Elapsed time between two events is greater for moving observer

Question

Lorentz's equations show the elapsed time between two events in moving frame is greater than rest frame. But moving observer's clock ticks slower. Isn't that an inconsistency?

Consider two observers, Earthman and Spaceman in two standard inertial frames S and S' respectively. Earthman creates two events, both at the origin of his coordinate system; the first event occurs at $t=0$; the second at $t=T$. Therefore Earthman perceives the elapsed time between the events is $T$.

Earthman perceives that due to Spaceman's relative velocity his light-clock ticks less often than Earthman's identical clock. Therefore, Earthman perceives that Spaceman's measure of elapsed time between the events, which corresponds to ticks on his clock between the events, will be less than T as shown below.

$$t'=T/\gamma$$

However, Earthman applies the following Lorentz Transformation to conclude that Spaceman's measure of the elapsed time between the two events is greater than $T$ as shown below.

$$t'= \gamma(t-xv/c^2) = \gamma(T-0v/c^2) = \gamma T$$

Answer 1

Earthman creates two events, both at the origin of his coordinate system; the first event occurs at $t=0$; the second at $t=T$. Therefore Earthman perceives the elapsed time between the events is T

Earthman observes that the elapsed time between the events is $T$. It's crucial to note that Earthman requires a single clock at relative rest to measure this elapsed time.

Earthman perceives that due to Spaceman's relative velocity his light-clock ticks less often than Earthman's identical clock. Therefore, Earthman perceives that Spaceman's measure of elapsed time between the events, which corresponds to ticks on his clock between the events, will be less than T

Here is where your reasoning fails. It's true that Earthman observes Spaceman's clocks to tick less often. But it's also true that the clock Spaceman uses to record the time of the first event is not the clock that is used to record the time of the second event (remember, the 2nd event is at a different location then the first event - according to Spaceman - so, two clocks are needed by Spaceman, one at rest at the location of the first event, and the other at rest at the location of the second event).

So Spaceman must subtract the time reading on one clock from the time reading on another clock in order to determine the elapsed time between the events.

According to Spaceman, the two different clocks are synchronized but Earthman observes that the Spaceman clock used to record the time of the second event is ahead of the Spaceman clock used to record the time of the first event!

According to Earthman, Spaceman calculates a longer elapsed time (due to the unsynchronized clocks) even though Earthman observes the two clocks run slower than his clock.

So it just isn't true that "Earthman perceives that Spaceman's measure of elapsed time between the events, which corresponds to ticks on his clock between the events, will be less than T"

Instead, Earthman observes that Spaceman has calculated the elapsed time with unsynchronized clocks that are running slower than his clock.

Answer 2

So, you have two events, $A:(x=0, t=0)$, and $B:(x=0, t=T)$.

However, Earthman applies the following Lorentz Transformation to conclude that Spaceman's measure of the elapsed time between the two events is greater than T as shown below.

Correct. Spaceman will measure the time difference between the events, $t'$, as being greater than $T$.

Earthman perceives that due to Spaceman's relative velocity his light-clock ticks less often than Earthman's identical clock. Therefore, Earthman perceives that Spaceman's measure of elapsed time between the events, which corresponds to ticks on his clock between the events, will be less than T as shown below.

Incorrect. What "Earthman perceives that due to Spaceman's relative velocity his light-clock ticks less often than Earthman's identical clock" means is that when Earthman looks at two events that have the same value of $x'$, $\Delta t'$ for those two events will be lower than $\Delta t$. Remember, Earthman and Spaceman disagree not just about time, but also space. When Spaceman looks at his clock, he sees it as moving through just time. Earthman sees it as moving through space as well, and thus sees it as moving through more time.

When Earthman and Spaceman look at two events that Spaceman says happen at the same location (such as a clock that is, in Spaceman's reference frame, stationary), Earthman measures there being more time between those events than Spaceman does. When they look at an event that Earthman says happen at the same location, Earthman measures there being less time than Spaceman does.

Answer 3

Lorentz's equations show the elapsed time between two events in moving frame is greater than rest frame. But moving observer's clock ticks slower. Isn't that an inconsistency?

I am assuming that the moving frame actually refers to the frame relative to the observer at rest, and the rest frame refers to the frame relative to the moving observer. In addition, I also believe that a better phrase than ‘elapsed time’ would be ‘perceived time’, since ‘elapsed time’ brings with it an idea that absolute time exists, when in fact it does not exist in special relativity.

I do not believe there is an inconsistency. While what you’ve said is true, that the moving observer’s clock ticks slower when observed by the observer at rest(in the moving frame), this is only true for the moving frame. Once you change frames, the perceived time would change. To the moving observer(who is in the rest frame), his clock would appear to tick normally while the stationary observer’s clock would appear to tick slower.

Therefore, Earthman perceives that Spaceman's measure of elapsed time between the events, which corresponds to ticks on his clock between the events, will be less than T as shown below.

𝑡′=𝑇/𝛾

I believe your equation may be wrong since the Lorentz factor 𝛾 = 1/(1+v^2/c^2). For t′ to be less than T, you should be multiplying by 𝛾 and not dividing by 𝛾. If so then there is no inconsistency with the Lorentz transformation.

Answer 4

I believe it is important to denote correctly, who is "at rest" and who is "moving". Quite often books and articles suppose, that if the events are "co-located", this observer is necessarily is "at rest".

There is a lot of confusion, who moves and who at rest, because according to this theory someone can be either "at rest" or "moving". An observer is at rest, but he is moving, for someone he is moving, but someone is moving according to him. This makes dizzy.

Let's someone traveled from point A to point B. There were two events: departure from A and arrival to B. These events were not co-located in the Earth frame, but co-located in the travelling observer's frame, because he had remained in the same car of the train.

An observer on the Earth (who is "at rest") measures interval of travel time with two spatially separated and synchronized clocks, an observer in the train measures travel time interval with a single clock. Hence, if the events for someone were co-located, this observer was the "moving one". His clock measured shorter interval of time, than two spatially separates clocks.

Wikipedia article about Motion claims: In physics, motion is the change in position of an object with respect to its surroundings in a given interval of time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, and speed. Motion of a body is observed by attaching a frame of reference to an observer and measuring the change in position of the body relative to that frame.

To move relatively to a single clock is not possible, because it "has no surroundings". Two spatially separated clocks create that very surroundings, and single clock moves relatively to them.

Single clock measures shorter time interval, than two spatially separated in Einstein – synchronized clocks. Two spatially separated clocks record greater time interval.

That means, that since single moving clock runs slower, time in the whole reference frame, in which this clock is moving, runs faster.

Time in the reference frame of “stationary” observer is the same in any point of his frame, it does not depend on coordinate, that goes out straight from Lorentz transformation for time.

Time Dilation Special Relativity