PDA

View Full Version : Relativity

Michael_FJS
2004-Jul-19, 10:15 PM
I was thinking about Einsteins theory of relativity and thought of somthing.

It&#39;s about that twin paradox thing, where one stays on Earth, and the otherone travels away in a space ship near to the speed of light, and when he comes back he has aged only 2 years where his brother has aged 100 (i.e he is dead)

This happends because as the spaceship approches lightspeed, time slows down, therefore the person on the ship ages less.

What I thought about, what because the spaceship is moving near lightspeed away relative to Earth, then wouldn&#39;t the Earth also be moving near lightspeed in the opposite direction relative to the spaceship? Therefore the Earth would undergo the same time dialation effect as the spaceship, and time would slow down by the same amount for both Earth and the spaceship, which would mean that when the ship returns to Earth, the twins would still be the same age.

Arramon
2004-Jul-19, 11:02 PM
I personally don&#39;t think that someone here on Earth would age more if another person &#39;zipped&#39; away at lightspeed and came back at the same rate... The time here on Earth neither stops nor slows or speeds up because person A moved away from person B, at ANY rate.

Wouldn&#39;t that be like the solar flares moving through our solar system?
Particles riding Solar winds travel at like 460+ km/s, which would take about 2 1/2 months to get from the Sun to the heliosheath (theoretically)... Several probes and satellites scattered throughout our system picked up the strongest solar flares of last October rushing towards the outer planets.. If someone were to move at the same speed, why would Time be affected? Or even a factor?

As far as we know, Time is a constant... where or when has Time ever slowed down? So if person B were to wear a watch and came back, would his watch reflect the Time that Person B should be in (a time affectively slowed down by the great speed)? Or would person B&#39;s time show that it is perfectly matched up with Person A&#39;s, if both were synchronized...

..a ship flashes from out of the interstellar medium, and slams into the heated region of the Centuari system.

pop&#33; ~ crackle&#33; "Hey&#33; Houston&#33; I don&#39;t believe it...&#33; I don&#39;t believe this&#33; It took me only 5 seconds to get to Alpha Centauri&#33;" ~ crackle&#33;

The subspace transmission beams itself across the ISM divide and back towards planet Earth. The message is heard.. and a reply is sent. 10 minutes later...

~ pop&#33; "..whoa&#33;&#33; Are you serious, Zipper1? We read you&#33; Loud and clear&#33; You say only 5 seconds...&#33; Repeat, 5 seconds???" ~ crackle&#33;

crackle&#33; ~ pop&#33; "..I arrived at 5 past 0500&#33;&#33; I left at 0500 exactly, Houston&#33; Do you copy?&#33; I arrived at 0505 hours..&#33;" ~hsssss...

Another ten minutes later... ~ crack&#33; hsss "...Zipper1, Zipper1, we have your arrival there at over 3 months from when you left Earth&#33;&#33; Do you read? The current date is July 19th.. you left on April 15th, Zipper1... repeat.. you left over 3 months ago... over..."...

So would it end up like this? Where Time is skewed/bent/warped/slowed down for one, and sped up for the other?

Or would the scenario play out like:

"yes&#33;&#33; We have you at only 5 seconds for time travelled from the Sol system to Alpha Centauri...&#33;"

..very interesting indeed...

blueshift
2004-Jul-19, 11:05 PM
Michael_FJS,

You are right that the traveler can claim that Earth is moving relative to the ship.

The ship, however, changes its motion while Earth does not. It turns around.
The Earth extends most of its energy through time since it does not change its location to the same degree that the ship does.

Separation in spacetime is called "interval".

The ship&#39;s interval between events is spacelike while the Earth&#39;s interval is more
timelike.

Spacetime for both will be equal. Remember Pythagorean&#39;s theorum..
Time^2 + space^2 = Spacetime^2
The two frames of reference share the hypotenuse, the spacetime, but differ in
each one&#39;s measurement of energy through space and energy through time.

I hope this didn&#39;t make matters worse.

Arramon,

Your post suggests that you do not have much of a background in relativity.

The effects of relativity do concern time and space but the basis of special relativity lies in motion...experiments that deal with motion and defining what "rest"
means. I would study up on Aristotle&#39;s views of motion and rest and see how
they differed from Galileo&#39;s..

Aristotle could not accept that the Earth moved because he didn&#39;t have relativity
around to explain why someone leaps off of a moving Earth and returns to the same spot..He wasn&#39;t insane. Experiments he conducted suggested that only Earth
had gravity. You drop something and it hits your foot or the ground. It doesn&#39;t fly off to the Sun..

Look it up and work your way to the present...slowly.

Now try a few simple experiments..and ask yourself a few questions..

If you play catch with your friend outside, do you have to take into account that
Earth is moving at a 30 km/sec clip in its orbit around the Sun and 220 km/sec
in its galactic orbit? Of course not. You go right on playing catch.

What about on a big cruise ship in a calm sea with no turbulence but moving steady, in uniform motion, not changing any direction or speed at all...Can you play catch there as well and still not take into acount the cruise ship&#39;s speed?

When you heat water in a microwave, do you take into account the Earth&#39;s motion
when you set the timer? Not at all..What if you brought the microwave with on the cruise ship? Do you set the timer to a different time because the ship is moving?

Not at all.

How about in a jet airliner going 600mph, a steady 600 mph and the seat belt sign has been switched off? Do you set the timer on the microwave to a different time?
Of course not.

Therefore the laws of motion and the laws of electromagnetism operate the same in all uniform motions..Whatever shares your motion defines your reference frame.

Put a flash bulb in a vacuum and let it flash just once.
After one second where are all the photons? ?

They would all form a circle with a radius from the bulb = 186,000 miles or
300,000 km.

Now, imagine a huge tower placed at the point where the flash bulb flashed.
The tower is 186,000 miles high and a mirror is placed at the top..

In the first experiment the tower sits still. The flash goes off. Where is the photon one second later? At the mirror..2 seconds later? Back to the floor on the bulb.

Add two observers to the scene. One inside the tower and one outside. They both agree that 2 seconds lapsed for the photon&#39;s journey up and down the tower.

Now put the tower in motion to your right at a very high speed and set off the flash
at the point of departure.

For the observer in the tower it goes straight up and straight down just as it did before..Time measured? 2 seconds.

But the observer on the ground sees something different. The path of the photon
travels on an angle, relative to the ground he is standing on and the photon&#39;s path
does not go straight up and down but slants up and down and makes the shape of a triangle when you add in the distance the tower traveled after two seconds..

The speed of the photon is the same for both. The same photon travels a different distance to each observer. Therefore they measure different times and both are correct in their claims to their measurement..Do a little math and you will see this is true..

Each uniform motion is just as valid as the other for doing these measurements.

It sounds crazy, but has been confirmed in experiments with muons, which decay at a precise rate. Muons in a lab will decay far more quickly than muons that are
moving near speed c. Experiments at Mt. Washington in New Hampshire in the 1950s confirmed this.

Time dialation does occur and has been confirmed by television engineers who must take relativity into effect when calculating the speed of electrons that barrel
down the CRT in your TV. They must shrink space enough so that the electrons cannot tumble or you wouldn&#39;t get a well organized scene on your screen.

The twin who leaves on the hypothetical space craft out to Andromeda doesn&#39;t feel any different than his twin at home does..But , in comparison, all the traveler&#39;s
brain cells will decay at a slower rate. Earthlings will measure his mass increasing
but he won&#39;t notice a thing..All the motions in his watch will experience the same thing..Earthlings will measure his watch increasing in mass while the watches back here remain the same...

For just &#036;55 you can get a whole video course on the subject that explains it very simple.
Send me an email and I&#39;ll give you the information

blueshift

antoniseb
2004-Jul-19, 11:06 PM
Hi Michael_FJS welcome to the forum.

First this should go into questions and answers, since it is a question, and since Relativity is part of the accepted package that Alternative theories are alternative to.

Second, the twin paradox is a tough question for exactly the reason you point out. There are a lot of web-sites with bogus and conflicting answers for it. Perhaps the clearest answer is that the way you described it, we change which frame of reference we use when we look at the problem. There are three. One on Earth, one that parallels the out-bound spaceship, and one that parallels the inbound spaceship. If you measure it from any of those three frames, you get a result that the twin in the spaceship ages less [by the same amount each time].

Another thing that often gets pointed out is that there is something special about the acceleration to change frames of reference that distinguishes the ages of the two brothers. I never fully understood this part of the explanation.

StarLab
2004-Jul-20, 12:14 AM
Lemme see if I can help out a bit, what with my sixteen-year-old mind: In this scenario, the earth is the control of the experiment. All of the measured velocities of the spaceship are relative to earth&#39;s. I hope this helps.

TheThorn
2004-Jul-20, 04:53 PM
StarLab, the point is that in Special Relativity, there is no preferred inertial frame of reference. So, as Michael pointed out, the spaceship is (apparently) just as valid a frame as the Earth. And, in that frame of reference, it is the Earth that is moving.

Michael, you have hit on exactly whay this is called a "paradox". Most people think that it is a paradox because one twin ages faster than the other - but that is just a straight-forward, if unexpected, result of relativity, not a paradox at all. What is a paradox is that depending on which frame of reference you use, each twin should age faster than the other. That&#39;s not possible, and therefore a paradox.

Like antoniseb, I don&#39;t really understand the resolution of the paradox, other than that it is contained in General Relativity (time dilation is a result of Special Relativity). Note that the spaceship is not an "inertial" frame of reference - it is accelerating (otherwise it could never return). That&#39;s the key here (and why I used the word "apparently" above).

Bottom line: the spaceship is not just as valid an inertial frame of reference as the Earth, because it is NOT an inertial frame of reference.

Sometimes people point out that from the spaceship, it looks like the Earth is accelerating (i.e. that maybe the spaceship is the inertial frame of reference), but this is incorrect. An observer on the Earth would not feel an acceleration, while an observer on the spaceship would, so it is quite easy to differentiate which frame is inertial (not accelerating) and which isn&#39;t.

StarLab
2004-Jul-20, 11:59 PM
That&#39;s almost precisely what I was saying, though not with the exact same wording; the spaceship is moving faster, and therefore earth is the preferred control/frame of reference.

blueshift
2004-Jul-21, 04:35 AM
The Earth is in the same reference frame as the distant galaxy or star that the ship is flying to..To synchronize clocks at both places a signal would have to sent from the midpoint to the star and Earth.

An "observer" is not someone who functions at one location. With the help of an
army of registering clocks, he operates all over the place at locations that do not move or change distances from each other between the star and Earth. A reference frame is not a location.

Our radio telescopes measure distant jets coming at us whose leads can be approaching us night by night. If we simply calulate the distance traveled as we
measure it in a 24 hour period, the jet can be approaching us at several times the speed of light..

This is erroneous. One day of our time is not one day of jet time..

To see how time dialates, pick up Richard Wolfson&#39;s "Simply Einstein" and he shows, with simple high school geometry how to derive and calculate it.
He put it in the appendix on page 242..It&#39;s very readable.

blueshift

StarLab
2004-Jul-21, 04:58 AM
Blueshift, got a scanner? Can you get that page on the internet via google?

blueshift
2004-Jul-21, 04:50 PM
Star Lab,

I don&#39;t own a scanner. I&#39;m still fairly new with computers..

We might need to exchange a few emails..

Maybe I can snail mail you either the whole book or I might just draw up a diagram with the accompanied formula..

Wolfson has a video course that I mentioned in an earlier post on this thread
that is excellent and geared toward non scientists..

If you comprehend Pythagorean&#39;s theorum ( a right triangle&#39;s base squared plus
the height squared = the hypotenuse squared ), you can derive the time dialation formula from it by combining it with the tower experiment I posted earlier on this thread..

Don&#39;t think of physics as some "deep" subject that takes great comprehension.

I was lucky. I have a large number of physicists in my astronomy club from Fermilab and Argonne who insist that physicists see things with shallow minds,
looking at the universe as if it is an onion, peeling back thin layers one at a time,
studying both their properties and how we uncover them and making simple
calculations to see what is missing and to search for some tracks that give us indications..

The only trouble is, that by unraveling these thin layers, we can create a wake in our path that looks like a confusing mess to someone who just walked up...We can
keep that person back there by insisting the journey is "hard" and that there is a lot of catching up to do.

This sounds crazy...but the slower you go, the faster you&#39;ll catch up..because you&#39;ll have a firmer understanding of basic principles that someone who merely absorbed or memorized the right answers to the right questions..

blueshift

David S
2004-Jul-26, 05:49 AM
So does anyone have an answer to Micheals question? What is the resolution for the twins "paradox"?

And I can&#39;t believe that it would rely on something as flimsy as the space ship has to "turn around", since that is a practical concern and this is a hypothetical senario (and in hypothetical senarios, we can just as easily have the earth rocket away from the spaceship).

Besides, there are all sorts of ways around that. You could have the spaceship not turn around and return to earth, and just communicate with radiowaves. You could have 2 spaceships. You could carry out the experiment in a centurfuge that is moving at close to the speed of light. You could have the spaceship fly close to a neutron star is such a way as to make the space do a complete 180&#39; and head back to earth without ever having to change course or accelerate in any way, ect.

rahuldandekar
2004-Jul-26, 09:22 AM
I think that Time slows down only in an accelerating frame of reference. The earth has only centripetal acceleration, while the spaceship has a linear acceleration. If the spaceship is moving with a constant velocity, then Time would not slow down.

Am I right?

antoniseb
2004-Jul-26, 10:38 AM
Originally posted by rahuldandekar@Jul 26 2004, 09:22 AM
Am I right?
If you mean that relativistic time dilation only happens while an object is changing velocity, then no you are not right. Just as an example, highly unstable particles and nuclei can travel thousands, and even millions of lightyears at near the speed of light without decaying. In the particle&#39;s frame of reference, it is because the space from where they originated to here has been forshortened. To us it is because time for them is moving slower.

bunny
2004-Jul-26, 01:03 PM
Hope this clears some confusion.

If the person on the rocket is moving away from the earth at relativistic speeds, the earth is the frame of reference and the person in the rocket experiences the dilation.

By the same account, if we then say the rocket is the frame of reference the earth appears to move away at the speed of light and so the person on earth experiences the elongation (not dilation remember).

For the twins to to be the same age again the rocket has to accelerate away from the earth then stop, then the earth has to accelerate towards the rocket. For this reason it has been said that the acceleration is the key to time dilation and not speed.

A force must be exerted on a object for it to have acceleration. A force is being exerted on the rocket and so the dilation occurs on it. No force is being exerted on the earth so it cannot feel any dilation (or elongation) regardless of which direction you are travelling -towards or away.
I have never felt this is a satisfactory answer, IMO time is universal and constant and Einstein (the plagarist) just keeps getting in the way of good ideas.

(I think I may have nailed my colours to the post with that last comment..&#33;)

David S
2004-Jul-26, 03:20 PM
The equation for determining time dialation is:

T = To / (square root (1-(V^2/C^2)))

T = reference frame of object in motion
To = reference fram of stationary observer
V= velocity difference between object in motion and observer
C= speed of light

For this equation acceleration is not one of the variables, and therefore cannot affect time dialation. It is only the velocity difference between the observer and object in motion that creates the time dialation.

I believe the "paradox" is created when each twin applies this equation, using their own reference frame as the staionary observer. There is something which determines that the twin who is in motion cannot view their reference frame as stationary, but I don&#39;t see how it&#39;s acceleration. I also don&#39;t see how it&#39;s velocity, since from the viewpoint of the twin in the spaceship he is standing still and the earth is rocketing away from him. In other words, the twin in the spaceship views his own velocity as zero.

So what is it that makes the the spaceship an invalid reference frame?

Relmuis
2004-Jul-26, 03:43 PM
There is one aspect that may throw some light on the question.

If a train passes me at great speed, a clock on that train will run slow relative to me. That is, if I look at the passing clock, and mentally correct for the fact that the light from this clock takes a certain (variable) time t reach my eyes, I will find that the clock on the train is moving slow.

However, if there are many clocks on the train, I will also find that they do not all give the same time (after correcting again for the time their light takes to reach me). The clocks on the rear of the train will be ahead of the clock near the driver&#39;s seat, in amounts that exactly correspond to their distance from the driver&#39;s seat.

Now, if I do not follow any particular clock, but amalgamate all the different clocks flashing by into one single clock, this amalgamated clock will run fast, even though every individual clock runs slow. In fact, the rate of my own, stationary clock will be the geometric mean of the speed of the individual moving clocks and the speed of the amalgamated clock.

This makes the situation symmetric, because someone on the train might look at a row of stationary clocks flashing by and amalgamate them too. Each stationary clock would seem to run slow, but the amalgamated clock would seem to run fast.

It means, however, that every state of non-accelerated motion has its own system of time zones.

If someone boards a train for Alpha Centauri, there changes trains to another train bound for Earth, and leaves the train there, he changes three times into another system. Two of these times, the local clocks may (just by coincidence) read the same time when they pass each other. But the third time they must then read different times.

Therefore, the traveller will need to adjust his watch at least once. And his stationary friend, who never leaves Earth, will of course not need to adjust his watch. Which means that they have not experienced the same amount of time.

Acceleration as such does not come into it, but the change from one state of motion (or velocity) to another state of motion (which may be instantaneous) does.

If you are sitting in a merry-go-round, you will experience a little bit of time less than someone who is merely looking at the merry-go-round from outside. That&#39;s because you are continually changing your state (speed and/or direction) of motion.

TheThorn
2004-Jul-26, 05:46 PM
OK, I admitted earlier that I didn&#39;t really understand the details of this, but that the answer was in the fact that the sapeceship had to accelerate to bring the twins back together again. All this discussion forced me to look up the details.

May I recommend that everyone go read the (several) explanations at This Page. (http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html) They are very good and I don&#39;t want to try to reproduce them here, &#39;cause I&#39;ll certainly screw up.

One side comment to Relmuis: You said "Acceleration as such does not come into it, but the change from one state of motion (or velocity) to another state of motion (which may be instantaneous) does." What do you think acceleration is, if it&#39;s not "a change from one state of motion (or velocity) to another". That&#39;s the very definition of velocity&#33;

Relmuis
2004-Jul-28, 09:15 AM

Acceleration is the rate of velocity change. That is: the amount of velocity change divided by the amount of time needed for the change.

Just like speed (or velocity) is the rate of place change. Distance (or vector distance) moved divided by the time needed for the move.

Acceleration does not come into my story, because the switch can be instantaneous. In fact, nobody really has to change trains. A person in the outbound train can synchronize his watch to the clock of Earth station flashing by. When at Alpha Centauri, he may hold his watch to the window, so that a person on an inbound train passing Alpha Centauri at the same instant may synchronize her watch with his. And when this second person passes Earth, she will find that her watch does not read the same time as the clock of Earth station (which was used to set the first watch). I.E. the three time intervals (Earth-Earth, Earth-Alpha Centauri and Alpha Centauri-Earth) do not add up. But no one and no thing has been accelerated in this version of the story.

rahuldandekar
2004-Jul-28, 11:27 AM
It&#39;s a surprise that though relativity is learnt by many, all have differences on it&#39;s interpretation. Funny wierd...

blueshift
2004-Jul-28, 10:50 PM
Here&#39;s how to derive the time dialation formula..

A float is passing by in a parade and all of us on this site are street side.
Two people on the float, each facing each other across the float with their backs to the sidewalk and perpendicular to the path of travel.

Both play catch with a ball, back and forth.

To them the ball travels in a straight segment, relative to the float.

To us bystanders on the ground it takes a diagonal path relative to the ground.
We see an isoscoles traingle for a path.

On the float time measured between two events ( the ball goes across and returns to the same spot ) are events that occur in the same place.

This frame is called "t prime" or written as t&#39;.

On the sidewalk time measured between two events occur at different places.

This frame is called t.

Substitute a photon for the ball.

The distance across the float = L

The distance across and back = 2L

Distance = speed x time..

Speed of a photon = c ...Therefore,

2L = ct&#39;

Divide both sides of the equation by c gives:

t&#39; = 2L/c

Now for the t frame of reference....The float is moving at speed "v" and we call "t"
as the time between departure and return of the photon across the float.

Since distance = speed x time, the distance the float travels = vt.
When the float travels half of this distance, the photon takes a diagonal path up to the apex of the triangle..The length of that photon&#39;s path forms the hypotenuse to a right triangle with the base = 1/2 vt and the height = L

From Pythagoras we get the hypotenuse squared = ( 1/2 vt )^2 + L^2 or,

((1/4 v^2 (t^2) + L^2))

The square root of the above gives the length of the hypotenuse

The total path covered is twice this distance.

Time = t

Speed of photon = c

Distance = speed x time gives

ct = 2 times the square root of ((1/4 v^2 ( t^2) + L^2 ))

We need to get "t" on one side of the equation and its tangled up in a square root.

Square both sides and we get:

C^2 ( t^2) = 4 (( 1/4 v^2 ( t^2 ) + L^2 ))...carrying out math on the right side gives:

c^2 (t^2) = v^2 ( t^2) + 4L^2

Subtract v^2 ( t^2) to isolate 4L^2 gives:

4L^2 = c^2 ( t^2) - v^2 (t^2)...Now divide both sides by c^2 and take the square root of both sides. The result is:

2L/c = t times ( the square root of 1- v^2/c^2 )

From earlier we learned that t&#39; = 2L/c....Therefore,:

t&#39; = t times ( the square root of 1 - v^2/c^2)

Voila&#33; Please print that. I&#39;m not going to repeat it on this keyboard again...

I dragged it right out of Richard Wolfson&#39;s " Simply Einstein"..

blueshift