Contraction effects at relativistic velocities

[asprin]

Just an issue I want to be absolutely clear about.

Scenario,

WE propell a rod of iron with a dimensions of 100 meters *1 meters through space with the 100 meter length perpedicular to the direction of velocity.

So the rod is traveling width edge forward. ( perpedicular to direction)

the question is :

Does the perpendicular length of our rod contract?

( if the velocity length is 1 meter and our width is 100 meters as per direction)

And also

Is there a table that is available to give contraction results vs veocity somewhere on the Net?

say contraction vs Earth meter at velocities like 0.1c, 0.2c, 0.3c etc etc

any help would be appreciated

 

[asprin]

another version of the same question

If we have two ships both traveling parallel to each other at relativistic velocities is the distance of separation subject to contraction?

 

[Ba]

I don't think so not with both ships at the same line perpendicular to motion, because they are not moving at all relative to each other and the horizontal distance will not shrink no matter how fast they go.

 

[Doc Al]

length contraction formula

WE propell a rod of iron with a dimensions of 100 meters *1 meters through space with the 100 meter length perpedicular to the direction of velocity.

So the rod is traveling width edge forward. ( perpedicular to direction)

the question is :

Does the perpendicular length of our rod contract?

Assuming you mean the length perpendicular to the motion (the "100 meter" length), then no. Length contraction only occurs in the direction of motion.

Is there a table that is available to give contraction results vs veocity somewhere on the Net?

No need for a table, the formula is simple. Let L_0 be the length of an object in its own (rest) frame; then L will be the length of the object measured from a frame in which the object is moving at speed v parallel to its length:
L = L_0 \sqrt{1 - v^2/c^2}

 

[selfAdjoint]

Assuming you mean the length perpendicular to the motion (the "100 meter" length), then no. Length contraction only occurs in the direction of motion.

No need for a table, the formula is simple. Let L_0 be the length of an object in its own (rest) frame; then L will be the length of the object measured from a frame in which the object is moving at speed v parallel to its length:
L = L_0 \sqrt{1 - v^2/c^2}

This is relative to the frame which sees the rod moving with velocity v.

 

[Doc Al]

This is relative to the frame which sees the rod moving with velocity v.

Thanks, selfAdjoint. I think my wording was a bit convoluted!

 

[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?

 

[Janus]

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'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.

 

[bino]

no i mean simultaneously according to the ship

 

[Janus]

no i mean simultaneously according to the ship

If they are fired simultaneously according to the ship, then the planet can't plan it out such that the ship is between the bullets when the ship flies past the planet. In that case, according to the planet, the guns do not fire simultaneously. For instance, if from the ship frame, the bullets hit the ship while being fired simultaneously, then from the planet frame, the guns will fire in a staggered order timed such that the bullets each hit the ship

 

[bino]

why is it not simultaneous from both frames? if it is simultaneous from the planet then the ship sees that farthest gun shooting off first. right?

 

[pmb_phy]

another version of the same question

If we have two ships both traveling parallel to each other at relativistic velocities is the distance of separation subject to contraction?

With regards to your first question - No. For the reason why please see the bottom section of - http://www.geocities.com/physics_world/sr/lorentz_contraction.htm

As per the question above - Assume that the rockets are identical and that they fire their engines at a constant rate as measured in an instantaneous rest frame. Let them start from rest each lying on the x-axis, one in front of the other. Then let an observer in the inertial frame measure the distance between them. The inertial observer will detect no change in distance between the two rockets. Not let either of the observers who are at rest in the rockets measure the distance. Each observer will measure the distance increasing. This is known as Bell's Spaceship Paradox. See - http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html

The reason has to do with gravitational time dilation and Lorentz contraction.

The observer in the trailing rocket ship can think of himself as being at rest in a uniform gravitational field. He's firing his engines so as to neither rise nor fall. This observer sees the the other rocket ship higher in the gravitational field. However, due to gravitational time dilation, the bottom observer will see the rocket as firing his engines at a rate which is faster than his and therefore he gains height.

Pete

 

[bino]

there is no rocket trailing the other. there are two guns, parallel to each other, and firing perpendicular to the line of movement of the ship.

l=bullets path
-=ships path
s=space

ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
-----------lShipl------------
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss
ssssssssssslsssslssssssssssss

 

[Doc Al]

if it is simultaneous from the planet then the ship sees that farthest gun shooting off first. right?

From the viewpoint of the ship, the gun in the rear fires first.

 

[Janus]

why is it not simultaneous from both frames? if it is simultaneous from the planet then the ship sees that farthest gun shooting off first. right?

Here, maybe this will help:

http://www.geocities.com/janus58.geo/simultaneous.html

 

[bino]

the animation did not work for me.

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

B would shoot first right?

 

[bino]

i got the animation to work. the ship that is moving in both animations looks like it is trying to out run the light. how can they see the explosion before they see the light of the flash. they could not possibly see the explosion until they see the light from the explosion.

 

[Janus]

i got the animation to work. the ship that is moving in both animations looks like it is trying to out run the light. how can they see the explosion before they see the light of the flash. they could not possibly see the explosion until they see the light from the explosion.

It's not that observer two sees the explosions happen before he sees the flashes, it is when observer 2 determines the explosions had to occur in order for him to see the flashes at the same time. Example: If a light flash originates 1 light sec from me, and I see that flash when my clock reads 12:00:01, I know that the flash started when my clock read 12:00:00

And if the flash originated 2 light secs from me and I see the flash at 12:00:01, then it originated when my clock read 11:59:59, Thus, if I see two flashes at the same time and one originated further from me, then the further explosion occurred before the closer.

Also:
Assume you have observers at the actuators in the same frame as observer two. They have clocks which are synchronized with each other and a clock carried by observer 2.

The blue observer stops his clock when he is opposite the other blue marker (We can assume that he passes so closely to that marker that we don't have to consider light signal delay.)

The green observer does the same when he passes the green marker.

Thus each observer has a clock that records when the explosion from his marker occured. If you were to bring the clocks together you will find that that the green observer's clock will read less than the blue observer's, and thus according to synchonized clocks sharing the same frame as observer 2, the explosions did not occur at the same time.

 

[bino]

would time dilation come into effect?

 

[Janus]

would time dilation come into effect?

Time dilation needs to be considered when we are comparing the clock rates between frames. Such as the rate that observer 1 would determine for frame 2's clocks, or the rate that frame 1's clocks would have as measured by observer 2. In this instance, we are only concerned with when the events occurred in each frame according to its own clocks.

 

[Doc Al]

would time dilation come into effect?

As Janus points out, it depends on what you wish to know. Referring to your example of the ship flying past the planet with the guns: If all you want to know is "According to the ship, which gun fires first" then you need only consider the desynchronization of the planet clocks as seen from the ship (the "relativity of simultaneity"). But if you want to know "According to the ship, how much time passes between the firing of the two guns" then you need to consider time dilation as well.

 

[bino]

ok then from the view of the ship, 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?

 

[bino]

because both side of the ship would be taking off at simualtanousley from the view of the planet.

 

[bino]

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

 

[Jake]

what causes this non-simultaneosness? Just the fact that light moves at a limited speed and therefore depending on what direction you are going the light has less space to travel, but more going in a diff. direction? Or something to that effect. Meaning, is the non-simultaneosness just the cause of a the way the information travels or is there something else to it?

 

[bino]

length contraction in the two different frames cause the non-simultaneosness in the frames. look at the animation on the first page. since each frame sees the other as being shorter they have different times as to when each pass each other.

 

[Jake]

But length contraction itself is just a cause of non-simultaneosness - I think. If not, then what's the cause of length contraction?

 

[bino]

http://www.physics.nyu.edu/hogg/sr/sr.pdf
read this its pretty good.

 

[bino]

but basically in order to have light to be constant for all observers we need to alter time and space to make that happen.

 

[quantumdude]

but basically in order to have light to be constant for all observers we need to alter time and space to make that happen.

No, "we" don't do it at all. It's just what happens.

 

[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.