Homogeneity of space and time in Landau & Lifshitz

[Mike_bb]

Hello!

I have a problem with understanding homogeneity of the space.

In the book Landau&Lifshitz it was written that two spacetime intervals differ by constant $a$: $ds'^2=ads^2$

The coefficient $a$ can't depend on coordinates or time of interval because this violated homogeneity of the space.

My questions are:
1. Could anyone explain on the examples what does "violated homogeneity of the space" mean in this case?
2. How does dependence of coordinates and time of the interval violate homogeneity of the space?

As hard as I tried I couldn't understand it many times.

Thanks.

 

[PeterDonis]

In the book Landau&Lifshitz it was written

Which book? Where? Please give a specific book/chapter/section/page reference.

 

[Mike_bb]

Which book? Where? Please give a specific book/chapter/section/page reference.

Landau & Lifshitz' Classical Theory of Fields (pp.14-15)

 

[PeterDonis]

Landau & Lifshitz' Classical Theory of Fields (pp.14-15)

Ok, so their $ds$ and $ds'$ are describing the same infinitesimal spacetime interval in flat spacetime, just in two different inertial frames.

The coefficient $a$ can't depend on coordinates or time of interval because this violated homogeneity of the space.

The homoegeneity of space and time, i.e., spacetime. That just means that the way two different inertial frames are related to each other has to be the same no matter what point in spacetime we look at.

 

[PeterDonis]

their $ds$ and $ds'$ and are describing the same infinitesimal spacetime interval in flat spacetime, just in two different inertial frames.

Btw, I personally would argue that the statement I just quoted above, all by itself, is sufficient to establish that we must have $ds = ds'$, because $ds$ is just the spacetime length of the interval, and that must be an invariant; it can't depend on which inertialf frame we choose. I think this is how a more modern textbook on GR would make the argument.

 

[Mike_bb]

The homoegeneity of space and time, i.e., spacetime.

In other source "homogeneity of space and time" means homogeneity of space and homogeneity of time. Not a spacetime.

 

[Matterwave]

Imagine I say "this piece of clay is homogeneous" what would that mean? We can look at the density $\rho$ and conclude that $\rho \neq \rho(\vec{x})$ i.e. that the density is not dependent on the spatial coordinates. L+L are just using this same concept but for spacetime.

In other source "homogeneity of space and time" means homogeneity of space and homogeneity of time. Not a spacetime.

Can you give more context so we know what this other source is talking about?

 

[Ibix]

In other source "homogeneity of space and time" means homogeneity of space and homogeneity of time. Not a spacetime.

Homogeneity of space means invariance under purely spatial translations. Thus if you have two events whose coordinates are $x_a$ ($a=0,1,2,3$) and $x_a+dx_a$, and two other events whose coordinates are $x'_a$ and $x'_a+dx_a$ ($x_0=x'_0$ for a spatial translation) then you want the $ds^2$ between the two pairs to depend only on $dx_a$, not $x_a$ or $x'_a$. Otherwise, it isn't invariant under translation.

You can make the same argument with $x_0\neq x'_0$, $x_i=x'_i$, $i\neq 0$ to insist that $ds^2$ can't depend on $t$.

I think that gets you to $ds^2=Adt^2-Bdx^2-Cdy^2-Ddz^2$. Isotropy (rotational invariance) tells you $B=C=D$, and requiring the frame invariance of $c$ gets you $A=c^2B$.

I think that's right.

 

[PeterDonis]

In other source

What source? Again, please give a specific reference.

"homogeneity of space and time" means homogeneity of space and homogeneity of time. Not a spacetime.

I'm not sure there's actually a difference, but without seeing the actual other source you mentioned and what it says, I have no way of knowing.

 

[PeterDonis]

Homogeneity of space means invariance under purely spatial translations.

By analogy, "homogeneity of time" would mean invariance under time translations. And "homogeneity of space and time", which is what Landau and Lifshitz actually say, would mean invariance under any kind of translation, even one that combined "space" and "time" components.

 

[Ibix]

By analogy, "homogeneity of time" would mean invariance under time translations. And "homogeneity of space and time", which is what Landau and Lifshitz actually say, would mean invariance under any kind of translation, even one that combined "space" and "time" components.

I agree. I suspect L&L simply haven't got as far as defining spacetime yet (note the use of "fictitious four-dimensional space" just after eq. 2.4 in the excerpt in #3) so are relying on pre-relativistic concepts.

 

[Mike_bb]

Homogeneity of space means invariance under purely spatial translations. Thus if you have two events whose coordinates are $x_a$ ($a=0,1,2,3$) and $x_a+dx_a$, and two other events whose coordinates are $x'_a$ and $x'_a+dx_a$ ($x_0=x'_0$ for a spatial translation) then you want the $ds^2$ between the two pairs to depend only on $dx_a$, not $x_a$ or $x'_a$. Otherwise, it isn't invariant under translation.

You can make the same argument with $x_0\neq x'_0$, $x_i=x'_i$, $i\neq 0$ to insist that $ds^2$ can't depend on $t$.

I think that gets you to $ds^2=Adt^2-Bdx^2-Cdy^2-Ddz^2$. Isotropy (rotational invariance) tells you $B=C=D$, and requiring the frame invariance of $c$ gets you $A=c^2B$.

I think that's right.

From this we have that $ds'^2=ds^2$, right?

 

[Ibix]

From this we have that $ds'^2=ds^2$, right?

I think so, yes.

 

[Mike_bb]

I think so, yes.

And then we prove that $ds'^2=ds_1^2=ds_2^2=....$ for other intervals in all inertial frames.

 

[Matterwave]

And then we prove that $ds'^2=ds_1^2=ds_2^2=....$ for other intervals in all inertial frames.

If you go with the L+L derivation you only need to prove $ds^2=ds'^2$ because $ds'$ is arbitrary. Note the quote "in any other system" at the bottom of the first page. So you're done.

 

[Mike_bb]

I think so, yes.

If it so then why do we need proof L&L?

 

[martinbn]

From this we have that $ds'^2=ds^2$, right?

How do you get this from what Ibix wrote?

 

[Ibix]

If it so then why do we need proof L&L?

Apologies, I've reused the prime for something and confused myself about what you were asking.

Restating my argument, consider two events with coordinates $x_a$ and $x_a+dx_a$ and two other events with coordinates $X_a$ and $X_a+dx_a$. As I argued, homogeneity of space and time, isotropy, and the requirement that $ds^2=0$ for light tells you that $ds^2=B(c^2dt^2-dx^2-dy^2-dz^2)$, where $B$ can't depend on the coordinates.

You can make the same argument in another frame and obtain $ds'^2=B'(c^2dt'^2-dx'^2-dy'^2-dz'^2$. I think then that the "infinitesimals of the same order" allows L&L to write $ds'^2=a\,ds^2$, where their $a=B/B'$

 

[Mike_bb]

How do you get this from what Ibix wrote?

I'm already starting to doubt that this is true.

But I thought following way: if we have $dx_1$ and $x_1+dx_1$; $dx_2$ and $x_2+dx_2$; and so on then $dx_1'=dx_1$;$dx_2'=dx_2$ and so on.

Now I'm not sure that this is correct reasoning.

 

[Mike_bb]

Ibix,

What do you think about following reasoning?

If the space is homogeneity and isotropy we can use the fact that arbitrary point can be translated or rotated to any point. From this we can conclude that coefficient $a$ between $ds'^2$ and $ds^2$ doesn't depend on coordinate (and time).

 

[Matterwave]

I'm already starting to doubt that this is true.

What Ibix wrote gets you to the first highlighted eqn in the image of your post #3. You have to follow the rest of L+L arguments to get to the final eqn 2.6.

$dx_1'=dx_1$;$dx_2'=dx_2$ and so on.

This is not true. Assuming you are following Ibix's conventions that the subscripts tell you which coordinate, it is certainly not true that the individual coordinates all equal each other between two frames. That would defeat the purpose of making a change of coordinates.

 

[Matterwave]

If the space is homogeneity and isotropy we can use the fact that arbitrary point can be translated or rotated to any point. From this we can conclude that coefficient $a$ between $ds'^2$ and $ds^2$ doesn't depend on coordinate (and time).

You need only homogeneity to show that $a$ does not depend on coordinates. You use isotropy to show that $a$ can only depend on the magnitude of the relative velocity ($V_1\equiv \sqrt{V^2_{1x}+V^2_{1y}+V^2_{1z}}$) between the two frames. These statements are made below the first highlighted eqn in your image in post #3.

 

[Mike_bb]

You need only homogeneity to show that $a$ does not depend on coordinates. You use isotropy to show that $a$ can only depend on the magnitude of the relative velocity ($V_1\equiv \sqrt{V^2_{1x}+V^2_{1y}+V^2_{1z}}$) between the two frames. These statements are made below the first highlighted eqn in your image in post #3.

Thx. Does my post #20 contain true reasoning?

 

[Matterwave]

Thx. Does my post #20 contain true reasoning?

It's a bit loose.

I would simply say: if space and time are homogeneous then we can conclude that $a$ in $ds^2=ads'^2$ does not depend on the coordinates (or time).

In other words, in post #20 you have to call out time homogeneity (not just space) and you can drop isotropy since it's not needed there (it's needed for other parts of the overall argument).