Adding 2 speeds close to c

[Kedas]

three objects

X=earth, Y=space station, Z=space craft

 

The space station Y travels at 0.866c away from earth X, at the space station they launch a space craft Z that is also moving away from Y at 0.866c.

 

now what does earth see:

Y at 0.866c

Z at somewhere between 0.866c and c

 

From earth point of view this means Z is barely moving away from Y but at Y they see Z moving very fast away in the same way that X sees Y.

 

So how can Z be fast and slow at the same time.

The only explanation is that due to the fact that time at the space station Z is going much slower it's only from their point of view that it's going that fast.

1 second on the space station is two seconds on earth. So the speed of Z in reference to Y of 0.866c is only 0.433c seen from earth.

But the sum of 0.866c and 0.433c is still faster than c and that's not possible.

 

Z from earth should be 0.9897c or 0.123c faster than Y (not 0.433c)

 

So what did I forget? space contraction?

[Janus]

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[Janus]

You are not including two additional effects: Length contraction and the relativity of simultaneity. The relativity of Simultaneity is important in the following way.

 

Assume you have second space station YY located some distance from Y in the direction from Y as Y is moving with respect to the Earth, and stationary with respect to Y. It has a clock which reads the same as the one on station Y as determined by station Y and YY.

Thus from Y and YY's view they look at what time the clock on Y reads when Z leaves, then what time YY reads when Z arrives, and knowing the distance between YY and Y compute that Z traveled at .866c.

 

Now from the Earth view, Z leaves Y when Y reads the same time as Y and YY sees, and arrives at YY when YY reads the same time as Y and YY sees, but from the Earth view, the Clocks on YY and Y do not read the same at the same time (this is after you factor out the light signal delay due to the fact that YY is a different distance from Earth than Y) . Clock YY will actually lag behind clock Y.

 

Thus when the Earth figures out how long it takes for Z to travel from Y to YY, taking into account how far apart Y and YY are due to length contraction, how fast their clocks are ticking with respect to its own, the fact that YY and Y's clocks are out of sync according to the Earth, and how fast Y and YY are moving with respect to the Earth, the Earth will determine that Z has to have a velocity of between .866 c and c (about .99c) to meet all of the above conditions

[CanadaAotS]

lol so confusing...

 

some kind of diagram would be helpful

[Kedas]

Why would Clock YY lag behind clock Y?

They are in the same reference frame so if it are two or just one big station it should be the same from earth.

 

I want to know how to get to 0.123c taking all effects in account.

I'm measuring their speeds with a laser on earth and Y is measuring the speed of Z with a laser on Y.

[Janus]

Why would Clock YY lag behind clock Y?

They are in the same reference frame so if it are two or just one big station it should be the same from earth.

 

Relativity of Simultaneity. Events that are simultaneous in a given reference frame are not simultaneous as determined by another reference frame when those events are separated by a distance measured along the line of relative motion between the frames. Before you can go any further with understanding Relativity, you will have to come to terms with this concept.

[Kedas]

OK I have two clocks in sync both will give a light flash at the same time.

I put them at both ends of a train.

The train moves fast the flashes go off, I see them go off at the same time and those at the train also see them go off at the same time. But since the train moved after the flashes went off the light doesn't meet in the midde of the train.

And on the train light also moves at c so the flashes do meet in the middle from there point of view.

 

But how does that effect my speed measurement from earth?

Even with length contraction I still get more than 0.123c

 

edit: the math

the difference is z.(1-y²)/(1+y.z) with z speed Z relative to Y and y speed of Y relative to X

y and z are factors of c

in my case it's 0.866*(1-0.866²)/(1+0.866²)=0.123 or 0.866/7

[Janus]

OK I have two clocks in sync both will give a light flash at the same time.

I put them at both ends of a train.

The train moves fast the flashes go off' date=' I see them go off at the same time and those at the train also see them go off at the same time. But since the train moved after the flashes went off the light doesn't meet in the midde of the train.

And on the train light also moves at c so the flashes do meet in the middle from there point of view.

 

[/quote']You've got this backward. If the flashes meet in the middle of the train in the frame of the Train, they must also meet in the middle of the train according to the frame of the embankment. What the embankment observer and the train observer will disagree about is whether the clocks are in sync and whether the two flashes originate at the same time

 

But how does that effect my speed measurement from earth?

Even with length contraction I still get more than 0.123c

 

 

Okay,

 

First we start in the frame of clock Y and clock YY; they are separated by 1 light sec distance in their frame. In order to make sure that the clocks are in sync, clock Y sends a radio signal to YY saying " at the moment I sent this, I read 0". Clock YY receives this signal, and knowing it was sent 1 sec ago (the time it took to travel the distance from Y to YY), set itself to 1 sec, and thus according to Y and YY the two clocks are in sync.

 

Now the same event in the frame of the Earth.

Clock Y sends it's signal which travels at c towards YY, but YY and Y are moving at .866 c, and thus YY is moving away from the point of emission and the signal must chase it down. Of course the YY starts out .5 light sec from Y as measured in the Earth frame and this takes .5/(c-.866c) = 3.73 secs by Earth's clock. Y is time dilated so it will accumulate 3.73/2 = 1.87 secs in the time it takes for the signal to reach YY. Again when the signal reaches YY, it sets itself to 1 sec, thus at this point, according to the Earth frame clock Y reads 1.87 sec and YY reads Y. IOW at this point and for every point hereafter, in the Earth frame YY lags Y by .87 sec.

 

Now back to the Frame of Y and YY.

 

Z leaves Y when Y reads 2 sec. It takes 1/.866c = 1.155 secs to reach YY and thus both Y and YY will read 3.155 sec when Z reaches YY

 

Earth frame.

Z leaves Y when Y reads 2 sec, It also arrives at YY when YY reads 3.155 sec. When YY reads 3.155 sec Y reads .87 sec more, or 4.03 secs. Thus while Z moves from Y to YY, Y advances from 2 to 4.03 sec and accumulates 2.03 seconds. Clock Y is time dilated which means that while 2.03 seconds pass on clock Y, 4.06 seconds pass on the Earth clock. IOW, it takes 4.06 seconds for Z to travel from Y to YY by the Earth clock.

 

The distance between Y and YY is .5 lightseconds, so to get Z's relative speed to Y and YY as measured by the Earth we take .5/4.06 = .123c

Which, when added to .866 c gives us a velocity of .989c for the relative velocity of Z to the Earth as measured from the Earth.

[Kedas]

yes, I got it backward. What happens stays the same it's when it happens that changes.

 

Thanks for the explanation

This is a much more understanding way of calculating it than the simple formula.

 

Test

When Z travels back from YY to Y at the same speed of 0.866c (according to Y&YY).

Then Earth will see Z at a speed of 0.5/((1.155-0.866)*2)=0.866c right?

This means that the speed of Z will be the same for Earth, Y and YY.

[Apeofman]

Hi,

You can find a geometric special relativity calculator at magen.co.uk . Study of it may help you to visualise how the Lorentz transform works, and the relationships between different observers.

It is an ongoing project and the documentation doesn't quite fit, but gives enough info. to anyone with a basic knowledge of the subject.

No maths are involved or presented. The solutions are from a geometric model/construction, driven by a Java geometry engine.

The calculator page can take more than five minutes to load using a 56k modem so some patience is required.

Ive been looking for a forum like this for ages. Maybe i'll pose one or two questions myself. The longer i study relativity the less certain i am that i really know it.