# Twin Paradox

chroot
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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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chroot
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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 lets 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 peice the entire thing together from start to finish.

chroot
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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

chroot
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Prague said:
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.
Okay, you should never mention miles again. Ever. If you ever even think of miles again, you're doing something wrong. Do not call c 186282 miles/second. Do not call c 300,000 km/sec either. Call c one light-year per year.

That's right -- c is just one in units where you use light-years and years as your units of length and time.

Now, gamma's easy to calculate, as I showed you. When you're considering velocities as fractions of c, then the c's cancel out. When you're considering 0.99c, for example, the term (0.99c)^2/c^2 can be simplified to just 0.99^2. Gamma is then just 1/(1-0.99^2), which you can do on any calculator. No miles. No unit conversions. If you ever attempt to use miles ever again, you're doing something wrong.

Now, gamma is dimensionless. It does not have units of time, or length, or anything. Gamma is just a number, a multiplicative factor.

When you want to use gamma in the time-dilation or length-contraction equations, leave the times in years, and the lengths in light-years.

If the twin is travelling three light-years at 0.99c, that three light-years will be contracted to 3/gamma light-years (about 0.423 light-years). Simple division. No miles.

The twin will cross those 3/gamma light-years in (3/gamma)/0.99c. Remember in our units c is one-light yer per year. Here are the units:

$$\frac{3 \,\textrm{light-years}}{\gamma} \, \cdot \, \frac{1}{0.99 \,\textrm{light-years per year}} = 0.427 \,\textrm{years}$$

Or, about 156 days.

- Warren

Ah, I understand this now. Thank you, hopefully this will get me by on my paper. I'll use this to prove there is a time difference, then I can say that the paradox is which twin is aging slowly because each twins reference is that the other twin is slowing down. And the simple answer is, the traveling twin ages slower because he leaves the the constant inertial rate of X, correct?

chroot
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You can use the time-dilation equations to show what the paradox means. The time-dilation equation is symmetrical, meaning that both the earth-bound and the spaceship-bound twin sees the others' clock as running slowly. You can show some numbers to make the concept of time dilation more concrete.

The resolution of the paradox comes from the realization that, for the twins to re-unite, one of them must turn around. The twin that turns around, of course, is the one on the spaceship. He has to push buttons on his console to fire his engines. He feels the forces of the engines. His coffee cup gets knocked over, etc. After he's turned around, he's no longer in the same frame of reference that he was when he was moving away from the Earth. The introduction of this new third frame breaks the symmetry of the paradox. When you solve the problem will all three frames considered, the paradox is no longer a paradox: All observers will see the spaceship-twin as aging less than the Earth-twin.

- Warren

Alright, thanks alot I understand everything now. Now on to writing.

Thank to everyone for the help.

chroot
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Good luck with the paper!

- Warren