On Einstein's explanation of the invariance of c

...and yet you still can't tell me why #912 is wrong.

Einstein's case relies on the relativity of simultaneity. #912 shows it to be false.

 

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I understand. I just gave an example with some inflated times evidently. The measured times will be correct. Are you saying the equation is wrong or that my times are unrealistic in the example?

 

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#912 is wrong for the same reasons all your other posts are wrong.

The only false is your belief that you understand the same problem Einstein found a solution for. You are the only person alive who can see why he (not you) made a big mistake. You no doubt also believe that nobody else in the last century has believed the same thing as you, and has been shown to be wrong. Einstein unfortunately for you, was right, you're the one who is wrong and you keep making the same basic errors in thinking, over and over.

You're nowhere near as 'intelligent' as he was, or as a lot of other people are or were. Why can't you just admit you should leave it to the bigger kids? What makes you think your basic to the point of being useless math says anything useful?

You have less than a coherent idea of what it is you're trying to do. The problem is nowhere near as simple as you appear to think. But then, you can't think. What you do is go around in the same closed circle.

Plenty of people have "tried" to explain to you where these continual mistakes are, but because you're a dumbass, you think it's just fine to ignore questions or explanations and stick to your ridiculous half-assed model. The one you can't seem to explain, you know, that one.

So stop posting shit like "you can't tell me why it's wrong", and the especially worn-out "don't you understand that". YOU don't understand, and the sooner you understand this the better (maybe we can all get some sleep).

 

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Instead of calling me names why don't you show me some of your superior intelligence and point out my mistakes in #912?

 

It could be because I'm intelligent enough to suspect strongly that it would be a complete waste of time, given the history of this thread.

Only one more to go and there will be 50 pages of crap. (a sciforums record?)

 

Neddy Bate, are you saying that my times in the example are unrealistic, or that my equation is incorrect?

 

No problem wasting time calling names, but you can't be bothered to explain my mistakes in #912?

 

No, the problem is with wasting time trying to explain all of your mistakes, there are too many and you ignore all the explanations.

That has to be the definition of a waste of time.

 

Arfa Brane: Not even close.

Paul Dixon has the record with his thread relating to Sol going supernova due to the Cern & other particle accelerators causing a transition into DeSitter (?spelling) space.

 

Neddy Bate, I acknowledge the light travel times in the other example were wrong. I've corrected the numbers in this example. Lets consider the train and the skate for a moment, with the skate traveling 90 degrees to the train.


Light traveled 29,979,245.8 meters in .1 seconds along the hypotenuse.

a^2+b^2=c^2

The distance light traveled in .1 seconds is c in the theorem, so c^2=898,755,178,736,817.64 meters


a^2=449,377,589,368,408.82 meters
b^2=449,377,589,368,408.82 meters

a=21,198,528.000038323887394410859085 meters
b=21,198,528.000038323887394410859085 meters

In the a and b frames (train and skate frames), light is measured to travel 21,198,528.000038323887394410859085 meters in .1 seconds.

In the a and b frames (train and skate frames), light travels at the velocity of 211,985,280.00038323887394410859085 m/s.

That means the train and skate are traveling an absolute velocity of 87,807,177.9996167611260558914092 m/s (299,792,458 - 211,985,280.00038323887394410859085 = 87,807,177.9996167611260558914092).

If two clocks are .21198528000038323887394410859085 meters apart from each other in the train and skate frames, it takes light .000000001 seconds to travel the distance between the clocks. That means in the train and skate frames, light travels at 211,985,280.00038323887394410859085 m/s. Why? Because the frames have an absolute velocity of 87,807,177.9996167611260558914092 m/s, so it takes light more time (in this case) to travel a length in that frame than it would if the frame had a zero velocity. Again, 211,985,280.00038323887394410859085 m/s + 87,807,177.9996167611260558914092 m/s = 299,792,458 m/s

So again, using v = (ct-l)/t

The distance between the clocks is .21198528000038323887394410859085 meters. It takes light .000000001 seconds to travel from one clock to the other clock. Of course c is 299,792,458 m/s.

The absolute velocity of the train and skate is 87,807,177.9996167611260558914092 m/s.

H = sqrt(.21198528000038323887394410859085^2 + .21198528000038323887394410859085^2)
H = sqrt(0.044937758936840881999999999999638 + 0.044937758936840881999999999999638)
H = sqrt(0.089875517873681763999999999998638)
H = 0.29979245799999999999999999999773

0.29979245799999999999999999999773 meters in .000000001 seconds is 299,792,458 m/s
Light travels at 299,792,458 m/s along the hypotenuse.

My equation is correct.

 

Oh, great! I'm glad you agree that there was a problem with those last numbers.

This is ok, I guess. You've got A and B equal to each other, but they don't have to be. What A really represents is how far the train traveled in 0.1 second. What B really represents is how far the skate traveled relative to the train, PLUS the length of the skate. Remember, the light ray is actually traveling down the edge of the skate, so the length of the skate is part of the B leg of the triangle.

No. The distance that the train traveled is A, as measured from the absolute frame. The train has no windows, so it does not know how far it moved until you complete your light test. The train can see the skate moving, and the distance the skate moves is B-L where L is the length of the skate.

But the skate is another closed box with no windows. Inside the skate box, the skater only knows that the light traveled down the skate, which is a distance of L. If your time is correct, then the skater also knows the light traveled traveled a distance of L in 0.1 seconds. This is where you are now. Keep gong, you're getting there!

 

a and b are the distance light travels along the x and y directions in reference to the train. In order to get to x,y (21,198,528.000038323887394410859085 ,21,198,528.000038323887394410859085), light had to travel the hypotenuse, which was 29,979,245.8 meters along the train.

The train and the skate each have absolute velocities of 87,807,177.9996167611260558914092 m/s.

So in .1 seconds, the train traveled 8,780,717.79996167611260558914092 meters in the x direction, and the skate traveled 8,780,717.79996167611260558914092 meters in the y direction.

Since the train traveled 8,780,717.79996167611260558914092 meters in the x direction in .1 seconds, and light traveled in the train in the x direction
21,198,528.000038323887394410859085 meters in the x direction in .1 seconds, the light traveled a total distance of 29,979,245.8 meters in .1 seconds which is the speed of light. But again, in the train frame, light only traveled 21,198,528.000038323887394410859085 meters in the x direction in .1 seconds, which means in the train frame light is measured to be 211,985,280.0003832388739441085908 m/s, as the train observers are unaware of the train's 87,807,177.9996167611260558914092 m/s absolute velocity.

 

So how long is the skate?

 

Sorry, I was editing the post when you typed.

 

Say I have a cube that is one lightsecond in length, per side. I place that cube at the origin point (0,0,0) of the absolute frame. Let the units of length be lightseconds, so that these points represent corners of the cube while it is at rest:

(1,0,0)
(0,1,0)
(0,0,1)

Next, I choose a random absolute velocity for the cube, and tell you that it takes light the following times:

1.201 seconds to reach the corner which used to be at (1,0,0).
1.336 seconds to reach the corner which used to be at (0,1,0).
1.482 seconds to reach the corner which used to be at (0,0,1).

Now all you have to do is tell me the absolute velocity of the cube. Hint: the absolute velocity has three components, because the cube is actually moving down the x, y, and z axes at the same time. Are you up for it, Motor Daddy?

 

The length of the skate is irrelevant. Think of the skate as a big box, traveling inside an even bigger box (the train.)

 

I told you I'm not a mathematician. Are you trying to wear me out?

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You know I'm right so can you show me how you would do it using my equation and methods?

 

The whole idea was to send a light signals down the skate! lol.

 

But the skate is just another box, and you are inside that box using light to determine the velocity of that box. In the train's frame the numbers will add up.

 

Motor Daddy,
Here is my calculation for the above problem. I think you will find that your equations are not capable of solving for these numbers. But feel free to try your equations, and see if you can make them work.

When the cube is at rest, the PRIMARY corner of the cube is located at (0,0,0). We will use the primary corner to describe the cube's location. Since the cube is one lightsecond long on each side, three other corners are located at these points when the cube is at rest:
(1,0,0)
(0,1,0)
(0,0,1)

A light sphere is emitted at point (0,0,0) at the same instant as the cube acquires constant absolute velocity through three dimensional space. I have chosen a velocity such that the three components are as follows:
x-component = 0.1c
y-component = 0.2c
z-component = 0.3c

The problem gives us three times. The first time is 1.201 seconds, so the PRIMARY corner of the cube would be located at (0.120, 0.240, 0.360) as you can see from these equations:
x = 1.201(0.1) = 0.120
y = 1.201(0.2) = 0.240
z = 1.201(0.3) = 0.360
And from that, it follows that the corner which used to be at (1,0,0) is now located at (1.120, 0.240, 0.360) because all I had to do was add one to the x-coordinate.

If I want to know exactly how far away that corner is from the origin, I can use this equation:
d = sqrt(1.120^2 + 0.240^2 + 0.360^2)
d = 1.201
And that happens to be how far the light sphere is from the origin, because the time is (1.201 seconds) and the light sphere expands at one lightsecond per second. So we know the light is just now reaching that corner of the cube.



The next time given by the problem is 1.336 seconds, so the PRIMARY corner of the cube would be located at (0.133, 0.267, 0.400) as you can see from these equations:
x = 1.336(0.1) = 0.133
y = 1.336(0.2) = 0.267
z = 1.336(0.3) = 0.400
And from that, it follows that the corner which used to be at (0,1,0) is now located at (0.133, 1.267, 0.400) because all I had to do was add one to the y-coordinate.

If I want to know exactly how far away that corner is from the origin, I can use this equation:
d = sqrt(0.133^2 + 1.267^2 + 0.400^2)
d = 1.336
And that happens to be how far the light sphere is from the origin, because the time is (1.336 seconds) and the light sphere expands at one lightsecond per second. So we know the light is just now reaching that corner of the cube.



The final time given by the problem is 1.482 seconds, so the PRIMARY corner of the cube would be located at (0.148, 0.296, 0.444) as you can see from these equations:
x = 1.482(0.1) = 0.148
y = 1.482(0.2) = 0.296
z = 1.482(0.3) = 0.444
And from that, it follows that the corner which used to be at (0,0,1) is now located at (0.148, 0.296, 1.444) because all I had to do was add one to the z-coordinate.

If I want to know exactly how far away that corner is from the origin, I can use this equation:
d = sqrt(0.148^2 + 0.296^2 + 1.444^2)
d = 1.482
And that happens to be how far the light sphere is from the origin, because the time is (1.482 seconds) and the light sphere expands at one lightsecond per second. So we know the light is just now reaching that corner of the cube.