Spaceship paradox

Question

I am having trouble with the first experiment proposed in this PBS SpaceTime video: Can We Break the Universe? (2021) (the description of the experiment starts at the moment I'm linking to).

A spaceship leaves Earth at close to speed of light $c$ towards the sun.

The video states

Special relativity tells us a clock on the spaceship will appear to tick more slowly than a clock on Earth from a point of view of an observer back on Earth.

Then it explains that from the point of view of the spaceship, it can be considered that Earth and the Sun are the ones moving at close to $c$, and thus "that means it sees clocks back on Earth ticking more slowly".

This is the aspect I'm having trouble with. How can both be true at the same time? How can a clock both appear to be ticking faster and slower than the other?

To be precise, say the spaceship left at $t=0$ s. When the spaceship reaches its destination, say its clock shows $t=t_{\text{ship}}$, and the Earth clock shows $t=t_{\text{Earth}}$.

$t_{\text{ship}}$ is either larger, equal or smaller than $t_{\text{Earth}}$. Depending on the answer, one has indeed been running faster or slower than the other, and I'm happy with that (other than I realize that would mean there is an absolute frame of reference, which of course would also be a problem), but I don't see how both can be true.

Answer 1

For the sake of proper comparison, let's assume the final destination is Earth, so the spaceship makes a round trip.

• Let us assume the spaceship is travelling with essentially a constant velocity throughout the entire journey, from beginning to end. At the end of the trip, in the spaceship's reference frame, $t_{\textrm{Earth}}>t_{\textrm{ship}}$. In the Earth's reference frame however, $t_{\textrm{ship}}>t_{\textrm{Earth}}$. There is no paradox because these are two different inertial reference frames. You only need to have consistent answers for $t_{\textrm{Earth}}\gtrless t_{\textrm{ship}}$ in the same inertial reference frame. It may sound ridiculous, but this is somewhat the same idea as asking two people whether object $A$ is spatially "in front" of object $B$. This of course depends on how each person defines their coordinate axes. If one person's coordinate system is rotated by $180^\circ$ relative to the other's, they could have different answers. The answer need only be the same if each person's coordinate systems are aligned appropriately.

• If we make the more realistic assumption that the spaceship begins in the reference frame of the Earth, accelerates to a speed close to that of light, then decelerates back to the velocity of Earth, we will be able to make fair comparisons. At the start of the journey, $t_{\textrm{Earth}}=t_{\textrm{ship}}=0$. At the end of the journey, due to the acceleration and deceleration of the spaceship, $t_{\textrm{Earth}}>t_{\textrm{ship}}$. This is essentially the famous twin paradox.

Answer 2

"Special relativity tells us a clock on the spaceship will appear to tick more slowly than a clock on Earth from a point of view of an observer back on Earth" To paraphrase Einstein, explanations should be made as simple as possible but no simpler. This explanation is simpler than possible in that it tries to condense a proper summary and make it sound simple, but a little thought shows the quoted sentence to be garbled.

The main thing missing is the concept of a frame of reference. This may be imagined as a framework of measuring tapes, and clocks all over the place, synchronised with each other, so we can pinpoint the position in both space and time of an event, in this frame of reference. For example, the Earth's frame of reference may be imagined to extend right out into space.

Instead of comparing the time interval, $T_0$, between two ticks of a clock on a spaceship as registered by that clock and ticks of a clock on earth, the relevant comparison is between $T_0$ and the time, $T$, between these ticks as registered by clocks in the Earth's frame of reference that are present at both events, that is clocks in the Earth's frame that the spaceship passes as its clock ticks.

We find that $T>T_0$, that is the time interval between two events (as measured by clocks local to the events) in a frame in which the events occur in different places is greater than the time between the same two events as measured by a clock in a frame in which the events occur at the same place.

Answer 3

There's a very simple analogy with people's everyday experience that explains the basic idea of how it works.

Consider two people walking across a field at the same speed but in slightly different directions. Because they're walking at a constant speed, their movement in the forwards direction is analogous to time, and distances sideways can be considered analogous to space.

However, each walker is facing in a different direction and so has a different definition of forwards and sideways - or time and space. When the second walker looks at the first, he sees his ruler tilted, and hence 'shorter' according to his own definition. He also sees the first walker gradually falling behind himself. Because the system is symmetric, the first walker sees the second walker in exactly the same way.

This property of both walkers seeing the other gradually falling behind themselves as they walk across the field, progressing more slowly in the 'forwards' direction, is exactly analogous to moving clocks both running slower than the other. The reason this is possible is that they are using different definitions of 'time', pointing in different directions in spacetime.

The geometry of spacetime combines space and time like the geometry of the field combines 'forwards' and 'sideways'. The main difference is that spacetime has a different rule for Pythagoras' Theorem, with a sign change. On the field, the rule is $d^2=f^2+s^2$: the distance along a diagonal squared is equal to the distance travelled forwards squared, plus the distance travelled sideways, squared. In spacetime, it would be $d^2=f^2-s^2$. Time and space contribute with opposite signs. This changes the geometric details somewhat, so be careful about taking the analogy too literally. However, it works quite well for many situations.

The time experienced by a moving clock is the length of its path through spacetime. Travelling from A to B by a straight path or a wiggly path gives different lengths, and hence each traveller can take a different amount of time to get there, even though the routes both start and end at the same times and places. We're already familar with thinking of lengths and distances behaving this way, it's not that odd - we're just not used to thinking of time working in the same way.