The Relativity of Simultaneity

[Neddy Bate]

Second, GPS has to be frequently updated for it to maintain any sort of accuracy, so don't pretend it maintains accuracy on its own!

"Frequently updated?" The GPS satellites are designed to go 180 days without any form of contact from the system. They just keep on transmitting whatever time the on-board atomic clock says, as well as the satellites current location relative to the surface of the earth.

If your theory were to be true, GPS probably would not work even if they updated the satellites constantly! How could they account for all the different speeds of light in all of the different directions? It would be a nightmare if it were even possible.

 

[James R]

Are you going to back up your statement that light can be measured to be c in all directions in the box in the example I posted? I am quite interested in seeing your numbers!

It's the same old train problem done over again, so there's little point.

You have a source half way between two walls of the cube. In the cube frame, light has equal distances to travel to get to the two walls. By Einstein's second postulate it travels at the same speed in the two directions. Therefore, the time taken in the cube frame is the same in each direction.

Done.

 

[James R]

Too boring? That's funny, James.

I'm glad you find it amusing.

Really, what's the point in going over the same thing again and again? There's nothing new to add.

Tell me in which direction you can measure light to be c in this box?

I already told you: in every direction.

Try to keep up.

 

[Motor Daddy]

I'm glad you find it amusing.

Really, what's the point in going over the same thing again and again? There's nothing new to add.



I already told you: in every direction.

Try to keep up.

You quit the other thread long before I posted this. You never worked this problem in 3D.

You can not measure the speed of light to be c in any direction in the box, PERIOD! This proves Einstein's second postulate is GARBAGE!

 

[Motor Daddy]

It's the same old train problem done over again, so there's little point.

You have a source half way between two walls of the cube. In the cube frame, light has equal distances to travel to get to the two walls. By Einstein's second postulate it travels at the same speed in the two directions. Therefore, the time taken in the cube frame is the same in each direction.

Done.

Wrong, it is .5 light seconds in length between the center of the cube (the source) and the receivers. It takes .65 seconds for light to travel from the center to the y receiver. Can you explain that?

 

[James R]

You quit the other thread long before I posted this. You never worked this problem in 3D.

Ok. I'll "work the problem".

The cube has side lengths of 1 light-second. The source is at the centre. The light travel time, in the cube frame, is therefore 0.5 seconds from the source to the centre of each wall. If you want the time taken for light to get to a corner of the cube or some other point, I can calculate that for you, too.

You can not measure the speed of light to be c in any direction in the box, PERIOD! This proves Einstein's second postulate is GARBAGE!

Where's the real-world evidence that proves your claim?

Oh, that's right, there isn't any. This proves your postulate is GARBAGE!

 

[Motor Daddy]

"Frequently updated?" The GPS satellites are designed to go 180 days without any form of contact from the system. They just keep on transmitting whatever time the on-board atomic clock says, as well as the satellites current location relative to the surface of the earth.

If your theory were to be true, GPS probably would not work even if they updated the satellites constantly! How could they account for all the different speeds of light in all of the different directions? It would be a nightmare if it were even possible.

What is the distance between the satellite and the surface of the Earth?

 

[Motor Daddy]

Ok. I'll "work the problem".

The cube has side lengths of 1 light-second. The source is at the centre. The light travel time, in the cube frame, is therefore 0.5 seconds from the source to the centre of each wall. If you want the time taken for light to get to a corner of the cube or some other point, I can calculate that for you, too.

You're wrong, James, it takes .65 seconds for light to get from the center of the cube to the y receiver.

 

[Neddy Bate]

It is not radically different.

How far away from the surface of the earth is a satellite?

GPS needs to maintain nanosecond accuracy in order to function as designed. That means the travel time of the signals really should not vary by more than about 0.000000001 seconds.

GPS satellites orbit the earth at about 20,200 km above the surface of the earth, but sometimes signals need to go farther than just straight down to the surface. It takes at least three satellite signals for a GPS receiver to triangulate your location, so at least two of those satellites are not going to be directly over your head.

But of course the earth's "absolute speed" is unknown, so I'll let you work out the maximum possible absolute speed of the earth from those numbers. This should be interesting!

 

[James R]

Wrong, it is .5 light seconds in length between the center of the cube (the source) and the receivers. It takes .65 seconds for light to travel from the center to the y receiver. Can you explain that?

Maybe.

In which frame of reference does it take 0.65 seconds?

Also, just let me clarify. Are you saying the cube moves or the source? In which direction does it move? Is it the x direction or the y direction?

Regardless of the answers to these questions, in the cube frame the travel times are 0.5 seconds to the walls in all directions.

 

[Motor Daddy]

Maybe.

In which frame of reference does it take 0.65 seconds?

Also, just let me clarify. Are you saying the cube moves or the source? In which direction does it move? Is it the x direction or the y direction?

In the box frame of reference inside the box it takes .65 seconds for light to get to the y receiver.

The source remains at the center of the cube at all times.

The cube is moving in the x direction and the velocity and time is noted.

Regardless of the answers to these questions, in the cube frame the travel times are 0.5 seconds to the walls in all directions.

No they are not. You show me your math if you think the times can be .5 seconds. Plus, you need to tell me the location of the x and y receivers.

 

[James R]

In the box frame of reference inside the box it takes .65 seconds for light to get to the y receiver.

The source remains at the center of the cube at all times.

The cube is moving in the x direction and the velocity and time is noted.

Where is this y receiver of yours, exactly? Are you saying it is in the exact centre of the cube wall and moves with the cube?

If so, then in the reference frame of the source, the y receiver moves in the x direction at speed v (the speed of the cube).

What is the speed, v, of the cube? Once you tell me that, I can calculate the time taken, as measured in the source frame. I have already given you the time in the cube frame.

You show me your math if you think the times can be .5 seconds. Plus, you need to tell me the location of the x and y receivers.

In the cube frame, the source is located at (x',y',z')=(0,0,0).
In the cube frame, the y receiver is located at (x',y',z')=(0, 0.5 light-seconds, 0).
In the cube frame, light travels at 1 light-second per second (Einstein's second postulate).
In the cube frame the light must travel a distance of 0.5 light-seconds, which takes 0.5 seconds.

In the source frame, the travel time will depend on the velocity of the cube relative to the source, so I need to know what v is.

 

[James R]

v(x)=.63897 c

I pulled this quote from a post of yours above. If I assume that this is the speed of the cube relative to the source then:

The distance in the source frame for light to go from source to y receiver is (in light-seconds, where c=1 light-second per second):

$$d = \sqrt{(0.5)^2 + (0.63897c)t} = ct$$

where I have used Einstein's 2nd postulate again to say that the speed of light is c in the source frame. Therefore:

$$(0.5)^2 + (0.63897c)t -c^2t^2 = 0$$

Solving this using the quadratic formula gives:

$$t = \frac{-0.63897c \pm \sqrt{(0.63897c)^2 + 4c^2(0.5)^2}}{-2c^2}$$

Putting c=1 in these units and solving gives:

t = 0.913 seconds

This is the light travel time in the source frame to get to the y detector. (I hope I haven't made a mistake in the calculation. It's possible.)

 

[arfa brane]

Again, nobody disputes the speed of light is constant.

There is at least one person who does dispute the constant speed of light, that person is you.

You have claimed, repeatedly, that the measured speed of light depends on the absolute velocity of the frame the measurement is made in. Using logic, that means you can't determine the actual speed of light by measuring it in any frame which is moving.

In that case, the "measured" speed of light is different to the "determined" speed of light, unless the determination is a result of measurement. Are you aware of a way to determine the speed of light without measuring the speed of light?

That would make you some kind of a genius, alright. Hell yeah.

 

[Neddy Bate]

Alright here are my calculations for the "maximum absolute speed" of the earth, according to MD's theory in regards to GPS:

For simplicity, I assume the GPS satellite signals are sent directly to the earth's surface, so the distance the signal travels is simply 20,200km. I also assume there is one GPS satellite oriented in such a way that the signal travels in the same direction as the earth's "absolute velocity". Then, 6 hours later, this same satellite would be oriented in such a way that the signal travels in the opposite direction to the earth's "absolute velocity".

Now, GPS requires that the maximum travel time fluctuation over that 6 hour period to be about 0.000000001 seconds, or else the accuracy of GPS starts to become unacceptable. Therefore, I have this equation:

0.000000001 = (20,200/(299,792-v)) - (20,200/(299,792+v))

Which tells me that the "absolute speed" of the earth can only be approximately v=0.003km/s maximum, according to MD's theory.

There you go, MD. It looks like GPS indicates that your theory can only be correct if the earth has an absolute speed of approximately 0.003km/s maximum. Does that sound reasonable to you? Hint: The speed of the earth relative to the sun is about 29km/s. D'oh!

 

[Neddy Bate]

I pulled this quote from a post of yours above. If I assume that this is the speed of the cube relative to the source then:

The distance in the source frame for light to go from source to y receiver is (in light-seconds, where c=1 light-second per second):

$$d = \sqrt{(0.5)^2 + (0.63897c)t} = ct$$

where I have used Einstein's 2nd postulate again to say that the speed of light is c in the source frame. Therefore:

$$(0.5)^2 + (0.63897c)t -c^2t^2 = 0$$

Solving this using the quadratic formula gives:

$$t = \frac{-0.63897c \pm \sqrt{(0.63897c)^2 + 4c^2(0.5)^2}}{-2c^2}$$

Putting c=1 in these units and solving gives:

t = 0.913 seconds

This is the light travel time in the source frame to get to the y detector. (I hope I haven't made a mistake in the calculation. It's possible.)

JamesR,

A while ago, I helped Motor Daddy figure out how to calculate the "absolute velocity" of a cube from inside the cube. We used MD's own version of the universe, which does not include length contraction or time dilation. The cube's velocity through the "absolute rest frame" was allowed to have a three-dimensional direction vector. These are the equations I came up with:

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

and

$$
(ct_y)^2 = (t_yv_x)^2 + (t_yv_y+L)^2 + (t_yv_z)^2
$$

and

$$
(ct_z)^2 = (t_zv_x)^2 + (t_zv_y)^2 + (t_zv_z+L)^2
$$

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

and $$t_y$$ is the measured signal transit time along the y-axis of the cube,

and $$t_z$$ is the measured signal transit time along the z-axis of the cube,

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

and $$v_y$$ is the component of the absolute velocity along the y-axis,

and $$v_z$$ is the component of the absolute velocity along the z-axis,

and $$L$$ is the length of the signal path relative to the cube.

I never could figure out how to solve these equations simultaneously...

 

[quantum_wave]

I am typing this very slowly in hopes that it will sink in this time. It is not theory it is actual measurement. The speed of light in a vacuum is always measured at the same speed regardless of the inertial frame.


Or we could use the special relativity (You know that robust theory that has been shown to be correct for 100 years) and realize that time and length contraction come into play. The wave front (speed of light) never changes speed in a vacuum.

It was very kind of you to type so slowly. Smart guys like you are needed to keep use dullards up to speed. Oh gosh, it is time dilation and length contraction. Wow, thank you. I’ll be able to sleep from now on.

I’ll certainly lose interest in the issue now that I know that there is such a robust theory out there. Oh wait, I knew that. I won’t concern my pea brain with how it all works, I’m just glad that the great minds out there know how it works. You do don’t you? You’re not just saying that it is geometry that causes the SR effects are you. No, that would be another issue I guess. Oh wait, why it is another issue; that is what this is all about IMHO.

Bottom line is that dopes like me don’t know how time dilation and length contraction work. They can be quantified mathematically but we don’t know the physics or the mechanics of how they work. Without them we have the mystery that is solved by time dilation and length contraction and about which you typed slowly to inform me. I don’t question the math but certainly you can give me a link that explains the mechanics and forces. Thanks in advance.

 

[Motor Daddy]

I pulled this quote from a post of yours above. If I assume that this is the speed of the cube relative to the source then:

The distance in the source frame for light to go from source to y receiver is (in light-seconds, where c=1 light-second per second):

$$d = \sqrt{(0.5)^2 + (0.63897c)t} = ct$$

where I have used Einstein's 2nd postulate again to say that the speed of light is c in the source frame. Therefore:

$$(0.5)^2 + (0.63897c)t -c^2t^2 = 0$$

Solving this using the quadratic formula gives:

$$t = \frac{-0.63897c \pm \sqrt{(0.63897c)^2 + 4c^2(0.5)^2}}{-2c^2}$$

Putting c=1 in these units and solving gives:

t = 0.913 seconds

This is the light travel time in the source frame to get to the y detector. (I hope I haven't made a mistake in the calculation. It's possible.)

Wrong again, James. There is no dispute that the y receiver is in a specific location at .65 seconds. What locations do you think the y receiver is at .65 seconds and .913 seconds?

You will not get your length contraction to work, even after I tell you the absolute velocity of the cube.

Try again!

 

[quantum_wave]

OK, well maybe you have a slightly different take on MD's argument. I don't think MD would have any problem with the idea that the speed of light would be measured to change when comparing different frames. For example, he says that the speed of light is 299,792 km/s in the absolute rest frame only. Thus, in a reference frame that has an absolute speed of 1,000 km/s, he says that the speed of light would be measured to be 298,792 km/s in one direction, and 301,792 km/s in the other direction.

On the other hand, it seems that you, quantum_wave, are concerned that if the speed of light could somehow be 299,792 km/s in all reference frames, that the wave front would have to change speed? Perhaps reading too much MD has got you all confused?




Yes, according to MD's postulate, the speed would be measured to be different.




Oh, okay, I think I see what you mean now. You are saying that, in order for Einstein's light postulate to hold true, the speed would have to change from c+v (for example) to c. That is true enough, and it can be explained. Relativity employs these three means to achieve that end:

1. time dilation
2. length contraction
3. relativity of simultaneity

You can derive all of these things from Einstein's two postulates.

When I got up and noticed all the posting activity I decided that this thread was not going to get the the issue as I saw it, that of how time dilation and length contraction work. Not how "mathematically" because they work if both the speed of light is constant in all frames and if the the principle of relativity is correct. The math works and fully supports both time dilation and length contraction. I hope that acknowledgement satisfies you.

 

[Motor Daddy]

JamesR,

A while ago, I helped Motor Daddy figure out how to calculate the "absolute velocity" of a cube from inside the cube. We used MD's own version of the universe, which does not include length contraction or time dilation. The cube's velocity through the "absolute rest frame" was allowed to have a three-dimensional direction vector. These are the equations I came up with:

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

and

$$
(ct_y)^2 = (t_yv_x)^2 + (t_yv_y+L)^2 + (t_yv_z)^2
$$

and

$$
(ct_z)^2 = (t_zv_x)^2 + (t_zv_y)^2 + (t_zv_z+L)^2
$$

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

and $$t_y$$ is the measured signal transit time along the y-axis of the cube,

and $$t_z$$ is the measured signal transit time along the z-axis of the cube,

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

and $$v_y$$ is the component of the absolute velocity along the y-axis,

and $$v_z$$ is the component of the absolute velocity along the z-axis,

and $$L$$ is the length of the signal path relative to the cube.

I never could figure out how to solve these equations simultaneously...

You didn't help me figure out the absolute velocity of the cube, you helped me figure out the time difference in each receiver, by teaching me how to use a coordinate system to tell exactly where each receiver is at an exact time, and I thank you for that. Knowing the exact coordinates of the receivers proves my point that light simply can't be measured to be c in any location of a box that has an absolute velocity greater than zero!

As you can see at the bottom of that post, my v=(ct-l)/t is correct, which I knew long ago, since I first told you the velocity of a country without first knowing the length of the country.