Relativistic rolling tank

[przyk]

Maybe MacM will have better luck with Hooke than martillo did with deBroglie.

I doubt it, for the exact reason you pointed out: Hooke's law is not a "fundamental" law. It results from the forces exerted between neighbouring atoms.

As I expect you already know, there's a stable equilibrium distance two atoms can be situated from one another without exerting a net force on one another. Push them closer and they repel; pull them apart and they attract. It turns out that the force varies roughly in a linear fashion near this equilibrium distance, which explains Hooke's law. The full law describing the inter-atomic forces can be derived using quantum mechanics, which does have a relativistic formulation.

 

[DaleSpam]

Is it really that complicated? If you describe your conveyorbelt or wheel, say by tracking the wordlines of all its constituent parts in one reference frame, you can use the Lorentz transformations to find the worldlines in any other frame.

I took your suggestion, or how I misunderstood your point anyway, and went ahead and did the math as far as I could. Now, I don't know the relativistic formulation of Hooke's law, so I did not do any calculations regarding the "stationary" geometry or any acceleration which would require considering the material properties. Here is my approach:

In the tank's rest frame the worldline of an arbitrary material point is
s = {ct, L/π asin[sin[π/L (v t + x0)]]} where v is the velocity of the tank relative to the ground, L is the length of the tread, and x0 is a "phase term" which gives the initial location and direction of the material point and distinguishes one part of the track from another.

Lorentz transforming into the ground frame gives
s' = {γ (ct + Lv/(cπ) asin[sin[π/L (v t + x0)]]), γ (vt + L/π asin[sin[π/L (v t + x0)]])}

Now, a given material point makes a complete circuit around the track every time increment of 2 L/v. So we can determine the spacetime separation of one full revolution by subtracting s and s' evaluated at t=0 from s and s' evaluated at t=2L/v. Doing that we get
Δs={2cL/v, 0}
Δs'={γ2cL/v, γ2L}
note that Δs' is simply the Lorentz transform of Δs, so far everything is consistent.

Taking the proper-time derivative of s we get the four-velocity
u = {γc, γv sign[cos[π/L (v t + x0)]]}

Lorentz transforming into the ground frame gives
u' = {γ²/c (c² + v² sign[cos[π/L (v t + x0)]]), γ² v (1 + sign[cos[π/L (v t + x0)]])}

In order to get the proper distance between two material points in the tread we use the following procedure. First we choose a given material point by specifying x0 and a specific event on its worldline by specifying t. Second, we determine the spacelike plane of simultaneity by solving for all events S = {T,X} such that u.(S-s) = 0. Third, we choose a second material point by specifying a different x0 and we find the intersection event between the plane S and the second worldline. Fourth, the proper distance between the two material points is equal to the spacetime interval between s and the intersection event.

The worldline and four-velocity functions are not analytically invertible as far as I know, so we will have to do this part numerically. Using c=1, L=2, and v=.6, and using x0=.5 for the first material point and x0=.6 for the second material point, and setting t=0 for specifying the event on the first point's worldline we get the following in the tank's rest frame:
s1 = {0, .5}
u1.(S-s1)=0 -> .3 + T = .6 X
s2 = {.094, .656}
|s2-s1| = .125

In the ground frame we get:
s1' = {.375, .625}
u1'.(S'-s1')=0 -> .176 + T' = .882 X'
s2' = {.609, .891}
|s2'-s1'| = .125

So both frames measure the same proper distance between the two material points. Therefore, even though I don't know the relativistic formulation of Hooke's law, since the two frames agree on the strain they must also agree on the stress and any failure characteristics.

-Dale

 

[Pete]

Well get it straight who brought the point up!

Like I said, I don't care. It's not a competition - you're not going to get any prizes for being the first one right.

Also are you completely unaware that he is actually taking the piss out of you as well as attempting to with me?

Again, I don't care. I just respond to the facts as posted.

Really, imaplanck, it's not a pissing contest. If you want to have a personal -"I'm smarter than you" competition with 2inq, go ahead... but do it somewhere else. Nobody else cares.

 

[imaplanck.]

Like I said, I don't care. It's not a competition - you're not going to get any prizes for being the first one right.

Again, I don't care. I just respond to the facts as posted.

Really, imaplanck, it's not a pissing contest. If you want to have a personal -"I'm smarter than you" competition with 2inq, go ahead... but do it somewhere else. Nobody else cares.

Look I dont give a shit who says it first, I just object to his dishonesty. He has clearly (several times now) twisted around what I have corrected him on and sayed he has corrected me on it. If he said I agree with me on it instead of claiming he thought of it I wouldn't mind. Look it's not about you I dont expect you to follow everything that is said, it is about his taking the piss.
Anyway why are you turning on me? Are you making an about turn into relativity being merely a mathematical theory like mac and 2zingger?

 

[DaleSpam]

I just object to his dishonesty.

I guess I didn't pay attention carefully to your current conversation with him, but I have never found 2inquisitive to be dishonest in the past when we have had pretty long discussions about reference frames. I am sure that whatever you are complaining about was not intentionally dishonest on his part. He is one of the few anti-SR people here with whom you can actually have a discussion. I suspect you are mis-interpreting his comments.

-Dale

 

[imaplanck.]

I guess I didn't pay attention carefully to your current conversation with him, but I have never found 2inquisitive to be dishonest in the past when we have had pretty long discussions about reference frames. I am sure that whatever you are complaining about was not intentionally dishonest on his part. He is one of the few anti-SR people here with whom you can actually have a discussion. I suspect you are mis-interpreting his comments.

-Dale

Well hes either dishonest or a complete retard, I dont care which one you believe and this discussion is pointless.
I dont care if you and Pete want to get in bed with this guy, Im fed up with these twonkers who cant even be bothered to put a little effort into studying and regarding experiment yet put tons of effort into damaging productivity and progress. Go ahead and nursemaid these Igneous andesites as if they have ears to listen, Im fucked if I care.

 

[DaleSpam]

If you don't think that arguing with the anti-SR crowd is fun then chances are you won't enjoy most of the threads on this forum. Obviously, other than simple entertainment value, these discussions are completely pointless, since nobody is going to change anybody else's mind. Just have fun.

-Dale

 

[2inquisitive]

imaplank,

Why are you focusing on the track being contracted in either just the top or the bottom of the track? The proper length of the track (overall length) will still be increased by a factor of 2. Why cant you get that?

Last edited by imaplanck. : 07-27-06 at 07:23 PM.

my response,

imaplanck, according to a ground observer, the bottom of the tread is stationary, the axels (and the body of the tank) are moving to his right at .866c, and the top of the tread is moving to his right at 2 x .866c. Can't you understand that the top of the tread is moving to the right of the ground observer at twice the speed of the tank body?

According to an observer inside the tank, the top of the tread moves forward at .866c and the bottom of the tread moves rearward at .866c. Both top and bottom would be contracted by the same amount, according to the tank observer.

The observer on the ground sees a different scenario. He sees the bottom of the tread at rest with the ground, the tank moving to the right at .866c, and the top of the tread moving to the right at twice the speed of the tank itself. The tank tread moves in different directions at the top and bottom. This creates the paradox in the ground frame. The ground observer does not see the top of the tread and the tank moving to the right at the same speed. The tread still moves relative to the tank body and its axels.

by imaplank'

“ Originally Posted by craniumbuoyancyaid
The observer on the ground sees a different scenario. He sees the bottom of the tread at rest with the ground, the tank moving to the right at .866c, and the top of the tread moving to the right at twice the speed of the tank itself. The tank tread moves in different directions at the top and bottom. This creates the paradox in the ground frame. The ground observer does not see the top of the tread and the tank moving to the right at the same speed. The tread still moves relative to the tank body and its axels ”


Yes again we had to tell you that(several times). Where is the paradox though????

Last edited by imaplanck. : 07-27-06 at 07:08 PM.

by imaplanck again,

IT APPLIES IN ALL FRAMES FUCKWIT. THERE IS NO NEED TO BE FRAME SPECIFIC IF IT APPLIES IN ALL FRAMES(ALL THOUGH THIS WAS ONLY A REITERATION OF A STATEMENT I MADE WHERE I DID ACTUALLY SPECIFY THAT IT APPLIES IN ALL FRAMES).
JESUS YOU ARE AN IMBECILE.

Last edited by imaplanck. : 07-27-06 at 07:28 PM.

imaplanck, you stated the tank tread 'stretched' by a factor of 2 in all frames of reference. I disagree with this, so how can you be the 'first' to bring up a point, as you say? What 'point' is it that you claim I 'stole' from you since we don't agree?

Let me state my viewpoint again. DaleSpam gave a correct explaination in his post if one begins the exercise in the TANK frame. I don't think anyone disagrees with Dale if you begin in the tank frame of reference, then transform to the ground frame. A tank observer 'sees' both the top tread and bottom tread contracted by identical amounts from his frame.

My scenario above, to which Pete responded, was from beginning in the ground frame, not the tank frame. Special Theory states there are no preferred frames, so I believe it is fine to begin from the ground frame, OK? Beginning in the ground frame, the top tread is moving to the right at 2 x .866c, for a contraction factor of 7. The tank, the axels, the ends of the track are all moving to the right at .866c, for a contraction factor of 2. The bottom of the tank tread is stationary wrt the ground, so the 'cleats' would not be contracted. It is when one begins in the ground frame of reference when odd things happen, not when one begins in the tank frame and transforms into the ground frame.
by imaplanck,

tHE bottom of the track is stretched x2 and the bottom of the track is stretched x 2(for reasons that I have tried to explain but I just cant at the moment), there is no concertinaing up on the top of the track compared to the bottom being overly stretched if you calculate correctly

.
post by 2inquisitive, starting in the ground frame:

There will be seven times as many cleats on the top of the track as on the bottom of the track in the ground frame. The cleats on top of the track will be 1/7 the distance apart as the cleats on the bottom, as measured from the ground frame.

The axels themselves correspond to the familiar 'relativistic cartwheel' in the gedankin. The 'cogs' on the axels are wider apart at the ground than they are after they have moved to the top of the axel. This explanation is consistent with Special Theory, I believe.

 

[DaleSpam]

It doesn't matter which frame you start in. As you say, it is fine to begin in the ground frame. It is only a matter of convenience to start in the tank frame since the worldline of the particle is easier to define there.

I can certainly repeat the exercise starting from the ground frame if you like, but what kind of "odd things" do you expect? Are you talking about normal SR "odd things" like length contraction or abnormal "odd things" like paradox?

-Dale

 

[2inquisitive]

Dale, when you begin in the ground frame, the distance between the cleats on the bottom tread is not contracted but the total length of the tread is contracted by a factor of two, along with the axels and tank. The TOP of the tread is contracted by a factor of seven because the top tread is moving twice the speed of the tank and its axels ( 2 x .866c) in the ground frame. Now, when you begin in the tank rest frame, both top and bottom treads are contracted by a factor of two, because both are moving at .866c relative to the tank, they are just moving in opposite directions. In your exercise, you used the time it takes for a point to complete a revolution around the tread in each frame. That method makes no distinction between top and bottom treads as seen by a ground observer.

Length contraction of the tread when beginning in the ground frame is what I am questioning. In the ground frame, both the total length of the bottom tread and the axels are contracted by a factor of two, but the top tread is contracted by a factor of seven. I don't see where either the top or bottom of the tread is 'stretched by a factor of two'. The top of the tread would have to stretch 3 1/2 times the axel spacing, or break. The spacing between the 'cleats' or measuring marks would have to be their proper rest length on the bottom and contracted by a factor of seven on the top, while the total length of the tread would be contracted by a factor of two on the bottom and seven on the top.

 

[DaleSpam]

That method makes no distinction between top and bottom treads as seen by a ground observer.

It most certainly does. Simply plot out s'. You will clearly see that there is a very great distinction between the top and the bottom treads as seen by a ground observer. My work is there out in the open for anyone to check. Please don't claim that I ignored such an obvious thing when I clearly did not.

Length contraction of the tread when beginning in the ground frame is what I am questioning. In the ground frame, both the total length of the bottom tread and the axels are contracted by a factor of two, but the top tread is contracted by a factor of seven. I don't see where either the top or bottom of the tread is 'stretched by a factor of two'. The top of the tread would have to stretch 3 1/2 times the axel spacing, or break. The spacing between the 'cleats' or measuring marks would have to be their proper rest length on the bottom and contracted by a factor of seven on the top, while the total length of the tread would be contracted by a factor of two on the bottom and seven on the top.

I know that I don't know the SR formulation of Hooke's law. If you know it then please post the reference, I would definitely like to learn it. If you do not, then you shouldn't assume that the "spacing between the 'cleats' or measuring marks would have to be their proper rest length on the bottom". For simplicity I would recommend sticking with the track already in motion and not trying to consider what it would look like at rest.

-Dale

 

[2inquisitive]

DaleSpam,

I know that I don't know the SR formulation of Hooke's law. If you know it then please post the reference, I would definitely like to learn it. If you do not, then you shouldn't assume that the "spacing between the 'cleats' or measuring marks would have to be their proper rest length on the bottom". For simplicity I would recommend sticking with the track already in motion and not trying to consider what it would look like at rest.

When beginning in the ground frame, why would Hooke's law have anything to do with proper rest length spacing of the cleats? The cleats are not moving relative to the ground, thus they will have their proper length spacing. The total length of the track would be contracted by a factor of two at the bottom and seven at the top because it is moving wrt the ground, but the cleats, or distance markings are not.

 

[Pete]

When beginning in the ground frame, why would Hooke's law have anything to do with proper rest length spacing of the cleats?

Because the spacing of the cleats is dependent on the stress on the track, according to Hooke's law.

 

[Pete]

Length contraction of the tread when beginning in the ground frame is what I am questioning. In the ground frame, both the total length of the bottom tread and the axels are contracted by a factor of two, but the top tread is contracted by a factor of seven.

These don't go together... You have either two factors of two, or a factor of one and a factor of seven, depending on exactly what you mean:

In the ground frame, both the track on top and the track on bottom are half the tank's proper length.
In the ground frame, the track on the bottom is not length contracted, and the track on top is length contracted by a factor of seven.

I don't see where either the top or bottom of the tread is 'stretched by a factor of two'.

If the track is stretchable and under tension, then it will be stretched, right? I suggest that it is, in fact, under tension.

The top of the tread would have to stretch 3 1/2 times the axel spacing, or break.

I don't think so... let's see why.

The spacing between the 'cleats' or measuring marks would have to be their proper rest length on the bottom...

Their proper stretched length, if the track is under tension.

... and contracted by a factor of seven on the top,

Contracted by a factor of seven from their stretched length, if the track is under tension.

while the total length of the tread would be contracted by a factor of two on the bottom and seven on the top.

The stretched track (if it is under tension) would be uncontracted on the bottom, and contracted by a factor of seven on the top.


Let's say the track's proper length when not under tension is 8 metres, with 8 cleats, one every metre. Say the wheels are very small, and the tank is 4 metres long.
So when the tank is sitting still, there are 4 cleats on the bottom and 4 and top, one every metre, right?

Now, let's look at the tank rolling at 0.866c in the ground frame. The question is Is the track stretched, and by what factor? Let's call this factor S. If the track is not stretched, then S will be one.

The proper length of the track will be 8S metres, with one cleat every S metres.

In the ground frame:
The tank will be 2 metres long.
There is 2 metres track on the ground, not length contracted. There will be 2/S cleats in this section.
There is 2 meters of track not on the ground, length contracted from 14 metres. There will be 14/S cleats in this section.

So, the total proper length of track (working from the ground frame) must be 16 metres, containing 2/S + 14/S cleats. This gives us to ways to work out S:

Total length of track = 8S metres = 16 metres, therefore S = 2.
Total number of cleats = 8 = 2/S + 14/S = 16/S, therefore S = 2.

So working from the ground frame, we find that the track must be under enough stress to stretch it by a factor of 2.

 

[2inquisitive]

Because the spacing of the cleats is dependent on the stress on the track, according to Hooke's law.

I thought the tank was in a non-accelerating (inertial) frame, so where would the stress come from that effects spacing of the cleats? I can understand the weight of the tank exerting a downward force, a stress, but why would there be a lateral stress when starting from the ground observer's frame of reference?

 

[Pete]

I thought the tank was in a non-accelerating (inertial) frame, so where would the stress come from that effects spacing of the cleats? I can understand the weight of the tank exerting a downward force, a stress, but why would there be a lateral stress when starting from the ground observer's frame of reference?

I don't precisely know, and this troubles me.

If no such stress can exist, then SR can not be valid.

I suspect that the stress comes from the track shrinking due to the extreme length contraction in the top track... but this doesn't sit completely comfortably.

Note that the proper spacing of the cleats is defined in the track frame, which is not inertial.

 

[2inquisitive]

Pete,

Careful... I know you mean that the measured length of track on the ground at any moment is half that in the tank frame, but some might think that you mean that the actual section of track on the ground is contracted.

Pete, I am beginning in the ground observer's frame of reference. In the ground observer's frame, both the distance between the axels and the tank itself will appear contracted by a factor of two because both are moving relative to the ground observer at .866c. I have not transformed to the tank's frame of reference yet.
Pete,

If the track is stretchable and under tension, then it will be stretched, right?

Again, beginning in the ground frame of reference, only the top portion of the tread will need to be stretched. The cleats are 'at rest' wrt the ground and the distance between the axels is contracted according to a ground observer.
Pete,
In the ground frame:

The tank will be 2 metres long.
There is 2 metres track on the ground, not length contracted. There will be 2/S cleats in this section.

OK with this so far.
Pete,

There is 2 meters of track not on the ground, length contracted from 14 metres. There will be 14/S cleats in this section.

So, the total proper length of track (working from the ground frame) must be 16 metres, containing 2/S + 14/S cleats.

Here is where I don't see any correlation with Special Theory. Why would the track be stretched to twice its proper length when the tank is moving in the ground frame? We have not considered the tank frame yet, only working from what we observer in the ground frame.

 

[2inquisitive]

Oops, I just noticed something. Here is your statement again:

There is 2 metres track on the ground, not length contracted. There will be 2/S cleats in this section.

The spacing between the cleats is not length contracted, but the end-to-end length of the tread is contracted from 4 meters to 2 meters. The tank, and its tread & axels, are rolling past the ground observer at .866c.

 

[Pete]

Oops, I just noticed something. Here is your statement again:

There is 2 metres track on the ground, not length contracted. There will be 2/S cleats in this section.

The spacing between the cleats is not length contracted, but the end-to-end length of the tread is contracted from 4 meters to 2 meters. The tank, and its tread & axels, are rolling past the ground observer at .866c.

The track on the ground is not length contracted.
There is only 2m on the ground in the ground frame because the tank is length ontracted, but this on-the ground track is stationary in the ground frame - it is not length contracted from 4m.

You could consider this as an effect of relative simultaneity - in the tank frame, one cleat leaves the ground at the same time as the 4th cleat along touches the ground. But in the ground frame, that same cleat leaves the ground well before the 4th cleat along touches the ground.

The point is that whatever way you think about it, there are 2/S cleats on the ground at any time in the ground frame, right?

 

[Pete]

Again, beginning in the ground frame of reference, only the top portion of the tread will need to be stretched. The cleats are 'at rest' wrt the ground and the distance between the axels is contracted according to a ground observer.

I suggest that the entire tread is stetched equally along its length. The numbers posted support that position.

Here is where I don't see any correlation with Special Theory. Why would the track be stretched to twice its proper length when the tank is moving in the ground frame?

That's what I just demonstrated. Using SR and working purely from the ground frame, we discovered that S=2.

We have not considered the tank frame yet, only working from what we observer in the ground frame.

And the stretch factor must work out to be the same in the tank frame. If the tread is physically stressed, that's a fact that's independent of reference frame.