Measuring the non-linearity residual between average and summed instantaneous velocities

[drumbo]

I have some questions, and it involves some speculation, but it is somewhat grounded within the currently accepted theory of relativity so I hope what I am proposing is not too speculative. Velocity is defined as displacement divided by time elapsed, and we can calculate the average velocity for an entire trip or the instantaneous velocity at a moment in time. The picture below illustrates this with the scalar portion:

Now let's do a thought experiment. Imagine that we are here on Earth, and John takes off in a space ship to head to a destination 10 light years away. In John's reference frame he records that it takes 11.11... years to complete the trip implying that the average velocity for the trip was 0.9 light years per year. Meanwhile back on Earth we have been observing John's spaceship. Due to John's high average velocity of 0.9c we experience time dilation which implies that by the time John reaches his destination more than 11.11 years will have passed here on Earth, and therefore if we calculate John's average velocity for the trip from our reference frame it will be greater than 0.9 light years per year. The point is that the average velocity of an object is not the same in all reference frames(?).

Now consider an alternative question. Is the instantaneous velocity of an object the same from the perspective all reference frames? If it is then the equation

where v(t) is velocity at time t and x(t) is position at time t is not true(?) for reference frames not fixed on the object whose velocity and displacement are being measured. In fact due to time dilation we must introduce the non-linearity residual (NLR)

since the instantaneous velocities do not(?) add linearly to give the average velocity. Note that NLR > 0.

So my questions are straightforward; is it true that the average velocity of an object is not the same in all reference frames, is it true that the instantaneous velocity of an object the same from the perspective all reference frames, and must NLR > 0?

I am asking this since if what I have conjectured is true, then could we not measure the NLR by sending probes into space? Send a probe in any direction really, and have it record the distance it travels along with the time elapsed. Simultaneously observe the probe from here on Earth recording the distance it travels along with the time elapsed, and then could we not calculate the NLR? Sorry if I am demonstrating a complete ignorance of relativity, I have not studied physics beyond some first year electives in university which focused on Newtonian mechanics and waves.

I am very intrigued by this since a calculation of the NLR, if it exists, over different paths in space could reveal some useful information.

Edit: I made a small error in the 2nd equation

Edited September 11, 2020 by drumbo

[Bufofrog]

1 hour ago, drumbo said:

John takes off in a space ship to head to a destination 10 light years away. In John's reference frame he records that it takes 11.11... years to complete the trip implying that the average velocity for the trip was 0.9 light years per year

That is not possible.  If he was going .9c his travel time would be WAY less than 10 years (too lazy to work it out).

[drumbo]

10 minutes ago, Bufofrog said:

That is not possible.  If he was going .9c his travel time would be WAY less than 10 years (too lazy to work it out).

I see, and if he was going 0.9c would his travel time from the Earth's reference frame be 11.11... years?

[Bufofrog]

2 hours ago, drumbo said:

I see, and if he was going 0.9c would his travel time from the Earth's reference frame be 11.11... years?

Yes.

[md65536]

10 hours ago, drumbo said:

I see, and if he was going 0.9c would his travel time from the Earth's reference frame be 11.11... years?

You said the trip was 10 light years, but didn't say what frame that's measured in. It sounds like you intended that to be in the Earth's reference frame, so Earth would measure the trip taking 11.11 years. In John's reference frame, the distance to the destination is length-contracted, so even though the destination approaches John at 0.9c, it arrives at John sooner than 11.11 years.

Gamma is ~ 2.29, the contracted travel distance is 10 LY/gamma = 4.36 LY, taking 4.84 years at .9c, measured by John.

12 hours ago, drumbo said:

In John's reference frame he records that it takes 11.11... years to complete the trip implying that the average velocity for the trip was 0.9 light years per year. Meanwhile back on Earth we have been observing John's spaceship. Due to John's high average velocity of 0.9c we experience time dilation which implies that by the time John reaches his destination more than 11.11 years will have passed here on Earth, and therefore if we calculate John's average velocity for the trip from our reference frame it will be greater than 0.9 light years per year.

This could be true (until the last sentence) if it was 10 LY as measured by John. If that were true, then it's true that he'd measure 11.11 years and that Earth would measure more (11.11 years * gamma = 25.49 years), but Earth would also measure the distance traveled as much greater too (10 LY * gamma = 22.94 LY), and the speed would still be .9 c.

Earth would not experience time dilation in those measurements, it would measure 25.49 years as normal. The only time dilation it would experience is that a moving clock ticks slower, and it agrees that John's clock ticks only 11.11 years during Earth's 25.49 years.

[swansont]

16 hours ago, drumbo said:

we must introduce the non-linearity residual (NLR)

Of what utility is it that we “must” introduce this? In what calculation would it be used?

 

16 hours ago, drumbo said:

since the instantaneous velocities do not(?) add linearly to give the average velocity. Note that NLR > 0.

So my questions are straightforward; is it true that the average velocity of an object is not the same in all reference frames, is it true that the instantaneous velocity of an object the same from the perspective all reference frames, and must NLR > 0?

You insist that NLR > 0, and then ask if it is. If you don’t know, don’t assert.

 

16 hours ago, drumbo said:

I am very intrigued by this since a calculation of the NLR, if it exists, over different paths in space could reveal some useful information.

What useful information do you expect?

 

 

[drumbo]

11 hours ago, md65536 said:

Earth would also measure the distance traveled as much greater too (10 LY * gamma = 22.94 LY), and the speed would still be .9 c

I don't quite follow this. How can the distance traveled not be the same in all reference frames? If we measure from Earth that the destination is 10 LY away then John will have overshot his destination if he travels 22.94 LY.

6 hours ago, swansont said:

You insist that NLR > 0, and then ask if it is. If you don’t know, don’t assert.

The assertion was contingent on the equation:

being incorrect which is the question, if the equation is incorrect the assertion must be true.

6 hours ago, swansont said:

What useful information do you expect?

Measuring the NLR may allow us to map out the convexity of space-time and discover its shape.

Edited September 12, 2020 by drumbo

[swansont]

11 minutes ago, drumbo said:

I don't quite follow this. How can the distance traveled not be the same in all reference frames? If we measure from Earth that the destination is 10 LY away then John will have overshot his destination if he travels 22.94 LY.

Distance is relative. It depends on the frame in which it is measured. 

 

11 minutes ago, drumbo said:

Measuring the NLR may allow us to map out the convexity of space-time and discover its shape.

You have not connected these. How does NLR relate to convexity?

[drumbo]

2 minutes ago, swansont said:

Distance is relative. It depends on the frame in which it is measured.

Ok. It appears I understand even less about relativity than I thought. I'll take a few days to read an introductory book.

[md65536]

3 hours ago, drumbo said:

Ok. It appears I understand even less about relativity than I thought. I'll take a few days to read an introductory book.

Sure! It's "length contraction". In case I wasn't clear, I was talking about two different cases, depending on what the 10 LY refers to. Either the 10 LY is the distance from Earth to destination as measured by Earth, and John measures that as length-contracted to a shorter distance. Or 10 LY is the distance that John measures, and Earth measures a longer distance that John measures as length-contracted to 10 LY.