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2nd thought experiment

05102009, 08:13 AM #1
2nd thought experiment
Here's Einstein's second experiment, and I've read different versions.
Again two observers, one on a train, one on the ground. The train observer drops a rock and watches it fall. He observes a rock dropping straight down as it is released, but it will hit the ground a few feet farther down the track.
The ground observer watches the same action. Some have stated that the ground observer will see a parabola, in which the rock will descend sideways as it descends downward.
However, even Einstein didn't say this. Instead, he drew a graph in which the space and time(movement in space) were stationary.
By making both space and time static, Einstein was able to trace a parabolic path across the graph. The "movement", combining space and time, resembled a "wave".
But, in actual experience, this will not be so from the observer's perspective.
Simply reverse the conditions, and instead of a train moving from left to right before the ground observer, suppose it is the ground that is made to move from right to left, and the train is stationary.
The person on the train then drops the stone. The ground observer will still see the same stone dropping down, and hitting the ground at the same position as before. The only difference is that the ground has been made to move at the same speed as the earlier experiment, while the train is now stationary. Instead of the train moving from left to right in front of the ground observer, the ground is moving right to left, making exactly the same results.
Einstein's graph will still create the same parabola by treating either position as static, with the path of the falling rock represented as a parabola or kind of half wave.
The path of the rock terminates at the "bottom" of the parabola.
This occurs, not by observation of actual experience, but by treating time and space(position and velocity) as static.

05102009, 03:44 PM #2
Re: 2nd thought experiment
Don't suppose you have a link to somewhere where this is explained with diagrams? I'm not sure I follow just from your explanation.

05102009, 06:47 PM #3
Re: 2nd thought experiment
Not really. Just the book written by Einstein.
I arrived at my conclusion about the fall of the rock from the train simply by imagining the train moving from left to right first, and then the ground moving under the train while the train was stationary, with the ground moving from right to left at the same speed. The rock would fall straight down in either case, landing at the exact spot assuming the train or ground was moving at the same speed in each thought experiment.
The thing that keeps eating at the back of my mind is your explanation of an electron traveling as a wave.
The more acceleration applied to the train(energy), the longer the slope of the parabola. The slower the movement of the train, the shorter the curve of the parabola.
But the parabola exists only if you treat both velocity and position as static in relation to each other, and draw the parabola on a graph as a combination of both position and velocity(position of stationary observer, velocity of moving observer).
While this is a kind of wave function(parabola), the position of the rock is selected by examining the trajectory of the parabola as a combination of both velocity and position.

05102009, 07:39 PM #4
Re: 2nd thought experiment
Ok, I think I see what you're getting at, but the path of the rock and the electron wavefunction are unrelated concepts. They just happen to represent each other superficially.
While the path that the rock takes is kind of wave like, this is very different to the sort of wave that we mean when we talk about electrons.
If you froze time at any point during the rock's motion, the rock would only occupy one place. It's either in my hand, or it's somewhere between my hand and the floor, or it's on the floor. No matter what, though, you can say that the rock is X meters from the floor, and Y meters from my hand, and Z meters from the carriage door etc...
Now, if you were to freeze time and look at a wave about to break on the beach, you can't say the same thing. It doesn't have any particular location, it's spread out across a large area, and if you imagined and infinitely long beach, then the wave would have an infinite expanse.
In the case of water breaking on the beach, the wave is spread out in only one direction, the direction parallel to the beach. An electron, though, is spread out in 3 dimensions, so if you were to ask "where is it", there's no definite answer. It could be anywhere, though some places are more statistically likely than others.
Anyway, the idea that an electron is a wave is a quantum mechanical one, and Einstein's thought experiments on the trains are relativistic ones. Quantum Mechanics and Relativity are difficult to combine into one consistent paradigm  it's not something you do until the final year of a physics degree, and some uni's don't cover it at all.

05112009, 06:34 AM #5

05112009, 06:53 AM #6
Re: 2nd thought experiment
While your explanation makes sense, and i agree as far as I can understand it, I'm still bothered by Einstein's idea of conflating space and time into a static graph to show a parabola.
If you envision a graph with the vertical side showing the fall of the rock, and the horizontal side showing the "direction" of that fall in time, you will see an arc, with the slope being sharp or gradual, depending on the acceleration of the train. If slow, the arc appears to fall nearly vertical, if very fast, the rock appears to fall in a sweeping arc toward the right, assuming the train is traveling in that direction.
So, while we can pinpoint the location of the rock at any point in its arc of fall, there is still a problem: the rock never fell in an arc. It fell straight down from its own referential frame.
This is shown simply by reversing the experiment from a moving train to a moving surface under the stationary train.
In fact, no "arc" or parabola has occurred. It is invented by describing an imaginary path that occurred by combining time and space into one graph.
This means that the "wave" appearance can only occur as a result of the combination of space and time, as a dimension imposed graphically on a physical event.
In reality, the rock never fell in an arcing motion. It simply fell down to a moving surface under it. In either case, the rock behaved as if the train were not moving at all.
A stationary observer would see this, the same as the train observer. No relative frames of reference.
Einstein "fudged" on this one.

05112009, 06:16 PM #7
Re: 2nd thought experiment
Oh, ok, I see. Did the rock fall straight down, or did it fall in an arc?
Both explanations are equally correct. Each person sees something different, and the whole point of the assumptions behind special relativity are that neither person's point of view is more valid than anothers. Both descriptions are true in their own frame of reference.
There is one thing that you mixed up though. The rock didn't fall straight down in it's own reference frame, it fell straight down in the reference frame of someone that is standing still with resepect to the ground.
In the rock's own reference frame, it didn't move at all. It stayed still while the ground rushed up towards it.
This is really getting into general relativity which is much more complicated, so I hope it doesn't get confusing. Perhaps I could rephrase the thought experiment in terms of an example that doesn't depend on gravity, but I need to get some work done right now, so when I've got some time...

05122009, 08:36 AM #8

05132009, 08:13 AM #9
Re: 2nd thought experiment
Theres one point that troubles me. The description of the rock falling in the pattern of an arc. This is incorrect. All falling objects(in a vacuum) fall in a parabola.
An arc would be a path equidistant from one point. A complete arc is a circle. If the rock were falling in an arc and the ground didn't exist, it would swing around and reach the same spot again.
The rock will have a constant x motion and a changing y motion, on earth 9.8 meters per second squared.
So, y = 9.8 x^2
Of course since you are holding the rock in your hand, the vertical motoin starts at zero. If you draw a parabola and turn it upside down, the rock will follow the path from the flat spot downward.
The person on the ground sees it stretched out very very wide, and the person on the train sees it squshed into infinity (squared acceleration). Both can still be expressed as parabola's, but the person on the train will not see movement on the horizontal.
Does this sound right?

05132009, 11:32 AM #10
Re: 2nd thought experiment
An arc would be a path equidistant from one point. A complete arc is a circle. If the rock were falling in an arc and the ground didn't exist, it would swing around and reach the same spot againLast edited by doojie; 05132009 at 11:34 AM.

05132009, 01:13 PM #11

05142009, 11:06 AM #12
Re: 2nd thought experiment
Are you saying all things fall in a parabola from a moving position?
If, at a much slower speed, the ground observer let his eyes follow the man on the train as he drops the rock, the man on the ground would simply see a rock falling straight down, but it would hit the ground a few feet up or down the track.
But if the ground observer were somehow moving and the train were standing still, the ground observer would still see exactly the same falling path of the rock, assuming the speed of the ground and everything were repeated.
The inertial frame of reference here is what is a bit confusing, since the train, moving at a steady rate, would provide an inertial reference, but the rock would have the force of gravity(F=MA) acting on it as it fell.
Nothing would affect the inertial frame. And gravity would act the same whether the train was moving or not.
I'm using what Kazza calls a "colloquial" reference, because there is no parabola except in the graph which combines train movement with gravity.
The fall of the rock on the graph would represent a series of "digital" positions as the train moved across the graph, but the "digital" progression of dots would be connected by a linear representation(analog) of the rock's fall. This would take the shape of a parabola.
It seems to me that the parabola is a kind of analog graphing of a digital process in time, a combination of time and space.

05192009, 08:12 AM #13
Re: 2nd thought experiment
[QUOTE=doojie;755925]Are you saying all things fall in a parabola from a moving position?
[QUOTE]
You are right, the fall from gravity is F=MA
Since the mass is constant, and the speed sideways is constant, then A will be 9.8 meters/second squared.(metric)
It's easier if you graph distance vs time, but any object which is accelerating at a square of time will move in a parabola. Since the horizontal motion will be constant, it is a nice analog to time passing, so if you trace the fall path, the horizontal movement will track evenly with time and the vertical movement will show the vertical displacement. The rock will travel in a parabola every time.
Of course all this has to happen in a vacuum or air friction gets into it... And the gravity will have to always come from the same direction, which is hard to do at light speed.
Also, any object that is thrown in the air shows more of a parabola. The peak of the throw will be the top of the parabola, and the other side should match the first exactly (assuming no air resistance)

05202009, 09:47 AM #14
Re: 2nd thought experiment
[quote=Lord_jag;757320][quote=doojie;755925]Are you saying all things fall in a parabola from a moving position?
You are right, the fall from gravity is F=MA
Since the mass is constant, and the speed sideways is constant, then A will be 9.8 meters/second squared.(metric)
It's easier if you graph distance vs time, but any object which is accelerating at a square of time will move in a parabola. Since the horizontal motion will be constant, it is a nice analog to time passing, so if you trace the fall path, the horizontal movement will track evenly with time and the vertical movement will show the vertical displacement. The rock will travel in a parabola every time.
Of course all this has to happen in a vacuum or air friction gets into it... And the gravity will have to always come from the same direction, which is hard to do at light speed.
Also, any object that is thrown in the air shows more of a parabola. The peak of the throw will be the top of the parabola, and the other side should match the first exactly (assuming no air resistance)
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