Explanation of Galilean transformations

[penroseandpaper]

Hi everyone,

We've just started special relativity and I'm just wondering if you'd mind clarifying something for me.

The transformation is described as x'=x-vt, where x' is moving relative to x. However, in the diagram I've attached, x' is ahead of x ; so why is the transformation described with -vt? Is this because we're calculating x' with regards to frame a, which is "traveling" in the opposite direction to x'and therefore x' sits to the negative of x=0?

I'm confused because if that is the case, why isn't the positive x direction given as that opposite to x'?

Any advice or examples would be greatly appreciated.

Penn

 

[PeroK]

Just put some numbers in. An event at $x = 100, t = 3$ with $v = 10$

The origin of the primed reference frame is at $x = 30$ at $t = 3$, so the primed coordinate for the event is $x' = 70$. That's the distance of the event from the primed origin. Which is the definition of the $x'$ coordinate for the event.

$x' = x - vt$

PS which is just what your diagram shows!

 

[RPinPA]

I'm confused because if that is the case, why isn't the positive x direction given as that opposite to x'?

Keep things simple. You have a pair of x and y axes. You have another pair of x and y axes. One of them is moving off to the right.

What you're doing in these transforms is asking what coordinates are assigned to any given event by two different observers.

Let's say B (the primed frame) is the one moving to the right relative to A, at velocity v, starting at x = 0 at t = 0.

How does A describe B's position? What equation does A use for x? Clearly it's x = vt, right?

Now, how does B describe his own position in primed coordinates? He says that from his point of view, the primed axes are not moving, because he's sitting at the origin of those axes. So he says x' = 0 for all time.

What if B is holding an object in front of him, at x' = a meters? It moves along with him, so A says that object is also moving at velocity v. So A says that object is at position x = a + vt and B says the object is at position x' = a. How do I get x' from x? x' = x - vt.

You're thinking in terms of x = x' + vt I think. Whatever x'(t) something is doing relative to the primed moving axes, x will add vt to that motion. But that makes x' = x - vt.

Suppose we have C moving at 20 m/s relative to A. So A describes C's motion as x = 20t.

Now suppose B is also moving at 20 m/s relative to A. Intuitively you understand that from B's point of view, C appears to be fixed, right? If you're in a fast moving car and you're next to another fast moving car with the same velocity, it stays fixed from your point of view. So A describes C as moving at speed 0, x' = 20t - 20t = 0.

Suppose C is moving at 20 m/s (relative to A) and B is moving at 21 m/s relative to A. You have no problem believing that from B's point of view, C will be receding at 1 m/s, right? A describes C's motion as x = 20t but B describes C's motion as x' = 20t - 21t = -1 t.

And suppose C is moving at 20 m/s and B is moving at 15 m/s. Again A says x = 20t for C, but B describes C's motion as x' = 20t - 15t = 5t.

Do you see that in every case, to get B's description of any motion from A's description, we subtract vt?