# Help with Galilean transforms?

1. Apr 1, 2014

### emmac

I am a little confused about the Galilean transforms.

If I have t'=$\gamma$t

Is t ALWAYS the frame of reference that we are in?
i.e. If a spacecraft is moving away from Earth, and I wanted to measure the time taken on the spacecraft to reach some distance, would t be the variable I should look for?

If yes, does the length contraction formula (l'=$\gamma$l) work the same way?
i.e. If I had two men at rest, a distance d from each other, and a spacecraft flies past and I want to measure the distance someone on the spacecraft travels - would l be the variable I wanted to find?

Help would be greatly appreciated!

2. Apr 1, 2014

3. Apr 1, 2014

### ghwellsjr

Yes, it appears you are confused. There's no gamma in the Galilean transforms. I think you mean the Lorentz transform.

Everybody and everything is in every frame of reference. I always use the unprimed variables as the defining reference frame and the primed frame as the one I want to transform all the coordinates of all the events to. You decide if you want the Earth to be the defining frame or not. You can do it anyway you want.

You really should learn to use the full transformation equations to figure out problems like this, they always work. The shortcut formulas can get you into trouble unless you know how to apply them.

However, I would suggest that you use units where c=1 and velocity is in terms of a ratio to the speed of light which we call beta, β. And then if you limit all motion to the x-axis, you don't have to consider the y- and z-axes and then you can draw a spacetime diagram for your defining scenario.

I have already drawn a great many diagrams depicting lots of different scenarios. If you do a search on my name for "diagram" you might find one that answers your questions.

4. Apr 2, 2014

### emmac

Ah, I see. The thing is, I can use the full Lorentz transformation eqns in certain situations - but I don't understand when the Lorentz transforms reduce down to the 'shortcut' formulae I posted earlier.

i.e. in x'=$\gamma$(x-ut)

In what situation is x = 0?

I can understand the situation when there are two events occuring in each frame of reference, then Δx is the distance between the two events in the unprimed frame - but I would really appreciate if you could explain to me when x=0 for a general situation. Thank you

5. Apr 2, 2014

### Staff: Mentor

$x$ is always zero at the origin of the unprimed coordinate system. If we say that I am at rest using the unprimed coordinates and that the origin of the unprimed coordinate system is where I happen to be at noon today, then:
At noon, the coordinates of my location in space-time are $(x=0,t=0)$. One second later, they will be $(x=0,t=1)$, a second after that, they will be $(x=0,t=2)$, and so forth.

It's worth taking a moment to think about two other people here: my brother who is also at rest relative to me but standing one light-second away; and you, traveling by by at speed $v$ and coincidentally passing me at exactly noon. The coordinates of my brother in the unprimed system at noon are $(x=1,t=0)$, one second later they will be $(x=1,t=1)$, a second after that they will be $(x=1,t=2)$, and so forth. Meanwhile, the coordinates of the ship in the unprimed system will at those three times will be $(x=0,t=0)$, $(x=v,t=1)$, and $(x=2v,t=2)$; in general they will be $(x=vt,t=t)$. (It's a good exercise to plug that general value into the Lorentz transforms to see what the coordinates of the location of the ship look like in the primed frame).

6. Apr 2, 2014

### emmac

Thank you very much for clearing that up - it makes a lot of sense. I hope you don't mind me asking another question, but in the formula:

$x'=$$\gamma$$(x-ut)$

Under what circumstances is t=0?

7. Apr 2, 2014

### Staff: Mentor

In my example above: Using coordinates in which I and my brother are at rest, $t$ is zero at noon everywhere. You'll note that my brother was at the point in spacetime with coordinates $(x=1, t=0)$ at noon, for example; and if he and I had synchronized our wristwatches, his watch will read noon at $(x=1, t=0)$ just as mine at $(x=0, t=0)$ does.

When two events have the same $t$ coordinate, whether it's zero or anything else we say that they are "simultaneous" or that they "happened at the same time". But you have to be careful with relativity of simultaneity here - two events that are simultaneous in using coordinates from one frame are not necessarily simultaneous using coordinates from another frame. For example, the two events "My watch reads noon" at $(x=0, t=0)$ and "the brother's watch reads noon" at $(x=1, t=0)$ are not simultaneous using the primed coordinates in which the spaceship is at rest. (Another good exercise: Calculate the $x'$ and $t'$ coordinates of these two events, verify that the $t'$ coordinates are not equal even though the $t$ coordinates are).

When two events have the same $x$ coordinate, whether it's zero or anything else, we say that they happened "at the same place".

(Once you have this down, you'll be able to derive the "shortcut" formulas for length contraction and time dilation, and that's by far the best of learning when you can get away with using them instead of the more general Lorentz transforms. But here's a hint: the length of an object is the distance between the point where one end is at a given time and the point where the other end is - at the same time).