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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.

Thanks!

joshuaronis
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2 Answers2

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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.

Douglas
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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.

Explicitly:

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

Simon Woods
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