Trouble understanding frames of reference

Moose

I am having trouble grasping why time for a spaceship leaving earth at near the speed of light passes more slowly than time on earth, rather than the other way around. I understand special relativity well enough to know that there is no absolute frame of reference and that, from the spaceship’s frame of reference, it is earth that is moving at near the speed of light.

The podcast I am listening to explains that, because the spaceship turns around and head back to earth, it is not in a state of uniform motion. But why does that matter? Is it not the case that less time passes for the spaceship even before it turns around, during which time it has experienced uniform motion?

topsquark

Forum Staff
I am having trouble grasping why time for a spaceship leaving earth at near the speed of light passes more slowly than time on earth, rather than the other way around. I understand special relativity well enough to know that there is no absolute frame of reference and that, from the spaceship’s frame of reference, it is earth that is moving at near the speed of light.

The podcast I am listening to explains that, because the spaceship turns around and head back to earth, it is not in a state of uniform motion. But why does that matter? Is it not the case that less time passes for the spaceship even before it turns around, during which time it has experienced uniform motion?
When the spaceship is turning around it is not in an inertial state compared to the Earth. By the Principle of Equivalence (General Relativity) we can know that the spaceship is the one that is moving and hence the relative frame concept breaks down. (Basically everyone can agree that it is the spaceship that is accelerating because the Earth isn't in this example. So we can't say that, relative to the spaceship that the Earth is accelerating.)

That is the cruicial point. I am not particularly good at explaining the details of the SR version of how to break the problem down, so if you still have questions I'll have to pass the ball.

-Dan

Woody

There is a way of defining a 4 dimensional space-time quantity that is an analogue to our familiar 3D distance.

S²=(t/c)²+X²+Y²+Z² (from imperfect memory, please check)

If you work out this 4D space-time distance between 2 events, it will be invariant regardless of reference frame.

For an accelerating body and an inertial body to start off and arrive at the same pair of events,
they must have travelled the same overall "distance" in 4D space-time.
Since the accelerating body must have travelled further in 3D space, it must have travelled less far in time.

2 people

HallsofIvy

I believe that should be S²= (t/c)²- X²- Y²- Z².

3 people

MikeFontenot

During both periods during his trip in which the rocket guy is NOT accelerating (i.e., doesn't have his rocket turned on), he will conclude that the home twin (she) is ageing more slowly than he is. And she (for the entire trip) concludes that he is ageing more slowly than she is. They disagree, but that doesn't result in any inconsistencies. But when the rocket guy uses his rocket to reverse course at the turnaround point, he IS accelerating, and he concludes that during that brief time (for him), the home twin gets MUCH older. In contrast, the home twin doesn't think that either of them ages during the sudden turnaround. When they are reunited, they will each agree about their respective ages then (as they of course must), but they will not agree about their respective ages at all other times during the trip.

2 people

Pmb

PHF Hall of Fame
See - Time Dilation

For the basics. The basic reason why that happens is that the symmetry of the scenario of broken why the spaceship slows down as it approaches the distant planet and the same thing happens on the way back.

1 person