# Relativistic rolling tank

2inquisitive said:
Pete, according to CANGAS's scenario, the length of the track is 20 feet, from centerline of the axel to centerline of the other axel. We can change this to 20 meters to make it easier for you.
Feet is fine. I'd like to focus on making it easy for CANGAS.

(1) the bottom of the tread will contract according to the contraction factor for 0 v; 1.0. Therefore the bottom of the tread will remain 20 meters long.
Not so.
In the ground frame, each part of the tread that is on the ground will be its proper length.
However, less of the tread is on the ground in he ground frame than in the tank frame at any one instant because of the relativity of simultaneity.

We've been through this in detail earlier in the thread, 2inq. -I can see I'm going to have to slow it down and put clocks in for you.

Pete,
Not so.
In the ground frame, each part of the tread that is on the ground will be its proper length.
However, less of the tread is on the ground in he ground frame than in the tank frame at any one instant because of the relativity of simultaneity.
All my clocks are on the ground on purpose. We don't care what the tank clock records, only the amount of time recorded on the synchronized ground clocks as the tank passes.

Fine. In the ground frame, the tank axles are 10 feet apart, so clearly there are 10 feet of track on the ground. This track is not length contracted, so 10 feet is its proper length.

I am using two synchronized clocks located on the ground to determine lengths. One clock will start recording time when a single cleat, while it is located below the centerline of the axel, trips a laser beam projected slightly above the ground and record until that same cleat lifts off the ground. The second clock will start recording time when an axel trips a laser beam as the first axel passes, and then stop when the second axel trips the laser beam a second time. The clocks are sitting on top of each other, and the laser beams are as well, one slightly above the ground and the other at axel-heigth. You are the mathematician, can you tell me how many nanoseconds will be recorded on each clock?
10 feet at 983.6 megafeet per second takes 117 picoseconds.
Same for both clocks.

We could add another laser beam and clock to record the number of cleats passing, and the total time taken, for the top of the tread. Each cleat would simply break the laser beam as it passed. Newton says this clock will also record 7.698 ns and would record 20 cleats, of course.
Same time and 39 cleats. Every cleat except the one on the ground will break this beam.

CANGAS said:
1.732c; contraction factor = HOLY MOLY! shorter than zero length
This didn't tip you off that you were doing something wrong? CANGAS your bias makes you look like an idiot.

-Dale

2inquisitive said:
You are the mathematician, can you tell me how many nanoseconds will be recorded on each clock? I get 7.698 nanoseconds recorded on each clock, assuming a speed of .866c for the tank and Newtonian physics. What does Special Theory predict?
2inquisitive, I have already sloved this starting in the ground frame, simply plug in whatever numbers you like:
DaleSpam said:
I finished the analysis starting from the ground frame. The most difficult part was figuring out a formula to use for the worldline of an arbitrary material point of the track in the ground frame. I finally found a reference on MathWorld that described the Fourier Series expansion of an asymmetric triangle wave. Using that I developed an analytical expression for the point's worldline in the ground frame:
s = {ct, vt + ∑(b sin[k(vt-x0)])}
where
b = 4 (-1)^n c^4 L sin[π n (c^2 + v^2)/(2c^2)]/(n^2 π^2 (c^4 - v^4))
k = n π/(L γ^2)
L is the length of the tank tread (axle spacing) in the ground frame
v is the velocity of the tank in the ground frame
x0 is a term indicating the initial position and direction
and the summation is from n=1 to n=infinity
-Dale

It seems to me that the disagreements are coming about primarily due to an oversimplification of the problem. Please excuse the lack of a complete description here, there are sill some parts of the problem I find too confusing.

Ignore for the moment that the three 'parts' of the tank are interconnected and review the problem, view them as three separate bodies. Neither the body of the tank, the upper track, nor the lower track is in the same relativistic frame of reference as the other two.

If one were to sit at the front of the tank and look backwards the length between the cleats on the top tread would appear to be shorter as they approach at some fraction of the speed of light. Meanwhile the cleats on the bottom track would appear to be further apart as they rush away from the observer at the same speed. But the length of the body of the take would appear as normal since the observer is moving at the same speed (fig 1). If we sit at the back of the tank and look forward we see the reverse effect (fig 2).

Strangest of all, if we sit exactly in the middle of the tank we see both effects on both tracks. In fact, if we viewed the top track the cleats would appear closer together but widen as they moved past the middle and away from us toward the front, appearing to be the normal distance apart at the moment the midpoint between the cleats passes us (fig 3). As we can see in this last example the length of the track would be exactly the same as the length of the tank body and the same length as the track at rest.

(fig 1)
=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=
[][][][][][][][][][][][][][][][][][][][][][][][][][][][][] o
==|==|==|==|==|==|==|==|==|==|==|==|==

(fig 2)
==|==|==|==|==|==|==|==|==|==|==|=
o [][][][][][][][][][][][][][][][][][][][][][][][][]
=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=

(fig 3)
=|=|=|=|=|=|=|=|=|=|=|==|==|==|==|==|==
[][][][][][][][][][][][][][]o[][][][][][][][][][][][][][]
==|==|==|==|==|==|=|=|=|=|=|=|=|=|=|=|=

But no matter the distance between the cleats the length of the track itself would continue to appear the same length as the body of the tank because the length of track we are observing is not moving relative to the body of the tank. Stepping back to the interconnected parts POV just for a moment, because of its shape the tack is, in effect, infinitely long. If we imagine an ideally smooth track with no observable differentiations this becomes more apparent (fig 4).

========================
[][][][][][][][][][]o[][][][][][][][][][][]
========================

Or if we imagine an infinitely long track (never mind the mechanics).

<=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|==|==|==|==|==|==|==|==|==|==|==|==|==|==|=>
----------------------------------------[][][][][][][][]o[][][][][][][][][]
<==|==|==|==|==|==|==|==|==|==|==|==|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=|=>

The next issue is the frame of reference of the ground. Although the tank body is moving at speed relative to the ground the track on the ground is not. In fact, the track on the ground is not moving at all relative to the ground's frame of reference.

Observe the point of view of an individual cleat. It contacts the ground at a specific point and remains there for the entire portion of its journey along the ground. This would also mean that the cleat marks would be the same distance as their resting positions would be.

The lower part of the track will be in the ground's frame of reference (the cleats touching the ground do not move at all relative to the ground), the body of the tank in another frame (moving at speed relative to the ground), and the top of the track in a third frame.

The reference frame of the cleat is also useful in thinking about strain., for while on the top or on the bottom the cleat would measure its distance as constant relative to the cleats before and aft. It is only at the axle, when the cleat is suddenly accelerated in a different direction that it would measure strain between it and the next cleat. Thus the strain is not a consequence of relativistic speed but of acceleration.

It seems to me the tricky part is when the track moves around the axle as the track then needs to accelerate at a different angle and reverse direction. And if we go back to the spinning disk each part of the disk is constantly being accelerated. I'm not sure how to handle that, though it seems to me that the strain would be perfectly distributed. The strain between point A and B might be immense but the strain between each division of that distance would be half the total. One thinks of neighboring atoms in a spinning disk that are being accelerated at almost exactly the same rate and direction.

Anyway, these are my inexpert thoughts. I'll leave it to those more intelligent.

~Raithere

It is worthy of note that simultaneously when a stationary observer in the station area and an observer who is a rider on the tank ponder the tank as whizzing on the road at 0.866c:

1. The stationary observer observes the tank to be moving at velocity v = 0.866c,
....The bottom track ( and the road ) to be to be stationary,
....And the top track to be moving at 2v = 1.732c

2. The rider on the tank observes the the tank to be stationary,
....The bottom track ( and the road ) to be moving rearward at v = 0.866c,
....And the top track to be moving forward at v = 0.866c

Therefore the stationary observer observes the top track to be moving at a velocity beyond that prohibited by the Special Relativity universal speed limit at the same time that the rider on the tank observes the top track to be comfortably under the speed limit.

Since the consequences of the speed limit violation are not trivial ( at 1.0c the top track has already contracted to infinite shortness ) the dichotomy between the two observations is not trivial. One observer observes the top track to be shorter than at rest while the other observes it to have vanished into Riemann hyperspace some time since.

Pete,
10 feet at 983.6 megafeet per second takes 117 picoseconds.
Same for both clocks.
Pete, I suggested you stick to meters to keep from getting confused. Seems you've got that tank going lickety-split again! Light itself will only travel 35.1 millimeters in 117 picoseconds. Light travels .3mm in one picosecond, .012 inch in one picosecond, or .001 feet in one picosecond. It takes 10,000 picoseconds for light to travel 10 feet. At .866c, it takes the tank 11547.34 picoseconds to travel 10 feet.

Pete,
Fine. In the ground frame, the tank axles are 10 feet apart, so clearly there are 10 feet of track on the ground. This track is not length contracted, so 10 feet is its proper length.
The axels were 20 feet apart before the tank began the pass. How did you arrive at '10 feet' from my clock ensemble? I did something similar to what GPS does to synchronize moving clocks. I eliminated the relativity of simultaneity of moving clocks. All three points of interest will break the three laser beams simultaneously, both at the beginning of the time cycle at at the end of the time cycle. In my measurement scheme, there will be no length contractions of any of the tank points moving past the clocks. It is a well known fact that length contraction is a product of Einstein clock synchronization and relativity of simultaneity. I eliminated both with my measurement scheme. If you want to do the measurements in feet and picoseconds, all three clocks will read 23094.68 picoseconds.

CANGAS said:
....And the top track to be moving at 2v = 1.732c ... Therefore the stationary observer observes the top track to be moving at a velocity beyond that prohibited by the Special Relativity universal speed limit
I can't believe that the 1.732c didn't clue you in to your glaring mistake; particularly since you are apparently aware that c is a universal speed limit. It is amazing how you think you have some sort of great insight that the rest of us are missing when you can't even get the basic concepts right.

-Dale

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2inquisitive said:
Pete,

Pete, I suggested you stick to meters to keep from getting confused. Seems you've got that tank going lickety-split again! Light itself will only travel 35.1 millimeters in 117 picoseconds.
Ha! I would've made the same mistake in metric!
I shifted a decimal point the wrong way. I was misled by a mistake in your post (10 metres at gamma 2 takes 38.5 ns, not 7.7 ns)
10 feet at a gamma factor of two takes 11.7 ns (11700 ps), not 117 ps.
Same for both clocks.

2inquisitive said:
The axels were 20 feet apart before the tank began the pass. How did you arrive at '10 feet' from my clock ensemble?
I deduced the numbers from the distance, the same as you did.
You said "...the distance between the axels is 10 meters according to the ground observer."

I did something similar to what GPS does to synchronize moving clocks. I eliminated the relativity of simultaneity of moving clocks.
Sure, you're only using clocks in the ground frame. As long as you don't consider simultaneity in any other frame, then you're fine.

All three points of interest will break the three laser beams simultaneously, both at the beginning of the time cycle at at the end of the time cycle. In my measurement scheme, there will be no length contractions of any of the tank points moving past the clocks.
In your dreams! You can't just turn off length contraction!
It is a well known fact that length contraction is a product of Einstein clock synchronization and relativity of simultaneity. I eliminated both with my measurement scheme.
You haven't eliminated the relativity of simultaneity, you've avoided considering by only using ground frame clocks.

“ Originally Posted by 2inquisitive
The axels were 20 feet apart before the tank began the pass. How did you arrive at '10 feet' from my clock ensemble? ”

Pete,
I deduced the numbers from the distance, the same as you did.
You said "...the distance between the axels is 10 meters according to the ground observer."
You have quoted me a little out of context. That statement was my acknowledgement of your position in your displayed animation. Is that a mistake? My position is stated in the next paragraph, that by using the clock scheme I proposed, the tank, tank tread, or axel distance would not change from their proper length, i.e. 20 meters for the tread.

Pete,
In your dreams! You can't just turn off length contraction!
You haven't eliminated the relativity of simultaneity, you've avoided considering by only using ground frame clocks.
Yes I can, by eliminating Einstein clock synchronization. Einstein clock synchronization relys on the one-way speed of light always being 1/2 the round trip travel time. It works OK in a true rest frame, but fails in moving frames. Einstein introduces relativity of simultaneity to cover the failure. There are other synchronization schemes used in physics in which relativity of simultaneity does not rear its head. I used one. GPS uses one. Astronomers that use the rest frame of the CMB use them. GPS is the best example that they do work accurately, that they are natural (reflect reality accurately). There are even formulations of Maxwell's electrodynamics that are in use to work with the frames. They use the (c-squared -1) parameter. There is one I have seen used repeatedly called Maxwell-Cern-Simons electrodynamics, but I don't know enough to know how, or if, it differs from the (c-squared -1) formulation.

2inq, you're way off track. You're effectively covering your eyes and saying "If I can't see a relative simultaneity problem, then it doesn't exist!"

The GPS and CMB schemes are specifically designed to synchronize the clocks is a particular reference frame. They don't ignore the relativity of simultaneity - they work around it by being careful about reference frames and not trying to synchronize clocks in more than one frame. It doesn't mean that those clocks are synchronized in all frames - it mens that they are synchronized in one well specified frame.

Your "synchronization scheme" is to specify that all your ground clocks are synchronized, and that's fine. But it doesn't mean that they're synchronized in all frames, and it doesn't mean that you've magically made SR disappear.

Pete, place clocks at the three measured points of the tank, (1) the tank tread on the ground, (2) the axel of the tank, and (3) the top of the tread. You will find that clock (1) beats at the same rate as the ground clock in my scheme. The clock measures the same time as the ground clock to move 20 meters (or feet). (2) the clock on the axel, or in the tank, beats at 1/2 the rate of the ground clock. It will accumilate 1/2 the time compared to the ground clock to cover the 20 meters. (3) the clock on the top tread will beat at 1/7 the rate of the ground clock. It will accumilate only 1/7 the time of the ground clock to pass the 20 meters.
If you 'contract' the lengths of the tank and tread in the ground frame, the clocks on the tank will read wrong. For instance, if you state the distance between the moving axels is only 10 meters in the ground frame, the tank clock will only accumilate 1/4 the time as the ground clock to travel the 10 meters. My clock scheme does away with relativity of simultaneity and length contraction, but not relativistic clock effects. Of course, it neither proves nor disproves that the tank clocks will actually accumilate less time than the ground clocks. It also does not distinquish between 'true' time dilation or just a 'clock effect' caused by motion through spacetime (aether?).

2inq, your "clock scheme" is three clocks at rest and in the same place in the same reference frame. Suggesting that this "does away with relativity of simultaneity and length contraction" is just stupid. If you want to just declare that the the tank isnt length contracted, go ahead... but trying to prove SR to be incorrect by assuming it to be incorrect is again just stupid.

You asked what SR predicts. Dale told you. I told you. If you didn't follow it, ask for clarification.

If you 'contract' the lengths of the tank and tread in the ground frame, the clocks on the tank will read wrong. For instance, if you state the distance between the moving axels is only 10 meters in the ground frame, the tank clock will only accumilate 1/4 the time as the ground clock to travel the 10 meters.
It's clear you are either unwilling or unable to work this through properly.
Here's what SR says happens with your three new clocks. I have worded these answers carefully for a reason, so be careful about what you read into them. In particular, be careful about simultaneity. Assuming very small wheel radius:

1) In the ground frame, a clock attached to the tread will be on the ground for 11.7ns, it will tick over 11.7ns in that time, and the tank will move foward 10 feet.
In the tank frame, it will be on the ground for 23.5ns, it will tick over 11.7ns in that time, and the ground will move back 20 feet.

2) In the ground frame, a clock on the axle will tick over 5.87ns in the 11.7ns it takes for the tank to move forward 10 feet.
In the tank frame, it will tick over 23.5ns in the 23.5ns it takes for the ground to move back 20 feet.

3) In the ground frame, a clock on the tread will be off the ground for 82.2ns, it will tick over 11.7ns in that time, move forward 80 feet, and the tank will move forward 70 feet.
In the tank frame, a clock on the tread will be off the ground for 23.5ns, it will tick over 11.7ns and move forward 20 feet in that time, and the ground will move back 20 feet.

If you are unsure how I derived these results, please let me know. I may have made a mistake. We can work through a Lorentz transform for each event of interest to find out.

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2inquisitive,

It is pretty amusing to watch you here. Do you realize that the only way you have found to generate a paradox here is to selectively ignore parts of SR? The only way you have found to get self-contradictory predictions is to occasionally use SR and occasionally not. Well, congratulations: you have invented and disproved some half-baked 2inq's-special pseudo-relativity theory. Not much of an accomplishment, MacM has been doing it for years.

If you want to prove SR wrong then you need to find a problem with SR, not your pseudo-relativity. But yes, I agree that your pseudo-relativity is wrong.

-Dale

PS Its not as difficult as you are pretending. Plug the numbers into the formulas I gave; I already did all of the hard work for you.

Pete,
1)
In the ground frame, ....and the tank will move foward 10 feet.
2) In the ground frame, ....it takes for the tank to move forward 10 feet.
3) In the ground frame,....and the tank will move forward 70 feet.
OK, to begin with, there is only one plane of reference on the ground. The laser beams located directly above each other, a single event takes place in spacetime as the various parts of the single tank ( bottom cleat, axel, and top cleat) all break the laser beams simultaneously in the ground frame. This event is when the front of the tank breaks the beams, starting all clocks, both on those on the various parts of the tank and those on the ground. question (1) Do you disagree that the beams were broken simultaneously and the clocks started recording time simultaneously? We don't care about the rates the various clocks beat at, only the simultaniety of the event, the breaking of the beams and the starting of the clocks.

As the rear of the tank (the bottom cleat, the rear axel, and a cleat that has rotated directly above the centerline of the rear axel, the last cleat that can trip the laser) passes by the laser beams a second time, all clocks stop recording time. You stated all three ground clocks will record 11.7 ns between events, the breaking of the laser beams twice. We measured the length of the tank tread and distance between the axels as 20 feet while the tank was stationary on the ground. A normal person would calculate that the tank passed by the laser beams at 1.73c. You say the axel distance and the bottom of the tread has physically shrunk to 10 feet. question (2) How does the ground observer, or the ground clocks, calculate this length? He has no idea how fast the tank is going, he is using his lasers to time the tanks passing. He doesn't know if the tank is travelling at 1000 miles per hour or .999c before he makes his timing measurements. He does know the tank measured 20 feet in length from his previous measurements.

question (3) You state the tank will travel 70 feet in 11.7 nanoseconds according to the ground clock and laser that times the passing of the top tread. The ground observer only knows that the tank took 11.7 ns to travel past him, timed the same by all three of his clocks. Why would he assume the tank was travelling at 6.06c when this clock recorded the same time as the other two clocks? Why would he assume a 70 foot long tank passed by him in 11.7 ns? He measured the tread length, and tank, as 20 feet.

2inq, you need to be clear about what clocks you're talking about.

In your last post, you said:
2inq said:
Pete, place clocks at the three measured points of the tank, (1) the tank tread on the ground, (2) the axel of the tank, and (3) the top of the tread.
And you went on to make calculations as those clocks were attached to the treads and tank structure.

Now, you're implying that these clocks are the same as your previous clocks, which are at rest in the ground frame.

It was quite clear in my post that the clocks I described are not in the same vertical plane for more than an instant. I don't think you even attempted to discern what I was talking about.

For example:
2inq said:
question (3) You state the tank will travel 70 feet in 11.7 nanoseconds according to the ground clock
This is crap. I said:
In the ground frame, a clock on the tread will be off the ground for 82.2ns, it will tick over 11.7ns in that time, move forward 80 feet, and the tank will move forward 70 feet.

Read it carefullly this time. Here it is again, reworded.
In 82.2ns in the ground frame, the tank moves 70 feet and a clock on the top tread will move from the back of the tank to the front (a distance of 80 feet in the ground frame). That clock will tick over 11.7ns in this time.