Is Velocity Addition in Special Relativity Contradictory?

[calebhoilday]

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?

 

[Austin0]

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?

Hi...interesting problem.

SO far it has occurred to me that if you replace the tracks with wheels then the lower section velocity never is zero in the rest frame. It is the reciprocal of the top velocity.

Perhaps the problem is centered on the idea that the lower tracks are not moving wrt the ground.

Looking at a single track section gives this impression.

Looking at the lower tracks as a whole they are clearly moving. The front segment descending and the rear segment ascending.

I don't know if any of this might be useful to you. It certainly hasn't cleared up the enigma for me yet.

 

[Ich]

The velocity they have according to a stationary observer is twice the speed of the tank.

You explicitly showed that this is not the case.

 

[calebhoilday]

Hi Ich,

If i was on the ground, and the tank has a speed of 0.45C according to my frame of reference, the top-side has a speed of 0.9C. When I wrote they, i mean't the two bottom-sides of the two tank tracks, not the top and the bottom sides of a tank track. Sorry if that was confusing.

Hi Austin0,

I still remain perplexed. I made it a tank so that for a considerable amount of time the surface of the tank that is in contact with the ground, would be stationary to it. A wheel is essentially the same, except that the surface in contact with the ground is smaller and it returns to velocity rather quickly, so it is difficult to imagine.

 

[Ken Natton]

Not entirely certain that I can add any thing to this, calebholiday, but in the case of the wheel, I recognise the concept you are mentioning, I have encountered it before, and it was presented as one of those illusory non-problems that more experienced students like to tease freshmen with. In the case of the wheel, the apparent problem arises because of an attempt to analyse angular motion as if it is linear motion. If one takes one’s reference point as the centre of the wheel, then clearly all points on the periphery of the wheel have exactly the same angular motion, hence it is not at all mysterious that the wheel does not rip itself to bits. It is an interesting twist to move it then to a tank track where the angular motion around the wheel at each end becomes linear motion between the wheels. But the solution is clearly a very similar point where the peripheral speed of the tank track is constant. And there is no particular need to define it in terms of a tank moving at a practically impossible fraction of the speed of light. The point works perfectly well if the tank is moving at 10 mph. If sufficient force is applied, a tank track being pulled at 10 mph in opposite directions would break apart, whereas the clear reality is that it doesn’t.

 

[calebhoilday]

Hi ken Natton,

It is only from the tank perspective that you would expect the tracks to fail, observations from the ground frame of reference, holds their to be no issue. So if the tracks fail, it would make no sense from the ground frame of reference.

I don't think that the tracks would break. But it is undeniable, if the subtraction formula is true then there will exist a difference between the top-side and bottom side speeds, in the reference frame of the tank. I don't know how one can explain how this difference wouldn't stop the tracks from breaking. To say that the top-side and the bottom side have the same speed in the tanks perspective would be to deny the velocity subtraction formula. The sides of the tracks have a well defined velocity in the ground frame of reference for a conceivable period of time, so using the velocity subtraction formula is appropriate in the situation.

 

[Austin0]

Hi ken Natton,

It is only from the tank perspective that you would expect the tracks to fail, observations from the ground frame of reference, holds their to be no issue. So if the tracks fail, it would make no sense from the ground frame of reference.

I don't think that the tracks would break. But it is undeniable, if the subtraction formula is true then there will exist a difference between the top-side and bottom side speeds, in the reference frame of the tank. I don't know how one can explain how this difference wouldn't stop the tracks from breaking. To say that the top-side and the bottom side have the same speed in the tanks perspective would be to deny the velocity subtraction formula. The sides of the tracks have a well defined velocity in the ground frame of reference for a conceivable period of time, so using the velocity subtraction formula is appropriate in the situation.

If you simply raise the tank off the ground and assume the same relative motion of all parts it would seem that relative to the ground the tracks velocities would be tank velocity plus or minus a reciprocal value .
Perhaps the problem is: The difference between a wheel and a tank track dissappears.
Tracking any particular track segment it is clearly traveling at the same essential velocity as the tank with simply a minor reciprocal motion +x and -x . I.e. Over a given interval of time it will have traveled essentially the same distance.

Maybe

 

[Ich]

If i was on the ground, and the tank has a speed of 0.45C according to my frame of reference, the top-side has a speed of 0.9C.

Listen:
The top side does not have a speed of 0.9 c in the ground frame.
Makes more sense now?

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.

From this (valid, btw) logic it is clear that the top side has v=0.45 and the bottom side has v=-.45.
From this, and the velocity addition formula, you get
"If i was on the ground, and the tank has a speed of 0.45C according to my frame of reference, the top-side has a speed of 0.74844C."
And then the subtraction formula works again.

 

[Ken Natton]

Now I’m in the position that while I was composing a reply to you calebholiday, Ich has already supplied a better answer. However, I still say different perspectives often help, so, for what it is worth, here’s what I wrote:


I don’t know for certain calebholiday, but I would still suggest to you that your problem stems from a mistaken analysis. Again, I actually find it easier to conceive in the case of the wheel. The faulty analysis says that if the car is traveling at 70mph then from the ground perspective, as you call it, top-dead-centre of the wheel is traveling at 140mph and bottom-dead-centre is stationary. That means that any point on the periphery of the tyre is constantly accelerating from stationary to 140 mph in one half a wheel spin and then back again to stationary in another half a wheel spin. This might lead you to believe that the wheel is subjected to intolerable forces and yet clearly it isn’t because the wheel survives. Undoubtedly a wheel is subject to some pretty large forces when the car is traveling at 70 mph, but nothing so dramatic as this analysis would suggest, and the whole mystery disappears when you view it as angular motion of the wheel, which recognises that the angular speed of the wheel is constant. In the case of your tank track, you’ve made it more involved but the point is really the same and there is no denial of the velocity addition formula.

I’m not sure if this is really related, other experts on this forum will explain it much better than me, but there certainly is an issue related to electrons and their orbital and spin motions. Again it is perhaps easier to conceive as an ordinary sphere rotating on its axis and moving through space. The net effect of its rotational and linear speeds cannot add up such that any point on the periphery of the sphere moves faster than the speed of light. So the faster the rotational speed, the more limited the linear speed. Now it is often pointed out that macro world spheres spinning and moving through space are a deeply flawed analogy for what happens in the weird and wonderful world of the sub-atomic particle, as I said, others might explain it better than me, but I know that it is an issue of consideration in quantum physics. Whether the explanation would impinge upon your struggle with your tank tracks moving through space I don’t know.

 

[calebhoilday]

If the tank has a speed of 0.45C and the speed of the bottom-side is 0, for the ground frame of reference, then the top side has a speed of 0.9C

http://www.youtube.com/watch?v=3I_KoCeAdEw&feature=related

Watch this and see the grip in the tracks advance forward, faster than the robot. What you are saying is that, the tracks move as fast on the top-side (0.45C) as the tank(0.45). This is clearly does not happen.

 

[Ich]

What you are saying is that, the tracks move as fast on the top-side (0.45C) as the tank(0.45). This is clearly does not happen.

What I've been trying to say is that - in the tank frame - the top side has v=0.45 and the bottom side has v=-.45.

What you're doing:

tank speed = 0.45,
top side speed relative to the tank = 0.45
so top side speed relative to the ground = 0.45+0.45 = 0.9.

What you should do instead:

tank speed = 0.45,
top side speed relative to the tank = 0.45
so top side speed relative to the ground = (0.45+0.45)/(1+0.45²) = 0.74844.

 

[calebhoilday]

The tanks speed relative to the ground must be half the top-side speed relative to the ground.

The top-side speed 0.45C relative to the tank, would only occur in classical relativity.

 

[Fredrik]

Caleb, you seem to be doing exactly the same thing that you're doing in your other threads, which is to completely ignore the answers you're getting. Ich knows what he's talking about, and you know that you don't, so why don't you listen to him?

 

[Austin0]

The tanks speed relative to the ground must be half the top-side speed relative to the ground.

The top-side speed 0.45C relative to the tank, would only occur in classical relativity.

I think that ich is applying the addition formula from the perspective of the tank and deriving the top speed relative to the ground from that,,, while you seem to be applying it from the ground to the tank and top.
But your velocity for the top in the ground frame is incorrect.

Unless I am totally misreading ich's analysis.

 

[Austin0]

Now I’m in the position that while I was composing a reply to you calebholiday, Ich has already supplied a better answer. However, I still say different perspectives often help, so, for what it is worth, here’s what I wrote:


I don’t know for certain calebholiday, but I would still suggest to you that your problem stems from a mistaken analysis. Again, I actually find it easier to conceive in the case of the wheel. The faulty analysis says that if the car is traveling at 70mph then from the ground perspective, as you call it, top-dead-centre of the wheel is traveling at 140mph and bottom-dead-centre is stationary. That means that any point on the periphery of the tyre is constantly accelerating from stationary to 140 mph in one half a wheel spin and then back again to stationary in another half a wheel spin. This might lead you to believe that the wheel is subjected to intolerable forces and yet clearly it isn’t because the wheel survives. Undoubtedly a wheel is subject to some pretty large forces when the car is traveling at 70 mph, but nothing so dramatic as this analysis would suggest, and the whole mystery disappears when you view it as angular motion of the wheel, which recognises that the angular speed of the wheel is constant. In the case of your tank track, you’ve made it more involved but the point is really the same and there is no denial of the velocity addition formula.

I’m not sure if this is really related, other experts on this forum will explain it much better than me, but there certainly is an issue related to electrons and their orbital and spin motions. Again it is perhaps easier to conceive as an ordinary sphere rotating on its axis and moving through space. The net effect of its rotational and linear speeds cannot add up such that any point on the periphery of the sphere moves faster than the speed of light. So the faster the rotational speed, the more limited the linear speed. Now it is often pointed out that macro world spheres spinning and moving through space are a deeply flawed analogy for what happens in the weird and wonderful world of the sub-atomic particle, as I said, others might explain it better than me, but I know that it is an issue of consideration in quantum physics. Whether the explanation would impinge upon your struggle with your tank tracks moving through space I don’t know.

I think this is a fallacy. AN illusion from looking at a single instant in time. The motionless arrow in flight.
The bottom point of the actual wheel is never motionless wrt the road outside of a durationless instant. And the top orf the wheel is never moving 140 wrt the road. As soon as time starts up the top is translating linear motiuon into angular with a decreased linear velocity wrt the road and the car.

The electron question is indeed intriguing. My conclusion was that electron orbits would be ellipses or ellipsoids as a result of the linear v limitation of approaching c.

 

[calebhoilday]

The topside speed relative to the tank is based on the measurements of 0.45C speed from a ground observer and the fact that the topside speed relative to the ground is double the tank speed.

It has to be this or it doen't operate in the ground frame.

tank speed = 0.45, (if the two others are right this is wrong)
top side speed relative to the tank = 0.45 (no this is affected by the high velocity)
so top side speed relative to the ground = (0.45+0.45)/(1+0.45²) = 0.74844.

Anyone reading draw a diagram of the tank from a side view and from the start use classical relativity from the ground frame, based on the speed 0.45C of the tank, the fact that 0 is the the bottom-side speed and the top-side speed must be double the tank speed.

Then work out the speeds relative to the tank using classical relativity. Top-side (0.9C-0.45C) bottom side (0-0.45C). Now multiply 1 / (1-(SV/C^2)) by the classical formulas, that I am sure everyone agrees with and get the special relativity answer.

The special relativity velocity addition formula and velocity subtraction formula inherently doesn't maintain proportion that a tank seems to require at first glance due to S or U being in the formula.

someone should be able to explain what is going on without dismissing the measurements and calculations.

 

[DrGreg]

...the fact that the topside speed relative to the ground is double the tank speed.

Several people in this thread have told you this is wrong, and given the reason why, and now I'm telling you too. Why do you think it is right? Sometimes in relativity the things that you think are "obviously true" turn out to be false.

Go back and re-read the reasons you have already been given why the topside speed is 0.74844c, not 0.9c.

 

[Fredrik]

Caleb, this is a tricky problem, so it's very easy to get confused here. You should probably focus on the more basic stuff until you understand them. One thing you're missing here (quite understandably) is that the topside is Lorentz contracted and forcefully stretched. This will mess up the arguments that you probably used to convince yourself of the wrong value of the topside speed in the ground frame.

 

[calebhoilday]

I know length contracts and I know from the ground the speed of the top side must be double the speed of the tank. I'll be unable to reply for a couple of days due to work commitments. This conclusion is unacceptable and is evading the problem. If anyone agrees they should make their voice heard.

 

[Fredrik]

Do you think we're all lying to you? Your claim implies that the velocities in the tank frame of the two sides are not v and -v. Do you really think that makes more sense than the topside speed being different from 2v in the ground frame, where the top is Lorentz contracted and the bottom isn't?

 

[Austin0]

I know length contracts and I know from the ground the speed of the top side must be double the speed of the tank. I'll be unable to reply for a couple of days due to work commitments. This conclusion is unacceptable and is evading the problem. If anyone agrees they should make their voice heard.

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.

Having thought it over more I have come to realize that it doesn't resolve the query really.

SOme thoughts: The question was what figure would Addition of v in the Earth frame provide?

1) The resolution takes an assumed answer to the question and inserts it as a premise. Then works the math from there.
Providing a number which conflicts with the expected number in the Earth frame.
Besides begging the question it also seems to indicate a preferred frame i.e. the tank.
That measurements in that frame or assumed measurements would be intrinsically more valid and should be inserted as valid measurements in another frame.

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.
The tank can be made as long as you want for convenience.
The distance and time are recorded with exactly the same ruler and clocks that were used to derive the velocity of the tank itself.
This would seem to pose a real question that seems to go beyond merely counter intuitive.

WHat principle of SR or physics would explain how this could be possible?

Fredrik's point about contraction is very relevant to the total question but regarding the simple measurement of distance traveled by a single point it doesn't seem to apply.



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.
This of course may be the answer to the engineering side of the question; I.e. a .45c tank is just not a vialble physical possibility.

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 .
Aside from the rapid ,to say the least, decceleration required turning from top to bottom it would seem to pose a question for the reciprocity of the L contraction to reconcile.

Perhaps this all just "proves" tanks don't make good relativistic vehicles?

 

[Ken Natton]

Hi Austin0,

I don’t question your tenacity on this point. Certainly your approach has a greater ring of rationality about it than Calebholiday’s approach. It is okay to be sceptical and to question, but Calebholiday seems to just blindly refuse to accept anything he / she is told.

Of course, I defer absolutely to Ich and Fredrik as contributors who have a much deeper understanding than my own. But I would deny just slavishly agreeing with them and effectively saying, ‘yeah, whatever they said’. I think I have an explanation that works in my own mind. So what I propose to do is to try to lay out the way I have rationalised it. It might be that this explanation is wrong, and if it is, I’m sure Ich and / or Fredrik will not hesitate to point up its flaws. Perhaps from that will emerge a better understanding for all of us.

As a side issue, I remain convinced that the correct explanation is absolutely no different for the tank track than it is for an ordinary wheel, or for a rotating sphere traveling in space, but since you seem to prefer the tank track version, let’s stick with that. Pedantic point I know, but for the record, only one wheel would be a drive wheel. The other would be a free-running carrier wheel. The problem isn’t just speed synchronising the two drives with sufficient accuracy, it is also nearly impossible to get them to load share properly. You would almost certainly find that one drive was taking nearly all of the load in any case. Easier to have one drive and one carrier wheel.

In any case, in the simple, vanilla flavoured version of the velocity addition formula, from a stationary reference frame, if one body is traveling at 0.45C and a second body is traveling at 0.45C relative to the first body, what speed is the second body traveling relative to the static reference frame? Newtonian mechanics would say 0.9C, relativity says 0.74844C. (I have taken the dangerous step of not doing the calculation for myself, I have assumed Ich’s mathematics are up to scratch.) I’m assuming nobody has a problem with that, not even Calebholiday.

But it is an important subtlety to grasp that the relativity value would raise an anomaly in terms of the distance traveled in the time taken, except for the length contraction and time dilation which must also be considered. All three (velocity addition, length contraction, time dilation) are driven by the Lorentz Transformation, thus it should not be a surprise that they balance out.

So, in our tank track situation, this explanation should work perfectly for the top side of the tank track. The body of the tank is our first body, and the top side track is our second body. Beyond that the explanation is identical.

The confusion seems to centre on the returning tank track because, in the reference frame of the tank body, the bottom side is traveling in the opposite direction making it a negative value. But the calculation is the same. It is still the same velocity addition formula, it is just that the second value is now negative. The same corresponding operations must be done to length contraction and time dilation to balance out the distance covered in the time, but my expectation is that the mathematics would work out. (I can see that I can’t just leave this to Ich. I am going to have to do the calculations myself to check that they do. I will be astonished if they don’t.)

But, of course, it is a critical point that these things only apply in the static reference frame. In the reference frame of being on board the tank the idea that the opposite directional length contractions and time dilations balance out should not be a surprise.

 

[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.
The tank can be made as long as you want for convenience.
The distance and time are recorded with exactly the same ruler and clocks that were used to derive the velocity of the tank itself.
This would seem to pose a real question that seems to go beyond merely counter intuitive.

WHat principle of SR or physics would explain how this could be possible?

Fredrik's point about contraction is very relevant to the total question but regarding the simple measurement of distance traveled by a single point it doesn't seem to apply.

Hi Austin0,

I don’t question your tenacity on this point. Certainly your approach has a greater ring of rationality about it than Calebholiday’s approach. It is okay to be sceptical and to question, but Calebholiday seems to just blindly refuse to accept anything he / she is told.

Hi Ken Natton
My post had nothing to do with Calebholiday. I hadn't paid any attention to his posts beyond the initial problem.
As I said I completely accepted Ich's solution and was ready to forget the question but the "tenacity" of my thought process kept examining the problem.

Of course, I defer absolutely to Ich and Fredrik as contributors who have a much deeper understanding than my own. It might be that my explanation is wrong, and if it is, I’m sure Ich and / or Fredrik will not hesitate to point up its flaws. Perhaps from that will emerge a better understanding for all of us.

I am also sure they wont

As a side issue, I remain convinced that the correct explanation is absolutely no different for the tank track than it is for an ordinary wheel, or for a rotating sphere traveling in space, but since you seem to prefer the tank track version, let’s stick with that.

It was not my preference but the conditions set for the problem.
If you think there is no essential difference between tracks and a wheel then I would have to disagree.

In any case, in the simple, vanilla flavoured version of the velocity addition formula, from a stationary reference frame, if one body is traveling at 0.45C and a second body is traveling at 0.45C relative to the first body, what speed is the second body traveling relative to the static reference frame? Newtonian mechanics would say 0.9C, relativity says 0.74844C. (I have taken the dangerous step of not doing the calculation for myself, I have assumed Ich’s mathematics are up to scratch.) I’m assuming nobody has a problem with that, not even Calebholiday.

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

But it is an important subtlety to grasp that the relativity value would raise an anomaly in terms of the distance traveled in the time taken, except for the length contraction and time dilation which must also be considered. All three (velocity addition, length contraction, time dilation) are driven by the Lorentz Transformation, thus it should not be a surprise that they balance out.

This the crux of my proposition. I wouldn't consider it a subtlety.

Without getting distracted with engineering, the mechanical physics and the other factors of contraction etc., you have this essential problem.

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 \pix 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.

...If there is some explanation why this would or could not be true ;great, But keep in mind it would have to apply in the track frame equally.

...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.

...Now you can run this around in circles (wheels) in your mind and if you come up with an explanation of how this is possible then I will be amazed and impressed .. :-)

The confusion seems to centre on the returning tank track because, in the reference frame of the tank body, the bottom side is traveling in the opposite direction making it a negative value. But the calculation is the same. It is still the same velocity addition formula, it is just that the second value is now negative. The same corresponding operations must be done to length contraction and time dilation to balance out the distance covered in the time, but my expectation is that the mathematics would work out. (I can see that I can’t just leave this to Ich. I am going to have to do the calculations myself to check that they do. I will be astonished if they don’t.)

[\QUOTE]
...I am sure you can take Ich's calculations to the bank but
if you are not familiar with it there is a fantastic on line calculator for additive velocities and all other Lorentzian transforms. Plus much else.
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

 

[Ken Natton]

Ah Austin0, and now I have to confess to you that I am struggling a little bit with the mathematics. I have a vision of Ich and Fredrik chuckling behind their hands. The difference between me and some other contributors to this forum is that I recognise that my struggle is rooted in my limitations, not in the limitations of physical science or, even more ridiculously, in some grand conspiracy by the physics establishment to mislead me.

I am trying to do it with some arbitrary actual values. I am sure that it can be done logically using an algebraic approach that would stand for all actual values, something more akin to your approach. I confess that I do not completely follow your logic, but I will spend a little longer going through it, to see if I can get what you are saying. And in due course, whether or not I get to a solution that satisfies me, I will present exactly what I have done, exposing all my inadequacies for the amusement of all, in the hope that one of the experts will come on and point out where I am going wrong. But I am not out of ideas yet, so just a little longer if you please…

 

[calebhoilday]

Im back... i think that as much as the velocities in one frame have to be balanced they have to be in the another. 0.9C topside from the ground frame required for the tank to have a velocity of 0.45. You may turn the problem around and find S rather than U based on the assumption of balance and say that the topside according to the tank is 0.45C but then your not going to get 0.9C as S which is required for the appropriate balance. I just have a preference for using the 0.9C assumption rather than the 0.45C assumption. Either way there this a problem.

 

[matheinste]

Several people in this thread have told you this is wrong, and given the reason why, and now I'm telling you too. Why do you think it is right? Sometimes in relativity the things that you think are "obviously true" turn out to be false.

Go back and re-read the reasons you have already been given why the topside speed is 0.74844c, not 0.9c.

Im back... i think that as much as the velocities in one frame have to be balanced they have to be in the another. 0.9C topside from the ground frame required for the tank to have a velocity of 0.45. You may turn the problem around and find S rather than U based on the assumption of balance and say that the topside according to the tank is 0.45C but then your not going to get 0.9C as S which is required for the appropriate balance. I just have a preference for using the 0.9C assumption rather than the 0.45C assumption. Either way there this a problem.

As DrGreg and others have told you, the speed of the top of the track relative to the ground is not 0.9c no matter what common sense may tell you.

Matheinste.

 

[Ken Natton]

Okay. I have some crisis of confidence in the value of this. I can only hope it leads to some kind of resolution. In any case, explicitly, the formulae I have used are:

Velocity Addition Formula: s=\frac{v+u}{1+\frac{vu}{c^2}}

Length Contraction Formula: x&#039;=x\sqrt{1-\frac{v^2}{c^2}}

Time Dilation Formula: t&#039;=\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}

I have not derived these formulae myself for this exercise. At some level, they are articles of faith on my part that they are accurate. Although I have previously seen the derivation of the Lorentz Transformation and know that it can be derived with nothing more complex than Pythagoras’ Theorem.

So:

c = 299 792 458 m/s

Tank body velocity = v = 0.45c = 134 906 606 m/s

Upper tank track velocity relative to the tank body = u = 0.45c = 134 906 606 m/s

Upper tank track velocity relative to the stationary reference frame = s = 224 376 892 m/s

\frac{s}{c} = 0.7484407 – Ich’s maths were correct! Surprise, surprise!

Since it has not previously been specified, I have decided to assign an arbitrary wheel base of 10m. That is 10 metres from top dead centre of the rear wheel to top dead centre of the front wheel. I am not considering motion of the track around the wheels, I am only interested in resolving the linear forward motion of the track with the linear rearward motion. If that resolves, the problem is solved, it seems to me.

By the length contraction formula, in the ground reference frame:

Wheel base = x’ = 8.93m

Time measured by a clock in the ground reference frame for any point on the track to cover the wheel base going forward = t=\frac{x’}{s}=3.98 \times 10^-8}s.

By the time dilation formula the same time period in the ground reference frame, measured by a clock on board the tank = t&#039;=4.457 \times 10^-8s.

To be honest, I’m not entirely certain that the last two assertions are correct. But this is where I really hit a difficulty. If I try to use the velocity addition formula for the rearward motion of the returning track, because v = u, then v – u = 0. Therefore the answer equates to zero. The suggestion is that in the ground reference frame the retuning tank track is actually static. Therefore, the time for any point on the returning tank track to get from the front wheel to the back wheel, is actually the time it takes the tank itself to move forward such that the rear wheel now occupies the position that the front wheel did at the start. Thus, that time measured by a clock in the ground reference frame is \frac{x’}{v}=6.62 \times 10^-8s, which by the time dilation formula becomes 7.413 \times 10^-8s in the ground reference frame, measured by a clock on board the tank.

Is this the anomaly Austin0 refers to? Or have I got something seriously wrong?


P.S. I'm trying Latex for the first time. Could not get the x 10^-8 notation to work whatever I tried! Very frustrating!

 

[yossell]

Ken - bracket the whole thing you want superscripted: 10^{-8}
10^{-8}

 

[Ken Natton]

Yes, thanks yossell, I thought I had tried that. I think there is an additional problem that if you edit the Latex configuration, the preview doesn’t necessarily update. I had aborted the whole post once and re-copied it to get around that.

In any case, never mind that. What about the maths yossell? I am not yet a convert to Austin0 and Calebholiday’s view, but I’m sliding down the ship’s deck into the shark’s mouth. Save me yossell! Save me…..!

 

[yossell]

Ken! What can I say? Relativity is a con and all the self-proclaimed experts have been lying through their teeth to keep it alive for years now. This is news?

But seriously...

I think the tank is an interesting situation to think about. But I think it's complicated: it involves changes in velocity - that means accelerations and that means forces. Why? Because the parts of the track change reference frame as they move from top to bottom. This makes a full analysis of how exactly the tracks spin round the wheel complex and, without further account of the forces acting on the track, the situation is probably under described. As Fredrik mentioned, since the bottom track is effectively stationary in the stationary reference frame, it will not be Lorentz contracted, while the top of the track will be extremely Lorentz contracted. From the stationary frame, it seems as though the bottom tracks will be much longer than the top of the track - and this complication may be important.

What I'm waiting for, though, is a clear explanation of what exactly the problem is supposed to be. Then we can analyse and see how the paradox plays out.

For my own part...I'm not at all sure about what you're doing with your wheel base formula. It looks as though you're only computing the distance between the axes of the wheels? But, because of the lorentz distortion, this does not seem to be the same as the length of the track at the base.

 

[Fredrik]

Ah Austin0, and now I have to confess to you that I am struggling a little bit with the mathematics. I have a vision of Ich and Fredrik chuckling behind their hands.

Well now I'm chuckling a bit , but I have to admit that I haven't read your discussion with Austin. I might have a look at it later, but I haven't yet.

Im back... i think that as much as the velocities in one frame have to be balanced they have to be in the another.

That sounds reasonable at first, but in SR, it's a logical contradiction. The fact that in the tank frame, the top and bottom are Lorentz contracted by the same factor, while in the ground frame they aren't, should give you enough to at least suspect that it's the ground frame that you're wrong about. The contradiction implies that you're wrong about one of them, and which is more likely, that you're wrong about the frame in which the description is simple, or about the frame in which it's hard?

 

[Ich]

Ok, I'm back, and I think it's worthwile to explain a bit more.

First, a comment:

Besides begging the question it also seems to indicate a preferred frame i.e. the tank.
That measurements in that frame or assumed measurements would be intrinsically more valid and should be inserted as valid measurements in another frame.

Yes, the tank frame is preferred: It's the only one where beginners have a nonzero chance of even guessing the right result. In the other frames, it's not the question whether or not you make a mistake. The question is how many mistakes you make, and if you at least manage to give a wrong answer, or end up in the "not even wrong" camp. That's nothing to do with your individual abilities (I'm talking about the general case). It's intrinsic to the learning process: The math is deceptively easy, but the underlying concepts need a huge amount of discipline and practice to be mastered. Even worse, as a layman, you're normally not even exposed to the concepts, but live necessarily in the pop-sci ETHER world.

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.

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.

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.
This is not really important for the problem at hand, but it shows that even the seemingly easiest case is trickier than one would think.

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 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.

If we observe the upper track, we see that it moves at v=0.74844, while the middle plane moves at v=0.45. We now invoke what has been dubbed the "closing speed" in recent threads. It is v1-v2=0.29844 for the upper track.
The closing speed for the lower track is 0-0.45 = -0.45.

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.

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.

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.).

 

[DrGreg]

I think there is an additional problem that if you edit the Latex configuration, the preview doesn’t necessarily update.

This is a known bug in the website. To work round it, use your internet browser's "reload" or "refresh" command after you press the preview button (and then resend when prompted to do so).

 

[calebhoilday]

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.

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) 45 of these must pass from the top to the bottom. The speed of these dots must overtake the speed of the tank from a ground perspective for any to be feed in at all. In classical relativity the speed of the dots would have to be 0.9C, but due to length contractions, these dots are closer together and so S should be slower than 0.9C. 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?

 

[calebhoilday]

P.S. This was just an adaption of a problem I was shown in a intro physics course in 07, it involved a wheel and it took a bit to understand (the tank is easier to explain). What should of been posted in this tread, was something similar to the above, which i think is still worded bad.

I was hoping that someone would give a good explanation, in the hope they could also explain the my true problem. I do not see how relativity of simultaneity is integrated into velocity formula's.

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.

 

[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 \pix 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 \pix 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.