Acceleration Challenge

All you need is Lawrence tranfirmation formulas, some relativity knowledge and a bit of integral calculus. It's a math challange, not an intellegence challange. Who has the time. [I'm sorry for the spelling mistakes, if any, I rarly read any of that stuff in English]

Not so fast, Mogad, not so fast! You are right with Lorentz transformation, but, once the Newtonian sum is made, the answer is, well… fairly obvious. No real relativity required, just a bit of intuition. The equation s = 0.5[e^(at)]/a is just a 'teaser' - not really required, but insight giving.

Did you mean the Lorentz-Transformation formulas?

That would apply to the Einstien Special Theory of Relativity, I think. For the first part of the question it is framed using classical physics and ignores relativity. At least that is the way I read it.

I'm gonna give your question a try. For Newton, there would be no cap on velocity, only the constant acceleration there so the trip would take about 1414 years and the peak velocity would be about 1414c or an impressive 1414 times the speed of light. Is this correct? I used x=1/2at^2 and v=at with that 1ly/y^2~g approximation.

Einstein would say the peak velocity would be near c but the ship would be a hell of a lot heavier. Is this right?

You're perfectly right with both velocities! The Newtonian trip would however take somewhat longer... The speed would not be 1414c for the whole trip!

As an (optional) extra, what about the on-board time that Einstein would have calculated? It's in the formula s = 0.5[e^(at)]/a.

Looks about 2 years, but that depends on how close you get to the speed of light. Small fractions of the speed of light can translate to HUGE differences when traveling 2 million light years!!!

No, 2 years are a bit optimistic! From your other reply (below), it seems that you think that there is a one year acceleration up to the speed of light and then a one year deceleration down to zero and you're there? Not quite. In relativity, you never reach the speed of light, but approach it in asymptotical fashion. The acceleration lasts for the whole time, +1g for 1st half and -1g for 2nd half.
With acceleration approximated to a = 1 ly/y^2, the answer lurks in the simplified equation s = 0.5e^t, where s = 1 million light years (half-way, giving half the time as well). The whole thing is explained with an engineering spin in a download available at linear acceleration.

Yes, you never reach the speed of light, but you can get very close. In classical physics you can accelerate at 9.81 m/s*s from a standing start. If you maintain that acceleration for almost 354 days you would theoretically reach the speed of light. Since you can only get very close, I simplified the problem and I assumed infinite propulsion and no issues dealing with the exponential increase in mass at relativistic speeds.

First, I assumed the time dilation only is significant at velocities very close to the speed of light (98-99%) since the dilation effect is exponential. Two, if it takes nearly a year to get close to the speed of light at 9.81 m/s*s, you could tweak your velocity such that the remaining 1,999,998 years traveling at that speed would essentially be a blink of the eye for the crew. Then you start the deceleration phase which lasts somewhat less than one year. Total shipboard time would appear to be about 2 years, give or take some days or weeks.

Of course the shipboard computers to regulate the craft's speed and thrust duration would need to be quite extraordinary, but this was simply an exercise in frivolity, at least at this stage of mankind's technological development.

As stated before, no, not 2 years! If the ship could move at the speed of light, yes, the "blink of an eye" would be valid (in fact, it would be far to long.) But according to the proven physical laws of the day, the ship cannot reach the speed of light. It simply approaches that speed asymptotically and the on-board time is many times more - look at the simplified equation again!

Incidentally, no "infinite propulsion" is needed, if by it is meant "infinite force". The misconception arises mostly out of the "outdated" concept that the mass of the spaceship will approach infinity at the speed of light. For the ship, the crew and the engine, the ship's mass will still be "normal". The total energy required for such a mission will however be enormous and is not practically available, or even possible.

If you can find the time, read the download on the web page referenced before: linear-acceleration. It will clear up a lot of issues.

I understand. However, is it not possible to get sufficiently close to the speed of light so that the time dilation would be so great as to be nearly instantaneous for the portion of the journey at that speed?

That is, you spend the first year getting the craft to 99.999999999999999999% of c at a rate of acceleration which is 1 g. Then you coast for a few million years and just before you arrive, you turn the ship around and begin deceleration at 1 G until it stops at the nearest rest stop or the Restaurant at the End of the Universe (wherever you are going).

This assumes you have the needed energy to do all of this.

I was under the impression that the effects of the time dilation would become close to infinite when you are infinitely close to the speed of light.

I'll reread that link you sent. Thanks for wonder website you set up, too!!!

Quote from your post #5280 "....you spend the first year getting the craft to 99.999999999999999999% of c at a rate of acceleration which is 1 g". Nope, this is not possible in relativity theory. At 1g, you will reach a speed of 99.99999999995% of c after 14.5 years on-board time, which happens to be at the halfway mark to Andromeda. Then you decelerate at 1g for another 14.5 years, for a total on-board travel time of 29 years.

If you stop your acceleration after only one year, your speed will be lower and the trip would take very much longer. How much longer? I will work it out, but give me some time - you know, still working at the engineering job and doing this in the evenings (GMT+2 zone). If you read that pdf file mentioned on the web page, you may actually work it out by yourself before I get round to it!

Okay. I assumed that you could use Newtonian physics to arrive at the amount of time to reach a speed just below c. I guess that calculation is not valid using v/a = t, even from an outside inertial frame.

I was just trying to pin down a number for the acceleration and deceleration phases using a simple process.

Thanks for the information.

Curiosity did get the better of me and I worked out your scenario. The simplified relativistic equation given before is not a good approximation for short periods. Using the full equation as in the referenced document, you will reach just 70.7% of c after one year at 1g acceleration. If you now switch off the propulsion and float to Andromeda, it will take 1.414 million years (on-board time) to get there! (The time dilation factor is only 0.707).

Even at a painful 10g for 1 year, you will reach 99.5% of c, but the on-bard time required is still almost 200 thousand years. There is only one way to (hopefully) survive the trip and that is continuous acceleration, pushing your speed as close to c as possible.

I would love to see the math. I used V = at as my starting point. Solving for t is:

t = v/a

t = 3.00 e8 / 9.81 = 3.06 e7 seconds

There are 365.242 days in a year, so 365.242 * 24 * 60 * 60 = 3.16 e7 seconds per year.

t = 3.06 e7 / 3.16 e7 = 0.97 years to reach a velocity of 3.00 e8 m/s (c). I know you can't reach c, but I am thinking that getting to a fraction of c would take almost the same time, so for all intents and purposes 0.97 years should be close enough, but...

I must be dense. What is wrong with the Newtonian math I am using? I think I am asking why doesn't constant acceleration yield predictable values of time (from an external inertial frame) for a target velocity?

Newtonian math is simply not valid when speed becomes a significant portion of c! As you will easily see in the (earlier) referenced pdf doc, a constant acceleration on-board the craft does not translate to constant acceleration in the external reference frame.
At any given instant, the on-board acceleration must be multiplied by [1-(v/c)^2]^(1.5) to get the instantaneous acceleration in the reference frame. This changing acceleration must be integrated to get speed.
If you want a constant acceleration in the reference frame, the on-board acceleration must increase exponentially – not a pleasant idea!

Ah! That makes sense! Thanks for the explanation.

I just realized the link you provided was your own website. The website looks great and seems to have a ton of information. Unfortunately now I'm going to subject you to tons of questions regarding relativity ;)

My first question is this. As I understand it, the closer an object gets towards the speed of light, the mass increases. If a star is spinning quickly enough the surface could be moving at relativistic speeds. Would this result in increased mass for the star or does the motion have to be linear? If it did result in a change in the mass of the star, could it result in the star becoming a black hole (if the effect was large enough)?

Roger, you are welcome to ask and I will try my engineering best to answer:-)

The idea of mass increase at high velocity is a bit blurred, because mass and energy are equivalent (E=mc^2). Obviously, at high velocity, the energy of an object increases, as measured by some fixed reference frame. An object with high rotational speed has more energy (mass) than an equivalent non-rotating object. There is however a limit to how fast any object can rotate before destroying itself – even black holes can only rotate at a certain maximum rate – the so-called "extreme Kerr black hole".

The short(?) answer to your question is: no, an object cannot rotate itself into a black hole without flying apart.

Just to let you know where this question is coming from. I read this article (See link - http://wwwa.britannica.com/eb/article-52857).

Basically it talks about a neutron star that spins at 600 times per second. The article mentions some spin faster. Unfortunately the article fails to mention the diameter of the star. I've read that some Neutron stars are around 20 km in diameter. If I use this value for the star spinning 600 times per second, I get a surface speed of 3.6 * 10^7 for the neutron star, or roughly .12c . So it seems spinning Neutron stars can spin at high speeds.

Doing a quick google search I also found this link (http://www.nrao.edu/pr/2006/mspulsar/) Which appears to discuss a neutron star spinning 716 times per second with a diameter (maybe) of 20 km. 4.5 x 10^7 m/s or roughly .15 c.

What would be considered a "relativistic speed"? I suspect it would have to be over .5 c at least. Let me read about this Kerr black hole. Thanks for the feedback. I have a keen interest in relativity so I'll be visiting your website.

As far as I know, there is no hard-and-fast definition for "relativistic speed", but it is loosely speaking when the ratio (v/c)^2 becomes "significant", whatever that may mean. It probably means that some effect of it becomes detectable, so 0.5c would be considered pretty relativistic! The time dilation factor would be sqrt[1-(0.5)^2] = 0.866. I tend to think even 0.01c is very relativistic.

About spinning neutron stars: every massive structure would have a maximum rotational speed before disintegrating - it depends, amongst other things, on the gravitational binding. I plan to write an "engineering piece" on Kerr black holes on my website in the not too distant future. Let me know what you find elsewhere.

For Newton there is no maximum velocity that you can't exceed, so classical physics should apply. That being said, the final velocity for an object is:

Vf = sqrt (2ax)

Where a = acceleration and x = distance.

If the speed of light is c = 299792458 m/s

and x is the distance light travels in 1 million years

and there are 31556909 seconds in a year (given 365.242 days per year)

We have 1,000,000,000 y * 31556909 s/y = 3.155690 e16 seconds.

I multiplied the number of meters light travels in one year by the total number of seconds light takes to go 1,000,000,000 light years and got:

3.155690 e16 s * 299792458 m/s = 9.460523 e24 meters

Plugging in the numbers above to the equation for final velocity:

Vf = sqrt (2 * 9.81 m/s*s * 9.460523 e 24 m) = 1.362407 e13 m/s

Calculate the number of times the speed of light divides into that number:

1.352407 e13 m/s / 299792458 m/s = 45,445 c

Someone check my math!

To calculate the time it takes to get to the half way point I used a derivation of x = (at*t)/2

t = sqrt( 2x/a) = sqrt ( 2*9.460423 e24 m / 9.81 m/s*s) = 1.928730 e24 seconds

To get years you divide that number by 31556909 seconds/year to get 6.12 e16 years.

That just seems wrong! I know you spend most of your time just accelerating toward your final velocity and things build geometrically, but that seems like a lot of reruns of "I Love Lucy" to me.

The theory of relativity states that the maximum velocity an object can travel is limited to something just less than the speed of light. At 1 G acceleration you would reach c long before the halfway point. The equation for the point where you reach the speed of light is (V*V) / (2 * a) = x.

(299792458 m/s * 299792458 m/s) / (2 * 9.81 m/s*s) = 4.58 e15 meters

The half way point is 9.46 e24 meters, so you reach the speed of light (4.58 e15 m / 9.46 e24 m) * 100% = 4.8 e-8 % (far less than 1% of the distance to the half-way point). Almost all your time is spent weightless! However, for those on board the time dilation effect would make that seem like a blink of an eye. So 99.9999999% of the distance you would be traveling at very close to the speed of light.

The next question is how long does it take to reach the speed of light at 1 G? If v = at, then t = v/a: t = 299792458 m/s / 9.81 m/s*s = 30,559,884 seconds

That is 30,559,884 s / 31,556,909 s/y = 0.968 years! Wow, just under one year to accelerate to the speed of light and one more to decelerate to zero again at 1 G. That makes the trip only 2 million years long.

So someone tell me where I went wrong, because the two possibilities don't agree with each other.

You made life very difficult for yourself, maths-wise. Rather accept 1g =~ 1 ly/y^2 (it's 1.03, actually) and the calcs are dead easy.

Yeah, but I wanted to torture myself and get there using the three basic equations of motion. I think I went too fast somewhere and something blew up numerically. ;-)

Yup, redid my math and I made a mistake early up the chain. I get 1,437 times the speed of light, which is close to your calculations.

The difference between the 1414 of Rodger Pink and mine is probably due to the length of a year. I used 365.242 days.