Does the Bottom of a Continuous Track Move

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• Special and General Relativity

Velocity subtraction problem

• Thread starter calebhoilday
• Start date Jul 19, 2010

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?

Answers and Replies

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

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.

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.

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.

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.

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

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.

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 travelling at 70mph then from the ground perspective, as you call it, top-dead-centre of the wheel is travelling 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 travelling 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.

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.

Last edited: Jul 19, 2010

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.

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.

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?

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.

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 travelling at 70mph then from the ground perspective, as you call it, top-dead-centre of the wheel is travelling 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 travelling 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.

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 im 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.

...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.

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.

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.

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?

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 prefered 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 dont make good relativistic vehicles???

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 travelling 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 travelling at 0.45C and a second body is travelling at 0.45C relative to the first body, what speed is the second body travelling 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 travelled 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 travelling 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.

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 travelling 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 travelling at 0.45C and a second body is travelling at 0.45C relative to the first body, what speed is the second body travelling 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 travelled 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 [tex]\pi[/tex]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.

...........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 travelling 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

Last edited: Jul 22, 2010

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…

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.

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.

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: [tex]s=\frac{v+u}{1+\frac{vu}{c^2}}[/tex]

Length Contraction Formula: [tex]x'=x\sqrt{1-\frac{v^2}{c^2}}[/tex]

Time Dilation Formula: [tex]t'=\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

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

[tex]\frac{s}{c}[/tex] = 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 = [tex]t=\frac{x'}{s}=3.98 \times 10^-8}[/tex]s.

By the time dilation formula the same time period in the ground reference frame, measured by a clock on board the tank = [tex]t'=4.457 \times 10^-8 [/tex]s.

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 [tex]\frac{x'}{v}=6.62 \times 10^-8[/tex]s, which by the time dilation formula becomes [tex]7.413 \times 10^-8[/tex]s 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!

Ken - bracket the whole thing you want superscripted: 10^{-8}
[tex]10^{-8}[/tex]

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…..!!!

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.

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?

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 prefered 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.).

Last edited: Jul 22, 2010

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).

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?

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.

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