Oh for GOD sakes...of course I understand GR and SR.
The evidence begs to differ.Since the velocity of c is 186,282 miles per second...the closing rate of the two craft is 372,526.74 mps....very close to twice the velocity of light.
Oh for GOD sakes...of course I understand GR and SR.
The evidence begs to differ.Since the velocity of c is 186,282 miles per second...the closing rate of the two craft is 372,526.74 mps....very close to twice the velocity of light.
I am talking about CLOSING DISTANCE RATE.The evidence begs to differ.
They will collide in 30 minutes.In who's reference frame?
At 10mph the answers will all be virtually identical. But at relativistic speeds they won't be.
, but if I use these people's formula for adding relativistic speeds,"CLOSING RATE OF 372,526.74 MILES PER SECOND."
186,263.37 + 186,263.37
V = ---------------------------- = 186,281.999 miles per second
1 + (186,263.37 * 186,263.37)
-------------------------
c^2
Dave...in answer...along with answering the other above members question...of course Light or Radar does not take on the rate of travel of the Broadcasting Craft...the craft is traveling at .9999c...the light leaving that craft travels at c...not c plus .9999c.SplitInfinity, I guess I don't understand SR.
You tell me the ships have a , but if I use these people's formula for adding relativistic speeds,
http://einstein.stanford.edu/content/relativity/q1021.html
the result looks more like this -
Can you explain what's the difference here?Rich (BB code):186,263.37 + 186,263.37 V = ---------------------------- = 186,281.999 miles per second 1 + (186,263.37 * 186,263.37) ------------------------- c^2
These statements are correct.The light leaves the space craft at C, but it also arrives to the other space craft at C. The speeds are identical. The frequency shifts, which is where the time dilation comes in.
Bill, with all due respect this statement only partially correct, but has little meaning.Like I said elsewhere, it is not a linear curve. You can forever approach C, but never exceed it from your reference point, or anyone elses for that matter.
Okay, but this has little bearing on the OP. No third observer is implied or required.If you see (as a 3rd party observer) two objects closing at FTL speeds, note that both ships are still under C as seen by you. From your relative viewpoint, they are not at C, nor will ever be.
Not sure what you are getting at here.If you change the reference to one of the other ships, neither you not the other ship is closing at FTL, it will always be under C.
Bill, there is nothing in SR that says things cannot be traveling with respect to one another at or beyond C in whole universe terms.Speeds are measure in reference to where you are measuring them from. And since the relationship is not linear, it is not intuitive in the slightest.
Bill...actually...your not getting it. I am not talking about or even infering there is a 3rd observer.You still don't seem to get it. The closing rate from a 3rd party point of view is an illusion. It is not real. The real closing rate is always from the objects approaching each other. In other words, relative to each other. That is why it is referred to as special relativity. It is also why Bill O said a 3rd party point of view complicates things, it allows spurious arguments to be presented that have no real basis in fact.
Nature is full of illusions.
Actually...the guy in this picture was wearing Black Leather and Studs about 15 minutes before this pic was taken as the guy was on a lunch break and has to go back to work.Hey Kermie, that's good of you to guard that sleeping horse with a big stick. Keep those graffiti artists from drawing more Chinese characters on his butt!
At the risk of flogging this dead horse one more time...Bill...actually...your not getting it. I am not talking about or even infering there is a 3rd observer.
Yes...Relativity...and SR.
My entire point that I have been attempting to make is based upon two crafts traveling at each other where there is a Closing Rate in the extreme...as there is with this example...not being able to use any Electronic Sensor Data...ie...Radar Send and Return...being useless as in that say the distance between each craft traveling at .9999c is 2 Light Seconds and at this distance both craft issue a Radar Send.
Now are you going to honestly tell me that the Radar Return Data for either craft will be of any use?
You are right of course...but regardless of whether Shroaty's Cat is dead or alive or both or in a non-determinate state...if the Astronauts on both craft were informed ahead of time that at some point their craft would be traveling right at the other craft...and this would be a good reason for them to be sending out a Radar Send...how well their sensor data would help them prevent an impact of their craft with the other would be very important.At the risk of flogging this dead horse one more time...
As soon as you mention that the two craft approach from the east and from the west, a 3rd observer is introduced. But this observer is completely irrelevant to the craft and what they see. If you put yourself on one of the craft, the situation becomes simpler and symmetrical. The only difference is the direction of the shuttle and the resultant transfer of mass.
I am impressed. A variation of this problem set was given to more than 100 students over 4 years and you are only the 2nd person to point this out. However, the ill conceived choice we made in describing direction was neither significant nor brought to bear at any point in the set beyond the introduction. But kudos for seeing this.As soon as you mention that the two craft approach from the east and from the west, a 3rd observer is introduced.