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A few relativistic questions.


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I did some maths and found that that at 99.999999% the speed of light, the distance to Alpha Centauri would decrease to 39 AU. 1. Does this affect take place in the traveller's reference frame? 2. Are you basically travelling 39 AU at 99.999999% the speed of light now from your frame of reference?

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That's correct, from the traveller's frame of reference they have only travelled 39 AU.  So, in their frame of reference, their trip only takes them about 13 days (ignoring acceleration/deceleration periods, and assuming my calculations are correct!).  Sounds perfect!  However, it's important to remember the concurrent time dilation.  The trip only takes about 13 days according to the traveller's clock, on their spaceship with them.  However, for a stationary observer watching them, the trip still takes just over four years, according to the stationary observer's clock.  Therefore, as with the twin paradox, although only 13 days has passed for the traveller, for anyone sitting back home on Earth, over four years have passed.

 

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8 minutes ago, slickfonzy said:

That's correct, from the traveller's frame of reference they have only travelled 39 AU.  So, in their frame of reference, their trip only takes them about 13 days (ignoring acceleration/deceleration periods, and assuming my calculations are correct!).  Sounds perfect!  However, it's important to remember the concurrent time dilation.  The trip only takes about 13 days according to the traveller's clock, on their spaceship with them.  However, for a stationary observer watching them, the trip still takes just over four years, according to the stationary observer's clock.  Therefore, as with the twin paradox, although only 13 days has passed for the traveller, for anyone sitting back home on Earth, over four years have passed.

 

Thanks! Would you mind just quickly showing how you calculated 13 days? It would be really helpful. I've obviously made a mistake in my workings.

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Just checked, seems I made the mistake :embarrassed: sorry! Running through the calculations again I get 5.4 hours, rather than 13 days - is that the same as you have worked out?

distance (traveller's frame) = s  = 39 AU = 5.834*(10^12) metres

c is 299,792,458 m/s, so traveller's speed = v = 0.99999999*c

time (traveller's frame) = s /v = [5.834*(10^12)] / 0.99999999*c

which gives a time of 19,460 seconds, so divide by 60 for minutes, 60 again for hours, gives 5.4 hours.

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5 minutes ago, slickfonzy said:

Just checked, seems I made the mistake :embarrassed: sorry! Running through the calculations again I get 5.4 hours, rather than 13 days - is that the same as you have worked out?

distance (traveller's frame) = s  = 39 AU = 5.834*(10^12) metres

c is 299,792,458 m/s, so traveller's speed = v = 0.99999999*c

time (traveller's frame) = s /v = [5.834*(10^12)] / 0.99999999*c

which gives a time of 19,460 seconds, so divide by 60 for minutes, 60 again for hours, gives 5.4 hours.

Yeah I got about 5.4 hours. Thanks for checking though!

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13 minutes ago, slickfonzy said:

Just checked, seems I made the mistake :embarrassed: sorry! Running through the calculations again I get 5.4 hours, rather than 13 days - is that the same as you have worked out?

distance (traveller's frame) = s  = 39 AU = 5.834*(10^12) metres

c is 299,792,458 m/s, so traveller's speed = v = 0.99999999*c

time (traveller's frame) = s /v = [5.834*(10^12)] / 0.99999999*c

which gives a time of 19,460 seconds, so divide by 60 for minutes, 60 again for hours, gives 5.4 hours.

Yeah I got about 5.4 hours. Thanks for checking though! I'm making a video on this. Make sure to check it out! It's on my new channel called Corkey no2. It wont be up for a couple of days.

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Would that be for an astronaut who jumps on to a rocket that whizzes by at 99.999999% of c  when it passes Earth and jumps off when it passes alpha Centauri? That is not as easy as it sounds. An astronaut would not survive jumping on and off a relativistic rocket.

Maybe this works better: Suppose you want to travel the 4.2 light years to alpha centauri. You want to start from rest and to arrive at rest. You would need a rocket that would accelerate to a maximum speed at midpoint, and then turn around and break with the same acceleration.

In this case, if you want the maximum speed to be 99.999999% of the speed of light, you would need a massive acceleration of 3262 g. The journey would be magnificent. Measured in Earth time, it would last 4.20059 years. Measured in ship time, it would last 0.000090 years, or 47.3 minutes. Of course, an astronaut would not survive a trip at 3262 g.

So for alpha Centasuri, 99.999999 % of c  is not feasible as it is such a short trip. About 95% of the speed of light would work fine though. At an acceleration of 1 g, the trip would last 5.82 years Earth time, 3.53 years ship time, and the maximum speed would be 0.949 times the speed of light. That is very comfortable especially since the 1 g acceleration would make for a perfect artificial gravity.

Below you find a small Excel workbook (Einstein_Rocket_v2.xlsx) which does these calculations. It is based on the program ROCKET.BAS published by Sky & Telescope. Apart from the examples above, it for instance tells you that a 1000 light years travel in a 1 g ship would last 1001.93611 years Earth time, 13.4519169 years ship time, and involve a maximum speed of 0.999998129 times the speed of light.

In Rocket.bas_v2.pdf you find the formulas used and the acceleration calculation for the 99.999999 % of c  trip.

 

Excel workbook: Einstein_Rocket_v2.xlsx       Pdf with formulas: Rocket.bas_v2.pdf

 

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On 10/5/2016 at 19:41, Ruud said:

Would that be for an astronaut who jumps on to a rocket that whizzes by at 99.999999% of c  when it passes Earth and jumps off when it passes alpha Centauri? That is not as easy as it sounds. An astronaut would not survive jumping on and off a relativistic rocket.

Maybe this works better: Suppose you want to travel the 4.2 light years to alpha centauri. You want to start from rest and to arrive at rest. You would need a rocket that would accelerate to a maximum speed at midpoint, and then turn around and break with the same acceleration.

In this case, if you want the maximum speed to be 99.999999% of the speed of light, you would need a massive acceleration of 3262 g. The journey would be magnificent. Measured in Earth time, it would last 4.20059 years. Measured in ship time, it would last 0.000090 years, or 47.3 minutes. Of course, an astronaut would not survive a trip at 3262 g.

So for alpha Centasuri, 99.999999 % of c  is not feasible as it is such a short trip. About 95% of the speed of light would work fine though. At an acceleration of 1 g, the trip would last 5.82 years Earth time, 3.53 years ship time, and the maximum speed would be 0.949 times the speed of light. That is very comfortable especially since the 1 g acceleration would make for a perfect artificial gravity.

Below you find a small Excel workbook (Einstein_Rocket_v2.xlsx) which does these calculations. It is based on the program ROCKET.BAS published by Sky & Telescope. Apart from the examples above, it for instance tells you that a 1000 light years travel in a 1 g ship would last 1001.93611 years Earth time, 13.4519169 years ship time, and involve a maximum speed of 0.999998129 times the speed of light.

In Rocket.bas_v2.pdf you find the formulas used and the acceleration calculation for the 99.999999 % of c  trip.

 

Excel workbook: Einstein_Rocket_v2.xlsx       Pdf with formulas: Rocket.bas_v2.pdf

 

Thanks very much!

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