anonymous
  • anonymous
check my work
Physics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • 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
  • IrishBoy123
SR, excellent! will look later... @Michele_Laino

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anonymous
  • anonymous
ok thanks:)
Michele_Laino
  • Michele_Laino
let's suppose the rod lies on reference \(S'\)
Michele_Laino
  • Michele_Laino
so its proper length is \(L'\)
Michele_Laino
  • Michele_Laino
namely I mean its real length is \(L'\)
Michele_Laino
  • 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
  • Michele_Laino
of course, reference \(S'\) is moving, with respect to \(S\), with speed \(V\)
Michele_Laino
  • 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
  • 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
  • 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
  • anonymous
Why have you used lorentz instead of inverse lorentz when u need to find length in frame S?
Michele_Laino
  • 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
  • Michele_Laino
same \(t\), not same \(t'\)
anonymous
  • 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
  • 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
  • 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
  • anonymous
is that right?
Michele_Laino
  • 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
  • Michele_Laino
\(\Delta t\)*
Michele_Laino
  • 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
  • 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
  • anonymous
So my derivations r correct !!
Michele_Laino
  • Michele_Laino
the second one is correct!
anonymous
  • anonymous
what's wrong with the first one ??
Michele_Laino
  • Michele_Laino
sorry, also the first derivation is correct!
anonymous
  • anonymous
can u help me with another question
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • 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
  • 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
  • Michele_Laino
oops..\(g=g'=1\)
anonymous
  • anonymous
ooh no idea what you talking about, I haven't read about tensors yet, is there an easier way to represent?
Michele_Laino
  • 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
  • anonymous
Nope not at all, I'm a freshman
anonymous
  • anonymous
I don't believe it is even in our 3 year course
Michele_Laino
  • 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
  • Michele_Laino
oops... \[\sqrt g d\tau = \sqrt g \int {dx{\kern 1pt} dy{\kern 1pt} dz{\kern 1pt} cdt} \]
Michele_Laino
  • Michele_Laino
it is a proposition from General Relativity
anonymous
  • 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
  • Michele_Laino
I think that it is correct!
anonymous
  • anonymous
The part I don't understand is why \[dx'=\gamma dx\]
Michele_Laino
  • Michele_Laino
here we have to specify the role of reference system \(S\) and \(S'\) first
anonymous
  • 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
  • Michele_Laino
let's suppose that \(S'\) is moving, with respect to \(S\) with a speed \(V\)
anonymous
  • anonymous
yep!
Michele_Laino
  • Michele_Laino
along a \(x-\)direction, for example
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
then we can write this: \[dx = \frac{{dx'}}{\gamma }\] which is the length contraction
Michele_Laino
  • Michele_Laino
furthermore, we can write this: \[dt = \gamma dt'\] which is the time dilation
Michele_Laino
  • Michele_Laino
as you can see the gamma's cancel
Michele_Laino
  • Michele_Laino
oops.. I meant gamma factors
anonymous
  • 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
  • Michele_Laino
I suppose that I'm in reference \(S\)
Michele_Laino
  • Michele_Laino
so: \(dx'\) is the real length, and \(dx\) is the measured length
Michele_Laino
  • Michele_Laino
similarly for \(dt',dt\)
anonymous
  • 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
  • anonymous
that would make sense then
Michele_Laino
  • Michele_Laino
I have used terms "observed" and "measured" with the same meaning
anonymous
  • 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
  • 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
  • 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
  • anonymous
Yeah, makes sense, i was doing opposite in length contraction, starting with dx and going to dx'
Michele_Laino
  • Michele_Laino
yes! I think so
anonymous
  • anonymous
good thread !! thx
Michele_Laino
  • Michele_Laino
:)
IrishBoy123
  • IrishBoy123
awesome thread!!! thank you both. @Abhisar @Astrophysics @arindameducationusc @iambatman @Radar @Robert136 @rvc @UnkleRhaukus
Abhisar
  • Abhisar
Wow!, that was some nice stuff!
Astrophysics
  • Astrophysics
Thanks for the tag @IrishBoy123 This is a great thread, thank you! Special relativity at work :)
rvc
  • rvc
Thanks @IrishBoy123 :)

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