Question

two observers, O and O' fixed relative to two coordinate system oxyz and O'x'y'z' respectively, observes the motion of a particle appear to have the same force acting on it, if and only if the two coordinate systems are moving at a constant velocity relative to each other. prove it.

Answer 1

ok

Answer 2

If the two observers are moving at constant velocity relative to each other, then they are not experiencing time dilation relative to each other. Right?
So if you drew a space time diagram of both O & O' they would both be standard cartesian plane systems.
Or at least these are my thoughts on the matter of Proving it :D
Cheers

Answer 3

No. There is time dilation going on.

Answer 4

Why would there be time dilation going on? If both observers are moving at constant velocity with respect to each-other then they are in the same intertial reference frame and therefore would not experience time dilation with respect to each-other.
Cheers :D

Answer 5

I see what you're thinking, but then you're using the terminology incorrectly. When two observers move with a constant velocity with respect to each other, we mean that they are moving apart. If they are in the same inertial frame, then they're not moving with respect to one another.

Answer 6

Congratulations Kfujioka! You win the prize for the most useless confrontation with another user who is actually trying to help a person solve a question they dont know the answer to.
"we mean that they are moving apart" <---no where in the problem statement.
"if and only if the two coordinate systems are moving at a constant velocity relative to each other. prove it." <--- stop nitpicking at me and start working on this. unless after all this type to me you still have nothing to add...
No more cheers for you.

Answer 7

so whats the answer???

Answer 8

i would like hear the answers from both point of views. you know that would make answer more detailed or just better to think about or something.
cheers for all observers B)

Answer 9

Thanks for keeping an open mind scrap. I've been pondering this one for a solid day now. I'm not fully satisfied with anything I come up with though. I will post again once I'm more confident. Whitey- I never meant to be confrontational with you, just trying to help scrap avoid something that I thought was misleading. I stand by my interpretation, "moving at a constant velocity relative to each other" means the observers are moving apart and are not in the same inertial frame, otherwise the problem would state something like, "at rest with respect to each other."

Answer 10

This approach is a bit dirty, but the principles should be there if you want to make it more rigorous.\[\vec{F} = \frac{d\vec{p}}{dt} = \frac{d}{dt}[m\vec{v}]=m\frac{d\vec{v}}{dt}= m\vec{a}\]\[\vec{F'}=\frac{d\vec{p'}}{dt'}=\frac{d}{dt'}[\gamma m \vec{v}]=\frac{d\gamma}{dt'}m\vec{v} + \gamma m \frac{d\vec{v}}{dt'}\]From the chain rule:\[\frac{d\vec{v}}{dt'}=\frac{d\vec{v}}{dt}\frac{dt}{dt'}=\vec{a}\frac{dt}{dt'}\]From the formula for time dilation:\[t' = \gamma t \rightarrow \frac{dt'}{dt'}=\gamma \frac{dt}{dt'} \rightarrow \frac{dt}{dt'}=\frac{1}{\gamma}\]Putting the pieces together we have, \[\vec{F'}=\frac{d\gamma}{dt'}m\vec{v} + \gamma m \frac{d\vec{v}}{dt'} = \frac{d\gamma}{dt'}m\vec{v} + m \vec{a} \]So, F = F' only if the time derivative on gamma is 0. That, in turn, can only be the case if the two frames are moving with a constant velocity relative to each other. If you dig a little yourself, just keep in mind that there's two different v's. The explicit "v" above is the velocity of a particle of mass m in the unprimed frame. The gamma contains a "V" of the velocity between the two frames.

Answer 11

cool.

Answer 12

This doesn't change the bottom line, but I just noticed an inconsistency. I didn't take the time derivative of gamma when working with the time dilation. When you do that, you get another term in the equation for F' that, again, has a time derivative of gamma. So the argument still holds. Check my work again though, maybe I'm being stupid somewhere else.