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Homework Help: Twin Paradox

  1. Apr 4, 2005 #1
    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.)

    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.
    Last edited: Apr 4, 2005
  2. jcsd
  3. Apr 4, 2005 #2
    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.
  4. Apr 4, 2005 #3
    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.
  5. Apr 4, 2005 #4
    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?
  6. Apr 4, 2005 #5
    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 wont have to compensate writing skills to express an equation in a paragraph.
  7. Apr 4, 2005 #6
    [tex] t_{dil} = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]

    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:

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

    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 [tex] t_0 [/tex] will give you the first equation I cited. [tex] \gamma [/tex] is always very close to 1.
  8. Apr 4, 2005 #7

    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.
  9. Apr 4, 2005 #8
    In this case your time dilation equation would be:

    [tex] t_{dil} = \frac{t_0}{\sqrt{1-\frac{(.99c)^2}{c^2}}} [/tex]

    Person A experiences time [tex]t_{dil}. [/tex]
    Person B experiences time [tex]t_0 [/tex]

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

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

    [itex] t_0 = \frac{40}{.99} = 39.6 [/itex]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:

    [tex] t_{dil} = \frac{t_0}{\sqrt{1-\frac{(.99c)^2}{c^2}}} [/tex]

    The denominator becomes [itex] \sqrt{1-(.99)^2} = \sqrt{0.0199} = 0.1411 [/itex]

    [tex] t_{dil} = \frac{t_0}{0.1411} = \frac{39.6}{0.1411} = 280.652 years [/tex]

    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.
  10. Apr 4, 2005 #9
    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
  11. Apr 4, 2005 #10
    You are right I think, but I'm not sure how that would incorporate into the above. I actually dont 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.
  12. Apr 4, 2005 #11
    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 wouldnt 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.
    Last edited: Apr 4, 2005
  13. Apr 4, 2005 #12
    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.

    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.

    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.

    [tex] F = \frac{dp}{dt} [/tex] As dt-> 0 F -> [tex]\infty [/tex]

    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.
  14. Apr 4, 2005 #13


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    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?
  15. Apr 5, 2005 #14
    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 [tex]t_0[/tex]
    B experiences time [tex] t_{dil}[/tex]
    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.

    [tex]t_0 = \frac{d}{v}[/tex]


    [tex]t_0 = \frac{1116000}{184140} = 6.060606061 [/tex]
    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

    [tex] t_{dil} = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]


    [tex] t_{dil} = \frac{t_0}{\sqrt{1-\frac{(184140)^2}{(186000)^2}}} = .1410673598[/tex]


    [tex]t_{dil} = \frac{t_0}{0.1410673598} = \frac{6.060606061}{0.1410673598} = 42.96249728 [/tex]
    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 arguement and answer it. I choose a physics questions becuase it's completely different then what we do in philosophy :bugeye:
    Last edited: Apr 5, 2005
  16. Apr 5, 2005 #15
    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 cant think of
  17. Apr 5, 2005 #16
    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 hes experiencing time as t_dil" even though hes 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.
  18. Apr 5, 2005 #17
    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?
  19. Apr 5, 2005 #18
    It just means this is the point we start counting from.
    [tex]t_0 = \frac{d}{v}[/tex]

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

    I dont 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 travelled 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.

    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, Im taking a logic in argment class now and I can't think of a decent argument you could shape up about this.
    Bolded: This doesnt sound right
    Underlined: You are killing your credibility by doubting your knowledge, dont include this in your paper, you shouldnt argue something you arent qualified to.

    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 travelling at those speeds under those conditions, the above time dilation effects will occur.
  20. Apr 5, 2005 #19
    The arguement is that B will be younger than A. It's a bad assignment, all he said was make a 1,000 word arguement 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?
    Last edited: Apr 5, 2005
  21. Apr 5, 2005 #20
    MKS is the standard system of units internationally used
    M stands for meters, K for kilograms, and S for seconds.
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