# I Galilean transformation paradox help

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1. Sep 16, 2016

### kevinki

I'm getting quite stuck on this problem here.
Galileo said that Xb = Xa - V*Ta.
(This follows from dv = dx/t --> Xa - Xb = t*dv --> the above formula)
Thus, it is concluded Xa = Xb + V*Ta, but why?
In my thought experiment the objects are moving relative to each other,
thus if A is moving away from B, then B is moving away from A.
(I also read that it is convention for the Doppler-effect that V is + for source and observer approaching each other and - for source and observer receding each other)
So should V not be positive or negative for both equations?

2. Sep 16, 2016

### Staff: Mentor

Moderator's note: moved to Classical Physics since it is about the Galilean transformation, not the Lorentz transformation.

3. Sep 16, 2016

### Ibix

Think about what the equations mean. I have a ruler at rest in front of me that shows what you've called Xa. I slide another ruler past it at a steady velocity V - that ruler shows Xb. If I decide to call the instant when the two rulers' zero points coincide time=0, then I can immediately see that the first expression tells me the Xb that corresponds to any given Xa at the time Ta - just think about where the origin of the second ruler (Xb=0) is in terms of Xa and Ta.

I can make the same argument starting with the second ruler. The only thing to note is that if the second ruler was moving in the +x direction according to the first then the first ruler must be moving in the -x direction according to the second - which is where the opposite sign comes from.

Different conventions can be used. The convention used above is that we pick a direction and call it +x, and velocities are positive in that direction. Note that V is the velocity measured in the "a" frame in both equations.

4. Sep 16, 2016

### Staff: Mentor

No. You are mixing up different sign conventions which are used for different purposes.

For the Doppler effect, the sign convention is + for approaching and - for receding, because we are looking for the frequency shift as seen by the receiver. So the key thing is the motion of the source relative to the receiver. So if we change receivers, we change the sign convention.

For the Galilean transformation, there are two frames involved--as you've labeled them they are the a frame and the b frame. The equations you wrote are transformations from one frame to the other, and the key condition they must satisfy is that each one is the other's inverse. To make that work, we must adopt a sign convention for V such that the motion of a relative to b is opposite in sign from the motion of b relative to a.

In other words, the + direction of V picks out a direction in space that is the same for both a and b; it doesn't shift around when we switch frames. Or, to put it another way, the transformation equations as you have written them assume that we have constructed the two frames so that the + direction of V is the same direction in space (for example, pointing towards the star Canopus) for both.

5. Sep 16, 2016

### kevinki

This makes it clearer

Yes, I did get it messed up.
I THINK to sum it up:
With the Doppler effect you assume that the receiver is static (therefore + and -)
With the Galilean Transformation you assume the environment is static (therefore fixed cartesian co-ordinate systems)

I think when you get caught up in special relativity, you forget there is such a thing as fixed directions in space.
It would seem more instinctual to use the + for approaching and - for receding convention, but with that you can not form the theory.
You would get Xb = Xa + V*Ta and Xa = Xb + V*Tb

6. Sep 16, 2016

### PeroK

Why not look at it graphically? With the usual x-axis, the formula assumes that a positive $V$ is motion of b to the right, and that he origins coincide at $t = 0$.

A point a distance $X_a$ from the a-origin at time $t$ will be $X_a - Vt$ from the b-origin, simply because the b-origin has moved $Vt$ to the right.

If you try to draw that, you may see what Galileo saw. (Assume $V$ is positive for simplicity.)

You can then look at it from b's perspective, with a moving to the the left at $V$. You should get the same formula.