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@CrazyScientist You refuse to admit to not understanding relativity, when you clearly don't. You make illogical assertions and stubbornly defend them. That is very troll-like behaviour. Are you going to start addressing the points that people are putting to you or will you continue to act like a troll?
No. Frame in which the receiver is moving is not the inertial frame of that receiver.
Quote from: CrazyScientist on 04/05/2021 16:28:18No. Frame in which the receiver is moving is not the inertial frame of that receiver.You are causing confusion by misusing a term (inertial frame) which is already defined in both Galilean and Special Relativity. You need a different term.In Galilean Relativity it is not possible to tell whether you are moving if your frame or motion is inertial.
Quote from: jeffreyH on 03/05/2021 11:21:33@CrazyScientist You refuse to admit to not understanding relativity, when you clearly don't. You make illogical assertions and stubbornly defend them. That is very troll-like behaviour. Are you going to start addressing the points that people are putting to you or will you continue to act like a troll?I'm trying to answer to all comments, but I might missed some of them. Tell me which post I didn't adress and I will do it next.
Quote from: CrazyScientist on 04/05/2021 16:29:32Quote from: jeffreyH on 03/05/2021 11:21:33@CrazyScientist You refuse to admit to not understanding relativity, when you clearly don't. You make illogical assertions and stubbornly defend them. That is very troll-like behaviour. Are you going to start addressing the points that people are putting to you or will you continue to act like a troll?I'm trying to answer to all comments, but I might missed some of them. Tell me which post I didn't adress and I will do it next.How about all of them
Ok, I think that I need to adress the Hafele-Keating experiment once again. As I showed in one of my previous posts, my predictions were in 100% consistent with the effects of time dilation predicted by SRT (to my own surprise) - and since my model suppose to be an alternative solution to the constancy of c in relative motion, it makes it kinda self-contradictory. This is why I probably have to explain the major differences between my model of relativity and the SRT.I will begin this subject with a question: Let's say, that 2 objects A and B are moving at 0,5c in relation to each other - which one of them will experience higher rate of time flow?According to my model, time will be flowing for both of them at the same rate. But according to SRT time will flow faster for the stationary object and it will flow slower for the moving one. But then which object is the stationary one, if both of them are moving in relation to each other?But to adress this issue in more details, I will try to explain, what in my model makes the difference between this scenario:And this one://www.youtube.com/watch?v=Tiu2BWpxtwsI've spent some time making a scene, which will include both scenarios - that means clocks, which remain on the Earth's surface and those onboard 2 planes, that move in opposite directions along the equator. In this scenario in the time of 2 full Earth's rotational cycles (2 days) each of 2 planes moves around Earth once, but in opposite directions. This results in a situation, where from the perspective of an observer, who remains suspended in a fixed position in the interplanetary space, Earth rotates 2 times, one plane moves around the Earth once, while the second plane moves around it 3 times:As I said earlier, in my model clocks, that remain on the surface will be always synchronized, despite them moving at different velocities due to their locations at different latitudes. Image below should show you why it is so:In shortcut, all 3 waveforms have different wavelenghts, but it is compensated by the differences of their velocities, so that all of them have the same frequency of cycles (24 per each rotation).However things will be different for the clocks onboard moving planes, as one of them (faster one - red color) will now count 48 cycles during 3 rotations around Earth and the second one (slower one - green color) will count 48 cycles during one rotation.If we look at the waveform for the clock onboard the faster plane (red one), we'll see, that it's wavelenght got bigger by 1/3 in relation to the clock placed on the surface at the equator:At the same time, wavelenght in the waveform representing the clock onboard the slower plane (green one), will get 2 times shorter, in relation to the clock at the equator:And since the rate of time flow is in my model defined by the frequency of cycles, time will flow 2 times faster onboard the green plane and 1/3 slower onboard the red plane in relation to clocks, which remain on the Earth's surface. Of course, those results are not representing the actual experiment, as velocities in my scenario are given in relation to constant c, while in real-life rotational velocity of Earh, just like the velocities of planes are MUCH slower than the light - so, the effects of time dilation observed in real-life will be MUCH smaller. And just in case, I will remind you, that the differences in the number of counted cycles onboard the planes are (for some reason) in 100% consistent with the time dilation, which is predicted in SRT...
Have you ever considered working for Disney? They have fairy tales all wrapped up.
Quote from: jeffreyH on 04/05/2021 21:16:01Have you ever considered working for Disney? They have fairy tales all wrapped up.I would love to - I'm sure they would pay me much more than I earn rght now.
Quote from: CrazyScientist on 04/05/2021 21:21:22Quote from: jeffreyH on 04/05/2021 21:16:01Have you ever considered working for Disney? They have fairy tales all wrapped up.I would love to - I'm sure they would pay me much more than I earn rght now.Then why not just let us have a sane conversation. You put a lot of effort into defending an untenable position. You could be actually learning some very interesting things, but you need to put your listening ears on.You might actually make some friends, instead of simply antagonising people. If you simply carry on the way you are then you lose that possibility.
No. A frame, where observer is moving is not his frame
Because you can't measure the speed of light in one-directional motion. This means, that you have to "ask" the receiver at which time the light reached it - and if that light was emitted 4ly away from that receiver, it wil "tell", that light reached it 4 years after emission
I didn't say it is constant in relation to all frames.
I said, that it's constant in relation to observer in his own inertial frame
Sure - you can show it to me. I did it, so I know, what it represents. Yes - I boosted the light according to the velocity of receiver (0,5c) and now 0,5c + 0,5c makes 1c, which is observed in the rest frame of that receiver
Quote from: CrazyScientist on 04/05/2021 16:23:57No. A frame, where observer is moving is not his frameI am not sure you have got this yet so I will address it. You can only be in your own frame, how could you be in a frame that is not your own? If I am moving at a constant velocity I am in an inertial frame. I am obviously in my own frame. If I pass by somebody not moving relative to me then they are in another inertial frame. They also are in there own frame. It is impossible to be in another frame than your own. It would be the same as saying you are moving at different speed than you are moving; it makes no sense.
Quote from: CrazyScientist on 04/05/2021 16:23:57Because you can't measure the speed of light in one-directional motion. This means, that you have to "ask" the receiver at which time the light reached it - and if that light was emitted 4ly away from that receiver, it wil "tell", that light reached it 4 years after emissionNot according to the space time diagram.
Quote from: CrazyScientist on 04/05/2021 16:23:57I didn't say it is constant in relation to all frames.That's to bad, because experimentation shows that the speed of light is c in ALL inertial frames.
Quote from: CrazyScientist on 04/05/2021 16:23:57 I said, that it's constant in relation to observer in his own inertial frameThat is not what you drew in your space time diagram! Wrong ship frame.jpg (30.78 kB . 600x450 - viewed 4289 times)This is a space time diagram drawn with the spaceship inertial frame at rest. You drew the green line which shows that from the frame of the spaceship the speed of light is 0.5c. That means the speed of light is not constant in the spaceships inertial frame. I'm sorry it that is not what you want but that is what the space time diagram is showing.
That means the speed of light is not constant in the spaceships inertial frame
I drew the space time diagram based on your postulate (the speed of light is constant in all inertial frames) and you get a travel time of 2.7 years.So in one diagram the travel time makes no sense and in the other diagram the speed makes no sense.The problem is in the underlying assumptions you made about the relativity - It doesn't work.
the speed of light is constant in all inertial frames
c is constant in relation to every observer in his own inertial frame
Quote from: CrazyScientist on 04/05/2021 16:23:57Sure - you can show it to me. I did it, so I know, what it represents. Yes - I boosted the light according to the velocity of receiver (0,5c) and now 0,5c + 0,5c makes 1c, which is observed in the rest frame of that receiverThis is not difficult! That means in the in the rest frame of the spaceships the speed of light is 0.5 c, which violates your postulate.
Let's pretend there was no receiver. Then you would draw the speed of light as c for the spaceship frame. If there is a receiver though, all the sudden the speed of light drops by half? Think about what you are saying, it makes no sense.
Let's say you are moving in relation to me. You're at rest in your rest frame (inertial as I said before) and I am at rest in my own rest frame. But at the same time you are in motion in my rest frame and I am at motion in your rest frame.
This is my postulate:QuoteQuotec is constant in relation to every observer in his own inertial frame.
c is constant in relation to every observer in his own inertial frame.
Please tell me which of these space time diagrams is from the spaceships rest frame (what you call it's inertial frame)
This one:
Quote from: CrazyScientist on 05/05/2021 00:36:22This one:Then the speed of light is not c in the rest frame of the spaceship. Which violates your postulates.
It is not c in relation to moving receiver - so it's in 100% consistent with my postulate.
Quote from: CrazyScientist on 05/05/2021 01:43:41It is not c in relation to moving receiver - so it's in 100% consistent with my postulate.You're unbelievable. You admit that the space time diagram is from the rest frame of the spaceship with the speed of light at .5c and then say it doesn't violate your postulate.
To remind you, you said "c is constant in relation to every observer in his own inertial frame".
To reflect from this obvious inconsistency you wave your arms about the speed of light relative to the receiver, which has nothing to do with the rest frame of the spaceship. Let me guess, if we remove the receiver from the diagram then the speed of light will snap back to c right?