The Relativity of Simultaneity

[James R]

Motor Daddy:

Once again you've shown you're not interested in the reality, just what you want to believe. By all means, stick with those beliefs if it makes you happy. Ignorance is in fact, bliss!!

In reality, objects can and do have motion in space. Only in Einstein's fantasy world does everyone get to assume to be at rest.

You keep talking about "reality" as if you had some kind of experimental evidence.

But we all know you have nothing, nada, zero, zilch, zip in terms of evidence from "reality".

So what is this "reality" of yours? Nothing but your imagination.

Don't talk to me about "reality" until you've gone and looked at it.

The only way it takes .5 seconds for light to get to all the receivers, simultaneously, is if the cube has an absolute zero velocity. If the cube has a velocity, it is impossible for light to arrive at each receiver simultaneously.

In the frame of the source, yes. In the frame of the cube, no.

It's all about reference frames, Motor Daddy. While you insist on only working in one and ignoring all others, you'll never make any progress.

Furthermore, if the cube has a velocity in space in the x direction, the time it takes light to reach the y and the z receivers is different, even though they have zero component velocities.

This is just wrong.

SR has no explanation, because:

1. SR assumes the speed of light is always measured to be c in all frames (which it is clearly NOT in this correct example)

2. SR assumes length contraction, which will prove to be incorrect in this example.

3. SR assumes the box is at rest, and that the box doesn't have a velocity, because Einstein doesn't acknowledge that the box could have an absolute velocity.

1. Yes, this is what SR assumes. You assume the wrong thing; SR assumes the right thing. That makes you wrong, not SR.

2. Length contraction is a direct consequence of the postulates of SR. You can't disprove it by working with different postulates. Of course different postulates will give you different answers. The only way to disprove length contraction is to do real-world experiments. How many have you done? None.

3. SR assumes the box is at rest in the box frame. SR assumes the box is moving in the source frame. SR can use any of a million different frames with no trouble at all. There is no absolute velocity.

We are inside the cube, and we sent a light to the y receiver .5 light seconds away from the source at the center of the cube. The light made it to the receiver in .65 seconds.

Show me your calculation of this. Questions:

1. What is the side length of the cube?
2. What is the velocity of the cube?
3. Was the source located at the centre of the cube at t=0 seconds?
4. Was the y receiver at the same x,z location as the source at t=0?
5. How far did the light travel from the source in .65 seconds?
6. How far from the source was the y receiver after 0.65 seconds?

There is no external frame to relate to, and we don't know if the box is in motion or not.

Depends what you mean by "external frame". There are certainly at least two frames in the problem: the rest frame of the source and the rest frame of the box.

 

[James R]

DonQuixote:

This meter-defined-by-light-speed seems very odd to me, but as far as I can tell you are right that this is how it is presently defined. .... Since relativists dominates the scientific community, they probably had a hand in the adoption of this definition. ....
I'm probably missing something here...

The metre used to be defined as the distance between two marks on a particular metal bar kept in Paris.

The speed of light had to be measured, and the result was a number in metres per second.

Over time, the accuracy of experimental methods to measure the speed of light increased dramatically, until a point was reached in which the experimental error in the measurements of the speed of light was far lower than the experimental error in measuring the length between the two lines on that metal bar in Paris.

At that point, it became more sensible to define the metre in terms of the speed of light, because the speed of light was known more accurately than the distance on the bar could be measured using any other method.

So, a new definition was adopted.

This definition of a metre in terms of the speed of light does not depend in any way on which theory of relativity you prefer, except that it is implicit in the definition that the metre in any reference frame is the measured distance that light will travel in exactly 1/299792458 seconds as measured in the same reference frame.

What the definition does NOT say is: the speed of light is 299792458 metres per second in ONLY one special "absolute" frame called the Motor Daddy imaginary fantasy reference frame.

 

[Motor Daddy]

Good. Have you done a graphic that shows this. Maybe start with a cube (square) with a point in the center as frame one. That would be the point of origin of the light wave. Then frame two shows the box moving away from the point of origin and shows the spherical (circular on the paper) wave expanding within the box. Then frame three could be the point where the wave front reaches the first receptor. BTW, as I think about it, the motion would cause the light wave to take longer to reach the front of the box than to reach the back of the box. It would reach the back of the box first wouldn't it, and then the sides, and finally the front.

I haven't done a graphic.

You understand my position well. The source remains fixed at the center of the cube at all times. The light sphere expands from the point in space it was emitted. If the source remains at the center of the light sphere, that also means the walls of the cube are not in motion, therefore the light sphere reaches all three receivers at the center of each the x, y, and z walls simultaneously at .5 seconds. You know the absolute velocity of the cube is zero because the times are all .5 seconds.

If the cube had an absolute velocity greater than zero, the light sphere would be emitted from the source, and the source would move away from the center of the light sphere, which means the walls also move, which means the light sphere will reach each receiver in a different amount of time than .5 seconds. From that a calculation can be done to know the absolute velocity of the cube in space.

One very important thing to note. If the cube is in motion in only the x direction (as in my example) the light sphere can not contact the y and z receivers in .5 seconds, because even though the y and z component velocities are zero, the receivers are moving in the x direction, and that means the light sphere can't reach them in the distance of .5 light seconds (.5 seconds). Light has to travel farther to reach them.

So in my example, if you were in the cube, motion unknown, you would measure light to take .65 seconds to travel the distance between the source and the y and z receivers which is .5 light seconds in the cube frame. It proves that light is not measured to be c in any direction in the cube frame at anything other than an absolute zero velocity, refuting Einstein's second postulate, which means his theory needs to be file 13'd!

 

[Motor Daddy]

How does this look, MD?

Yup, that's it.

 

[Motor Daddy]

When someone says that there is no acceleration due to gravity because a beach ball does not have an increasing velocity through solid ground that is just ignorance.

Being a laymen does not make you stupid or irrational, but it sure doesn't preclude it!

If you want to respond to that, respond in the proper thread, we have enough going in this thread already.

I never said there was no acceleration due to gravity. You said the acceleration at the SURFACE of the earth was 9.81 m/s^2. Well, I'm at the surface, and my velocity is not increasing or decreasing, so the rate of change of velocity (acceleration) is 0 m/s^2.

You seem to want to pretend that since I'm at the surface of the earth my acceleration is 9.81 m/s^2??? Prove it!!! Tell me how much time it will take for me to reach the center of the earth, if I start at a 0 m/s velocity towards the center, and have an acceleration rate of 9.81 m/s^2???

 

[Motor Daddy]

Motor Daddy:



Apparently, there is, because I get a different answer to you.

Let's go back a step. What is the speed of the cube? Was I correct in using 0.63897c? That's a strange number to pick for an example, but I'm happy to use it if you want to.

Or is the speed of the cube some other number (in which case I need to re-do my calculation)?

The ONLY way I know the absolute velocity of the cube is to first measure the times to the receivers. You don't plan on telling me that those are not the actual times after I tell you the velocity, do you? That would be absurd, James. I measure times, I calculate the velocity, and then you use that velocity to tell me the times are wrong, that they are actually ...???

The times are measured first to each receiver in the cube, and from those times the velocity is known. At that point there is no changing the times or the velocity. It is already recorded in the history books!

If the speed of the cube is 0.63897c in the x direction, then at t=0.65 seconds the y receiver is at:

$$(x,y,z)=(0.63897c \times 0.65 metres,0.5 light-seconds, 0)$$

Again, I'm working on the assumption that the y receiver was at (0,0.5 light seconds, 0) at time t=0. Correct me if that's wrong.

That's correct, the y receiver was at (0,.5,0) at t=0.

 

[Motor Daddy]

In the frame of the source, yes. In the frame of the cube, no.

The source is fixed at the center of the cube, and remains at the center of the cube at all times.

This is just wrong.

No it is not wrong. If the cube has a motion in the x direction, even though the y and z component velocities are zero, the y and z receivers are moving in the x direction. So it is impossible for the light sphere to reach them in .5 seconds. If you think otherwise, tell me what the coordinates are at t=.65 for each the y and z receivers? I will then show you your error.

Show me your calculation of this. Questions:

1. What is the side length of the cube?
2. What is the velocity of the cube?
3. Was the source located at the centre of the cube at t=0 seconds?
4. Was the y receiver at the same x,z location as the source at t=0?
5. How far did the light travel from the source in .65 seconds?
6. How far from the source was the y receiver after 0.65 seconds?

1. 1 light second
2. TBD
3. Yes
4. The source location at t=0 is (0,0,0) and the y receiver location at t=0 is (0,.5,0)
5. Light traveled .65 light seconds from the point in space that the source was when it emitted the light (0,0,0).
6. .5 light seconds.

Depends what you mean by "external frame". There are certainly at least two frames in the problem: the rest frame of the source and the rest frame of the box.

The source is in the same frame as the cube. The source is fixed to the center of the cube and remains there at all times.

 

[quantum_wave]

I have added the motion and some text from Einstein to the graphic. Anything is subject to change, MD, so let me know what you want done. Can you copy this and work with it in Paint? Or do you want additions made and if so specify them.

 

[Motor Daddy]

My art work may not be to scale.

The red dots represent location (0,0,0) where the light sphere was originated at t=0.

The first frame is t=0

The second frame is t=.65 seconds when the light sphere contacts the y and z receivers in my example.

The third frame is t=1.384930 seconds when the light sphere contacts the x receiver.

As you can clearly see, the light can not contact the y and z receivers in .5 seconds if the box has an absolute velocity which it clearly does in this pic.

 

[quantum_wave]

My art work may not be to scale.

The red dots represent location (0,0,0) where the light sphere was originated at t=0.

The first frame is t=0

The second frame is t=.65 when the light sphere contacts the y and z receivers in my example.

The third frame is t=1.384930 when the light sphere contacts the x receiver.

As you can clearly see, the light can not contact the y and z receivers in .5 seconds if the box has an absolute velocity which it clearly does in this pic.

I think you did a good job on that graphic.

Essentially you are talking about the addition of velocities as discussed in Chapter 6, do you agree? There is nothing wrong with that and it marks a point in Einstein's development of SR. Your theory uses the addition of velocities concept in the moving box, and the constant speed of the light sphere from the point of origin, right.

 

[Motor Daddy]

Not really the addition of velocities. You can see that the y and z component velocities are zero, and yet it takes light .65 seconds to travel the distance of .5 light seconds in the cube frame.

If you were in the cube, you would not see the hypotenuse that the light traveled along to get to the receiver, you would only be able to measure that light traveled at the rate of .5/.65=.77c along the distance between the source and the y receiver in the cube frame. In the cube frame, there really is no "hypotenuse."

This is my equation based on the Pythagorean theorem and is correct to find the x component velocity and the x component time, when the y and z component velocities are zero, and their times are known.

v(x)=sqrt(t(y)^2-l(y)^2)/t(y)
t(x)=l(x)/(c-v(x))

 

[quantum_wave]

Not really the addition of velocities. You can see that the y and z component velocities are zero, and yet it takes light .65 seconds to travel the distance of .5 light seconds in the cube frame.

If you consider the point of origin of the light to be fixed, like the embankment is fixed in chapter 6, then the motion of the box equates to the motion of the person walking along the embankment as the light sphere expands. Agree?

If you were in the cube, you would not see the hypotenuse that the light traveled along to get to the receiver, you would only be able to measure that light traveled at the rate of .5/.65=.77c along the distance between the source and the y receiver in the cube frame. In the cube frame, there really is no "hypotenuse."

This is my equation based on the Pythagorean theorem and is correct to find the x component velocity and the x component time, when the y and z component velocities are zero, and their times are known.

v(x)=sqrt(t(y)^2-l(y)^2)/t(y)
t(x)=l(x)/(c-v(x))

Can you show the numbers in your equations that equal .77c; can you show the equations and numbers for the motion of the box?
Then it would be nice to plug them into the graphic .

 

[Motor Daddy]

If you consider the point of origin of the light to be fixed, like the embankment is fixed in chapter 6, then the motion of the box equates to the motion of the person walking along the embankment as the light sphere expands. Agree?

It's not about "considering" the embankment to be at rest, you have to measure the embankment's velocity. Sure you could say the person traveled 20 meters along the embankment in 5 seconds, but that doesn't mean he had an absolute velocity of 4 m/s unless you know the embankment had an absolute zero velocity. For all you know the embankment may have had an absolute velocity of 300 m/s in the opposite direction, and the person traveling along the embankment had a 296 m/s velocity in the same direction.

In order to measure using light you have to measure properly. Assuming the embankment to be at rest, and then using light to measure the person is just plain wrong! That's where Einstein goes astray. He assumes the embankment to be at rest, because he doesn't have a way to measure the absolute velocity of the embankment. So he creates a world of illusions.

Can you show the numbers in your equations that equal .77c; can you show the equations and numbers for the motion of the box?
Then it would be nice to plug them into the graphic .

The times are measured, they are not calculated. Once the times are known you need to figure out each component velocity separately, as I don't have a simple equation to do that. The 3d scenario is very complicated because the cube is traveling and the sphere is growing, so you need to find the exact point on the sphere that contacts the receiver at an exact time. It is VERY VERY complicated, and I am not a mathematician. Maybe someone that can help with the math would be willing to tell us an equation that would work??

 

[quantum_wave]

It's not about "considering" the embankment to be at rest, you have to measure the embankment's velocity. Sure you could say the person traveled 20 meters along the embankment in 5 seconds, but that doesn't mean he had an absolute velocity of 4 m/s unless you know the embankment had an absolute zero velocity. For all you know the embankment may have had an absolute velocity of 300 m/s in the opposite direction, and the person traveling along the embankment had a 296 m/s velocity in the same direction.

In order to measure using light you have to measure properly. Assuming the embankment to be at rest, and then using light to measure the person is just plain wrong! That's where Einstein goes astray. He assumes the embankment to be at rest, because he doesn't have a way to measure the absolute velocity of the embankment. So he creates a world of illusions.

Einstein specifies that the embankment may very will have velocity but can be considered at rest in regard to a given event, like the flash of light, because the origin of that event is fixed relative to the embankment. In your approach it is OK to consider the point of origin fixed relative to the expanding light sphere isn't it? Do you agree that the fixed point of origin could be in motion relative to an observer in the box?

The times are measured, they are not calculated. Once the times are known you need to figure out each component velocity separately, as I don't have a simple equation to do that. The 3d scenario is very complicated because the cube is traveling and the sphere is growing, so you need to find the exact point on the sphere that contacts the receiver at an exact time. It is VERY VERY complicated, and I am not a mathematician. Maybe someone that can help with the math would be willing to tell us an equation that would work??

I'm with you on that and am certain that we can accomplish the calculations.

I am wondering why you gave me the equations for v(x) and t(x) if you aren't using them for calculations?

 

[Motor Daddy]

Einstein specifies that the embankment may very will have velocity but can be considered at rest in regard to a given event, like the flash of light, because the origin of that event is fixed relative to the embankment.

Again, that is like saying the origin of light in the cube is fixed to the source, which it is clearly not. The origin of light is fixed to the location in space that it was emitted (0,0,0) at t=0. The origin of light has nothing to do with being fixed to any object which may or may not have had an absolute velocity.

In your approach it is OK to consider the point of origin fixed relative to the expanding light sphere isn't it? Do you agree that the fixed point of origin could be in motion relative to an observer in the box?

The light sphere expands from the point in space it was emitted, regardless of what the source does after it was emitted.

Again, the fixed point is not in motion, it is a point in space (0,0,0) at t=0. That point is not in motion, it is simply a point in space. It is not an object, it is a location at t=0.

I am wondering why you gave me the equations for v(x) and t(x) if you aren't using them for calculations?

If you simply know the y and z times of .65 seconds, as measured, then you can use that formula to arrive at the x time and absolute velocity in the x direction.

 

[quantum_wave]

Again, that is like saying the origin of light in the cube is fixed to the source, which it is clearly not. The origin of light is fixed to the location in space that it was emitted (0,0,0) at t=0. The origin of light has nothing to do with being fixed to any object which may or may not have had an absolute velocity.

The light sphere expands from the point in space it was emitted, regardless of what the source does after it was emitted.

That is true. Again it is my choice of words that has me going astray. The light sphere expands from the point in space it was emitted, which I was awkwardly referring to the point of origin. I see the distinction I think. When I say the point of origin it could be construed to be the light source which is clearly not fixed. Am I getting it straight yet?

Again, the fixed point is not in motion, it is a point in space (0,0,0) at t=0. That point is not in motion, it is simply a point in space. It is not an object, it is a location at t=0.

That I can understand and I don't think I have mis-characterized the fixed point in space.

If you simply know the y and z times of .65 seconds, as measured, then you can use that formula to arrive at the x time and absolute velocity in the x direction.

You're the boss . Now what lines or times are measured? Can you show them on the graphic and assign the .65 value to them on the graphic for clarity? Fill in anything else you can in the graphic if you want us to be able to refer to it in the future.

 

[Motor Daddy]

You can fill in the times in the graph you just posted with the sphere radius lines. Those radius lines are .65 light seconds in length, and 1.384930 light seconds in length, respectively.

The .65 line represents the time it took for light to contact the y and z receivers.

The 1.384930 line represents the time it took to contact the x receiver.


So it took .65 seconds for light to contact the y receiver, which is .5 light seconds away from the source in the cube frame, and it took 1.384930 seconds for light to contact the x receiver, which is .5 light seconds away from the source in the cube frame.

So the speed of light is measured to be .5/.65=.77c from the source to the y and z receivers in the cube frame, and the speed of light is measured to be .5/1.384930=.361c from the source to the x receiver in the cube frame.

Those are what is measured in the cube frame. In the absolute frame light travels at c, clearly, as the radius lines of the light sphere are .65/.65=1c, and 1.384930/1.384930=1c. Light always travels at c, but it is measured to be different in different frames due to the velocity of the frame.

 

[quantum_wave]

You can fill in the times in the graph you just posted with the sphere radius lines. Those radius lines are .65 light seconds in length, and 1.384930 light seconds in length, respectively.

The .65 line represents the time it took for light to contact the y and z receivers.

The 1.384930 line represents the time it took to contact the x receiver.


So it took .65 seconds for light to contact the y receiver, which is .5 light seconds away from the source in the cube frame, and it took 1.384930 seconds for light to contact the x receiver, which is .5 seconds light seconds away from the source in the cube frame.

So the speed of light is measured to be .5/.65=.77c from the source to the y and z receivers in the cube frame, and the speed of light is measured to be .5/1.384930=.361c from the source to the x receiver in the cube frame.

Those are what is measured in the cube frame. In the absolute frame light travels at c, clearly, as the radius lines of the light sphere are .65/.65=1c, and 1.384930/1.384930=1c. Light always travels at c, but it is measured to be different in different frames due to the velocity of the frame.

OK, I'll send it to my staff when she gets home, lol. (I'm pretty sure I'll have to do it.)

A question comes to mind about the .5 light seconds that the y receiver is away from the source in the cube frame. Let me test my understanding of that statement. If there was an observer in the center of the box frame, that observer would expect it to take .5 light seconds for light to travel from the source that is at the center of the box to the x, y and z receptors. However, when that observer gets feedback from the receptors he has the data from the receptors that says .65, 1.384930, .65 respectively, right.

So he/she can calculate the motion of the box relative to the fixed point at the center of the expanding light sphere? And that is the figure that James R gave us, 0.63897c, correct?

 

[Motor Daddy]

If there was an observer in the center of the box frame, that observer would expect it to take .5 light seconds for light to travel from the source that is at the center of the box to the x, y and z receptors. However, when that observer gets feedback from the receptors he has the data from the receptors that says .65, 1.384930, .65 respectively, right.

Correct.

So he/she can calculate the motion of the box relative to the fixed point at the center of the expanding light sphere? And that is the figure that James R gave us, 0.63897c, correct?

Correct. James got that velocity from my example. You can calculate it too with the known times of .65 second to the y and z receivers using my equation that I previously posted:

v(x)=sqrt(t(y)^2-l(y)^2)/t(y)
t(x)=l(x)/(c-v(x))


v(x)=sqrt(.65^2-.5^2)/.65
v(x)=sqrt(.4225-.25)/.65
v(x)=sqrt(.1725)/.65
v(x)=0.41533119314590374262921313724537/.65
v(x)=0.63897106637831345019878944191596 c


t(x)=l(x)/(c-v(x))
t(x)=.5/0.36102893362168654980121055808404
t(x)=1.384930551089674865417977078419 seconds

 

[quantum_wave]

Correct.

Correct. James got that velocity from my example. You can calculate it too with the known times of .65 second to the y and z receivers using my equation that I previously posted:

v(x)=sqrt(t(y)^2-l(y)^2)/t(y)
t(x)=l(x)/(c-v(x))


v(x)=sqrt(.65^2-.5^2)/.65
v(x)=sqrt(.4225-.25)/.65
v(x)=sqrt(.1725)/.65
v(x)=0.41533119314590374262921313724537/.65
v(x)=0.63897106637831345019878944191596 c


t(x)=l(x)/(c-v(x))
t(x)=.5/0.36102893362168654980121055808404
t(x)=1.384930551089674865417977078419 seconds

Excellent. I think my staff (lol) can put that all on one graphic.

I want to move on while we wait for the staff to catch up.

You have suffered a barrage of criticism for not acknowledging the postulate of SR that says light speed is the same in all frames. Your graphic takes one light sphere emitted from within a box, and the box is in motion relative to the point the light sphere was emitted from in space (not from the source as such). And you are getting different speeds of light relative to the motion of the box, but constant speeds when measured from the fixed point in space where the light was emitted. This is the MD view of reality, right?

One example used by others to show you are wrong deals with light spheres emitted from a star and the speed of that light is measured from two locations in Earth's orbit around the sun, one while the Earth is moving away from the star and one while the Earth is moving toward the star. I assume the wave length of the measured light is slightly red/blue shifted depending on the Earth's relative motion to the star, but they are saying that the speed of the light was measured to be the same in both cases which confirms SR.

I suggest that there is an all new coordinate system within which your scenario is playing out. It would be a coordinate system where measurements made in the SR example of light from a distant star would be subjected to mathematical equations that would transform their measurements so they agreed with what your measurements would show, if you were to measure the speed of light from that distant star using your logic. Do you follow this reasoning?

To move on to that analysis, can you envision a graphic that would lay out how to measure the same light sphere coming from a distant star from two perspectives, moving toward that star, and moving away from that star, using your logic? I think you are saying that the speed of that light will be c from the point in space where the light sphere you measure was emitted (not from where the star is), but that the measurement from the two different Earth boxes will show the speed of the star light to be different when we are moving toward the light sphere compared to when we are moving away from the light sphere. Is that correct? Then if we apply the MD transformation equations to their measurement they would agree with yours.

This doesn't mean you have reality on your side, lol. It just means that it is possible to use SR measurements to determine MD measurements, and vice versa. The difference between the two will then quantify the different "realities" represented by each coordinate system.