Combined speeds at relativistic velocities

[Moontanman]

4 minutes ago, swansont said:

If two rockets each approached the space station at 0.5c, from opposite directions, they would be approaching each other at 0.8c

(.5 + .5)/(1 + 0.5*0.5)

(Galilean addition would give you c)

Thank you, exactly what I was wondering but I didn't realize the station was necessary. 

 

[pzkpfw]

It's always about who is measuring what relative to what.

[joigus]

2 hours ago, Moontanman said:

Thank you, exactly what I was wondering but I didn't realize the station was necessary. 

 

It was, because otherwise one rocket's relative speed to the other one is equal and opposite to the converse, no matter what relativity principle you use (Galilean or Einsteinian).

It's exactly as Swansont said with 0.5, 0.5, giving 0.8 (in units of c)

It's perhaps an illuminating exercise to do it with 

0.99999 and 0.99999. It gives (0.99999+0.99999)/(1+0.99999*0.99999) = 0.9999999999

(in units of c) which is practically just c.

But, and here's what interesting, with small velocities as compared to c.

0.00001, 0.00001, it gives (0.00001+0.00001)/(1+0.00001*0.00001)=0.00002000000000

which is so close to the simple addition of velocities that nobody could tell the difference. That's why our intuition tells us velocities are additive.

[pzkpfw]

42 minutes ago, joigus said:

It was, because otherwise one rocket's relative speed to the other one is equal and opposite to the converse, no matter what relativity principle you use (Galilean or Einsteinian).

It's exactly as Swansont said with 0.5, 0.5, giving 0.8 (in units of c)

It's perhaps an illuminating exercise to do it with 

0.99999 and 0.99999. It gives (0.99999+0.99999)/(1+0.99999*0.99999) = 0.9999999999

(in units of c) which is practically just c.

But, and here's what interesting, with small velocities as compared to c.

0.00001, 0.00001, it gives (0.00001+0.00001)/(1+0.00001*0.00001)=0.00002000000000

which is so close to the simple addition of velocities that nobody could tell the difference. That's why our intuition tells us velocities are additive.

Just to link it up for Moontanman, the bits I bolded above are worked examples (reverse order) of what I described in the 3rd post:

Quote

 

... Where v and u' are small fractions of c, this ends up being a division by almost 1, i.e. at regular day to day speeds, it hardly makes a difference.

But where v and u' are large fractions of c, that becomes close to a division by two, i.e. you can add two speeds close to c, and still get a speed close to c. ...

 

 

Edited May 2 by pzkpfw
Fixed a typo in my quote of myself

[exchemist]

11 hours ago, Moontanman said:

Thank you, exactly what I was wondering but I didn't realize the station was necessary. 

 

It's simply that velocity always has to be stated relative to something in order to have any meaning. 2 cars in adjoining lanes on a motorway may be travelling at 5mph relative to each other but at 75 and 80mph relative to a policeman by the side of a road with a speed camera, and at 20 and 25mph relative to a truck which it itself travelling at 55mph relative to said policeman. The problem is we unthinkingly assume "the ground" is our reference frame in daily speech, treating it a bit as if it were an absolute frame of reference, when in reality there is no such thing.  (This should however be a bit more obvious in space.)

So you can't say you have a spacecraft "travelling at 0.999% of c" without saying with respect to what. Hence I proposed a space station to provide a reference, so that these speeds have a meaning. 

Edited May 2 by exchemist

[Halc]

12 hours ago, Moontanman said:

Thank you, exactly what I was wondering but I didn't realize the station was necessary. 

The station isn't necessary.  All that is needed is a frame reference. One can say that each rocket is moving at .9999c relative to frame arbitrary abstract inertial frame S which happens to have nothing stationary in it.  The lack of a stationary object in S would make it difficult for any rocket guy to directly measure that speed, but it can still be computed. A frame is, after all, an abstraction, not a physical thing.

So if X is moving west relative to S at 0.9999c and Y is moving east relative to S at 0.9999c, then X is moving at about 0.999999995c relative to Y and V-V.

(.9999 + .9999)/(1 + 0.9999*0.9999) = 0.999999995

 

12 hours ago, Moontanman said:

The observer is on each rocket, the rocket judges its own speed by its departure point

In that case the velocity of the two respective departure points relative to each other needs to be specified. Without that, there is no way to compare the rocket speeds relative to each other since there can be no common frame.

Usually, in the absence of an explicitly specified frame, an observer on any object (a rocket say  judges his own speed to be zero.

 

[Moontanman]

I do appreciate everyone answering this question, I honestly had no idea I was opening such a can of worms! While I am aware of Relativity and how frames of reference worked I was genuinely unaware of how fundamental the idea of reference frames was in this scenario. Thank you very much for answering my question in detail, while I am sure I am stilling missing much of the concept due to not speaking mathematics I am much closer than I was to understanding.  

[Ghideon]

I'm a little late, I think all aspects are sorted out already. But I had started drawing a picture of different aspects of the relative velocities in the scenario. Posting in case anyone finds it helpful:

Rocket A is traveling from space station a at velocity V_Aa (Aa denotes that the velocity is the relative velocity between a and A)

Rocket B travels from station b at velocity V_Bb . The rockets will meet head on and V_Aa=-W_Bb

Equations:
1: a and b (the space stations) are at rest relative one another, in the same inertial frame of reference  
2: When each rocket's crew measures the velocity of the other approaching rocket, relative to their own rocked, the velocity does not match the sum of velocities relative their starting points. (Assuming the correct sign is used since velocity is vector)
3: for low speeds the sum of velocities is a reasonable approximation
4: Rocket A will measure its speed relative its starting point to be the same (with negative sign) as the rocket B will measure relative it's starting point
5: Rocket A will measure leaving a with the same velocity that it is approaching b (minus the sign)
6: In any case the relative velocities between the rockets, measured by crew on board A, is greater than the rocket's velocity relative it's starting point a (and symmetrical for rocket B and crew)

7 (now shown) observers on a or b will agree that the rockets will meet half way between a and b. The sum of distance traveled by A plus B can be more than a light signal could travel in the same amount of time.


(did this in a bit of a hurry between other things, please point at errors)