Everything we perceive is in the past due to finite speed of light
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Motor Daddy
Valued Senior Member
Given how I've been treated over this diagram and the utter disrespect that people have shown me without even understanding the diagram, I'm sure you'll excuse me for saying, Fuck Off! I am helping you and you are treating me like shit, so I am returning the favor! Continue living in your fantasy land, by all means. It's amusing to me knowing that you don't know shit, you're just an ass kissing Einstein groupie that's been told what to think!! You paid to learn someone's bullshit! How does that make you feel??
Motor Daddy
Valued Senior Member
Oh, I'm treating you much much better than I've been treated by your peers in the scientific community so I have no sympathy for you!
It's good being paid to learn about real things, I must say. Can I please calculate $$c_y$$ now?
Motor Daddy
Valued Senior Member
Either y or z makes no difference. If you want to go 3D here:
Measured times to receivers:
x time: .761972 seconds
y time: .761972 seconds
z time: .92 seconds
Component velocities:
v(x) = .2300c
v(y) = .2300c
v(z) = .4022c
Coordinates of source at center of cube at .761972 seconds: (0.17525356,0.17525356,0.3064651384)
x .761972(.2300c) 0.17525356
y .761972(.2300c) 0.17525356
z .761972(.4022c) 0.3064651384
Coordinates of source at center of cube at .92 seconds: (.2116,.2116,.370)
x .92(.2300c) .2116
y .92(.2300c) .2116
z .92(.4022c) .370
Coordinates of source at center of cube at 1.0 seconds: (.23,.23,.4022)
x 1.0(.2300c) .23
y 1.0(.2300c) .23
z 1.0(.4022c) .4022
Distance center of cube traveled from start coordinates (0,0,0) to (.23,.23,.4022) in 1 second is 155,072,655.74 meters, so the absolute velocity of the center of the cube is 155,072,655.74 m/s.
Component velocities
x .23c=68,952,,265.34 m/s
y .23c=68,952,,265.34 m/s
z .4022c= 120,576,526.6076 m/s
I understand my model, but I am placing control factors on your BS, and in order to eliminate your (Einstein's) BS I incorporated the z time, which is the 2D problem. I can go 3D if you choose, although it takes me considerable time and effort to do that. I created that 2D equation in order to dispel your BS, as you will soon find out!!!
You are getting quite agitated, you need to get back on your meds.
Motor Daddy: No I don't want to go into your 3D case, because I can get all the results I need just by picking a convenient coordinate system which reduces it to a 2D problem. I would pick my x-axis to align with the movement of the box and get the exact same thing. If you can disprove my reasoning at the end here, then we can look at other examples.
Calculating $$c_y$$ in Motor Daddy's system: The beam traverses a distance on the y-axis of 0.5 lightseconds, which Motor Daddy says the train observer detects at z at a time of 0.65 seconds. So $$c_y=\frac{0.5}{0.65}c=0.76923c$$, this of course is only for light reaching z, not x. With metre sticks and clocks placed onboard the train, the observer can measure this velocity component on the train without needing to know about anything that's happening outside. We can thus use this info to calculate the train's velocity: $$v_x=c_x=\sqrt{c^2-c_y^2}=0.63897c$$
Well golly, whadda ya know.
So then Motor Daddy, I have shown you how you can use clocks and metre sticks inside the train to measure the speeds of the light beams reaching z and x relative to the train, and use that info to calculate the postulated speed of the train through "space", getting the same value both times. Do you agree?
Calculating $$c_y$$ in Motor Daddy's system: The beam traverses a distance on the y-axis of 0.5 lightseconds, which Motor Daddy says the train observer detects at z at a time of 0.65 seconds. So $$c_y=\frac{0.5}{0.65}c=0.76923c$$, this of course is only for light reaching z, not x. With metre sticks and clocks placed onboard the train, the observer can measure this velocity component on the train without needing to know about anything that's happening outside. We can thus use this info to calculate the train's velocity: $$v_x=c_x=\sqrt{c^2-c_y^2}=0.63897c$$
Well golly, whadda ya know.
So then Motor Daddy, I have shown you how you can use clocks and metre sticks inside the train to measure the speeds of the light beams reaching z and x relative to the train, and use that info to calculate the postulated speed of the train through "space", getting the same value both times. Do you agree?
Mr. Motor Daddy, Oh, I've played your game before. When you cannot "reel someone in" to you inane "non-theory of relativity", you resort to name calling and hissy-fits and then you"cut bait" and run off.
Earlier, when I mentioned my association with Sylvania/DARCOM/TECOM/GPS, you claimed that I was "part of the problem" and referred to GPS constantly updating as proof of the failure of the currently accepted "theory of relativity".
When I responded with an intelligent and truthful response that shot down any argument that you could bait me into - you ignored it and moved along.
Motor Daddy, I do not mind being the dmoe - you fail miserably at even beginning to understand what dmoe actually means - not nearly as much as you fail to understand "relativity", I must add.
Heck, from everything of yours I have read (unlike you, I do actually read and even re-read posts!) and re-read, I am not entirely convinced that you even fully understand the word "relative".
I am sorry, but open discussions and the sharing of knowledge, is not what you are offering - and you always eventually show just that.
If you ever own up to that fact, your understanding of everything will only increase, and then maybe, just maybe YOU will be able to really ADD SOMETHING to any discussion - even possibly one on relativity.
Until then, I am more than happy to "commentate play by play".
Still, dmoe
Motor Daddy
Valued Senior Member
Oh golly gee! I tell you how to do it, and then you all of a sudden calculate the velocity of the box from within the box, which nobody has ever done before! You're amazing! Praise you for being the first person in the history of the world to calculate the velocity of the box from within the box!!!! Praise the Lord!! You did it!!!! (rolls eyes!)
You didn't tell me anything. I followed your basic, simplistic postulates to their logical conclusion. All I needed was the same starting point you took, which was to assume that the pulse reaches detector z at 0.65 seconds. I wanted to make sure you followed and agreed with each step so I wasn't wasting my time.
Now we can agree then, that by your logic, in principle I could bring velocity detectors onboard the train and place them at locations z and x, measure the effective velocity of the light relative to the train, and thereby determine my supposed velocity through "space". I have indeed just shown how to do that. Correct?
Motor Daddy
Valued Senior Member
If I would think it was incorrect I wouldn't have posted it! BY DEFINITION IT IS CORRECT, and if later you come back and show me different, then you have just violated the definition of the meter!
Motor Daddy must not have any more bait prepared.
Once again he has chosen to belittle his chosen foe - even going so far as to bring a "Lord" into his childish ploy!
It is really heating up in here, is ignorance, once again going to rear its ugly head ?
Oh, I forgot, Motor Daddy has already played the "ignorance card" already, with more than one posting or poster, in this thread.
Odds are, Motor Daddy will soon "cut and run", again.
Once again he has chosen to belittle his chosen foe - even going so far as to bring a "Lord" into his childish ploy!
It is really heating up in here, is ignorance, once again going to rear its ugly head ?
Oh, I forgot, Motor Daddy has already played the "ignorance card" already, with more than one posting or poster, in this thread.
Odds are, Motor Daddy will soon "cut and run", again.
Motor Daddy
Valued Senior Member
Odds are I'll be banned before I "cut and run." That's what happens when my foes run out of arguments and they have nothing left but to either admit defeat or ban me. They don't admit defeat, so there is one choice left!
Ok. So you agree that by your presumptions, if I place a velocity detector at x or along the path to x, the observer on the train will measure light in that direction to have an effective speed of $$c_1=0.361029c$$, travelling along the x-axis. The observer on the train will similarly place a velocity detector along the path to z, and measure light reaching z to appear to be travelling along the z-axis with velocity $$c_2=0.76923c$$.
I can then calculate the speed of the train through space by either of the following two methods:
Using the speed measured along the path to x: $$v_x=c-c_1=0.63897c$$
Using the speed measured along the path to z: $$v_x=\sqrt{c^2-c_2^2}=0.63897c$$
Would you say that this is valid reasoning, Motor Daddy? Do you believe this method will work as stated?
Motor Daddy
Valued Senior Member
Moron: The time to the z receiver is what it is and is not an indication of the x receiver's time. You don't take the speed of light in the direction of the x axis and use that to measure the time to the z axis! Look at the freakin' diagram and the 2D equation!
CptBork, Motor Daddy.
If I may again interpose a 'circuit breaker' for the growing 'tension' between you two as the discussion has proceeded to this point...
I ask that you two please reply to the series of naive questions I posed to you both (see post #81) in relation to my own illustrative scenario therein designed to 'tease out' your respective takes on this 'lightspheres' behaviour etc.
Only if you both give me your respective answers as requested can I then compare the logics and physics for consistency internally and externally (based on the respective theoretical/empirical basis from which your respective answers will show where each of you 'is coming from').
Your respective answers will automatically imply to me the logics and physics which your respective views bring to the 'lightsphere' scenarios of all sorts.
I need your answers if I am to enhance my naive understandings of both your perspectives and how both compare with my own naive understandings already in this context. Thanks!
If I may again interpose a 'circuit breaker' for the growing 'tension' between you two as the discussion has proceeded to this point...
I ask that you two please reply to the series of naive questions I posed to you both (see post #81) in relation to my own illustrative scenario therein designed to 'tease out' your respective takes on this 'lightspheres' behaviour etc.
Only if you both give me your respective answers as requested can I then compare the logics and physics for consistency internally and externally (based on the respective theoretical/empirical basis from which your respective answers will show where each of you 'is coming from').
Your respective answers will automatically imply to me the logics and physics which your respective views bring to the 'lightsphere' scenarios of all sorts.
I need your answers if I am to enhance my naive understandings of both your perspectives and how both compare with my own naive understandings already in this context. Thanks!
In the central observer's frame, there's only one light sphere. The observers moving to the left and right would be seen moving to the left and right of the light sphere's center, respectively. Note that the central observer could synchronize a bunch of clocks and place them around the surface of this sphere, so they all get hit at the same time. From the other two observers' points of view, these clocks would all give the same time readout when they're hit, but they'd get hit at different times- in the other two observers' frames, the central observers' clocks are no longer synchronized.
Each of the two "moving" observers will also see a light sphere centered around themselves; the observer moving to the right sees the other two observers moving towards the left edge of the sphere. The observer moving to the right sees the other moving to the right edge of the sphere. It all comes down to synchronization being a relative concept just like time and distance intervals.
I think you may find answers to these questions in my answers to the others, as they're related questions.
If I use velocity detectors instead of time clocks, by your logic I can still determine the train's velocity through "space", yes? These velocity detectors are still based on standardized clocks and meter sticks, no assumptions whatsoever about light. Will this work, or will it not work?
Motor Daddy
Valued Senior Member
They are NOT "velocity detectors" they are TIME DETECTORS!