Is Velocity Addition in Special Relativity Contradictory?

[Austin0]

Like everyone here , when I saw Ich's solution I thought it was a clever approach to the question and concluded it was a done deal.

2) Even if this assumption should be correct it still ignores the physics and measurement in the Earth frame.

In that frame it is very simple:
...A point on the track is chosen located at the top of the rear drive wheel ,as the tank moves forward that point is tracked until it is located on top of the front drive wheel.

If you apply the distance traveled according to the .74844 figure and then apply contraction on top of this you get a track point (segment) that has not traveled as far as the wheel base and gearing. The contraction figure for the base is .8930 and for the track is .6632.
SO either something is amiss or the track should decompose.

There are the additional questions; in the tank frame both the top and bottom are moving and so would be equally contracted.
In the Earth frame the top would be contracted but not the bottom .

Given: ...Ich's velocity figure is totally accurate.

....The other effects are in operation, the tank and track is contracted etc.

....We now are measuring the track velocity.

....We start with a coordinate time and position measurement at the rear point on the track where it is equivalent to the rear wheel base.

...We take another measurement at the point where it is colocated with the topmost point of the front wheel (and base)

...From these two points we derive velocity for the track point and the wheel base.

...The distance traveled by the track is simply dx=dx_t

...for the wheel base it is dx-(wheel base...dx)=dx_b

...I think it is safe to say that geometrically dx_t=2 dx_b

...Using round geared wheels and geared track the forward motion of the base is $$\pi$$x 2r(radius of wheel) per revolution. The top track must advance by twice this value to maintain continuity with the wheel base and gear teeth.[/B
...Velocity .For the track this is simply dx_t/dt=v_t

For the base it is ...dx_b/dt=v_b

If we assume that v_t <v_b then either

...1)....dx_t<2 dx_b

or ...2)...dt_t >dt_b

,,,,,,,,,,,,,Considering there are only 2 measurements.. t₁-t₀=dt

,,,,,,,,,,,,Number 2) is not not very viable.

This is pure kinematics . What is measured after the physics is already in effect.

First, a comment:

Yes, the tank frame is preferred: It's the only one where beginners have a nonzero chance of even guessing the right result. .

As a pedagogical preference I wouldn't disagree but there is the 1st P to consider also.

The most important thing in learning relativity is to tell the difference between logic and preconception. In the rest frame, many preconceptions happen to coincide with what the logic actually tells us. So if you get different results, most likely both are wrong, but the one derived in the moving frame is certainly wrong.

COmplete agreement regarding preconception i.e. intuitive logic as it applies to learning SR.

But there is the logic that applies within SR and also applies to SR as a logical theoretical structure.
Specifically testing its derivative applications regarding consistency with the fundamental structure.
Please do not try to twist this around into some kind of questioning of SR.

My world view is founded on both the logical consistency of the fundamental structure as elucidated by A Einstein as well as its total validity as a description of the real world.
Rock Solid

The term moving frame is normally simply an operative term. With the implicit understanding that both frames are "moving" dependiong on perspective.

Now for the math.

The tank is moving at v=0.45 (I use c=1, that's easier). That's not a question, that's the setup.

In the tank frame:
The ground has v=-.45 by some basic principles, and since the lower track is at rest wrt the ground it has v=-.45, too.
Now imagine a virtual vertical plane separating the rear part from the front part of the tank. Call it the middle plane.
Now there's an equation of continuity that says that, in any given timespan, as many track segments have to cross said plane from back to forth as in the other direction, lest the segments accumulate somewhere.
From symmetry principles it is evident that v=0.45 for the upper track fulfills the continuity condition. It can be shown mathematically that this is the only solution.

Agreed ;within the tank frame.

This is not entirely trivial, however.
For example, the segments all like to be Lorentz-contracted. If you have a fixed number of (well-fitting) segments and spin up the tank, this cannot be achieved. Either the track will break, or the segments all get stretched. You could also add more segments until the track fits again.

.

...A point on the track is chosen located at the top of the rear drive wheel ,as the tank moves forward that point is tracked until it is located on top of the front drive wheel.

If you apply the distance traveled according to the .74844 figure and then apply contraction on top of this you get a track point (segment) that has not traveled as far as the wheel base and gearing. The contraction figure for the base is .8930 and for the track is .6632.
SO either something is amiss or the track should decompose.

Ok, now we've established v_upper = 0.45 in the tank frame, and v_tank = 0.45 in the ground frame, we have that formula that tells us that v_upper in the ground frame is 0.74844. This is not up to debate. That's dictated by logic, and if you have some logic that says otherwise: it's wrong.

I am not sure what you are referring to when you say "is not up for debate" or what "logic "you are referring to.

If you mean the formula given those v's,, returns 0.74844 I certainly never questioned it.

If you mean the "logic" that was behind the velocity in the tank frame then I never questioned that either

I have absolutely no question that Ich's math was accurate as was the logic of Newtonian mechanics he applied in the tank frame.

If you mean the logically derived conclusion is not up to debate then I would have questions.

As far as I know the domain of applicability of the Addition of Velocities equation is limited to ;
Independant inertial frames or objects.

a) The track as a whole or as segments would not seem to be inertial.
It is under constant acceleration through force.

b) The track is not independant. It is physically connected to both the tank and the ground.
I don't just mean in contact but that the physics of both frames have direct causal effects.

SO it would seem there is doubt if it would apply to this situation at all.

This would seem to then raise some doubt whether the derived figure can be taken at its normally , factual value , without further ocnsideration.

I don't know where all of you went wrong in the "moving" (=ground) frame, but here's a description that at least gives the correct result. It's a convoluted argument, but that's always the case if you think in terms of the ether: time dilation and length contraction.

Since I never made any assertion at all as to the quantitative relative velocity from the ground frame I am unsure how I could have gone "wrong"

Here's the point where some of you see the continuity equation violated, and claim that logic dictates both closing speeds be the same (up to the sign). It's easy to show how this is wrong.

How many segments ("n") do actually move from back to forth and otherwise in, say, 1 second?
Use some algebra to find that n = v/l * 1 second, where v is the closing speed and l the length of a segment.
The lower track is at rest in the ground frame. The're no relativity to be obeyed, and for the sake of simplicity we set the length of a segment in that frame to be 1 second. (Remember, c=1. If you want to use the usual dimensions: that means l = 1 lightsecond = ~300000 km.)
Then n_lower = 0.45/1 s = -0.45/s.

The upper segments are slower (lower absolute closing velocity), but contracted and thus more densely packed. From the Lorentz contraction formula, we get: l_upper = 0.6632 s.
And we find n_upper = 0.29844/0.6632 s = 0.45/s.

...Using round geared wheels and geared track the forward motion of the base is $$\pi$$x 2r(radius of wheel) per revolution. The top track must advance by twice this value to maintain continuity with the wheel base and gear teeth.

I absolutely never made any such claiim that both speeds must be the same.

Given your contraction figures which I am sure are accurate:

This would mean that not only the links were contracted but, if we assume gearing, then the number of links between wheels would be fixed and so the distance between the tops of the wheels would also have to be contracted relative to the ,bottom of the track, distance between the bottoms of the wheels. Yes??

WOuldn't you assume this would create intolerable stress on the wheels??

And wheelbase??

Would you maintain that this was a workable reality by the same criteria you applied to the tank frame??

What a surprise, if it works in the rest frame, the math somehow magically conspires to make it work in a moving frame, too.
That's not a coincidence, it's a mathematically proven property of relativity. Whenever you find it violated, you got something wrong. To repeat: you got something wrong, not relativity and not the international conspiracy of pysicists.
Whenever it doesn't work, seek you error, and don't take your flawed logic for granted and insist the the others have to be wrong.

.

...I am sure you can take Ich's calculations to the bank

I have absolutely no question that Ich's math was accurate as was the logic of Newtonian mechanics he applied in the tank frame.

Edit: The last sentences appear somehow harsh, and they are not aimed at you, Ken Natton, and only partially at Austin0 (more from the experience in a different thread, I'd say you, Austin0, can make use of this advice.).

Which part of the advice? If you mean your international conspiracy; please spare me.

If you mean my flawed logic please point to a specific point of my logic and show me.

Not this blanket assumption and ad hominem assertion.

As applied to me this whole post is directed at claims I never made .

This also applies to the other thread you referred to. I never came to conclusions or made assertions regarding the problem. I never said everyone was wrong outside of a final logical argument with someone. I made an error, not of logic, but of perspective and recognized it in the thread. As the majority of the thread was not regarding the actual problem but simply dickering over the boundary conditions which I acceeded to , I fail to see how any of the above applies. Once again if you think there are flawed points of logic or any assumptions not consistent with the principles of SR fine. Bring them forth. I am ready to learn.

 

[Ken Natton]

Oh dear guys. Listen, it was me that first mentioned this silly notion of conspiracies, and I am sure you understand that I intended to be ironic. I certainly didn’t intend to make the conversation take this turn.

Calebholiday, I would essentially agree with you that it is not enough to simply say, ‘that is the way it is’. But I’m not entirely in accord with your assertion that ‘things should be explained adequately’. The onus is on us, the laymen, to draw the telling insights from the experts. It is not enough to sit there passively and say ‘explain it to me’. It requires a deal more effort on your part. Some of that effort requires you to do a little reading, mainly to help you to ask the right questions. But also, you need to attend closely to the replies to pick up the hints about which are the right follow-up questions. And some actually quite minor changes to your approach and your phraseology might tend to engender slightly more constructive conversations with the experts. I accept that sometimes, if a conversation has a certain edge, it makes it more compelling to follow. But if it tips beyond a certain point the result is unlikely to be helpful to any of us.

For my part, I think that Ich’s post #32 does contain a great deal of insight, but I’m not certain that it does resolve completely the anomaly we are struggling with here. I need to ponder it a little longer to fully understand it (Ich’s post). My failure to prove what I was trying to prove with a little mathematics only proves that the mathematics required are a deal more involved than those I was trying to use, and that I do not possesses the mathematical ability to do what I attempted. I do not accept that I am in any way being the credulous layman if I continue to assert my certainty that it can be done in essentially the that way I was trying to do it. And I continue to believe, Austin0, that in its essentials, the true solution is the same for a wheel as it is for a tank track. The key point is that the peripheral speed of the wheel or the tank track is constant throughout its revolution and the track is under no tension whatever other than that imparted by the drive wheel. I also reassert my belief that the essential problem is no different at 10mph than it is at 0.45c. Unfortunately, I possesses not the ability to prove it.

 

[phyti]

Caleb;

v: speed of tank
vb: speed of bottom tread
vt: speed of top tread
c = 1

The ground observer measures:
v = .45, vb = 0, vt = .9

The driver is moving at .45, so his speed affects whatever calculations he makes.
The moving observer uses the difference formula which you did, but only on the top.
He measures:
vt = (.90-.45)/(1-.9*.45) = .45/.595 = .756
vb= (-.45-.45)/(1+.45*.45) = -.90/1.203 = -.748
What's missing is the acceleration around the wheels, which should make the values equal.

If we correct the top speed to the ground (0-speed) frame using the addition formula we get
ut = (.45+.756)/(1+.45*.756) = 1.206/1.340 = .90

A general note: speed calculations are independent of time dilation and it's consequent length contraction.

 

[matheinste]

Caleb;

v: speed of tank
vb: speed of bottom tread
vt: speed of top tread
c = 1

The ground observer measures:
v=0.5c vb=0 vt = 0.9c

.

If this is relative to the ground then why .9c?

Matheinste

 

[Fredrik]

the Velocity that S or the topside to the relative observer must be, one where the ground observer would observe 0.45 light-seconds of track being feed from the topside to the bottom.

I can't make sense of what you're saying.

Imagine that the track has little dots placed on it every 0.1 of a light second (have to imagine a big tank in this instance)

This doesn't tell us anything, because you left out essential information about frames and velocities.

The speed of these dots must overtake the speed of the tank from a ground perspective for any to be feed in at all.

True.

In classical relativity

I don't know what that means. Special relativity is a classical theory. Are you talking about the pre-relativistic (Galilean) concept of spacetime?

the speed of the dots would have to be 0.9C

In Galilean spacetime, yes.

, but due to length contractions, these dots are closer together and so S should be slower than 0.9C.

You're forgetting that the track is being forcefully stretched. Imagine a tank floating in space when you switch on the engine. The track speeds up (in the inertial frame comoving with the tank) and gets Lorentz contracted, but it can't get shorter because it's wrapped around something that doesn't change shape. So it has to get forcefully stretched. If the distance between two adjacent dots was 0.1 ls in the tank's frame before the engine was switched on. It's now gamma*0.1 ls in an inertial frame that's co-moving with those two dots. It has to get stretched by a factor of gamma to fit around the thing it's wrapped around.

If in another situation and the speed of S and V are different with S being the speed which produces a 50% contraction, the speed of S would only need to be 50% faster than V for it to feed the bottom-side adequately.

Do you agree or disagree with this?

I don't understand what you're saying. You need to be more careful when you explain things. Make sure that it's clear what symbols such as S and V are representing, and never specify a time, distance or velocity without making it clear what object and what coordinate system you're talking about.

It's not enough to have someone tell you something is the way it is. Things should be explained adequately, so people don't have to take your word for it.

You could also learn about Lorentz transformations and work through the derivations of length contraction, time dilation, and velocity addition like the rest of us. It's a bit naive to expect that a difficult problem has a simple explanation that you can understand when you still haven't studied the basics. You should stop assuming that we're all feeding you BS, and start working on understanding the basics. Do you understand Lorentz transformations yet? Do you know how to use them to derive the velocity addition rule? Why not? Don't you think exercises like that would help you understand problems like this one?

 

[phyti]

If this is relative to the ground then why .9c?

Matheinste

The top track moves twice as fast as the tank.

 

[Austin0]

The top track moves twice as fast as the tank.

Hi phyti I would certainly agree with you as far as the determination of velocity is purely kinematic,,a point moving from A to B with the assumption that all Lorentzian effects, contraction etc are already taking place.

SO the determination you have arrived at is an abstraction with no regard for whether or not the physics of contraction would make it possible for part of the system to actually operate at that velocity.

Basically doing what Ich did but from the gound perspective?

Or are you assuming that contraction has no physical meaning ? Is an artifact of measurement due to motion?

Do you think an actual mechanism could function with the relative degrees of contraction between the top and bottom and the top and the tanks itself??

 

[matheinste]

The top track moves twice as fast as the tank.

Relative to the ground the bottom track is at rest when in contact with the ground. The speed of the bottom track relative to the tank is -0.45c when in contact with the ground and the speed top track relative to the tank is +0.45c when parallel to the ground assuming the tank is traveling at top speed. The other points on the track are more difficult to analyse because they are not moving inertially.

Take the rest frame of the ground, which is also the rest frame of the bottom track when in contact with the ground, and we have the tank traveling at 045c relative to that frame, that is a given, assuming the tank is going at top whack and the top speedof the tank is quoted as being relative to the ground in the proposed scenario. Now the top track is going at 0.45 relative to the rest frame of the tank. So we need to use the relativistic velocity addition formula to arrive at the speed of the top track relative to the ground. This does not give 0.9c as many previous posters have shown.

Using the relative speed of the top track as 0.9c and plugging into other equations propagates the error. The only way 0.9c crops up is as the closing speed of the top and bottom tracks relative to tanks rest frame.

Matheinste.

 

[phyti]

Using the relative speed of the top track as 0.9c and plugging into other equations propagates the error. The only way 0.9c crops up is as the closing speed of the top and bottom tracks relative to tanks rest frame.

Matheinste.

The ground observer does not use his speed (.90), but that from the driver.
The driver calculates vt as .756c.
Using the addition formula, the ground observer adjusts this
downward to .45 relative to the driver/tank, as shown.
Moving observers calculate relative speeds too large (due to simultaneity convention).
The compostion formula adjusts them to the fixed (0c) frame.
If you solve v=(b-a)/(1-ab) for b, you get the addition
equation.
If the approx. .75c is acceptable for one track, it must also be for the other.

 

[matheinste]

The ground observer does not use his speed (.90), but that from the driver.
The driver calculates vt as .756c.
Using the addition formula, the ground observer adjusts this
downward to .45 relative to the driver/tank, as shown.
Moving observers calculate relative speeds too large (due to simultaneity convention).
The compostion formula adjusts them to the fixed (0c) frame.
If you solve v=(b-a)/(1-ab) for b, you get the addition
equation.
If the approx. .75c is acceptable for one track, it must also be for the other.

I'm sorry but i cannot see where the 0.9c comes from.

The statement "the top track moves at twice the speed of the tank" or "the top track has a speed of 0.9c relative to the ground" are the statement I am unhappy with.

Matheinste.

 

[Fredrik]

If the approx. .75c is acceptable for one track, it must also be for the other.

Wouldn't you also say that if 0.45c is acceptable for one track, it must also be for the other? This is what we're saying in the tank frame. Surely this can't be any less obvious than what you said? And yet, by the velocity addition formula, these two statements contradict each other. Matheinste is right. Ich posted the correct solution earlier in this thread.

 

[Ich]

Specifically testing its derivative applications regarding consistency with the fundamental structure.
Please do not try to twist this around into some kind of questioning of SR.

It is good to repeatedly find out how everything is consistent. The problem is, you are not in a position to do so. You'll need another few years of learning and getting accustomed. At least I needed a few years to come to an understanding of every single claimed paradox I came across.

As far as I know the domain of applicability of the Addition of Velocities equation is limited to ; Independant inertial frames or objects.

It is limited to inertial frames. "Independent inertial frames" makes no sense, as I can't imagine what a dependent inertial frame would be.

a) The track as a whole or as segments would not seem to be inertial.
It is under constant acceleration through force.

Each segment is inertial while it's moving linearly.

b) The track is not independant. It is physically connected to both the tank and the ground.
I don't just mean in contact but that the physics of both frames have direct causal effects.

"Physical connectedness" is no criteria for something being inertial or not. If it's moving with constant velocity (speed and direction), it is inertial. Period.

This would mean that not only the links were contracted but, if we assume gearing, then the number of links between wheels would be fixed and so the distance between the tops of the wheels would also have to be contracted relative to the ,bottom of the track, distance between the bottoms of the wheels. Yes??

No..
This is a splendid example of logic vs preconception. Try to check what you can really be sure of, and what not. For example, the bottoms (and tops) of the wheels are stationary in the tank frame. They are points - as opposed to events. You know that distances between stationary points are Lorentz contracted. Both by the same factor, of course.
By similar arguments you should know that the top segments undergo contraction wrt the bottom segments.
So where does your logic fail and turn into preconception?

There's a single answer in more than 99% of all cases where "logic" clashes with reality: You forgot the relativity of simultaneity.


Counting the number of segments between wheels contains an assumption concerning the simultaneity of spatially separated events.

Once again if you think there are flawed points of logic or any assumptions not consistent with the principles of SR fine. Bring them forth.

That's what I'm doing.
And all I was saying in the post you took offence at: Your basic principle must be that you are wrong if you spot an inconsistency (in fact, you provably are wrong). If people are telling you where - specifically - you went wrong, there's no use defending that point. Instead, try help to nail the error down.

Oh dear guys. Listen, it was me that first mentioned this silly notion of conspiracies, and I am sure you understand that I intended to be ironic.

I understood, and my mentioning of it was metallic, too.
The words were aimed at calebhoilday. Re-read the thread and you'll find them totally appropriate. If there's some collateral damage, I can live with it. At least I spared those who noticeably seek understanding. I'm sure Austin0 does so, too, but unfortunately I failed to notice the determination behind his/her efforts. This may be totally my bad.

@phyti: Please don't!

 

[phyti]

From post #1;

Bottom-side velocity according to the tank
U = (0 - 0.45C) / (1-(0C*0.45C/C^2))
= -0.45C / 1
= -0.45C

Top-side velocity according to the tank
U = (0.9 - 0.45C) / (1-(0.9*0.45/C^2))
= 0.45C / (1-0.405)
= 0.45C / 0.595
= 0.7563C

The bottom track speed is 0 relative to the ground.
The bottom track speed is -.45c relative to the tank.
That's the posters mistake.

The driver/tank must see the same amount of track moving forward as that moving backward, since the amount of track is constant.

If the tank moves at .45c in the ground frame,
and the top track moves 2x the speed of the tank to function correctly,
then the top track moves at .90c in the ground frame.

 

[yossell]

Phyti:
What's the reasoning that gets you from here:

The driver/tank must see the same amount of track moving forward as that moving backward, since the amount of track is constant.

to here:

If the tank moves at .45c in the ground frame,
and the top track moves 2x the speed of the tank to function correctly,
then the top track moves at .90c in the ground frame.

In particular, I want to understand how you justify `the top track moves 2x the speed of the tank to function correctly', particularly in the light of the fact that, in the ground frame, the upper track is heavily lorentz contracted while the lower tread is not.

 

[Austin0]

Phyti:
What's the reasoning that gets you from here:[/QUOTE
___________________________
Originally Posted by phyti

The driver/tank must see the same amount of track moving forward as that moving backward, since the amount of track is constant.

In particular
to here:

___________________________
Originally Posted by phyti

If the tank moves at .45c in the ground frame,
and the top track moves 2x the speed of the tank to function correctly,
then the top track moves at .90c in the ground frame.

In particular, I want to understand how you justify `the top track moves 2x the speed of the tank to function correctly', particularly in the light of the fact that, in the ground frame, the upper track is heavily lorentz contracted while the lower tread is not.

Maybe the meaning is better understood as "functioning at all"

The contraction itself raises questions and problems.
In the tank frame both the upper and lower are contracted, while in the ground frame only the upper is. So in the tank frame there is not the same problem of physical possibility of mechanics as the ground frame .
Even though the track contracting relative to the wheel base in the tank frame would seem to pose some problem regarding stress etc. unless you assume Fredrik's stretching track.

 

[Austin0]

I do not accept that I am in any way being the credulous layman if I continue to assert my certainty that it can be done in essentially the that way I was trying to do it. And I continue to believe, Austin0, that in its essentials, the true solution is the same for a wheel as it is for a tank track. The key point is that the peripheral speed of the wheel or the tank track is constant throughout its revolution and the track is under no tension whatever other than that imparted by the drive wheel. I also reassert my belief that the essential problem is no different at 10mph than it is at 0.45c. Unfortunately, I possesses not the ability to prove it.

I think you might rethink thiese points.IMHO

The track is under constant tension and acceleration throughout.
The force, i.e. momentum of the drive wheel is distributed throughout the whole wheel. Otherwise the tank does not move. The Earth is also accelerating the track by providing counter force otherwise the tank would not move.

This not only means tension between all the links but means it is an accelerating system, even the motionless bottom of the track. In the same way that a coordinate observer in gravity ,hovering through thrust and thus motionless wrt the erath is still accelerating.

If you mean the abstract problem of calculation you may be right. But there is an obvious difference between .45 c and 45 mph as far as physical possibility.
It may in fact be impossible to physically accelerate a wheel /disk in any frme to relativistically significant velocities.
Not because of relativistic affects but simply due to the counter relationship of inertial forces and the binding strngth of intermolecular tensile forces.

 

[yossell]

O Austin0!
I'm just want to see a precise argument for the 0.9c claim. In the hope that: when we clearly see the assumptions that go into the argument, and the chain of reasoning, we'll be able to see where it goes wrong and where and how intuition led us astray and then there'll be enlightenment for all.

If it turns out there is no error, then we've found a contradiction in SR and this thread will ascend into immortality, and it's Nobel Prizes all round. Just a precise, step be step, argument, with premises, logical chains, `this, then this, therefore that' - you know the kind of thing...

-oh, wait.

 

[matheinste]

I assume the original question was posed as an exercise in relativity to show an apparent paradox and for the reader to resolve this. I don't think the setter of the question was after a full understanding of the mechanical engineering aspects of the scenario as the speeds involved are pretty unrealistic in the first place.

Although length contaction is involved it does not need to be invoked specifically. I see the problem as a straightforward basic exercise in reasoning with reference frames and applying the relativistic velocity addition fromula.

As far as 0.9c is concerned this can only be arrived at for the top track relative to the ground using the Galilean transformation. But again earlier posters have already said this.

So forget all the complications and see the problem for what it probably began as--a simple teaching example.

Matheinste

 

[Austin0]

Wouldn't you also say that if 0.45c is acceptable for one track, it must also be for the other? This is what we're saying in the tank frame. Surely this can't be any less obvious than what you said? And yet, by the velocity addition formula, these two statements contradict each other. Matheinste is right. Ich posted the correct solution earlier in this thread.

Both frames agree on a relative v between the tank and ground. Both frames agree the bottom track is motionless in Earth frame and moving -v wrt the tank.
The only contention is the top.

Both Ich's and phyti's logic is equally valid in terms of Newtonian mechanics and physical systems. "Surely this can't be any less obvious than what you said?" is certainly true but it is equally true that it is no more less obvious either.

Just that this disagreement (alternative valid applications of the math) has arisen at all is a clear sign that the OP is not a usual dilemma. The contradiction is not in the v addition formula it is inherent in the problem .

My opinion is that you can apply relative simultaneity, which being reciprocal is not going to change anything, you can argue over contraction etc. but in the end there will be no resolution between the self evident validity of the physics in both frames which leads to contradiction.
phyti and Ich are presenting equally valid perspectives and any choice between them is purely arbitrary and personal.
In one frame the top v must equal bottom -v or the track will disrupt. SO you adjust for this and then in the Earth frame the track disrupts but it is the same track so it appears that both frames agree the track will disrupt

 

[Austin0]

I assume the original question was posed as an exercise in relativity to show an apparent paradox and for the reader to resolve this. I don't think the setter of the question was after a full understanding of the mechanical engineering aspects of the scenario as the speeds involved are pretty unrealistic in the first place.

Although length contaction is involved it does not need to be invoked specifically. I see the problem as a straightforward basic exercise in reasoning with reference frames and applying the relativistic velocity addition fromula.

As far as 0.9c is concerned this can only be arrived at for the top track relative to the ground using the Galilean transformation. But again earlier posters have already said this.

So forget all the complications and see the problem for what it probably began as--a simple teaching example.

Matheinste

How is the tank frame any different? The relative speed of the track is derived through Gallilean transform and Newtonian mechanics, no?

 

[yossell]

Although length contaction is involved it does not need to be invoked specifically.

Ah - interesting. You're probably right. Still, in this case, I think it may need to be invoked - though it depends on what the worry is. I think that there's a *intuitive worry* about how, from the stationary observer's point of view, the tank can function over a period of time without the track getting hopelessly messed up because of the discrepancy of the velocities of the two sides. Because of the discrepancy between velocities, it *seems* as though the upper part of the tread is getting fed more track at the rear of the tank than it is revolving below at the front end of the tank (hence phyti's talk of `amount' of track). It *seems* this cannot go on indefinitely. Yet, from the tank's point of view, there is no problem - upper and lower treads are balanced. So we *seem* to have a puzzle. This is why the tank example may seem more puzzling than just the familiar non-Newtonian addition of velocities.

Once there is distortion/stretching, though, velocities can differ while amount stays the same - and we see how this tank puzzle can be solved.

But until there's a very clear outline of exactly why people think the upper speed should be 0.9c, something as clear as one of Ich's posts, this is just speculation.

 

[matheinste]

How is the tank frame any different? The relative speed of the track is derived through Gallilean transform and Newtonian mechanics, no?

How so! The original speed in the problem given and how it was first derived is irrelevant. Any relative velocities must be arrived at using relativistic callculations. It is after all posed as a relativity problem.

The closing speed of the tracks viewed from the tank rest frame is determined using Galilean methods because we are using two speeds in the same frame viewed from a second frame and this is indeed 0.9c but the figure is irrelevant to the solution.

There is nothing mysterious or fundamentally challenging in this problem. Don't over complicate it.

Matheinste.

 

[matheinste]

Ah - interesting. You're probably right. Still, in this case, I think it may need to be invoked - though it depends on what the worry is. I think that there's a *intuitive worry* about how, from the stationary observer's point of view, the tank can function over a period of time without the track getting hopelessly messed up because of the discrepancy of the velocities of the two sides. Because of the discrepancy between velocities, it *seems* as though the upper part of the tread is getting fed more track at the rear of the tank than it is revolving below at the front end of the tank (hence phyti's talk of `amount' of track). It *seems* this cannot go on indefinitely. Yet, from the tank's point of view, there is no problem - upper and lower treads are balanced. So we *seem* to have a puzzle. This is why the tank example may seem more puzzling than just the familiar non-Newtonian addition of velocities.

Once there is distortion/stretching, though, velocities can differ while amount stays the same - and we see how this tank puzzle can be solved.

But until there's a very clear outline of exactly why people think the upper speed should be 0.9c, something as clear as one of Ich's posts, this is just speculation.

Relativity of simultaneity takes care of the tracks?

Matheinste.

 

[yossell]

Relativity of simultaneity takes care of the tracks?
Matheinste.

Maybe - but the `amount' (pseudo)argument is generated just from the point of view of the stationary observer - from naive calculations, the upper track isn't moving fast enough to get rid of the amount of track from below that's being fed in. Eventually, something must give. It's hard to see how relativity of simultaneity can help here, as he's not changing frame. If, however, because of relativity the upper track has been stretched or is just contracted, then I can see how the amount argument fails.

 

[matheinste]

Maybe - but the `amount' (pseudo)argument is generated just from the point of view of the stationary observer - from naive calculations, the upper track isn't moving fast enough to get rid of the amount of track from below that's being fed in. Eventually, something must give. It's hard to see how relativity of simultaneity can help here, as he's not changing frame. If, however, because of relativity the upper track has been stretched or is just contracted, then I can see how the amount argument fails.

Let's try a simple logic argument as the relativity of simultaneity in this case, although an easy concept is, for me, hard to explain rigorously in a few words in this case. Viewing things in the rest frame of the body of the tank the driving mechanism either stands the strain or it does not and flies apart. You seem to be saying that there is no problem from the point of view of the tank. Let us assume that apart from the ridiculously high speeds that this is so. If the mechanism remains intact in one frame then it remains intact in all frames and of course the converse is also true.

Matheinste

 

[Austin0]

O Austin0!
I'm just want to see a precise argument for the 0.9c claim. In the hope that: when we clearly see the assumptions that go into the argument, and the chain of reasoning, we'll be able to see where it goes wrong and where and how intuition led us astray and then there'll be enlightenment for all.

I have no doubt that you are fully capable of providing a precies argument for the .9 conclusion yourself if you so desired ,using the same logic and mechanics that Ich employed to derive the .45 c velocity in the tank frame.

If it turns out there is no error, then we've found a contradiction in SR and this thread will ascend into immortality, and it's Nobel Prizes all round. Just a precise, step be step, argument, with premises, logical chains, `this, then this, therefore that' - you know the kind of thing...

-oh, wait.

AH! you mean the same kind of of precise step by step argument that Ich presented.
??
As for our impending Nobels I think I won't bother clearing space on the mantel yet. Wait I don't have a mantle.
As for contradiction in SR I don't see that possibility as this question is strictly speaking outside the applicability of the Addition of Velocities equation. WHich is just an equation which does not incorporate physical logic outside of inertial systems.
BTW what was the meaning of your final "-oh, wait.".....? Anticipation of your Nobel or was the coffee ready and you needed a break?

 

[Austin0]

Let's try a simple logic argument as the relativity of simultaneity in this case, although an easy concept is, for me, hard to explain rigorously in a few words in this case. Viewing things in the rest frame of the body of the tank the driving mechanism either stands the strain or it does not and flies apart. You seem to be saying that there is no problem from the point of view of the tank. Let us assume that apart from the ridiculously high speeds that this is so. If the mechanism remains intact in one frame then it remains intact in all frames and of course the converse is also true.

Matheinste

ExaCTLY

 

[Austin0]

Maybe - but the `amount' (pseudo)argument is generated just from the point of view of the stationary observer - from naive calculations, the upper track isn't moving fast enough to get rid of the amount of track from below that's being fed in. Eventually, something must give. It's hard to see how relativity of simultaneity can help here, as he's not changing frame. If, however, because of relativity the upper track has been stretched or is just contracted, then I can see how the amount argument fails.

Isn't the stretched idea itself completely outside physics as we know it?

If you simply assume gearing between the track and both wheels then the "amount" has a precise geometric and quantitive evasluation doesn't it?

If not then it would appear incumbent on you to provide a reasonable alternative view, no?

 

[matheinste]

I have no doubt that you are fully capable of providing a precies argument for the .9 conclusion yourself if you so desired ,using the same logic and mechanics that Ich employed to derive the .45 c velocity in the tank frame.



AH! you mean the same kind of of precise step by step argument that Ich presented.
??
As for our impending Nobels I think I won't bother clearing space on the mantel yet. Wait I don't have a mantle.
As for contradiction in SR I don't see that possibility as this question is strictly speaking outside the applicability of the Addition of Velocities equation. WHich is just an equation which does not incorporate physical logic outside of inertial systems.
BTW what was the meaning of your final "-oh, wait.".....? Anticipation of your Nobel or was the coffee ready and you needed a break?

My argument is against the 0.9c value of the relative speed of the top track relative to the ground. In other words the top track is not moving at twice the speed of the tank relative to the ground.

The use of the relativity velocity addition formula is applicable as I and others have implied throughout the thread by using, or in my case mentioning, the relativity formula for velocities. It is no different from the usual case of a spaceship flying at 0.45c realtive to the Earth and a second one flying at 0.45c relative to the first. The second spaceship is not traveling at twice the speed of the first relative to the eatrth

Matheinste.

 

[Fredrik]

If the tank moves at .45c in the ground frame,
and the top track moves 2x the speed of the tank to function correctly,
then the top track moves at .90c in the ground frame.

Let's assume that this is correct, just to see where it takes us. What you said implies that the bottom track is comoving with the ground. In the ground frame, the tank's velocity is v=0.45 (in units such that c=1). By symmetry (the principle of relativity), or by definition of the inertial frame we associate with the motion of an object, this implies that the velocity of the ground in the tank's frame is -v. So...relative to the tank, the bottom track has velocity -v=-0.45, and relative to the bottom track, the top track has velocity u=0.9 (according to your assumptions). Now the velocity addition formula tells us that the velocity of the top track relative to the tank is (-v+u)/(1+(-v)u))=(-0.45+0.9)/(1-0.45*0.9)=0.45/0.595≈0.756, while the velocity of the bottom track relative to the tank is -0.45. This contradicts your other claim:

The driver/tank must see the same amount of track moving forward as that moving backward, since the amount of track is constant.

So your assumptions can't all be correct.

 

[DrGreg]

O Austin0!
I'm just want to see a precise argument for the 0.9c claim. In the hope that: when we clearly see the assumptions that go into the argument, and the chain of reasoning, we'll be able to see where it goes wrong and where and how intuition led us astray and then there'll be enlightenment for all.

I have no doubt that you are fully capable of providing a precies argument for the .9 conclusion yourself if you so desired ,using the same logic and mechanics that Ich employed to derive the .45 c velocity in the tank frame.

I think you misunderstand the point yossell is making. As far as I can see, unless I've missed it, nobody in this thread has given any reason for the 0.9c velocity in the ground frame that amounts to much more than "it must be", which is no good reason at all. The people who mistakenly believe this to be true need to sit down and think about it and produce a detailed step-by-step proof, then the rest of us can examine the "proof" to find out where the error is.

In the extremely unlikely event that your "proof" has no error, you will go down in history as the person who disproved relativity.

 

[starthaus]

I would like to understand velocity addition and subtraction in special relativity, more than I currently do. It would be greatly appreciated if one could comment on the outcome of the following thought experiment.

Imagine ‘the super-tank’ a tank capable of speeds of 0.45 C is being designed. Someone on the design team, raises a potential problem. When considering the tanks tracks, the tracks that are in-contact with the ground or the bottom-side tracks, have no velocity until the tank moves over them and pulls them to the top-side. The velocity they have according to a stationary observer is twice the speed of the tank.

The member of the design team states that if you treat the tank as the stationary observer, then what ever speed the top-side has the bottom-side needs to have, just in the opposite direction. If this is not the case then the tank tracks would rip apart, as either the top-side or bottom-side tracks would not feed enough track to the other.

It is then shown that the tank will not have the same but opposing velocity for its tracks, in the reference frame of the tank, based on the velocity of the bottom-side tank relative to the ground observer being 0 and the top-side 0.9C.

U = (S-V) / (1-(SV/C^2))
U: The velocity of the tracks according to the tanks frame of reference.
S: The velocity of the tracks according to the ground frame of reference.
V: The velocity of the tank according to the ground frame of reference.
C: the speed of light in a vacuum.

Bottom-side velocity according to the tank
U = (0 - 0.45C) / (1-(0C*0.45C/C^2))
= -0.45C / 1
= -0.45C

Top-side velocity according to the tank
U = (0.9 - 0.45C) / (1-(0.9*0.45/C^2))
= 0.45C / (1-0.405)
= 0.45C / 0.595
= 0.7563C

How can this difference exist considering the concerns of the designer?

The tank body advances at $v=0.45c$ wrt the ground.
Since the tank rolls over the bottom track without slipping, then, in the tank frame

-the bottom of the track moves at $v_b=-0.45c$
-since the track doesn't stretch, the top of the track moves at $v_t=+0.45c$

In the gound frame the speeds are different:

$$v'_t=\frac{v_t+v}{1+vv_t/c^2}=.748c$$
(not 0.9c as you keep claiming)

$$v'_b=\frac{v_b+v}{1+vv_b/c^2}=0$$

$$v_{tank}=v=0.45c$$

This is a simple problem, it does not merit so many posts.

 

[Fredrik]

In the extremely unlikely event that your "proof" has no error, you will go down in history as the person who disproved relativity.

I'd like to add that since we're just talking about Minkowski spacetime, which is just the set [tex]\mathbb R^2[/itex] with a specific bilinear form defined on it, and since the set and the bilinear form both exist by the ZFC axioms of set theory, it would disprove ZFC as well. So we would have to throw out all of mathematics and start over from square 1...if we can assume that the concept of "1" is still valid.

This is a simple problem, it does not merit so many posts.

Ich posted your solution twice on the first page. Caleb et al are simply ignoring it, because the speed of the top track "must" be twice the speed of the tank. (That's their argument, not mine).

 

[starthaus]

Ich posted your solution twice on the first page. Caleb et al are simply ignoring it, because the speed of the top track "must" be twice the speed of the tank. (That's their argument, not mine).

I've been scooped
:LOL:

 

[Ich]

phyti and Ich are presenting equally valid perspectives and any choice between them is purely arbitrary and personal.
In one frame the top v must equal bottom -v or the track will disrupt. SO you adjust for this and then in the Earth frame the track disrupts but it is the same track so it appears that both frames agree the track will disrupt

Sorry, but as long as you don't take the time to read what I've written, it'd be a waste of time to write more. We're exactly where we've started.