally not related distances and times we can calculate and confirm that speed of fast aircraft on angled trajectory is max speed (c)
jv1 Posted May 16 Author Posted May 16 I came up with one more thought experiment - this time there is 3 observers but instead of light clock that have flash lights and stop watches. The speed of light is the only speed we observe. The observed distance- trajectory is changing . speed is always constant . The time change is directly proportionate to change of distance - always . And there is no time change ,or speed change without change in distance (different trajectory) -2
jv1 Posted May 16 Author Posted May 16 I came up with one more thought experiment - this time there is 3 observers but instead of light clock that have flash lights and stop watches. The speed of light is the only speed we observe. The observed distance- trajectory is changing . speed is always constant . The time change is directly proportionate to change of distance - always . And there is no time change ,or speed change without change in distance (different trajectory) First picture - on the top shows that mirrors of light clock 2 moved to the right 1 m and the light dot of clock 2 is observed to be late (light dot is not in a position at mirror B of second light clock). The light dot of clock one - observed is to be touching mirror B of light clock one). That is why the clock 2 is observed to show slower time. And again this happens because trajectory Of light dot in clock 2 - distance increased. -1
jv1 Posted May 20 Author Posted May 20 On previous drawing light clocks were not perpendicular (movement direction of light dot) to direction of movement of mirrors (v2). Here is explanation for light clocks (movement of light dot inside the clock - system / event 1) moving perpendicular to v2(movement of mirrors). The observed length ( orange do) in non moving (t=0)and moving (t1=1 s) of light dot trajectory of clock 2 which is observed by observer 1 id the same when mirrors are at rest and when mirrors are moving at speed v2(clock 2). In both cases speed of light dot of clock 2 is c and time travelled is 1 second. The difference is that when t=0 (time explains that two clocks 1 and 2 and their observers 1 and 2 do not move realtive to each other . Both light dots are moving - they are event 1 - reference event. They both travel at the speed of light c and travel the same distance 3x10e8m This is constant - this never changes. Observed trajectory direction is changing but the distance light travels in 1 second never changes. That is the hole and only purpose of light clock in time dilation experiment - reference light signal. From known light speed of light dots(vertical sytem/event 1) and observed angle Alfa between do- orange observed distance and direction of horizontal speed (system /event 2) we can calculate relative speed v2 between any two objects anywhere in universe . But instead using all of this - why do not we just make xyz greed for time and apply classic physics and get the total time . Than we make xyz greed for distance . Than from formula v=d/t find the speed . ? -1
jv1 Posted May 21 Author Posted May 21 Here is a bit better drawing for light clocks positioned in line with speed v2. I think this is easier way to see that light for travels at speed of light in both clocks (stationary and moving) and it travels for same time 1 sec. Mirrors are moving simultaneously for one second in the Same direction and the trajectory of light dot is observed shorter. This is direct consequence of two events happaning at the same time . Movement of light dot and movement of mirrors . Again the movement of light dot is reference movement and it will happen simultaneously with any event we observe. That is purpose of light clock- reference signal .
Mordred Posted May 21 Posted May 21 (edited) You know you really should use the transformations for Lorentz. They are easy to use and work regardless of velocity of the signals. Others have already mentioned that Edited May 21 by Mordred
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