[CrazyScientist (OP)]

@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.

[Colin2B]

No. 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.

[CrazyScientist (OP)]

No. 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.

Thanks! Then maybe I will use the term "rest frame" instead. Will it be ok? What i ment by "inertial frame" is a frame, where the object of interest appears to be at rest.

[jeffreyH]

@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

[CrazyScientist (OP)]

@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

I'm doing my best

[CrazyScientist (OP)]

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:

I'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...

[jeffreyH]

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:

I'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.

[CrazyScientist (OP)]

Have 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.

[jeffreyH]

Have 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.

[CrazyScientist (OP)]

Have 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.

I'm trying to. Look for example at my response to Colin2B couple posts above. He had a CONCRETE objection and I thanked him for it.

[Origin]

No. A frame, where observer is moving is not his frame

I 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.

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

Not according to the space time diagram.

I 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.

I said, that it's constant in relation to observer in his own inertial frame

That is not what you drew in your space time diagram!

Wrong ship frame.jpg (30.78 kB . 600x450 - viewed 4467 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. 

ship frame.jpg (21.29 kB . 600x450 - viewed 2695 times)
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.

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

This 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.

[CrazyScientist (OP)]

No. A frame, where observer is moving is not his frame

I 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.

Ok. First of all, let's follow Colin2B advise and stop using the term "inertial frame" - it is causing confusion, when discussing Galilean relativity. What I ment by this term, is a frame, where the object of interest (e.g. the observer) appears to be at rests, so from now on, I will now only use the term "rest frame".

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. Those things suppose to be well known by someone, who professionally deals with physics..

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

Not according to the space time diagram.

Yes - according to those both diagrams light emitted 4ly away from the sensor (blue line) is reaching it 4 years after the emission at t=0

Maybe you're reading all those diagrams incorrectly....

I 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.

"That's to bad, because experimentation shows that the speed of light is c in ALL inertial frames"

It is impossible to measure the speed of light in a moving frame - at least not in one-directional motion of light...
I won't even count, how many times I said it before...

I said, that it's constant in relation to observer in his own inertial frame

That is not what you drew in your space time diagram!

Wrong ship frame.jpg (30.78 kB . 600x450 - viewed 4467 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. 

Of course it does - why wouldn't it? You say:

That means the speed of light is not constant in the spaceships inertial frame

And I didn't say it has to. I said that c is NOT constant as long, as light moves in relation to a MOVING frame - and the receiver is here in a MOVING 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.

This:

the speed of light is constant in all inertial frames

is not my postulate

This is my postulate:

c is constant in relation to every observer in his own inertial frame

You are just not capable to comprehend the difference

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

This 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.

It is in 100% consistent with my postulate ("c is constant in relation to every observer in his own inertial frame"). Speed of light in relation to a moving frame is NOT constant.

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.

Without the receiver speed of light is immeasurable...

[CrazyScientist (OP)]

Maybe this will help:

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.

Light moves at constant c in relation to me in my own rest frame and it moves at constant c in relation to you in your own rest frame. Light DOESN'T move at constant c in relation to you in my own rest frame and it DOESN'T move at constant c in relation to me in your own rest frame...

It really can't be explained more clearly. If you still won't be able to understand it, then it won't look good for your intelligence...

[Origin]

You are not arguing in good faith.

This is my postulate:
Quote

c is constant in relation to every observer in his own inertial frame.

OK.

Wrong ship frame.jpg (30.78 kB . 600x450 - viewed 5871 times)

Laser frame.jpg (22.46 kB . 600x450 - viewed 4242 times)quote


Please tell me which of these space time diagrams is from the spaceships rest frame (what you call it's inertial frame).

[CrazyScientist (OP)]

Please tell me which of these space time diagrams is from the spaceships rest frame (what you call it's inertial frame)

This one:

And I won't call it "inertial frame" anymore. Now I'll call it only as the "rest fame" - space ship is at rest

[Origin]

This one:

Then the speed of light is not c in the rest frame of the spaceship.  Which violates your postulates.

[CrazyScientist (OP)]

This 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.
Time at which light is being recorded by the receiver is measured in the rest frame of that receiver - and in that frame light is moving at constant c in relation to receiver at rest

[CrazyScientist (OP)]

here let me repeat this once again:

Let's say you are moving in relation to me. You're at rest in your rest frame 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.

Light moves at constant c in relation to me in my own rest frame and it moves at constant c in relation to you in your own rest frame. Light DOESN'T move at constant c in relation to you in my rest frame and it DOESN'T move at constant c in relation to me in your rest frame

[Origin]

It 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 deflect 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?

[CrazyScientist (OP)]

It 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.

Yes

To remind you, you said "c is constant in relation to every observer in his own inertial frame".

Yes

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?

No. It will "snap" back to c if the receiver is at rest in relation to spaceship - but it isn't...