check my work

- anonymous

check my work

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- schrodinger

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- anonymous

Consider a rod kept on the ground, it is at rest with respect to a frame of reference S
of an observer standing at a distance of x1 from the rod(since both r moving along with the earth, they are at rest wrt the earth). He thus measure it's length (L) to be x2-x1, it is the rod's length in a frame of reference respect to which is at rest, so it is the rod's proper length
|dw:1444042534166:dw|
\[L=x_{2}-x_{1}\]
Now consider the observer to be standing in the same place, but the rod is now kept in a vehicle that is moving with a velocity v in the positive x direction w.r.t. the observer
The rod is now at rest with respect to the frame of reference of the bus, and it's length now measured by an observer in the bus is given as
\[L'=x_{2}'-x_{1}'\]|dw:1444043022456:dw|
But for the observer on ground, the coordinates are given by using inverse lorentz transformation equation(we use inverse as we want to find coordinates in S frame)
\[x_{1}^{new}=\frac{x_{1}'+vt'}{\sqrt{1-\frac{v^2}{c^2}}} \space \space , \space \space x_{2}^{new}=\frac{x_{2}'+vt'}{\sqrt{1-\frac{v^2}{c^2}}}\]
\[L^{new}=x_{2}^{new}-x_{1}^{new}=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}(x_{2}'+vt'-x_{1}'-vt')=\gamma(x_{2}'-x_{1}')=\gamma L'\]
But the new length observed by the observer on the ground should be same as before, as he has not moved from his original position
so we have
\[L=L^{new} \space \space \space ; \space \space \space L=\gamma L' \implies L'=\sqrt{1-\frac{v^2}{c^2}}L\]
@IrishBoy123

- anonymous

Another assumption that is made is that the coordinates of both the end points of the rod are measured simultaneously in both frames

- IrishBoy123

SR, excellent!
will look later...
@Michele_Laino

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## More answers

- anonymous

ok thanks:)

- Michele_Laino

let's suppose the rod lies on reference \(S'\)

- Michele_Laino

so its proper length is \(L'\)

- Michele_Laino

namely I mean its real length is \(L'\)

- Michele_Laino

now, I have to measure its length in reference \(S\), and I have to make the measure of both ends of the rod at the same time \(t\)

- Michele_Laino

of course, reference \(S'\) is moving, with respect to \(S\), with speed \(V\)

- Michele_Laino

so, using \(Lorentz\; transformation\), I can write this:
\[\begin{gathered}
x{'_2} = \gamma \left( {{x_2} - Vt} \right) \hfill \\
x{'_1} = \gamma \left( {{x_1} - Vt} \right) \hfill \\
\end{gathered} \]
and subtracring the second equation from the first one, I can write this:
\[\begin{gathered}
x{'_2} - x{'_1} = \gamma \left( {{x_2} - {x_1}} \right) \hfill \\
{L_O} = \gamma {L_{meas}} \hfill \\
\end{gathered} \]

- Michele_Laino

where \(L_O\) is the real length and \(L_{meas}\) is the measured length. So, finally, we get:
\[{L_{meas}} = \frac{{{L_O}}}{\gamma }\]
namely the so called \(Length\; contraction\), being \(\gamma>1\)

- anonymous

Suppose 2 events are recorded in the same position in their respective frames by 2 observers, one frame at rest w.r.t. the earth(S) and the other frame moving at a velocity v in the positive x direction w.r.t. the earth(S')
The time interval between the events occurring at time t1 and t2 for the S frame and time t1' and t2' for frame S' are thus given as
\[\Delta t=t_{2}-t_{1}\]\[\Delta t'=t_{2}'-t_{1}'\]
But for an observer in frame S he observes the times t1 and t2 for the events occurring in frame S' according to inverse lorentz transformation as
\[t_{2}=\gamma(t_{2}'+\frac{vx'}{c^2}) \space \space ; \space \space t_{1}=\gamma(t_{1}'+\frac{vx'}{c^2})\]
\[\therefore \Delta t=\gamma(t_{2}'+\frac{vx}{c^2}-t_{1}'-\frac{vx}{c^2})=\gamma(t_{2}'-t_{1}')=\gamma \Delta t'\]
Thus
\[\Delta t'=\frac{1}{\gamma} \Delta t\]

- anonymous

Why have you used lorentz instead of inverse lorentz when u need to find length in frame S?

- Michele_Laino

because I have to measure the length in \(S\) and I have to make the measure of position of the ends of the rod at the same tame \(t\)

- Michele_Laino

same \(t\), not same \(t'\)

- anonymous

Ohk, since we are measuring in frame S, we require the same time in frame S. This point was always bothering me, that why do we use the lorentz instead of inverse, if we are calculating for frame S

- Michele_Laino

since \(x'_2\) is a function of \(x_2,t\), similarly for \(x'_1\), which is a function of \(x_1,t\)

- anonymous

Then my derivation for time dilation wrong....If I correct my error I should arrive at the result
\[\Delta t'=\gamma \Delta t\]

- anonymous

is that right?

- Michele_Laino

let's see:
if the real interval time is \(t'_2-t'_1=\Delta t'\), then I have to measure \(Delta t\) in reference \(S\) at the same position in \(S'\), namely \(x'_2=x'_1\)

- Michele_Laino

\(\Delta t\)*

- Michele_Laino

so we can write this:
\[\begin{gathered}
{t_2} = \gamma \left( {t{'_2} - \frac{V}{{{c^2}}}x'} \right) \hfill \\
{t_1} = \gamma \left( {t{'_1} - \frac{V}{{{c^2}}}x'} \right) \hfill \\
\end{gathered} \]
since the two events occurred at the same position \(x'_2=x'_1=x'\)

- Michele_Laino

again, subtracting the first equation, from the second one, we get:
\[\Delta t = \gamma \Delta t'\]
namely the so called \(time\;dilation\), being \(\gamma >1\)

- anonymous

So my derivations r correct !!

- Michele_Laino

the second one is correct!

- anonymous

what's wrong with the first one ??

- Michele_Laino

sorry, also the first derivation is correct!

- anonymous

can u help me with another question

- Michele_Laino

ok!

- anonymous

Prove that the 4-dimensional volume element
\[dxdydzdt\]
Is invariant under Lorentz transformations
From our previous derivations we have
\[L=\gamma L'\]
\[\implies \Delta x=\gamma \Delta x'\]
Taking the limits
\[\Delta x \rightarrow 0 \space \space \space ; \space \space \space \Delta x' \rightarrow 0\]
\[dx=\gamma dx'\]\[\implies dx'=\frac{1}{\gamma }dx\]
Obviously
\[dy'=dy \space \space \space ; \space \space \space dz'=dz\]
and from our time dilation we get
\[\Delta t'=\frac{1}{\gamma}\Delta t\]
Again taking the limits
\[\Delta t \rightarrow 0 \space \space \space ; \space \space \space \Delta t' \rightarrow 0\]\[dt'=\frac{1}{\gamma}dt\]
thus we get
\[dx'dy'dz'dt'=\frac{1}{\gamma^2}dxdydzdt\]
Which is not the same

- Michele_Laino

the natural way to express the invariance of volume, is to consider the fundamental tensor \(g_{ik}\), namely we have to show the truth of a more general statement:
\[\sqrt {g'} dV' = \sqrt g dV\]
where \(\sqrt g\)is the square root of the determinant of the tensor \(g_{ik}\)
Since in special relativity, we have \(g=g'=-1\), then we have showed your statement, namely: \(dV'=dV\)

- Michele_Laino

oops..\(g=g'=1\)

- anonymous

ooh no idea what you talking about, I haven't read about tensors yet, is there an easier way to represent?

- Michele_Laino

I'm sorry, it is the only way, which I know, in order to solve your exercise. Are you familiar with tensor calculus?

- anonymous

Nope not at all, I'm a freshman

- anonymous

I don't believe it is even in our 3 year course

- Michele_Laino

since, what is invariant is the volume of the 4-dimensional space, namely what is invariant is the subsequent quantity:
\[d\tau = \int {dx{\kern 1pt} dy{\kern 1pt} dz{\kern 1pt} cdt} \]

- Michele_Laino

oops...
\[\sqrt g d\tau = \sqrt g \int {dx{\kern 1pt} dy{\kern 1pt} dz{\kern 1pt} cdt} \]

- Michele_Laino

it is a proposition from General Relativity

- anonymous

If you don't know an easier way, that's alright, I can leave this exercise for later, I doubt it will even come in the examination, mostly they want us to do the derivations and a numerical related with it,however I was curious but in the end this problem was only confusing me more to be honest. In the book they have taken
\[dx'=\gamma dx\]\[dy'=dy\]\[dz'=dz\]\[dt'=\frac{dt}{\gamma}\]
if you multiply that out the gamma cancels

- Michele_Laino

I think that it is correct!

- anonymous

The part I don't understand is why
\[dx'=\gamma dx\]

- Michele_Laino

here we have to specify the role of reference system \(S\) and \(S'\) first

- anonymous

Physically I can realize it as the contraction of the length exactly cancelling out the dilation of time, but mathematically I can't interpret the equation of dx'

- Michele_Laino

let's suppose that \(S'\) is moving, with respect to \(S\) with a speed \(V\)

- anonymous

yep!

- Michele_Laino

along a \(x-\)direction, for example

- anonymous

ok

- Michele_Laino

then we can write this:
\[dx = \frac{{dx'}}{\gamma }\]
which is the length contraction

- Michele_Laino

furthermore, we can write this:
\[dt = \gamma dt'\] which is the time dilation

- Michele_Laino

as you can see the gamma's cancel

- Michele_Laino

oops.. I meant gamma factors

- anonymous

shouldn't it be
\[dx'=\frac{dx}{\gamma}\]
Since a length in frame S' will be smaller than S, we observer the length in S' to be a fraction of the length in S

- Michele_Laino

I suppose that I'm in reference \(S\)

- Michele_Laino

so: \(dx'\) is the real length, and \(dx\) is the measured length

- Michele_Laino

similarly for \(dt',dt\)

- anonymous

Then why is the expression for length contraction has the primes interchanged?Is it because I've measured in S and observed in S'? where as if for time im measuring in t' and observing for t, I should respectively measure length in S' and then observe in S?

- anonymous

that would make sense then

- Michele_Laino

I have used terms "observed" and "measured" with the same meaning

- anonymous

Actually I've measured the proper length in frame S in my derivation, that is why my length element is coming different, I should measure the proper length in frame S' because im measuring proper time in frame S'

- Michele_Laino

I have supposed to be in reference \(S\) and then to measure the quantities \(dx',dt'\) in reference \(S\)If you are in reference \S\), then you have to measure length and time in \(S\), namely you have to write what you measure when you are in \(S\)

- Michele_Laino

more precisely: \(dx',dt'\) are the given quantities, and you have to write what are \(dx,dt\) when you are in \(S\)

- anonymous

Yeah, makes sense, i was doing opposite in length contraction, starting with dx and going to dx'

- Michele_Laino

yes! I think so

- anonymous

good thread !! thx

- Michele_Laino

:)

- IrishBoy123

awesome thread!!! thank you both.
@Abhisar
@Astrophysics
@arindameducationusc
@iambatman
@Radar
@Robert136
@rvc
@UnkleRhaukus

- Abhisar

Wow!, that was some nice stuff!

- Astrophysics

Thanks for the tag @IrishBoy123
This is a great thread, thank you!
Special relativity at work :)

- rvc

Thanks @IrishBoy123 :)

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