Lorentz transformation independence of axis orthogonal to velocity

[center o bass]

Apriori -- before taking any of the postulates of special relativity into account -- we might say that the lorentz transformations between two frames K and K', where K' is moving w. speed v along the x-axis of K, is given by
$$\vec{x}' = F(\vec x, t)$$
and
$$t' = G(\vec x, t).$$
Now, i want to conclude that $y' = y$ and $z'=z$ as well as $x' = F(x,t)$. I believe this follows form homogeniety of space and time, but I do now know exactly how to make the argument.

 

[ChrisVer]

in general since you wanted to put a vector on x, you should have put a vector on F too.

the argument you are trying to use, if I understand your question correctly, is that you are supposing a transformation which only changes the x-axis [or in general you are free to choose the axis system so that everything happens along the x-axis]. That is a result of isotropy of space (which is contained in homogeneity), where you are free to choose whichever orientation and build a xyz system on it.

 

[Meir Achuz]

I think x'=x comes from requiring the inverse LT relating x and x' to give the same answer as the direct LT relating x' and x.

 

[nrqed]

Apriori -- before taking any of the postulates of special relativity into account -- we might say that the lorentz transformations between two frames K and K', where K' is moving w. speed v along the x-axis of K, is given by
$$\vec{x}' = F(\vec x, t)$$
and
$$t' = G(\vec x, t).$$
Now, i want to conclude that $y' = y$ and $z'=z$ as well as $x' = F(x,t)$. I believe this follows form homogeniety of space and time, but I do now know exactly how to make the argument.

Here is one thought experiment that shows this. Consider the y axis. We agree that by symmetry, if $y \neq y'$, there should be no way to tell which frame is moving. For example, if from the point of view of S one observes that $y < y'$ then we must have that from the point of view of S' $y' < y$ so that there is no way to break the symmetry between the two frames.

But it is then clear that this is not possible.
To see this,
let's say there is tree painted on a wall at rest in S and the tree has a height y. And let's say someone in Y' is holding a paint brush at a height y' and is leaving a horizontal painted line on the wall as S' is moving with respect to S. Then there are only three possibilities: the brush leaves a line on the wall that crosses the tree (in which case y' < y), the brush misses the tree by a certain distance (in which case y' > y ) or leaves a mark exactly at the same height as the tree (y=y').

The key point is that the conclusion tag is reached is agreed by everyone, in all frames (whether the painted line goes over the tree or not is not frame dependent!). And the only situation that is consistent with symmetry between the two frames is y=y'. It is clear that the same reasoning leads to z=z'.

 

[center o bass]

Here is one thought experiment that shows this. Consider the y axis. We agree that by symmetry, if $y \neq y'$, there should be no way to tell which frame is moving. For example, if from the point of view of S one observes that $y < y'$ then we must have that from the point of view of S' $y' < y$ so that there is no way to break the symmetry between the two frames.

But it is then clear that this is not possible.
To see this,
let's say there is tree painted on a wall at rest in S and the tree has a height y. And let's say someone in Y' is holding a paint brush at a height y' and is leaving a horizontal painted line on the wall as S' is moving with respect to S. Then there are only three possibilities: the brush leaves a line on the wall that crosses the tree (in which case y' < y), the brush misses the tree by a certain distance (in which case y' > y ) or leaves a mark exactly at the same height as the tree (y=y').

The key point is that the conclusion tag is reached is agreed by everyone, in all frames (whether the painted line goes over the tree or not is not frame dependent!). And the only situation that is consistent with symmetry between the two frames is y=y'. It is clear that the same reasoning leads to z=z'.

I agree with your reasoning, but can you invoke some principle that can convince me that if "the painted line goes over the tree or not is not frame dependent"?

I know that it would be quite absurd that this were the case, but before relativity was invented it was quite absurd for people to accept length contraction and time dilation as well. So do you have a simple, principled reason for that statement?

Anyway, were the two later paragraphs really necessary? Can't we just say: Either $y'=y$ or $y'\neq y$ and let's assume the latter. Then one possibility is that $y'>y$, but by the principle for relativity this must imply that $y > y'$ for otherwise one could tell which frame were moving by communicating ones y-coordinates. But clearly $y> y'$ and $y<y'$ are incompatible for any logically consistent theory as these are numbers. Thus, $y\neq y'$ is ruled out, and the only possibility left is $y'=y$.

Edit: I'm trying to find out why I can't conclude that $x'=x$ by the same argument. Any tips?

Edit-Edit: I've left out time. Clearly $y> y'$ and $y<y'$ are possible at different times. So I'm back to asking you for a fundamental principle that can rule out the absurd situation where the tree get painted or not depending on the observer.

 

[center o bass]

I think x'=x comes from requiring the inverse LT relating x and x' to give the same answer as the direct LT relating x' and x.

Could you outline a formal argument on how its done? Or could you refer me to one?

 

[nrqed]

I agree with your reasoning, but can you invoke some principle that can convince me that if "the painted line goes over the tree or not is not frame dependent"?

Well, where paint mark is above or below the tip of the tree cannot be a frame independent concept! We just have to take a picture of it! Or we can travel back to the wall, slow down and look at it :-)

 

[strangerep]

Apriori -- before taking any of the postulates of special relativity into account -- we might say that the lorentz transformations between two frames K and K', where K' is moving w. speed v along the x-axis of K, is given by
$$\vec{x}' = F(\vec x, t)$$
and
$$t' = G(\vec x, t).$$
Now, i want to conclude that $y' = y$ and $z'=z$ as well as $x' = F(x,t)$. I believe this follows form homogeniety of space and time, but I do now know exactly how to make the argument.

If you're "not taking any of the postulates of special relativity into account", I'm not sure you can derive anything much.

If you adopt the relativity principle, and spatial isotropy, (but not homogeneity) one can derive a generalized (fractional--linear) transformation which does change the transverse coordinates by a scale factor. If you invoke homogeneity, that forces you back to linear transformations of standard Lorentz form wherein the transverse coordinates are unaffected.

Rindler's textbook contains some of the derivations.