How time dilation occurs

[Uclock]

HydrogenBond

 

Time dilation takes energy to occur. In the case of special relativity, the energy comes from the kinetic energy in velocity. If we were to give off a light signal from a rocket ship traveling near C, moving perpendicualr to a stationary earth reference, the signal would expand as it goes from our zone of distance contraction, into the low distance contraction on the earth reference. The perpendicular motion factors out relative motion to or from us. It would be like light coming out of extreme gravity. It would red shift as space expands as we move away from the gravity zone. In the rocketship's case the energy added to get the motion creates a distance-time contraction shroud around the ship.

 

Can you explain what you mean by ‘distance contraction’?

 

 

If you look at EM energy, it is wavelength times frequency. Notice how the blue light compares to the red light; it has a smaller wavelength (distance contracts) and faster frequency (reciprical of frequency is time, with faster frequency meaning longer time or time dilation). With all these photons in the ship time dilated (blue shifted) and energy needed to measure time, time measurement becomes time dilated.

 

I thought EM energy is planks constant times frequency not wavelength times frequency. Wavelength time frequency only gives you the velocity which is c.

 

Tony

[Jay-qu]

Im not saying it.. read my post more carefully. I was equating it to doppler shift.

 

Imagine the ship clock sends a pulse of light every 1 sec (ship time), then on the way out the pulses will be spread and on the way back they will bunch up.

[CraigD]

Because the distance traveled by the light with which each clock is “seen” is different for each indicated time of each clock.

Sorry, you have completely lost me. The distance light has to travel between the two clocks is exactly the same or are you saying the distance is different?

The distance between the ship and earth when a particular clock indicates a particular time (eg: the clock on the ship says “1.0 years”) is different than when it indicates another (eg: 2.0).

 

Consider this example:

• Let the velocity of the ship be a convenient [math]\sqrt{\frac34} \dot= .866 c[/math], so the time dilation ratio is 1:2, the same as in the example in post #25
  
• Visualize the 2 clocks as fancy cuckoo clocks. Twice a year, a door opens and a little mechanical bird emerges holding a brightly lit sign with the numeral “0.5”, “1.0”, “1.5”, etc., then bird and sign disappear back into the clock for another year. The clocks also have a small higher-precision display visible only a nearby observer
  
• After 1 years according to the earth clock, the Earth cuckoo flashes its “1.0” sign, while at the same instant (as viewed by a distant observer equally distant from both clocks), the ship’s cuckoo (which is advancing at 1/2 the rate) flashes its “0.5” sign
  
• At that instant (as viewed by the distant observer), the ship is about .866 light years from Earth
  
• An observer on Earth, then, sees the ship’s “0.5”, and glancing at the Earth clocks high-precision display, notes that it says “1.866” (1 + the time it took the light from the “0.5” to reach Earth)
  
• An observer on the ship sees Earth’s “1.0”, and glancing at the ship clock’s high-precision, notes that it says “3.732”. Because the ship is moving, the light from the Earth’s “1.0” doesn’t reach it as soon (as viewed by the distant observer). The formula for how many years the distant observer sees the "1.0"s light take to reach the ship (t) is [math]1 t = d \dot= .866 + .866 t[/math], which we solve to find [math]t \dot= \frac{.866}{1-.866} \dot= 6.464[/math]. Adding this to the 1 year already on the clock when the “1.0” was flashed, and dividing by the ships time dilation factor gives [math]\frac{6.462+1}2 \dot= 3.732 [/math]
  

Note that, based on the data they receive, both the Earth observer and the ship observer conclude that the other’s clock is running slower – the Earth observer [math]\frac{.5}{1.866} \dot= .2679[/math], the ship observer [math]\frac{1}{3.731} \dot= .2679[/math]. To the distant observer, however, the Earth clock appears to be running at normal speed, the ship’s at [math]\frac12[/math] normal speed.

 

You can consider an example at any time in the trip. The formulae necessary are different for the return leg of the trip than for the outbound. If you are more visually that numerically inclined, you may find it more useful to get slightly less precise values using paper, pencil, and a ruler. Of special interest is the time interval where the ship has reversed direction, but Earth is still receiving light it emitted when it was outbound.

Why would the rates suddenly change? Do you mean this change in the beat of the clocks happens when the ship is accelerating or when it is in uniform motion?

The sudden change in rate is due to the ships sudden change in direction. It occurs immediately after this change in direction, which in my simplified example, is assumed to occur so quickly that it need no clock-watching occurs during its period of very high acceleration.

But a change in direction requires acceleration, so are you saying that the acceleration is causing time to dilate?

Only indirectly.

 

By definition, acceleration causes change in velocity ([math]a= \frac{\Delta v}{\Delta t}[/math]). The acceleration caused the ship's velocity to change. The change in velocity caused a change in the ship’s time dilation factor. The time dilation factor stays the same until the magnitude of the ship’s velocity changes, which requires acceleration. However, an acceleration that changes only the direction, not the magnitude, of the ship’s velocity doesn’t affect its time dilation factor due to special relativity. It does change it’s time dilation due to general relativity, but that’s a different subject, and one I’ve taken care to make insignificant to the examples in this thread.

[jungjedi]

I think you’re thinking of an episode of Star Trek, jung padawan. ;) Specifically, the “time-traveling solar slingshot maneuver” from ”Tomorrow Is Yesterday which was repeated in the movie Star Trek IV: The Voyage Home.

 

Though pretty cool special effects for their day, the physics of this is pure fiction. The solar slingshot maneuver is performed more times than astronomers can count, by major and minor solar bodies in cometary orbits. They reach modest velocities of at most a few thousand m/s, (.00001 c), producing insignificant time dilation factors of (around .99999999995 maximum), far from the effect depicted in Star Trek. And, special relativity only allows time travel at an increased rate in the usual direction, into the future, not the past, as in STrek.

 

then what of the concept of frame dragging proposed by einstien.around a gravastar or nuetron star the effect must be phenomenal

[CraigD]

then what of the concept of frame dragging proposed by einstien.around a gravastar or nuetron star the effect must be phenomenal

AFAIK, these ideas are variations of the Tipler Cylinder Proposed in the 1970s by Frank Tipler. In principle, such a region of spacetime could allow a traveler to visit any timelike point of it, meaning that it would be a time machine that could take a traveler to any time after the machine was created and before it will be destroyed.

 

In Tipler’s original, mathematically rigorous work, the cylinder needs to be not only very massive and fast-spinning, but infinitely long. Later, he and supporters of the idea argued in a speculative way the possibility that a sort of “quantum physical information loss” might allow a cylinder of finite length to cause the same effect, if no possible experiment at some region near the cylinder could determine that the cylinder was not infinitely long.

 

Most theorists appear to disagree with this argument, or have other objections to frame-dragging time machines. Personally, I lack the skills to have an well-informed opinion on the subject. It’s an interesting subject, though.

[jungjedi]

AFAIK, these ideas are variations of the Tipler Cylinder Proposed in the 1970s by Frank Tipler. In principle, such a region of spacetime could allow a traveler to visit any timelike point of it, meaning that it would be a time machine that could take a traveler to any time after the machine was created and before it will be destroyed.

 

In Tipler’s original, mathematically rigorous work, the cylinder needs to be not only very massive and fast-spinning, but infinitely long. Later, he and supporters of the idea argued in a speculative way the possibility that a sort of “quantum physical information loss” might allow a cylinder of finite length to cause the same effect, if no possible experiment at some region near the cylinder could determine that the cylinder was not infinitely long.

 

Most theorists appear to disagree with this argument, or have other objections to frame-dragging time machines. Personally, I lack the skills to have an well-informed opinion on the subject. It’s an interesting subject, though.

 

this is the interesting part(A limitation of the Tipler cylinder is that it is only possible to travel to times (and places) in which the cylinder already exists. Thus, one could not travel backwards further than the date that the cylinder was activated)-from wiki

[Uclock]

CraigD

 

The distance between the ship and earth when a particular clock indicates a particular time (eg: the clock on the ship says “1.0 years”) is different than when it indicates another (eg: 2.0).

 

Yes, that of course is logical.

 

 

Consider this example:

• Let the velocity of the ship be a convenient , so the time dilation ratio is 1:2, the same as in the example in post #25

 

And the same must be said for the velocity of the Earth from the ships frame.

 

•

Visualize the 2 clocks as fancy cuckoo clocks. Twice a year, a door opens and a little mechanical bird emerges holding a brightly lit sign with the numeral “0.5”, “1.0”, “1.5”, etc., then bird and sign disappear back into the clock for another year. The clocks also have a small higher-precision display visible only a nearby observer

I hope the bird doesn’t get travel sick.

 

• After 1 years according to the earth clock, the Earth cuckoo flashes its “1.0” sign, while at the same instant (as viewed by a distant observer equally distant from both clocks), the ship’s cuckoo (which is advancing at 1/2 the rate) flashes its “0.5” sign

But the observer must have a relative velocity to the Earth in order to remain at an equal distance from the ship and the Earth. After one year passes the ship will also see the Earth clock flash up 0.5 because the Earth has relative velocity to the ship.

 

• At that instant (as viewed by the distant observer), the ship is about .866 light years from Earth

• An observer on Earth, then, sees the ship’s “0.5”, and glancing at the Earth clocks high-precision display, notes that it says “1.866” (1 + the time it took the light from the “0.5” to reach Earth)

But the same must be said of both frames because the distance and the relative velocity is the same between both frames. You are treating one frame differently to the other. Either velocity is relative or it isn’t.

 

• An observer on the ship sees Earth’s “1.0”, and glancing at the ship clock’s high-precision, notes that it says “3.732”. Because the ship is moving, the light from the Earth’s “1.0” doesn’t reach it as soon (as viewed by the distant observer). The formula for how many years the distant observer sees the "1.0"s light take to reach the ship (t) is , which we solve to find . Adding this to the 1 year already on the clock when the “1.0” was flashed, and dividing by the ships time dilation factor gives

Note that, based on the data they receive, both the Earth observer and the ship observer conclude that the other’s clock is running slower – the Earth observer , the ship observer . To the distant observer, however, the Earth clock appears to be running at normal speed, the ship’s at normal speed.

 

Yes I did a similar task when learning relativity some years ago but try doing exactly the same thing without using a third frame. You will find you cannot logically say velocity dilates time if velocity is relative. There is however always a time delay because of the speed of light which I have seen used to try and solve the paradox but that also comes up short.

 

 

You can consider an example at any time in the trip. The formulae necessary are different for the return leg of the trip than for the outbound. If you are more visually that numerically inclined, you may find it more useful to get slightly less precise values using paper, pencil, and a ruler. Of special interest is the time interval where the ship has reversed direction, but Earth is still receiving light it emitted when it was outbound.

 

And so is the ship because the distance the light has to travel is the same.

 

But a change in direction requires acceleration, so are you saying that the acceleration is causing time to dilate?

 

Only indirectly.

 

By definition, acceleration causes change in velocity ( ). The acceleration caused the ship's velocity to change. The change in velocity caused a change in the ship’s time dilation factor. The time dilation factor stays the same until the magnitude of the ship’s velocity changes, which requires acceleration. However, an acceleration that changes only the direction, not the magnitude, of the ship’s velocity doesn’t affect its time dilation factor due to special relativity. It does change it’s time dilation due to general relativity, but that’s a different subject, and one I’ve taken care to make insignificant to the examples in this thread.

 

I am sorry CraigD but you have failed to convince me that this explanation solves the twin paradox. Can you try again using just two frames, that of the observer on the Earth and the one on the ship?

 

Tony

[arkain101]

I have a question to toss in here.

 

Conisder we have a object leaving directly away from earth and I have a nifty camera that can zoom or bend the light enough to see the object as it passes by hypothetical markers in space that are at rest relative to us on earth.

Simple illustration:

| is a marker

o> is the flying object

 

(earth)----------|----------|----------|o>---------|----------|

 

 

Each marker will represent 1 light minute distance.

Such as, 1min,2min,3min,4min,5min..

We will have to pretend earth is also at rest.

 

Here is the question.

 

When I take a picture of the object when I see it at the 3min marker, and my super duper camera takes a picture with almost instantanious shutter speed, and displays a picture on a screen for me to see it at this position.

 

Do I say.

a)No this object is truly not where the picture says it is because the light is 3mins old from its origin point?

b)This object is more or less where the picture shows where it is?

[Jay-qu]

Even if you know that the object is 3 light mins away you cant possibly know where the object is at that time, the ship would lie somewhere within a 3 light min sphere of the point where you took the picture of.

[Guest chendoh]

You can not say either

 

You are pretending the earth is at rest

[arkain101]

Hmm both replies were confusing lol :lol:

[Guest chendoh]

If 100,000 spectators at a nascar race watch 43 drivers drive at an average speed of 195.000 on a two mile track.........

 

Are the 43 Drivers younger than the spectators at the end of the race?

[Guest chendoh]

Why confusing?.....In Astrometrics One can not Presume what you prepose.

[CraigD]

Conisder we have a object leaving directly away from earth and I have a nifty camera that can zoom or bend the light enough to see the object as it passes by hypothetical markers in space that are at rest relative to us on earth. …

Well, a camera – the receiving end of a light observation – can’t much bend or otherwise effect light far away from it. The scenario you arkain describes, though, isn’t difficult to realize using simpler techniques – for example, each marker could have a proximity detection system set to trigger a bright “flash” when the object passed it.

When I take a picture of the object when I see it at the 3min marker, and my super duper camera takes a picture with almost instantanious shutter speed, and displays a picture on a screen for me to see it at this position.

The shutter on a camera only controls when it stops and starts detecting light. This is useful for avoiding blurring from a stream of photons from a moving object, but not necessary if the light to be detected is already a short-duration pulse, like the flashing markers described above, or the brightly-lit sign bearing cuckoos in post #37.

Do I say.

a)No this object is truly not where the picture says it is because the light is 3mins old from its origin point?

b)This object is more or less where the picture shows where it is?

Unless the object has a very low speed relative to Earth, I’d pick choice “a”. You are seeing the object where it was 3 minutes ago, not where it “really” is now.

 

When discussing observations over distances where the travel times of light are significant (within the precision of our timing devices), concepts like “where it really is now” and “where the picture says it is” need to be treated with special care. It’s best to qualify every description of an observation with the observer’s position and velocity relative to the observed. With this information, and the rigorous application of fairly simple math, geometry, and the postulate that the speed of light is constant for all observers, practically all of the “paradoxes” associated with such observations can be made to disappear.

[CraigD]

If 100,000 spectators at a nascar race watch 43 drivers drive at an average speed of 195.000 on a two mile track.........

 

Are the 43 Drivers younger than the spectators at the end of the race?

Yes.

 

For the calculations to be easy, we’ll have to make a minor adjustment: stop or slow the Earth’s rotation so that it’s much smaller than the speed of the cars. Otherwise, all sorts of complicated East/West factors enter into them.

 

That done, for a 500 mile race at a constant speed of 195 mile/hr (not something ever seen at a NASCAR race ;)), the drivers will be at least [math]\frac{500}{195} \left (1-\sqrt{1-\frac{195}{670616629}} \right )[/math] = .00000074 hours younger than the spectators.

 

500 such races, and they’ll be a whole second younger. Not a very effective life-extension technique. :D

[CraigD]

Consider this example:

• Let the velocity of the ship be a convenient [about .866 c], so the time dilation ratio is 1:2, the same as in the example in post #25

And the same must be said for the velocity of the Earth from the ships frame.

No, it must not. This point is crucial to the whole demonstration. The idea that, because each observer observes the others clock to run slower (on the outbound trip) and faster (on the inbound) by exactly the same ratio (about .2679 outbound, 3.732 inbound), neither knows who is “moving” and who is “stationary”, is incorrect. In the Earth-ship system, both observers can in principle (and almost certainly in practice) detect that the ship, not the Earth, changes velocity. Spring scale, LASER interferometers, and any other kind of accelerometer on the ship show this, while ones on Earth show an absence of acceleration. A distant observer with constant angular velocity notes a change in the angular position of the ship, not the Earth. The ability to distinguish which body has been accelerated, and which has not, is critical to all the subsequent discussion, and all discussions of relativity.

 

It is not important if either observer is capable of measuring acceleration, or distinguishing which body accelerates, or if they or a distant observer actually exist. The described phenomena are objectively real, and occur whether observed or not.

• At that instant (as viewed by the distant observer), the ship is about .866 light years from Earth

• An observer on Earth, then, sees the ship’s “0.5”, and glancing at the Earth clocks high-precision display, notes that it says “1.866” (1 + the time it took the light from the “0.5” to reach Earth)

But the same must be said of both frames because the distance and the relative velocity is the same between both frames.

No, it must not be.

 

The first reason is, due to length contraction, which has only been briefly mentioned in this thread, but is an essential prediction of special relativity, the distance between earth and the moving observer, or any length measured in the direction of the ship’s velocity is not the same as measured from Earth

 

The next can be illustrated by a simple example involving everyday objects:

• 2 practiced marble-roller, Alice and Bob, (or unpracticed ones with simple marble-rolling devices) have a couple of smooth-rolling marbles, and a smooth floor, such that the marbles role at a constant velocity as far as the experiment requires
  
• From a distance of 2 meters, both simultaneously role their marble toward the other with a speed of 1/4 m/s
  
• At the instant of release, Bob begins moving away from Alice at a speed of 1/12 m/s
  
• At 8 seconds after release, Alice catches Bob’s marble
  
• At 12 seconds after release, Bob catches Alice’s
  

So, even though Alice and Bob are the same distance apart when the marbles are released, they don’t catch the marbles at the same instant, because Bob moved.

 

A third observer, Carol, sees the Alice-Bob marble exchange in essentially the same way the distant observer sees the Earth-Ship exchanges, absent the observation of a noticeable time dilation of Bob.

 

It is the distance between where the photon (or marble) is released, and where it is received, that determines its travel time, not the distance between where it is released and where whatever receives it is when it is released. This distance is different for a moving/moved observer than for a stationary one.

You are treating one frame differently to the other. Either velocity is relative or it isn’t.

Yes, I am, for the reasons above. I can treat the frames differently, because they are measurably different.

Yes I did a similar task [to the time dilation examples in this thread] when learning relativity some years ago but try doing exactly the same thing without using a third frame. You will find you cannot logically say velocity dilates time if velocity is relative.

I only included discussion of a third, distant observer for clarity. The actual calculations in post #25 and post #37 are all in terms of the unaccelerated frame, Earth.

 

Though I’ve tried presenting a resolution of the “twins paradox” in an original manner, such presentations have been around as long as special relativity, and are all fairly similar. The twins paradox was not intended to be and is not a “real” paradox illustrating a fundamental difficulty or incompleteness with any theory, but an educational example. The best advice I can offer skeptics of it is to work though it with the best geometric rigor you can manage – the “paradox” really does “go away”.

 

Some variations of the standard paradox can be educational, for example:

• The ship never turns around and returns
  
• The ship stops (accelerates to 0 velocity relative to earth) but does not return
  
• The Earth observer travels by spaceship to join the observer on the ship, whether the original ship turns around and returns, stops, or neither.
  

[kalexia]

Hi. I am confused about travelling near the speed of light. If a star is, say, 10 light years away, then it obviously takes light 10 years to travel from the star to us. Why then does it take someone travelling at, say, half the speed of light, less than 10 years to get there? I just can't figure it out.