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Suppose a rod is moving at speed $v$ relative to me along its length. $L_0 = {}$length of the rod in the frame in which the rod is at rest $L = {}$length of the rod in my frame

Then $$L = L_0 \sqrt{1-\frac{v^2}{c^2}} $$ Let us now consider another scenario. I am moving towards a star at speed $v$. Then the distance between me and the star is given by $$L = L_0 \sqrt{1-\frac{v^2}{c^2}} $$ My question is: What is the meaning of $L_0$? I understand the meaning of a rod being at rest in some frame. But what is the meaning of distance between me and the star being at rest in any frame? Can space between two points be at rest in any frame?

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There is no frame in which "space" is at rest.

Using the length contraction formula $$ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} $$ for the distance to the star is slightly misleading. The distance $L$ in your frame corresponds to two points, $x_0 = 0$ for you and $x_1 = L$ for the star, that you are observing simultaneously, say at time $t_0$. In the star's frame the same locations will have coordinates $$ x'_0 = - \frac{vt_0}{\sqrt{1 - \frac{v^2}{c^2}}}\\ x'_1 = \frac{L - vt_0}{\sqrt{1 - \frac{v^2}{c^2}}} $$ and you'll be tempted to say that the distance in the star's frame is $$ L_0 = x'_1 - x'_0 = \frac{L}{\sqrt{1 - \frac{v^2}{c^2}}} $$ or that $$ L = L_0 \sqrt{1 - \frac{v^2}{c^2}} $$ But in the star's frame, locations $x'_0$ and $x'_1$ are actually observed at different times, reading $$ t'_0 = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}\\ t'_1 = \frac{t_0 - v x_1 /c^2}{\sqrt{1 - \frac{v^2}{c^2}}} < t'_0 $$ So this $L_0$ is not a distance in the proper sense of the word.

The important thing to learn here is that this is a result of relativity of simultaneity: events/locations/measurements that are simultaneous in one frame will not be simultaneous in another.

In fact the same thing happens for the rod, but at least in that case we do have a meaningful "rest length" in the rod's frame.

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  • $\begingroup$ When I move towards a star then the distance between me and the star gets contracted. But, to use the formula of length contraction, I need to know the "rest distance". But there is no meaningful "rest distance". So how to use the length contraction formula? $\endgroup$
    – Animesh
    Commented Jun 26, 2016 at 9:55
  • $\begingroup$ Is the following correct? I am in some inertial frame. I am moving with speed v towards the star according to some other inertial frame S'. In my frame light will take time t to arrive from the star to me. In frame S', light will take time t' to arrive from the star to me. Then we can use the length contraction formula by putting L = c t and L' = c t'. Is this correct? $\endgroup$
    – Animesh
    Commented Jun 26, 2016 at 9:59
  • $\begingroup$ No, this is not correct. It's also not incorrect. It's completely meaningless. "The time light will take to arrive from the star to me" depends entirely on when the light leaves the star. (Don't forget that the star keeps getting closer!). What light beam's travel time are you talking about? One that has just arrived at your ship? One that has just left the star according to you? One that has just left the star according to an observer stationary with respect to the star? $\endgroup$
    – WillO
    Commented Jun 26, 2016 at 13:00
  • $\begingroup$ It is certainly true that if you say a given light beam has traveled for a time $t$ then you must say it has traveled a distance $L=ct$. But this is not at all the same thing as the distance from you to the star. $\endgroup$
    – WillO
    Commented Jun 26, 2016 at 13:00
  • $\begingroup$ @Animesh If you are trying to relate relativity of simultaneity to communication at the speed of light, the answer is no, it does not arise due to a propagation delay. You can check why by calculating the "propagating speed" of the delay using the $|x'_1 - x'_0|$ and $|t'_1 - t'_0|$ as in the answer: $|x'_1 - x'_0|/|t'_1 - t'_0| = c^2/v > c$, it would have to be superluminal. So it is not a "removable" relativity of simultaneity, it is much stronger. $\endgroup$
    – udrv
    Commented Jun 26, 2016 at 19:28
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First: You've chosen to express "the distance from me to the star" as some formula involving a quantity called $L_0$ --- and then you've asked us to tell you what $L_0$ means. But you're the person who wrote down this expression, so only you can know what you meant by it.

Second: It's very hard to tell what you're actually asking, but I am pretty sure that I can identify the source of your fundamental confusion. Namely: The length of the rod (as defined in your frame, the rod's frame, or any other frame) does not change over time. The distance from you to the star does change over time. You seem to be trying to treat the length of the rod and the distance to the star as perfectly analogous, but they are not analogous at all.

Third: If two observers are both present at an event $E$, and if you know the coordinates that the first observer assigns to an event $F$, you can Lorentz-transform those coordinates to find the coordinates that the second observer assigns to event $F$. You seem to be imagining two observers, one at an event $E$ on the spaceship and one at a different event $E'$ on the star, and trying to Lorentz transform one set of coordinates to the other by blindly applying a formula.

So start over: You are on the ship. Your clock strikes 1:00. Call this event $A$. At that moment (according to you), a clock on the star strikes 1:00. Call this event $X$. At that moment, according to an observer on the star, your clock is striking 2:00. Call this event $B$.

"The distance from you to the star" depends on both the observer and the time when the observation is made. To you, at 1:00 by your clock, "the distance between you and the star" means the distance, in your coordinates, from $A$ to $X$. To you, at 2:00 by your clock, "the distance between you and the star" means something else. To your friend on the star, at 1:00 by his clock, "the distance between you and the star" means the distance, in his coordinates, from $X$ to $B$. Before you can start comparing one of these distances to another, you have to decide which two you're trying to compare.

Finally, with regard to the time it takes light to get from the star to you --- remember that the light currently arriving at your ship left the star at a time when the star was further away than it is now. Light currently leaving the star will arrive at your ship at a time when the star is closer than it is now. So if you want to talk about "the time it takes light to get from the star to your ship", you need to decide not just who's measuring, but which light beam's travel times you're talking about. (Or perhaps you mean some other travel time.)

Bottom line: Different questions have different answers. Questions like "How far does the guy on the star say he is from me?" have many possible meanings, each of which yields a different answer. The key thing you are missing is that you are not being clear (certainly not with us, and I suspect not with yourself) about what you're trying to ask.

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  • $\begingroup$ My confusion is: As I understand, Lo in the Lorentz length contraction formula is the rest-length of a rod. But suppose we are not dealing with a rod. Rather, we are dealing with the distance from Earth to a star. If I am moving towards the star, does distance between me and the star contracts (according to my frame) because of my movement? If yes, then what should I treat as "rest distance"? I can understand the meaning of a rod being at rest in some inertial frame. But what is the meaning of distance being at rest in any frame? $\endgroup$
    – Animesh
    Commented Jun 26, 2016 at 14:44
  • $\begingroup$ And what I keep telling you is that "the distance between you and the star" is, by itself, an entirely meaningless phrase. Before you can ask whether it contracts, you have to decide what it means. And nobody other than Animesh can answer the question: "What does Animesh mean?" $\endgroup$
    – WillO
    Commented Jun 26, 2016 at 15:46
  • $\begingroup$ (Also: If $L_0$ is the length of some rod, what is it doing in a formula that has nothing to do with rods?) $\endgroup$
    – WillO
    Commented Jun 26, 2016 at 15:47
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Distance contraction is a case of length contraction. The length of a rod in its rest frame is contracted for observers moving near light speed. Distance contraction is working in an analog way: The length of the rod corresponds to the distance between a mass object at the point of departure and a mass object at an end point.

For this purpose we must suppose that both points are belonging to the same frame, that means that the relative velocity between the two mass objects at the starting point and at the end point is zero (or at least non-relativistic). For practical purposes, this definition is in most cases sufficient, for example if you want to define a distance between Earth and an exoplanet. An astronaut moving from Earth to the exoplanet will have to take into account the distance contraction. As a result, he may reach exoplanets within his lifetime which he could not have reached without the phenomenon of distance contraction.

In contrast, the astronaut traveling near light speed and the exoplanet moving at non-relativistic velocity with respect to Earth cannot be considered as one and the same reference frame because their relative velocity is relativistic. That means that your example does not work: if you are approaching a star at relativistic relative velocity, you are not in the same reference frame as the star. No rod can be held between you and the star.

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  • $\begingroup$ if the Earth and the star are at rest relative to each other, then does it mean that the length of a rod from Earth to the star is analogous to rest-length? $\endgroup$
    – Animesh
    Commented Jun 26, 2016 at 15:01
  • $\begingroup$ Yes, the distance is taking the place of the rod. Any astronaut traveling with relativistic velocity between Earth and star would be an observer, observing the distance contraction. $\endgroup$
    – Moonraker
    Commented Jun 26, 2016 at 16:16

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