schroder said:
You are changing the rules of the game, again Vanesch.
The claim of Galilean relativity is that when on the treadmill, the tread is moving with respect to the stationary air while when in the wind the air is moving with respect to the stationary ground and the frames are equivocal.
Yes, so the treadmill in one setup is the equivalent of the ground in the other.
And the observation made on the treadmill is that the cart, by virtue of advancing in the opposite direction to the motion of the tread, is going faster than the tread.
No, of course not. It is not "going faster than the thread". It is going faster WITH RESPECT TO THE TREAD, than the air is WITH RESPECT TO THE TREAD.
Then, by the equivalence of reference frames, the claim is being made that the cart, when in the wind, will go faster than the wind.
Yes, because here also, the cart, WITH RESPECT TO THE GROUND is going faster than the air, WITH RESPECT TO THE GROUND.
It would be a a bit silly to say that the cart is going faster than the ground, wouldn't it ?
What I am saying is: I can show that the cart on the treadmill is actually not going faster than the tread,
Sure, but nobody is claiming that.
with respect to the tread but is going slower.
You cannot have a thing X which goes slower WRT to a thing Y, than the thing Y ITSELF is going WRT to itself, can you ?
Because a thing doesn't move wrt itself. So it is hard to go slower than a thing that doesn't move, right ?
And, once I show this, I want to know if everyone will then accept that when in the wind, the cart will move slower than the wind.
Nope. You compare the wrong things.
What you show is that the cart is going slower WRT the air than the ground is WRT the air.
What people claim is that the cart is going FASTER wrt the ground than the air is WRT the ground.
For that, you have to understand negative numbers of course.
What you show is: | C - A | < | G - A|
What is claimed is that |C - G| > |A - G| = | G - A |
What you show doesn't invalidate the second claim.
Take as an example: C = -2, A = 0, G = 10.
You show: | -2 - 0 | < | 10 - 0| OK
Nevertheless, we have: | C - G | = | -2 - 10| = 12 > |G - A| = |10 - 0| = 10
(yes, 12 > 10)
Note: as in the above inequalities, we only have differences, it doesn't matter if we add or subtract a common number to all of them, right ?
That's the mathematical expression of saying that relative velocities can be calculated in any frame, BTW.