about the principle of relativity

ENDLESSYOU

When we dealing with questions like "S'-frame moves with respect to S-frame at a velocity u in x direction" then we also assume that "S-frame also moves with respect to S'-frame at a velocity -u in x direction"
Why this assumption is correct?

tiny-tim

Hi ENDLESSYOU! Welcome to PF!

We have to assume that the observers in S and S' are using units (of distance and time) which make their speeds equal …

if S was using metres, and S' was using feet, it wouldn't work!

(btw, this has nothing to do with einstein, it works for galileo also)

ghwellsjr

I think you are asking why the velocity changes sign, correct? Well, isn't it obvious if there is a relative speed between us and I see you moving east with respect to me, then you will see me moving west with respect to you? It really doesn't matter which one or both of us is actually moving, we're talking about relative speed. Doesn't this make sense to you?

ENDLESSYOU

Thank you! But what I really mean is that if they are using the same units ,then why u+u'=0?
Because of symmetry?

ENDLESSYOU

Thank you but I'm sorry my presentation is so poor.My question is why |u|=|u'|? Though it's obvious..

tiny-tim

Because they are also using lined-up direction-names.

S's North is S''s North.

If S's North was S''s South, then u' would be u, not -u !

ENDLESSYOU

My confusion is not the direction but the magnitude of them. Why |u|=|-u|?

ghwellsjr

Are you wondering why the magnitudes of the speeds are the same when one (or both) of them is experiencing time dilation and length contraction?

ENDLESSYOU

No.It's about two frames ,not two particles in two frames.So even in the sense of Galileo,the question still exits.

tiny-tim

I thought you were happy with my previous answer …

ghwellsjr

Would a good answer for you be that when you take any event in any frame and transform it to any other frame moving at u with respect to the first frame and then you transform again from that second frame with -u, you get the same event you started with in the first frame?

ENDLESSYOU

All right.I still can't persuade myself.But thank you all the same.Please give me a minute. Let me think about it .

HallsofIvy

Let's suppose that u sees u' moving east at, say, 10 m/s. If they are next to one another at t= 0 then u will see u' move 100 m in 10 seconds. From u' point of view, u is moving west but he will still measure 100 m between them after 10 seconds. That's why u' sees u moving at 10m/s also but in the opposite direction: [itex]v_u= -v_{u'}[/itex]

ENDLESSYOU

Thank you! But what you explain isn't the point I'm confused.Let me think..

ENDLESSYOU

I think what you say is something like symmetry.Let me think.

Chestermiller

This is not the only relative geometric orientation of the S and S' frames of reference that can be used (obviously), although it does greatly simplify the mathematical form of the Lorentz Transformation. This particular relative orientation is sometimes referred to as Standard Configuration.
The relative velocity of the S' frame of reference with respect to the S frame of reference can generally be in any spatial direction (as reckoned from the S frame of reference), and the relative velocity of the S frame of reference with respect to the S ' frame of reference can be generally in any spatial direction (as reckoned from the S' frame of reference). The only constraint is that the magnitudes of these 3d velocity vectors must be equal to one another.
Incidentally, there is a slight error in your description. You should replace the words "at a velocity -u in x direction" with the words "at a velocity -u in x' direction". The x and x' axes are not pointing in the same directions in 4D spacetime, even for Standard Configuration. Each has a component in the time direction of the other's frame of reference.

Chet