The Symmetric Twin Paradox
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From: train on 20 Jun 2010 06:12 On Jun 20, 2:13 pm, "Inertial" <relativ...(a)rest.com> wrote: > "train" <gehan.ameresek...(a)gmail.com> wrote in message > > news:2ec5e817-9bb9-43f7-b407-57a440ababdb(a)a9g2000prd.googlegroups.com... > > > > > > > On Jun 19, 6:06 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> "train" <gehan.ameresek...(a)gmail.com> wrote in message > > >>news:8848af8d-c47c-4d28-b572-cfd072537de9(a)s4g2000prh.googlegroups.com.... > > >> > On Jun 19, 1:11 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> >> "colp" <c...(a)solder.ath.cx> wrote in message > > >> >>news:81c945d7-ef5d-4905-bc17-ff691d4025fd(a)z15g2000prh.googlegroups.com... > > >> >> > On Jun 19, 7:34 pm, "Inertial" <relativ...(a)rest.com> wrote: > >> >> >> "colp" <c...(a)solder.ath.cx> wrote in message > > >> >> >>news:3f27a5b2-6fe9-4f52-9d45-033de8e4f473(a)g39g2000pri.googlegroups.com... > > >> >> >> > On Jun 19, 3:27 am, Tom Roberts <tjroberts...(a)sbcglobal.net> > >> >> >> > wrote: > >> >> >> >> colp wrote: > >> >> >> >> > It is not necessary for me to showing you the math in order > >> >> >> >> > for > >> >> >> >> > you > >> >> >> >> > to > >> >> >> >> > identify the errors in the article. > > >> >> >> >> The basic error in that article is that they DID NOT use the > >> >> >> >> math > >> >> >> >> of > >> >> >> >> SR. > > >> >> >> > That isn't necessarily an error. Can you show how their math > >> >> >> > resulted > >> >> >> > in coming to an incorrect conclusion? > > >> >> >> >> Instead > >> >> >> >> they used a comic-book description of SR such as "moving clocks > >> >> >> >> run > >> >> >> >> slow" -- SR > >> >> >> >> does NOT say that; > > >> >> >> > The truth is not determined by what SR says > > >> >> >> The truth about what SR says IS determinets by what SR says > > >> >> > Circular reasoning. > > >> >> Nope. Your LACK of reasoning. You say the truth of SR is determined > >> >> by > >> >> what SR does NOT say. SR does NOT say the twins are less than each > >> >> other > >> >> over the whole experiment .. it says they have the same ages. > > >> > Even if the twins both show the same age after te symmetric travel, I > >> > made the point earlier that twin A has moved relative to twin B and > >> > twin B has moved relative to twin A. > > >> Of course they have > > >> > When realtive motion occurs, time > >> > dilation occurs. > > >> Yeup > > >> > The additional paradox is > > >> > how can both twins show the same age when relative movement between > >> > them has occurred? > > >> Same way as one twin can be younger than the other in the usual twins > >> paradox. > > > My point is that the assertion that the twins show the same age is in > > contradiction with the fact that the twins have moved relative to each > > other. I cannot make it more simple than that > > Wrong > > > The moving clock runs slow > > Yes it does .. as measured by the 'stationary' clock > > > The stay at home twin ages faster means that all stay at home twins > > age faster than the all traveling twins that follow the exact same > > flight profile. > > > But a particular traveling twin`s clock is always in motion with > > respect to another traveling twins clock if they move in paths 90 > > degrees to each other for example. > > > Is this not true? > > Only half the story. > > > oh maybe as AE said we have to give up common sense. And reason? > > Nope. Only intuition As far as I know a statment can be true or not true . Half true equals what?
From: train on 20 Jun 2010 06:28 On Jun 20, 2:39 pm, "Androcles" <Headmas...(a)Hogwarts.physics_z> wrote: > "harald" <h...(a)swissonline.ch> wrote in message > > news:1febcc23-2ef4-4331-bcde-20d305fc7873(a)r27g2000yqb.googlegroups.com... > On Jun 20, 2:07 am, train <gehan.ameresek...(a)gmail.com> wrote: > [..] > > > > >> >> >> Instead > > > >> >> >> they used a comic-book description of SR such as "moving clocks > > > >> >> >> run > > > >> >> >> slow" -- SR does NOT say that; > > Einstein wrote that in a scientific paper in 1905; > ======================================= > One man's "scientific paper" is another's man's comedy act, and you are > a fxxxxg joke. OK so mathematics is supposed to settle this sort of thing before coming to blows... Ok so for a spaceship named creatively as A and a spaceship named as B. Both are together then accelerate for n seconds directly away from each other. Bring out the applicable formula please.. dt = dt'/sqrt(1-v^2/c^2) Are we all agreed on this?
From: harald on 20 Jun 2010 06:35 On Jun 20, 12:28 pm, train <gehan.ameresek...(a)gmail.com> wrote: [..] > OK so mathematics is supposed to settle this sort of thing before > coming to blows... The correct *application* of mathematics; as I pointed out to you: "Einstein's introduction sentence of his 1905 paper is misleading: not relative motion between any objects, but motion relative to inertial systems is used for the calculations." > Ok so for a spaceship named creatively as A and a spaceship named as > B. > > Both are together then accelerate for n seconds directly away from > each other. > > Bring out the applicable formula please.. > > dt = dt'/sqrt(1-v^2/c^2) > > Are we all agreed on this? Good start. ;-) Harald
From: Inertial on 20 Jun 2010 07:29 "train" <gehan.ameresekere(a)gmail.com> wrote in message news:ee0f98c2-eef6-4c6f-9b56-513976e112ae(a)k25g2000prh.googlegroups.com... > On Jun 20, 2:13 pm, "Inertial" <relativ...(a)rest.com> wrote: >> "train" <gehan.ameresek...(a)gmail.com> wrote in message >> >> news:2ec5e817-9bb9-43f7-b407-57a440ababdb(a)a9g2000prd.googlegroups.com... >> >> >> >> >> >> > On Jun 19, 6:06 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> "train" <gehan.ameresek...(a)gmail.com> wrote in message >> >> >>news:8848af8d-c47c-4d28-b572-cfd072537de9(a)s4g2000prh.googlegroups.com... >> >> >> > On Jun 19, 1:11 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> "colp" <c...(a)solder.ath.cx> wrote in message >> >> >> >>news:81c945d7-ef5d-4905-bc17-ff691d4025fd(a)z15g2000prh.googlegroups.com... >> >> >> >> > On Jun 19, 7:34 pm, "Inertial" <relativ...(a)rest.com> wrote: >> >> >> >> "colp" <c...(a)solder.ath.cx> wrote in message >> >> >> >> >>news:3f27a5b2-6fe9-4f52-9d45-033de8e4f473(a)g39g2000pri.googlegroups.com... >> >> >> >> >> > On Jun 19, 3:27 am, Tom Roberts <tjroberts...(a)sbcglobal.net> >> >> >> >> > wrote: >> >> >> >> >> colp wrote: >> >> >> >> >> > It is not necessary for me to showing you the math in order >> >> >> >> >> > for >> >> >> >> >> > you >> >> >> >> >> > to >> >> >> >> >> > identify the errors in the article. >> >> >> >> >> >> The basic error in that article is that they DID NOT use the >> >> >> >> >> math >> >> >> >> >> of >> >> >> >> >> SR. >> >> >> >> >> > That isn't necessarily an error. Can you show how their math >> >> >> >> > resulted >> >> >> >> > in coming to an incorrect conclusion? >> >> >> >> >> >> Instead >> >> >> >> >> they used a comic-book description of SR such as "moving >> >> >> >> >> clocks >> >> >> >> >> run >> >> >> >> >> slow" -- SR >> >> >> >> >> does NOT say that; >> >> >> >> >> > The truth is not determined by what SR says >> >> >> >> >> The truth about what SR says IS determinets by what SR says >> >> >> >> > Circular reasoning. >> >> >> >> Nope. Your LACK of reasoning. You say the truth of SR is >> >> >> determined >> >> >> by >> >> >> what SR does NOT say. SR does NOT say the twins are less than each >> >> >> other >> >> >> over the whole experiment .. it says they have the same ages. >> >> >> > Even if the twins both show the same age after te symmetric travel, >> >> > I >> >> > made the point earlier that twin A has moved relative to twin B and >> >> > twin B has moved relative to twin A. >> >> >> Of course they have >> >> >> > When realtive motion occurs, time >> >> > dilation occurs. >> >> >> Yeup >> >> >> > The additional paradox is >> >> >> > how can both twins show the same age when relative movement between >> >> > them has occurred? >> >> >> Same way as one twin can be younger than the other in the usual twins >> >> paradox. >> >> > My point is that the assertion that the twins show the same age is in >> > contradiction with the fact that the twins have moved relative to each >> > other. I cannot make it more simple than that >> >> Wrong >> >> > The moving clock runs slow >> >> Yes it does .. as measured by the 'stationary' clock >> >> > The stay at home twin ages faster means that all stay at home twins >> > age faster than the all traveling twins that follow the exact same >> > flight profile. >> >> > But a particular traveling twin`s clock is always in motion with >> > respect to another traveling twins clock if they move in paths 90 >> > degrees to each other for example. >> >> > Is this not true? >> >> Only half the story. >> >> > oh maybe as AE said we have to give up common sense. And reason? >> >> Nope. Only intuition > > As far as I know a statment can be true or not true . Half true equals > what? There is more than one statement to make. Stating only half the facts is only half of the truth
From: Daryl McCullough on 20 Jun 2010 09:10
colp says... >On Jun 20, 1:11=A0am, Tom Roberts <tjroberts...(a)sbcglobal.net> wrote: >> You MUST learn what SR ACTUALLY says. > >What do you think that SR actually says about the symmetric twin >thought experiment? What SR says is that the laws of mechanics and electromagnetism are invariant under the Lorentz transformations. That's the physical content of SR. This was already true of Maxwell's equations, but not of Newtonian mechanics. So part of the development of SR involved finding a replacement to Newton's equations of motion that were invariant under the Lorentz transformations. I know that's not what you meant---you don't actually want to know about real physics, you want to know about thought experiments involving clocks and twins and rockets. What SR says for those sorts of thought experiments is that, as measured in any inertial coordinate system, 1. Light travels at constant velocity, at speed c, in all directions, independent of the motion of the source. 2. An ideal clock traveling at speed v for time period t will show an elapsed time of T = t square-root(1-(v/c)^2). 3. An extended object traveling at speed v will, after reaching its equilibrium shape, be contracted in the direction of motion by a factor of square-root(1-(v/c)^2). 4. An object not under the influence of any forces will move at constant speed. Those are all true in any inertial coordinate system. The final claim, which is actually derivable from the first 4, (alternatively, 2&3 can be derived from 5) is: 5. If C is an inertial coordinate system, and C' is obtained from C by a translation, rotation or Lorentz transformation, then C' is an inertial coordinate system. So anything you want to derive about twins moving around, sending light signals, carrying clocks, etc. is derivable from 1-5. Pick any inertial coordinate system you like to analyze the situation, and compute away using 1-5. The usual twin paradox: A twin hops in a rocket at time t=0, speeds away to a distance L, turns around rapidly, and speeds back at speed v. How much older is he when he returns? By rule 2 above, it's pretty simple: the elapsed time on the rocket for the full trip will be 2 (L/v) square-root(1-(v/c)^2). In contrast, the stay-at-home twin ages by an amount 2L/v. Done. The traveling twin is younger. Now, if you want to compute the results from the point of view of some other frame, that's fine. Go ahead, and use 1-5. Let's compute it from the outgoing frame of the traveling twin. First, we have to figure out what happens in this frame, since the description above was from the point of view of the stay-at-home twin. To get the corresponding description for the new frame, we use the Lorentz transformations. There are three major events involved: E1: The two twins depart. In the stay-at-home frame, this has coordinates (x=0, t=0). By the Lorentz transformations, in the new frame, it has the same coordinates: (x'=0, t'=0). E2: The traveling twin turns around. In the stay-at-home frame, this has coordinates (x=L, t=L/v). By the Lorentz transformations, in the new frame, it has the coordinates: x'=0, t' = gamma (t-vx/c^2) = gamma (L/v - vL/c^2) = L/v gamma (1-v^2/c^2) = L/v 1/gamma (since gamma = 1/square-root(1-(v/c)^2). E3: The twins reunite. In the stay-at-home frame, this has coordinates (x=0, t=2L/v). In the new frame, it has coordinates x' = gamma (x-vt) = -gamma 2L t' = gamma (t - vx/c^2) = gamma 2L/v Okay, so in the new frame, the story looks like this: The stay-at-home twin travels at constant speed v in the -x direction the whole time. He moves from x'=0 to x'=-gamma 2L. The total time for the trip is gamma 2L/v. So using our rule 2 above, if the stay-at-home twin travels at speed v for a length of time t, then he will age by an amount t square-root(1-(v/c)^2). In this case, the trip took time gamma 2L/v as measured in this frame, so the stay-at-home twin ages by an amount: gamma 2L/v * square-root(1-(v/c)^2) = 2L/v (since gamma = 1/square-root(1-(v/c)^2) So, in this frame, the stay-at-home twin is *also* 2L/v, just like in the first frame. Now, let's calculate the age of the traveling twin in this frame: As viewed in this frame, the traveling twin stays put at x'=0 between the times t'=0 and t'= L/v 1/gamma. His velocity during this period is 0 in this new frame, so during the first part of the trip, he ages by an amount L/v 1/gamma. What about the second part of the trip (after turnaround)? During this part of the trip, the traveling twin goes from x'=0 at time t'=L/v 1/gamma to x'=-gamma 2L at time t' = gamma 2L/v. So let's figure out what the traveling twin's speed was during this part of the trip. The total elapsed time was: t_end - t_start = gamma 2L/v - L/v 1/gamma = gamma L/v (2 - 1/gamma^2) = gamma L/v (1 + (v/c)^2) (where I used 1/gamma^2 = 1 - (v/c)^2) The total distance traveled was: |x_end - x_start| = gamma 2L So the velocity during this part of the journey is v' = gamma 2L/(gamma L/v (1 + (v/c)^2)) = 2v/(1+(v/c)^2) (This is what would have been obtained using the velocity addition formula) Now, let's calculate square-root(1-(v'/c)^2) for this part of the trip: (v'/c)^2 = 4(v/c)^2/(1 + 2(v/c)^2 + (v/c)^4) 1 - (v'/c)^2 = (1 - 2(v/c)^2 + (v/c)^4)/(1+2(v/c)^2 + (v/c)^4) square-root(1-(v'/c)^2) = (1-(v/c)^2)/(1+(v/c)^2) To find out how much the traveling twin ages during the last part of the journey, we multiply this factor by the elapsed time, which we calculated as gamma L/v (1 + (v/c)^2) Aging during last part = gamma L/v (1 + (v/c)^2) * (1-(v/c)^2)/(1+(v/c)^2) = gamma L/v (1-(v/c)^2) = L/v 1/gamma (where again I used gamma = 1/square-root(1-(v/c)^2). So the total aging done by the traveling twin turns out to be L/v 1/gamma during both legs of the journey. So when the journey is over, he has aged by 2L/v 1/gamma while the stay-at-home twin has aged by 2L/v So once again, the traveling twin is younger when they get back together. So if you use the actual equations of SR (as opposed to guessing), then the results are completely consistent: In *every* inertial coordinate system, the answer is the same: The stay-at-home twin ages more than the traveling twin. We could go through the symmetric case, if you want, but the results will be this: During the outward trip, each twin ages by an amount L/v square-root(1-(v/c)^2). During the return trip, each twin ages by the same amount. So when the twins get back together, they have both aged by an amount 2L/v square-root(1-(v/c)^2). They are the same age when they get back together. Every inertial coordinate system agrees on this answer. There is no paradox, in the sense of two different ways to compute things leading to two different results for the ages. The paradox comes up when you *fail* to correctly use SR. -- Daryl McCullough Ithaca, NY |