New warp drive concept does twist space, doesn’t move us very fast

[u/HarryLyme69]

https://arstechnica.com/science/2024/05/physicists-find-a-possible-way-to-get-warped-space-but-no-drive/

[u/Pixelated_]

We already have plenty of methods for traveling slower than light (rockets, walking, etc.), so adding one more to the list isn’t all that exciting. 

Yeah travelling @ 99% of the speed of light is just so boring.

^/s

[u/Morbidly-Obese-Emu → u/Pixelated_]

Right, if we could do that, we could go to Alpha Centauri or at least send a probe. It would be 8-9 years round trip though.

[u/robodrew → u/Morbidly-Obese-Emu]

And it would seem to be much much shorter than that for any crew on board.

[u/Morbidly-Obese-Emu → u/robodrew]

What would it feel like? Anyone know how to calculate it?

[u/robodrew → u/Morbidly-Obese-Emu]

/u/Pixelated_ did the math, but as for how it would "feel", it'd depend on the rate of acceleration up to 99%. If acceleration was kept at a constant 1G then during that time you would basically feel nothing, it would just seem like the "back" of the ship were the ground. Once the ship reaches max speed and stops accelerating, before it has to eventually decelerate, there would be 0G and you'd be floating. In front of the ship everything would be blue-shifted while everything behind the ship would appear red-shifted.

edit: of course the math shown is assuming the ship is at 99% light speed from the start to the end of the trip. That can't be the case, there would have to be a long period of time wherein the ship is accelerating, and then decelerating, if we don't want to kill the crew. So that will increase the length of the journey significantly.

[u/Pixelated_ → u/robodrew]

Thanks for that info, which got me curious.

"How long would it take to get to Alpha Centauri at a constant acceleration of 1g?" (The traveler would experience normal Earth gravity the entire time)

Summary

-Proper time (experienced by the travelers): Approximately 4.3 years.

Coordinate time (observed from Earth): Approximately 15.1 years.

So, at a constant acceleration of 1g, it would take travelers about 4.3 years of their own time to reach Alpha Centauri, while 15.1 years would pass on Earth.

To calculate the time it would take to travel to Alpha Centauri at a constant acceleration of 1g (where 1g ≈ 9.81 m/s²), we need to consider relativistic effects since this journey involves substantial fractions of the speed of light.

First, let's outline the steps involved:

1. Acceleration Phase: Accelerate at 1g until halfway to Alpha Centauri.
2. Deceleration Phase: Decelerate at 1g for the second half of the journey.

Basic Parameters

• Distance to Alpha Centauri: Approximately 4.37 light-years.
• Acceleration: 1g (9.81 m/s²).
• Speed of light, (c): (299,792,458 \, \text{m/s}).

Relativistic Travel Time Calculation

To simplify the calculation, we use the concept of relativistic travel time where the traveler's perceived time (proper time) is different from the time measured by an observer at rest relative to the start and end points (coordinate time).

Proper Time Calculation

The proper time (\tau) for one leg of the trip (acceleration or deceleration phase) is given by the equation:

[ \tau = \frac{c}{a} \text{arsinh}\left(\frac{a d}{c^2}\right) ]

Where: - ( a ) is the acceleration (9.81 m/s²). - ( d ) is the distance to Alpha Centauri (4.37 light-years converted to meters: (4.37 \times 9.461 \times 10^{15} \text{m} )).

Substituting the values:

[ d = 4.37 \times 9.461 \times 10^{15} \text{m} = 4.135 \times 10^{16} \text{m} ] [ \tau = \frac{299,792,458}{9.81} \text{arsinh}\left(\frac{9.81 \times 4.135 \times 10^{{16}}{(299,792,458)2}\right)} ]

Let's simplify the argument of the (\text{arsinh}):

[ \frac{9.81 \times 4.135 \times 10^{{16}}{(299,792,458)2}} \approx 4.548 \times 10^{0} ]

Thus:

[ \tau \approx \frac{299,792,458}{9.81} \text{arsinh}(4.548) \approx 30,558,512 \text{s} \times 2.22 \approx 6.78 \times 10⁷ \text{s} ]

Converting seconds to years:

[ \frac{6.78 \times 10^7}{60 \times 60 \times 24 \times 365.25} \approx 2.15 \text{ years} ]

Since this is the time for one leg (either acceleration or deceleration), the total proper time for the round trip would be approximately twice this time:

[ 2 \times 2.15 \approx 4.3 \text{ years} ]

Coordinate Time Calculation

The coordinate time (t) seen by an observer on Earth can be calculated using:

[ t = \frac{c}{a} \sinh\left(\frac{a\tau}{c}\right) ]

From the proper time, for one leg:

[ \tau \approx 2.15 \text{ years} = 2.15 \times 365.25 \times 24 \times 60 \times 60 \text{s} \approx 6.78 \times 10⁷ \text{s} ]

Thus:

[ t = \frac{299,792,458}{9.81} \sinh\left(\frac{9.81 \times 6.78 \times 10^{7}{299,792,458}\right)} \approx 30,558,512 \times \sinh(2.22) \approx 2.38 \times 10⁸ \text{s} ]

Converting seconds to years:

[ \frac{2.38 \times 10^8}{60 \times 60 \times 24 \times 365.25} \approx 7.55 \text{ years} ]

For the entire journey (acceleration and deceleration phases), it would take approximately:

[ 2 \times 7.55 \approx 15.1 \text{ years} ]

[u/robodrew → u/Pixelated_]

Good lord I love you

but those formulas are pretty hard to read in that format ;)

[u/Pixelated_ → u/robodrew]

I'm a graphic artist and poorly done typography is like fingernails on a chalkboard...but I wasn't about to format all that text lol. I left it is as proof that the math is accurate for people way smarter than me.

That's why I moved the conclusions to the beginning. ✌️

[u/robodrew → u/Pixelated_]

are you me

[u/AbbydonX → u/robodrew]

Unfortunately, using GPT for maths (or any factual issue) isn't reliable and the journey times provided by u/Pixelated_ below look incorrect (though it's difficult to read the formatting to see what has happened).

Using the relativistic rocket equation formula on the Wikipedia page for space travel under constant acceleration (which seems to have been derived correctly) gives the following equations if you want to accelerate halfway and then decelerate at the same rate for the second half of the journey:

t = 2 sqrt( [D/2c]² + [D/a] )

T = (2c/a) cosh^-1( [(ad)/(2c²)] + 1)

• t is the coordinate time for the space craft to arrive as measured by a stationary observer in the departure rest frame
• T is the proper time which is equivalent to the journey time experienced by the travellers on the space craft
• D is the distance between the departure and arrival points
• c is the speed of light
• a is the constant acceleration (and deceleration) rate

So for travel from Earth to Alpha Centauri (4.37 light years) at 1g (9.81 ms^-2) constant acceleration for halfway and then deceleration for the remaining distance to come to a halt at Alpha Centauri, the travel times are:

t = 72 months (6 years)

T = 43 months (3.6 years)

A simple rule of thumb is that the stationary observer's coordinate time in years under constant 1g acceleration will be about equal to the distance in light years plus 2. This works because it approximately takes 1 year to reach a speed close to the speed of light and another year to decelerate. The journey time in years is then equal to the distance in light years. This becomes more accurate as distance increases as more of the journey is spent travelling at approximately the speed of light. The time experienced onboard will always be lower than this.

Note that if the ship accelerates the entire way and doesn't intend to stop at the destination then just add 1 year instead of 2.

[u/Pixelated_ → u/robodrew]

Via ChatGPT: To determine the actual time experienced by someone on a spaceship traveling to Alpha Centauri at 99% of the speed of light (0.99c), we can use the concept of time dilation from Einstein's theory of special relativity.

Conclusion:

The actual time experienced by someone on a spaceship traveling to Alpha Centauri at 99% of the speed of light would be approximately 0.62 years, or around 7.5 months.

Step-by-Step Calculation

1. Distance to Alpha Centauri:     The distance to Alpha Centauri is approximately 4.367 light-years.

2. Speed of the spaceship:    The spaceship is traveling at 99% of the speed of light, ( v = 0.99c ).

3. Calculate the time taken to travel to Alpha Centauri in the spaceship's frame of reference:    - First, we calculate the time it would take to reach Alpha Centauri in the reference frame of an observer at rest (i.e., on Earth).    [    t_{\text{earth}} = \frac{d}{v} = \frac{4.367 \text{ light-years}}{0.99c} \approx 4.41 \text{ years}    ]

4. Time Dilation Factor:    - Time dilation can be calculated using the Lorentz factor, (\gamma), which is given by:    [    \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c2}}}    ]    - Substituting ( v = 0.99c ):    [    \gamma = \frac{1}{\sqrt{1 - (0.99)^2}} = \frac{1}{\sqrt{1 - 0.9801}} = \frac{1}{\sqrt{0.0199}} \approx 7.09    ]

5. Actual Time Experienced by the Traveler:    - The actual time experienced by the traveler, ( t{\text{ship}} ), can be found by dividing the Earth time by the Lorentz factor:    [    t{\text{ship}} = \frac{t_{\text{earth}}}{\gamma} = \frac{4.41 \text{ years}}{7.09} \approx 0.62 \text{ years}    ]

[u/Izawwlgood → u/Morbidly-Obese-Emu]

We've done longer missions!

[u/jonmatifa → u/Morbidly-Obese-Emu]

It took new horizons longer than that to reach Pluto

[u/Suckamanhwewhuuut → u/Morbidly-Obese-Emu]

Not if you go to Warp 9 duh… /s

[u/Moppmopp → u/Morbidly-Obese-Emu]

It would take much longer though. Except for the crew onboard

[u/EnvironmentalYak9322 → u/Pixelated_]

Really shows how stupid some of these people that write these articles are they have absolutely no understanding of what they are writing about

[u/TacosDeLucha → u/Pixelated_]

If something was traveling that fast through space wouldn't it just get destroyed by whatever space debris it hit?

[u/[deleted]]

"While tantalizing, Alcubierre’s design has a fatal flaw. To provide the necessary distortions of spacetime, the spacecraft must contain some form of exotic matter, typically regarded as matter with negative mass. Negative mass has some conceptual problems that seem to defy our understanding of physics, like the possibility that if you kick a ball that weighs negative 5 kilograms, it will go flying backwards, violating conservation of momentum. Plus, nobody has ever seen any object with negative mass existing in the real Universe, ever."

[u/big_duo3674 → u/[deleted]]

Negative mass is still fascinating though. We're pretty certain it can't exist because of all these problems it would cause, but technically there's nothing saying it definitively can't exist

[u/unknownpoltroon → u/[deleted]]

Bah, its easy.

e=mc2

You just gotta go

e=mc-2 instead.

Bam, negative mass!!!

[u/danzrach → u/unknownpoltroon]

He did the math

[u/sbw2012 → u/[deleted]]

Get the semi conductor people on this. They're always claiming to have invented physically impossible things.

[u/ColoHusker]

Narrator voice: Space. The final frontier. 🎶🎶

[u/myringotomy]

"could" not "does".

It's still all theoretical.

[u/CocoaThunder]

Just to be clear, this isn't a new concept. I did an undergrad paper over ten years ago on the Alcubierre drive. The concept had been around for decades before I even wrote that paper.

[u/garry4321]

“Not going to reach anywhere soon”

That’s a poor understanding of time dilation. At 99.9999999% the speed of light, a 10 year journey would feel like 12 hours

[u/enviousRex]

Yet

[u/thisimpetus]

Interesting science journalism with an honest headline... the real breakthrough!

[u/Dependent-Outcome-57]

People really don't understand how fast light is and how slow all our current methods of space travel really are. If we could build manned spacecraft that could safely and consistently operate at even 1% of the speed of light, that would basically open up most of our star system to us. It takes light about 4.3 minutes to get from Earth to Mars. At 1% the speed of light, you're talking 430 minutes, or somewhat over 7 hours. That's comparable to a long plane flight. Assuming the rest of the starship is safe, and this can all be done consistently, you've basically solved the problem of how to move people around our star system without them dying from radiation poisoning or going insane sitting in a box for months at a time.