[math-fun] More GR fun
After watching Susskind's lectures, I think I've finally understood one problem that has bothered me since undergraduate days almost 50 years ago. When studying special relativity, we learned about 'time dilation', where things moving faster than us have clocks that _appear_ to be going slower. We also learned about getting on a star cruiser after fathering a child & later coming home younger than your child (the so-called "twin paradox"). The question for me was 'what is the cause of the time dilation?' -- is it the speed itself, or is it the acceleration? It appears that the time dilation is related to the acceleration, so -- in effect -- all time dilation is caused by acceleration, whether it be 'real' acceleration due to rocket thrust, or to gravitational acceleration due to some mass in the neighborhood. I.e., the father/child paradox requires the acceleration to get up to speed, as well as the deceleration necessary to get back to meet with the child.
On 10/21/2013 4:48 PM, Henry Baker wrote:
After watching Susskind's lectures, I think I've finally understood one problem that has bothered me since undergraduate days almost 50 years ago.
When studying special relativity, we learned about 'time dilation', where things moving faster than us have clocks that _appear_ to be going slower.
We also learned about getting on a star cruiser after fathering a child & later coming home younger than your child (the so-called "twin paradox").
The question for me was 'what is the cause of the time dilation?' -- is it the speed itself, or is it the acceleration?
It appears that the time dilation is related to the acceleration,
Only in the sense that it's due to the path through spacetime not being straight. Suppose you pass by Earth at high speed in your spaceship and set your clock to an Earth clock as you pass. Then, continuing at constant velocity for a few years, you cross paths with another spaceship which is heading back to Earth at high speed. As you pass by, he sets his clock to yours. These three clocks are identical atomic clocks. None of them have accelerated. But when the inbound spaceship gets to Earth his clock will be behind the Earth bound clock. Why? Because in special relativity a straight line is the *longest* duration between two events. The two spaceship clocks measured the duration of a broken line path between the two events of passing Earth (one outbound one inbound), while the Earth clock measured the straight path between the events. It's no different Euclidean geometry in which a un-straight line between two points is longer than a straight one; except for the minus sign in the metric. Brent
so -- in effect -- all time dilation is caused by acceleration, whether it be 'real' acceleration due to rocket thrust, or to gravitational acceleration due to some mass in the neighborhood.
I.e., the father/child paradox requires the acceleration to get up to speed, as well as the deceleration necessary to get back to meet with the child.
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Another interesting special rel. "paradox" is a spaceship of rest-length L, travelling at a significant fraction of c, approaches a berth, or dock, of rest-length L. When the nose of the spaceship touches the far end of the berth, the spaceship stops. Just before that instant, a person standing on the ship sees the berth as shorter than the ship (the ship is sticking out); a person on the dock sees the ship as shorter than the dock (the ship is well inside the dock). What is going on? -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Henry Baker Sent: Monday, October 21, 2013 5:48 PM To: math-fun@mailman.xmission.com Subject: [EXTERNAL] [math-fun] More GR fun After watching Susskind's lectures, I think I've finally understood one problem that has bothered me since undergraduate days almost 50 years ago. When studying special relativity, we learned about 'time dilation', where things moving faster than us have clocks that _appear_ to be going slower. We also learned about getting on a star cruiser after fathering a child & later coming home younger than your child (the so-called "twin paradox"). The question for me was 'what is the cause of the time dilation?' -- is it the speed itself, or is it the acceleration? It appears that the time dilation is related to the acceleration, so -- in effect -- all time dilation is caused by acceleration, whether it be 'real' acceleration due to rocket thrust, or to gravitational acceleration due to some mass in the neighborhood. I.e., the father/child paradox requires the acceleration to get up to speed, as well as the deceleration necessary to get back to meet with the child. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On 10/28/2013 6:23 AM, Cordwell, William R wrote:
Another interesting special rel. "paradox" is a spaceship of rest-length L, travelling at a significant fraction of c, approaches a berth, or dock, of rest-length L. When the nose of the spaceship touches the far end of the berth, the spaceship stops. Just before that instant, a person standing on the ship sees the berth as shorter than the ship (the ship is sticking out); a person on the dock sees the ship as shorter than the dock (the ship is well inside the dock). What is going on?
Aside from infinite deceleration that requires sonic velocity in the ship structure exceeding c? Aside from that there's the fact that where the back end of the ship is at the same instant the nose touches the berth and where the back of the dock is at that instant are both frame dependent. Brent
Yes, ignore rigidity problems. The interesting part is "What does the person on the dock see?" and "What does the person on the ship see?" -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of meekerdb Sent: Monday, October 28, 2013 12:00 PM To: math-fun Subject: Re: [math-fun] [EXTERNAL] More GR fun On 10/28/2013 6:23 AM, Cordwell, William R wrote:
Another interesting special rel. "paradox" is a spaceship of rest-length L, travelling at a significant fraction of c, approaches a berth, or dock, of rest-length L. When the nose of the spaceship touches the far end of the berth, the spaceship stops. Just before that instant, a person standing on the ship sees the berth as shorter than the ship (the ship is sticking out); a person on the dock sees the ship as shorter than the dock (the ship is well inside the dock). What is going on?
Aside from infinite deceleration that requires sonic velocity in the ship structure exceeding c? Aside from that there's the fact that where the back end of the ship is at the same instant the nose touches the berth and where the back of the dock is at that instant are both frame dependent. Brent _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
http://math.ucr.edu/home/baez/physics/Relativity/SR/barn_pole.html On Mon, Oct 28, 2013 at 12:06 PM, Cordwell, William R <wrcordw@sandia.gov> wrote:
Yes, ignore rigidity problems. The interesting part is "What does the person on the dock see?" and "What does the person on the ship see?"
-----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of meekerdb Sent: Monday, October 28, 2013 12:00 PM To: math-fun Subject: Re: [math-fun] [EXTERNAL] More GR fun
On 10/28/2013 6:23 AM, Cordwell, William R wrote:
Another interesting special rel. "paradox" is a spaceship of rest-length L, travelling at a significant fraction of c, approaches a berth, or dock, of rest-length L. When the nose of the spaceship touches the far end of the berth, the spaceship stops. Just before that instant, a person standing on the ship sees the berth as shorter than the ship (the ship is sticking out); a person on the dock sees the ship as shorter than the dock (the ship is well inside the dock). What is going on?
Aside from infinite deceleration that requires sonic velocity in the ship structure exceeding c? Aside from that there's the fact that where the back end of the ship is at the same instant the nose touches the berth and where the back of the dock is at that instant are both frame dependent.
Brent _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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participants (4)
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Cordwell, William R -
Henry Baker -
meekerdb -
Mike Stay