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A frame $F_b$ is being translated with a speed $v$ according to another frame $F_a$.

A point $A$ is attached to $F_a$. $B$ is attached to $F_b$.

$D_a$ is the distance $AB$ viewed from $Fa$ and $D_b$ is the distance $AB$ viewed from $F_b$.

What is the relation between $D_a$ and $D_b$ (in special relativity)?

(Is there something like $D_a=+vt_a$ and $D_b=-vt_b$?)

aayyachi
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  • We do have standards here, we don't just let any question remotely about physics be asked here. In its original form, your question seemed very much like a homework assignment (considered off-topic for this site, hence its closure). It appears you are looking for a derivation of the Lorentz contraction, which can be found in any reference book on SR (including Wikipedia's article on the matter) and would effectively be a duplicate of this question that John Rennie mentions. – Kyle Kanos Oct 29 '15 at 19:48
  • aayyachi: "frame $F_a$ is being translated with a speed $v$ according to another frame F_b$" -- These frames are thus two inertial systems; where (mutually) the members of one determine the speed of each member of the other as the same value $v$. "point $A$ is attached to $F_a$. $B$ is attached to $F_b$" -- Alright. (Instead of "is attached to" I'd say "was and remained a member of"). "$D_a$ is the distance $AB$ [...]" -- No: distance values are attributed to pairs of members of the same IS. The relation between $A$ and $B$ is characterized not by some distance, but by speed $v$. – user12262 Oct 29 '15 at 21:45
  • @Kanos, All human knowledge can be found somewhere in books. If sometimes one couldn't understand something, one ask others, that's what one can do here. Otherwhise, this site is not only for doing scientific research for 'new knowledge' – aayyachi Oct 30 '15 at 01:03
  • How could you remove my comment? This is not a homework and not off-topic. Also, Mr. Curious is down voting all my questions. – aayyachi Oct 30 '15 at 01:12
  • The only thing I add to the question is (...). Even without adding it, the question has its importance. – aayyachi Oct 30 '15 at 01:14

1 Answers1

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I would guess you are thinking about deriving the Lorentz contraction, but if so this isn't the way to go about it.

Suppose we put the point $A$ at the origin of $F_a$ and the point $B$ at the origin of $F_b$, and for convenience we take the zero time to be when the two origins coincide i.e. $A$ and $B$ are at the same point.

An observer at the origin in $F_a$ sees $B$ moving with velocity $+v$, so the distance to point $B$ is:

$$ D_a = vt $$

An observer at the origin in $F_b$ sees $A$ moving with velocity $-v$, so the distance to point $A$ is:

$$ D_b = -vt $$

So for both observers the distance between the points is the same.

If you're interested in deriving the Lorentz contraction see my answer to How do I derive the Lorentz contraction from the invariant interval?.

Community
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John Rennie
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  • If a distance (length of an object) is attached to $F_b$, it will be seen shorter from a viewer attached to $F_a$. That's why I was wondering about distance related to both frames. – aayyachi Oct 29 '15 at 18:54
  • don't $D_a = vt_a$ and $D_b = -vt_b$ ? – aayyachi Oct 29 '15 at 19:02