Malorie said:
Sorry JesseM,
I was going to deconstruct your whole explanation but it was getting to be to much chopping up and not enough input from me.
I'll just agree with the whole thing and add a question.
You used the words "proper time" throughout your post. Could you clarify this for me? Maybe it's just a terminology issue for me.
The "proper time" along a given worldline is just the time that would be measured by a clock moving along that worldline. It's a coordinate-invariant quantity which is directly analogous to the length along a given path in 2D space.
Malorie said:
One issue I'm having here is that I tend to be more of a visual person and verbal/literal discussions take a bit to sink in. A picture is definitely worth a thousand words for me.
Are you familiar with spacetime diagrams, like these one for the twin paradox?
http://www.csupomona.edu/~ajm/materials/twinparadox/twins.jpg
The diagram is taken from http://www.csupomona.edu/~ajm/materials/twinparadox.html ...on the left side we see a diagram drawn from the perspective of a frame where the Earth is at rest, on the right is one where the traveler was at rest during the outbound phase of the journey (up until t=4) but not afterwards. In both cases, the frame's time coordinate is the vertical axis and the frame's space coordinate is the horizontal axis. You can see that regardless of which frame you use, the Earth's worldline is a straight line going from the event of the traveler departing at the bottom to the event of the traveler returning at the top, while the traveler's worldline is has a bend in it (at t=5 in the left diagram, and t=4 in the right diagram). These sorts of diagrams make the geometry of the situation more apparent. Note that in the left diagram, if you just focus on the traveler's outbound leg, the \Delta x is 3 and \Delta t is 5, whereas if you look at the traveler's outbound leg in the right diagram, \Delta x is 0 and \Delta t is 4, which means in both cases you get the same value for the proper time along the outbound leg with the formula \sqrt{\Delta t^2 - \Delta x^2 } (which as I mentioned is analogous to the Pythagorean theorem that gives distance along a straight line in a Cartesian coordinate system).
Malorie said:
Let me try it this way:
Both twins start out in the same inertial frame.
The traveling twin changes their inertial frame by accelerating up to travel speed. Are we still together here?
After the traveling twin finishes accelerating and is again at rest in a now moving frame of refrence, both twins are now opposed in what they see (both see the other as moving through time slowly from their own perspecive). This state is no different than if the twins were moving away from each other instead of just one of them moving. Are we still together here?
Now the traveling twin decelerates and accelerates back toward the other twin. Changing their frame of reference from departing to aproaching.
You have to take into account the fact the
relativity of simultaneity here--different inertial frames can disagree about whether two events at different locations in space are simultaneous or not. Suppose as in the example above in the graphic above, the traveling twin's speed relative to the Earth is 0.6c, and the traveling twin turns around after 4 years of proper time have passed on his own clock. In the frame of the Earth, the event of the Earth clock showing that 5 years have passed is simultaneous with the event of the traveling twin's clock reading 4 years, which is when the traveling twin turns around--so in the Earth's frame the traveling twin's clock was slowed down by a factor of 0.8. In the frame where the traveling twin was at rest during the outbound leg of the trip, it was the Earth's clock that was slowed down by a factor of 0.8, so the event of the traveling twin's clock reading 4 years was simultaneous with the event of the Earth's clock reading 4*0.8 = 3.2 years. But if we then look at a third frame where the traveling twin was at rest during the
inbound leg of the trip, in this frame the event of the traveling twin's clock reading 4 years was simultaneous with the event of the Earth's clock reading 6.8 years. So you can't just jump from the outbound rest frame to the inbound rest frame without taking into account that they have different definitions of simultaneity and thus different opinions about the time on the Earth clock at the moment the traveling twin turned around.
All of this has parallels once again to the 2D analogy--if you have two paths between points in space, one straight and the other a bent path consisting of two segments at different angles (so together the two paths form a triangle), then you could pick one Cartesian coordinate system with its x-axis parallel to the bottom segment of the bent path, and see what point on the straight path has the same x-coordinate as the bend in the bent path; then you could pick a different Cartesian coordinate system with its x-axis parallel to the top segment of the bent path, and see what point on the straight path has the same x-coordinate as the bend in the bent path. This would yield two totally
different points on the straight path, as you can should be able to see if you draw a diagram. Does that lead you to conclude that all the extra distance along the bent path accumulated at the moment of the bend, so that if you drove a car along the bent path the odometer would suddenly jump when you reach the bend? Of course not--the greater length of the bent path has to do with its overall shape and the fact that neither of the two straight segments is parallel to the straight path, the bend itself can be made arbitrarily short so it contributes almost nothing to the overall length of the bent path.
Malorie said:
Once the traveling twin reaches their approach speed, again the twins are opposed in their view of the other (again they both see the other moving slowly though time from their own perspective). Again we reached a relative speed state that would be equivalent to both twins moving towards each other instead of just one of them moving. How am I doing, have I lost it yet?
Now the traveling twin decelerates on arrival back at the non-travelling twin.
Now where did the difference in time dilation happen? It wasn't while they were both at rest in their own reference frame. It was during the acceleration deceleration phases of the journey.
So where am I loosing it?
The problem here is that you are not sticking to a single inertial frame throughout the journey, you are thinking of one inertial frame for the outbound leg and another for the inbound leg. If you stick to a single inertial frame, you will find that there is no sudden change in either twin's age at the moment of acceleration, you can find how much each twin has aged in total just by looking at how much coordinate time went by on each leg of the journey, and figuring out how much each twin's clock would be slowed down relative to coordinate time based on their velocity during that leg. For example, in the right hand diagram above, drawn from the perspective of the inertial frame where the traveling twin was at rest during the outbound leg, the outbound leg lasted 4 years of coordinate time while the inbound leg lasted 8.5 years of coordinate time. In this frame the traveling twin was at rest during the outbound leg so his clock was keeping pace with coordinate time, meaning he aged 4 years of proper time during the outbound leg; then the traveling twin was traveling at 15/17c during the inbound leg, so during this leg his clock was slowed down by a factor of \sqrt{1 - (15/17)^2} = 0.470588, meaning during the 8.5 years the inbound leg lasted the traveling twin aged 0.470588*8.5 = 4 years of proper time. This means that if we add the traveling twin's proper time during the outbound and inbound legs we find he has aged a total of 4 + 4 = 8 years between leaving Earth and returning. Meanwhile in this frame the Earth had a constant velocity of 3/5c during both the outbound and inbound leg, so its clock was slowed down by a factor of \sqrt{1 - (3/5)^2} = 0.8, so during the 4 + 8.5 = 12.5 years of both legs combined the Earth's clock only ticked forward by 0.8*12.5 = 10 years of proper time. So here we have calculated how much each twin aged throughout the journey using only their velocities in each phase of the trip, as I said there were no jumps during the acceleration. You could do exactly the same thing from the perspective of a different inertial frame and you'd still conclude the traveling twin aged 8 years while the Earth twin aged 10.