Less elapsed time: Lorentz Transformation or Spacetime Interval ?

In summary, the 8 year elapsed time on the rocket clock is because it traveled a lesser distance in spacetime, S=8, with a clock that mechanically ran at the same rate as the stationary clock.
  • #1
morrobay
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With one clock stationary at the common origin for 10 time units and the other rocket clock
traveling out and back at v=.6c. When both clocks are compared at the origin the elapsed
time on the traveling clock will show less time by a factor of (.8) from the Lorentz Transformation.
The spacetime interval for the traveling clock will be S=8, sqrt( 10^2-6^2)
Compared to the stationary clocks spacetime interval S=10.
Note, it is understood if the rocket clocks coordinates were transformed to the inertial frame
by: t=gamma(1.25) (t' + (v)(x')/cc and x=gamma(1.25)(x'+(v)(t') then S=8
=sqrt(17^2-15^2).
So the question: Is the 8 units of elapsed time in the rocket clock because the physical mechanism was moving at a slower rate (time between ticks) in the rockets frame.
Or in the case of the spacetime interval, was the mechanism in the rockets clock ticking at the same rate (in rocket frame) as the rate of ticking in stationary frame, but the rocket clock just traveled a lesser distance in spacetime, S=8
 
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  • #2
morrobay said:
So the question: Is the 8 units of elapsed time in the rocket clock because the physical mechanism was moving at a slower rate (time between ticks) in the rockets frame.
A clock is never running slow in its own inertial rest frame...but the rocket must accelerate at some point to turn around, so what do you mean when you say "the rockets frame"? If we assume the acceleration to turn around is instantaneous and that the rocket moves inertially before the turnaround and afterwards, then the rocket has two different rest frames: 1) the frame where the rocket is at rest during the journey away from the origin, and 2) the frame where the rocket is at rest during the journey back towards the origin. If you analyze things from frame 1, then the rocket's clock is ticking normally during the journey out but running slow after the turnaround when it's no longer at rest in this frame, and likewise if you analyze things from frame 2, then the rocket wasn't at rest before the turnaround so its clock was running slow then.
 
  • #3
My question is not between the views of the rockets outbound and inbound reference frames
or the relativity of simultaneity between other frames. It is based on the overall conclusion when the two clocks are compared at the common origin after the rocket clock has traveled
outbound and inbound at v=.6c for a 10 year period from the reference frame of the earth.
That the elapsed time on the stationary clock is 10 yrs and elapsed time on the rockets clock is 8 years.
So is the 8 year elapsed time on the rocket because it was mechanically running slower(time between ticks) than the Earth clock - from the Lorentz transformation.
Or did the rockets clock record a proper time of 8 years because it traveled a lesser distance in spacetime, S=8, with a clock that mechanically ran at the same rate as the
stationary clock ?
 
  • #4
morrobay said:
Or did the rockets clock record a proper time of 8 years because it traveled a lesser distance in spacetime, S=8, with a clock that mechanically ran at the same rate as the stationary clock ?
This is the interpretation I prefer.
 
  • #5
morrobay said:
So is the 8 year elapsed time on the rocket because it was mechanically running slower(time between ticks) than the Earth clock - from the Lorentz transformation.
But the Lorentz transform doesn't really say this--you can pick a frame where the rocket clock was ticking faster than the Earth clock for at least part of the journey, it's only if you look at the average rate of ticking over the entire trip that all frames agree the rocket clock was ticking slower on average.
morrobay said:
Or did the rockets clock record a proper time of 8 years because it traveled a lesser distance in spacetime, S=8, with a clock that mechanically ran at the same rate as the
stationary clock ?
The distinction between these two options doesn't seem physically meaningful--they don't make any distinct predictions about any empirical results. Also, what does it even mean to ask whether a clock is "mechanically running slower" or if it "mechanically ran at the same rate" if you aren't asking this question from the perspective of a particular coordinate system? You can't compare rates of ticking without use of a coordinate system, your question is analogous to looking at two lines on a plane and asking whether they "physically have different slopes" or "physically have the same slope" without specifying a coordinate system to define "slope" in terms of dy/dx. The "slope" of a line on a 2D plane is an intrinsically coordinate-dependent notion, and I would say the same thing about the "rate of ticking" of a clock in 4D spacetime.
 

Related to Less elapsed time: Lorentz Transformation or Spacetime Interval ?

1. What is the Lorentz Transformation?

The Lorentz Transformation is a mathematical formula that describes how measurements of space and time change for an observer in one reference frame when compared to an observer in a different reference frame moving at a constant velocity relative to the first observer.

2. How does the Lorentz Transformation relate to the concept of spacetime interval?

The Lorentz Transformation is one of the key components of Einstein's theory of special relativity, which describes how the laws of physics are the same in all inertial reference frames. It is used to calculate the spacetime interval, which is the distance between two events in spacetime, taking into account both the spatial and temporal components.

3. What is the significance of less elapsed time in the Lorentz Transformation?

Less elapsed time in the Lorentz Transformation refers to the phenomenon of time dilation, where time appears to pass more slowly for an observer who is moving at a high velocity relative to another observer. This is a consequence of the constant speed of light in all reference frames, as predicted by Einstein's theory of special relativity.

4. Can the Lorentz Transformation be applied to all objects and situations?

Yes, the Lorentz Transformation is a fundamental principle of special relativity and can be applied to all objects and situations, as long as they are moving at a constant velocity relative to each other. It is used in a wide range of fields, including physics, engineering, and astronomy, to make accurate predictions about the behavior of objects in motion.

5. Are there any practical applications of the Lorentz Transformation?

Yes, the Lorentz Transformation has many practical applications in modern technology. It is used in the development of GPS systems, which rely on the precise measurement of time and the effects of time dilation in order to provide accurate location information. It also has practical applications in particle accelerators, where precise calculations of velocity and time are crucial for the success of experiments.

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