- #36
Dale
Mentor
- 35,358
- 13,591
You can look right at the EOM directly $$\ddot y= -2gy$$ It is inhomogenous since the acceleration depends on position.
If by deriving you mean solving the Newton equation, no, this is not the end goal. To see that the system is not homogeneous, you don't need to actually solve the Newton equation. There is a much simpler way to do it.gionole said:but I think end goal must be deriving y′ and comparing it to y. I think this question goes to you @Demystifier
Hm, ##x(0)## and ##x'(0)## for sure are different. Didn't we say that ##x(0)coswt## and ##x'(0)coswt## are non-homogeneous ? they behave differently and are not spatially shifted functions. Even if the we had gotten the exact same solution of ##x'(t)=x'(0)coswt##, it wouldn't be homogeneous. In reply #32, this is what you told me:If the system was homogeneous, in analogy with the homogeneous case above one would expect that it is
##x'(t) = x'(0)coswt##
3. Are you sure when you got ##x' = x+a##, you can write: ##L = \frac{m\dot x^2}{2} - V(x'+a)## ? shouldn'it be : ##L = \frac{m\dot x^2}{2} - V(x' - a)## since ##x = x' - a## ?I really have no idea why do you think so. As @Dale told you, they are not the same, and the initial condition is not the only difference between them.
Basically you are saying that ##Acoswt## and ##(A+k)coswt## are the same functions. But they are not. If you don't see that, then your source of confusion is at some very elementary level.
That's wrong. The correct Newton equation is ##\ddot y' = -2g(y'-k)##. Since the force is ##F=-2gy##, and since ##y=y'-k##, it follows that the force is ##F=-2g(y'-k)##. When you correct this, is everything resolved?gionole said:Passive Transformation
We can still write newton as: ##\ddot y' = -2gy'##(I didn't replace ##y'## by ##y+k##)
Sorry, it was a typo, now I've corrected it.gionole said:3. Are you sure when you got ##x' = x+a##, you can write: ##L = \frac{m\dot x^2}{2} - V(x'+a)## ? shouldn'it be : ##L = \frac{m\dot x^2}{2} - V(x' - a)## since ##x = x' - a## ?
Read again what I told you. I said they are different functions, but I didn't say that it means non-homogeneous.gionole said:2. Hm, ##x(0)## and ##x'(0)## for sure are different. Didn't we say that ##x(0)coswt## and ##x'(0)coswt## are non-homogeneous ? they behave differently and are not spatially shifted functions. Even if the we had gotten the exact same solution of ##x'(t)=x'(0)coswt##, it wouldn't be homogeneous. In reply #32, this is what you told me:
I think, you're right. So what you're saying is if system was homogeneous, 2 things must happen. E.O.M of ##x'(t)## definitely must be exact same form such as ##x'(t) = x(0)coswt## - (if not, clearly not homogeneous), but that's not enough, because even if it would have ##x'(t) = x(0)coswt##, it would still be non-homogeneous. So sum up is: if eoms are different, it clearly means system is non-homogeneous and no further need to check anything else. If eoms are the same though, it still doesn't mean homogeneity and we need to look at the trajectory solutions. Correct ?Read again what I told you. I said they are different functions, but I didn't say that it means non-homogeneous.
This doesn't make sense. The ##x'(t) = x(0)coswt## is not an EOM, it is a solution of an EOM.gionole said:E.O.M of ##x'(t)## definitely must be exact same form such as ##x'(t) = x(0)coswt##
No, same eoms are enough.gionole said:If eoms are the same though, it still doesn't mean homogeneity and we need to look at the trajectory solutions. Correct ?
Just because his coordinate is called ##y'## does not imply that the force is ##-2gy'##. The symbol ##y'## is not an alternative name for the same coordinate. It is really a different coordinate, ##y'\neq y##.gionole said:Question 2: Now, I think we're at a center point of my confusion. [/B]
I said: ##\ddot y' = -2gy'## in ##y'## frame. Why is this wrong ? the ball in the ##y'## frame has ##y'## coordinates, what makes you say that this is wrong ?
OK.gionole said:I understand why you're doing the following: ##\ddot y' = -2g(y'-k)## (You want to write equation of the ball not in ##y## coordinates, but in ##y'## coordinates, and that's clear,
The initial assumption ##\ddot y' = -2gy'## is wrong, hence the final result ##\ddot y = -2g(y+k)## is wrong. Indeed, the final result must be wrong because we know that in fact ##\ddot y = -2gy##. It is not consistent that both ##\ddot y = -2gy## and ##\ddot y = -2g(y+k)## are true, unless ##k=0##.gionole said:but I don't know what makes the following wrong: ##\ddot y' = -2gy' => \ddot y = -2g(y+k)## (maybe this one is correct as well, it's just doesn't give us ##y'(t)## which is what we need.)
Interesting. When we use active transformation method(##y## frame only), for each ball(since ball is dropped from some height and in 2nd experiment, it's dropped from higher location), we said that each ball would have ##\ddot y = -2gy##.No, same eoms are enough.
I should have been more precise. If the eom is ##m\ddot x = F(x)## and if ##F(x)## is the same for all ##x##, then the system is homogeneous. If ##F(x)## is not the same for all ##x##, then the system is not homogeneous. That's all.gionole said:Interesting.
Did you miss my post #36? Since ##-2gy## is not a constant function of ##y## the system is non-homogenous.gionole said:we said that each ball would have ##\ddot y = -2gy##.
1st ball - eom is ##\ddot y = -2gy##.
2nd ball - eom is ##\ddot y = -2gy##
You see, eoms are the same, so with what you say, the system has to be homogeneous.
In our explanations so far on this thread, when we did active translation method, we only changed ball's initial location to be dropped from and in passive transformation method, we just look at it from different frames, but everything in space stays at the same place. What does the quote talk about then ? Does it mean checking homogeneity on bigger scale ? I understand that homogeneity of space is scale-dependent, so I think that's what the quote means but for our thread discussion, I don't think that's necessary to do stuff like that since we're curious whether space is homogeneous on smaller scale(just to us being on earth and using ##mgy^2##). Correct (2) ?Set up an experiment. Find the result.
Move the experiment somewhere else and run it again. You will get the same result.
To do this, you must move all the important parts of the experiment. For example, if you drop a rock on earth, it falls. If you move the experiment out into space, it just floats. To do it right, you would have to move the earth too.
The results would not be exactly the same because gravity from the moon and the sun have a small effect. So you really need to move them too. And if you really get precise, everything in the universe has a small effect. So you need to move the whole universe. And if you do that, how do you know you have moved anything? It gets confusing.
A simple answer is, because passive and active transformation can be written in the same mathematical way.gionole said:What would be your explanation why passive transformation still gives a solid way of checking homogeneity ?
You say that the solution doesn't seem to match, but in fact you didn't write the solution explicitly. And if you try to find the solution, you will see that the solution for ##x'(t)## depends on ##y'(t)##, which is not something you want. That's because you have two unknowns, ##x'(t)## and ##y'(t)##, while one equation. You need two equations, but the second equation, the one involving ##\ddot y##, "disappeared" because you decided to do it separately for ##x## and ##y##. You need to turn the second equation back, because it is really the system of two equations for two unknowns. Or more geometrically, your particle moves in two dimensions, so a general motion in two dimensions cannot be described by one equation, you need two of them.gionole said:but what if we do it separately for x and y axis.
in ##x,y## frame, ##ma_x = -2kx => \ddot x = \frac{-2kx}{m}##, Now we can plug ##x'cosa - y'sina## instead of ##x## and we get: ##\ddot x'cosa - \ddot y'sina = \frac{-2k(x'cosa - y'sina)}{m}##. From this, solution ##x'(t)## doesn't seem to match with ##x(t)## from ##\ddot x = \frac{-2kx}{m}##. Any idea why ?
If you are interested in the laws of physics for the motion of the ball, you move the ball. If you are interested in the laws of physics for the motion of all balls in the Universe, in principle you should move all balls in the Universe. If you are interested in the laws of physics for the whole Universe, in principle you should move the whole Universe. Obviously, only the first thing can be done in practice, experimentally. The other two things can only be done mentally, in theory. So in reality we only make actual experiments of the first type, and then extrapolate the results by assuming that similar results would be also valid in thought experiments of the other two types. This whole philosophy can also be summarized by the proverb - Think globally, act locally.gionole said:While checking homogeneity in active transformation method, all we move is the object's initial location(i.e ball dropped from 10 meters is now dropped from 12 meters) and this really doesn't suggest that we move all points of space, we just move the object's initial location. What's your thoughts on this ?
@vanhees71 So in terms of ball,earth example, here is what I think active transformation would be like:vanhees71 said:Active transformation means you move the entire system under consideration around, while passive transformation means you move your reference frame and describe the same system from this other reference frame. Mathematically at the end you describe both as transformations of some coordinates, building together a group.