From one space dimension to two space dimensions in special relativity

[bernhard.rothenstein]

Consider that you have derived the Lorentz transformations in one space diomensions
x=g(x'+Vt')
t=g(t'+Vx'/c^2).
In order to extend the problem to two space dimensions many Authors add
y=y'.
Please tell me if that is the best approach.
Thanks.

 

[A.T.]

Are you looking for the relativistic vector addition?
http://en.wikipedia.org/wiki/Velocity-addition_formula#Vector_notation

 

[bernhard.rothenstein]

LT in two space dimensions

Are you looking for the relativistic vector addition?
http://en.wikipedia.org/wiki/Velocity-addition_formula#Vector_notation

Thanks. An alert student could ask if the transformation equation for the time coordinate remain correct if we go from y=0 to y different from zero.

 

[HallsofIvy]

??y= 0 is a purely arbitrary choice. Changing to y= any other value cannot change the physical situation at all.

The fact that y'= y (not y= 0) comes from the fact that the Lorentz contraction applies only in the direction of motion. Here x is chosen to be in the direction of motion.

 

[bernhard.rothenstein]

two space dimensions SR

??y= 0 is a purely arbitrary choice. Changing to y= any other value cannot change the physical situation at all.

The fact that y'= y (not y= 0) comes from the fact that the Lorentz contraction applies only in the direction of motion. Here x is chosen to be in the direction of motion.

Thank you. Is there an explanation via clock synchronization i.e. is there a way to show that all the clocks C(x=const,y different) display the same time?

 

[yuiop]

Thank you. Is there an explanation via clock synchronization i.e. is there a way to show that all the clocks C(x=const,y different) display the same time?

An informal explanation:

Your reference frame consists of all points that are stationary with respect to you. If an objects is moving with velocity v with respect to you we usually choose as a convenience an x-axis that is parallel to the velocity vector of the object (x') and a y-axis that is orthoganal to x. The y-axis in the moving frame (y') is orthoganal to the x' axis of the moving frame and parallel to your y axis. Since y' is parallel to y all points with the same y' coordinate in the moving frame (x' = constant) are the same distance from your y-axis and have the same velocity relative to your reference frame, so there is no reason that clocks with same y' coordinates should display different times.

 

[bernhard.rothenstein]

lorentz transformation in two space dimensions

An informal explanation:

Your reference frame consists of all points that are stationary with respect to you. If an objects is moving with velocity v with respect to you we usually choose as a convenience an x-axis that is parallel to the velocity vector of the object (x') and a y-axis that is orthoganal to x. The y-axis in the moving frame (y') is orthoganal to the x' axis of the moving frame and parallel to your y axis. Since y' is parallel to y all points with the same y' coordinate in the moving frame (x' = constant) are the same distance from your y-axis and have the same velocity relative to your reference frame, so there is no reason that clocks with same y' coordinates should display different times.

Thanks. Please have a look at the followings


Consider the clocks 1(0,0) and 2(x=rcosi,y=rsini) both at rest in I the first located at its origin O the second somewhere in the XOY plane. The first ticks whereas the second is stopped and fixed to display t=r/c. When 1 reads t=0 a light signal is emitted from O along a direction that makes an angle i with the positive direction of the OX axis. Let c(x)=ccosi be its projection on the OX axis. Arriving at the location of clock 2 the light signal starts it both clocks reading t=r/c. Let 3(x,y=0) be a clock located on the OX axis. It is synchronized by the OX component of the same light signal reading t=x/c(x).
Because t=r/c=rcosi/ccosi=x/c(x) the result is that after synchronization clocks 1 and 3 read the same running time and so
t'=g(t+Vx/c^2)
holds even for different values of y=y' as long as x is the same.
Please tell me is you find some flow there.

 

[yuiop]

Thanks. Please have a look at the followings


Consider the clocks 1(0,0) and 2(x=rcosi,y=rsini) both at rest in I the first located at its origin O the second somewhere in the XOY plane. The first ticks whereas the second is stopped and fixed to display t=r/c. When 1 reads t=0 a light signal is emitted from O along a direction that makes an angle i with the positive direction of the OX axis. Let c(x)=ccosi be its projection on the OX axis. Arriving at the location of clock 2 the light signal starts it both clocks reading t=r/c. Let 3(x,y=0) be a clock located on the OX axis. It is synchronized by the OX component of the same light signal reading t=x/c(x).
Because t=r/c=rcosi/ccosi=x/c(x) the result is that after synchronization clocks 1 and 3 read the same running time ...

That is one way of syncronising the clocks. Another way would be to preset clock 3 to x/c and have it started by the same omnidirectional signal that started clock 2 (which was preset to r/c). Certainly all 3 clocks would be showing the same running time in the rest frame, but so would any clocks (not having the same x coordinate) in the rest frame show the same running time. You have not shown why only the clocks with same x coordinate would appear to have the same running time to an observer moving relative to the clocks. For example clock 1 at the origin and clock 3 at some coordinate with x not equal to zero, would not show the same time in the moving reference frame.

and so t'=g(t+Vx/c^2) holds even for different values of y=y' as long as x is the same.
Please tell me is you find some flow there.

I think the best way to demonstrate that t' is independent of y is to derive the Lorentz transformation in one dimension and show that the same formulas are obtained when the Lorentz transformations are derived from a two dimensional situation such as the Michelson-Morley experiment.

It may also be worth noting that it is very easy to demonstrate that from a spatial point of view y must be equal to y'. Imagine two rings of the same radius moving on a common axis towards each other. If y did not equal y' then from the point of view of one moving observer, ring 1 may pass inside ring 2 while from the point of view of another observer with a different velocity ring 1 might pass outside ring 2. Clearly that cannot happen.

Hope that helps.

 

[country boy]

Consider that you have derived the Lorentz transformations in one space diomensions
x=g(x'+Vt')
t=g(t'+Vx'/c^2).
In order to extend the problem to two space dimensions many Authors add
y=y'.
Please tell me if that is the best approach.
Thanks.

Instead of x and x', it might be better to start with a fixed y-axis and a moving y'-axis (both axes are perpendicular to the direction of motion). Consider two meter sticks arranged parallel to y and y'. Let the moving meter stick (attached to y') travel past the stationary meter stick (attached to y) so that the ends are coincident as they pass. If the clocks along y and y' are synchronized in their respective frames, the moving and stationary observers must agree that the two coincident events (the passing at both ends) happened at the same time in each frame. Therefore, the moving and stationary meter sticks must be the same length.

This agreement now permits the moving and stationary observers to build light clocks along y and y' in the two frames that have the same mirror separation. With these clocks the transformation of time can be derived. Then, timing of events along the direction of motion can be used to derive the transformation between x and x'.