Special Relativity. Lorentz Boosts.

[binbagsss]

The question is given a straight rod parallel to the x-axis moving in the y direction at speed u, to find the angle between the rod and x'axis in R'. (the chosen method to solve being to write equations for the endpoints and transform them to R')

I am looking at my book's solution and don't understand some parts.

( I think these are stemming from my physical intuition as to what is producing a non-zero angle in R' *, essentially how the y components are transformed , and how the principle of measuring the endpoints simultaneously should be tied in , and in which frame R or R' should measuring the endpoints be simultaneous. )


Here's my method:

So if i say the rod has length l, the endpoints in R are given by (x1,y1,z1)= (0,ut1,0) and (x2,y2,z2)= (l,ut2,0), and in R' are given by (x1',y2',z2')=(0',ut1,0) and(l',ut2,0).

Where 0'= γ(0-vt1)=-vγt1 [1] and l'= γ(l-vt2) [2], so this length contraction part is fine.

Okay and t1'=γ(t1-0)= γt1 and t2'=(t2-vl/c^2)

y1'=ut1= [3]
y2'=ut2 [4]

So in terms of t':

t1=t1'/γ
t2=t2'/γ +vl/c^2

So using these in [1],[2],[3], and [4] :

θ=(U/γ(t2'-t1')+uvl/c^2 )/(-v(t2'+t1')+l/γ)

Now the book then gives the final answer as uvγ/c^2, which implies that t1'=t2'-(so I conclude the measurement is simultaneous in R')

But my question is the rod is parallel to the x-axis in R, so θ=0, => Δy=0=u(t2-t1) s.t t2=t1 (measurement also simultaneous in R). But if I simply look at Δy' = u(t2-t1) also as frame is moving in the x direction Δy=Δy', so why doesn't this instantly imply θ'=0?

Last Question
I don't really understand why Δy=Δy'? * So the argument is that because motion is in the x direction, so I understand if the y co-ordinates were not time dependent. But Δy=uΔt , so I would have thought that Δy'=uΔt' , where Δt' is given by the normal Lorentz Boost - physically the reason being that time runs differently in both frames...?

Many Thanks to anyone who can shed some light on this.

 

[voko]

You have not explained what R and R' are.

 

[vanhees71]

R and R' are simply the two inertial reference frames.

To understand the Lorentz-transformation properties of the rod, it's better to work with four-vectors, e.g., the endpoints in R are described by the four vectors
[tex]x_1=\begin{pmatrix}c t \\ 0 \\ u t \\ 0\end{pmatrix}, \quad x_2=\begin{pmatrix}c t \\ l \\ u t \\ 0 \end{pmatrix}.[/tex]
These you transform to R' using a Lorentz boost in [itex]x[/itex] direction. An observer in R' again measures all distances and angles by reading off the coordinates of the end points at the same time t' in his frame of reference. This gives rise to both the length contraction and the non-zero angle wrt. to the [itex]x'[/itex] axis for the observer in R'.

 

[binbagsss]

Okay so in frame R' the coordinates are read off at the same time t', but in R at t.
So I think my confusion lies in the ,perhaps, notation used by the book for t.
As motion is in x direction y remains unchanged, the book denotes these as ut=y1=y2=y1'=y2'.
So simply by looking at θ'=Δy=Δx=y2'-y1'/x2'-x1', from the above would you not immediately deduce θ'=0.
Looking at the rest of the solution I can see that when we express t in terms of t', a difference in t' for each end point arises as t' in the lorentz boost contains x in the formula, which is 0 and l respectively.
So this is why I would go for t1' and t2' instead?

Thanks in advance.

 

[Nickie]

I can understand your confusion and I will try my best to provide a clear explanation. Special relativity is a fundamental theory in physics that describes the relationship between space and time. It is based on two postulates: the laws of physics are the same for all observers in uniform motion, and the speed of light in a vacuum is constant for all observers regardless of their relative motion.

In this scenario, we have two frames of reference: R and R'. R is the rest frame, where the rod is at rest and R' is the moving frame, where the rod is moving at a constant velocity u in the y direction. The goal is to find the angle between the rod and the x' axis in R'.

To do this, we need to use the Lorentz transformation equations, which relate the coordinates and time in one frame to the coordinates and time in another frame. In this case, we are transforming from R to R'. The equations you have used in your method are correct, but I believe the confusion lies in the interpretation of these equations.

In the Lorentz transformation equations, t1 and t2 represent the time coordinates in the R frame, while t1' and t2' represent the time coordinates in the R' frame. So when you say t1 = t1'/γ and t2 = t2'/γ + vl/c^2, it means that t1 and t2 are measured at the same time in the R frame, but the corresponding times in the R' frame are t1' and t2' which are not necessarily simultaneous. This is because time runs differently in the two frames, as you have mentioned.

Now, let's look at the expression for θ. As you correctly pointed out, the y coordinates in R and R' are equal, so Δy = Δy'. However, the time coordinates in R and R' are not equal, as they are measured at different times. This is why we cannot simply say that Δy = Δy' implies θ' = 0. Instead, we need to use the Lorentz transformation equations for time to get the correct expression for θ'.

To address your last question, Δy = uΔt and Δy' = uΔt' are both correct, but they represent different things. Δy = uΔt means that in the R frame, the change in y coordinate is equal