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anonymous
 one year ago
check my work
anonymous
 one year ago
check my work

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Consider 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 x2x1, 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
 one year ago
Best ResponseYou've already chosen the best response.0Another assumption that is made is that the coordinates of both the end points of the rod are measured simultaneously in both frames

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0SR, excellent! will look later... @Michele_Laino

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3let's suppose the rod lies on reference \(S'\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so its proper length is \(L'\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3namely I mean its real length is \(L'\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3now, 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
 one year ago
Best ResponseYou've already chosen the best response.3of course, reference \(S'\) is moving, with respect to \(S\), with speed \(V\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so, 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
 one year ago
Best ResponseYou've already chosen the best response.3where \(L_O\) 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
 one year ago
Best ResponseYou've already chosen the best response.0Suppose 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
 one year ago
Best ResponseYou've already chosen the best response.0Why have you used lorentz instead of inverse lorentz when u need to find length in frame S?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3because 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
 one year ago
Best ResponseYou've already chosen the best response.3same \(t\), not same \(t'\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohk, 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
 one year ago
Best ResponseYou've already chosen the best response.3since \(x'_2\) is a function of \(x_2,t\), similarly for \(x'_1\), which is a function of \(x_1,t\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then my derivation for time dilation wrong....If I correct my error I should arrive at the result \[\Delta t'=\gamma \Delta t\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3let's see: if the real interval time is \(t'_2t'_1=\Delta t'\), then I have to measure \(Delta t\) in reference \(S\) at the same position in \(S'\), namely \(x'_2=x'_1\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so 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
 one year ago
Best ResponseYou've already chosen the best response.3again, 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
 one year ago
Best ResponseYou've already chosen the best response.0So my derivations r correct !!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3the second one is correct!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what's wrong with the first one ??

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3sorry, also the first derivation is correct!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can u help me with another question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Prove that the 4dimensional 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
 one year ago
Best ResponseYou've already chosen the best response.3the 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
 one year ago
Best ResponseYou've already chosen the best response.3oops..\(g=g'=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ooh no idea what you talking about, I haven't read about tensors yet, is there an easier way to represent?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I'm sorry, it is the only way, which I know, in order to solve your exercise. Are you familiar with tensor calculus?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nope not at all, I'm a freshman

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't believe it is even in our 3 year course

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3since, what is invariant is the volume of the 4dimensional space, namely what is invariant is the subsequent quantity: \[d\tau = \int {dx{\kern 1pt} dy{\kern 1pt} dz{\kern 1pt} cdt} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3oops... \[\sqrt g d\tau = \sqrt g \int {dx{\kern 1pt} dy{\kern 1pt} dz{\kern 1pt} cdt} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3it is a proposition from General Relativity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.3I think that it is correct!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The part I don't understand is why \[dx'=\gamma dx\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3here we have to specify the role of reference system \(S\) and \(S'\) first

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Physically 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
 one year ago
Best ResponseYou've already chosen the best response.3let's suppose that \(S'\) is moving, with respect to \(S\) with a speed \(V\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3along a \(x\)direction, for example

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3then we can write this: \[dx = \frac{{dx'}}{\gamma }\] which is the length contraction

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3furthermore, we can write this: \[dt = \gamma dt'\] which is the time dilation

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3as you can see the gamma's cancel

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3oops.. I meant gamma factors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0shouldn'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
 one year ago
Best ResponseYou've already chosen the best response.3I suppose that I'm in reference \(S\)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so: \(dx'\) is the real length, and \(dx\) is the measured length

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3similarly for \(dt',dt\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Then 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
 one year ago
Best ResponseYou've already chosen the best response.0that would make sense then

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I have used terms "observed" and "measured" with the same meaning

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Actually 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
 one year ago
Best ResponseYou've already chosen the best response.3I 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
 one year ago
Best ResponseYou've already chosen the best response.3more precisely: \(dx',dt'\) are the given quantities, and you have to write what are \(dx,dt\) when you are in \(S\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, makes sense, i was doing opposite in length contraction, starting with dx and going to dx'

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0awesome thread!!! thank you both. @Abhisar @Astrophysics @arindameducationusc @iambatman @Radar @Robert136 @rvc @UnkleRhaukus

Abhisar
 one year ago
Best ResponseYou've already chosen the best response.0Wow!, that was some nice stuff!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the tag @IrishBoy123 This is a great thread, thank you! Special relativity at work :)
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