Understanding Frames of Reference in Relativity

[venomxx]

Im trying to understand frames of reference, I am very new to relativity so sorry if I am being silly!

Suppose you have a 2 spacecraft one traveling relative to the sun, and the other traveling relative to the first spacecraft in a perpendicualar direction. i know the velocitys of spacecraft A relative to the sun and the velocity of B relative to A.

Am i right in saying there's three frames of reference?

What I am trying to consider is relating spacecraft B to the sun...so am i right in saying i relate A to the sun with lorentz transformations and B to A and relate the two?

Im just a little confused since there's the aditional moving object!
Any explanation would help!

 

[tiny-tim]

Welcome to PF!

Hi venomxx! Welcome to PF!

Yes, each velocity has its own frame of reference.

Since A B and S have different velocities, each can only regard itself as stationary if it uses its "own" frame of reference.

To transform between coordinates in different frames of reference, you use a Lorentz transformation (which may be a boost or a rotation or a combination).

You can represent a Lorentz transformation by a 4x4 matrix.

To convert from B's coordinates to A's, and then to S's, you multiply the transformations (or the matrices).

The only complication is that if (as usually happens) B to A and A to S are boosts (with no rotation), then the combination B to S is a combination: a boost and a rotation.

 

[venomxx]

Cheers for the quick response! Iv a better understanding of the whole problem...i read into the lorentz boosts and it does help my understanding!:)

Im using a 4x4 matrix to transform from A to B to start and its a very long process, ill post again if iv any problems! Makes me wish i joined this earlier...:p

 

[Naty1]

Am i right in saying there's three frames of reference?

not exactly...you might say there are three "convenient" frames...but the Earth might be a logical fourth frame...

and if you were on any other of billions of planets, each of those would also be a convenient frame...

Another way to think of "frames of reference" is with regard to black holes...one convenient frame is in free fall towards the black hole, another convenient one is stationary at an infinite distance...both give different views of phenomena...

 

[atyy]

not exactly...you might say there are three "convenient" frames...but the Earth might be a logical fourth frame...

No. I agree with tiny-tim. A's velocity is specified relative to the sun, and B's velocity is specified relative to A (that's 2 frames we can throw in one for good measure )

 

[DrGreg]

No. I agree with tiny-tim. A's velocity is specified relative to the sun, and B's velocity is specified relative to A (that's 2 frames we can throw in one for good measure )

I think Naty1's point was that there are an infinite number of frames you could choose if you wanted to, but only three of any practical use in this problem. All these frames exist whether you choose to use them or not.

 

[Naty1]

DrGreg..thanks...that was exactly what I was pointing out...perhaps a minor point BUT...

it might not be obvious to someone starting out in GR that picking reference frames is an art form...I'm only beginning to understand how important such selection is after several years of casual study...it's one thing in special relativity,without gravity, implied in this post, far more subtle in general relativity with all the extremes of gravity:

I'm now reading Kip Thornes BLACK HOLES AND TIME WARPS and he points out how it took from the 1930's to the 1960's for physicists to understand the implosion of massive stars to a singularity was compulsory...The absolute horizon and the event horizon in black holes were subjects of study by eminent physicsts for years before they were understood...

And perhaps the craziest frames of all: either inside or outside an event horizon surrounding a black hole...still dazzles: Inside you can't see out; just on the horizon, time stops! And freefalling toward a black hole: yet another frame with significant insights...

 

[venomxx]

i understand where your coming from how i could choose any of infinite reference frames...does this mean i can pick my own reference frame to be me standing on the Earth watching the whole thing, then relate the sun, craft A and B to me?

the way i went was very long and tedious...whats the most efficient set of frames to use? The best i can think of is tinytims answer!

 

[JesseM]

venomxx, do you know about the relativistic velocity addition formula? This is probably the simplest approach to the problem, although you'll get the same answer if you use the full Lorentz transformation. If your ship B has velocity v relative to ship A (in A's rest frame), and ship A has velocity u (in the same direction) relative to the Sun (in the Sun's rest frame), then the velocity of ship B relative to the Sun is given by (u + v)/(1 + uv/c^2).

 

[tiny-tim]

If your ship B has velocity v relative to ship A (in A's rest frame), and ship A has velocity u (in the same direction) relative to the Sun (in the Sun's rest frame), then the velocity of ship B relative to the Sun is given by (u + v)/(1 + uv/c^2).

Hi JesseM!

But venomxx said …

… the other traveling relative to the first spacecraft in a perpendicualar direction.

(I think he meant "perpendiculiar" )

 

[JesseM]

Ah, I missed that. In this case, instead of doing matrix algebra, a simpler method might be to just pick two events on the worldline of B as seen in the A frame, and then figure out the coordinates of these events in the Sun frame. For example, in the A frame if ship B is moving in the y' direction at speed v, then if it's at the x'=0 and y'=0 at time t'=0, that means at time t'=t'₁ it'll be at position x'=0 and y'=vt'₁. Then if the A frame is moving at speed u in the x direction in the Sun frame, with the origins coinciding so the first event has coordinates x=0,y=0,t=0 in the Sun frame, according to the Lorentz transformation the second event has coordinates:

x = gamma*(x' + u*t') = gamma*u*t'₁
y = y' = vt'₁
t = gamma*(t' + ux'/c^2) = gamma*t'₁

where gamma = 1/sqrt(1 - u^2/c^2). So using the pythagorean theorem, in the Sun frame the ship has traveled a distance of sqrt(gamma^2*u^2*t'₁^2 + v^2*t'₁^2) = t'₁*sqrt(gamma^2*u^2 + v^2). Divide this by the time of gamma*t'₁ and you get a speed of sqrt(gamma^2*u^2 + v^2)/gamma. Of course I may have made an algebraic mistake so venomxx shouldn't take my word for it...