# I Velocity addition example, another confusing one from Morin

1. Dec 26, 2016

### Oz123

Ok, so here's another example Morin did in his book:

Ok, so the thing that confuses me is when he used the velocity addition equation:

Note:4c/5 and 3c/5 are their speeds relative to the ground.
Where v2 is the velocity of the S' frame, and v1 is the velocity of the object moving in the S' frame.
For the B's point of view, he used v1 as the velocity of A, and v2 as the velocity of B...Now, the thing that confuses me is that the velocity of B in B's frame should be zero, so why did he used v2 as non zero in B's frame? He seemed to just plug in numbers and don't explain why he used those numbers, can anyone explain to me why he used those numbers in? I would understand a bit if it's in the ground frame, v2 wouldn't be zero, but then v1 should be the velocity of A with respect to B not with respect to the ground.

2. Dec 26, 2016

### Ibix

$v_1$ and $v_2$ are the velocities of the trains in the ground frame. $u$ is the velocity measured in the frame where the train moving at $v_1$ is at rest.

One way to see what's going on is to take the non-relativistic limit, $v_1,v_2 <<c$. Then the denominator is very close to one and can be disregarded. Then $u=v_1-v_2$. So if you're a pedestrian and you see two cars doing 60mph in opposite directions in your frame, you know that one is stationary ($u=60-60$) while the other is doing 120mph ($u=60-(-60)$) seen from a frame doing 60mph.

The only change for the relativistic case is that velocities add in a slightly more complicated way (please do work out my car example using the full relativistic formula - I think the difference is on the 14th significant figure).

Note that Morin's 10.29 and your version of the velocity addition formula appear to be using opposite sign conventions for $v_2$. Not sure if this is simply a mistake, or indicative of some deeper misunderstanding.

Last edited: Dec 26, 2016
3. Dec 26, 2016

### Oz123

Thanks a lot!
Ya, I don't know why he used opposite signs, both the trains are going in the same direction, but the velocity addition formula right there is the one derived by Morin.

But v1, which is the train A is not at rest on any of the frame besides the train A itself, in the first calculation, he used B's frame as the rest frame, so I guess u is the velocity measured in the frame where the train moving at v2 (train B) is at rest?
From what I understand, suppose we use this instead:
$u = \frac {u' + v} {{1 + \frac {v u'} {c^2}}}$
(Morin's notation is quite confusing), but here I define u' as the velocity of thing A in the moving frame B (so v of A with respect to B), v is the velocity of the moving frame with respect to some other frame C (So v of B with respect to C). So u is the velocity of A with respect to C.
So what he used as u' is the velocity of the train A with respect to C (ground), and he used v as the velocity of the train B with respect to C. So from this, u must be the velocity of A with respect to B, so it makes sense, but why the negative sign? u and v are with respect to C, and in C's frame, they are both going in the same direction.

Last edited: Dec 26, 2016
4. Dec 26, 2016

### Oz123

Oh, I think I understand now why there is a negative sign, in non relativistic velocity, Vac=Vbc+Vba...So Vab=Vac-Vbc where Vab is velocity of a with respect to b etc. So in things like this, we need to look at the non relativistic limit if the vector addition makes sense, right? So if that happens, we also flip the signs in the denominator?

Last edited: Dec 26, 2016
5. Dec 26, 2016

### Ibix

Think about every day speeds. If you see a train, A, doing 60mph and another train, B, doing 70mph then you know that someone looking out of train A's window will see the window frame having velocity 0, train B passing at 10mph, and the landscape going by at -60mph. You are using $u=v_1-v_2$, where $v_2$ is the velocity (in the ground frame) of the person whose perspective you want, $v_1$ is the velocity (in the ground frame) that you wish to transform, and $u$ is the transformed velocity.

The only difference with what Morin is doing is that he's using a more precise calculation of the transformed speeds because $u=v_1-v_2$ is not a valid approximation. So he works out how fast a train going at 0.8c will be travelling, viewed from a train travelling at 0.6c. Then he figures out how long the overtake requires viewed from this frame.

6. Dec 26, 2016

### Oz123

Ya I think I get now, thanks a lot!!

7. Dec 26, 2016

### Oz123

Oh, also if we can think of the rest frame as B, so C is the moving frame with respect to B which is moving the other way, that's where the negative sign goes. Thanks for explaining :)

8. Dec 26, 2016

### Ibix

All that's happening here is a choice of sign convention. The smart thing to do is to declare that all velocities are positive moving to the right (for example). Then $$u=\frac {v_1-v_2}{1-v_1v_2/c^2}$$But some people like to think of it as a velocity addition formula and consider speeds in opposite directions to be both positive, and speeds in the same direction to be one negative one positive. In that case $$u=\frac {v_1+V_2}{1+v_1V_2/c^2}$$ I think that's just confusing, but ultimately the formulae are the same. Everyone agrees that $v_1=-V_1$, they just argue about whether it's easier to think about $v_1$ or $V_1$. I like the first convention because then everything is measured in the same sense and the maths keeps track of which way everything is pointing.
Relativity is a better model of reality than Newtonian physics, so I would be uncomfortable agreeing with this. That said, you can check your working by thinking: if these were highway speeds and I could use thr Newtonian approximation, would the signs come out the way I expect? Better to aim to become comfortable with the relativistic formulae, however.

9. Dec 26, 2016

### Oz123

Yup, thank you! Noted everything :)