# Basic Special Relativity Question

#### Fednis48

Registered Senior Member
This problem has been extracted from this thread because Tach refuses to address it unless it's in a new topic. The question is thus:

In a relativistically moving train, a metal rod is dropped from the ceiling, parallel to the floor. To stay in the realm of special relativity, let's say it's moving down at constant velocity rather than falling under gravity. In the reference frame of a stationary platform watching the train zoom by, does the rod stay parallel to the ground as it falls? The following calculation is Pete's attempt at a solution, and it concludes that the rod is not parallel to the ground in the platform frame.

$$S =$$ rest frame of the rod
$$S' =$$ rest frame of the train
$$S'' =$$ rest frame of the platform

$$v = (0, u)$$ is the velocity of $$S'$$ relative to $$S$$
$$v' = (V, 0)$$ is the velocity of $$S''$$ relative to $$S'$$
$$\gamma = 1/\sqrt{1-u^2/c^2}$$
$$\gamma' = 1/\sqrt{1-V^2/c^2}$$

The rod has two ends, A and B. In the rod frame, we place them at (x,y) = (0,0) and (1,0) respectively.

In $$S$$:
\begin{align} A &= (x_A, \ y_A) \\ &= (0, \ 0) \end{align}
\begin{align} B &= (x_B, \ y_B) \\ &= (1, \ 0) \end{align}
The angle of the rod with the x-axis at time t is:
\begin{align} \tan(\theta) &= \frac{y_B-y_A}{x_B-x_A} \\ &= 0 \end{align}​

In $$S'$$:
\begin{align} A' &= (x_A'(t'), \ y_A'(t')) \\ &= (x_A, \ \frac{y_A}{\gamma} - ut') \\ &= (0, \ -ut') \end{align}
\begin{align} B' &= (x_B'(t'), \ y_B'(t')) \\ &= (x_B, \ \frac{y_B}{\gamma} - ut') \\ &= (1, \ -ut') \end{align}
The angle of the rod with the x-axis at time t' is:
\begin{align} \tan(\theta') &= \frac{y_B'(t') - y_A'(t')}{x_B'(t')-x_A'(t')} \\ &= 0 \end{align}​

In $$S''$$:
\begin{align} A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\ &= \left(\frac{x_A'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \\ &= \left(-Vt'', \ \frac{-ut''}{\gamma'} \right) \end{align}
\begin{align} B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\ &= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \right) \end{align}
The angle of the rod with the x-axis at time t'' is:
\begin{align} \tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\ &= \frac{uV\gamma' / \gamma c^2}{1 / \gamma'} \\ &= \frac{uV\gamma'^2}{\gamma c^2} \end{align}​

Thanks Fednis48.
Tach, I'd particularly like to know whether you think there are any mistakes with the transformation of $$A'$$ to $$A''$$ and $$B'$$ to $$B''$$.

The following calculation is Pete's attempt at a solution, and it concludes that the rod is not parallel to the ground in the platform frame.

Another way to look at it is to consider that if the rod is parallel to the floor, then both ends of the rod must impact the floor simultaneously. And conversely, that if the ends of the rod do not impact the floor simultaneously, then the rod must not have been parallel to the floor.

In the train frame, let both ends of the rod impact the floor simultaneously at time t'. Let one end of the rod impact the floor at location x', and let the other end of the rod impact the floor at location x'+L where L is the proper length of the rod.

By the Lorentz transformation, the platform frame measures the time of impact of one end of the rod to be t_1:
$$t_1 = \gamma(t' + \frac{vx'}{c^2})$$

and the platform frame also measures the time of impact of the other end of the rod to be t_2:
$$t_2 = \gamma(t' + \frac{v(x'+L)}{c^2})$$

In order for the rod to be parallel to the floor, both ends of the rod must rod impact the floor simultaneously, thus t_1 must equal t_2:
$$t_1 = t_2$$

which lets us set the right hand sides of the first wo equation equal to each other:
$$\gamma(t' + \frac{vx'}{c^2}) = \gamma(t' + \frac{v(x'+L)}{c^2})$$

which simplifies to:
$$x' = x'+L$$

which simplifies to:
$$L = 0$$

Demonstrating that, according to the platform frame, the ends of the rod can only impact the floor simultaneously if the length of the rod is zero L=0. Thus, for any rod of non-zero length, the platform frame says the ends of the rod do not impact the floor simultaneously, and therefore such a rod cannot be parallel to the floor, according to the platform frame.

Right, the essence of the problem is in two questions Tach refused to answer in the other thread:

Do the ends of the rod hit the floor simultaneously in the train frame?
Do the ends of the rod hit the floor simultaneously in the platform frame?

Thanks Fednis48.
Tach, I'd particularly like to know whether you think there are any mistakes with the transformation of $$A'$$ to $$A''$$ and $$B'$$ to $$B''$$.

I think it looks solid. The step where you write y'' as $$-ut'$$ is a little odd, since t' isn't well-defined in the platform frame. So there's probably a more precise way to write it. But the conclusion looks right.

I think it looks solid. The step where you write y'' as $$-ut'$$ is a little odd, since t' isn't well-defined in the platform frame. So there's probably a more precise way to write it. But the conclusion looks right.

Yeah, I wasn't really happy with that either. I considered writing t'(x'', t'') to indicate that in S'', t' depends on x'' and t'', but it looked cluttered.

Yeah, I wasn't really happy with that either. I considered writing t'(x'', t'') to indicate that in S'', t' depends on x'' and t'', but it looked cluttered.

Ahh, gotcha. That would look cluttered; the mere fact that you went through it in that more precise way satisfies my puzzlement.

This problem has been extracted from this thread because Tach refuses to address it unless it's in a new topic. The question is thus:

In a relativistically moving train, a metal rod is dropped from the ceiling, parallel to the floor. To stay in the realm of special relativity, let's say it's moving down at constant velocity rather than falling under gravity. In the reference frame of a stationary platform watching the train zoom by, does the rod stay parallel to the ground as it falls? The following calculation is Pete's attempt at a solution, and it concludes that the rod is not parallel to the ground in the platform frame.

$$S =$$ rest frame of the rod
$$S' =$$ rest frame of the train
$$S'' =$$ rest frame of the platform

$$v = (0, u)$$ is the velocity of $$S'$$ relative to $$S$$
$$v' = (V, 0)$$ is the velocity of $$S''$$ relative to $$S'$$
$$\gamma = 1/\sqrt{1-u^2/c^2}$$
$$\gamma' = 1/\sqrt{1-V^2/c^2}$$

The rod has two ends, A and B. In the rod frame, we place them at (x,y) = (0,0) and (1,0) respectively.

In $$S$$:
\begin{align} A &= (x_A, \ y_A) \\ &= (0, \ 0) \end{align}
\begin{align} B &= (x_B, \ y_B) \\ &= (1, \ 0) \end{align}
The angle of the rod with the x-axis at time t is:
\begin{align} \tan(\theta) &= \frac{y_B-y_A}{x_B-x_A} \\ &= 0 \end{align}​

In $$S'$$:
\begin{align} A' &= (x_A'(t'), \ y_A'(t')) \\ &= (x_A, \ \frac{y_A}{\gamma} - ut') \\ &= (0, \ -ut') \end{align}
\begin{align} B' &= (x_B'(t'), \ y_B'(t')) \\ &= (x_B, \ \frac{y_B}{\gamma} - ut') \\ &= (1, \ -ut') \end{align}
The angle of the rod with the x-axis at time t' is:
\begin{align} \tan(\theta') &= \frac{y_B'(t') - y_A'(t')}{x_B'(t')-x_A'(t')} \\ &= 0 \end{align}​

In $$S''$$:
\begin{align} A'' &= \left(x_A''(t''), \ y_A''(t'') \right)\\ &= \left(\frac{x_A'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(-Vt'', \ -u\gamma'(t'' + \frac{Vx_A''}{c^2}) \right ) \\ &= \left(-Vt'', \ \frac{-ut''}{\gamma'} \right) \end{align}
\begin{align} B'' &= \left(x_B''(t''), \ y_B''(t'') \right) \\ &= \left(\frac{x_B'}{\gamma'} - Vt'', \ -ut' \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ -u\gamma'(t'' + \frac{Vx_B''}{c^2}) \right) \\ &= \left(\frac{1}{\gamma'} - Vt'', \ \frac{-ut''}{\gamma'} + \frac{uV\gamma'}{\gamma c^2} \right) \end{align}
The angle of the rod with the x-axis at time t'' is:
\begin{align} \tan(\theta'') &= \frac{y_B''(t'') - y_A''(t'')}{x_B''(t'')-x_A''(t'')} \\ &= \frac{uV\gamma' / \gamma c^2}{1 / \gamma'} \\ &= \frac{uV\gamma'^2}{\gamma c^2} \end{align}​

Basic algebra would give the result $$-\frac{uV\gamma'}{c^2}$$. (not that the physics treatment is correct, either).
You may want to check your math. The fact that Neddy Bate and Fednis48 agree with you result speaks volumes.
Once you figure out where your new error is , we can talk about the rest.

Another way to look at it is to consider that if the rod is parallel to the floor, then both ends of the rod must impact the floor simultaneously. And conversely, that if the ends of the rod do not impact the floor simultaneously, then the rod must not have been parallel to the floor.

In the train frame, let both ends of the rod impact the floor simultaneously at time t'. Let one end of the rod impact the floor at location x', and let the other end of the rod impact the floor at location x'+L where L is the proper length of the rod.

By the Lorentz transformation, the platform frame measures the time of impact of one end of the rod to be t_1:
$$t_1 = \gamma(t' + \frac{vx'}{c^2})$$

and the platform frame also measures the time of impact of the other end of the rod to be t_2:
$$t_2 = \gamma(t' + \frac{v(x'+L)}{c^2})$$

In order for the rod to be parallel to the floor, both ends of the rod must rod impact the floor simultaneously, thus t_1 must equal t_2:
$$t_1 = t_2$$

which lets us set the right hand sides of the first wo equation equal to each other:
$$\gamma(t' + \frac{vx'}{c^2}) = \gamma(t' + \frac{v(x'+L)}{c^2})$$

which simplifies to:
$$x' = x'+L$$

which simplifies to:
$$L = 0$$

Demonstrating that, according to the platform frame, the ends of the rod can only impact the floor simultaneously if the length of the rod is zero L=0. Thus, for any rod of non-zero length, the platform frame says the ends of the rod do not impact the floor simultaneously, and therefore such a rod cannot be parallel to the floor, according to the platform frame.

You can maybe simplify matters somewhat by imagining two separated light sources A and B which are turned on at the same time according to the stationary observer in the middle. However the moving observer who is instantaneously in the middle heading from A to B will claim that B was turned on before A. Put A and B on the ends of a falling rod, and the moving observer will claim that the B end hits the ground first. Flipping it around and putting the rod on the moving train then tells you that the rod is not parallel to the ground in the platform frame.

I'm very happy that you are now on record calling the Lorentz transformations "garbage." Speaks volumes about you.

The only thing I "conceded" in the other thread was that I was neglected gravity. I'm glad you popped into this thread to go on record that you still think gravity must be considered, even though gravity is not part of the problem in this thread.

Yeah! Go Tach!

I'm very happy that you are now on record calling the Lorentz transformations "garbage." Speaks volumes about you.

The only thing I "conceded" in the other thread was that I was neglected gravity. I'm glad you popped into this thread to go on record that you still think gravity must be considered, even though gravity is not part of the problem in this thread.

Yeah! Go Tach!

Careful...Hell hath no fury like a Tach scorned.

Basic algebra would give the result $$-\frac{uV\gamma'}{c^2}$$. (not that the physics treatment is correct, either).
You may want to check your math. The fact that Neddy Bate and Fednis48 agree with you result speaks volumes.
Once you figure out where your new error is , we can talk about the rest.

Edit: My bad, that's not right! There is a $$\gamma'^2$$ in the numerator and a $$\gamma$$ in the denominator. One is primed and the other is not, so they don't cancel. Pete's algebra is right, as far as I can tell. We specifically spun this thread off at your request to get you to point out errors, so if you see any, please identify them precisely ("basic algebra" is not precise enough, because it doesn't indicate which step(s) in the math are mistaken).

Original Post: Good point. Since the question of this thread is basically "does $$\theta=0$$", I was just focused on the numerator of the fraction, but the denominator turns into $$1/\gamma$$, which Pete failed to include. Dunno where you got that overall minus sign, though; I find the result to be $$\frac{uV\gamma'}{c^2}$$. Of course, both of our expressions are still nonzero.

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Careful...Hell hath no fury like a Tach scorned.

LOL, his standard is to spew venom with every post, so how much worse can it get?

LOL, his standard is to spew venom with every post, so how much worse can it get?

I don't wanna find out lol

Edit: My bad, that's not right! There is a $$\gamma'^2$$ in the numerator and a $$\gamma$$ in the denominator.

I understand that you desperately want to debate but your basic algebra skills get in the way. This is very basic math.....

One is primed and the other is not, so they don't cancel. Pete's algebra is right, as far as I can tell.

As far you can tell...you are wrong. Want to try again?

LOL, his standard is to spew venom with every post, so how much worse can it get?

how about you helped the other two guys, they don't get it. Neither do you.

I'm very happy that you are now on record calling the Lorentz transformations "garbage."

No, I called your reasoning, garbage, let's be precise.

Neddy Bate said:
LOL, his standard is to spew venom with every post, so how much worse can it get?

I don't wanna find out lol

See his two posts above this. Not too bad, because he doesn't even bother to demonstrate his claims.

Hi Tach,
Please point out any errors you see, so we can fix them. I'd also like to know what problems you see with the physical treatment.
I don't want to waste time, I just want to find the right answers.