Time dilation question with a clock and sped of light?
830077
I think it would be around 75%\n\nSo for in 1 hour earth time the clock would of done 30 minutes.
101086
One example to measure the speed of red light is to use equipment that includes a LED (light-emitting diode) that emits a regular series of pulses of red light that are only 20 nanoseconds in duration.\n\n    (A nanosecond is one-billionth of a second or 1/1,000,000,000 second. It is also written as 10-9 seconds, 10^-9 or 1e-9, where the -9 indicates the number of zeros in the denominator.)\n\nThat means the pulse of light blinks on for 20 billionths of a second or 20/1,000,000,000 second and blinks 40,000 times per second. Having such a short pulse allows the scientist to measure the difference in time it takes to travel two different paths. If the duration was longer, the distance traveled would have to be longer.\n\nBy splitting the light beam with a half-silvered mirror, one beam travels to a mirror 10 meters away and then back to a photodiode detector. The other beam is reflected off a mirror only a few centimeters away. The time difference for the two beams is about 67 nanoseconds, which can be displayed on a regular dual beam oscilloscope.\n\nThe total distance the light travels is 20 meters, which equals 0.02 kilometer (20/1000).\n\nThe speed of light is then:\n\nc = 0.02 kilometers / 67*10-9 seconds = 298,500 kilometers per second\n\nThis is a fairly accurate reading and pretty close to the actual speed of 299,792 km/s.
1002813
The key point to this whole question is if you could travel faster than light.  Since at our present knowledge of physics says no, and our current tech is not able to get anywhere remotely near the speed of light let alone faster than it, we will assume this hypothetical question implies these two things don't matter.\n\nIf so, then yes you could see into the past if you had an adequate viewing device which could discern the light you would receive at the end of your journey--light which you left behind by going faster than it.  But to see exaclty the number of years in the past as the light year distance you travelled, you woould have to move instantaneously.  If you had any non-relativistic travel time, you would have to consider time of your journey to determine how far in past you were seeing.  If you travelled 1.1 times the speed of light, you would only get to see .1*light year distance in years past.
789176
I'll take a simpler interpretation of the question and say one second is "calculated" to be 1/86400th of a day, where 86400 is the product of 24 hours times 60 minutes/hour times 60 seconds/minute. Problem is that the day (say from local noon to local noon by measuring the sun's highest point in the sky) is not constant throughout the year, so it was decided to use the average day looking over the whole year. \nFor very precise timings of experiments in the laboratory, one needs a very good pendulum or something else that oscillates or vibrates at a very constant frequency; then you can work back to what fraction of a day or how many seconds something takes. \nIn modern times cesium atoms provided the constant well defined measurable frequency, and scientists agreed to "define" one second as some number of these oscillations (9,192,631,770). Now when the earth decides to rotate at a different rate, scientists add leap seconds instead of redefining the second every year. \nLately the reliability of the cesium atom has been questioned, and since the speed of light (in vacuum) is defined as a fundamental "constant", the second can indeed be defined as how long it takes light to go some distance (299,792,458 meters), but that seems circular to me because I see a meter defined as how far light goes in 1/299,792,458 of a second.
12211
Velocity distorts time.  The faster you go, the slower time goes for you.  This effect is negligable unless you are going very close to the speed of light.  In fact the formula is:\nt'=t/sqrt(1-v^2/c^2\nwhere t' is the time that has passed from a stationary observer, t is the time that has passed for you, sqrt means square root and ^2 means squared.  As you can notice, when v is small, since c^2 is so huge, you get a miniscule decimal, and 1 minus that number is basically as small as 1.0000000000001 or something, and so time is not really affected.  As v gets larger(it can never be the speed of light, then the denomonator would be zero) that fraction becomes closer to 1, so when you subtract it from one, the denomonator gets smaller and smaller, meaning t gets smaller(in order for t' to remain constant) this means that time slows down for the object moving at such velocities!  This formula is correct, it is part of Einstein's theory of relativity, and I have had to derive it myself from ordinary mechanics formulas.  \n\nAccording to the same factor, length also contracts.  It has been observed in certain tiny particles coming to the earth's surface.  These particles have a certain lifespan before they become unstable and decay.  They are also moving at a certain speed. (which is very fast; close to the speed of light.) You should simply be able to calculate v=d/t (velocity=distance/time) to find out how far they would travel.  However, they somehow travel farther than that.  When this calculated distance is put into that formula, we come up with the correct distance they actually travel!
798878
Yes time "dilation".\n\nt.sub.b = t.sub.a/sqrt(1-(v/c)^2)\nEinsteins Special Theory of Relativity, and earlier references.\nwhere .sub.  means following character is a subscript\nsqrt(   ) means square root()\nx^2  means x squared, v is objects velocity and c is speed of light.\nand t (should actually be delta t) = (t.sub.1 - t.sub.0) or "duration"\n\nThe time duration of person "b" traveling at a speed of "v" observed by (relative to) person "a" is the above function of the speed and the duration.\n\nas "v" gets faster and faster, v/c gets closer and closer to 1 and the denominator gets closer and closer to zero  (1 minus v^2/c^2), see?  Which means  the time duration person "b" seems to experience as observed by person "a" stretches out (it dilates) or slows down.  Things seem to take longer when "b" is doing them than when "a" does them, according to person "a".  \n\nExample:  two fire crackers with exactly the same length fuses which will explode exactly 5 seconds after they are lit.  One is floating outside person "a"'s spaceship and the other is about to zoom by (being launched by a friend using a mass driver cannon.... )  Person "a" powers up her laser and takes the shot as the fire cracker zooms by and miracle of miracles! lights both with one shot!  The one floating there goes off exactly 5 seconds later.  Question when did the other one go off?  Answer: depends on how fast is zoomed by!  if v/c=0.8 then (v/c)^2=.64  and the square root of (1-.64) = sqrt(.36)=0.6 so 5/0.6= 8.33 seconds or 3.33 seconds after the other one. \nThe mind boggling thing is that if the friend was zooming along with the speeding fire cracker, he would see that his went off exactly 5 seconds after the laser hit the fuse but the one floating outside his friends space craft took another 3.3 seconds to explode.
351868
Time Dilation: Time dilation is a consequence of Albert Einstein's theories of relativity. It occurs when one observer finds a clock which is measuring proper time ticking slow with respect to the relative time defined by his own clock (which is also measuring proper time). In special relativity, this occurs for clocks which are moving with respect to an inertial observer. In general relativity this occurs for clocks at lower potentials in a gravitational field, and is called gravitational time dilation.
53950
Your "authority", if they was refering to relativity and time dilation, was wrong.\n\nThe special theory of relativity, including time dilation, is a well established experimental fact.  It has been tested and verified in many different ways (particle acceleators, cosmic rays, satellites, etc.).
1062755
That question goes in line with my latest thought on einstein theory of relativity.  If something approaches the speed of light (or at the speed of light) time slows for it but not the observer, so how do we measure it if the light itself is time compressed?
1102952
Gravitational time dilation caused by a spherically symmetric, nonrotating massive body can be computed exactly by using the Schwarzschild solution to the Einstein field equations. This is also a good approximation to a slowly rotating, roughly spherical body like the Earth. The time dilation given is:\n\nt0=t1√(1-2GM/(rc²)) where:\n\nt0 is the time between two events for an observer within the gravitational field\nt1 is the time between those same events for an observer well outside the gravitational field\nG is Newton's  gravitational constant\nM is the mass of the object creating the gravitational field\nr is the distance of the observer in the gravitational field from the center of said field\nc is the speed of light
469813
is the phenomenon where the observed time rate of an observer's reference frame is different from that of a different reference frame. In Albert Einstein's theories of relativity the effect is manifested in two ways:\n\n    * In special relativity, clocks that are moving with speed v with respect to an inertial system of observation are found to be running slow. This effect is embodied in the Lorentz transformation.\n\n    * In general relativity, clocks at lower potentials in a gravitational field are found to be running slower. This gravitational time dilation is only briefly mentioned in this article but is described elsewhere (see also gravitational red shift).\n\nThe formula for determining time dilation in special relativity is:\n\n    &#92;Delta T = &#92;gamma &#92;Delta T_0 &#92;!\n\n        where\n\n            ΔT is the time period as measured by a stationary observer,\n            ΔT0 is the time period of the moving object as seen by the stationary observer,\n            &#92;gamma &#92;equiv &#92;frac{1}{&#92;sqrt{1 - &#92;frac{u^2}{c^2}}} is the Lorentz factor,\n            u is the relative velocity between the observer and the object, and\n            c is the speed of light.\n\n\nNote that this formula assumes that both object and observer are in the same spot, and that their +x direction is the same. The lorentz transforms can be used for more general cases.\n\n\nThis effect is extremely small at ordinary speeds, and can then be safely ignored. It is only when an object approaches speeds on the order of 30,000 km/s (1/10 of the speed of light) that it becomes important.\n\nTime dilation by the Lorentz factor was predicted by Joseph Larmor (1897), at least for electrons orbiting a nucleus. Thus "... individual electrons describe corresponding parts of their orbits in times shorter for the [rest] system in the ratio :&#92;sqrt{1 - v^2/c^2}" (Larmor 1897). Time dilation of magnitude corresponding to this (Lorentz) factor has been experimentally confirmed.
63805
Actually there is no speed of dark.\nBut according to me it should be the same as the speed of light because as soon as the light goes the dark appears.So it must be as the speed of light i.e. 3* 10^8 meters/seconds
191791
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/tdil.html\nThe equation is true and this first link has his time dilation equation along with some others.\nWe can justify it by experimentation. The following link describes the Ives-Stilwell experiment.  http://www.mpi-hd.mpg.de/ion-storage/Lorentz/relativity.html
269338
The speed of light is 3 x 10^10 cm/sec or 3 x 10^8 m/sec or 3 x 10^5 km/sec.\n\nWavelength = speed of light / frequency\n\nBe sure you use the right units.
333385
According to Einstein, all time is relative, actually.  The simple act of moving through space changes time relative to everyone.  It's only near the speed of light, however, that the effect really becomes noticeable.  As you approach the speed of light, time slows to a stop.  You wouldn't directly notice it, but the rest of the universe would seem to be slowing to a stop, in your perception.  I could explain it here, but Wikipedia sums it up rather nicely:
506771
Actually by what Einstein explained, da light will still be in  front of da car, it won't go behind u no matter how fast u r going even if u r faster dan da speed of light. Weird? rit. I too thought so, It was explained in discovery channel. Well I'll try 2 explain it 2 u, But it wud hav been easier if I cud actually show it by som animations. When u r travellin at da speed of light, time stops. Ya it does. Time is somthin like per user setting, it is unique for u & each & every one. Think dat ur carrying an Einstein's clock, which is a simple devise with 2 mirrors facing each other with a photon bouncin of the mirror. Each time it bounces from a mirror u'll hear a tic sound, that will b 1 second for u. Say u travel at a speed little less dan light u r movin forward but da photon has 2 move a lot more distance than it initially had been, which was in a straight line, but now it's motion will b som what like a hypotenuse, so it has 2 travell more distance dependin on ur speed. So ur 1 sec will not be da same as 1 sec is for som 1 on earth. So actually ur headlight will be actually movin faster dan u no matter what
958698
If t is the time that has been measured in the clock's reference frame, then t/2 is the time that has been measured by the observer.\n\nNow t = (t/2)[1-(v^2)/(c^2)]^(1/2)\n\nthe t's cancel and you are left with 1 = (1/2)[1-(v^2)/(c^2)]^(1/2)\nor [1-(v^2)/(c^2)]^(1/2) = 1/2\n\nsquare both sides to get 1 - (v^2)/(c^2) = 1/4\n\nsubtract one from each side to get -(v^2)/(c^2) = -3/4\n\nor (v^2)/(c^2) = 3/4\n\nthis means that v^2 = (3/4)(c^2)\n\ntaking the square root of both sides gives the final answer:\n\nv = [3^(1/2)]c/2.\n\nNot the most intuitive answer in the world.