I Distance Contraction: Is D or D*sqrt(1-(v/c)^2) Traveled?

[Kairos]

Two planets are immobile one with respect to the other and distant of D. A rocket passes from one to the other at high speed v. Is the distance really traveled by the rocket between these planets D or only D*sqrt(1-(v/c)^2) ?

 

[robphy]

Draw a Spacetime diagram.

 

[Orodruin]

Is the distance really traveled by the rocket between these planets D or only D*sqrt(1-(v/c)^2) ?

In what frame?

 

[Kairos]

In what frame?

in the frame of the rocket's odometer

 

[Orodruin]

The rocket travels a distance of exactly zero in its own rest frame.

 

[Kairos]

okay I answered your question wrong! So my question is what distance does the rocket counter indicate between the two planets

 

[Orodruin]

What is a rocket counter? Are you asking what the distance between the planets are in the rocket frame?

 

[Kairos]

yes I guess that's my question.

 

[Orodruin]

Then yes. The distance between the planets in the rocket frame is $D \sqrt{1 - v^2/c^2}$. When the first planet passes the rocket, the other is a distance $D \sqrt{1 - v^2/c^2}$ away.

 

[Kairos]

Thank you very much!

 

[Sagittarius A-Star]

yes I guess that's my question.

Call the rest frame of the two planets $S$ (unprimed frame) and the rest frame of the rocket $S'$ (primed frame) and use the Lorentz transformation.

The spatial distance between the starting event on planet A and the arrival event on planet B in the two planet's rest frame is $\Delta x = D$.

From ...

The rocket travels a distance of exactly zero in its own rest frame.

... follows:

The spatial distance between the starting event and the arrival event in the rocket's rest frame is:

$\Delta x' = 0$.

inverse Lorentz transformation:

$\Delta x = \gamma (\Delta x' + v \Delta t')$

$D = \Delta x = \gamma (0 + v \Delta t')$

The distance, which planet B is moving towards the rocket in the rocket's rest frame while $\Delta t'$ is:

$v\Delta t' = D/\gamma = D \sqrt{1-v^2/c^2}$.

 

[Kairos]

and in the frame of an inhabitant of one of the planets, the traveled distance of the rocket between the two planets is also $D \sqrt{1-v^2/c^2}$ or $D$?

 

[Orodruin]

and in the frame of an inhabitant of one of the planets, the traveled distance of the rocket between the two planets is also $D \sqrt{1-v^2/c^2}$ or $D$?

$D$

 

[Kairos]

Thanks, I was confused because in the case of uniform relative motion, I thought we might as well assume that it is the planet that is moving relative to the rocket, so the situation is symmetric..

 

[Sagittarius A-Star]

so the situation is symmetric..

The situation is not symmetric:

Imagine a very long rod with one end at planet A and the other end at planet B.

In frame $S$ the rod is at rest (=> not length contracted), in frame $S'$ the rod is moving (=> length contracted).

 

[Kairos]

Yes I understand this reasoning. But let us imagine now in the frame of the astronaut S', a long rod whose length corresponds to the distance between the crossing points of the planets. In the S frame, this rod must also be shortened. When two intervals cross each other at constant speed, each one keeps its value seen internally and sees the other one contracted, right?

 

[Sagittarius A-Star]

Yes I understand this reasoning. But let us imagine now in the frame of the astronaut S', a long rod whose length corresponds to the distance between the crossing points of the planets.

In frame S' the "crossing points" are at the same x'-location. This is the fixed location, where the rocket is at rest. So the length of this rod would be zero, according you description.

Of course, you could define the length of this rod differently, to have a finite length >0.

In the S frame, this rod must also be shortened. When two intervals cross each other at constant speed, each one keeps its value seen internally and sees the other one contracted, right?

Yes.

 

[PeroK]

Thanks, I was confused because in the case of uniform relative motion, I thought we might as well assume that it is the planet that is moving relative to the rocket, so the situation is symmetric..

... the rocket is length contracted in the planet frame. If the rocket had rest length $D$, then there's the symmetry.

 

[Nugatory]

When two intervals cross each other at constant speed, each one keeps its value seen internally and sees the other one contracted, right?

Yes, but don’t overlook the relativity of simultaneity here. If the two ends of the rod line up with the two planets at the same time using one frame, they will not in the other.

You might want to spend some time on the “pole-barn paradox” - it’s the starting point for understanding how we can consider either to be length-contracted without contradiction.

 

[Kairos]

In frame S' the "crossing points" are at the same x'-location. This is the fixed location, where the rocket is at rest. So the length of this rod would be zero, according you description.

Indeed! and thank you for all your answers. The attached diagram may be clearer than my questions. C1 and C2 are the crossing points with the two planets, on the left in the frame of an inhabitant of planet 1 and on the right in the frame of the astronaut. Is the distance marked D the same for both points of view or not?

 

[Nugatory]

Indeed! and thank you for all your answers. The attached diagram may be clearer than my questions. C1 and C2 are the crossing points with the two planets, on the left in the frame of an inhabitant of planet 1 and on the right in the frame of the astronaut. Is the distance marked D the same for both points of view or not?

(The angled lines in the right-hand diagram should slant the other way, southeast to northwest).

In the left-hand diagram D is the distance between the planets using the frame in which they are at rest. In the right-hand diagram D is the rest length of the rod. There is no particular reason for those to be the same unless we chose the rod or placed the planets to make it so. Neither D is related to any contracted length.

 

[robphy]

Here's the spacetime diagram that I suggested in #2.

In the rest frame of the planets,
their separation (the spatial-distance between the red worldlines along the red dot-dash line) D is 5 light-years. (There is a hyperbola [analogous to a circle] representing 5 spacelike-units.)

An inertial astronaut (Bob) traveling at velocity (3/5)c measures a
a smaller separation (the spatial-distance between the red worldlines along the blue dot-dash line)... smaller because the dot-dash segment doesn't meet that 5-light-year hyperbola.
This separation for the case with numbers given here turns out to be only 4 light-years.

Numbers are less important.
First should be on the fact that
the measured separation is smaller
(for relative-motion along the extent of a long-meterstick between the planets).

What is key is how the inertial astronauts define simultaneity
(by the tangency to the hyperbola at the point-event where the astronaut worldline meets the [future-timelike] hyperbola [analogous to a circle] representing (here) a spacetime-interval for 4 timelike-units).

It might be helpful to consider the Euclidean version of this construction.

 

[Sagittarius A-Star]

Indeed! and thank you for all your answers. The attached diagram may be clearer than my questions. C1 and C2 are the crossing points with the two planets, on the left in the frame of an inhabitant of planet 1 and on the right in the frame of the astronaut. Is the distance marked D the same for both points of view or not?

No. In addition to what the others answered:

In frame S' (rest frame of the rocket), the (length contracted) distance between the planets is $D \sqrt{1-v^2/c^2}$.

You assume that the rod at rest in frame S' has the same length $D \sqrt{1-v^2/c^2}$. The two ends of this rod each meet a planet at the same time $t'$.

This rod is even shorter by length contraction in frame S: $D (1-v^2/c^2)$. But in frame S the two ends of this rod meet planets at different times $t$. So in frame S the length of this (moving) rod is not equal to the distance of the planets.

Source:
https://commons.wikimedia.org/wiki/File:Animated_Spacetime_Diagram_-_Length_Contraction.gif

 

[Kairos]

understood, thank you

 

[Kairos]

Yes, but don’t overlook the relativity of simultaneity here. If the two ends of the rod line up with the two planets at the same time using one frame, they will not in the other.

You might want to spend some time on the “pole-barn paradox” - it’s the starting point for understanding how we can consider either to be length-contracted without contradiction.

Thank you, I am indeed intrigued by this point. Imagine that the rocket is followed by a second one, rocket2, in its rest frame at a distance D and that the rockets are in the same alignment as the planets. I assumed that perceived from the planets rest frame, the distance traveled from the planet 1 by the rocket 1 when the rocket 2 reaches the planet 1 is also $D \sqrt{1 - v^2/c^2}$ You don't think so?

 

[Nugatory]

Thank you, I am indeed intrigued by this point. Imagine that the rocket is followed by a second one, rocket2, in its rest frame at a distance D and that the rockets are in the same alignment as the planets. I assumed that perceived from the planets rest frame, the distance traveled from the planet 1 by the rocket 1 when the rocket 2 reaches the planet 1 is also $D \sqrt{1 - v^2/c^2}$ You don't think so?

I’m sorry but I have no idea what you’re trying to say here. But I think it is a mistake to try to further complicate the problem before you understand the less complicated case with one rocket and one distance between two planets.

We have a rocket with rest length $D$; that is, using the frame in which the rocket is at rest the length of the rocket is $D$. We have two planets at rest relative to one another; using the frame in which both are at rest they are separated by a distance $D$.

Using the frame in which the planets are at rest:
- The length of the rocket is $D/\sqrt{1-v^2/c^2}$.
- The nose of the rocket passes planet 2 after the tail of the rocket reaches planet 1

Using the frame in which the rocket is at rest:
- The distance between the planets is $D/\sqrt{1-v^2/c^2}$.
- The nose of the rocket passes planet 2 before the tail of the rocket reaches planet 1

 

[Kairos]

Indeed, I expressed this poorly. I postulated two rockets separated by D instead of a very very long rocket but I think there is agreement on the principle. That's okay for me, thanks a lot

 

[robphy]

Thank you, I am indeed intrigued by this point. Imagine that the rocket is followed by a second one, rocket2, in its rest frame at a distance D and that the rockets are in the same alignment as the planets. I assumed that perceived from the planets rest frame, the distance traveled from the planet 1 by the rocket 1 when the rocket 2 reaches the planet 1 is also $D \sqrt{1 - v^2/c^2}$ You don't think so?

Here are some more descriptive spacetime diagrams
using "light-clock diamonds", which have equal areas
(in accordance with Lorentz invariance).

Note that when $v=(3/5)c$, then $\gamma=\frac{1}{\sqrt{1-(v/c)^2}}=\frac{5}{4}$.
More important for drawing these diagrams is that the Doppler factor $k=\sqrt{ \frac{1+(v/c)}{1-(v/c)}}=2$, which are the stretching and shrinking factors that relate the blue diamonds to the red diamonds.

The following diagram can be constructed by hand from the previous one.
But, with practice, the information in this diagram can be read off the previous one.