What is the discrepancy between proper time and observed time in this scenario?

[s0ft]

This is a conceptual problem, right out of a book.
Say Mavis is moving in a spaceship at 0.6c relative to Stanley on earth. When Mavis just passes earth, both of them start their clocks. When Mavis reads 0.4s, what does Stanley read on his?
Now here's the problem. I think 0.4s being proper time, the time interval will seem longer from Stanley's reference frame. So, it should be γτ where τ=0.4s.
But the book would have me believe otherwise.
What's wrong here?

 

[Matterwave]

The problem is ill posed because you are still assuming that the statement "when Mavis reads .4s" can be applied to Stanley's reference frame without caveat. So are you asking what Stanley's clock will read when STANLEY reads Mavis's clock as .4s? If so, do you take into account the time delay in the light signal Stanley must use? Or does Stanley account for this?

 

[s0ft]

I'm sorry if I was unclear. It is : ...Mavis reads 0.4s on her timer ...

 

[DiracPool]

But the book would have me believe otherwise.

What does the book say?

They are both going to read their own proper time on their own clocks.

Stanley will read a shorter time than 0.4s while looking at Mavis's clock because Stanley is in the "laboratory frame" and Mavis is in the primed or proper time frame.

T=t'√(1-v^2/c^2)

 

[stevendaryl]

This is a conceptual problem, right out of a book.
Say Mavis is moving in a spaceship at 0.6c relative to Stanley on earth. When Mavis just passes earth, both of them start their clocks. When Mavis reads 0.4s, what does Stanley read on his?
Now here's the problem. I think 0.4s being proper time, the time interval will seem longer from Stanley's reference frame. So, it should be γτ where τ=0.4s.
But the book would have me believe otherwise.
What's wrong here?

When you ask a question such as "When Mavis's clock reads 0.4s, what time does Stanley's clock read?", you are really asking a question about simultaneity of events. You are asking "What is happening at Stanley's clock at the same time that Mavis's clock reads 0.4s?" The problem with that question is that the notion of two distant events happening at the "same time" (or "simultaneously") is relative (which is why it's call "relativity"). Simultaneity of distant events is relative.

So let's name a few events:

e_1 = the event where Mavis' clock reads 0.4s/γ
e_2 = the event where Mavis' clock reads 0.4s
e_3 = the event where Mavis' clock reads γ 0.4s
e_4 = the event where Stanley's clock reads 0.4s/γ
e_5 = the event where Stanley's clock reads 0.4s
e_6 = the event where Stanley's clock reads γ 0.4s

In Mavis's frame,

• e_1 happens first.
• e_2 happens at the same time as e_4
• e_3 happens at the same time as e_5

In Stanley's frame,

• e_4 happens first.
• e_5 happens at the same time as e_1
• e_6 happens at the same time as e_2

So, according to Mavis, Stanley's clock is running slow, and according to Stanley, Mavis' clock is running slow.

 

[Matterwave]

I'm sorry if I was unclear. It is : ...Mavis reads 0.4s on her timer ...

But this act of Mavis reading .4s on her timer does not describe a universal time, that's the whole point of relativity. This "Mavis reads .4s on her timer" happens at different times according to different people, so are you asking at what time Stanley would say Mavis reads .4s on her timer?

 

[s0ft]

What does the book say?

The book says Stanley will measure a shorter time of 0.32s.
I'll directly quote the book:

It is tempting--but wrong-- to answer that Stanley's timer reads 0.5s. We are considering a different pair of events, the starting and the reading of Stanley's timer, that both occur at the same point in Stanley's Earth frame. These two events occur at different positions in Mavis's frame, so the time interval of 0.4s that she measures between these events is equal to t.(In her frame, Stanley passes her at time zero and is a distance behind her of (1.8 x 10^8 m/s)(0.4 s) = 7.2 x 10^7 m at time 0.4 s.) The time on Stanley's timer is now the proper time:
t_o = t x √(1-(u/c)^2) = 0.4 x √(1-0.6^2) = 0.32 s

 

[DiracPool]

The book says Stanley will measure a shorter time of 0.32s.
I'll directly quote the book:

Well, that sounds about right to me, without doing the actual calculation..

It's shorter than 0.4s.

 

[stevendaryl]

So, let me put some numbers to the events, using γ=0.8

• e_1 = Mavis' clock shows time t'=0.32s
• e_2 = Mavis' clock shows time t'=0.4s
• e_3 = Mavis' clock shows time t'=0.5s
• e_4 = Stanley's clock shows time t=0.32s
• e_5 = Stanley's clock shows time t=0.4s
• e_6 = Stanley's clock shows time t=0.5s

In Mavis' rest frame, e_2 and e_4 are simultaneous.
In Stanley's rest frame, e_2 and e_6 are simultaneous.

 

[DiracPool]

So, let me put some numbers to the events, using γ=0.8

• e_1 = Mavis' clock shows time t'=0.32s
• e_2 = Mavis' clock shows time t'=0.4s
• e_3 = Mavis' clock shows time t'=0.5s
• e_4 = Stanley's clock shows time t=0.32s
• e_5 = Stanley's clock shows time t=0.4s
• e_6 = Stanley's clock shows time t=0.5s

In Mavis' rest frame, e_2 and e_4 are simultaneous.
In Stanley's rest frame, e_2 and e_6 are simultaneous.

I like the way you set up that table, stevendaryl, kind of just spells it all out there.

 

[s0ft]

The table cleared it up. So, what was actually being asked was the time from Mavis's perspective. I hadn't realized that.
Thank you!

 

[ghwellsjr]

This is a conceptual problem, right out of a book.
Say Mavis is moving in a spaceship at 0.6c relative to Stanley on earth. When Mavis just passes earth, both of them start their clocks. When Mavis reads 0.4s, what does Stanley read on his?
Now here's the problem. I think 0.4s being proper time, the time interval will seem longer from Stanley's reference frame. So, it should be γτ where τ=0.4s.
But the book would have me believe otherwise.
What's wrong here?

I am in total agreement with you.

The phrase "When Mavis" doesn't have to imply Mavis's frame as the authors insist. They previously used the phrase in the statement "When Mavis just passes the earth" which is clearly not Mavis's frame or it would have been stated as "When the Earth just passes Mavis". In fact everything in the problem is stated as specified according to Stanley's Earth frame so there is no reason to assume that the question has jumped to Mavis's frame.

So if the authors insist that there is only one correct answer (which there isn't), then it must be according to the one and only frame that they have been talking about all along.