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## Main Question or Discussion Point

First off I apologize for lengthiness, but I want to be thorough... please read the entire post

I was just reading up on time dilation on wiki, and saw that the formula for determining time dilation in special relativity is essentially delta t times the Lawrentz factor.

in other words for time dilation,

t / [(1 - v^2 / c^2)^1/2]

The lawrentz factor describes things like time dilation, length contraction and relative mass, but I think there may be a problem with the application of it. The lawrentz factor is

c / [(c^2 - v^2)^1/2]

but it is commonly reduced to

1 / [(1 - B^2)^1/2]

where B = v / c

this reduction is done by simply pulling out a c^2 from under the square root, and then canceling it with the top c.

Here is the problem.

If we are dealing with a vector problem, as most time dilation theories do, (i.e. traveling out at a significant fraction of the speed of light, and coming back at the same speed therefore reducing the aging process) the reduced equation essentially assumes absolute values of the vectors. Here is an example.

NO ASSUMPTIONS PLEASE KEEP READING

Let us assume for the problem a reference point of (0,0,0) coordinates in space on an X,Y,Z axis.

The direction we will travel from point (0,0,0) to planet A will be in a positive Z direction at a rate of 2.0E8 meters per second with no X or Y components. Assume the distance is such that it takes 5 days on a spacecraft to get there.

Using the unreduced formula, the Lawrentz factor for the trip there will be

3.0E8 / [((9.0E16)-(4.0E16))^1/2] = 3.0E8 / 2.24E8 = 1.34

on the trip back we have

-3.0E8 / [((9.0E16)-(4.0E16))^1/2] = -3.0E8 / 2.24E8 = -1.34

the -3.08 on top must be negative because the speed of light and the velocity we determined are relative to a positive z vector, but now we are headed in a negative z direction... so a factor describing relative time and speed to a stationary point must maintain that relativity. The -3.0E8 and -2.0E8 on the bottom were both squared according to the equation, and thus lost the signs attributed to them, leaving them at 9.0E16 and 4.0E16 (i know this is basic, but i must point it out to avoid argument)

if we add the lawrentz factors, the total difference in relative time is 0

5days(1.34) +5days(-1.34) = 0

Finally, If we did this same equation with the reduced formula (where c had been canceled out), we would get 1.7 days fewer on both the trip there and the trip back, leaving us 3.4 days "younger" than our brothers at point (0,0,0)

How has nobody noticed this flaw...

To me, this is similar to the Doppler effect. If you hear an ambulance going away from you the pitch decreases. When it comes closer to you the pitch increases

If you had a telescope and you were able to watch the spaceship fly away from point (0,0,0) it would take you 1.7 days to observe 1 day of behavior on the ship... the light is stretched out... but if you watch the way back, it will only take you about .746 days to observe one day of behavior, because the light would be compacted and closer together... regardless, the entire trip could be observed in exactly 10 days. It would just appear to take 6.7 days for them to get there and 3.3 days for them to get back...

can anyone echo if I am on to something here? and why it hasn't been noticed before?

I was just reading up on time dilation on wiki, and saw that the formula for determining time dilation in special relativity is essentially delta t times the Lawrentz factor.

in other words for time dilation,

t / [(1 - v^2 / c^2)^1/2]

The lawrentz factor describes things like time dilation, length contraction and relative mass, but I think there may be a problem with the application of it. The lawrentz factor is

c / [(c^2 - v^2)^1/2]

but it is commonly reduced to

1 / [(1 - B^2)^1/2]

where B = v / c

this reduction is done by simply pulling out a c^2 from under the square root, and then canceling it with the top c.

Here is the problem.

If we are dealing with a vector problem, as most time dilation theories do, (i.e. traveling out at a significant fraction of the speed of light, and coming back at the same speed therefore reducing the aging process) the reduced equation essentially assumes absolute values of the vectors. Here is an example.

NO ASSUMPTIONS PLEASE KEEP READING

Let us assume for the problem a reference point of (0,0,0) coordinates in space on an X,Y,Z axis.

The direction we will travel from point (0,0,0) to planet A will be in a positive Z direction at a rate of 2.0E8 meters per second with no X or Y components. Assume the distance is such that it takes 5 days on a spacecraft to get there.

Using the unreduced formula, the Lawrentz factor for the trip there will be

3.0E8 / [((9.0E16)-(4.0E16))^1/2] = 3.0E8 / 2.24E8 = 1.34

on the trip back we have

-3.0E8 / [((9.0E16)-(4.0E16))^1/2] = -3.0E8 / 2.24E8 = -1.34

the -3.08 on top must be negative because the speed of light and the velocity we determined are relative to a positive z vector, but now we are headed in a negative z direction... so a factor describing relative time and speed to a stationary point must maintain that relativity. The -3.0E8 and -2.0E8 on the bottom were both squared according to the equation, and thus lost the signs attributed to them, leaving them at 9.0E16 and 4.0E16 (i know this is basic, but i must point it out to avoid argument)

if we add the lawrentz factors, the total difference in relative time is 0

5days(1.34) +5days(-1.34) = 0

Finally, If we did this same equation with the reduced formula (where c had been canceled out), we would get 1.7 days fewer on both the trip there and the trip back, leaving us 3.4 days "younger" than our brothers at point (0,0,0)

How has nobody noticed this flaw...

To me, this is similar to the Doppler effect. If you hear an ambulance going away from you the pitch decreases. When it comes closer to you the pitch increases

If you had a telescope and you were able to watch the spaceship fly away from point (0,0,0) it would take you 1.7 days to observe 1 day of behavior on the ship... the light is stretched out... but if you watch the way back, it will only take you about .746 days to observe one day of behavior, because the light would be compacted and closer together... regardless, the entire trip could be observed in exactly 10 days. It would just appear to take 6.7 days for them to get there and 3.3 days for them to get back...

can anyone echo if I am on to something here? and why it hasn't been noticed before?

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