# Exploring the Twin Paradox: A 1,000 Word Essay

• Prague
In summary, the conversation is about the Twin Paradox, a theory in physics that explains how time can be affected by speed. The conversation involves discussing the concept of the paradox and finding simple solutions to understand it, as well as looking for resources to help explain the mathematical equations involved. The conversation also includes a summary of the theory and a request for help in solving a problem related to the theory.
Prague
Hey, I am instructed to argue a theory in a 1,000 word essay (can exceed by 250 words). Anyways, I am going to write about the Twin Paradox. I myself am not good with this theory, I just "understood" it, well yesterday when I got a few good answers in the Special & General Relativity forum. Anyways, I have to have my abstract in tomorrow how does this sound for the intro (I have a week and two days to finish the actual paper, so I figure I'll do extensive research and my little understanding or the math/history behind it will improve as I research.)

If there are two life forms consisting of the same senescence inhabiting a life supporting mass with a constant inertial rate, I argue that if one life form were to leave traveling at close to the speed of light, and then turn back and travel back to the life supporting mass, the life form that did not travel will in fact have aged significantly more than that of which had traveled. This theory is commonly referred to as the "Twin Paradox," which was solved by Albert Einstein in 1911. For simplification matters I will create a story line similar to usual depictions of the "Twin Paradox." For the most part I will convey the logic behind this theory through words, however certain ineffable situations occur, which will be expressed through certain mathematical equations stated in Einstein's theory of Special Relativity, as well as other related theories.

Also, anyone know of any sites that show step by step mathematical equations etc... to explain and solve the twin paradox? I figure I'll make my own storyline up, I just want to make sure I account for the stuff used to solve this paradox.

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That first line was a killer to understand, but your facts are right only if you say that he has aged in comparison to the twin that is not travelling. Also knock off that very last comma, and 'For simplication matters' kinda sounds funny.

Try wikipedia.org , its a general science encyclopedia and is great for things like this, just search for relativity and navigate through to find the mathematical concepts.

thanks alot, yea I do have to shorten that first sentence, bit long. I'm going to try to get a start on the paper now, so i'll be researching the equations, i'll post little later to check if I get things right. Thanks.

Ok, I've been reading information from published works, each seem to always delve into the paradoxal aspect of the 'twin paradox' (clearly), to either prove or disprove that it is in fact a paradox. I however, will blindly accept in my paper that it is not a paradox; in fact I'll be omitting that completely. I simply want to prove that Twin B ages slower than Twin A. And I will be using the most straightforward method. Now, I just need some help separating what is the most straightforward method. Can anyone help me out?

My situation will be just like the other ones.

Two life forms of the same senescence live on a life supporting mass. This mass has a constant inertia. We will label the two life forms A and B and the life supporting mass X. B travels away from X traveling at 90c. At the point of departure A and B are considered to be at the age of 0. After B travels 20 light years, B turns around and travels back to X. Upon returning B finds that A is much older than B. How is this possible?

Well, it will be something like that but structure better. Anyways, any help on simple solutions?

It would make ALOT more sense if you wrote it mathematically. What class is this for? If you write it mathematically you'll probably get a better grade, because you won't have to compensate writing skills to express an equation in a paragraph.

$$t_{dil} = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

T_dil is the time dilation factor.
T_0 is called 'proper time', or in this case will be time as experienced by your non-travelling twin. The RHS of the equation will give you the time the traveller will experience. It is basically this:

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

In relativistic terms, this equation will give you the effects of relativistic motion on the entity you are trying to explain, in this time, multiplying both sides by $$t_0$$ will give you the first equation I cited. $$\gamma$$ is always very close to 1.

whozum said:
$$t_{dil} = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

T_dil is the time dilation factor.
T_0 is called 'proper time', or in this case will be time as experienced by your non-travelling twin. The RHS of the equation will give you the time the traveller will experience. It is basically this:

$$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

In relativistic terms, this equation will give you the effects of relativistic motion on the entity you are trying to explain, in this time, multiplying both sides by $$t_0$$ will give you the first equation I cited. $$\gamma$$ is always very close to 1.

Hmm, I am finding this somewhat hard to follow now. When I discussed this problem in the SR forum, I understood the answer of why they aged differently through just words. I really want to understand this math but it seems everywhere I go there are different equations popping up.

OK, so perhaps you can help me out since I wish to solve this mathematically. Here is what I state in my paper.

Two twins, A and B are born on the planet X, which has a constant inertial rate. At the time of B's departure from X both A and B had an age of 0. Twin B travels 20 light years at a speed of 99c. Once reaching the destination of 20 light years, twin B returns to X again at 99c. What age will A and B be and why are the ages different?

well it will be something like that but I just need help solving it now. Thanks alot.

In this case your time dilation equation would be:

$$t_{dil} = \frac{t_0}{\sqrt{1-\frac{(.99c)^2}{c^2}}}$$

Person A experiences time $$t_{dil}.$$
Person B experiences time $$t_0$$

Person A is stationary and has age t(0) = 0
Person B is traveling at .99c for 40 light years.

To find the time he is gone, divide distance by velocity:
If he was traveling at c, it would take him 40 years to get there.
Since he was traveling at .99c, it would take him slightly less time:

$t_0 = \frac{40}{.99} = 39.6$years. This is the time Person B will claim he was gone for. Notice he does not feel any effects of time dilation. This is because the physics of any constant inertial frame are the same.

To find the time Person B was gone from A's point of view, we plug into our time equation to find the effects of time dilation:

$$t_{dil} = \frac{t_0}{\sqrt{1-\frac{(.99c)^2}{c^2}}}$$

The denominator becomes $\sqrt{1-(.99)^2} = \sqrt{0.0199} = 0.1411$

$$t_{dil} = \frac{t_0}{0.1411} = \frac{39.6}{0.1411} = 280.652 years$$

As you will see he will come back to meet his twin's great great great great grandkids.

Hope this helps you understand. If you need any explanation let me know, and if I made any errors (which I probably did) I'm sure someone will jump in and say something.

don't you have to take into account the fact that the traveling twin is not in an inertial reference frame, but instead an accelerated one when he turns around to go back? it would require an infinite acceleration for him to slow down, turn around, and speed back up in 0 time. i believe that the resolution to the paradox has something to do with the physics that occurs when the traveling twin is in the accelerated reference frame? i could be wrong though

You are right I think, but I'm not sure how that would incorporate into the above. I actually don't know about this, but if he was to travel 40 years and somehow compare his clock to Person A's then he would find that they had gone through a time lapse of ~240 years. The complications arising in this comparison are beyond me.

Wait, so is the work you did above correct? Or is there a part missing? Also, why does it take him slightly less time to get to reach his destination if he is traveling at .99c? Isn't .99c 99% of c? Thus, he would be traveling at 184,140 miles per second opposed to 186,000 miles per second (speed of light)?

Also, why would it require an infinite amount of acceleration for him to slow down? If he is traveling .99c wouldn't he just need -.99c to stop, then he would need .99c acc to gain speed again. This works if you make acceleration from 0 to .99c 0 seconds (or 1 second [whatever it is]).

Also it seems that 20 light years to each point is far too long, I think I am going to change it to 3 light years for a total of 6 light years travel.

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Prague said:
Wait, so is the work you did above correct? Or is there a part missing?
The work is correct, it just does not take into account the requirements of returning to Person A. Recall the laws of relativity only apply to frames that have constant inertias, any time there is an acceleration, these laws fumble.

Also, why does it take him slightly less time to get to reach his destination if he is traveling at .99c? Isn't .99c 99% of c? Thus, he would be traveling at 184,140 miles per second opposed to 186,000 miles per second (speed of light)?

You're right, I wrote 40/.99 but did 40*.99 in my calculator. The correct figure for t_0 = 40.4, instead of 39.6.

Also, why would it require an infinite amount of acceleration for him to slow down? If he is traveling .99c wouldn't he just need -.99c to stop, then he would need .99c acc to gain speed again. This works if you make acceleration from 0 to .99c 0 seconds (or 1 second [whatever it is]).

It would take infinite acceleration to impart a change of velocity almost equal to 2c in t = 0. You can't apply a force in 0 seconds, it always takes SOME amount of time.

$$F = \frac{dp}{dt}$$ As dt-> 0 F -> $$\infty$$

Also it seems that 20 light years to each point is far too long, I think I am going to change it to 3 light years for a total of 6 light years travel.

Just plug in 6 wherever you saw 20, or 12 wherever you saw 40, and do the math yourself. You'll experience a significant drop in time dilation.

Am I missing something here? Why is this a paradox? Seems pretty cut and dried to me.

Or is it called a paradox because it appears to be a paradox to people not "in the know"?

What kind of class is this for?

Two twins, A and B are born on the planet X, which has a constant inertial rate. At the time of B's departure from X both A and B had an age of 0. Twin B travels 3 light years at a speed of 99c. Once reaching the destination of 3 light years, twin B returns to X again at 99c. What age will A and B be and why are the ages different?

Ok, so here goes.

A experiences time $$t_0$$
B experiences time $$t_{dil}$$
B travels at .99c (184140 mps)
A is stationary and has an age of t(0) = 0 (not sure what this means).

To find how long B is gone we take the distance and divide it by velocity.

$$t_0 = \frac{d}{v}$$

thus,

$$t_0 = \frac{1116000}{184140} = 6.060606061$$
is this light years becuase it involves twin B who is traveling in light years? And if he only felt 6 light years, is that really 6 years for him even though they are light years?

To find how long B was gone in reference to A we must find the time dilation for the equation

$$t_{dil} = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

therefore,

$$t_{dil} = \frac{t_0}{\sqrt{1-\frac{(184140)^2}{(186000)^2}}} = .1410673598$$

then,

$$t_{dil} = \frac{t_0}{0.1410673598} = \frac{6.060606061}{0.1410673598} = 42.96249728$$
this is years correct? because this involves twin A who is stationary so its years?

Thus, we conclude that if A and B were to meet again A = 42.96249728 years of age and B = 6.060606061 years of age.

----------------

I added little pointers in small text below concepts I really didn't understand. So if you could answer those, that would help. Also, does this look right?

As for why this is a paradox, well this particular way of solving it is not a paradox. The paradox lies in when you look at the clock from the view point of twin A and the view point of Twin B, both seem to slow down so both A and B would say that the other twin would be older/younger. Its something like that, I am not to clear on it, I didn't look into it because I didn't understand the basics anyways.

This is for a philosophy class, we have to come up with a specific argument and answer it. I choose a physics questions becuase it's completely different then what we do in philosophy

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ek said:
Am I missing something here? Why is this a paradox? Seems pretty cut and dried to me.

Or is it called a paradox because it appears to be a paradox to people not "in the know"?

What kind of class is this for?

Special and General Relativity come to play largely during Quantum mechanics and analysis on the very microscale.

Its a paradox because a) Its not possible to return the clocks to the same position without accelerating, and the effects of acceleration have an unknown effect on this.

Other reasons I can't think of

The problem in the paradox is that each person will claim that they are experiencing proper time with no effects of time dilation, this is because one can never claim that they are the one moving. There is no frame of reference that includes the motion of both people at a constant rate. So Person B will say "im experiencing time as t_0, and he's experiencing time as t_dil" even though he's flying thru spsace at .99c, Person A is EQUALLY as correct in making the exact same claim.

Absolute velocity can not be determined unless an acceleration is present.

well, i did state that the planet A stays on is at constant inertia. So therefore, wouldn't B need to acc. anyways to get to the speed of light right off the planet?

Prague said:
Two twins, A and B are born on the planet X, which has a constant inertial rate. At the time of B's departure from X both A and B had an age of 0. Twin B travels 3 light years at a speed of 99c. Once reaching the destination of 3 light years, twin B returns to X again at 99c. What age will A and B be and why are the ages different?

Ok, so here goes.

A experiences time $$t_0$$
B experiences time $$t_{dil}$$
B travels at .99c (184140 mps)
A is stationary and has an age of t(0) = 0 (not sure what this means).
It just means this is the point we start counting from.
To find how long B is gone we take the distance and divide it by velocity. Also try to stick to SI units, MKS

$$t_{dil} = \frac{d}{v}$$
$$t_0 = \frac{d}{v}$$

If he is timing himself, then the time he finds for the distance he travels will be proper to him.

thus,

$$t_{dil} = \frac{1116000}{184140} = 6.060606061$$
is this light years becuase it involves twin B who is traveling in light years? And if he only felt 6 light years, is that really 6 years for him even though they are light years?
I don't know what these numbers are, but it looks like your putting miles into this, like I said above, all units should be MKS. Notice that 1 light year is a distance. It is the distance light travels in one year. He traveled a distance of 6 light years at speed .99c = 6.06years of travel in his perspective, to find the years of travel through the other person's perspective you must use the time dilation equation.

To find how long B was gone in reference to A we must find the time dilation for the equation

$$t_{dil} = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

therefore,

$$t_{dil} = \frac{t_0}{\sqrt{1-\frac{(184140)^2}{(186000)^2}}} = .1410673598$$

then,

$$t_{dil} = \frac{t_0}{0.1410673598} = \frac{6.060606061}{0.1410673598} = 42.96249728$$
this is years correct? because this involves twin A who is stationary so its years?

Thus, we conclude that if A and B were to meet again A = 42.96249728 years of age and B = 6.060606061 years of age.
Notice you got the same time dilation factor as I did. The time dilation for a 6 lightyear trip is ~40yrs but for a 40 lightyear trip is ~240yrs.

Also The units you plug in will be the units you get out, this is why units are so important, we have to be consistent in order to get a consistent result.
Just a future note, stick with MKS units. Also, for a philosophy paper, I don't think this would be hte best argument to pose, you are merely stating scientific facts, I am taking a logic in argment class now and I can't think of a decent argument you could shape up about this.
----------------
I added little pointers in small text below concepts I really didn't understand. So if you could answer those, that would help. Also, does this look right?

As for why this is a paradox, well this particular way of solving it is not a paradox. The paradox lies in when you look at the clock from the view point of twin A and the view point of Twin B, both seem to slow down so both A and B would say that the other twin would be older/younger. Its something like that, I am not to clear on it, I didn't look into it because I didn't understand the basics anyways.
Bolded: This doesn't sound right
Underlined: You are killing your credibility by doubting your knowledge, don't include this in your paper, you shouldn't argue something you arent qualified to.

This is for a philosophy class, we have to come up with a specific argument and answer it. I choose a physics questions becuase it's completely different then what we do in philosophy

well, i did state that the planet A stays on is at constant inertia. So therefore, wouldn't B need to acc. anyways to get to the speed of light right off the planet?
This is why this is physically impossible, it would atke an infinite amount of force to accelerate something ot speed of light. However, we are not considering getting to these speeds in these formulas. These formulas merely state that if you were to be traveling at those speeds under those conditions, the above time dilation effects will occur.

I don't know what these numbers are, but it looks like your putting miles into this, like I said above, all units should be MKS. Notice that 1 light year is a distance. It is the distance light travels in one year. He traveled a distance of 6 light years at speed .99c = 6.06years of travel in his perspective, to find the years of travel through the other person's perspective you must use the time dilation equation.

The argument is that B will be younger than A. It's a bad assignment, all he said was make a 1,000 word argument about anything you want. All you need is 6 published references (which I found) and an arguement. I figured I'd have some fun with it and actually learn something in the process.

-As for units, I just found that out. I just spent awhile converting everything into the correct units but using miles and seconds etc and to my surpirse, I got the same answer. So, lesson learned.

- I am not familiar with MKS, any good reference sites?

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MKS is the standard system of units internationally used
M stands for meters, K for kilograms, and S for seconds.

Oh wow. There's a whole lot of bad math and misinformation in this thread. Whozum, I am -again- going to ask that you stay out of threads on topics you do not yet completely understand. I'm sorry, but you are doing nothing but hopelessly confusing people. I'm serious, this cannot continue. I'm sure you don't realize how often you're wrong, but you really do not seem to be educated well enough to be answering these kinds of threads. Please stop.

Let me see if I can straighten this out a bit.

The equation originally posted by whozum,

$$t = \gamma t_0$$

where

$$\gamma \equiv \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

is often called "the time dilation equation." It's a direct result of using the Lorentz transform to transform coordinates between two observers (say, twins) in relative motion.

On the right-hand side, $t_0$ is the time elapsed on one of the twin's wristwatches. On the left-hand side, $t$ is the amount of time elapsed on the other twin's wristwatch in the same period of time.

The factor $\gamma$ is always equal to or greater than one. When there is no relative velocity (the twins are at rest with respect to each other), $\gamma$ is one. When the twins are traveling at 0.99c with respect to each other, it's about seven.

What this means is that when one twin observes one second elapsing on his watch, the other twin observes about seven. If one twin is flashing a laser beam every second (according to his watch), the other twin will observe the flashes coming only once every seven seconds.

Everything is symmetrical -- if both twins are flashing laser beams at each other, each will see the other's flashes as running slowly.

Now, not only does high relative motion cause time to dilate, lengths also contract. In other words, when you're at rest with respect to both the Sun and the star Alpha Centauri, the two stars appear to be about 4.3 light-years apart. When you're traveling at 0.99c from the Sun to Alpha Centauri, however, that distance appears to be contracted by the familiar factor, $\gamma$:

$$L = \frac{L_0}{\gamma}$$

If you're going 0.99c, the distance that appeared to be 4.3 light-years while at rest will only appear to be about six-tenths of a light-year. According to the twin aboard the starship, the trip to Alpha Centauri would only take about 223 days, less than a year. If the twin traveled even faster, it would appear to him to take even less time. At 0.999999c, the trip would take him only two days, according to his watch. To the twin sitting back on Earth, the trip will take the other twin nearly 4.3 years.

The twin "paradox" is this: why can't the twin aboard the spaceship claim himself at rest, while the Earth-bound twin is really the one traveling at 0.99c? The answer is that the traveling twin has to turn around at some point in his journey, and by turning around, he is leaving one frame of reference and entering another. You cannot analyze the entire problem without considering an additional, third frame of reference occupied by the twin on his return trip.

The "paradox" is not really a paradox at all if you understand how the special theory of relativity works.

Here's an excellent page that will explain the twin paradox is any many ways as you could possibly wish:

- Warren

Thanks alot, I'll take a look at that page and report back if I have any other questions.

Hey, I seem to be confused now. I can't seem to mathematically solve my problem anymore with the correct way. Can anyone help me?

Actually I did it, but I did it like this.

$$\gamma \equiv \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

where,

$$\gamma \equiv \frac{1}{\sqrt{1-\frac{(.99*184140)^2}{186000^2}}}$$

thus,

$$t = \gamma t_0$$

is,

this is where I get stuck. How do I figure out the time that has gone by for twin A and B. If I do it this way,

$$t = 7.08881205*6 = 42.5328723$$

then the answer doesn't seem right. It means that twin A is about 42 and twin B is about 7 and those age differences seem a little high for 6 light years of travel.

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I don't know what you're doing with all that 184140 stuff. You're making it harder than it should be.

Look at the expression (0.99 c)^2 / c^2. The c's cancel, leaving you with just 0.99^2.

Gamma for v = 0.99c is thus

One twin's watch ticks off one second. The other twin measures the same time interval as roughly seven seconds on his own watch.

- Warren

Ok so I take the twin that ticks just one second will be twin B (who traveled away) and the twin that ticks 7.08881205 is the one on Earth (twin a)

Alright now how do i figure out how long twin A is away for in B's reference and how long twin B is away for in A's reference?

Prague said:
Ok so I take the twin that ticks just one second will be twin B (who traveled away) and the twin that ticks 7.08881205 is the one on Earth (twin a)
No. During the trip out, both twins will measure the others' clocks as running seven times slower than their own.
Alright now how do i figure out how long twin A is away for in B's reference and how long twin B is away for in A's reference?
Please give me the numbers you're working with. Is the trip 40 light-years as measured by the Earth-bound observer, or what?

- Warren

Oh sorry, Twin A and B live on X blah blah blah when B departs their age set to 0. Twin B travels 3 light years at .99c and then returns at the same speed.

so what would those two final equations look like?

Also, $$t_0$$ is the time experienced by A (homebound twin) and t is the time experienced by B (traveler) right?

Thanks, I was confused at first with all the different ways I was told to solve it but I am getting it now.

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Prague said:
so what would those two final equations look like?
I've already given you all the equations.
Also, $$t_0$$ is the time experienced by A (homebound twin) and t is the time experienced by B (traveler) right?
$t_0$ is the time experienced by either of the twins, as measured by his own wristwatch. t is the time experienced by the other for the same interval.

- Warren

chroot said:
I've already given you all the equations.

I know, but from what I gather in the google.com equation is this.

$$3 light years * \frac{ \sqrt{1- \frac{.99c^2}{c^2}}}{.99*c}$$

now, if this is correct I understand that it will equal the 156 days. However, what I do not understand is...

Why you are dividing $$\gamma$$ by $$.99 * c$$.
Also, when I try to convert 3 light years to actual numbers and then change the .99 * c to 184140 the equation doesn't work. So I don't know what 3 light years or even c represents if it doesn't work.

Basically what I am trying to say is, if I were to do this on an actual calculator, not google.com how would I write it (with numbers included). I can't put a c so I assume its 186000, I can't write 3 light years so I assume its 558000, but putting these in they don't calculate to the 156 days.

I hope you understand what I am getting at.

chroot said:
Prague said:
Originally Posted by Prague
Ok so I take the twin that ticks just one second will be twin B (who traveled away) and the twin that ticks 7.08881205 is the one on Earth (twin a)
No. During the trip out, both twins will measure the others' clocks as running seven times slower than their own.
Alright now how do i figure out how long twin A is away for in B's reference and how long twin B is away for in A's reference?
Please give me the numbers you're working with. Is the trip 40 light-years as measured by the Earth-bound observer, or what?

- Warren

Are you sure this is right Warren? On the out bound trip each measure the other time as slower? This only works if the universe started off with both of them in a initial frame, which is a contrived, if not impossible scenario.

When the traveling twin accelerates for his/her relativistic journey he is no longer on equal footing with his twin. So the traveling twin ages slower on both legs of the trip.

What the traveling twin can know and what he can see are two different things. If measure means what the traveling twin can see then yes he measures his counterparts clock as going slow on the outbound trip, but that does not mean he can realistically expect it to be moving slower.

Prague said:
chroot said:
I've already given you all the equations.

I know, but from what I gather in the google.com equation is this.

$$3 light years * \frac{ \sqrt{1- \frac{.99c^2}{c^2}}}{.99*c}$$

now, if this is correct I understand that it will equal the 156 days. However, what I do not understand is...

Why you are dividing $$\gamma$$ by $$.99 * c$$.
Also, when I try to convert 3 light years to actual numbers and then change the .99 * c to 184140 the equation doesn't work. So I don't know what 3 light years or even c represents if it doesn't work.

Basically what I am trying to say is, if I were to do this on an actual calculator, not google.com how would I write it (with numbers included). I can't put a c so I assume its 186000, I can't write 3 light years so I assume its 558000, but putting these in they don't calculate to the 156 days.

I hope you understand what I am getting at.

He was useing:
$$L = \frac{L_0}{\gamma}$$

L is the distance that the traveling twin experiences and .99c is the speed he is going so:
$$\frac{L}{v} =\frac{\frac{L}{\gamma}}{v} = \frac{\frac{L_0}{\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}}}{v} = \frac{L_0 \sqrt{1-\frac{v^2}{c^2}}}{v}$$

Hmm, alright I see now. So let's see,

$$\gamma \equiv \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

gives me the seconds for the twin opposite the observer. Which is 7.08881205

$$t = \gamma t_0 (not sure what this one is used for. and [tex] L = \frac{L_0}{\gamma}$$

gives me the days/year experienced by the observes/traveler.

and

$$\frac{L}{v} =\frac{\frac{L}{\gamma}}{v} = \frac{\frac{L_0}{\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}}}{v} = \frac{L_0 \sqrt{1-\frac{v^2}{c^2}}}{v}$$

is the actual equations. Sorry for asking so much, I really want to understand this. I can't seem to grasp how to plug all the actual numbers in. What would $$L_0$$ be and L is 3 (or 6 if I do full trip)? Is it ok just to put 3 in? Or do I have to convert the 3 into light years or miles etc...

I am just trying to piece the entire thing together from start to finish.

Davorak said:
When the traveling twin accelerates for his/her relativistic journey he is no longer on equal footing with his twin. So the traveling twin ages slower on both legs of the trip.
I'm trying quite hard to avoid the technicalities here; I understand my treatment is not entirely complete (I provided links to Baez' treatment, which is more complete than anything I could write here). The original poster is trying to write an essay on the topic, and first must just understand time dilation and length contraction, so I'm pressing that bit first.

- Warren

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