Question

http://assets.openstudy.com/updates/attachments/570a1b81e4b03d42c805090a-ganeshie8-1460280209242-zz.png

Answer 1

i'd say proper time is as measured in ship and proper length is as measured on planet moon system

so \(L_{pm} = \gamma L_{ship}\)
\(\Delta t_{pm} = \gamma \Delta t_{ship } \)

or is that too simplistic?

Answer 2

I am extremely confused. Is the word "Earth" in this question used in two different sense? The planet "Earth" and as an adjective to describe that something is owned by people on Earth?

Answer 3

@thomas5267 yeah the wording is very confusing. Here is how I interpret :

A starship has been sent from Earth to check another planet which has a moon. I think the two frames of reference that can be considered in this problem are :
1) starship frame
2) planet-moon frame

So we don't need to consider Earth in this problem

Answer 4

@IrishBoy123
here \(\gamma \approx 5\), so would the answers be
\(L_{pm} = 5*4*10^8m\)
\(\Delta t_{pm} = 5*1.1 s\)

?

Answer 5

That makes sense. The length should be contracted in the spaceship frame relative to the planet frame and hence the time it takes for the microwave burst to reach outpost should be longer than observed in the spaceship frame as well.

Answer 6

Exactly! we three are on same page here. But the textbook solution says otherwise and its solution is as confusing as the problem statement itself. Wanna take a look at the solution ?

Answer 7

What is the answer in textbook then? I supose the Earth outpost is meant to be interpret as the outpost *of* Earth not the outpost *on* Earth.

Answer 8

Here I am attaching the solution of textbook. I really don't get how the distance between `moon and planet` in moon-planet frame could be anything other than \(\gamma *4*10^8\).

Answer 9

The short answer is the transformation formula given in the book is wrong. The long answer is I have to restart my computer so I will give the answer 10 minutes later.

Answer 10

Ahh okay, take your time... I have all night :)

Answer 11

Actually the transformation provided by the book is right lol.

Answer 12

The Lorentz boost in the x direction is
\[
\begin{pmatrix}
x'\\
ct'
\end{pmatrix}
=\begin{pmatrix}
\gamma&-\gamma v/c\\
-\gamma v/c&\gamma
\end{pmatrix}
\begin{pmatrix}
x\\
ct
\end{pmatrix}
\]
Let the UNprimed frame be the spaceship frame and let the primed frame the planet frame. Suppose the microwave burst is observed at \((x_1,t_1)\) and the bang on outpost of Earth is at \((x_2,t_2)\). Then\[
\begin{pmatrix}
x_2'\\
ct_2'
\end{pmatrix}
-
\begin{pmatrix}
x_1'\\
ct_1'
\end{pmatrix}
=\begin{pmatrix}
\gamma&-\gamma v/c\\
-\gamma v/c&\gamma
\end{pmatrix}
\begin{pmatrix}
x_2-x_1\\
ct_2-ct_1
\end{pmatrix}
\]
The problem with the length contraction equation is that it assumes the two events are observed at the same time in the observer frame. Since the two events (the microwave burst and the bang) are observed at different times, the length contraction equation needs to be modified.

Answer 13

sorry for my question but in the text of this exercise i dont get anywhere the word ,,EARTH" just planet and moon and star-ship - or has a different text again what not was posted here ?

Answer 14

Still I have issues with interpreting \(\Delta x'\) as the distance between moon and planet

Answer 15

The Earth is in the question because the outpost is owned by Earthlings. The planet Earth has nothing to do with the scenario. XD

Answer 16

\(\Delta x' = x_2'-x_1'\) is the distance between the moon and the planet in the inertial frame of the planet, or in other words, the rest distance.

Answer 17

then that is the proper length and any measurement made from any other frame of reference must be less than that proper length right ?

Answer 18

Here \(4*10^8\) is the distance between `moon and planet` as measured in ship's frame

So this must be less than the corresponding length in moon-planet frame, which is the proper length.

Answer 19

I still don't see why we can't use length contraction thingy here..

Answer 20

The problem is that the standard length contraction assumes that the two events are "measured" at the same time, i.e. \(\Delta t =0\). In this case the bang happens after the burst and the time difference must be compensated for.

Answer 21

In ship's frame, we cannot have two different lengths for the distance between two objects

Answer 22

Am I wrong in interpreting that the given 4*10^8 is the distance between moon and planet in ship's frame ?

Answer 23

By definition, the phrase "measured distance between two objects" assumes that the measurements at the endpoints are made simultaneoulsy in the measured frame right ?

Answer 24

Suppose the burst is detected at time t in spaceship frame (t unprimed). Then the bang is detected 1.1 seconds later. The time difference between the burst and the bang needs to be compensated for I think.

Answer 25

In ship's frame,
the moon-planet frame moves a distance of v*t.
But this movement shouldn't change the measured distance between moon and planet as it depends only on v and not on t hmm idk

Answer 26

"By definition, the phrase "measured distance between two objects" assumes that the measurements at the endpoints are made simultaneoulsy in the measured frame right ?"

Say the object to be measured is a ruler. The standard derivation of length contraction assumes that the measurement of both ends occur at the same time in the *observer* frame. Suppose the measurement happens at the same time in the *ruler* frame, the result is actually length expansion if I got the math right. Since two different methods have two different result, then it is clear that we have to define measurement more rigorously and it is a nightmare. Why I know it is a nightmare? Because I sent a email to my chemistry teacher asking about why different measuring method can have different results and it was a nightmare to type.

Answer 27

When 1.1 s later the bang arrives at the ship the ship would have moved (a lot in fact) and this has to be compensated for.

Answer 28

Are you saying the distance between moon and planet in the ships frame changes based on when the bang occurs ?

Answer 29

BTW, the answer to the email is *drum rolls*

He reckons that special relativity is wrong and points me to http://www.conspiracyoflight.com/

Answer 30

Maybe I am interpreting the given question incorrectly. I thought that the "measured distance between moon and planet" and the bang bang events are not related.

Answer 31

By not related, I meant to say :

In ship's frame, they measured the distance between moon and planet on some different day. And the bang happened some other day.

Answer 32

The distance between moon and planet of course is a constant (at least it is implied by the question). However, since there is a 1.1 s difference between the microwave burst and the bang as observed by the spaceship in the spaceship frame, the movement of the spaceship must be compensated for. The moon and the planet will be in different position when the bang is observed compared to when the microwave burst is observed in the planet-moon frame.

Answer 33

The last sentence should be:
The spaceship will be in different position when the bang is observed by the spaceship compared to when the microwave burst is observed by the spaceship in the planet-moon frame.

Answer 34

that is fine, but why am i not allowed to use length contraction formula here ?

Answer 35

The derivation of length contraction assumes the both ends of the ruler (or whatever) is observed at the same time in the observer frame. The burst and the bang are not observed at the same time in the observer frame in this case, they are separated by 1.1 s.

Answer 36

One of the reason why that email is so hard to type is for every event or "thing" that is measured, a frame of reference has to be specified or else there might be ambiguity and the result might vary depend on which frame of reference is chosen.

Answer 37

Again, what has burst and bang anything to do with the numbner 4*10^8 ?

Answer 38

i originally though forget about planet and just imagine a big flying ruler connecting them. there the assumption is that you are looking at both ends at the same time.

but here i think maybe they see the ends at different times, so you have to use Lorentz diagram and plot them etc etc

Answer 39

I see what you mean now. The question is now unambiguously BAD.

Answer 40

soz: "forget about planet **and moon** and just imagine a big flying ruler connecting them"

Answer 41

Or unambiguously ambiguous if you fancy words.

Answer 42

this is from memory so it might not be too right bit i recall the task of deriving the formula for length contraction from a lorentz diagram to be really unintuitive and tricky. cos you gotta fix the ends at the same time in the moving frame. that might shed some light on why this is what it is.

Answer 43

\(\Delta x' = \gamma(\Delta x - v\Delta t)\)

Let \(\Delta x'\) be the proper length.
If we now let\(\Delta t=0\), then \(\Delta x\) is the difference in the measurements made simultaneously.

Maybe I'll pause on this problem for now. I'll get back with fresh mind tomorrow. Not able to comprehend properly...

Answer 44

i'll see if i can find better explanation of what i mean by that....

Answer 45

Yes you comprehended the question correctly. The question is at fault. The problem is how do you measure the distance between the moon and the planet. Different measuring methods can be shown to have different results.

Answer 46

Consider two ends of the ruler. Let the primed be the ruler frame.
\[
\begin{pmatrix}
x'\\
ct'
\end{pmatrix}
=
\begin{pmatrix}
\gamma&-\gamma v/c\\
-\gamma v/c&\gamma
\end{pmatrix}
\begin{pmatrix}
x\\
ct
\end{pmatrix}\\
\Delta x'=\gamma\Delta x\\
\Delta x= \frac{1}{\gamma}\Delta x',\,\frac{1}{\gamma}<1
\]
Suppose the measurement is done simultaneously in the unprimed observer frame. Then.
\[
\begin{pmatrix}
x_2'-x_1'\\
ct_2'-ct_1'
\end{pmatrix}
=
\begin{pmatrix}
\gamma&-\gamma v/c\\
-\gamma v/c&\gamma
\end{pmatrix}
\begin{pmatrix}
x_2-x_1\\
0
\end{pmatrix}\\
\begin{pmatrix}
x_2'-x_1'\\
ct_2'-ct_1'
\end{pmatrix}
=
\begin{pmatrix}
\gamma(x_2-x_1)\\
-\gamma v(x_2-x_1)/c
\end{pmatrix}
\]
That's the standard derivation. However suppose that the ruler is measured simultaneously in the ruler frame by the observer, then
\[
\begin{align*}
\begin{pmatrix}
x_2'-x_1'\\
0
\end{pmatrix}
&=
\begin{pmatrix}
\gamma&-\gamma v/c\\
-\gamma v/c&\gamma
\end{pmatrix}
\begin{pmatrix}
x_2-x_1\\
ct_2-ct_1
\end{pmatrix}\\
\begin{pmatrix}
x_2'-x_1'\\
0
\end{pmatrix}
&=
\begin{pmatrix}
\gamma(x_2-x_1-v(ct_2-ct_1)/c)\\
-\gamma( v(x_2-x_1)/c+ct_2-ct_1)
\end{pmatrix}\\
\begin{pmatrix}
x_2'-x_1'\\
0
\end{pmatrix}
&=
\begin{pmatrix}
\gamma(x_2-x_1-v(t_2-t_1))\\
-\gamma( v(x_2-x_1)/c+ct_2-ct_1)
\end{pmatrix}
\end{align*}
\]\[
\begin{align*}
0&=-\gamma( v(x_2-x_1)/c+ct_2-ct_1)\\
ct_2-ct_1&=v(x_2-x_1)/c\\
t_2-t_1&=v(x_2-x_1)/c^2
\end{align*}
\]\[
\begin{align*}
x_2'-x_1'&=\gamma(x_2-x_1-v(t_2-t_1))\\
&=\gamma(x_2-x_1-v^2(x_2-x_1)/c^2)\\
&=\gamma(x_2-x_1)(1-v^2/c^2)\\
&=\frac{1}{\gamma}(x_2-x_1)\\
\Delta x'&=\frac{1}{\gamma}\Delta x\\
\Delta x&=\gamma \Delta x'
\end{align*}
\]
Remember, \(\Delta x'\) is the rest length. So depending on the measuring method, you can get a length contraction or a length expansion, which needless to say is BAD! This prompts the question "what is a measurement?"

Answer 47

I put the result of the standard derivation of length contraction in the wrong place. It should be in the second math block not the first.

Answer 48

Consider two ends of a ruler. Let the primed frame be the ruler frame.
\[
\begin{pmatrix}
x'\\
ct'
\end{pmatrix}
=
\begin{pmatrix}
\gamma&-\gamma v/c\\
-\gamma v/c&\gamma
\end{pmatrix}
\begin{pmatrix}
x\\
ct
\end{pmatrix}\\
\]Suppose the measurements are done simultaneously in the unprimed observer frame. Then\[
\begin{align*}
\begin{pmatrix}
x_2'-x_1'\\
ct_2'-ct_1'
\end{pmatrix}
&=
\begin{pmatrix}
\gamma&-\gamma v/c\\
-\gamma v/c&\gamma
\end{pmatrix}
\begin{pmatrix}
x_2-x_1\\
0
\end{pmatrix}\\
\begin{pmatrix}
x_2'-x_1'\\
ct_2'-ct_1'
\end{pmatrix}
&=
\begin{pmatrix}
\gamma(x_2-x_1)\\
-\gamma v(x_2-x_1)/c
\end{pmatrix}\\
\Delta x'&=\gamma\Delta x\\
\Delta x&= \frac{1}{\gamma}\Delta x',\,\frac{1}{\gamma}<1
\end{align*}
\]That's the standard derivation of length contraction. However suppose that both ends of the ruler is measured simultaneously, according to an observer in the ruler frame, by the observer in the unprimed observer frame*, then\[
\begin{align*}
\begin{pmatrix}
x_2'-x_1'\\
0
\end{pmatrix}
&=
\begin{pmatrix}
\gamma&-\gamma v/c\\
-\gamma v/c&\gamma
\end{pmatrix}
\begin{pmatrix}
x_2-x_1\\
ct_2-ct_1
\end{pmatrix}\\
\begin{pmatrix}
x_2'-x_1'\\
0
\end{pmatrix}
&=
\begin{pmatrix}
\gamma(x_2-x_1-v(ct_2-ct_1)/c)\\
\gamma( -v(x_2-x_1)/c+ct_2-ct_1)
\end{pmatrix}\\
\begin{pmatrix}
x_2'-x_1'\\
0
\end{pmatrix}
&=
\begin{pmatrix}
\gamma(x_2-x_1-v(t_2-t_1))\\
\gamma( -v(x_2-x_1)/c+ct_2-ct_1)
\end{pmatrix}
\end{align*}
\]\[
\begin{align*}
0&=\gamma( -v(x_2-x_1)/c+ct_2-ct_1)\\
ct_2-ct_1&=v(x_2-x_1)/c\\
t_2-t_1&=v(x_2-x_1)/c^2
\end{align*}
\]\[
\begin{align*}
x_2'-x_1'&=\gamma(x_2-x_1-v(t_2-t_1))\\
&=\gamma(x_2-x_1-v^2(x_2-x_1)/c^2)\\
&=\gamma(x_2-x_1)(1-v^2/c^2)\\
&=\frac{1}{\sqrt{1-v^2/c^2}}(x_2-x_1)(1-v^2/c^2)\\
&=\sqrt{1-v^2/c^2}(x_2-x_1)\\
&=\frac{1}{\gamma}(x_2-x_1)\\
\Delta x'&=\frac{1}{\gamma}\Delta x\\
\Delta x&=\gamma \Delta x'
\end{align*}
\]
We now have two results. If both ends of ruler are measured simultaneously in observer's frame, you get length contraction of \(\Delta x = \dfrac{1}{\gamma}\Delta x'\). However, if both ends of the ruler is measured simutaneously, according to an observer in the ruler's frame, by an observer in the observer frame, you get length expansion of \(\Delta x = \gamma \Delta x'\). This is clearly BAD and raises the question "What is a measurement?"

*"Both ends of the ruler is measured simultaneously, according to an observer in the ruler frame, by the observer in the unprimed observer frame" means the measurement is done simutaneously according to the definition of simutaneity of an observer in the ruler frame, but the measurement itself is done by the observer in the observer frame. In other words, the ruler "feels" the measurement at the same time in the ruler frame, but the measurement is done by the observer in the observer frame.