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Lets say There is a moving board moving to the right in the speed of $V = 0.8c\,\hat x$.

We know that the angle that the pole creates with the Y axis of our system is 31 degrees $(\alpha = \tan^{-1}(0.6) = 31^\circ)$.

I want to find that angle from the system of the board $\Rightarrow\alpha'$

Which Lorentz equation do I use?

(Let's say I don't want to use shortcuts like the length contraction equation, and I want to get to it by using the basic Lorentz transformation).

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Let me see if I understand your question correctly: you have an object (say a rod or board) inclined at some angle $\alpha$ in some reference frame $S$, and you would like the find the angle $\alpha^\prime$ in a reference frame $S^\prime$. Since you've mentioned $\alpha^\prime$ as the angle from the rod's frame, I'm using the convention that the rest frame (i.e., the frame in which the rod is at rest) is $S^\prime$. Let me write down the Lorentz Transformations:

\begin{equation} \begin{aligned} &\text{(A)}\quad\Delta x^\prime = \gamma \left(\Delta x - v \Delta t\right)\\ &\text{(B)}\quad \Delta t^\prime = \gamma \left( \Delta t - \frac{v}{c^2}\Delta x\right)\\ \\ &\text{(C)}\quad\Delta x = \gamma \left(\Delta x^\prime + v \Delta t^\prime \right)\\ &\text{(D)}\quad \Delta t = \gamma \left( \Delta t^\prime + \frac{v}{c^2}\Delta x^\prime \right)\\ \end{aligned} \label{LT} \end{equation}

The question as to which Lorentz Transformation(s) should be used is a good one, I feel, as it's very easy to get confused and use the wrong ones. (Suddenly, you'll find lengths expanding in stead of contracting and so on! Take a look at my answer here.)

As I point out in the above answer,

For an observer sitting in $S^\prime$, since the object is at rest with respect to him, its length $L^\prime$ is simply the difference in the coordinates, irrespective of when $x_B^\prime$ and $x_A^\prime$ are measured. He could measure $x_B^\prime$, have a coffee, and then measure $x_A^\prime$ and the difference would give him the length. However, for an observer sitting in $S$, since the object is moving with respect to her, both the endpoints $x_B$ and $x_A$ need to be measured simultaneously in her frame of reference ($S$) in order for the difference to be the length $L$. (In other words, if she has a coffee between measuring $x_B$ and $x_A$, the object would have moved between measurements!) So, we have $$L^\prime = x_B^\prime - x_A^\prime |_\text{ for any $\Delta t^\prime$}$$ $$L = x_B - x_A |_\text{ only when $\Delta t=0$}$$

Now, let's try to answer your question. You are interested in relating the angle $\alpha^\prime$ with the angle $\alpha$. From trigonometry, it's clear that

$$\tan{\alpha} = \frac{L_x}{L_y} \quad\quad \text{ and } \quad \tan{\alpha^\prime} = \frac{L_x^\prime}{L_y^\prime}.$$

Of course, since the direction of motion is only along $x$, $L_y = L_y^\prime$.

Now, all we need to do is to relate the lengths $L_x$ and $L_x^\prime$. As I've already pointed out, this means that we need to find a relation between $\Delta x$ and $ \Delta x^\prime$, when $\Delta t=0$, since $L_x = \Delta x$ if (and only if) $\Delta t =0$, since the rod is moving in the frame $S$ and therefore its endpoints have to be measured simultaneously.

So, we ask ourselves, which Lorentz Transformation relates $\Delta x, \Delta x^\prime$, and $\Delta t$? The answer is, of course, (A) Remember, while $\Delta t=0$, we are saying nothing about $\Delta t^\prime$. It turns out that $\Delta t^\prime$ is not zero! This is why it isn't helpful to use (C) for example, since we'd first have to find $\Delta t^\prime$. So,

\begin{equation*} \begin{aligned} \Delta x^\prime &= \gamma \left(\Delta x - v \Delta t\right)\\ \Delta x^\prime|_{\Delta t = 0} &= \gamma \left(\Delta x|_{\Delta t =0} - v \Delta t|_{\Delta t = 0}\right)\\ \\ L_x^\prime &= \gamma L_x \end{aligned} \end{equation*}

Plugging this into our trigonometric identity, we see that

$$\tan{\alpha^\prime} = \frac{L_x^\prime}{L_y^\prime} = \frac{\gamma L_x}{L_y} = \gamma \tan{\alpha}.$$

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  • $\begingroup$ Your answer was very helpful! thank you for that, but now I have another question, which relates to the first one. Lets say that now everything is dark around us, and our reference frame is the camera. We send a flash of light at any time t1. The light gets to the rod when it's left edge (of the rod) correlates with the camera's left edge, and then the light gets back to the camera. What is the length of the "stain" (the picture of the rod) that we get from the camera? $\endgroup$ Commented Jan 6, 2020 at 12:51
  • $\begingroup$ Sounds like quite a different question, I think I was given almost exactly the same question in an examination (only it involved a photographic plate ;) ). I'd suggest you open a new question, and show what you've managed to do and where you're getting stuck :) $\endgroup$
    – Philip
    Commented Jan 6, 2020 at 14:26

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