# Velocity subtraction problem

1. Jul 19, 2010

### 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?

2. Jul 19, 2010

### Austin0

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.

3. Jul 19, 2010

### Ich

You explicitly showed that this is not the case.

4. Jul 19, 2010

### 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.

5. Jul 19, 2010

### 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.

6. Jul 19, 2010

### 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.

7. Jul 19, 2010

### Austin0

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

8. Jul 19, 2010

### Ich

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

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.

9. Jul 19, 2010

### 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 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.

10. Jul 19, 2010

### 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

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
11. Jul 19, 2010

### Ich

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.

12. Jul 19, 2010

### 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.

13. Jul 19, 2010

### Fredrik

Staff Emeritus
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?

14. Jul 19, 2010

### Austin0

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.

15. Jul 19, 2010

### Austin0

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.

16. Jul 19, 2010

### 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 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.

17. Jul 19, 2010

### DrGreg

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.

18. Jul 19, 2010

### Fredrik

Staff Emeritus
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.

19. Jul 20, 2010

### 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.

20. Jul 20, 2010

### Fredrik

Staff Emeritus
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?