0
$\begingroup$

On the Wikipedia page for the Twin Paradox, the example lays out the perspective of each twin in turn. Both twins are portrayed as understanding the ship's velocity as v, and the travelling twin's sense of time is then explained by saying that the earth-distant star system, being in effect one giant object, undergoes length contraction ie the twin travels less distance, ie less time went by for the twin. But if the entire rest of the twin's system, ie. the twin's whole sense of distance, has shrunk, shouldn't the twin's sense of velocity shrink too? How can they agree on velocity but not distance, when v = d/t ? The v is relative to what?

Or are we saying that our usual sense of reality is wrong, when we think of an object's location at a given time (absolute or otherwise) as an intrinsic truth upon which we define velocity as Δlocation/unit time, when in fact it's the other way around and an object's velocity is intrinsic. Are we saying velocity actually IS absolute even though time and distance aren't?

$\endgroup$
3
$\begingroup$

Suppose the stationary twin occupies the origin of our coordinates, $x=0$, and observes that the velocity of the traveling twin is $v$, in their frame. In other words, suppose that the traveling twin covers a certain distance $d$ in a certain time $t$, such that $v=d/t$. Let's find out what the travelling twin measures the stationary twin's velocity to be, by directly applying a Lorentz transformation, boosting to a frame with velocity $v$:

$$x'=\gamma(x-vt)$$

$$t'=\gamma(t-vx/c^2)$$

The stationary twin is at the origin of our original coordinates, so plugging in $x=0$, we get the coordinates of the stationary twin in the traveling twin's frame:

$$x'=-\gamma vt$$

$$t'=\gamma t$$

Now let's calculate the velocity of the stationary twin in this frame:

$$v'=\frac{x'}{t'}=\frac{-\gamma v t}{\gamma t}=-v$$

In other words, the velocities observed by each twin for the other are equal and opposite. The effects of time dilation and length contraction exactly cancel out.

$\endgroup$
1
  • $\begingroup$ Right. So even though observers in different frames can't agree on time and distance, they can agree on velocity. I just wanted to make sure I had that right because I find it cool. $\endgroup$ – Justin Poirier Aug 14 '20 at 16:25
1
$\begingroup$

To answer your title question: no. The twin's paradox resolution is that the traveling twin has to change inertial reference frames to come back home, so the scenario is not symmetric between the two twins. Just because there is a different outcome for each twin does not mean there is a universal time and special relativity is wrong.

To get to the body of your question, both twins agree on the relative velocity between their own frames. As a hand-wavy argument, you seem to be focusing on how they disagree about distances, but they will also disagree about time intervals as well. Since velocity depends on both distances and time intervals, these effects essentially "cancel out" when looking at relative velocities. A more precise handling of this can be found on probably_someone's answer.

This doesn't mean velocity is absolute; velocities are still relative. You always have to specify a velocity with respect to something. It is just that if I am considering my velocity relative to you, and you are considering your velocity relative to me, then we will find these velocities to be equal and opposite. However, your velocity relative to some other observer does not need to be the same as your velocity relative to me. In general it won't be, unless that third observer happens to be at rest relative to me.

$\endgroup$
7
  • $\begingroup$ My point is that we are saying they have the same relative velocity and THEN saying given that, and given their measured distances between each other, their difference in perceived time is x. Normally we think of velocity as a function of time, not the other way around. This is at the very least fascinating, and my intention was to confirm and share my fascination, not discredit relativity with hand-waving. $\endgroup$ – Justin Poirier Aug 14 '20 at 16:04
  • $\begingroup$ @JustinPoirier You don't have to assume that they have the same relative velocity. You can prove it with a Lorentz transformation, as you can see in my answer. $\endgroup$ – probably_someone Aug 14 '20 at 16:05
  • $\begingroup$ @probably_someone Thanks. Putting a link in my answer just in case $\endgroup$ – BioPhysicist Aug 14 '20 at 16:05
  • $\begingroup$ @probably_someone that's technically true, but it's essentially just proving an axiom of lorentz transformations. They were, after all, derived from the fact that the relative velocities are equal and opposite. It's a bit like using $F=ma$ instead of $F=dp/dt$ and then showing that the mass must be constant in that case. You want to show that relative velocities are equal? Use velocity addition to boost from $v=0$ to $v'$, since velocity is 0 in one's own frame of reference $\endgroup$ – Jim Aug 14 '20 at 17:15
  • 1
    $\begingroup$ @Jim What I did here was demonstrate the self-consistency of special relativity, which does, in fact, have value, just like demonstrating that using $F=dp/dt$ should give you the same predictions as using $F=ma$. $\endgroup$ – probably_someone Aug 14 '20 at 17:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.