Understanding Time Dilation: How Fast Do You Have to Go?

In summary, time dilation is always applicable but may not be noticeable at everyday speeds. Relativistic effects become noticeable at speeds around 0.001 times the speed of light. The formula for time dilation can be used to see how it depends on the speed of the clock compared to the speed of light. A relativity calculator can be used to calculate the Relativistic Change Factor, which indicates how much longer a time interval on the spaceship will appear on Earth. When the spaceship makes a round trip, the effects of time dilation become more clear.
  • #1
afbla
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Hi I was wondering about how fast do you have to go before time dilation comes into effect

P.S. I'am no Qauntum physics professor so I don't know a lot about relativity
 
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  • #2
This has nothing to do with quantum physics, so I'm moving it out of the Quantum Physics Forum.

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

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
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
Time Dilation Formula

The measured rate of a moving clock is given by the following formula, where [itex]\Delta T_0[/itex] is the time interval according to the moving clock itself and [itex]\Delta T[/itex] is the time interval as measured by clocks in the "stationary" frame ([itex]v[/itex] is the speed of the clock according to the stationary frame):
[tex]\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 ([itex]v[/itex]) compared to the speed of light ([itex]c[/itex]).
 
  • #5
afbla, there’s a nice little relativity calculator http://www.1728.com/reltivty.htm?b0=299792 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
Vast said:
afbla, there’s a nice little relativity calculator http://www.1728.com/reltivty.htm?b0=299792 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.

Does "3.944 years on earth" mean the Earth planet has gone round the sun for 3.944 times?
 
  • #7
Sam Woole said:
Does "3.944 years on earth" mean the Earth planet has gone round the sun for 3.944 times?
Yes, that's what it means.
 
  • #8
Sam Woole said:
Does "3.944 years on earth" mean the Earth planet has gone round the sun for 3.944 times?

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
Doc Al said:
Yes, that's what it means.

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
Sam Woole said:
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."
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.

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?
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!

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.
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
Sam Woole said:
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.
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 traveling twin's clock will say 1 year has passed. The two twins don't disagree about what the other twin's clock reads--the traveling 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 traveling 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 traveling 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
JesseM said:
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 traveling twin's clock will say 1 year has passed. The two twins don't disagree about what the other twin's clock reads--the traveling 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 traveling 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 traveling 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.

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
Sam Woole said:
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.
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
Doc Al said:
Both twins agree that only 3.944 years have elapsed on earth.

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
Sam Woole said:
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.
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 traveling 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 traveling twin's clock." So if they departed at the age of n, this sentence tells you that the traveling twin would agree that the earth-twin was n+3.944, and the earth-twin would agree that the traveling twin was n+1.
 
  • #16
Doc Al said:
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!

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
Sam Woole said:
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.
All you can deduce is that relativity does not agree with your preconceptions about time.

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?
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 traveling 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 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 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.

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.
It's not that they work differently, it's that they were moved differently. One accelerates; the other doesn't.
 
  • #18
Doc Al said:
It's not that they work differently, it's that they were moved differently. One accelerates; the other doesn't.

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
Sam Woole said:
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.

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
Sam Woole said:
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.
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.


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.
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.
In such a contradictory confusion people like me certainly cannot learn relativity. I do not understand how could some have.
It seems to me that you much prefer the comfort of your preconceptions.

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.
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
Sam Woole said:
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.
No. As ZapperZ points out, only inertial (constant velocity) motion is relative, acceleration is absolute in SR.
Sam Woole said:
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.
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 traveling 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.
Sam Woole said:
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.
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
Sam Woole said:
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?
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 traveling twin's quartz clock (or his heart, or any oscillator that is traveling with him) will cycle at a different rate (from a previous rate) as the traveling 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 traveling 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 traveling oscillators might be influenced by.).

One manifestation of this is that the periods of any oscillators moving with the traveling 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 traveling twin's previous state of motion at rest wrt the earth-twin system.

To keep things simple just assume that the traveling 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 traveling twin will count the same number of Earth rotations or earth-sun revolutions as the earthbound twin for the trip interval, but the traveling 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 traveling 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.
 
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  • #23
Sherlock said:
The physical reason for this is because the traveling 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 traveling oscillators might be influenced by.).

One manifestation of this is that the periods of any oscillators moving with the traveling 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 traveling twin's previous state of motion at rest wrt the earth-twin system.
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}[/tex]. So if two twins are moving apart at constant velocity, you can analyze things from the earth-twin's frame, in which the traveling twin's clock will be slowed down; or you can analyze things from the traveling 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 traveling 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 traveling 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 traveling twin's clock and the earth-twins' clock read at the moment they reunite, and the answer will always be that the traveling twin's clock is behind, assuming the traveling twin accelerated and the earth-twin did not (on the other hand, if the traveling twin was moving away from the Earth at constant velocity, then you strapped huge rockets on the Earth and accelerated it towards the traveling twin, then the earth-twin's clock would be behind when they reunited).
 
  • #24
Sherlock said:
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 traveling twin's quartz clock (or his heart, or any oscillator that is traveling with him) will cycle at a different rate (from a previous rate) as the traveling 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 traveling 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 traveling oscillators might be influenced by.).
This is completely wrong. There is no "interaction" with anything that accounts for time dilation.
One manifestation of this is that the periods of any oscillators moving with the traveling 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 traveling twin's previous state of motion at rest wrt the earth-twin system.

To keep things simple just assume that the traveling 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 traveling twin will count the same number of Earth rotations or earth-sun revolutions as the earthbound twin for the trip interval, but the traveling 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.
No, not for the reasons 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.

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
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?
 
  • #26
Xargoth said:
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?

I'm not so sure the answers will help the other posters much, but here they are

There is no absolute reference point in the universe.

To add velocities, one has to use the relativistic velocity addition formula. Velocities measured in the same frame add as vectors - velocities measured in different frames have to be transformed to the same frame before they can be added. This transformation and addition process can be and usually is incorporated into a single formula as follows:

If person B is passing person A at a velocity v1 as measured in the A frame, and person C is passing person B in the same direction at a velocity V2 in the B frame, as per the following diagram


A
B------> v1 (relative to A)
C-----> v2 (relative to B)

The velocity of C relative to A is given by the formula

v_tot = (v1+v2)/(1+v1*v2/c^2)

NOT (v1+v2).
 
  • #27
To go back to something that is probably slightly more likely to help other posters, the main problem in relativity is forgetting things that one has learned that do not apply.

Specifically, one has to forget the idea of absolute simultaneity. One of the best routes, IMO, is to start with the most general possible view, which does not consider simultaneity at all, but only what individual clocks measure.

One takes it as a basic fact that two clocks can be syncrhonized when they are at the same point in space and time, but that if they follow different paths, they do not necessarily stay synchronized.

This is obviously weird, but it is very helpful if one can accept this as a given and work out the logical consequences, rather than insisting that this can't happen.

The twin paradox then becomes a total non-issue, we expect clocks to disagree when they meet up again as a matter of course.

It is by far the simplest to consider the case of special relativity, where space-time is flat, and there is no gravity. This will be the default assumption for the following remarks.

With such a general viewpoint, one wonders whether one can say anything at all about the behavior of clocks. There is in fact something very interesting one can say, which is this.

If one considers the set of all possible paths joining two events (points in space-time), there is one and only one path (given the assumption of flat space-time) that maximizes the value of a clock traversing it. This is the path taken by an object that moves naturally, the path taken by an object that does not experience any forces.

Any other path will have a shorter elapsed time than this natural path, called a "geodesic" path.
 
  • #28
pervect said:
A
B------> v1 (relative to A)
C-----> v2 (relative to B)

The velocity of C relative to A is given by the formula

v_tot = (v1+v2)/(1+v1*v2/c^2)

i gotto tell i find these equations funny, for example i can just say;

insert lower V here / insert higher V here, V can not be lower than 1..

if both V is equal than the answer is undefined

And now try going over 1..No offense;I am simply curious, is the equation formed by the idea that you can't exceed c, or the equation came with test results in favour of the constant c?

And i really would appreciate if i can get a link to the journal that this equation was published and was it Einstein himself? I don't even know who published what yet..I am quite new to this topic, and need to learn a lot before i become convinced that the instruments i am using are reliable..I have quite a problem with information you see, i never trust anything :redface:
 
  • #29
Xargoth said:
i gotto tell i find these equations funny, for example i can just say;

insert lower V here / insert higher V here, V can not be lower than 1..
You mean v_tot can't be lower than 1? And by "1", do you mean 1c? Actually, if both v1 and v2 are lower than c, then v_tot will always be lower than c as well. And if either v1 or v2 are equal to c, then v_tot is equal to c.
Xargoth said:
if both V is equal than the answer is undefined
That's not true, you must be using the equation wrong. For example, let v1=v2=0.5c, then v_tot = (0.5c + 0.5c)/(1 + 0.5c*0.5c/c^2) = 1c/(1 + 0.25)
= 0.8c.
Xargoth said:
No offense;I am simply curious, is the equation formed by the idea that you can't exceed c, or the equation came with test results in favour of the constant c?
The equation can be derived from the Lorentz transformation, which tells you how two different coordinate systems are related to each other. If I use coordinates (x,y,z,t) and you use coordinates (x',y',z',t'), and in my rest frame you are moving at velocity v along my x-axis, and the origins of our coordinate systems coincided at time t=t'=0, then any event which has coordinates x,y,z,t in my rest frame will have the following coordinates in your rest frame:

[tex]x' = \gamma (x - vt)[/tex]
[tex]y' = y[/tex]
[tex]z' = z[/tex]
[tex]t' = \gamma (t - vx/c^2)[/tex]

where [tex]\gamma = 1/\sqrt{1 - v^2/c^2}[/tex]

This coordinate transform is itself derived from the assumption that each observer uses a network of physical rulers and clocks at rest with respect to themselves to assign coordinates to events, and that each observer synchronizes different clocks in his system using the assumption that light always travels at c in his own rest frame. If you also make the assumption that the laws of physics will look the same in each observer's rest frame, you get this coordinate transformation.
Xargoth said:
And i really would appreciate if i can get a link to the journal that this equation was published and was it Einstein himself? I don't even know who published what yet..I am quite new to this topic, and need to learn a lot before i become convinced that the instruments i am using are reliable..I have quite a problem with information you see, i never trust anything :redface:
The equation can be found in section 5 of Einstein's original 1905 paper on relativity (the Lorentz transformation equations appear at the end of section 3).
 
  • #30
JesseM said:
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, per se, doesn't say it. But, I think modern physics (eg., interaction with zero point fields;the idea of dark energy; the idea of gravitational fields as quantizable wave fields of varying complexity and intensity; etc.) is moving in this direction, or at least is not opposed to the general idea that alterations in the periods of oscillators which are moving anomalously wrt previous states of systems which encompass them are due to corresponding differences in their interactions with various wave phenomena which pervade and permeate the system -- maybe on a hierarchy of scales.

I don't think anything that I wrote contradicts the kinematics or the standard geometrical interpretations of relativity. It's just that when I considered these as real physical effects and was given a hint as to the immediate physical cause of why the period of an oscillator in some state of motion, v, is altered by an amount that's proportional to the Lorentz transformation when it's in some state of motion > v or < v, then the idea of a cumulative effect (eg., differential aging, the twin clock effect) was easier to accept -- and coupled with the idea that deeper physical explanations will be dealt with (maybe, eventually) using wave mechanics, then I could get on with learning the nuances of Relativity as a calculational tool rather than getting bogged down puzzling over it's apparent physical non-intuitiveness.

JesseM said:
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 . So if two twins are moving apart at constant velocity, you can analyze things from the earth-twin's frame, in which the traveling twin's clock will be slowed down; or you can analyze things from the traveling 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.
Time dilation is a symmetrical effect. Ok. I don't think I've written anything contrary to this, so I don't understand your point here.

Differential aging implies anomalous motion. But anomalous motion itself wouldn't account for the observed cumulative effects unless that motion was causing some real physical changes (idealized partly as length contraction and mass increase) in the oscillators that undergo the motion and index the changes during the intervals of their anomalous motion.
And physical change implies interaction.
 
  • #31
Janus said:
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.
What's pre-Relativistic about my notion of time? Time is what you read on a clock. A clock is an oscillator of one sort or another, and an accumulator that indexes (counts) the oscillations. The time of an event (in Relativity) is the reading on a clock next to the event. And from Relativity we know that as an oscillator's state of motion changes, then it's period and rate of timekeeping changes.

Is it the idea that so called 'empty space' isn't empty that you disagree with? This idea is part and parcel of standard modern physics, afaik.
 
  • #32
Sherlock said:
Is it the idea that so called 'empty space' isn't empty that you disagree with? This idea is part and parcel of standard modern physics, afaik.

That's true, but it's not used as an "explanation" for time dilation in mainstream modern physics. If you have some non-mainstream ideas that you'd like to develop, then according to PF's rules you need to submit them to the "Independent Research" forum.

https://www.physicsforums.com/forumdisplay.php?f=146

Advocacy of non-mainstream theories is off-topic in the other forums (such as this one).
 
  • #33
Sherlock said:
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. :-))
Thank you, Sherlock. You inspired me on the idea of natural clock. As a result I felt the time dilation idea cannot be properly justified.
 
  • #34
ZapperZ said:
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.

ZapperZ, maybe you were talking about experiments in earth, or in a gravitational field. If you were in a space where gravity is zero, can you detect you are accelerating?

Even if I got it wrong with this acceleration, which was considered as absolute as JesseM pointed out, do we know which motion, relative or absolute, will cause time dilation?
 
  • #35
Sam Woole said:
ZapperZ, maybe you were talking about experiments in earth, or in a gravitational field. If you were in a space where gravity is zero, can you detect you are accelerating?
Yes, you will experience G-forces when you accelerate in space, just like how when you're in a car that's accelerating you feel yourself pushed back into the seat. From the point of view of an inertial frame, this isn't a true "force" like gravity (it's sometimes called a http://www.hcc.hawaii.edu/~rickb/SciColumns/FictForce.04Feb96.html for this reason), it's just that the car seat is accelerating and it has to overcome the inertia of your body to accelerate it to the same speed. But from the point of view of your own non-inertial frame, it feels just like a force is pulling you backwards.
Sam Woole said:
Even if I got it wrong with this acceleration, which was considered as absolute as JesseM pointed out, do we know which motion, relative or absolute, will cause time dilation?
According to relativity time dilation is a function of velocity rather than acceleration. But if two clocks are moving apart at constant velocity, then in each clock's own rest frame it will be the other clock that is running slower--the only situation where you get an objective answer to which clock is "really" behind is the one where one clock turns around (accelerates) and moves back towards the first clock, so they can meet at a single location in space and see which clock is behind when they meet.
 
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