What does Lorentz Factor really imply, if time Dilation is symmetric?

Question

Let us imagine a spaceship traveling to a planet around a star 4.0 light years away exactly at 99 percent the speed of light (Lorentz factor of around ~7) relative to earth. For simplicity let us assume the distance between earth and this exoplanet is constant at exactly 4.0 ly (earth exoplanet forms a inertial reference frame) Further we assume that there is no change in gravitational potential such as moving away from the gravity well of our solar system into the exoplanet star system,importantly no acceleration and deceleration is involved

According to my calculations from the perspective of the space ship it would be traveling 4 ly/7= 0.57 light year, and the time elapsed when it passes by the exoplanet from its perspective would be 0.57 ly/0.99 c= 0.5757..years or about 6.91 months. From the perspective of earth when the space craft lands on the exoplanets (4 ly/0.99c) or 4.0404..years would have elapsed..and spacecraft traveled exactly 4 ly. Are the calculations based on the stated assumptions correct so far?

What I think a Lorentz factor of 7 means, for every second on spacecraft clock, 7 seconds would elapse on the spacecraft, and vice versa (that is one second by earth's clock 7 second would transpire on the spaceship) Is this interpretation of Gamma of 7 correct?

Let us consider two events A clock on spaceship reads that it leaves earth and B-clock on spaceship records the instant the rocket passes by exoplanet at 0.99c relative to it and earth. The ship records A-B interval as 0.5757 years (this is the proper time). According to set of clock rest with earth this interval would be Delta T 7*0.5757=4.04 years. Let us reverse scenario: Event C (clock on earth records spacecraft leaving, this is coordinated wtih A), and Event D (clock on earth records time it receives confirmation spacecraft has passed exoplanet..this is 8.04 years later, 4.04 years for the spacecraft planet+4 years to get the signal back of succesful passby).... Would hte interval C-D be measured as 56.28 years by the ships clock? Its the symmetry/reciprocity of time dilation in special relativity I am having trouble understanding.

Further let us consider an event E also measured in Earth proper time where 2 years after the departure of the spaceship a hurricane destroys the launch pad. The proper time by Earth clock for this is 2 years, but would the time elapsed between C-E according to a set of clocks moving at the same velocity (i.e at rest relative to) the spaceship (which is still traveling at 0.99c into the galaxy relative to earth after passing exoplanet) be measured at 7*2=14 years? That is according to the ship Event E transpired AFTER the ship passed the exoplanet, while according to earth it was BEFORE the ship passed exoplanet (i.e the order of event is reversed according to the frame of reference).

TLDR, Clarify what would the event C-D and C-E be measured by a set of clocks moving with the same velocity as spaceship, and would the order of events C-E be reversed in Earth, and spaceship's reference frame? Are the questions clear?

Answer 1

"what Lorentz factor of 7 means, for every second on earth clock, 7 seconds would elapse on the spacecraft, and vice versa"

This has to be stated more carefully in order to be clear. At the moment it could be correct or incorrect depending on how it is interpreted.

The way to gain clarity is, I suggest, to introduce the concept of event. You specify two events, and then report the time elapsed between those events according to one set of clocks or another.

For example: let event A be "clock on spaceship reads noon" and event B be "clock on spaceship reads 1:00pm". The time elapsed between these events, according to a set of clocks at rest relative to Earth, is 7 hours.

Now how about the other way around. If we take those same two events, the answers will remain what we just said: 1 hour for spaceship, 7 hours for Earth. To look at the other scenario we need to pick a different pair of events. For example, consider event C which is "clock on Earth reads noon" and event D which is "clock on Earth reads 1:00pm". The time interval between C and D, according to a set of clocks moving with the same velocity as the spaceship, is 7 hours.

In the above scenario one can arrange that A and C are the same event, but then B and D will be different events. Or one could arrange that B and D are the same events, but then A and C will be different. And so on.

The time dilation factor is equal to the Lorentz factor $\gamma$ when we compare time in a reference frame to proper time. Proper time is the time interval between two events for the inertial frame where those events are at the same location. Or, another way to say it: if there is a clock with inertial motion (i.e. constant velocity or at rest) present at both events then that clock gives the proper time $\tau$ between the events. Some other clock, moving at velocity $v$ relative to the one present at both events, will register $\Delta t = \gamma \tau$ as the time interval.

In the scenario described in the question, the spaceship clock gives the proper time between events "leave Earth" and "arrive at exoplanet". A clock fixed to Earth could be used to find the proper time between events such as "spaceship leaves launch pad" and "ground control crew have a party on the launch pad after receiving news that all is well after one year."

Answer 2

Your mistake is to assume that for every second that passes on the ship seven seconds would pass on Earth. Time dilation doesn't work that way. What time dilation means is that time in the frame of the ship is out of synch with time in the frame of the Earth.

If the Earth takes a second (in Earth time) to move between two clocks in the frame of the ship, then the difference in the readings of the two clocks the Earth passes will be seven seconds. So the Earth clock seems to be running slow by comparison to the readings on two separate clocks in the ship frame.

Conversely, if a clock in the ship frame takes a second (according to itself) to move between two clocks in the Earth frame, then the time difference between the two Earth clocks will be seven seconds, so the ship clock will seem to be running slow when compared to two clocks in different places in the earth frame.

So the key point here is that you never directly compare one clock with another- instead you are comparing the elapsed time on one clock with the difference between the times shown on two separate clocks in two different places in the other frame. In reality all the clocks tick at one second per second.