Could time-dilation be used to create faster super computers?

[PJay_A]

The greater the gravity, the slower the time. Or is it the reverse? Whatever it is, my idea is to take advantage of the time differention to create faster super computers. For instance, if a computer would appear to be processing faster in  orbit than say the same machine on the ground, let's launch a network of super computer satellites with laser communications. Or if the opposite is true, build a series of super computers and house them on centrifuges to simulate greater gravity (or figure out a way to protect them from the high pressures of Jupiter's atmophere and send them on permanent baloons).

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The greater the gravity, the slower the time. Or is it the reverse? Whatever it is, my idea is to take advantage of the time differention to create faster super computers. For instance, if a computer would appear to be processing faster in&nbsp; orbit than say the same machine on the ground, let's launch a network of super computer satellites with laser communications. Or if the opposite is true, build a series of super computers and house them on centrifuges to simulate greater gravity (or figure out a way to protect them from the high pressures of Jupiter's atmophere and send them on permanent baloons). <br />Posted by PJay_A</DIV></p><p>You have the order of effects correct, clocks in a large gravitational field appear to rujn slowly to an observer in a lower gravitatinal field.&nbsp; But the effect is VERY small except for VERY large gravitational fields.&nbsp; At practical levels there is not enough effect to be important for the application that you have in mind.&nbsp; For purposes of speed of computation the difference between the gravitational field of the Earth and zero gravity is not important.</p><p><br /><br />&nbsp;</p> <div class="Discussion_UserSignature"> </div>

 

[R1]

<p><font size="2">Or you could put a computer in an accelerator?</font></p><p><font size="2">I know, it's also absolutely impractical and totally inefficient.&nbsp; Not a good idea.</font></p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>

 

[R1]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'><font size="2" color="#000080">Or you could put a computer in an accelerator?I know, it's also absolutely impractical and totally inefficient.&nbsp; Not a good idea.</font>&nbsp; <br />Posted by john1r</DIV><br /><br /><font size="2">No.&nbsp; I was wrong.&nbsp; Accelerating it does slow its time, but it slows down everything, including its processing speed.</font></p><p><font size="2">The first idea of putting the computer in an unmoving state away from a gravitational field would process slightly faster, but nothing practically useful unless it is running calculations or programs&nbsp;that require billions of years of time.</font></p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p>&nbsp;</p><p>GPS satallites only gain about 40 milliseconds per day.&nbsp; Interesting idea, but not very practicle.&nbsp; Besides, we have quantum computers on the horizon.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[R1]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;GPS satallites only gain about 40 milliseconds per day.&nbsp; Interesting idea, but not very practicle.&nbsp; Besides, we have quantum computers on the horizon.&nbsp; <br />Posted by derekmcd</DIV><br /><br /><font size="2">Interesting indeed.&nbsp; At 40 ms /day,&nbsp; this is almost 200 days per million yrs., or about 450 yrs. per billion yrs.&nbsp; </font></p><p><font size="2">It's not so much that orbital spacetime gains time, it's more like at sea level we are lagging so far back in time relative to the space that the earth is in.</font></p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Interesting indeed.&nbsp; At 40 ms /day,&nbsp; this is almost 200 days per million yrs., or about 450 yrs. per billion yrs.&nbsp; It's not so much that orbital spacetime gains time, it's more like at sea level we are lagging so far back in time relative to the space that the earth is in.&nbsp; <br /> Posted by john1r</DIV></p><p>I though your numbers looked a little large, then I realized the mistake was mine.&nbsp; I meant to say <strong><em>nano</em></strong>seconds... not milliseconds.&nbsp; A wee bit of a difference... <img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-embarassed.gif" border="0" alt="Embarassed" title="Embarassed" /></p><p>It's mistakes like that that make things crash.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[R1]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I though your numbers looked a little large, then I realized the mistake was mine.&nbsp; I meant to say nanoseconds... not milliseconds.&nbsp; A wee bit of a difference... It's mistakes like that that make things crash.&nbsp; <br />Posted by derekmcd</DIV><br /><br /><font size="2">&nbsp;</font></p><p><font size="2">it's alright.&nbsp; even at 40ns/day&nbsp; I get approximately 20 days per 100 million yrs.</font></p><p><font size="2">or almost 5 yrs. per 10 Billion years.</font></p><p><font size="2">I'm mostly trying to illustrate how proximity to a gravitational field causes things to lag far in the past, over large scale time durations.&nbsp; In proximity of stars, black holes, etc. the time lag is even larger, of course.&nbsp; </font></p><p><font size="2">When they flew an atomic clock on a plane and compared it to the one on the ground, it may not have even been a full day flight.&nbsp; The planes altitude is also far lower than the satellite's.&nbsp; </font></p><p><font size="2">(If its orbit is geostationary, does the satellite essentially have&nbsp;<u>no</u> kinematic time reduction? )</font></p> <div class="Discussion_UserSignature"> </div>

 

[SpeedFreek]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>(If its orbit is geostationary, does the satellite essentially have&nbsp;no kinematic time reduction? ) <br /> Posted by john1r</DIV></p><p>The orbit is not geostationary and the difference is 38 microseconds or 38 millionths of a second. Microseconds come in between nanoseconds and milliseconds. </p><p>From <font size="2">Real-World Relativity: The GPS Navigation System </font></p><p>"The current GPS configuration consists of a network of 24 satellites in high orbits around the Earth. Each satellite in the GPS constellation orbits at an altitude of about 20,000 km from the ground, and has an orbital speed of about 14,000 km/hour (the orbital period is roughly 12 hours - contrary to popular belief, GPS satellites are not in geosynchronous or geostationary orbits).</p><p>Because an observer on the ground sees the satellites in motion relative to them, Special Relativity predicts that we should see their clocks ticking more slowly. Special Relativity predicts that the on-board atomic clocks on the satellites should fall behind clocks on the ground by about 7 microseconds per day because of the slower ticking rate due to the time dilation effect of their relative motion</p><p> Further, the satellites are in orbits high above the Earth, where the curvature of spacetime due to the Earth's mass is less than it is at the Earth's surface. A prediction of General Relativity is that clocks closer to a massive object will seem to tick more slowly than those located further away. As such, when viewed from the surface of the Earth, the clocks on the satellites appear to be ticking <em>faster</em> than identical clocks on the ground. A calculation using General Relativity predicts that the clocks in each GPS satellite should get ahead of ground-based clocks by 45 microseconds per day.</p><p>The combination of these two relativitic effects means that the clocks on-board each satellite should tick faster than identical clocks on the ground by about 38 microseconds per day (45-7=38)! This sounds small, but the high-precision required of the GPS system requires nanosecond accuracy, and 38 microseconds is 38,000 nanoseconds. If these effects were not properly taken into account, a navigational fix based on the GPS constellation would be false after only 2 minutes, and errors in global positions would continue to accumulate at a rate of about 10 kilometers each day."</p><p>&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <p><font color="#ff0000">_______________________________________________<br /></font><font size="2"><em>SpeedFreek</em></font> </p> </div>

 

[MeteorWayne]

Replying to:<BR/><DIV CLASS='Discussion_PostQuote'> This sounds small, but the high-precision required of the GPS system requires nanosecond accuracy, and 38 microseconds is 38,000 nanoseconds. If these effects were not properly taken into account, a navigational fix based on the GPS constellation would be false after only 2 minutes, and errors in global positions would continue to accumulate at a rate of about 10 kilometers each day."&nbsp;&nbsp; <br />Posted by SpeedFreek</DIV><br /><br />Thus providing exquisite proof of Einstein's theories! <div class="Discussion_UserSignature"> <p><font color="#000080"><em><font color="#000000">But the Krell forgot one thing John. Monsters. Monsters from the Id.</font></em> </font></p><p><font color="#000080">I really, really, really, really miss the "first unread post" function</font><font color="#000080"> </font></p> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Thus providing exquisite proof of Einstein's theories! <br />Posted by MeteorWayne</DIV></p><p>Precisely.&nbsp; And laying waste to the various nut cases who dote on showing that Einstein was wrong because those theories don't meet the test of their personal intuition or "common sense".</p> <div class="Discussion_UserSignature"> </div>

 

[R1]

<p><br /><font size="2">True.&nbsp; </font><font size="2">speedfreek thanks for your post too.&nbsp; As it turned out I used microsecseconds the second time.</font></p><p><font size="2">Is the center of the earth under the influence of gravitational time dilation&nbsp; (as determined by the radius of the earth) even though it may be in a kinetically balanced (zero force) state?&nbsp;</font></p> <div class="Discussion_UserSignature"> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>True.&nbsp; speedfreek thanks for your post too.&nbsp; As it turned out I used microsecseconds the second time.Is the center of the earth under the influence of gravitational time dilation&nbsp; (as determined by the radius of the earth) even though it may be in a kinetically balanced (zero force) state?&nbsp; <br />Posted by john1r</DIV></p><p>Since there is zero gravitational field in the center of a homogeneous sphere there should be no time dilation effect.</p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The orbit is not geostationary and the difference is 38 microseconds or 38 millionths of a second. Microseconds come in between nanoseconds and milliseconds.<br /> Posted by SpeedFreek</DIV></p><p>Thanks again, speedfreek.&nbsp; My memory must be failing me.&nbsp; Five seconds worth of googling could've bailed me out on that one.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Since there is zero gravitational field in the center of a homogeneous sphere there should be no time dilation effect. <br /> Posted by DrRocket</DIV></p><p>I'm not so sure this is correct.&nbsp; Gravitational time dilation is dependent on the gravitational potential.&nbsp; The only time that is zero is at infinity.&nbsp; I would think that time dilation would be at its maximum at the center of the gravity well.&nbsp; I don't believe it is related to the "g" force.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[UncertainH]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I'm not so sure this is correct.&nbsp; Gravitational time dilation is dependent on the gravitational potential.&nbsp; The only time that is zero is at infinity.&nbsp; I would think that time dilation would be at its maximum at the center of the gravity well.&nbsp; I don't believe it is related to the "g" force.&nbsp; <br />Posted by derekmcd</DIV></p><p>&nbsp;</p><p>Gravitational field strength is indeed 0 at the center of a sphere, the potential is -3GM/2r at the center of a sphere of radius r. I'm not sure how that translates to time dilation but it wouldn't be infinity that's for sure<br /></p>

 

[exoscientist]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The greater the gravity, the slower the time. Or is it the reverse? Whatever it is, my idea is to take advantage of the time differention to create faster super computers. For instance, if a computer would appear to be processing faster in&nbsp; orbit than say the same machine on the ground, let's launch a network of super computer satellites with laser communications. Or if the opposite is true, build a series of super computers and house them on centrifuges to simulate greater gravity (or figure out a way to protect them from the high pressures of Jupiter's atmophere and send them on permanent baloons). <br /> Posted by PJay_A</DIV></p><p>&nbsp;Some scientists are investigating the possibility that the light speed barrier might be broken at ultra high energies as happens with some cosmic rays. If so, then you could make the calculations go as fast as you want by using suffienctly energetic particles.</p><p>Testing Lorentz Violation Using Propagating UHECRs.<br />http://arxiv.org/abs/0805.1275 </p><p>&nbsp;</p><p>&nbsp; Bob Clark </p> <div class="Discussion_UserSignature"> </div>

 

[R1]

<p><font size="2">alright.&nbsp; well dilation itself is additive, even regardless of whether it is kinematic or gravitational, as in the previous experiment with the satellite.&nbsp; So&nbsp;would it&nbsp;appear that dilation should zero out at a sphere's center? &nbsp; Conceiving this mathematically is somewhat strange, it appears that the sphere has an infinite number of radii.&nbsp; Each radius representing a line segment of 'mass' exerting gravity on the sphere's center.&nbsp; The net result, however, should probably be based on the net result of all radii.</font></p><p><font size="2">It&nbsp;calls for&nbsp;a strange-looking well.&nbsp; A planet's gravity well (and dilation diagram similarly) is not a plain and simple curve or hyperbola, it appears it should be a strange looking map-like diagram.&nbsp; </font></p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>&nbsp;Gravitational field strength is indeed 0 at the center of a sphere, the potential is -3GM/2r at the center of a sphere of radius r. I'm not sure how that translates to time dilation but it wouldn't be infinity that's for sure <br /> Posted by UncertainH</DIV></p><p>While the field strength may be zero, the potential is not as represented by -3GM/2r at the center of a sphere.&nbsp; The only time the potential is at zero is when the radius is extended to infinity.&nbsp; The potential will alway be negative and gravitatioinal time dilation is based on the difference between the potentials.&nbsp; I did not mean that at the center of the Earth, time dilation is infinite.&nbsp; I only meant that at the center of any sphere, time dilation is at it's maximum.&nbsp; Your clock ticks the slowest at the center.</p><p>I has to do with how much work is required to climb out of the gravity well.&nbsp; It has nothing to do with gravitational forces.&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[UncertainH]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>While the field strength may be zero, the potential is not as represented by -3GM/2r at the center of a sphere.&nbsp; The only time the potential is at zero is when the radius is extended to infinity.&nbsp; The potential will alway be negative and gravitatioinal time dilation is based on the difference between the potentials.&nbsp; I did not mean that at the center of the Earth, time dilation is infinite.&nbsp; I only meant that at the center of any sphere, time dilation is at it's maximum.&nbsp; Your clock ticks the slowest at the center.I has to do with how much work is required to climb out of the gravity well.&nbsp; It has nothing to do with gravitational forces.&nbsp;&nbsp; <br />Posted by derekmcd</DIV></p><p>Perhaps you misread that post. r is the radius of the sphere not the distance from the center of the sphere. The full equation for the gravitational potential inside a sphere of radius r where x is the distance of a point from the center of the sphere is -(GM/2r3)*(3r2-x2). As x approaches 0 (the center) the equation becomes -3GM/2r as x approaches r (the surface) the equation becomes -GM/r. Thus the potential has its largest negative value at the center and you are right the clocks tick the slowest. I think that would correctly be described as the minimum time dilation. The maximum dilation is as the distance from the sphere itself approaches infinity and the potential approaches 0 and the clocks tick the fastest.</p><p>&nbsp;</p><p><br /><br />&nbsp;</p>

 

[SpeedFreek]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Thus the potential has its largest negative value at the center and you are right the clocks tick the slowest. I think that would correctly be described as the minimum time dilation. The maximum dilation is as the distance from the sphere itself approaches infinity and the potential approaches 0 and the clocks tick the fastest.&nbsp;&nbsp; <br /> Posted by UncertainH</DIV></p><p>I would have it the other way round, if you want to think of it in terms of the clock that ticks the fastest or slowest.</p><p>The "fastest" clock would be a clock in free-fall in deep space, the furthest from any massive object. The "slowest" clock would be a clock at the event horizon of a black hole. The conditions on the surface of the Earth, or even at its centre, are far closer to the "fastest" clock than they are to the "slowest" clock.</p><p>The <em>arrow of time</em> should define the any "direction" we want to frame around time-dilation. Time moves in only one direction - forwards.</p><p>Only from the frame of reference where the clock is "fastest", where <em>every other</em> clock in the universe seems to be running slower than themselves, is there no time-dilation. Time runs at its baseline rate in empty space. Introduce mass or energy into that space, and you start dilating time when compared to that baseline!</p><p>(This post is not meant in any way to imply any notion of absolute time (which might only be possible in a truly empty universe) and ignores time-dilation due to 1: the expansion of the universe. 2: Having to move a clock at relativistic speeds to get it out of the Virgo Cluster. Also, all clocks tick at the same rate, a second per second. The difference in any <em>apparent</em> clock rate between yours and someone elses frame of reference is due to the difference in the frames of reference and their notions of simultaneity, which is ultimately due to their respective paths through space-time).</p><p><img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-wink.gif" border="0" alt="Wink" title="Wink" /> </p> <div class="Discussion_UserSignature"> <p><font color="#ff0000">_______________________________________________<br /></font><font size="2"><em>SpeedFreek</em></font> </p> </div>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>While the field strength may be zero, the potential is not as represented by -3GM/2r at the center of a sphere.&nbsp; The only time the potential is at zero is when the radius is extended to infinity.&nbsp; The potential will alway be negative and gravitatioinal time dilation is based on the difference between the potentials.&nbsp; I did not mean that at the center of the Earth, time dilation is infinite.&nbsp; I only meant that at the center of any sphere, time dilation is at it's maximum.&nbsp; Your clock ticks the slowest at the center.I has to do with how much work is required to climb out of the gravity well.&nbsp; It has nothing to do with gravitational forces.&nbsp;&nbsp; <br />Posted by derekmcd</DIV></p><p>I don't think so.&nbsp; At the very center of a uniform non-rotatint sphere the effect is as if the sphere were not there at all.&nbsp; The gravitational time dilation should be zero, the minimum value within the sphere.&nbsp; Here is a Wiki article that seems to be consistent with this reasoning.</p><p>http://en.wikipedia.org/wiki/Gravitational_time_dilation<br /></p> <div class="Discussion_UserSignature"> </div>