# Time Dialation Question

1. Sep 28, 2005

### afbla

Hi I was wondering about how fast do you have to go before time dialation comes into effect

P.S. I'am no Qauntum physics professor so I don't know alot about relativity

2. Sep 29, 2005

### Tom Mattson

Staff Emeritus
This has nothing to do with quantum physics, so I'm moving it out of the Quantum Physics Forum.

It's not as though relativistic effects "kick in" at some critical speed. They are always applicable. It's just that they aren't noticeable at everyday speeds.

Now, if you want to know when relativistic effects are noticeable, that would depend on how sensitive your instruments are.

3. Sep 29, 2005

### pervect

Staff Emeritus
To give a very rough idea, at .001 times the speed of light, time dilation effects are about half a part per million. That's 3*10^5 meters/second.

Cesium clocks have accuracies on the order of a few parts per 10^14, so they can detect time dilation at speeds on the order of slightly over 10^-7 c, say 50 meters/sec, which is only around a hunderd miles/hour.

This comes from gamma = 1/sqrt(1-(v/c)^2) ~ 1 + (1/2)*(v/c)^2 for small v via a taylor series expansion.

4. Sep 29, 2005

### Staff: Mentor

Time Dilation Formula

The measured rate of a moving clock is given by the following formula, where $\Delta T_0$ is the time interval according to the moving clock itself and $\Delta T$ is the time interval as measured by clocks in the "stationary" frame ($v$ is the speed of the clock according to the stationary frame):
$$\Delta T = \frac{\Delta T_0}{\sqrt{1 - \frac{v^2}{c^2}}}[/itex] Thus a time interval measured on the moving clock is observed to take longer according to the stationary frame; this is the meaning of time dilation and the statement "moving clocks run slow". You can use this formula to see how time dilation depends on the speed of the clock ($v$) compared to the speed of light ($c$). 5. Sep 29, 2005 ### Vast afbla, there’s a nice little relativity calculator here especially for time dilation. Where it says “input” just put in a number close to the speed of light such as 290000 km/s and click on km/second. You’ll notice it gives a Relativistic Change Factor, which means, if it’s 3.944802246249386 for the above input of 290000 km/s, 1 year of your time on a spaceship traveling at such a speed would be 3.944 years on earth. Also notice that for significant time dilation to take affect the speed needs to be at least 0.9 the speed of light. 6. Oct 3, 2005 ### Sam Woole Does "3.944 years on earth" mean the earth planet has gone round the sun for 3.944 times? 7. Oct 3, 2005 ### Doc Al ### Staff: Mentor Yes, that's what it means. 8. Oct 3, 2005 ### pervect Staff Emeritus More or less, though picking nits, there are a number of different definitons of "year", depending on how one measures the earth going around the sun. One usual definition is the time between vernal equinoxes, however there are a number of subtle isues here. To a level of 4 significant figures, though, the statement will be correct by most any of the possible subtly different definitions of "year". As far as timekeeping goes, though, our atomic clocks keep much more accuarate time than astronomical motions do, and have become our primary time standard. Atronomical motions will also not be very useful for a hypothetical space traveler as far as measuring time - he will not base his units of time by the astronomical motions of a distant planet around a distant sun (he won't even be able to observe them in real time). He will base his time on the atomic clocks that he caries with him. 9. Oct 4, 2005 ### Sam Woole Thank you, Doc Al. Maybe you knew that my question has something to do with time dilation. The whole sentence reads: "1 year of your time on a spaceship traveling at such a speed would be 3.944 years on earth." When the twin on the spaceship met his twin brother on earth, it means both twins have spent an equal time interval t, from departure on earth to meeting on earth. There could not have been two time intervals. Then a question has arisen. In this same time interval t, what has the planet earth done? Has it gone round the sun 1 time, or 3.944 times? Which is right, 1 or 3.944? Undoubtly there can be only one right, the 3.944. The 1 year on the spaceship is wrong. Does this mean that time dilation is nothing but false? It cannot be justified any way we try. 10. Oct 4, 2005 ### Doc Al ### Staff: Mentor This means that folks on earth would measure 3.944 years passing on earth while the folks on the spaceship only experienced 1 year. Note that this is according to the earth. Things get interesting--and unambiguous--when the spaceship is able to make a round trip. Again you have to realize that time is not an absolute, it really does depend on the relative motion of the frame doing the measuring. When the ship returns to earth, the two brothers will really be different ages! If you are really interested in learning about time dilation and relativity, stick around and ask questions. (You aren't ready to understand the traveling twins quite yet--you first need to understand the relativity of simultaneity!) But please don't start up again with the accusations of lying, cheating, and claiming that relativity is "nothing but false". It's tiresome. 11. Oct 4, 2005 ### JesseM Time dilation is based on what would be read by clocks moving along with each observer--in this case, the earth-twin's clock will say that 3.944 years have passed when they reunite, while the travelling twin's clock will say 1 year has passed. The two twins don't disagree about what the other twin's clock reads--the travelling twin agrees that 3.944 years have passed on the earth-twin's clock, and the earth-twin agrees that 1 year has passed on the travelling twin's clock. The earth going around the sun is just like another type of "clock" that stays at rest relative to the earth-twin, so of course everyone agrees it elapses 3.944 years as well. If the travelling twin carried a copy of the earth and sun along with him on the trip, then the duplicate earth would have only completed 1 orbit when the two twins reunited, and both twins would agree that this was true. 12. Oct 5, 2005 ### Sam Woole JesseM, my understanding of your words above is: if both twins departed at the age of n, when they united both twins agreed they were both (n + 3.944) years old, the same age according to the clock kept by the earth twin. On the other hand, according to the clock kept by the traveled twin, both agreed that both were (n + 1) years old, the same age. That is to say, whichever way we looked at it, there is no differential aging, no time dilation. 13. Oct 5, 2005 ### Doc Al ### Staff: Mentor Both twins agree that only 3.944 years have elapsed on earth. But what counts as far as aging goes is the time elapsed on the clocks that move along with each twin. Both twins will agree that the traveling twin is physically younger than the stay at home twin. The twins themselves are biological clocks. To make the difference more apparent, increase the speed so that 100 years go by on earth while only a year passes on the ship. When the twin returns to earth, he'll find his brother long dead. 14. Oct 5, 2005 ### jtbell ### Staff: Mentor And both twins agree that one year has elapsed on on the traveling twin's spaceship. (just to make this point clear) For a worked-out numeric example that demonstrates how the twins can arrive at this agreement, see posting #3 in this thread. 15. Oct 5, 2005 ### JesseM I don't know how you got that conclusion from my words. What I said was: "The two twins don't disagree about what the other twin's clock reads--the travelling twin agrees that 3.944 years have passed on the earth-twin's clock, and the earth-twin agrees that 1 year has passed on the travelling twin's clock." So if they departed at the age of n, this sentence tells you that the travelling twin would agree that the earth-twin was n+3.944, and the earth-twin would agree that the travelling twin was n+1. 16. Oct 5, 2005 ### Sam Woole While I do apologize for my offensive charges, on the other hand I do believe that there must be something wrong with the relativity theory, as can be deduced from your words above. When the traveling twin left, he was in his inertia frame. The clock he carried was at rest in his frame. His clock therefore would work exactly like any other clocks in inertia frames such as the one carried by the earthbound twin. Don't you agree? If you do, then both clocks would registered one identical departure time D when the twins departed; then both clocks would register one identical arrival time A when they met again. (A - D) would give one time interval, which means both twins are of the same age. If you don't agree, then show me why the clock carried by one twin would work differently from the clock carried by the other twin. 17. Oct 5, 2005 ### Doc Al ### Staff: Mentor All you can deduce is that relativity does not agree with your preconceptions about time. All clocks do work the same way. They just don't work the way you think they do! As measured from any inertial frame, the rate at which a moving clock operates depends on its speed. As long as the two clocks remain in their single inertial frames, they both can equally claim that the other's clock runs slow--perfect symmetry. But if the travelling twin makes a return trip to earth he cannot possibly remain in a single inertial frame; he must accelerate and thus change frames. (To really understand this you'll have to learn some relativity.) It would be a problem if the motion of the twins were perfectly symmetric and yet their clocks read different times when they reunited. But their motion is not symmetric! One remains in an inertial frame; the other accelerates. If you like, you can arrange for the two clocks to read the same at the start of the trip. But once you do, you'll find that they don't read the same when they are reunited. The only way you can deduce that the clocks would read the same is if you ignored what relativity has taught us about how moving clocks--and time itself--actually work and just assumed that time flows at the same rate for everyone, regardless of relative motion. It's not that they work differently, it's that they were moved differently. One accelerates; the other doesn't. 18. Oct 7, 2005 ### Sam Woole I believe you were contradicting not only yourselves but also Einstein. Here you said "One accelearates; the other doesn't" This directly contradicted the principle of relativity, which means (to me at least) all motions are relative. If the spaceship accelerates relative to earth, earth is also accelerating relative to the spaceship. Based on this known principle, the earthbound twin would see the clock on the spaceship to have registered 1 year while his own has done 3.944 years. Similarly the spaceship twin would see the clock on earth has registered 1 year while his own has done 3.944 years. Your words : "As long as the two clocks remain in their single inertial frames, they both can equally claim that the other's clock runs slow--perfect symmetry. " Namely, it is always the other guy's clock running slow, not mine, acceleration or not. The fact that it is the other guy's clock running slow also agrees with the physical phenomenon as we know it, that light needs a time interval to reach the earth from the spaceship, say 2.944 years, and vise versa. When the two clocks come together, united on earth, both certainly will read the samething, no time dilation, no differential aging. Not only you appeared contradicting yourselves, but also you were using languages that don't agree with convention. JesseM said the space twin carried a copy of the sun-earth. He also used the word "duplicate". If it was a duplicate, it must duplicate the number of orbits the earth has done. If it did not duplicate the number of orbits, then it was not a copy of the solar system. It was completely a different system, completely a different kind of clock, not the clock of identical construction specified by Einstein. In such a contradictory confusion people like me certainly cannot learn relativity. I do not understand how could some have. If it were true that speed would make us younger, it must also be true that life forms on earth will never die, because the earth is always moving at c relative to light. But we are all dying. It is obvious to me that there is no time dilation. 19. Oct 7, 2005 ### ZapperZ Staff Emeritus This is wrong. You can ALWAYS do an experiment to detect that you are accelerating. You cannot do an experiment to detect if you're moving without using another frame as a reference. The accelerating frame can always tell that it is accelerating, and can tell that another frame isn't. Zz. 20. Oct 7, 2005 ### Doc Al ### Staff: Mentor You need to learn to distinguish an inertial frame (non-accelerating) from an non-inertial frame (accelerating). The principle of special relativity can be written as "All inertial frames are equivalent for all experiments; no experiment can measure absolute velocity". If one frame accelerates, things are very different. Wrong again. A solar system like sun-earth works exactly like any other clock. You just mistakenly think that the operation of a clock is independent of its speed with respect to the frame observing it. It seems to me that you much prefer the comfort of your preconceptions. Wrong again. This is exactly the opposite of what relativity actually says. The principle of relativity says that I age at the usual rate according to my clocks, and that it would be a violation of physics if it were any other way. And you (on that uniformly moving spaceship) age at the usual rate according to your clocks. You can't tell that you are "really" moving (at least not by how your own clocks work). 21. Oct 7, 2005 ### JesseM No. As ZapperZ points out, only inertial (constant velocity) motion is relative, acceleration is absolute in SR. That's like saying "if it was a duplicate, it must have the same velocity and position as the original sun-earth". That's not the standard meaning of "duplicate"--a duplicate doesn't share every property in common with the original (certainly not the exact same spatial location, which would be impossible), it just means that if you exchanged the places of the duplicate and the original, nothing would be changed, it would be impossible to tell the difference. If the travelling twin (the one who accelerates in turning around) drags the original earth/sun system along with him while making the trip, and the stay-at-home twin (the one who does not accelerate) keeps the duplicate sun/earth in his own location, then it will be the original earth that will have only experienced 1 orbit around the original sun when they meet up, while the duplicate earth will have experienced 3.944 orbits around the duplicate sun. There is no valid inertial reference frame where light is at rest and we are moving at c--this is because of the postulate which says that the laws of physics must work the same way in every inertial reference frame, and since light waves move at c in some frames, they must move at c in every valid frame. 22. Oct 7, 2005 ### Sherlock No, not exactly. (And, let me say that I had the same problem grasping this as you seem to be having. I should also say that I'm not a physicist, so don't take anything I write as being necessarily correct -- of course, I'm sure the mentors will be on top of it. :-)) I've learned to think about it like this. The travelling twin's quartz clock (or his heart, or any oscillator that is travelling with him) will cycle at a different rate (from a previous rate) as the travelling twin accelerates away from the earth-twin system (or inertial frame) and eventually assumes a different uniform velocity relative to the earth-twin system (that is, different from his previous velocity relative to the earth-twin system prior to take off). The physical reason for this is because the travelling twin is interacting with the physical stuff (wave phenomena and interactions on a level that's not directly amenable to our sensory perception) that is presumed to pervade and permeate the empty (to our normal senses) space surrounding the earth-twin system (Lets assume the traveller doesn't leave our solar system -- which can be viewed as a complex interacting wave system where only the most intense regions -- ponderable bodies -- of interaction are amenable to our senses. Add to this the fact that our solar system is itself part of a local star system, which is part of the Milky Way galactic system, which is part of a galactic group, and so on. And also figuring in whatever is happening to space on a universal scale, eg. expansion due to kinetic energy imparted via the big bang, and there's really quite a lot of wave activity that the travelling oscillators might be influenced by.). One manifestation of this is that the periods of any oscillators moving with the travelling twin will increase (proportional to length contraction wrt the direction of motion), and therefore the rate at which they accumulate will decrease relative to the travelling twin's previous state of motion at rest wrt the earth-twin system. To keep things simple just assume that the travelling twin's clock and the earthbound twin's clock keep time at exactly the same rate when they're both on earth. During his round-trip, the travelling twin will count the same number of earth rotations or earth-sun revolutions as the earthbound twin for the trip interval, but the travelling twin's clock (and his heart and other oscillators moving with him) will accumulate fewer total oscillations for the trip than the earthbound twin's clock -- for the physical reason(s) given above. The Lorentz time transformation provides a way to calculate the difference between the accumulated cycles of two previously synchronized (or, preferably, equal wrt some common rest frame of reference -- like when they're both on earth) osclillators (clocks) that are moving wrt each other. Now, admittedly, the rationale that's been offered above is pretty hazy. But, I think it makes sense as a general approach to understanding that the effects of acceleration and velocity relative to a previous state of motion have to do with real physical interactions that produce real physical changes. When the traveller returns to earth his oscillators assume the periods that are normal for them in that state of motion (at rest wrt the earth). But, while he is travelling at high speed throughout the solar system they change in proportion to his velocity relative to his earthbound state. I guess I should add that the calculation might be complicated by how close the traveller gets to the centers of strong gravitational fields. I'm not sure how moving uniformly toward the center of a gravitational field is related to accelerating away from it. Last edited: Oct 7, 2005 23. Oct 7, 2005 ### JesseM This explanation does not quite correspond to what modern physics says about the reason one twin's clock is behind the other when they reunite--relativity does not say that clocks slow down because their velocity is higher relative to some "physical stuff" that is filling space, in fact that sounds like an aether theory, which relativity displaced. Relativity says any situation can be analyzed from any inertial frame you like, and whichever frame you use in analyzing the problem, clocks at rest in that frame will tick at a normal rate and clocks moving at velocity v relative to that frame will be slowed down by a factor of [tex]\sqrt{1 - v^2/c^2}$$. So if two twins are moving apart at constant velocity, you can analyze things from the earth-twin's frame, in which the travelling twin's clock will be slowed down; or you can analyze things from the travelling twin's frame, in which the earth-twin's clock will be slowed down. There is no objective truth about which clock is "really" running slower, according to relativity. However, if the travelling twin accelerates to turn around and return to earth, then no matter which inertial frame you choose, his clock will be running slower than the earth-twin's in at least part of his journey (for example, if you analyze things from the frame where the travelling twin is at rest during the outbound leg of his trip, then during the inbound leg his velocity will be even greater in this frame than the earth-twin's velocity, hence his clock will be running even slower during this leg). It works out so that no matter which frame you choose, you will always get the same prediction about what the travelling twin's clock and the earth-twins' clock read at the moment they reunite, and the answer will always be that the travelling twin's clock is behind, assuming the travelling twin accelerated and the earth-twin did not (on the other hand, if the travelling twin was moving away from the earth at constant velocity, then you strapped huge rockets on the earth and accelerated it towards the travelling twin, then the earth-twin's clock would be behind when they reunited).

24. Oct 7, 2005

### Janus

Staff Emeritus
This is completely wrong. There is no "interaction" with anything that accounts for time dilation.
No, not for the reasons given above.
The rationale above is completely wrong in terms of Relativity. The problem is that you are still trying to hold on to a Pre-Relativistic notion of "time", and this is incompatable with how Relativity treats time.

25. Oct 8, 2005

### Xargoth

Two simple questions then, i am sure the answers to these will clear the confusion here;

-While the universe is not STATIC, then what is the absolute reference point. Is there an absolute referance point?

-What prevents me to move relatively higher than "c"? I can have a vectoral velocity higher than speed of light for the reference point x? or Does relativity work some other way?