This has been bugging me

[between3and26characterslon]

There was a discussion some months back about travelling to our nearest star, Proxima Centauri, and how long it would take to get there if you accelerated to the speed of light at 1g. A 1g acceleration would be a reasonable acceleration to be comfortable to those who are travelling.

 

Some of the more mathematicaly inclined and knowledgeable proffered the suggestion it would take about 7 years (if I remember correctly) whereas I said I had heard it would take some 245 years.

 

So my question is this; if our Sun and Proxima Centauri are rigidly attached to an inertial frame A with time t and our mode of transport has its own frame B with time t' then is it correct to say that an acceleration of 1g in A = 1g in B. In other words if time for B t'=t/2 an accereration of 1g in A would = 2gin B.

 

This means to maintain a 1g acceleration aboard our transport the acceleration as measured from the intertial frame would decrease. If this is correct how long would it take, both in the inertial frame and our transport's frame, to accelerate to c at 1g and travel for the necessary distance to allow us to turn round and decellerate at 1g and arrive at Proxima Centauri?

[Tres Juicy]

 

This means to maintain a 1g acceleration aboard our transport the acceleration as measured from the intertial frame would decrease. If this is correct how long would it take, both in the inertial frame and our transport's frame, to accelerate to c at 1g and travel for the necessary distance to allow us to turn round and decellerate at 1g and arrive at Proxima Centauri?

 

I think you've got it back to front as it were.

 

To maintain a steady 1g onboard you would have to Increase acceleration as you travelled

[D H]

This means to maintain a 1g acceleration aboard our transport the acceleration as measured from the intertial frame would decrease. If this is correct how long would it take, both in the inertial frame and our transport's frame, to accelerate to c at 1g and travel for the necessary distance to allow us to turn round and decellerate at 1g and arrive at Proxima Centauri?

Assuming a distance of 4.3 light years, an acceleration of 1g, and a turnaround at the halfway point, it would take 5.93 years from the perspective of the Earth, 3.56 years from the perspective of the space travelers.

 

I think you've got it back to front as it were.

 

To maintain a steady 1g onboard you would have to Increase acceleration as you travelled

Nope. between3and26characterslon had it right. From the perspective of an inertial observer, the spaceship's velocity will approach c. The acceleration as computed by this inertial observer necessarily decreases as the ship gets closer and closer to the speed of light.

[Tres Juicy]

Nope. between3and26characterslon had it right. From the perspective of an inertial observer, the spaceship's velocity will approach c. The acceleration as computed by this inertial observer necessarily decreases as the ship gets closer and closer to the speed of light.

 

 

But in order to maintain 1g on board due to acceleration surely it must increase?

[D H]

But in order to maintain 1g on board due to acceleration surely it must increase?

Surely not. The spaceship is accelerating at a=1g as measured by sensors on the spaceship (e.g., an accelerometer). Assume the spaceship accelerates at this constant acceleration (spaceship frame) starting with a velocity of 0 at time t=0. The velocity of the spaceship as calculated by our inertial observer at some later time t is given by

 

[math]

v(t) = \frac{at}{\sqrt{1+\left(\frac{at}{c}\right)^2}}

= at\left(1+\left(\frac{at}c\right)^2\right)^{-1/2}[/math]

 

At some finite time, the calculated acceleration dv/dt is given by

 

[math]\frac{dv(t)}{dt} = a\left(1+\left(\frac{at}c\right)^2\right)^{-3/2}[/math]

 

For small t, the radical is nearly equal to 1, so initially the spaceship's acceleration from the perspective of an inertial observer is a. However, as the spaceship builds up speed, the term [math]1+(at/c)^2[/math] becomes larger and larger. The computed acceleration decreases.

 

It has to decrease. Note that v→c as t→∞. Thus dv/dt→0 as t→∞.

[Tres Juicy]

Surely not. The spaceship is accelerating at a=1g as measured by sensors on the spaceship (e.g., an accelerometer). Assume the spaceship accelerates at this constant acceleration (spaceship frame) starting with a velocity of 0 at time t=0. The velocity of the spaceship as calculated by our inertial observer at some later time t is given by

 

[math]v(t) = \frac{at}{\sqrt{1+\left(\frac{at}{c}\right)^2}}= at\left(1+\left(\frac{at}c\right)^2\right)^{-1/2}[/math]

 

At some finite time, the calculated acceleration dv/dt is given by

 

[math]\frac{dv(t)}{dt} = a\left(1+\left(\frac{at}c\right)^2\right)^{-3/2}[/math]

 

For small t, the radical is nearly equal to 1, so initially the spaceship's acceleration from the perspective of an inertial observer is a. However, as the spaceship builds up speed, the term [math]1+(at/c)^2[/math] becomes larger and larger. The computed acceleration decreases.

 

It has to decrease. Note that v→c as t→∞. Thus dv/dt→0 as t→∞.

 

My fault, I'm not thinking...

 

The speed must increase not the acceleration