Um… Okay, I got the link to work. Pretty! Kind of fun, although it goes by a bit too quickly for me to see all the numbers.
But here’s the problem
When you turn the diagram 90 degrees, you introduce the acceleration of gravity.
It isn’t the same experiment any longer. It’s different. The behavior of the two objects changes.
Pool balls on a level pool table behave differently from pool balls on a vertical pool table!
This has been going on for this entire thread! You change the terms of experiments, and then try to claim “It’s the same.” You turned Einstein’s “moving train” on its side, making it accelerate toward a gravitational source, and then tried to claim you hadn’t changed it at all.
I really don’t know how to go forward from here. You continue to treat zero-gravity situations and gravitational situations as if they are the same, and you don’t even seem to be conscious of this.
Actually the rotation of the two objects is irrelevant. If either object had a mass equal to the Earth gravity would play its part. The important point is that the example is an example of conservation of momentum only. This is what I mentioned in my last post. ** Tomo** is apparently unable to differentiate between the limited interactions in a theoretical scenario, and the vastly greater number of actions in a real world occurence. This inability of his is why he feels relativity is wrong. He is unable to make the leap from everyday observations to the physics of the immense.
Oh and I’d like a shot at your challenge regarding the Sagnac effect, but before I do I want to make sure I know the rules you made.
I can use 3 frames of reference - Transmitter Tx, receiver Rx and an independant reference (let’s call it my Aunt Emily, or Ae)
I can state that the interferometer is rotating WRT Ae.
I cannot make any relationship between the speed of the beams of light WRT Ae.
I can define the speed of the light beam WRT each other, Tx and Rx.
Is that correct, or are there any goalposts acting reletavistically?
Sorry for the double-post. Forgot to add this and edit window timed out.
[Quote=Bunny meant to say]
In the specific example above, look at the difference in speed. The objects approach each other with a combined velocity of 5(whatever units you like). And they separate with a velocity of 5(wuyl). If you were an observer on earth when the ball was dropped you would therefore see it bounce at exactly the same speed it hit. And this is in a perfectly elasctic collision. This is in complete contrast to the flubber experiment detailed previously. Congratulations Tomo, you’ve found a wiki entry that proves you wrong.
I should also have quoted the part where he says, “. . . make the cart on the left the size and mass of the Earth . . .”
(Now I’m having fun envisioning 3-D zero-G pool!)
Yep… This is why the bowling pin can shoot away from the bowling ball, much faster than the bowling ball was travelling.
It’s a classic “two equations in two unknowns,” and the results are compelled by the familiar conservation laws. Like much else in physics, the results are not necessarily going to accord with our intuitions.
(Scientific American had a lovely article on “Naive Physics,” and how we have basic expectations of how things behave that differ from how they really do. For instance, the idea of “impetus” makes sense to our minds. When you throw a baseball upward, it has impetus, which slowly drains away, so that, when it’s gone, the ball falls straight down. The actual parabolic arc is not intuitive!)
When we bounce basketballs on a floor, we get to simplify the equations remarkably, by taking the mass of the earth as effectively infinite. Since we can subtract this from both sides, it “goes away” nicely, and we get wonderfully symmetric behavior, like an (ideal) basketball bouncing upward with the same speed as it fell, and thus bouncing to the same height from which it was dropped.
Things get a little harder when the objects are of different, but rather similar, masses. Suppose someone gets cute, and substitutes a cue ball on your billiard table that has twice the mass of the other balls. Shoot this “heavy” cue at another ball, and the results are…not necessarily intuitive! The other ball can, in fact, rocket away from the cue ball faster than the cue ball was moving.
(Again, we see this every day with bowling balls and bowling pins, or golf balls and golf clubs.)
(And I keep typing “gold balls” when I mean “golf balls.” But, say, wouldn’t that be nice! Transmutation of elements! The darn things won’t drive for any distance, but they’re worth several thousand dollars each!)
Anyway, since I think we actually agree, I’m merely preaching to the preacher here, or trying to!
I’ve been away from the board for a while and so haven’t followed the thread, but i did do a search and it seems no one has made reference to Julian Barbour (The End of Time) and his idea that relativity isn’t sufficiently relative. He claims that Einstein had attempted to formulate relativity with a gridless (or Machian as he calls it) system but didn’t have the juice to do it and so he copped out and used space-time as the grid - a particularly amusing choice if one considers the growing contingent of physicists who seem to believe that time has no basis in reality.
And this shows how little you understand the basic elements of Newtonian mechanics. The equations apply in the (ideal) case of a perfectly elastic collision, i.e. when no energy is lost to any deformations of either object. As you said previously:
This is exactly the case in which these equations apply; in all other cases, some energy is lost (due to deformations, sound waves being emitted, friction, etc.), so the overall energy after the collision available for moving the carts is less than before it. This is just the same as in the case of the ball dropped from some height: it can bounce back, ideally, to its original height if no energy is lost; in any other case, it will fail to reach its prior height.
So let me walk you through the calculations. All we’ll need is the conservation of energy, and the conservation of momentum. Let’s call the incoming cart 1, and the (originally stationary) one 2. The incoming cart, prior to the collision, has mass m[sub]1[/sub], and velocity u[sub]1[/sub], and thus, a momentum of p[sub]1,before[/sub] = m[sub]1[/sub]u[sub]1[/sub] and an energy of E[sub]1,before[/sub] = m[sub]1[/sub]u[sub]1[/sub][sup]2[/sup]/2. Since the velocity of the stationary cart before the collision is 0, it has neither kinetic energy nor momentum, so the energy and momentum of cart 1 before the collision are the total energy and momentum.
After the collision, cart 1 still has some residual momentum and energy, p[sub]1,after[/sub] = m[sub]1[/sub]v[sub]1[/sub] and E[sub]1,after[/sub] = m[sub]1[/sub]v[sub]1[/sub][sup]2[/sup]/2; and cart 2 will have picked up energy and momentum of its own, p[sub]2,after[/sub] = m[sub]2[/sub]v[sub]2[/sub] and E[sub]2,after[/sub] = m[sub]2[/sub]v[sub]2[/sub][sup]2[/sup]/2. Now we know that both energy and momentum are conserved, such that the total energy and momentum before the collision must equal the total energy and momentum after the collision:
p[sub]1,before[/sub] = p[sub]1,after[/sub] + p[sub]2,after[/sub]
E[sub]1,before[/sub] = E[sub]1,after[/sub] + E[sub]2,after[/sub]
or:
m[sub]1[/sub]u[sub]1[/sub] = m[sub]1[/sub]v[sub]1[/sub] + m[sub]2[/sub]v[sub]2[/sub]
m[sub]1[/sub]u[sub]1[/sub][sup]2[/sup]/2 = m[sub]1[/sub]v[sub]1[/sub][sup]2[/sup]/2 + m[sub]2[/sub]v[sub]2[/sub][sup]2[/sup]/2
This is a system of two equations with two unknowns (v[sub]1[/sub] and v[sub]2[/sub]), which can be solved by simple arithmetic to get:
v[sub]1[/sub] = u[sub]1[/sub]*(m[sub]1[/sub] - m[sub]2[/sub])/(m[sub]1[/sub] + m[sub]2[/sub])
v[sub]2[/sub] = 2u[sub]1[/sub]*m[sub]1[/sub]/(m[sub]1[/sub] + m[sub]2[/sub])
Which, if m[sub]2[/sub] is sufficiently small compared to m[sub]1[/sub] that it can be neglected without making a significant error, yields just v[sub]1[/sub] = u[sub]1[/sub] and v[sub]2[/sub] = 2u[sub]1[/sub], meaning that the second cart goes twice as fast as the first one – the push force – originally did, as is (approximately) the case in the wiki example.
Now you may want to argue that this somehow isn’t applicable, but it’s what your own stipulations yield; and even if you want to include effects of elasticity etc., those will only slow down the speed of the second cart after collision by some margin. This isn’t subject to discussion – it’s simple textbook Newtonian mechanics.
The transmitter and receiver are in the same (rotating) frame. The light reaches both at different times because it travels a longer distance if it travels around the setup in the direction of rotation than if it travels against the direction of the rotation (because its target will move away, resp. towards, it during its time of propagation). Wiki explains this nicely. (It even explains how the result would be the same referenced to a co-rotating observer, i.e. one standing on the axis of rotation of the setup – there’s no ‘ether’ invoked here.)
Again, this is just a basic misunderstanding. If, in the setup, the gravity of the Earth (first cart) is relevant, then it will impart an acceleration on the second cart; if it starts without any kinetic energy, then it will ‘bounce’ at most as high as it was when released; if it has kinetic energy (or the Earth does), then of course it will ‘bounce higher’, as a ball does if you throw it hard towards the ground. If we ‘turn off’ Earth’s gravity, then the analysis is just the same as above: the cart is ‘hit’ with the Earth, and will zip off (if it was originally stationary) at about twice the velocity of the Earth hitting it.
Well, Einstein’s relativity does not depend on the ‘grid’, i.e. system of coordinates, in any way – it is generally covariant or diffeomorphism invariant, meaning that the physical quantities expressed by the theory are completely independent on whatever ‘grid’ one might choose. Barbour has an alternative theory, called shape dynamics, which is generally covariant only with respect to the three spatial dimensions, but includes a different symmetry, called conformal symmetry (basically, an invariance under arbitrary angle-preserving transformations) to make up for this; so it does in a sense ‘break’ the symmetry between space and time inherent in relativity. As far as I know, his theory is equivalent to general relativity at least with respect to physically relevant solutions, but I’m no expert on it. (It’s certainly fascinating and deeply original work, however.)
Man, this is certainly an old thread. I have another question to add to it. I haven’t read all the thread, but I’m about to go to sleep and will forget it if I don’t ask. I really doubt it’s already been asked.
It’s my understanding that there’s nothing about relativity that makes a curved universe (one in which you could travel in a straight line and end up where you started) impossible. In fact, it’s been a theory among scientists that this is how the universe is, though it’s looking less likely the more we test for it.
Let’s assume that the universe is indeed curved in such a manner and let’s pretend that it isn’t expanding or shrinking so that the second time you pass the same object, it’s at the same velocity. So what happens if you pass Earth twice at near light speed in your spaceship without accelerating? From the perspective of someone on Earth, the ship’s clock will be running much more slowly, and vice versa. However, both the Earth and the ship remain in inertial frames of reference between both encounters.
I thought maybe length contraction would solve it, but it only seems to make matters worse. Let’s say the universe is a million light years wide from Earth’s perspective. It would perceive the space ship as taking over a million years to fly across the universe and return but would expect a lot less time to have passed on the ship’s clock. On the other hand, the person on the ship might perceive the universe to be about a light-year across (in the direction of travel) so it would take a little over a year to pass Earth again (not conflicting with Earth’s perspective) but would expect maybe a few minutes to have passed on Earth (definitely conflicting).
So how can what seems to be a paradox be resolved? And how can the universe seem to be a different size from different perspectives? That would seem to suggest there is an absolute zero velocity: the one in which the universe is largest in all directions.
You have pretty much answered the question yourself! As you point out, in a compact space there is, in fact, a preferred frame of reference. Here is a paper discussing these issues.
Okay, I read all that, but I don’t understand how person S has a preferred frame of reference over person H. (Admittedly some of the math was over my head.) Since they’re both inertial, each would see the other move across the universe and then approach from the opposite direction. Are they saying that in a finite space, things aren’t entirely relative and that there is an absolute velocity? I’m not sure what they mean when they say the preferred frame is the one in which the length along a given side is shortest.
I am going to put this to bed right now.
If I drop an object onto a flat immovable surface (if you want to you can use a trampoline which easily deforms and reforms, ie it is springy, but you cannot lift the trampoline up as the object hits it), then the dropped object cannot, under any circumstances, bounce higher than the height it was dropped from. This also means that it cannot rebound from the surface any faster than it hit the surface, and I used the example of the trampoline to stop you pouncing on that. I am including any deformation and reformation in the contact speed. If it did bounce higher we would have perpetual motion.
The same applies to the Wiki example. If the smaller mass bounced off the larger faster than the contact speed, we would have perpetual motion.
So all of you who think it is possible, build a perpetual motion machine and I will believe you.
I asked for an explanation of the Sagnac effect according to relativity.
According to Einstein the light is traveling at c WRT the observer/receiver, so it is also traveling at c WRT the whole interferometer as the receiver is an integral part of it. Therefore it cannot travel a longer distance going with the direction of rotation any more than it can travel a shorter distance going against the direction of rotation. The distance between Tx and Rx is fixed. You have just described the aether theory explanation.
Try to keep sarcasm out of the discussion please. These rules are not made by me, but by Einstein. Einstein said that c is WRT the observer/receiver, so the Sagnac effect has to be explained internally. If you refer the speed of light back to a third FR, then that frame becomes the aether.
This was posted by me earlier.
There are two frames used in SRT, the transmitter and the receiver. The speed of light is the same for all observers (receivers). All other frames are irrelevant for working out the speed between transmitter and receiver which is by SR’s definition c. As there is no relative motion between Tx and Rx, the two light beams should arrive at the receiver at the same time no matter which direction they are going round the rotating path. They do not. If you use a third FR, and relate the speed of light to that, that FR becomes the aether. Please explain the Sagnac effect without recourse to a third reference frame. You are of course allowed to say that WRT this arbitrary frame, the Sagnac Interferometer is rotating. What you are not allowed to do is to refer the speed of light in the interferometer to this FR. The light is generated and transmitted in the interferometer, and is received in it also. It never goes outside it.
There are plenty of easy-to-observe real-world collisions where small masses emerge traveling faster than the contact speed – golf balls and baseballs are two obvious examples.
This is (literally) elementary freshman physics. If you don’t even understand the law of conservation of momentum, you’re in no position to critique relativity.
Here is a better paper you might try reading, but it may be behind a pay wall. What they mean about “the preferred frame is the one in which the length along a given side is shortest” is that the preferred frame is the one in which the size of the universe is smallest. There is another way in which this frame is preferred: it is the frame where if you send a light signal to the right and to the left, they both arrive back at the starting point at the same time. This is related to another way in which the frame is preferred: only in the preferred frame is clock synchronization possible, because there is no ambiguity created once the synchronization signal loops back to where it started. A lot of this in unintuitive and fairly advanced (it isn’t stuff you learn when you get a degree in physics), so it might be best to get ahold of the above paper and really sink your teeth into it. It does a nice job of solving the compact universe analog of the “barn paradox”, a javeline thrower whose javeline appears larger than the size of the universe in some reference frames
When a bowling ball moving at 5 m/s strikes a bowling pin, the bowling pin can rebound from the bowling ball at nearly 10 m/s from the point of view of an observer stationary wrt the bowling pin before the collision.** However, and this is the part which tomh4040 gets right, the relative speed of two objects after a collision is always less than or equal to the relative speed before the collision.** The bowling pin flies away at almost 10 m/s to the right, but the bowling ball is still traveling at nearly 5 m/s to the right, so the relative speed between the two is 5 m/s before and and a hair under 5 m/s after the collision.
tomh404 is right in that the rebound speed (with respect to each other) of two colliding objects is less than or equal to the approach speed.
Everyone else is right that a light object can rebound from a collision with a much heavier object with a speed almost twice that of the heavier object** from the perspective of an outside observer in an inertial reference frame.**
I think this isn’t correct, because, under idealized Newtonian situations, you can time-reverse a collision. You can run the film backwards, and not be able to tell which way is “real.” Thus, “before and after” don’t have unique meanings, and so, for what you say to be correct, the relative speeds would always have to be equal (since they could never be “less” without also being “greater.”
I’m not sure of this, but I think my logic is sound…
Of course, the equations I provided can’t lead to perpetual motion, since they derive from conservation of energy and thus, can’t violate it. And indeed, they completely and correctly describe the case of a small mass bouncing off a large one. Reproducing the solution for the velocities for your benefit:
v[sub]1[/sub] = u[sub]1[/sub]*(m[sub]1[/sub] - m[sub]2[/sub])/(m[sub]1[/sub] + m[sub]2[/sub])
v[sub]2[/sub] = 2u[sub]1[/sub]*m[sub]1[/sub]/(m[sub]1[/sub] + m[sub]2[/sub])
It’s easy to see that if m[sub]2[/sub] is much bigger than m[sub]1[/sub], such that m[sub]1[/sub] can be neglected, as is certainly the case for a ball bouncing off of the surface of the Earth, what we get as a solution is v[sub]1[/sub] = -u[sub]1[/sub], and v[sub]2[/sub] = 0, i.e. the smaller mass moving in the opposite direction at the speed it had before the collision (provided there are no friction/deformation losses), and the larger mass not moving at all. You can plug in realistic values if you like, but I don’t think any calculator will be able to show the difference to this ideal calculation.
Here’s a Java applet you can play around with in order to get some intuition for collisions.
Yes, the distance is fixed, but the interferometer rotates. So while the light is travelling – at its fixed speed c – the receiver will move towards it in the one direction, and away from it in the other. It’s the same as if you send a pulse of light after a car that’s moving away, or towards one that’s coming at you. Because the distance changes during the time of flight of the light pulse, the light will take longer to catch up with the car moving away than it does to reach the car closing in. Not sure what your problem is here.
That’s indeed right. The general solution for the velocities after collision (allowing for the second cart having a velocity prior to the collision of u[sub]2[/sub]) is:
v[sub]1[/sub] = (m[sub]1[/sub]u[sub]1[/sub] - m[sub]2[/sub]u[sub]1[/sub] + 2m[sub]2[/sub]u[sub]2[/sub])/(m[sub]1[/sub] + m[sub]2[/sub])
v[sub]2[/sub] = (m[sub]2[/sub]u[sub]2[/sub] - m[sub]1[/sub]u[sub]2[/sub] + 2m[sub]1[/sub]u[sub]1[/sub])/(m[sub]1[/sub] + m[sub]2[/sub])
Subtracting these, it’s immediate to see that |v[sub]1[/sub] - v[sub]2[/sub]| = |u[sub]1[/sub] - u[sub]2[/sub]|.
