# Reference frames in terms of homogenity/isotropy

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• gionole
gionole
Question 1: in the non-inertial frame, space is non-isotropic. If we're in an accelerated train frame, and we face forward(the same direction where train is accelerating) and drop a ball, ball moves backward. If we face backward and repeat the experiment, dropped ball moves forward to us. So we got different results depending on where we were facing - hence non-isotropic, but I wonder, when we say physics don't behave the same in different directions(as It happened in our case), shouldn't the experiments yield the different equations of motions ? It seems to me that if our ##x## axis is in the same direction as train is accelerating to, in both of our experiments, the equation of motion is ##x(t) = -\frac{at^2}{2}##. Shouldn't it yield different equations of motion ? I'm asking this because it's important in the Lagrangian case where logic is that in the inertial frame, ##L## can't depend on direction, which is fine, but i wanted to bring counter argument that if we used non-inertial frame and it depended on direction, we would get the different equations of motions and break things, but turns out I get the same thing for my experiments.

Question 2: In the non-inertial frame, space is said to be non-homogeneous. Can you give a good example(not about rotating example), why space is non-homogeneous in such frame ? Mainly, what we do is we do experiment when object is at position ##x##, then repeat the experiment where object is at ##x+\delta x## and if physics doesn't change(equation of motion) stays the same, we say it's homogeneous. Would you be able to give an example in the accelerated frame, such as object's physics is different at ##x## and ##x+dx##(i.e equation of motion is different ? more like a really world use case example. Just very simple one where you don't come up with pendulum or anything. I tried to use just ball being in an accelerated train, but i don't get different results while dropping the ball at ##x## and ##x+dx##.

The reason why I mention non-inertial frame is non-isotropic and non-homogeneous, are shown in the attached image taken from Landau's book at page: 5

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gionole said:
in the non-inertial frame, space is non-isotropic.
I don't agree. Your choice of coordinates is anisotropic - space is what it is. An isotropic space allows an inertial coordinate system to be defined; it doesn't require that you use it.

Ibix said:
I don't agree. Your choice of coordinates is anisotropic - space is what it is. An isotropic space allows an inertial coordinate system to be defined; it doesn't require that you use it.
can you check the attached image ? it's from Landau's book.

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Then I guess Landau and I gave different ideas of what is meant by "space". I'd see his point in relativity, but in classical mechanics, not so much.

gionole said:
can you check the attached image ? it's from Landau's book.
You should always quote more the what you want so that the context is understood. Also provide the reference (including location) so people can look it up themselves.

Ibix said:
Then I guess Landau and I gave different ideas of what is meant by "space". I'd see his point in relativity, but in classical mechanics, not so much.
well, he is basing his logic on gallileo's relativity principle. So I think my 2 questions above are valid.

Ibix said:
Then I guess Landau and I gave different ideas of what is meant by "space". I'd see his point in relativity, but in classical mechanics, not so much.
Ironically, though, the posted quote does come from Landau & Lifshitz's book about classical mechanics: Mechanics (Third Edition), section 3. Galileo's Relativity Principle.

You are misreading Landau. He says that an arbitrary (not non-inertial) frame is not homogenous or isotropic.

Frabjous said:
You are misreading Landau. He says that an arbitrary (not non-inertial) frame is not homogenous or isotropic.
Are there any other frames other than non inertial and inertial ? What is its name then ? What arbitrary means ?

gionole said:
Are there any other frames other than non inertial and inertial ? What is its name then ? What arbitrary means ?
An arbitrary frame essentially means any frame. I am guessing that one can show that the numbers of frames that do not show any of the symmetries are much greater than the number of frames that show at least one symmetry. So if you pick one frame, the probability that it has no symmetries is essentially one.

Take a frame with a constant acceleration. If you translate a particle to a different location, it will have the same laws of motion at that location. So space is homogenous. This contradicts your non-inertial means non-spatially homegenous.

Frabjous said:
An arbitrary frame essentially means any frame. I am guessing that one can show that the numbers of frames that do not show any of the symmetries are much greater than the number of frames that show at least one symmetry. So if you pick one frame, the probability that it has no symmetries is essentially one.

Take a frame with a constant acceleration. If you translate a particle to a different location, it will have the same laws of motion. So space is homogenous. This contradicts your non-inertial means non-spatially homegenous.
If you got a frame with constant acceleration, that frame is definitely non inertial which I think you agree. In this frame, yes, space is homogeneous.

Thats what made me confused but now I am asking can you give an example of a frame which is neither inertial nor non inertial ? I dont think there is such a frame. It is either inertial or non inertial and we just showed that space is homogeneous in non inertial frame and it is also easy to show that space is homogeneous in inertial frame as well. So what frame is Landau talking about ? Maybe could you provide an example ?

We showed that space is homogenous for a specific non-inertial frame, not in general.

A material (lagrangian) frame of reference in fluid dynamics does not have to be spatially homogenous.

Question 1: I mean if we are talking about a ball in an accelerated train, you can choose inertial frame(outside train) or non inertial frame(inside train). I guess for this case, we cant choose material frame? If we can what would it be ?

If we cant, landaus argument is for the very different types of cases. Have to dig on material frame of reference but still wanted to ask the question just to know your opinion.

Question 2: in the non-inertial frame, space is non-isotropic. If we're in an accelerated train frame, and we face forward(the same direction where train is accelerating) and drop a ball, ball moves backward. If we face backward and repeat the experiment, dropped ball moves forward to us. So we got different results depending on where we were facing - hence non-isotropic, but I wonder, when we say physics don't behave the same in different directions(as It happened in our case), shouldn't the experiments yield the different equations of motions ? It seems to me that if our ##x## axis is in the same direction as train is accelerating to, in both of our experiments, the equation of motion is ##x(t) = -\frac{at^2}{2}##. Shouldn't it yield different equations of motion ? I'm asking this because it's important in the Lagrangian case where logic is that in the inertial frame, ##L## can't depend on direction, which is fine, but i wanted to bring counter argument that if we used non-inertial frame and it depended on direction, we would get the different equations of motions and break things, but turns out I get the same thing for my experiments.

gionole said:
Question 1: I mean if we are talking about a ball in an accelerated train, you can choose inertial frame(outside train) or non inertial frame(inside train). I guess for this case, we cant choose material frame? If we can what would it be ?

If we cant, landaus argument is for the very different types of cases. Have to dig on material frame of reference but still wanted to ask the question just to know your opinion.

Question 2: in the non-inertial frame, space is non-isotropic. If we're in an accelerated train frame, and we face forward(the same direction where train is accelerating) and drop a ball, ball moves backward. If we face backward and repeat the experiment, dropped ball moves forward to us. So we got different results depending on where we were facing - hence non-isotropic, but I wonder, when we say physics don't behave the same in different directions(as It happened in our case), shouldn't the experiments yield the different equations of motions ? It seems to me that if our ##x## axis is in the same direction as train is accelerating to, in both of our experiments, the equation of motion is ##x(t) = -\frac{at^2}{2}##. Shouldn't it yield different equations of motion ? I'm asking this because it's important in the Lagrangian case where logic is that in the inertial frame, ##L## can't depend on direction, which is fine, but i wanted to bring counter argument that if we used non-inertial frame and it depended on direction, we would get the different equations of motions and break things, but turns out I get the same thing for my experiments.
Q1: If we are inside the train, we use the non-inertial frame. If we are outside the train, we use the inertial frame. Experiments give different results at the two locations. Space is not homogenous.
Q2: We determine a pseudo force in the direction toward the rear of the train. There is no such force in directions transverse to the train. Space is not isotropic.

1. If material frame is the same thing as inside being the train, yeah, in our train ball case, in it, space is still homogeneous. So fluids case is the only thing where space is non homogeneous ? Dont we have any good better example such as train ball case ? And to be honest, in my opinion, you should not be comparing experiments results of being in train and outside of train to say that space is non homogeneous. You should be showing that space is non homogeneous in the frame you are(not stepping out of it).

2. Well, I get that different physics, but does the equation of motion differ in those 2 experiments ? When we rotate to face the back of the train, are you rotating x axis ? If you dont rotate it, the equation of motion in both cases are exactly the same. If you rotate, yeah, signs are different but I dont know why you rotate x axis. You shouldnt.

1. Think about performing experiments on the surface of the ocean where the frame is the surface current (assume the currents do not change). Or a roller coaster. Different results in different places. Non-homogenous is space.
You need something to vary spatially.
2. Space has a preferred direction (transverse vs longitudinal). That is want non-isotropic means. (FYI, I quickly deleted my original reply, so you might not have read my final one)

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gionole said:
can you check the attached image ? it's from Landau's book.
Of course choosing coordinates can apparently break symmetries, because you have to introduce a reference frame. Take, e.g., our usual Euclidean 3D affine manifold, which is homogeneous and isotropic. Now introduce spherical coordinates. To do so you need to introduce a "polar axis" and an arbitrary fixed point ("the origin"). The latter of course breaks translation invariance by hand, because now you introduce this special point. With this distinguished point there's still the istropy round this point, but the polar axis also breaks this, i.e., what's manifest in polar coordinates is only the symmetry wrt. rotations around the polar axis.

This is also immediately clar, when you write down free Lagrangian for a single particle,
$$L=\frac{m}{2} \dot{\vec{x}}^2=\frac{m}{2} (\dot{r}^2 + r^2 \dot{\vartheta}^2 + r^2 \dot{\varphi}^2 \sin^2 \vartheta).$$
The only "cyclic coordinate" is ##\varphi##, i.e., only the rotations around the polar axis (##3##-axis in the standard Cartesian coordinate system used to introduce the spherical coordinates) is manifest in spherical coordinates, and you get only the canonical momentum conserved, which is the ##3##-component of angular momentum, but of course you know that there's full rotational symmetry around the origin, i.e., the full angularmomentum vector is conserved. This you can only figure out with the adequate symmetry analysis, using the full rotation group (which however is easier to express in Cartesian than in spherical coordinates).

gionole
vanhees71 said:
Of course choosing coordinates can apparently break symmetries, because you have to introduce a reference frame. Take, e.g., our usual Euclidean 3D affine manifold, which is homogeneous and isotropic. Now introduce spherical coordinates. To do so you need to introduce a "polar axis" and an arbitrary fixed point ("the origin"). The latter of course breaks translation invariance by hand, because now you introduce this special point. With this distinguished point there's still the istropy round this point, but the polar axis also breaks this, i.e., what's manifest in polar coordinates is only the symmetry wrt. rotations around the polar axis.

This is also immediately clar, when you write down free Lagrangian for a single particle,
$$L=\frac{m}{2} \dot{\vec{x}}^2=\frac{m}{2} (\dot{r}^2 + r^2 \dot{\vartheta}^2 + r^2 \dot{\varphi}^2 \sin^2 \vartheta).$$
The only "cyclic coordinate" is ##\varphi##, i.e., only the rotations around the polar axis (##3##-axis in the standard Cartesian coordinate system used to introduce the spherical coordinates) is manifest in spherical coordinates, and you get only the canonical momentum conserved, which is the ##3##-component of angular momentum, but of course you know that there's full rotational symmetry around the origin, i.e., the full angularmomentum vector is conserved. This you can only figure out with the adequate symmetry analysis, using the full rotation group (which however is easier to express in Cartesian than in spherical coordinates).
Thanks for the great answer. I'd also love to take your opinion on the following.

It's said that in the non-homogeneous space, physics change while doing the experiment at two different points. Let's say we got an accelerated train where at some parts(location of the train ##d_1##, train's acceleration is ##a_1## and at ##d_2##, it's ##a_2##) - It's like some parts of train is on a cliff, while the rest is already on the ground. If you're in train frame, thrown ball in the train to the train's direction from ##d_1## has ##x(t) = v_bt - \frac{a_1t^2}{2}## while from ##d_2##, it's ##x(t) = v_bt - \frac{a_2t^2}{2}## .

Since we say space is non-homogeneous(i.e physics change), that must mean that EOM's form changes. In our case, the difference between the EOM's is in one case, we got ##a_1## in there, in second case, we got ##a_2##. Ok, so far so good.

Let's look at the same experiments from ground frame. Ball is thrown from ##d_1## inside train. ##x(t) = (v_b + v_1)t##. And when it's thrown from ##d_2##, we got ##x(t) = (v_b + v_2)t##. ##v_1## and ##v_2## are the velocities that ball had at the exact moment before it was thrown. We know that space is homogeneous in inertial frame, so physics must be the same - i.e EOM's should not change, but we clearly see that they differ by ##v_2## and ##v_1##. How is it that we don't call this EOM's form change and we do call it form change in the above case in the train frame ?

vanhees71
If you describe your example from the point of view of the inertial frame nothing special happens. If you neglect the gravitation of the Earth (which you seem to do in this example), the ball will move with constant velocity wrt. this frame. If you work in the train's rest frame, of course you have this apparent asymmetry. It's again an asymmetry introduced by the choice of the reference frame.

gionole

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