Twin Paradox, Paradox?

[Erasmus00]

Relativity says this can not happen for the orbiting/circling twin because his clock must run faster than ours on earth if he's traveling at a significant fraction of c even at a constant speed (ie, no acceleration... granted he had to accelerate for an instant to gain speed, but assume we began the signal after that moment and he traveled at a constant rate thereafter).

 

Bolding the relevant phrase. In your experiment the person IS accelerating, constantly. The only way to maintain circular motion is a constant inward acceleration. In your example, the astronaut is constantly accelerating, there isn't one instant where he isn't. This is why your simple time dilation argument fails.

-Will

[xersan]

Bolding the relevant phrase. In your experiment the person IS accelerating, constantly. The only way to maintain circular motion is a constant inward acceleration. In your example, the astronaut is constantly accelerating, there isn't one instant where he isn't. This is why your simple time dilation argument fails.

-Will

This paradox is became affirmative for hundred years and the theory is sanctified in 2005. I think, the human is still influenced by mystisism. But, why don't the science and scientists solve this paradox ? Paradoxes are the warning about inappropriateness. I have some ideas the solution

[EWright]

Bolding the relevant phrase. In your experiment the person IS accelerating, constantly. The only way to maintain circular motion is a constant inward acceleration. In your example, the astronaut is constantly accelerating, there isn't one instant where he isn't. This is why your simple time dilation argument fails.

-Will

If you can't answer the question, just say so. Fine then, allow him to accelerate and explain the time dialation that he will see. Oh, and let's accelerate him at such a rate that he maintains a Geosynchronous orbit. We can slow the spaceship down if necessary and discuss the time experienced in nanoseconds if need be. But my question remains unanswered.

[Erasmus00]

If you can't answer the question, just say so. Fine then, allow him to accelerate and explain the time dialation that he will see. Oh, and let's accelerate him at such a rate that he maintains a Geosynchronous orbit. We can slow the spaceship down if necessary and discuss the time experienced in nanoseconds if need be. But my question remains unanswered.

 

I will ignore gravitational effects and just do an SR calculation. Also, note I've outlined this calculation before, you have ignored it.

 

Note, for variables I use common ones. x,y,z for space, c for speed of light, t for time in seconds of a stationary observer, r for radius, theta shall be written as O and is the polar angle, omega shall be written w as the angular velocity of the orbit. T will be for the proper time of our rotating astronaut.

 

Start with the Minkowski metric for flat spacetime. ds^2= c^2dt^2-dx^2-dy^2-dz^2.

 

Now, coordinate change into a rotating reference frame, assuming a geosynchronus orbit, w orbit= w earth = w. x = r cos(O +wt), y= r sin(O+wt). You get the rotating metric for flat space

ds^2 = (c^2-r^2 w^2)dt^2 - dr^2 -r^2dO^2 - 2r^2 wdtdO - dz^2.

 

Now, we compare the time the astronaut notice as compared to a non rotational observer. (we could also compare to an observer corotating but at a shorter radius. This is just a trivial extension of what I'm about to do).

 

Now, for a ciruclar orbit r is constant, so

 

ds = sqrt[(c+2rw)(c-2rw)]dt.

 

Now, ds = cdT, so,

 

dT= 1/c *sqrt[(c+2rw)(c-2rw)]dt

 

And that is our time dilation formual. Now, to extend this to an observer on Earth, simply use the formula above to figure out the proper time of the astronaut and the proper time of the observer on Earth, and compare them.

 

I apologize for any mistakes, I'm doing this all by hand and its getting late.

-Will

[EWright]

I will ignore gravitational effects and just do an SR calculation. Also, note I've outlined this calculation before, you have ignored it.

 

Note, for variables I use common ones. x,y,z for space, c for speed of light, t for time in seconds of a stationary observer, r for radius, theta shall be written as O and is the polar angle, omega shall be written w as the angular velocity of the orbit. T will be for the proper time of our rotating astronaut.

 

Start with the Minkowski metric for flat spacetime. ds^2= c^2dt^2-dx^2-dy^2-dz^2.

 

Now, coordinate change into a rotating reference frame, assuming a geosynchronus orbit, w orbit= w earth = w. x = r cos(O +wt), y= r sin(O+wt). You get the rotating metric for flat space

ds^2 = (c^2-r^2 w^2)dt^2 - dr^2 -r^2dO^2 - 2r^2 wdtdO - dz^2.

 

Now, we compare the time the astronaut notice as compared to a non rotational observer. (we could also compare to an observer corotating but at a shorter radius. This is just a trivial extension of what I'm about to do).

 

Now, for a ciruclar orbit r is constant, so

 

ds = sqrt[(c+2rw)(c-2rw)]dt.

 

Now, ds = cdT, so,

 

dT= 1/c *sqrt[(c+2rw)(c-2rw)]dt

 

And that is our time dilation formual. Now, to extend this to an observer on Earth, simply use the formula above to figure out the proper time of the astronaut and the proper time of the observer on Earth, and compare them.

 

I apologize for any mistakes, I'm doing this all by hand and its getting late.

-Will

:wave: You're so lucky I don't speak calculus, or I might have to cuss you out with variables... squared even! :shrug: Your answer just shows how wrapped up science is in Einstein's box. Can you really not describe the result's in plain english? I'm sure your point is that I should "learn the math" but perhaps you could learn to express your answers in lay terms. It is a straight forward enough question after all.

[Erasmus00]

:wave: You're so lucky I don't speak calculus, or I might have to cuss you out with variables... squared even! :shrug: Your answer just shows how wrapped up science is in Einstein's box. Can you really not describe the result's in plain english? I'm sure your point is that I should "learn the math" but perhaps you could learn to express your answers in lay terms. It is a straight forward enough question after all.

 

Straightforward answers often elude straightforward questions. Before you accuse me of being trapped in "Einstein's box" I'd point out that phyiscs is written in math. It is not a science of "plain English." For instance, I'm yet to hear a good "plain English" description of gyroscopes, though any physics student can do the math.

 

And your question is not straightforward. Straight forward thought experiments involve inertial frames. Motion that isn't accelerated. The traditional twin paradox is often seen as subtle because of the difficulty of handling one accleration in SR. Yours contains motion that is accelerated not once, but constantly which is a tricky thing for SR to handle.

 

Finally, I answered your question and derived the time dilation for you. What more could you want?

-Will

[EWright]

Straightforward answers often elude straightforward questions. Before you accuse me of being trapped in "Einstein's box" I'd point out that phyiscs is written in math. It is not a science of "plain English." For instance, I'm yet to hear a good "plain English" description of gyroscopes, though any physics student can do the math.

 

And your question is not straightforward. Straight forward thought experiments involve inertial frames. Motion that isn't accelerated. The traditional twin paradox is often seen as subtle because of the difficulty of handling one accleration in SR. Yours contains motion that is accelerated not once, but constantly which is a tricky thing for SR to handle.

 

Finally, I answered your question and derived the time dilation for you. What more could you want?

-Will

For you to answer whether the orbiting twin will see the time-coded light signal as running faster, slower or the same as it is seen on earth, not considering the delay time of about a second in reaching him. And if it varies, how that can be given as light must travel at a fixed speed. In english, not numbers. Can you not extrapolate that from the calculus you figured out?

[Erasmus00]

For you to answer whether the orbiting twin will see the time-coded light signal as running faster, slower or the same as it is seen on earth, not considering the delay time of about a second in reaching him. And if it varies, how that can be given as light must travel at a fixed speed. In english, not numbers. Can you not extrapolate that from the calculus you figured out?

 

Yes, he will see the light at a different frequency then the observer who emits it. The rotational frame red shift is exactly analogous to the gravitational red shift.

 

This shift occurs because time is dilated between an observer on Earth and an observer on the orbiting ship. The exact dilation factor I already worked out for you. Consider a period t between light pulses on Earth. The observer on the ship will see a period T between light pulses, because of time dilation, where t is related to T by the dilation factor. This has nothing to do with the signals "traveling" although you could work out the dilation factor using the equivence principle.

-Will