I asked this question earlier today:

Does Special Relativity Imply Multiple Realities?

And I'm still confused about the answers. However, I now have another scenario (which is definitely yet another scenario formulated by me incorrectly interpreting special relativity once again), but I think if this confusion is cleared up, I'll be one step closer to understanding everything!

Say we have Alice and Bob. Alice is in her space-suit in space, and Bob is whizzing by on a rocket ship past her at a velocity of $v$. Right now, they're right next to one another.

Additionally, there's a point $D$ a distance of $L$ away from Alice, that's stationary relative to her. From her point of view, Bob is traveling directly towards it.

Say that Alice measures that it takes a time of $t$ for Bob to reach the point $D$. That is, she measures that it takes a time of $t$ for Bob to travel the distance $L$.

She would admit that on Bob's clock, only $t\sqrt{1-\frac{v^2}{c^2}}$ would've passed.

Additionally, she concludes that since less time passed on Bob's clock, he must've seen the space between her and point $D$ contract. She concludes that he must've measured that he traveled a distance of $L\sqrt{1-\frac{v^2}{c^2}}$.

Now, we go to Bob. From his point of view, $D$ is moving towards him at a velocity of $V$.

(From this point on is where my interpretation changes from the linked question, and I'm not sure if the things I say from here on are correct...)

Lets say he actually DOES measure the length he travels to be $L\sqrt{1-\frac{v^2}{c^2}}$, where $L$ is the length Alice measured. Additionally, he actually DOES measure the time he travelled for to be $t\sqrt{1-\frac{v^2}{c^2}}$, where $t$ is the time Alice measured. (And again, I started this part with "lets say" because I'm not sure if he actually does).

So far, they agree on everything.

However, from Bob's point of view, Alice was travelling at a velocity of $v$ away from him. Which means that for whatever time he measured that passed $t'$, he must've measured that $t'\sqrt{1-\frac{v^2}{c^2}}$ passed for Alice...

Which means that he would've measured $t\sqrt{1-\frac{v^2}{c^2}}\sqrt{1-\frac{v^2}{c^2}}=t(1-\frac{v^2}{c^2})$ passed for Alice...

....I know this is wrong, but I'm not sure exactly where...

As I said in my other question, this is my first day learning about relativity, so the more thorough the explanation the better.


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    $\begingroup$ Good questions! You are falling into the trap that, actually, the majority of people coming to us with special relativity questions have. You've learned about length contraction and time dilation (which are very easy to glean off the internet), but you haven't learned about loss of simultaneity (which is more subtle, but just as important). Without additionally accounting for loss of simultaneity, you will run into all kinds of paradoxes and contradictions. $\endgroup$
    – knzhou
    Commented Dec 20, 2019 at 2:13
  • 1
    $\begingroup$ Besides linking to the hundreds of previous questions along these lines, all I can say is, find any good book (absolutely not anything popular and internet-based, these are all oversimplified, and they generally make less sense the smarter you are). I promise they will explain literally everything about all the questions you have asked so far; you are going down a very well-trodden road. $\endgroup$
    – knzhou
    Commented Dec 20, 2019 at 2:14
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    $\begingroup$ Examples of good books include Spacetime Physics, Special Relativity: For the Enthusiastic Beginner, and _ An Illustrated Guide to Relativity_. Actually, just about any book is good, as long as it's a real book. $\endgroup$
    – knzhou
    Commented Dec 20, 2019 at 2:16
  • $\begingroup$ Try reading story in the link below. It describes a fictitious world where the speed of light is only 10 km/hr. The story was written by George Gamow - a prominent 20th century theoretical physicist. asc.ohio-state.edu/durkin.2/phys1201/MrTompkinshort.pdf $\endgroup$
    – hermit
    Commented Dec 20, 2019 at 2:25
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    $\begingroup$ Yup, that's the right way to see it! That's the first big "jump" people always have to make when learning relativity. Now you, too, can answer a thousand special relativity questions on this site. $\endgroup$
    – knzhou
    Commented Dec 30, 2019 at 2:30

2 Answers 2


Consider the following events

1 Bob passes Alice, Bob starts his clock and Alice starts hers. (These events are absolutely simultaneous as they occur at the same time and place.)

2 Bob passes point D and stops his clock. (Again these events are absolutely simultaneous as they occur at the same time and place.)

3 Alice stops her clock.

In Alice's frame, event 3 is simultaneous with event 2 - she stops her clock at the same instant that Bob stops his. But crucially these events are not co-located, which means that in Bob's frame they are not simultaneous. According to Bob, Alice doesn't stop her clock until some time after he reaches point D.


In Alice's frame

  • 1 occurs at time $0$ (Bob's clock also reads $0$)
  • 2 & 3 occurs at time $t$ (Bob's clock reads $t/\gamma$)

In Bob's frame

  • 1 occurs at time $0$ (Alice's clock also reads $0$)
  • 2 occurs at time $t/\gamma$ (Alice's clock reads $t/\gamma^2$)
  • 3 occurs at time $t\gamma$ (Alice's clock reads $t$)

$\gamma$ is $1/\sqrt{1-\frac{v^2}{c^2}}$

  • $\begingroup$ Got you Simon. Thanks! Can you read my comment under the original question and verify that I understood it correctly? $\endgroup$ Commented Dec 30, 2019 at 1:44
  • $\begingroup$ @JoshuaRonis, it sounds like you've got it, but don't be surprised if you go through several more cycles of confusion followed by clarity before it really sinks in :-) $\endgroup$ Commented Dec 30, 2019 at 22:28
  • $\begingroup$ That's exactly whats happening to me right now Simon! Thank you! $\endgroup$ Commented Dec 30, 2019 at 23:33

Bob's perspective:

  1. I've reached point D.
  2. At the point in time that I consider to be "now", I calculate that Alice thinks I actually have not reached point D yet.

This discrepancy comes from the fact that "now" does not have a frame-independent definition unless it is paired with a location. To reason about this properly, we need points not in space or in time, but in spacetime - the combination of both location and time considered together.

So, Bob reaches location D. The time that this happens, at the location that this happens, is spacetime point X. Now, "now" may not have a frame-independent definition separate from location, but if we do specify a frame then we get an objective definition after all.

The time Bob reaches location D, in Bob's frame, but at Alice's location, is spacetime point Y.

The time of spacetime point Y, in Alice's frame, but at location D, is spacetime point Z.

Spacetime points X and Z are both at spatial location D, but they are not the same point - they have different times. More specifically, Z is before X. Considering point Y from each person's reference frame:

Alice's frame: At spacetime point Y, the current time of location D is at point Z. Bob has not reached it yet. Bob's frame: At spacetime point Y, the current time of location D is at point X. He is right that moment at location D.

When Bob, at point X, considers Alice's viewpoint, he can calculate that in her frame she'd be seeing point Z instead.

This discrepancy does not cause any issues because the combination of location and time - a spacetime coordinate - is an absolute frame-independent point. If two people, or one person and a message from another, meet, then the process of each of them arriving at the meeting's spacetime coordinate will involve a combination of frame changes and travel time that resolves any such discrepancies.

  • $\begingroup$ Thanks a lot! I think I understood it - I wrote a comment under the original question...hopefully it makes sense! $\endgroup$ Commented Dec 30, 2019 at 1:45

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