The Relativity of Simultaneity

Neddy Bate said:
Nice diagram, Motor Daddy!

In between frame 1 and frame 2, there is a time when the light hits the left wall of the cube. By my calculations, that happens at T=0.305069 seconds. To calculate that time, I used this equation:

$$
(ct_x)^2 = (t_xv_x-L)^2 + (t_xv_y)^2 + (t_xv_z)^2
$$

Where $$t_x$$ is the measured signal transit time along the negative x-axis of the cube,

and $$v_x$$ is the component of the absolute velocity along the positive x-axis,

and $$L$$ is the length of the signal path relative to the cube.
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That time is correct, it took .305069 seconds for light to reach that wall's receiver.
 
quantum_wave said:
Ned, is this the frame you are talking about?

Nedframe1.jpg
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No, it looks more like this:

attachment.php
 
Motor Daddy said:
Can relativity tell me the absolute velocity of the box?

Does relativity acknowledge that it takes .65 seconds for light to travel from the source to the y and z receivers?
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Relativity can tell you the velocity of this box. Your diagram takes place entirely in a reference frame through which the box is moving.
 
Neddy Bate said:
Relativity can tell you the velocity of this box. Your diagram takes place entirely in a reference frame through which the box is moving.
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Oh, good.

So the times of the y and z receivers are .832 seconds. What is the time at the x receiver and what is the absolute velocity of the box?
 
Motor Daddy said:
Oh, good.

So the times of the y and z receivers are .832 seconds. What is the time at the x receiver and what is the absolute velocity of the box?
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The hypothetical box-guy should be able to measure all three times. If you tell me the x, y and z times, I should be able to tell you the velocity.
 
Neddy Bate said:
The hypothetical box-guy should be able to measure all three times. If you tell me the x, y and z times, I should be able to tell you the velocity.
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Well the x component clock's battery died, so all he has is the y and z times.

Can relativity tell me the absolute velocity of the box and the time to the x receiver?
 
Motor Daddy said:
Are you implying the times at y and z could be the same and have different component velocities?

Yes, the y and z component velocities are zero.
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If the y and z component velocities were not zero, there might be more than one answer.


Motor Daddy said:
Well the x component clock's battery died, so all he has is the y and z times.

Can relativity tell me the absolute velocity of the box and the time to the x receiver?
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Since we know the y and z component velocities are zero, then the x component of the velocity is about 0.8c, and the time to reach the x receiver would be about 2.5 seconds.
 
Neddy Bate said:
If the y and z component velocities were not zero, there might be more than one answer.
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If the y and z receiver times are the same, the y and z component velocities are zero, regardless if they both read .5 seconds or .9345 seconds.




Neddy Bate said:
Since we know the y and z component velocities are zero, then the x component of the velocity is about 0.8c, and the time to reach the x receiver would be about 2.5 seconds.
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About? Give me exact numbers and show me the math according to Einstein's relativity. Show me the length between the source and the receiver in the cube frame according to relativity. Tell me the coordinates of all three receivers when the light reaches the x component receiver.

How does relativity resolve the issue of the light taking .832 seconds to travel from the source to the y and z receivers, which is .5 light seconds in length in the cube frame?
 
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Motor Daddy said:
If the y and z receiver times are the same, the y and z component velocities are zero, regardless if they both read .5 seconds or .9345 seconds.
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Wrong. If the cube has an absolute velocity with equal y and z components, and no x component, then the y and z receiver times would be the same. You need to learn your own theory.


Motor Daddy said:
About? Give me exact numbers and show me the math according to Einstein's relativity. Show me the length between the source and the receiver in the cube frame according to relativity. Tell me the coordinates of all three receivers when the light reaches the x component.
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I gave you my answers, which are accurate to three decimal places. I could make them more accurate if I wanted to, but I know your 0.832 times are only accurate to three decimal places, right?

LOL.

You could at at least admit my answers were correct before asking me to solve more problems.


Motor Daddy said:
How does relativity resolve the issue of the light taking .832 seconds to travel from the source to the y and z receivers, which is .5 light seconds in length in the cube frame?
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Easy. None of these measurements take place in the cube frame. They all take place in the absolute frame through which the cube is moving. Period! Prove otherwise. ;)
 
Neddy Bate said:
Wrong. If the cube has an absolute velocity with equal y and z components, and no x component, then the y and z receiver times would be the same. You need to learn your own theory.
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I said the y and z component velocities were zero and the times were .832 seconds.

Neddy Bate said:
I gave you my answers, which are accurate to three decimal places. I could make them more accurate if I wanted to, but I know your 0.832 times are only accurate to three decimal places, right?

LOL.

You could at at least admit my answers were correct before asking me to solve more problems.
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What are the coordinates of the x, y, and z receivers when the light reaches the x receiver? Quit trying to wiggle and squirm and tell me the coordinates!! :D

Neddy Bate said:
Easy. None of these measurements take place in the cube frame. They all take place in the absolute frame through which the cube is moving. Period! Prove otherwise. ;)
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Uh, maybe you don't understand. I am in the cube. I sent a light from the source in the cube to the receivers in the cube. The receivers show me the times. I am measuring the time it takes for light to travel from the source to the receivers in the cube.
 
Motor Daddy said:
I said the y and z component velocities were zero and the times were .832 seconds.
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You also said, "If the y and z receiver times are the same, the y and z component velocities are zero," which is not correct. That is why 3D equations are so important. I can tell you are still using 1D equations.


Motor Daddy said:
What are the coordinates of the y, y, and z receivers when the light reaches the x receiver? Quit trying to wiggle and squirm and tell me the coordinates!! :D
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You haven't even congratulated me on my correct answers to your previous questions. Why would I answer more questions? :)


Motor Daddy said:
Uh, maybe you don't understand. I am in the cube. I sent a light from the source in the cube to the receivers in the cube. The receivers show me the times. I am measuring the time it takes for light to travel from the source to the receivers in the cube.
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Your diagram takes place in the absolute reference frame. How do you know what is happening in the cube frame? Unless ... You Are the Cube??? :p
 
Neddy Bate said:
You also said, "If the y and z receiver times are the same, the y and z component velocities are zero," which is not correct. That is why 3D equations are so important. I can tell you are still using 1D equations.
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Correct, if the x component velocity is zero, there is a scenario where the y and z components have component velocities greater than zero and equal to each other, with the same times. I did say that the y and z components have zero velocities, though.




Neddy Bate said:
You haven't even congratulated me on my correct answers to your previous questions. Why would I answer more questions? :)
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You haven't given me coordinates.

Neddy Bate said:
Your diagram takes place in the absolute reference frame. How do you know what is happening in the cube frame? Unless ... You Are the Cube??? :p
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I am in the cube. I measured the times. I calculated the absolute velocity of the box, and I made a diagram to show how I can resolve the issue of the constancy of the speed of light remaining true, while at the same time light being measured to be traveling at different speeds in my cube. I concluded that the cube must have an absolute velocity, even though I have nothing to relate to, nor can I feel any acceleration or motion.
 
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There is an absolute point of reference and a box moving relative to that point in space. What is being invoked here is the invariant speed of light and how light that traverses space from that absolute point will be measured in a moving light box. Ned, you have pointed out before that the box can have motion other than rectilinear. Why don't we do a graphic with some other form of motion along two axes?
 
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Motor Daddy,


This was your question to me:


Motor Daddy said:
Oh, good.

So the times of the y and z receivers are .832 seconds. What is the time at the x receiver and what is the absolute velocity of the box?
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My answer:


Neddy Bate said:
Since we know the y and z component velocities are zero, then the x component of the velocity is about 0.8c, and the time to reach the x receiver would be about 2.5 seconds.
Click to expand...


Do you agree I was correct?
 
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