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.