Contraction effects at relativistic velocities

[RandallB]

100 meters still won't fit inside 99 meters

I'm going to assume that you mean for the guns to fire simultaneously according to the planet.

The answer is that the bullets will miss the ship, according to both the ship and the planet. From the ships view, the guns do not fire simultaneously.

This makes no sense even if only one of the 100's of guns fired there should be some chance it might hit. Spaced at 101m firing at a 100m ship there would only be 1 chance in 101 for a complete miss!

A better plan for the defending planet is to set the guns 99m apart fired at the same time from the planet there would be no chance of a miss.

Measure from either view -- time dialation, distance contraction, not even non-simultaneosness will save the ship cause they all have to taken into account. Once the numbers are worked all the way, the ship would not fit between any of the 99m spaces.

RB

 

[Doc Al]

But 44 meters easily fits inside 99 meters

A better plan for the defending planet is to set the guns 99m apart fired at the same time from the planet there would be no chance of a miss.

Measure from either view -- time dialation, distance contraction, not even non-simultaneosness will save the ship cause they all have to taken into account. Once the numbers are worked all the way, the ship would not fit between any of the 99m spaces.

You might want to redo your calculations.

Since the 100 m ship moving at 0.9c is contracted to about 44 m (according to the planet), it should fit easily between the 99 m spacing of the guns.

 

[RandallB]

High Noon - Showdown

You might want to redo your calculations.

Since the 100 m ship moving at 0.9c is contracted to about 44 m (according to the planet), it should fit easily between the 99 m spacing of the guns.

So your quick calculation of a LENGTH of 44 m has embolden your War department to attack our fair planet at it's only vulnerability - The dreaded 100 meter weapon delivered within a narrow window of time just after HIGH NOON !
You claim the length contraction to 44 m as the attacking ships come in at varied spacing that over half of them should find a space to get by our cannons spaced at 99m to cover the entire vulnerable opening.

Your demand we surrender based on your 44 m contraction data you have provided us is rejected!
Gary Cooper in my defense department has played Poker with your Scotty in Engineering before and this 44m in "Length" thing is just another bluff attempting to scare us with incomplete data!

Our defense department offers you the following telemetry data from our cannon defense test. Note our cannon A shows the nose of test 100 m test ship passing at exactly HIGH NOON + 103ns (nano-seconds).
The cannon B camera shows the nose reaching it at exactly the synchronized firing time for A B and all cannons as they fire simultaneously. With a near miss less than one half meter ahead of the nose. The nose camera agrees with cannon B's data that both of their clocks at this point are perfectly synchronized at HIGH NOON + 400 ns and the near miss. The Elapsed time of 297 Confirming matching your speed of 0.9c (c=0.3m/ns)

Gary Cooper is in a fearless mood and advises you have Scotty forward us
(and detail to you) the following:
1) the clock time on the nose of the ship when it passed cannon A. observed by both Cannon A and the nose.

2) the clock times observed by cannon A of the cannon and the tail clock when the tail crosses cannon A.

3) the confirming tail observations of cannon A time & Tail time at crossing.
Note the tail clock is of course synchronized with it own nose clock.

Then we can talk terms of your surrender as you've already destroyed your ability to get back to your home galaxy.

Randall B

...Woe unto those that abuse the rules of relativity
...without understanding the difference between length and distance.
...Bring it on - our cannons are ready!

 

[Doc Al]

loose cannons?

You claim the length contraction to 44 m as the attacking ships come in at varied spacing that over half of them should find a space to get by our cannons spaced at 99m to cover the entire vulnerable opening.

The ship has a proper length of 100 m and travels at 0.9c with respect to the planet. I claim that if the planet simultaneously fires two guns spaced 99 m apart at the precise moment that the ship's midpoint passes the midpoint of the guns--then the bullets will miss the ship. (Ignoring, of course, the vertical distance between the guns and the ship, and the thickness of the ship: assume both are zero.)

...Woe unto those that abuse the rules of relativity

My sentiments exactly!

 

[bino]

ok let's say there is a ship 100ft long moving at .90c. a planet perpendicular has guns 101ft apart at rest. the guns line of fire are parrallel to each other. the planet plans it out so that the ship will be in between the bullets when the ship flies past the planet. would the bullets hit the ship or not?

i meant that the ship was 100 when it was moving. not 100 when it was sitting still.

 

[RandallB]

Victory is Mine

The length of the ships are always 100m you have to measure that on board.
It doesn’t matter how fast it goes!

I changed the spacing of the cannons so it would be impossible for 100 m ships to get by them.

As you an see from above Doc sent his ships to there doom!

I understand as I watched from B cannon as Doc went by in the nose of his ship. The tail of the 100m in length ship is only 44m away while my cannon is only now firing at 100m away! HOW CAN I HOPE TO SURVIVE !
I will not know until the light from my cannon has the time to reach me at Cannon B.

But I do know the at the moment of firing Cannons only the nose is in my time and space.
It is the only part of his ship the where things happen simultaneous with me at cannon B due to being in the same place & space.

His tail 44m away is traveling IN MY FUTURE! And is already crumbling! And will not “sync” (as his clock is running slow) with me until the light from the tail and what remains of it, come blowing by me. About the same time as the light from cannon A arrives here showing A KILL!

DOC should have done the work to calculate all the time sequences as well, not just the distance calculations alone. But the destruction of his ship will move up to him in the nose of the ship in just a few nano sec.

RB

 

[Doc Al]

ok let's say there is a ship 100ft long moving at .90c. a planet perpendicular has guns 101ft apart at rest. the guns line of fire are parrallel to each other. the planet plans it out so that the ship will be in between the bullets when the ship flies past the planet. would the bullets hit the ship or not?

i meant that the ship was 100 when it was moving. not 100 when it was sitting still.

Kind of changes things, now doesn't it? (Why bother specifying the speed?) Nonetheless, if the length of the moving ship as measured in the planet frame is less than the spacing of the guns, and the guns are fired simultaneously according to the planet frame, then the bullets will miss the ship.

 

[Doc Al]

another misfire!

The length of the ships are always 100m you have to measure that on board.
It doesn’t matter how fast it goes!

The length of the ship measured in its own frame (aka, the proper length) will always be 100m, but its length measured in other frames depends on the speed of the ship.

DOC should have done the work to calculate all the time sequences as well, not just the distance calculations alone. But the destruction of his ship will move up to him in the nose of the ship in just a few nano sec.

You've managed to confuse yourself pretty good with all that bogus "time sequence" stuff. Doc Al's prescription: Run to the nearest relativity book, bury yourself in it, and don't come out until you understand Lorentz contraction!

(For extra credit: explain how the ship's pilot, well aware of relativistic effects, will agree that the bullets miss his ship---even though he measures the separation of the guns to be much less than the length of his ship.)

 

[bino]

if we set a ship off from the planet instead of guns then the ship we just launched would look like it took off crooked?

no one answered this

 

[bino]

what would happen from both views if we did not have the lorenz contraction?

or this one

 

[RandallB]

--- bogus "time sequence" stuff. Doc Al's prescription: Run to the nearest relativity book, bury yourself in it, and don't come out until you understand Lorentz contraction!

(For extra credit: ---------?)

You have got to be kidding!
I need to get buryed in a book and you haven't even tried to find the 3 sets of clock times I suggested! It cann't be that hard.

Let me give you an EZ one for extra credit:
We will send an offical observer to watch in a ship traveling by at 0.45c
Now you do agree the contraction from his view will be the same for both my cannons at 99m appart and your 100 m ship.

Or will it only be true in his view that 100m ships can not fit between 99 m spaced cannons - but in another view they can make it.
Different things happen depending on who looks at them? I don't think so.
I'll need to see some real numbers on that one before believing getting the time of events absolutly correct is not important!

RB

 

[zefram_c]

We will send an offical observer to watch in a ship traveling by at 0.45c Now you do agree the contraction from his view will be the same for both my cannons at 99m appart and your 100 m ship.

The contraction will be the same. But he will not observe the cannons firing simultaneously, and that will be enough to skew the outcome in favor of the ships.

Different things happen depending on who looks at them? I don't think so.

It depends on what you have in mind. Do you mean an 'event' happens in one frame but not another, eg the explosion of the ship? This should not happen in a reasonable physical theory, and indeed will not happen in SR, despite the best thought experiments one can conceive. Reaching such a conclusion means that SR was improperly applied. On the other hand, if a situation involves multiple events, it may be the case that different observers will not agree on the ordering or separation between them, and if something is defined by comparing such events (eg simultaneity), it is entirely possible, and perfectly consistent, for different observers to have different opinions on the subject.

 

[Doc Al]

You have got to be kidding!
I need to get buryed in a book and you haven't even tried to find the 3 sets of clock times I suggested! It cann't be that hard.

Not hard, just irrelevant. Now if you wanted to figure out how the ship sees the situation (the extra credit problem I proposed) then you'll need to figure out the sequence of clock readings according to the ship. (Of course, the ship agrees that the bullets miss!) But from the planet's view, it's trivial. I kid you not!

Let me give you an EZ one for extra credit:
We will send an offical observer to watch in a ship traveling by at 0.45c
Now you do agree the contraction from his view will be the same for both my cannons at 99m appart and your 100 m ship.

Well... no. If the observer travels at 0.45c with respect to the planet, then the observer will see the planet and ship moving at different speeds with respect to himself. (Perhaps you've forgotten the relativistic addition of velocity?) I think you mean that if the observer were to travel at a speed such that both planet and ship had the same speed with respect to him (about 0.63c, I believe), then the observer will see each contracted by the same factor. Sure... so what?

Length contraction is only one effect. To find out what "really happens" you must consider time dilation and desynchronization as well. Of course, from the planet frame things are simple since the guns fire simultaneously.

Or will it only be true in his view that 100m ships can not fit between 99 m spaced cannons - but in another view they can make it.
Different things happen depending on who looks at them? I don't think so.

As zefram_c points out, it depends on what you mean by "different things". Surely, whether the bullets hit (or miss) the ship--observers in any frame must agree on this. And they do! But as far as measurements of distances and times--those are frame dependent.

 

[RandallB]

You WIN

I still disagree on clock times and synchronization.
They are not irrelevant or trivial!
They are why YOU defeated me! KILLED me bad! And Gary Cooper was so good in High Noon! Bummer - please take the women and children from our Planet.
We have already sent the correct solution to the other planets in our system.

I'll explain your VICTORY.

Setting the whole battle field!
Clearly one cannon with a lucky shot could hit. But rather than luck. My planet covered the opening by placing 102 cannons 99m apart with the first and last at the outer limits of the danger window.

You of course wisely sent your ships down the required corridor nose to tail with just 1 m spacing between them. You are assured that at least one will get a near miss at the nose & the cannon behind to miss by -- I'll do the real math on that later.

So looking from the planet. The 100 ships are in front of well less than 50 of my cannon! I can only hope to hit less than EVERY OTHER ONE! And I cannot put them any closer together they are very large!

But looking from the first ship - Contraction has 100 guns firing away at only the lead ships and less than 50 of them at that! The ONES UP FRONT will all be hit most of them twice. But all the rear ships will get through!

That was one of my "Different views" - Here is one more:
Checking with the observer he's seeing a one to one hit on each and every ship as lengths are the contracted the same from this view.

Only one can be the true reality!
The planet view best shows how I incorrectly set the timing of firing at the same time on the planets clocks.
The other planets in our system are now setting the timing such that they fire cannon synchronized to the same time as seen by the hypothetical observer ship. The stagger this will require on the planet will adjust for not being able to pull the cannons closer together.

I called the bluff - when Scotty wasn't bluffing!
At least we went down gaining knowledge!

Thanks RB

 

[bino]

what the heck is he talking about?

 

[zefram_c]

Well... no. If the observer travels at 0.45c with respect to the planet, then the observer will see the planet and ship moving at different speeds with respect to himself. (Perhaps you've forgotten the relativistic addition of velocity?)

Actually I did, so I stand corrected. I was thinking of this case:

I think you mean that if the observer were to travel at a speed such that both planet and ship had the same speed with respect to him (about 0.63c, I believe), then the observer will see each contracted by the same factor.

As seen by such an observer, the only factor involved is de-synchronization of the clocks, and the bullets miss the ship, as in all other frames.

what the heck is he talking about?

Beats me, but since the "real math" is supposedly on its way, I will monitor this thread for a while.

 

[bino]

if we set a ship off from the planet instead of guns then the ship we just launched would look like it took off crooked? what would happen from both views if we did not have the lorenz contraction?

still getting no answers here?

 

[RandallB]

What are you talking about " the real math"?? - Like you said the math and figuring the times is easy, you don't need me to show you how to find that cannon B has to fire 184.8 nano sec BEFORE cannon A. And cannon C 184.8 nsec BEFORE B etc.

It was zefram c that was on target with the explanation that made the difference:

The contraction will be the same. But he will not observe the cannons firing simultaneously, and that will be enough to skew the outcome in favor of the ships.

zefram c is referring to the observer ship. Where we both overlooked the point that one half of .9c is not .45c. But his on target point that the observer was not seeing simultaneous firing cannons was the key.

For the observer to see simultaneous firings, The planet must fire B before A and every cannon out to the end must be advanced by an additional 184.8 nsec.

Now from the planet view the last ship is hit first by the 102nd cannon furthest from Cannon A.

The observer would see simultaneous firing at hits from all cannons.

And the Ships would see the cannons making their hits on 184.8 intervals also, BUT in the opposite order with the first Ship being hit first.

So yes all three views are seeing something different: first to last; last to first; or all at the same time. But now reality of what happens, all ships are hit and which two of the 102 cannons miss, still matches with all views.

Simple enough and easy math once I've got the "Relativity Spectacles" adjusted.

Thanks zefram c for helping me tweak mine.

Randall B

PS:
On "still getting no answers here?" without 'the lorenz contraction?' being in effect.
I'd expect the speeds of .63c plus .63c to give the ships a total speed of 1.26c resulting in "Sonic Light Booms" so bizarre I wouldn't know how to describe them!

 

[Doc Al]

What are you talking about " the real math"?? - Like you said the math and figuring the times is easy, you don't need me to show you how to find that cannon B has to fire 184.8 nano sec BEFORE cannon A. And cannon C 184.8 nsec BEFORE B etc.

Keep it simple. Two cannons; one ship. (See post #34) The planet's cannons are 99m apart; the ship's proper length is 100m and it travels at 0.9c with respect to the planet.

From the planet's view: When the ship is at the midpoint of the cannons, both cannons fire simultaneously. Of course, the bullets miss the ship, which has a length of only \Delta x/\gamma = about 44m.

From the ship's view: The two cannons do not fire simultaneously. The second cannon fires \gamma (\Delta x v/c^2) = about 680 nanoseconds before the first cannon, long before the ship is anywhere close. Of course, the bullets still miss the ship.

 

[bino]

randallb your still not making sense. that does not anwser my questions. nothing is moving so that it would be moving faster then c if there were no lorenz contraction.

 

[bino]

ssssssssss^ssss^sssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
-----------lShipl------------>
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssAssssBsssssssssss

B would shoot first right?

 

[zefram_c]

What are you talking about " the real math"??

I apologize for a poor choice of words, I simply referred to this statement:

You are assured that at least one will get a near miss at the nose & the cannon behind to miss by -- I'll do the real math on that later.

I was thrown off track by all the references to who beat who, but the follow-up was much clearer.

For the observer to see simultaneous firings, The planet must fire B before A and every cannon out to the end must be advanced by an additional 184.8 nsec... the Ships would see the cannons making their hits on 184.8 intervals also, BUT in the opposite order with the first Ship being hit first.

This is incorrect. De-synchronization is a frame-dependent effect, but not in the manner you derive. Since the observer is moving in the same direction as the ship, the order of the firing will be the same for him and the moving ship. In fact the moving ship will measure a larger delay. For an observer to see the same offset of 184.8ns but in the opposite direction, he would have to move at a velocity opposite to the first observer (ie. if the first observer travels at 0.63c towards the planet so he sees the planet approach him in front and the ships from behind at .63c each, the other observer would be moving at .63c away from the planet with a .9c velocity relative to the first one, and a greater velocity relative to the ships.)

So yes all three views are seeing something different: first to last; last to first; or all at the same time.

This is indeed true. Since the firings of the cannons are events with a spacelike separation (they are simultaneous and spaced out in the planet frame), there always exists one frame in which they are simultaneous, and two frames in which either occurs first. This is a property of all spacelike separated events.

But now reality of what happens, all ships are hit and which two of the 102 cannons miss, still matches with all views.

Agreed. If you time the bullets such that the ships are hit, all frames will agree which bullet hit what. Edit: This thread discussed different situations, so I make no claim that any particular situation will result in a hit.

I'd expect the speeds of .63c plus .63c to give the ships a total speed of 1.26c resulting in "Sonic Light Booms" so bizarre I wouldn't know how to describe them!

Sounds like someone's forgetting relativistic addition of velocities again

 

[RandallB]

For an observer to see the same offset of 184.8ns but in the opposite direction, he would have to move at a velocity opposite to the first observer (ie. if the first observer travels at 0.63c towards the planet so he sees the planet approach him in front and the ships from behind at .63c each, the other observer would be moving at .63c away from the planet with a .9c velocity relative to the first one, and a greater velocity relative to the ships.)

Didn't said the observer would see and offset. I said "For the observer to see simultaneous firings, The planet must fire B before A "
simultaneous OK
And I have no idea where your second 'observer' is.
We have a planet, one observer, and the ships.

Sounds like someone's forgetting relativistic addition of velocities again

Sounds like someone diden't read the question.
SO you think you can give an answer to post #47 ??
You tell me without lorenz contraction how are you going to keep "relativistic addition of velocities" ?

Randall B

 

[RandallB]

randallb your still not making sense. that does not anwser my questions. nothing is moving so that it would be moving faster then c if there were no lorenz contraction.

Of course things would move faster than c if you DUMP contraction!
Without contraction you can not just assume "relativistic addition of velocities" it comes from the fact that contraction exists.

RB

 

[Doc Al]

I'd expect the speeds of .63c plus .63c to give the ships a total speed of 1.26c resulting in "Sonic Light Booms" so bizarre I wouldn't know how to describe them!

Sounds like someone's forgetting relativistic addition of velocities again

I believe RandallB was giving an answer for the case without Lorentz contraction. Of course, in such a case the velocities of 0.63c would never arise. (0.63c came up because RandallB wanted an observer who sees both ship and planet moving at the same speed. Without relativity, that speed would just be 0.45c with respect to the planet. Of course, without relativity there is no reason to introduce a third observer.)

 

[RandallB]

ssssssssss^ssss^sssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
-----------lShipl------------>
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssAssssBsssssssssss

B would shoot first right?

No - not unless you turn the ships around and send them toward the planet!

Try this:

Ships <<<< ...1. 2. 3. 4. 5. 6
Observer <<
Planet - - ...A...B...C...D...E
(B2) occurs before (A1)

Ships << ....1..2..3..4..5
Observer - -
Planet >> ...A..B..C..D..E
(A1) & B2) same time

Ships - - .....1...2...3...4...5
Observer >>
Planet >>>> ......A.B.C.D.E
(B2) occurs after (A1)

So you B needs to fire before A to score the hits
Observer sees simultaneous hits
Ships see A firing before B

Timing and spacing of Cannon on the planet view must both be changed to achieve simultaneous firings and hits from the view of the ships.

Randall B

 

[bino]

ok let's say there is a ship 100ft long moving at .90c. a planet perpendicular has guns 101ft apart at rest. the guns line of fire are parrallel to each other. the planet plans it out so that the ship will be in between the bullets when the ship flies past the planet. would the bullets hit the ship or not?

this was my original question. there is only one ship with a length of 100 when its going .90c. there is two guns separated from each other by 101.A and B are guns they shoot one bullet each the lines are the paths the ship and the bullets go. so it looks like this:
ssssssssss^ssss^sssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
-----------lShipl------------>
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssAssssBsssssssssss

now my questions are:
would A or B shoot first?
if we set a ship off from the planet instead of guns then the ship we just launched would look like it took off crooked?
what would happen from both views if we did not have the lorenz contraction?

 

[RandallB]

I'm done here

Asprin got it started into a lot of varied questions and ideas here.

Seems like it's gone full circle. I've learned a bit!
Using a half relative speed observer was a good insite I'll use again.

See you in another thread.

RB

 

[bino]

would A or B shoot first?
if we set a ship off from the planet instead of guns then the ship we just launched would look like it took off crooked?
what would happen from both views if we did not have the lorenz contraction?

 

[Doc Al]

If you are serious about getting those questions answered you need to state them precisely. Define each scenario completely.