Galilean invariance question


by Skids
Tags: galilean, invariance
Skids
Skids is offline
#1
Dec6-09, 05:00 AM
P: 4
Hi, I've been reading Vic. Stenger's book "The comprehensible Cosmos" and have a question about the example he gives for Galilean invariance. In his example, Galileo drops a weight from the tower of Pisa and to a person standing near the tower (and thus in the same inertial? frame of the tower) the weight falls in a straight line. For an observer moving at velocity on a river near the tower, the weight falls in a parabolic shape. I can't visualize this. I've been on boats and watched things move on land and can't say I've noticed anything like what Stenger talks of. I'm not even sure how it relates to Galilean invariance. The diagram in Stenger's book doesn't really help. I'm probably asking a total noob question, but if anybody could help or point me to a better description or diagrams that'd be great.
Thanks in advance.

P.S. I did a search for Galilean invariance before posting but didn't find anything that appeared relevant. Hopefully I haven't done anything flame-worthy.
Phys.Org News Partner Physics news on Phys.org
Physicists design quantum switches which can be activated by single photons
'Dressed' laser aimed at clouds may be key to inducing rain, lightning
Higher-order nonlinear optical processes observed using the SACLA X-ray free-electron laser
arildno
arildno is offline
#2
Dec6-09, 05:20 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
The point Stenger is trying to make, is that relative to the boat passengers' REST FRAME, the Tower of Pisa AND the falling rock will have a horizontal velocity component (i.e, the negative velocity with respect to that velocity the boat is observed to have from the rest frame of the Tower of Pisa).

Since the rock's horizontal x-displacement component will be proportional to "time", whereas its vertical y-displacement component will be proportional to "time squared", y will be proportional to "x squared", i.e, the traced out curve will be PARABOLIC.


Now, this is compounded by the difficulty that, WHEN PERCEIVING, we are adjusted to assigning OURSELVES any and every relative motion that may exist between ourselves and the ground. (That's why, in earth quake situations, that so many people report the UNNATURALNESS of the feeling that the ground itself moves; it is so totally contrary to everyday experiences that we simply cannot regard the Earth itself as stationary)

THEREFORE, we instinctively merely measure the rock's relative position with respect to the Tower of Pisa, NOT its relative position to ourselves (i.e, its position relative to our rest frame).
Stenger is saying that relative to OURSELVES, regarding US as the stationary object, the rock will go in a parabolic curve, but that perspective is not naturally geared into our brain, which is, instead, wired into regarding the ground as the absolute rest frame.

Therefore, you need skill to regard YOURSELF as stationary, and strictly measure how the rock moves relative to YOU.
The easiest way is to use a mindless camera to register the movement of the rock, and indeed, the movement of the Tower of Pisa itself.
hamster143
hamster143 is offline
#3
Dec6-09, 05:40 AM
P: 986
That's probably a misleading example. The weight drops parallel to the tower, and the person observing from the boat is very likely to make a subconscious mental adjustment that makes it appear as if the objects falls in a straight line.

Would you agree that an object dropped from an airplane falls in a parabolic shape? What would the trajectory of the same object look like if you looked from an airplane that flies parallel and at the same velocity as the original airplane?

Skids
Skids is offline
#4
Dec6-09, 06:23 AM
P: 4

Galilean invariance question


Hi, thanks for your responses. I'm getting the idea I think. It is hard to imagine the ground as moving and the dude on the boat as stationary. I'm was sort of imagining the rock falling in a linear fashion, not parabolic. Something like y proportional to -x. But then, because Gravity accelerates the rock it would fall more slowly at the beggining of its descent and thus relative to the inertial frame of the person on the boat the rock would appear to cover more x relative to y than later when having accelerated a bit it would seem to fall almost straight down. I think that was my error. Well apart from not having the guy on the boat as stationary. A falling object accelerates at 9.8m/s/s so when released it starts off with 0 m/s velocity but accelerates to 9.8 m/s at t = 1s and so on. But in the first few milliseconds it hasn't fallen much at all and if the boat were travelling quite fast, it would appear to the guy on the boat that the rock has a fair amount of horizontal motion (the negative of the guy's velocity multiplied by the time) I suppose, but later as it had accelerated, horizontal motion would be swamped by vertical motion. Or something similar.

Regarding the plane. From an observer on the plane. If a ball were dropped it would decelerated relative to the plane and disappear downwards and backwards. Would be hard to observe. From someone on the ground with sufficient eyesite or measuring device. The ball would fall slowly at first because it would start with the jets velocity, but would decelerate horizontally and loose the horizontal component but accelerate vertically due to gravity. That would look like half a parabola (y = cx^2). That's easier to visualize as the ground is appears stationary in everyday experience (constant velocity). To an observer on another jet or plane travelling at the same velocity (same inertial frame?) the ball would fall in a (half) parabolic shape I suppose but with reverse direction as it would fall behind at first slowly then loose the horizontal motion as it gained vertical motion under gravity. Is that right?

Thanks again. Just typing out this response to your posts has helped me work it out a bit.
YellowTaxi
YellowTaxi is offline
#5
Dec6-09, 08:51 AM
P: 196
Regarding the plane. From an observer on the plane. If a ball were dropped it would decelerated relative to the plane and disappear downwards and backwards. Would be hard to observe. From someone on the ground with sufficient eyesite or measuring device. The ball would fall slowly at first because it would start with the jets velocity, but would decelerate horizontally and loose the horizontal component but accelerate vertically due to gravity. That would look like half a parabola (y = cx^2). That's easier to visualize as the ground is appears stationary in everyday experience (constant velocity). To an observer on another jet or plane travelling at the same velocity (same inertial frame?) the ball would fall in a (half) parabolic shape I suppose but with reverse direction as it would fall behind at first slowly then loose the horizontal motion as it gained vertical motion under gravity. Is that right?
In all of physics & mechanics it's generally assumed that air resistance is zero (unless you're dealing with aerodynamics or whatever). And I don't think you'll ever find a case where relativity and air resistance are considered together. In fact it was by visualising a situation were air resistance is zero, that Galileo was first able to unlock the laws of motion for us. So aeroplanes are not a good place to be sitting if you want to think about relativity. Not unless you're just aiming to confuse yourself...
arildno
arildno is offline
#6
Dec6-09, 09:42 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
A falling object accelerates at 9.8m/s/s so when released it starts off with 0 m/s velocity but accelerates to 9.8 m/s at t = 1s and so on. But in the first few milliseconds it hasn't fallen much at all and if the boat were travelling quite fast, it would appear to the guy on the boat that the rock has a fair amount of horizontal motion (the negative of the guy's velocity multiplied by the time) I suppose, but later as it had accelerated, horizontal motion would be swamped by vertical motion. Or something similar.
It will always have the same horizontal velocity component, but its vertical velocity component will as time goes on be a greater fraction of the rock's total speed.
So, it will seem to get a more and more vertical motion, just as the parabola does as you move outwards.
TurtleMeister
TurtleMeister is offline
#7
Dec6-09, 10:51 AM
P: 735
This video is relevant to your question and may help you better understand. It's old, but explains reference frames quite well. (It's in four parts)
http://www.youtube.com/watch?v=Y75kEf8xLxI
Skids
Skids is offline
#8
Dec6-09, 02:33 PM
P: 4
Thanks again. I see your point about ignoring air resistance. The motion is idealized. It's a model with some simplifications. So basically we start out with horizontal motion, which remains, but there is increasing vertical motion contributing to the change in direction and velocity. Which is the definition of acceleration (change in direction/change in velocity). Which gives a half parabolic shape. That makes sense.

I'll have to check out the video.

Thanks again.
Skids
Skids is offline
#9
Dec6-09, 02:47 PM
P: 4
Just looked at that video. That was helpful.

Regarding the plane. If an airliner is travelling along at say 800km/h at 10,000m and releases a rock, veiwed from another airliner travelling in the same direction and velocity then according to the evidence of that video I reckon the rock would appear to fall vertically if we ignore friction/air resistance. Though from the frame of reference of the Earth it would be be our parabolic path. This is because the frame of reference of the airliners have
s moved on by a certain distance from the release of the rock till strikes the ground. Correct?


Register to reply

Related Discussions
Galilean invariance and conserved quantities Classical Physics 8
Lorentz Invariance and Non-Galilean Invariance of Maxwell's Equations Classical Physics 4
Tensors: Lorentz vs Galilean invariance Special & General Relativity 1
Galilean invariance (and Maxwell's equations) Classical Physics 7
Schroedinger Equation - Galilean Invariance Quantum Physics 5