Consider the question:

Alice is in a spaceship moving at a velocity of 3/5c with respect to Bob on earth. When she passes earth both of their clocks read t = 0. a) When observers in Bob’s frame find Alice has moved 9×10^7m past the earth, what does Alice’s clock read. b) When Alice’s clock reads 0.4s, what does Bob’s clock read?

Upon trying to answer this question I have stumbled across something which is confusing me:

For part a) we can calculate the time as Alice's clock as 0.4 seconds.

However now we do part b) and by applying the formula for time dilation and taking 0.4 seconds in Alices frame as proper time Bobs clock now reads 0.32 seconds.

How is this possible as in part a) when Alices clock reads 0.4 seconds Bobs will read 0.5 (as 9×10^7 / 0.6c = 0.5) but then in part b) we see that Alices clock will read 0.4 yet Bobs will now read 0.32?

Any help would be appreciated.

  • $\begingroup$ Each observer sees the other's clock running slower. "When" (or "now") means a different thing for Bob than for Alice. $\endgroup$ – user126422 May 24 '17 at 1:19
  • $\begingroup$ insti.physics.sunysb.edu/~siegel/sr.html $\endgroup$ – Count Iblis May 24 '17 at 1:31

A spacetime diagram might help elaborate on the comment and answer you have already received.

I am going to use a spacetime diagram on rotated graph paper so that we can visualize the time and space intervals.

spacetime diagram on rotated graph paper

Let each light-clock diamond represent "0.1 sec".

Using Minkowski-right triangle $OPQ$, Alice has velocity $v_{Alice}=PQ/OP=(6/10)c=(3/5)c$ with respect to B. The time-dilation factor $\gamma=\frac{1}{\sqrt{1-v^2}}=5/4$. [In terms of rapidity $\theta$ (the Minkowski analogue of angle, where $v_{Alice}=\tanh\theta$, we have $\gamma=\cosh\theta=\cosh({\rm arctanh}(v))$.]

For part a)...
Alice arrives at event E.
According to Bob, event B on his worldline is simultaneous with E.
(BE is parallel to the spacelike diagonal of Bob's light-clock diamonds.)
So, when Bob says "Alice is $9\times10^{7}\rm\ m$ away", we have $BE=9\times10^{7}\rm\ m=9\times10^{7}\rm\ m\left(\frac{c}{3\times10^{8}\rm{\ m/s}}\right)=0.3\rm{\ light-sec}$, and thus $OB=BE/v_{Alice}=(0.3)/(3/5)=0.5$.
Then, since time-dilation in triangle OBE with hypotenuse OE implies $\gamma=\cosh\theta=\frac{ADJ}{HYP}=\frac{OB}{OE}$, we have $OE=OB/\gamma=(0.5)/(5/4)=0.4$, as you said.

For part (b)...
I will rephrase "When Alice’s clock reads 0.4s, what does Bob’s clock read?"
as "When Alice’s clock reads 0.4s, what does Alice say Bob’s clock reads [at the event E' on Bob's worldline that Alice says is simultaneous with event E on her worldline]?"

According to Alice, event E on her worldline is simultaneous with E' on Bob's worldline.
(EE' is parallel to the spacelike diagonal of Alice's light-clock diamonds.)

Since time-dilation in triangle OEE' with hypotenuse OE' implies $\gamma=\cosh\theta=\frac{ADJ}{HYP}=\frac{OE}{OE'}$, we have $OE'=OE/\gamma=(0.4)/(5/4)=0.32$, as you said.
[Note that $|EE'|\neq |BE|$.]

So, as others have implied, Alice and Bob disagree about which event on Bob's worldline is simultaneous with event E on Alice's worldline.

The spacetime diagram [on rotated graph paper] and its geometric/trigonometric interpretation hopefully makes this clearer [as compared to merely using a formula... without recognizing these interpretations].


Because in special relativity there is no concept of absolute simultaneity, so even if in Bob frame the event of his clock reading 0.5 is simultaneous with the event of Alice clock reading 0.4 this is not true for Alice. Notice that posing your doubt you are implicitly supposing absolute simultaneity since you are thinking that it is impossible that two different observers do not concur wether two events are happening in the same time or not. I would say more, you should have expect this result because otherwise the relativity principle would have been broken: the system is completely symmetric so if Alice time is dilated in Bob reference then Bob time must be dilated in Alice reference.


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