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Author Topic: How long do you have to accelerate at 1g to reach the speed of light?  (Read 12249 times)

Offline thedoc

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Jack Stott  asked the Naked Scientists:
   Dear Dr Chris.

I have plucked up the courage to write to you with a question from a colleague of mine which I can't answer.

I have no idea why he wants to know this ( I think he reads too many science fiction comics or watches too much 'Star Treck' ) but here goes :-

If an object of negligible size & mass is launched from a standing start in a vacuum, and is subjected to an acceleration force of 1 g - how long will it take to reach the speed of light.

Hope you are able to provide an answer, or even a formula to calculate an approximate result when and if you have the time.

Best Regards

Jack Stott BSc(Hon) Elec Eng Science
What do you think?
« Last Edit: 04/07/2013 22:48:46 by chris »


 

Offline graham.d

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It will never reach the speed of light. It would take an infinite amount of energy. This results from Einstein's theory of Special Relativity. However, from the perspective of anyone travelling on this craft, they will experience a contraction in the distances in their direction of travel. So although they can never get to 186,000 miles per second they can (theoretically), nonetheless, achieve a speed such that they can travel a distance that they may initially have measured (before accelerating) as 186,000 miles in less that 1 second as measured on their clocks. If I remember correctly this turns out to be the same time (to get to this speed) as would be calculated by Newtonian mechanics - I would need to check this with some maths to be sure. There are some unfortunate consequences of travelling this fast resulting from time dilation so that should return at some point you would find the earth you left having aged considerably compared to yourself.
 

Offline syhprum

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There are many unfortunate consequences of travelling near the speed of light, microscopic specks of dust would be like express trains when you hit them and the CMBR would be blue shifted up to Gamma rays.
 

Offline graham.d

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Yes. Forward Shields to maximum please Mr Scott :-)
 

Offline distimpson

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Jack Stott  asked the Naked Scientists:
   Dear Dr Chris.

I have plucked up the courage to write to you with a question from a colleague of mine which I can't answer.

I have no idea why he wants to know this ( I think he reads too many science fiction comics or watches too much 'Star Treck' ) but here goes :-

If an object of negligible size & mass is launched from a standing start in a vacuum, and is subjected to an acceleration force of 1 G - how long will it take to reach the speed of light.

Hope you are able to provide an answer, or even a formula to calculate an approximate result when and if you have the time.

Best Regards

Jack Stott BSc(Hon) Elec Eng Science
What do you think?

Wonderful question, the answer depends on the details, if "negligible" mass means zero then the particle is already moving at c and can not be at a standing start in any (inertial) reference frame. If not, then the velocity can be computed from the equations of Einstein's special theory of relativity, rather than repeat, please refer to this link for the details: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

Since we are assuming an acceleration of 1g, the size and mass does not enter into the velocity calculation, it will matter in terms of the energy required to accelerate the particle. So, after 1 year at 1g, 0.77 of the speed of light, 2 years, 0.97c, 12 years to get to 0.99999999996, pretty close to c but not close enough for a physicist :-)

Hope this helps, a simplified answer to a sophisticated question. Science fiction sometimes becomes science fact, keep the wonder!
« Last Edit: 01/05/2013 15:15:42 by distimpson »
 

Offline lightarrow

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Given a proper acceleration "a" (the acceleration felt by the astronaut inside his ship) and denoted with "T" the proper time (the time measured by the astronaut's wristwatch), the spaceship' speed u, as measured from Earth, is given by:

u(T) = c [exp(aT/c) - exp(-aT/c)] / [exp(aT/c) + exp(-aT/c)]

Example 1:
a = 1g ~= 10m/s2
T = 347.22 days ~= 30,000,000 s
c ~= 300,000,000 m/s
--> u = c [e - 1/e] / [e + 1/e] ~= 0.76 c

Example 2:
a = 1g ~= 10m/s2
T = 9.5 years ~= 300,000,000 s
c ~= 300,000,000 m/s
--> u = c [e10 - (1/e)10] / [e10 + (1/e)10] ~=  0,999999996c.

As others have pointed out, u becomes = c only when T = +oo.
« Last Edit: 01/05/2013 17:02:02 by lightarrow »
 

Offline Pmb

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As stated above a particle subjected to uniform accleration as measured in its own frame of reference wil never travel at the speed of light. It will only come closer and closer to it. On the other hand its impossible to force a particle to accelerate at a constant coordinate acceleration, i.e. as measured in a particular inertial frame of reference. E.g. place a charged particle in a uniform electric field then the force on the charge would be constant but its acceleration would decrease with time. You can think of this as the (inertial/relativistic) mass increasing with speed and thus with time.

I worked out this derivation here
http://home.comcast.net/~peter.m.brown/sr/uniform_accel.htm

There is a nice spacetime diagram which illustrates the particle's position as a function of time. It should make all this clearer.
 

Offline yor_on

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Very nice Pete.
 

John B

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« Reply #8 on: 27/06/2013 04:08:56 »
All that relativity garbage is just that. Garbage. As you aproach the speed of light nothing changes. It doesn't take any longer to accelerate the last half of the way to the speed of light as it does the first half. When you acheive the speed of light nothing happens , nothing changes. If you keep accelerating at 1G twice as long as that then you will be going twice the speed of light. Specks of dust should be destroying the space station by now because for all we know the solar system and the entire galaxy could be moving away or toward one another at 100s of times the speed of light. Unless your Christopher Colombuses Queen of Spain and you beleive the ocean continues on to infinity and where your sitting is the center of the universe, the sun goes around the earth and this is ground zero and you are perfectly still. Did you measure our speed in the universe by the warehouse walls at the edge morons? What I want to know is where these idioitic concepts like Vaporizing or going back in time come from? Since you haven't invented a spacecraft that can accelerate constantly at a 1G acceleration you automatically asume from some erouneous candle light and horse era mathimatical scribble that you will vaporize at the speed of light which if you recall they also tried to sell us about breaking the SOUND barrier before it was possible. Why do you mathematical types persist at this nonsense. It takes 380 something days to accelerate to and PAST the speed of light as closely as I had calculated at one time but I cannot ever get a straight answer on this very pertinent question because of all the geeknoid scientific daydreaming about these infinite forces barring coming near the speed of light GETTING IN THE WAY. Shut up all ready you have no proof for all that hogwash.
 

John B

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« Reply #9 on: 27/06/2013 04:30:47 »
Not worded quite correctly the post meant to compare the solar system and this galaxy moving toward or away from other galaxies (not within our own)at possibly 100s of times the speed of light.
 

Offline RD

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« Last Edit: 27/06/2013 04:51:54 by RD »
 

Offline galaxysim

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There is some extra weirdness that happens close to the speed of light. I read an article recently about two ships traveling close to the speed of light attached by a rope....does it break or no. Length contraction and other effects.

I didn't quite understand it, it was a bit involved...just thought id add that in.

Bell's spaceship paradox feel free to knock yourself out on this one, lol
http://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox [nofollow]

From the speedy travelers perspective there appears to be no upper bound or speed limit. You keep on going faster and faster.  Time dilation effects, you can travel anywhere in the observable universe in a short space of time, by your watch anyway.

I think a number of unpleasant side effects start to accumulate. This might mean that traveling very close to the speed of light might be impractical even if energy for propulsion wasn't an issue.



Shielding solution.
Imagine a giant cargo net full of ice floating just ahead of your speeding ship. This should take care of small dust grains and EM radiation. Space is cold so machine gunning that snowball and lobbing in the odd artillery shell isn't to much of a worry so long as the ice pack can dissipate the heat fast enough. Quite a few layers of steeply angled metal Armour also help.

A very powerful forward facing laser might also help. Heating a corridor of space directly in your path. This will tend to make the dust particles scatter out of the way. Assuming no weird ionization effects etc cause the opposite to happen.




« Last Edit: 04/07/2013 03:00:49 by galaxysim »
 

Offline Pmb

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Quote from: galaxysim
There is some weirdness that happens close to the speed of light. I read an article recently about two ships traveling close to the speed of light attached by a rope....does it break or no. Length contraction and other effects.
As you said, it's called Bell's Spaceship Paradox. I know the problem well. See
http://en.wikipedia.org/wiki/Bell's_spaceship_paradox
http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html

Quote from: galaxysim
I didn't quite understand it, it was a bit involved...just thought id add that in.
It's easy to understand once it's properly explained. Consider two spaceships which start out from rest in the inertial frame S. Connect them with a piece of string which is strung taught. Let each of them accelerate at the exact same rate so that as measured in S they maintain the exact same distance apart. Since the string also undergoes acceleration its speed is constantly increasing. Therefore its length must undergo a Lorentz contraction. Eventually the spring will break. Consider this from the Spaceship's point of view. Each astronaut sees the other one initially at rest with respect to each other. Their accelerations are the same so they must be at the same distance apart, right? Therefore the string doesnít break.  That's the paradox.

The resolution to the paradox is to consider the spaceship which is trailing behind to be at rest in a uniform gravitational field and the other ship being above him in the gravitational field with his rocket's turned on with the exact same amount of thrust. Since the local acceleration in a uniform gravitational field decreases as one goes higher in the gravitational field the astronaut in the higher position, who has his rocket engines doing the same thing as the one lower in the field, must be accelerating away from the spaceship below him. Since he's moving away the string connecting them breaks. Thatís the resolution to the paradox.

Make sense?

Quote from: galaxysim
From the speedy travelers perspective there appears to be no upper bound or speed limit. You keep on going faster and faster.  Time dilation effects, you can travel anywhere in the observable universe in a short space of time, by your watch anyway.
If the traveler feels the same acceleration in his frame of reference then his acceleration as measured from an inertial frame of reference will decrease with speed at such a rate that his speed approaches the speed of light but never gets there.
 

Offline lightarrow

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It's easy to understand once it's properly explained. Consider two spaceships which start out from rest in the inertial frame S. Connect them with a piece of string which is strung taught. Let each of them accelerate at the exact same rate so that as measured in S they maintain the exact same distance apart. Since the string also undergoes acceleration its speed is constantly increasing. Therefore its length must undergo a Lorentz contraction.
Lorentz contraction in which frame of reference? Not in S, since "as measured in S they maintain the exact same distance apart".
 

Offline Pmb

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Quote from: John B
All that relativity garbage is just that. Garbage.
Wrong. Youíre basing that on the way you imagine that it should be which is grounded in every day experience. But everyday experience does not give you any sense of what happens at high speed. If you were to base your assumptions on what really happens rather than what you think happens youíd be singing another tune.
Quote from: John B
As you aproach the speed of light nothing changes. It doesn't take any longer to accelerate the last half of the way to the speed of light as it does the first half. When you acheive the speed of light nothing happens , nothing changes. If you keep accelerating at 1G twice as long as that then you will be going twice the speed of light.
Thatís been proven to be wrong. We know how things work under constant proper acceleration because it happens in particle accelerator labs all the time. Theory and experiment correspond to exactly what is predicted.

Quote from: John B
Since you haven't invented a spacecraft that can accelerate constantly at a 1G acceleration you automatically assumeÖ
Wrong. We can accelerate things at constant proper acceleration. Itís done with particles all the time. E.g. if you place a charged particle in an uniform electric field then the force on it is constant and in a frame of reference which is momentarily at rest with respect to it the accelerate has a constant value. In the lab frame the acceleration decreases with time and the particle gets closer and closer to the speed of light in a manner predicted exactly (to experimental accuracy) by the theory of relativity

Quote from: John B
Why do you mathematical types persist at this nonsense.
Weíre not mathematical types. Weíre physicist types who use math to describe the physics. And because thatís whatís observed in the laboratory. Why do you non-mathematical types refuse to learn what actually happens rather than what you think happens? And why do people like you refuse to learn the physics?

Quote from: John B
Shut up all ready you have no proof for all that hogwash.
Why do you think being rude will help you convince people who know what actually happens that it doesnít actually happen?
 

Offline Pmb

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Quote from: lightarrow
Lorentz contraction in which frame of reference? Not in S, since "as measured in S they maintain the exact same distance apart".
Lorentz contracted in the frame in which the string is moving. When the string tries to Lorentz contract it breals because in its own istantaneous frame the spaceships are moving apart. I explained all that above.
 

Offline yor_on

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It does, as long as we give a 'inertial' preference to the ship under Pete. But what happens if we define the ship above as being 'inertial' too, at rest in a gravitational field? And if we assume the ship above to be at rest, with the ship under going for it at , ahem, full throttle (one constant uniform G acceleration) it becomes really intriguing :)

Then again, what about the contraction you observe, relative a distance measured in front of you. The ship 'under' should also see the space contracted in front of it, including the space between the ships as I think? What about the ship above, looking back, at the ship 'below', would they agree to it being a same distance between them?

As they are defined as being 'at rest' with each other? What would 'at rest' mean there?
Different constant distances?
 

Offline yor_on

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Assuming a length contraction to be a symmetry in both directions (backward as well as forward of the ships motion) you can define them as being at rest, and that should make the best sense, possibly :) but you have a acceleration of matter to consider too, as well as the rope. A little like the (not infinitely) rigid pole you move, poking at the moon, with the 'motion propagating' in the rod at a approximate speed of sound, compressing and decompressing itself. Here you have particles of that matter constantly compressing (and decompressing?) relative each other as the 'force' of one G 'propagates in the matter, or should I assume that they are in a constant uniform state of compression? What I mean is that I find it tough to see that matter as being in a equilibrium as it is accelerated.
« Last Edit: 04/07/2013 16:30:18 by yor_on »
 

Offline Pmb

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It does, as long as we give a 'inertial' preference to the ship under Pete. But what happens if we define the ship above as being 'inertial' too, at rest in a gravitational field? And if we assume the ship above to be at rest, with the ship under going for it at , ahem, full throttle (one constant uniform G acceleration) it becomes really intriguing :)

Then again, what about the contraction you observe, relative a distance measured in front of you. The ship 'under' should also see the space contracted in front of it, including the space between the ships as I think? What about the ship above, looking back, at the ship 'below', would they agree to it being a same distance between them?

As they are defined as being 'at rest' with each other? What would 'at rest' mean there?
Different constant distances?
There is a difference between the ships. One ship is lower in the gravitational field that the accelerating frame mimics while the other is higher up. This is what breaks the symmetry. The clock in the higher up (i.e. leading ship) has its clock running faster than the one lower in the field (i.e. the trailing ship).

Did you understand everything I was trying to explain when it came to looking at this all from the perspective of a gravitational field?
 

Offline yor_on

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Think so :)

You're defining the ship 'under' as being 'at rest', presenting a uniform gravitational field, aka a 'inertial object' (loosely expressed a planet of some mass). You then go to define the other ship as constantly accelerating in that gravitational field. It being at a higher altitude versus that 'field', so giving it a 'faster clock' relative the 'planetary gravity' represented by the ship under. The problem is, to me, that it seems that I can use any of the ships for such a definition, or assume both to be in a uniform gravitational field there?

And doing so, instead defining the one 'above' as presenting this uniform gravity, you get a result in where the ship 'under' is uniformly constantly accelerating into that field? Although it still is a symmetric solution, as you have one ship at a different 'height' relative that field, no matter how you define them, it becomes slightly confusing to me.
=

I don't think one can define the ship 'above' as 'hovering' at a constant height, using the first example of yours? As that wouldn't give me a equivalence with the situation of two ships constantly uniformly accelerating, as you have a ever growing expenditure of energy as the ships continue to accelerate?

I'm not sure there?
« Last Edit: 04/07/2013 18:09:20 by yor_on »
 

Offline Pmb

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Think so :)

You're defining the ship 'under' as being 'at rest', presenting a uniform gravitational field, aka a 'inertial object' (loosely expressed a planet of some mass). You then go to define the other ship as constantly accelerating in that gravitational field.
No. My neck hurts from typing or so long so I'll explain later.
 

Offline yor_on

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Ok Pete, I'll read it with interest.
 

Offline yor_on

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Is this solution using two exactly equivalent ships Pete? Everything the exact same for both ships? As mass, acceleration, etc? If I assume them them being 'exact replicas' of each other in every aspect, accelerating equivalently, in a equivalently 'flat space', then they also should be able to be described as belonging to a same frame of reference I think? Wouldn't such a definition make their 'clock-ticks' equivalent too? And if they are, how do I from that get to different heights?
 

Offline lightarrow

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Quote from: lightarrow
Lorentz contraction in which frame of reference? Not in S, since "as measured in S they maintain the exact same distance apart".
Lorentz contracted in the frame in which the string is moving. When the string tries to Lorentz contract it breals because in its own istantaneous frame the spaceships are moving apart. I explained all that above.
Quoting from the wiki page that you have linked:
<<In the inertial frame S, a delicate string or thread hangs between two identically accelerating spaceships.>>
Where does it say that "as measured in S they maintain the exact same distance apart"? If they maintained the same distance apart, the string wouldn't broke.
 

Offline Pmb

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Quote from: lightarrow
Where does it say that "as measured in S they maintain the exact same distance apart"?
The fact that the ships accelerate at the same rate and thus remain the same distance apart is the entire point of the paradox. In fact thatís what the paradox is all about.

Please read it again because it states
Quote
The distance between the spaceships does not undergo Lorentz contraction with respect to the distance at the start, because in S it is effectively defined to remain the same, due to the same acceleration of both spaceships.
I donít understand what the problem your having is. Itís very simple. As measured in an inertial frame S (in flat spacetime) if two ships accelerate at the exact same rate then they must remain the exact same distance apart at all times as measured in frame S. If you donít know that they you donít know the problem at hand because this is what the problem is all about.

Quote from: lightarrow
If they maintained the same distance apart, the string wouldn't broke.
That's quite wrong. Consider a string whose proper is L0. The proper length of an object is defined to be the length of the object as measured in the frame  in which the object is at rest.

In the inertial frame S string is moving parallel to its length. In S the string has undergone a Lorentz contraction and as such it will be shortened to the length L = L0*sqrt(1 - v2/c2). Since the ships accelerate at exactly the same rate the remain the same distance apart in the frame S. Therefore after the acceleration has ended and theyíre moving at constant speed the distance between them as measured in S will be greater than they were before they started moving. This means that they will be further apart than when they started. Since the properties of the string never changed the string had to break because the ends were moved apart.

QED

Any questions now?
 

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