More relativiy thought experiments

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The evidence begs to differ.
I am talking about CLOSING DISTANCE RATE.

No matter what...if two objects are traveling toward each other in realively uneffected by Gravity Space/Time...at a specific velocity...let's say...A is traveling toward B...A's velocity is 10mph...B's velocity is 10mph.

A and B are 10 miles apart. Both are traveling at 10mph. Both are traveling directly at each other in a straight line.

How long will it take the two to collide?

Split Infinity
 

WBahn

Joined Mar 31, 2012
29,979
In who's reference frame?

At 10mph the answers will all be virtually identical. But at relativistic speeds they won't be.
 
In who's reference frame?

At 10mph the answers will all be virtually identical. But at relativistic speeds they won't be.
They will collide in 30 minutes.

Now since the two craft in the other example are traveling at .9999c...thus BELOW the speed of light that being 186,282 mph....but they are traveling at 186,263.37 mph EACH....at each other in a straight line...the CLOSING DISTANT RATE is 372,526.74 MILES IN ONE SECOND.

Now the Closing Distance Rate just reflects the distance that will be traveled by both ships before impact and how long that will take.

Since the craft are traveling below the velocity of light...and let's be honest....9999c is IMPOSSIBLE to obtain using any form of propulsion or Electromagnetic Particle Interaction and could only be obtained by Folding Space/Time...but for arguement sake....the two craft are traveling right at each other at .9999c.

There is NOTHING in GR or SR that will prevent the fact that these two craft are going to impact each other. If one craft was standing still and the other craft headed toward it at .9999c...and the distance between the two craft was .9999 Light Second....then that is how long it will take for one craft to impact the craft standing still.... .9999 Second.

Since both are traveling at .9999c at each other and both are at this velocity at a distance of 1.9998 Light Seconds from each other...they will impact each other in .9999 Second.

Split Infinity
 

Thread Starter

BillO

Joined Nov 24, 2008
999
But who is measuring the distance between the two ships?

Besides, it does not matter. If SR is correctly applied, each observer will agree.

I'd advise that you keep it simple and do not try to bring a third observer into the argument. It will make the situation more complex.

But just to be sure you are aware, there are many, many observable objects in this universe traveling at (and beyond, but let's not get into that now) or close to C with respect to each other. The light still makes it to us. Communications can occur. You might be making the common mistake of assuming that somehow, like a ball tossed from a moving train, that light takes on some momentum from the velocity of the emitter. This is not the case.

If we disregard any 3rd observer for now and just concentrate on A and B, at some mutually agreeable moment in their respective time, they will see the distance between them as being particular amount. They will also see that they are moving toward each other at less than C. If they try to communicate, let say using light, they will both see the light traverse that distance at C. The light will cover the distance between them quicker than they will, so they will be able to communicate.

in any case, I am renowned at not being abel to teach, so let me refer you to someone that was revered for his ability in the respect. Go Here. You will find what remains of the recordings of Richard Feynman's famous lectures on physics which he gave at Caltech in the 60s. Of particular interest to this topic would be 16 and 17. If you look hard enough on the internet, you can also find PDFs for these lectures too, with the diagrams, text and math in them.
 

davebee

Joined Oct 22, 2008
540
SplitInfinity, I guess I don't understand SR.

You tell me the ships have a
"CLOSING RATE OF 372,526.74 MILES PER SECOND."
, 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 -

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
Can you explain what's the difference here?
 
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 -

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
Can you explain what's the difference here?
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.

A craft at a certain Velocity and in particular...very high velocities away from a Celestial Gravity Well...will incur Time Dialasion. Still...this HAS NO EFFECT upon CLOSING RATE of the two craft in mention.

Both are at a velocity of .9999c and headed right at each other. Thus Closing Rate is 1.9998c and Closing Distance is 1.9998 Light Seconds.

That means no matter how you cut it...a distance of 1.9998 Light Seconds exist between the two craft and at their given velocities of .9999c....they will crash into one another in .9999 seconds.

Now if you want to start getting into passage of time of the observer say on Earth or the Time Dilation that would exist due to the crafts velocity of .9999c each...that's another thing.

But as far as this whole question goes...that being...I am trying to send a signal or effect electromagnetically or use a sensor system from one craft to another....due to the velocities the two crafts are at...due to the fact that any useful data that one would need in an attempt to navigate or effect or contact one or the other craft....say a radar signal is sent from one craft to the other at a distance between them of One Light Year...by the time the Radar got to the other craft and returned almost the entire approx. 6 Trillion Miles that was between them when the Radar signal was sent will have been traversed by both closing craft thus any data recieved WILL BE USELESS.

This has been what I have been trying to say this whole time.

I am not debating GR or SR or any and all effects such as Time Dilation...etc...or the calculation due to Relativistic Speeds...YES...I am aware...I believe it was Bill or another that talked about this well at the beginning of this topic...but fact remains no matter how time is experienced for those watching or inside the craft...in normal space/time...uneffected by large gravitational effects....craft A has x as a velocity...craft B also is traveling at x. D=Distance...L=Closing Rate...

If at minimum 95% of D can be covered during L...and in this case it is .002% of D being covered...there is NO WAY any Sensors using anything EM are going to give you data worth anything.

Split Infinity
 

mikeleeson

Joined Aug 22, 2012
26
The original post simply describes two craft that are heading towards each other. The external observer measures the velocities as close to the speed of light - this indicates that the subsequent calculation should include relativistic effects.

The external observer has no other involvement in the problem. [I think this is what BillO is saying].

The two craft see a similar situation: another craft approaching at a velocity close to the speed of light. We can easily measure things from distant objects that are travelling close to the speed of light. The velocity of the distant object has no impact on the performance of our equipment. [I think this is what WBahn is saying]

Of course, we have to take relativity into account when we interpret the data from our equipment: the Doppler shift from a radar reflection (for example) would need to be calculated using relativistic equations.

Option 1
If craft B launches the doctor to the west, energy will be required. Craft B will see the doctor accelerate and gain kinetic energy; that energy must come from craft B. Captain B is right that this option will cost him energy.

Option 2
If craft B just separates the doctor's shuttle, no (additional) energy will be required. Craft B will have lost some kinetic energy because it has lost some mass (the doctor) but still has the same velocity. The doctor gains the kinetic energy (in fact the doctor already had this KE but is now separate from craft B).

Craft A now has to collect the doctor as the shuttle is passing. Craft A captures the shuttle and absorbs the kinetic energy because the 'collision' is not elastic. The momentum of craft A is reduced by the 'collision' with the shuttle since the total momentum is conserved. As the mass has increased, the velocity must decrease: craft A will lose speed.

My opinion
The problem is about transfer of mass and the effect on energy and momentum. The doctor exists in any frame of reference; relativity is only required when converting the speeds and times and accelerations of the doctor between the two different frames.
 

Wendy

Joined Mar 24, 2008
23,415
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.

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.

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.

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.

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.
 

Thread Starter

BillO

Joined Nov 24, 2008
999
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.
These statements are correct.

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.
Bill, with all due respect this statement only partially correct, but has little meaning.

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.
Okay, but this has little bearing on the OP. No third observer is implied or required.

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.
Not sure what you are getting at here.

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, there is nothing in SR that says things cannot be traveling with respect to one another at or beyond C in whole universe terms.

My OP was part of a problem set given out to physics students back in the 80's. This is was only one small part of the questions asked around the same situation. One of the key takeaways from it was that one observer's "I don't have the energy to accelerate any massive object to the speed of light" problem is another observer's "Oh goody, I can get all the energy I want from decelerating that object down from the speed of light" bonus. In other words, saying that you cannot accelerate something to the speed of light has no real meaning because there will exist another observation point that will see that as exactly the opposite situation.

As to the FTL thing, using relativity calculations and the concept of comoving coordinates, it has been determined that there are observable objects in this universe that are traveling faster, with respect to us, than the speed of light. This is due to expansion of the universe.

It should be noted that the mathematics of SR are not applicable over very large distances. Specifically, those distances where the expansion rate of the universe is at all significant GR needs to be used. In GR speed, or more correctly velocity, is a 'local' phenomenon.
 
BillO posted in reply to Bill M....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.

THANK YOU! I have been trying to get this point across in my several posts on this topic but YOU stated it both directly and correctly!

CLOSING RATE...the total distance that will be traversed by two objects or particles either both in motion or only one from a specific moment in linear time to a future moment in linear time when such objects or particles will intersect or collide....DIVIDED BY...the combined velocity of both and if occuring acceleration curve of one or both either increasing or declining....using a common measurement of time for both velocity and acceleration or decell curve for both objects or particles.

Thus two objects at 100 miles apart that will intersect due to vector with one objects velocity at 25 mph and another at 75 mph then adding these two velocities gets 100 mph thus divide and we get a closing rate of velocity of 100 mph and a closing rate of time of 1 Hour every 100 Miles.

SR and GR have nothing to do with the Closing Rate of objects or particles and as in the topics example...the closing rate is 1.9998c.

This does not mean ANYTHING is traveling at 1.9998c...it just means that the two craft have a Closing Rate of 1.9998c.

Split Infinity
 

Wendy

Joined Mar 24, 2008
23,415
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.
 
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.
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?

Split Infinity
 

Thread Starter

BillO

Joined Nov 24, 2008
999
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!:D
 
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!:D
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.

The horse has the day off from pulling a carriage with tourist couples around Central Park so the horse will Sleep In for the rest of the afternoon at the Ritz!

But the horse has another scheduled appointment with a client that night at the Hellfire Club and since the horse doesn't want to have to go near the park and possibly see his Boss in an attempt to get another stick...the horse is telling the guy holding the stick..."HEY! You can smell that stick as much as you want but if you want to take it to the office with you...It will cost you another $100 Bucks!" LOL!

Split Infinity
 

mikeleeson

Joined Aug 22, 2012
26
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?
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.
 
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.
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.

Split Infinity
 

mikeleeson

Joined Aug 22, 2012
26
If the two craft were informed ahead of time that they would need a doctor, maybe they would have taken more care with their general health.

Cats and radars are making this more hypothetical than it needs to be. It was an interesting discussion - thanks to BillO for starting it off. :)
 

Thread Starter

BillO

Joined Nov 24, 2008
999
As soon as you mention that the two craft approach from the east and from the west, a 3rd observer is introduced.
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.
 
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