Basic Special Relativity Question

I will.

 

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Good luck.

 

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Thanks, you too.

 

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Oh, are we talking about the train frame? My mistake. I agree that in the train frame, the rod is parallel to the floor of the train car, so it will fall through the slit and land parallel with the platform. In the platform frame, the rod is at an angle to the train floor, so by extension, it is at an angle to the slit and will land at an angle to the platform.

While we're addressing this question, let's not forgot the two other questions to which I am eagerly waiting your reply:

1. What is the correct Lorentz transform from \(S'\) to \(S''\)?

2. Without referencing \(S''\) or Pete's/my incorrect use of the Lorentz transform, what is unphysical about \(S'\)?

 

Took you a while.

No, it isn't.

You can keep waiting, you are using the standard crackpot approach, you ask a question, you get the answer repeatedly, it disagrees with your prejudices and you pretend that you didn't get an answer.

 

I have kept my word. I am looking forward to seeing Tach's solution to the puzzle.

 

I disagree. Prove me wrong, with math.

Quote any sentence in this thread answering either of my questions. You say I got the answer "repeatedly", so it should be really easy for you to find one answer to each question. To reiterate, the questions are:

1. What is the correct Lorentz transform from \(S'\) to \(S''\)?

2. Without referencing or Pete's/my incorrect use of the Lorentz transform, what is unphysical about \(S'\)?

 

The height of the train car is \(h\)
The diameter of the wheels is \(d\)
The rod "disengages" from the roof of the train car at \(\tau=0\) in the proper frame of the rod, \(S\)
The rod passes through the slit at \(\tau=\frac{h}{u}\), all points simultaneously.
The rod hits the platform at \(\tau=\frac{h+d}{u}\), all points simultaneously.
The rod collides at a zero angle (all points make contact simultaneously) with both the car floor and the platform.

 

Yes you had, you'll have to wait a little longer , both Fednis48 and Pete are still struggling with accepting that the rod makes a zero angle with both the platform and the car floor. On the upside, if you continue to keep quiet, I promise that the solution covers the most general case, the one that includes the gravitational field and the original release mechanism, i.e. it is a full GR problem, as stated by the OP in the other thread, not the dumbed down version "invented" in this thread.

 

You state that "The rod hits the platform at \(\tau=\frac{h+d}{u}\), all points simultaneously." with no justification or explanation. It does not follow, mathematically or logically, from any of the previous steps.

Also, until you either answer my questions or quote any post in this thread where you answered them, I will continue to post them to keep them from getting lost.

1. What is the correct Lorentz transform from \(S'\) to \(S''\)?

2. Without referencing or Pete's/my incorrect use of the Lorentz transform, what is unphysical about \(S'\)?

 

The rod makes a zero angle with the train floor in the rod frame. Therefore the rod makes a zero angle with the train floor in the train car frame.
The rod makes a zero angle with the platform in the rod frame. Therefore the rod make a zero angle with the platform in the platform frame.

 

LOL, you are lowering yourself with every step. How much further do you plan to take this charade?

 

I've written a response to this argument, but I'd like to reach an agreement on your first argument before addressing your second.

Are we agreed that the OP description of the rod's motion in \(S'\) is physically valid?
Are we agreed that a proper application of the Lorentz Transform to the \(S'\) equations of motion will give the correct equations of motion in \(S''\)?

 

Sorry for editing my post, Tach. The moment I submitted the post, I realized my reasoning implied something weird, and I wanted to check my work before I committed to anything. I've since done a sanity check, so I'm confident enough to move forward again. In doing so, I reiterate my original statement:

As you state, this calculation proves that the rod hits the ground horizontally in its own frame. I will also concede that the rod hits the ground horizontally in the train frame. It does not follow that the rod hits the ground horizontally in the platform frame.

For a change, you make a compelling reply, which mirrors my own uncertainties:

It does seem like, even if parallel-ness isn't universally conserved, two parallel vectors should be parallel in each other's reference frames. But this turns out to be untrue. To see why in a little more detail, check the section labeled "calculation" at the end of this post. But for now, suffice it to say that the rod is parallel with the platform in the rod frame, but the rod is not parallel with the platform in the platform frame. As weird as that is, it's not a contradiction in special relativity.

1. What is the correct Lorentz transform from \(S'\) to \(S''\)?

2. Without referencing or Pete's/my incorrect use of the Lorentz transform, what is unphysical about \(S'\)?

Calculation:

Consider four points, with starting frame coordinates in the form \((t,x,y)\) as follows:

\(\begin{align}A&=(t,-1,10-10t)\\ \alpha&=(t,1,10-10t)\\ B&=(t,9-10t,0)\\ \beta&=(t,11-10t,0)\end{align}\)

In words, there are two pairs of points, each separated horizontally by a distance of \(2\). \(A\) and \(\alpha\) start at \(y=10\) and approach the origin vertically, while \(B\) and \(\beta\) start at \(x=10\) and approach the origin horizontally. In the original frame, \(A\) meets \(B\) and \(\alpha\) meets \(\beta\) at \(t=1\). Calculating the coordinates in the reference frames that move with \(A\) and \(B\) is an interesting exercise, but for now I'll just give the quick version that summarizes the results relevant here.

For a Lorentz boost in the \(x(y)\) direction, \(t'\) depends on both \(t\) and \(x(y)\). When we boost into the \(A\) frame, we are moving vertically. The only difference between \(A(B)\) and \(\alpha(\beta)\) is \(x\), which doesn't affect anything in a vertically boosted frame, so \(A(B)\) and \(\alpha(\beta)\) are still at the same \(y\). Since both pairs of points still have equal \(y\), they are still parallel.

When we boost into the \(B\) frame, we are moving horizontally, so time in the boosted frame depends on both \(t\) and \(x\) in the original frame. Both pairs of points are separated in \(x\), so \(t'\) for \(A(B)\) are different from \(t'\) for \(\alpha(\beta)\). For \(B\) and \(\beta\), \(y=0\) independent of time, so there is still no difference. But for \(A\) and \(\alpha\), \(y\) depends on \(t\), so when we substitute in the \(x\)-dependent expression for \(t'\), the result is an \(x\)-dependent expression for \(y'\). This means the vector from \(A\) to \(\alpha\) at a fixed \(t'\) is no longer horizontal. In other words, the two vectors are parallel in one of the vectors' rest frames, but not the other! Sorry if this post is needlessly long and/or obvious, but I thought it was a cool result.

 

The rod makes a zero angle with the platform in the rod frame. Therefore the rod make a zero angle with the platform in the platform frame. PoR teaches you that the two frames are interchangeable, you cannot get differing answers depending on frame.
There cannot be contradiction between the results in the two frames. If you get a contradicting result in your calculation , your calculation is contradicting the principle of relativity (PoR), so, try finding your mistake. (Actually, you do not have a calculation, you have a long winded story that attempts to rehash Pete's calculation). The fact that your reasoning, if true, contradicts PoR, is the kiss of death to your conclusion.

 

I was surprised by my result myself. If you can find a reference saying that "If vectors \(A\) and \(B\) are parallel in the rest frame of \(A\), they are parallel in the rest frame of \(B\)", I would love to see it. Not only would it confirm that I made a mistake, I'm sure it would help me find it, because it would probably include some math showing the correct way to do Lorentz transforms between the two vectors' rest frames.

But do you know what else would help me find my mistake? You explaining mathematically where you think the mistake is. In fact, you could directly refute my point just by answering a question I've asked five times now:

1. What is the correct Lorentz transform from \(S'\) to \(S''\)?

The second question isn't directly relevant to the question of whether the rod and platform are parallel, but I'll repost it just the same because I would love an answer:

2. Without referencing \(S''\) or Pete's/my incorrect use of the Lorentz transform, what is unphysical about \(S'\)?

 

You should be surprised by your result since it violates the principle of relativity, you are getting two frames to disagree on a null scalar (the zero angle).
What happens is the following:

Imagine that you set a string of synchronized clocks on the falling rod, another string on the floor of the car and a third one on the platform.
The rod passes through the slit at a zero angle, both the clocks on the rod and the ones on the car floor show simultaneous contact. This is because there is no relative motion between the rod and the car floor in the x direction.
The rod contacts the platform with all points at the same time in the frame of the rod. But in the frame of the platform, all the clocks in the string show different time of contact (because there is desynchronization in the x direction, the direction of relative motion between the rod and the platform). The fact that the platform clocks show different coordinate time does NOT change the physics of the contact, the contact is the same in both frames (otherwise you get your PoR violation).

The above underlines the pitfalls of using coordinate-dependent time (in place of the proper time of the rod frame). Every time you use coordinate-dependent solutions, as you and Pete have stubbornly stuck to, you open yourselves to unphysical results.

I think that you can stop trolling now, you have received your answer.

 

I think you're proving my point, actually. Like you said, because of desynchronization, the clocks in the platform frame \(S''\) will never agree with each other at a given \(t''\). But experiments have to produce the same results in all reference frames, so the clocks should still give the same time of impact. The only way to reconcile these statements is that the clocks hit the platform at different \(t''\).

You are lying. You have never answered either question. Prove me wrong by quoting an answer to either question.

 

Any white flags yet?

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Unfortunately, there won't be any. The loser will just slowly slink away.


It's part of the Sf culture.

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