How does mutual time dilation work with one twin aging faster?
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From: Mathal on 8 Aug 2010 22:48 On Aug 8, 7:27 am, Gc <gcut...(a)hotmail.com> wrote: > On 8 elo, 17:18, Mathal <mathmusi...(a)gmail.com> wrote: > > > On Aug 8, 6:53 am, Gc <gcut...(a)hotmail.com> wrote:> On 8 elo, 08:06, BURT <macromi...(a)yahoo.com> wrote: > > > > > If you watched a clock that you are passing at high speed; if you are > > > > the one aging slower how can you see that it is aging more than you > > > > but ticking slower than you at the same time? > > > > All the important stuff in the twin paradox happens when the twin in > > > the spacecraft feels acceleration (it has to turn at some point if it > > > comes back to earth). The "aging difference effect" happens just when > > > the acceleration does. > > > No. The aging difference effect is due to a difference in relative > > velocity. If you get up and go for a walk you will be younger than the > > 'you' who decided instead to just sit in one place. Not much younger, > > but younger. > > No, your proper time is of course different then and that is solely > because you _felt more acceralation_. No. When the ship decelerates and returns, while the ship is motionless WRT the other twin, the twins experience the same rate of time and again when the ship with the travelling twin returns and stops in the frame of the stationary twin they share the same rate of time. At all other moments when the relative velocity of the frames differs, the difference in the time-span that occurs in the two frames increases. It is necessarily the travelling twin's clock (his frame) that is moving slower (time-wise) that the stationary twin. It is the velocity, not the acceleration. I'll give you this, you can't get from an initial shared state of rest to two frames moving WRT each other without acceleration, but there deosn't need to be any more than the initial aceleration i.e. the velocity can be constant for the majority of the trip. Mathal By the way, when did the Linus blaket term 'proper time' come back into vogue?
From: BURT on 8 Aug 2010 22:57 On Aug 8, 7:48 pm, Mathal <mathmusi...(a)gmail.com> wrote: > > > > > If you watched a clock that you are passing at high speed; if you are > > > > > the one aging slower how can you see that it is aging more than you > > > > > but ticking slower than you at the same time? > > > > > All the important stuff in the twin paradox happens when the twin in > > > > the spacecraft feels acceleration (it has to turn at some point if it > > > > comes back to earth). The "aging difference effect" happens just when > > > > the acceleration does. > > > > No. The aging difference effect is due to a difference in relative > > > velocity. If you get up and go for a walk you will be younger than the > > > 'you' who decided instead to just sit in one place. Not much younger, > > > but younger. > > > No, your proper time is of course different then and that is solely > > because you _felt more acceralation_. > > No. When the ship decelerates and returns, while the ship is > motionless WRT the other twin, Please explain how deceleration is motionless. > the twins experience the same rate of > time and again when the ship with the travelling twin returns and > stops in the frame of the stationary twin they share the same rate of > time. At all other moments when the relative velocity of the frames > differs, the difference in the time-span that occurs in the two frames > increases. It is necessarily the travelling twin's clock (his frame) > that is moving slower (time-wise) that the stationary twin. > > It is the velocity, not the acceleration. I'll give you this, you > can't get from an initial shared state of rest to two frames moving > WRT each other without acceleration, but there deosn't need to be any > more than the initial aceleration i.e. the velocity can be constant > for the majority of the trip. > > Mathal > By the way, when did the Linus blaket term 'proper time' come back > into vogue?- Hide quoted text - > > - Show quoted text - In order to accelerate energy must pass through every speed inbetween until ariving at an end speed. It is the changing of speed or acceleration that slows time. But only one twin accelerated and that one's clock runs slower. It had weightedness and the other did not. It ages slower. If at the end one is older they both cannot be going slower than the other as in the appearence of relativity. Lost time can be eliminated in the argument for a twin passing at near light speed. Mitch Raemsch
From: kenseto on 9 Aug 2010 10:07 On Aug 8, 10:48 pm, Mathal <mathmusi...(a)gmail.com> wrote: > On Aug 8, 7:27 am, Gc <gcut...(a)hotmail.com> wrote: > > > > > > > On 8 elo, 17:18, Mathal <mathmusi...(a)gmail.com> wrote: > > > > On Aug 8, 6:53 am, Gc <gcut...(a)hotmail.com> wrote:> On 8 elo, 08:06, BURT <macromi...(a)yahoo.com> wrote: > > > > > > If you watched a clock that you are passing at high speed; if you are > > > > > the one aging slower how can you see that it is aging more than you > > > > > but ticking slower than you at the same time? > > > > > All the important stuff in the twin paradox happens when the twin in > > > > the spacecraft feels acceleration (it has to turn at some point if it > > > > comes back to earth). The "aging difference effect" happens just when > > > > the acceleration does. > > > > No. The aging difference effect is due to a difference in relative > > > velocity. If you get up and go for a walk you will be younger than the > > > 'you' who decided instead to just sit in one place. Not much younger, > > > but younger. > > > No, your proper time is of course different then and that is solely > > because you _felt more acceralation_. > > No. When the ship decelerates and returns, while the ship is > motionless WRT the other twin, the twins experience the same rate of > time and again when the ship with the travelling twin returns and > stops in the frame of the stationary twin they share the same rate of > time. At all other moments when the relative velocity of the frames > differs, the difference in the time-span that occurs in the two frames > increases. It is necessarily the travelling twin's clock (his frame) > that is moving slower (time-wise) that the stationary twin. > > It is the velocity, not the acceleration. I'll give you this, you > can't get from an initial shared state of rest to two frames moving > WRT each other without acceleration, but there deosn't need to be any > more than the initial aceleration i.e. the velocity can be constant > for the majority of the trip. The acceleration of the traveling clock increases its state of absolute motion and thus makes it runs slower than the stay at home clock. Ken Seto > > Mathal > By the way, when did the Linus blaket term 'proper time' come back > into vogue?- Hide quoted text - > > - Show quoted text -
From: kenseto on 9 Aug 2010 10:36 On Aug 8, 1:26 pm, stevendaryl3...(a)yahoo.com (Daryl McCullough) wrote: > It's not so mysterious. Think about traveling along the x-axis. > If you travel at a constant speed for 1 hour elapsed time, you'll > end up a certain distance from your starting point. If you travel > twice as fast, you'll end up twice as far from your starting > point. There's nothing mysterious about that. > > Now, people are always traveling at a nonzero velocity in the t-direction.. > If you travel for one hour, you'll end up at a different time then when > you started. If you travel for one hour at twice at twice the velocity > in the t-direction, you'll end up twice as far along the t-axis. This is false....you assumed that the passage of time is velocity dependent. It is not. The problem here is that you assume that a clock second is a universal interval of time. It is not....a clock second in different frame will contain a different amount of absolute time. This is the reason why the speed of light is a constant math ratio in all frames as follows: Light path length of ruler(299,792,458 m long physically)/the absolute time content for a clco second co-moving with the ruler. > > The 4-D view of SR is that every object has a velocity in the x-direction, > the y-direction, the z-direction and the t-direction. Different travelers > have different velocities in the t-direction, so they travel different > distances along the t-axis. The passage of absolute time in constant in all frames....it is independent on the state of motion of the object. > > Rather than thinking one twin ages 1 hour while the other ages 1/2 hour, > instead you think that one twin takes a full hour to go from 12:00 to 1:00, > while the other twin only takes 1/2 hour to go the same distance along > the t-axis. Here you are assuming that an hour in one frame contain the same amount of time as an hour in a different frame. It is not. Before you can make a comparison you must convert the 1/2 hour of the traveling twin into the stay at home twin's clock by a factor of gamma. When you do that you will see that both experienced the same ampount of absolute time. Ken Seto > > -- > Daryl McCullough > Ithaca, NY
From: PD on 9 Aug 2010 11:58
On Aug 8, 12:06 am, BURT <macromi...(a)yahoo.com> wrote: > If you watched a clock that you are passing at high speed; if you are > the one aging slower how can you see that it is aging more than you > but ticking slower than you at the same time? > > If time dilation is mutual then one twin cannot age any different than > the other. But one does. Time dilation is mutual only for two clocks in *inertial* motion. In the twin puzzle and in the GPS case, this is not true. > > Please prove that time only appears slower. I say to you that you will > see the station's clock always running faster and mutual is an excuse; > If you are the one that accelerated as the station did not. The > difference is you felt weight at acceleration that the station doesn't > know about. > > Mitch Raemsch |