Strange Claims About Relativistic Time Split From A Alternative Theories Thread

[xinhangshen]

A-wal, I already told you that the image of clock A takes 10 minutes of the time of clock A to reach observer B and then it takes another 10 minutes of the time of clock A for the image of clock B to travel from B to A. The total time lapse of clock A is 20 minutes. Therefore, at 3:20 of clock A, observer A receives the image of clock B. How can you get 3:45 of the time of clock A when observer A receives the image of clock B?

Edited September 1, 2016 by xinhangshen

[A-wal]

Because they're moving away from each other so it will take longer for the second signal to travel the greater distance. 3:45 was arbitrary.

 

It's easier if they simultaneously (as observed by a third object sitting equal distantly between them) send each other signals so you can see the it's symmetric and there's no contradiction.

[xinhangshen]

Because they're moving away from each other so it will take longer for the second signal to travel the greater distance. 3:45 was arbitrary.

 

It's easier if they simultaneously (as observed by a third object sitting equal distantly between them) send each other signals so you can see the it's symmetric and there's no contradiction.

A-wal, you just admitted in a previous post that the path for the image of clock A traveling from A to B is exactly the same as the path for the image of clock B traveling from B to A observed in the reference frame of observer A. 

Edited September 2, 2016 by xinhangshen

[A-wal]

Yes! That's the whole point!

 

In your scenario you have A sending a signal to B first and then B sending a signal back once they've received A's signal so the distance between them will be greater when the signal is traveling from B to A than it was when it was traveling from A to B.

 

Just have them sending signals simultaneously to each other so you can see it from both frames at the same time (using the frame of a C observer equally between them). You get the same information but you get it from both sides.

 

Like I said, you're just isolated individual signals when you could simply have them looking at each other so that there's a continuous stream of signals and each sees the other as time dilated. That doesn't change if you use individual instances of signals between them.

[xinhangshen]

Yes! That's the whole point!

 

In your scenario you have A sending a signal to B first and then B sending a signal back once they've received A's signal so the distance between them will be greater when the signal is traveling from B to A than it was when it was traveling from A to B.

 

Just have them sending signals simultaneously to each other so you can see it from both frames at the same time (using the frame of a C observer equally between them). You get the same information but you get it from both sides.

 

Like I said, you're just isolated individual signals when you could simply have them looking at each other so that there's a continuous stream of signals and each sees the other as time dilated. That doesn't change if you use individual instances of signals between them.

A-wal, in the reference frame of observer A, the image of clock A reaching observer B and the image of clock B leaving observer B happen at exact the same location and same time, and the speed of observer B does not have any role here. The distance from A to B is the same as the distance frame B to A. The speed of light is also the same in both directions. Therefore, the times spent on the way for the images are exactly the same.

[A-wal]

Event 1: A sends a signal towards B.

Event 2: B receives A's signal and sends a signal back.

Event 3: A receives B's signal.

 

The distance between them is greater at event 2 than at event 1 and greater at event 3 than at event 2, because they're continuously moving or they would be in different frames.

[xinhangshen]

Event 1: A sends a signal towards B.

Event 2: B receives A's signal and sends a signal back.

Event 3: A receives B's signal.

 

The distance between them is greater at event 2 than at event 1 and greater at event 3 than at event 2, because they're continuously moving or they would be in different frames.

You just don't understand what an event is. An event is the coordinates of a point in the four-dimensional spacetime, which does not have distance. Distance can only exist between events.

 

In the reference frame of observer A, the distance between event 1 and event 2 is exactly the same as the distance between 2 and event 3 because location A is fixed in the reference frame of observer A.

Edited September 2, 2016 by xinhangshen

[A-wal]

Use the third observer who's an equal distance between them.

 

When A sends the signal there's less distance between A and B than there is when B sends the signal.

 

You just don't understand what coordinates are. You could just as easily say that the reference frame of observer B is fixed, it's completely arbitrary. That's why a third observer between them is handy because you can have A and B on an equal footing, they're both moving away from C at the same velocity.

Edited September 2, 2016 by A-wal

[xinhangshen]

Use the third observer who's an equal distance between them.

 

When A sends the signal there's less distance between A and B than there is when B sends the signal.

 

You just don't understand what coordinates are. You could just as easily say that the reference frame of observer B is fixed, it's completely arbitrary. That's why a third observer between them is handy because you can have A and B on an equal footing, they're both moving away from C at the same velocity.

You are just muddling the debate. Please discuss my thought experiment which is to create a contradiction for special relativity.  

[sluggo]

It seems there is a need to further explain the problem of special relativity in my thought experiment: 

 

Suppose there are two observers: A and B. They have a constant speed relative to each other. Each person has a clock with him, and thus the position of each observer is clearly known by the other observer in his reference frame. Now observer A wants to know the time of clock B. For example, at 3 o'clock of clock A, observer A sends the image of clock A to observer B. When observer B receives the image, he will immediately send the image of clock B back to observe A. According to special relativity, observer B will see that the time shown on the image of clock A (3 o'clock) will be earlier than the time of clock B (for example 3:15) minus the time for the image to travel from A to B (for example 10 minutes calculated by the position of clock B in the reference frame of observer A divided by constant c, the speed of light) because of time dilation. Now what will observer A see when the image of clock B arrives at observer A? Observer A will see that the time on the image of clock B (3:15) will be later than the time of clock A (3:20 calculated by the time he sent the image to B plus the time for the images traveling from A to B and from B to A) minus the time for the image traveling from B to A (10 minutes). That is, observer A will see that clock B is faster than his clock, contradict time dilation as predicted by special relativity.

 

One may argue that the time for an image traveling from A to B and the time for an image traveling from B to A are different observed in the frame of observer A and in the frame of observer B. OK, just do whatever modification as you want. After your modification, then what time will be shown on the image of clock B received by observer A? You will find, no matter how you choose the time on the image of clock B, the faster clock will always be faster observed in both reference frames. If they are identical, then there is only one answer: they have the same time observed in both reference frames (i.e. the time shown on the image of clock B must be 3:10 here).

A sends t at 3:00

B receives t at 3:15

B is aware he is ahead of A by 3:15-3:10=5 min.

B sets his clock to 3:10 and sends t

A receives t at 3:20

A and B know their clocks are synchronized.

 

No one has to adjust for light transit time.

They only need the clock readings at send and receive, and assign one of them to make a clock adjustment, B in this case.

 

The great advantage of the clock synchronization method is, you don't need to know your absolute speed in space, which you can't determine by any known methods.

[xinhangshen]

A sends t at 3:00

B receives t at 3:15

B is aware he is ahead of A by 3:15-3:10=5 min.

B sets his clock to 3:10 and sends t

A receives t at 3:20

A and B know their clocks are synchronized.

 

No one has to adjust for light transit time.

They only need the clock readings at send and receive, and assign one of them to make a clock adjustment, B in this case.

 

The great advantage of the clock synchronization method is, you don't need to know your absolute speed in space, which you can't determine by any known methods.

The situation is that the two clocks are identical relative to their own reference frame. We don't need to synchronize them. We just want to know the time shown on the received image of clock B. If clock B has not been adjusted, the image should show 3:15. Therefore, observer A will also see that clock A runs slower than clock B. That is, clock B is faster than clock A observed by both observers, that contradicts special relativity. Therefore, special relativity is wrong.

Edited September 3, 2016 by xinhangshen

[sluggo]

 

You have to compare intervals for time dilation.
In the drwg:
A interval is 0, .68
B assigns to 0, 1.00
B concludes A clock running slow
B interval is 0, .68
A assigns to 0, 1.00
A concludes B clock running slow
clocks synchronized at 0, not necessary but eliminates any time calculations!
This is still doppler effects.

[xinhangshen]

Now I have realized that there is no logical contradiction in the kinematics of special relativity if we strictly use relativistic time and length everywhere, just as use Newton's kinematics with Galilean time and length everywhere.

 

In fact all inertial reference frames can be described by both coordinate systems: one is based on Lorentz Transformation (called relativistic system) and the other is based on Galilean Transformation (called Galilean system). Relativistic system is mathematically equivalent to Galilean system. But the two coordinate systems result in two different interpretations of time, length and the speed of light: for relativistic system, time and length are relative but the speed of light is absolute, and for Galilean system, time and length are absolute but the speed of light is relative. Both interpretations are logically intact. That is, to describe the motions of objects (i.e. in kinematics), using relativistic system will produce exactly the same results as using Galilean system if we use strictly their respective definitions of time and length.

 

The major advantage of using special relativity is that it creates a set of electromagnetic equations invariant in inertial reference frames. If these equations are not correct, then special relativity is meaningless because of the difficulty in handling its relative time. According to special relativity, times of clocks with relative velocities such as those on GPS satellites flow at different rates and can never been universally synchronized. Because the time is relative in special relativity, there is no clear sequence of time flow in the universe so that we will never be able to correctly record history in relativistic time.

 

The advantage of Galilean coordinate system is that length and time are invariant in all inertial reference frames. Thus, time keeping becomes straightforward. For example, all clocks on GPS satellites show Galilean times that are synchronized not only relative to earth but also relative to each other no matter what relative velocities between them though in the synchronization process, the calculation of the speed of light should strictly follows Newton's velocity addition formula. With the absolute Galilean time, we have a clear sequence of time so that all historic events in the universe can be recorded in order. In the Galilean coordinate system, many modern physics theories are not developed yet, though so called relativistic theories of modern physics may completely wrong.

 

Up to now, all clocks we use in the world are universally synchronized. Thus, we are all using absolute Galilean time. It is a mistake to use our current universally synchronized time as relativistic time with relativistic theories to describe physical phenomena.

[A-wal]

In fact all inertial reference frames can be described by both coordinate systems: one is based on Lorentz Transformation (called relativistic system) and the other is based on Galilean Transformation (called Galilean system).

There's no test that can determine whether or not an object is in motion without comparing it to other objects because motion's true definition is a change in the amount of space between objects over time. That's also true in Galilean relativity but it that the speed of light isn't constant. The fact that we know that its velocity is independent of the motion of the source and so the same in every inertial frame means that SR is the correct transformation.

 

It may be harder to understand but that's just because we're not used to dealing with relatively high velocities.

[xinhangshen]

There's no test that can determine whether or not an object is in motion without comparing it to other objects because motion's true definition is a change in the amount of space between objects over time. That's also true in Galilean relativity but it that the speed of light isn't constant. The fact that we know that its velocity is independent of the motion of the source and so the same in every inertial frame means that SR is the correct transformation.

 

It may be harder to understand but that's just because we're not used to dealing with relatively high velocities.

 I have found the mathematical relations between Galilean spacetime and relativistic spacetime (see my papers available through the links in my previous posts). Every relativistic spacetime point within light reach can be uniquely mapped to a point of Galilean spacetime, and every point of Galilean spacetime within light reach can be uniquely mapped to a relativistic spacetime point. They don't produce any differences in kinematics. That is, the result of any kinematic problem calculated based on relativistic spacetime can be finally converted to the result in Galilean spacetime with the mathematical relationship I presented on the papers, which will be exactly the same as the result directly calculated based on Galilean kinematics, and vice versa. 

[A-wal]

Galilean transformations use a variable speed of light and that's been proven not to be the case. A constant speed of light requires sr.

 

Neither version uses an aether.

[xinhangshen]

Galilean transformations use a variable speed of light and that's been proven not to be the case. A constant speed of light requires sr.

 

Neither version uses an aether.

Do you know that the speed of light can be defined by both Galilean system and relativistic system? The Galilean version of the speed of light is not a constant while the relativistic version is constant. You can use the relationships between Galilean system and relativistic system I presented on my papers to do the conversions. The two definitions of the speed of light are mathematically equivalent.

 

Both Galilean and relativistic systems are logically intact. The clocks on GPS satellites are adjusted to Galilean time (faster than the corresponding relativistic times on individual satellites) because they are all synchronized to the ground clocks while relativistic time can never be used to synchronize clocks with relative speeds. We are all using Galilean time and relativistic time is just an artificial variable introduced to produce an artificial constant speed of light.

Edited September 9, 2016 by xinhangshen