L2 transformation reduces to the L1 transformation

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The discussion centers on the relationship between L2 and L1 transformations in the context of special relativity, specifically when two reference frames are in standard configuration. The L2 transformations are defined as r' = r + γv^[(1 - 1/γ)(r.v^) - βct] and ct' = γ(ct - r.β), while the L1 transformations are given by x' = γ(x - βct), y' = y, z' = z, and ct' = γ(ct - βx). The term "standard configuration" refers to both reference frames moving along the x-axis, which is crucial for the transformation equations to hold true. The user confirms the correct representation of the position vector as r = xi + yj + zk, emphasizing the importance of coordinate systems in these transformations.

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Students and professionals in physics, particularly those studying special relativity, as well as educators seeking to clarify the concepts of L1 and L2 transformations and their applications in different reference frames.

Pyrokenesis
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Hello.

I am having trouble answering the following question:

"Show that the L2 transformation reduces to the L1 transformation when the two reference frames are in standard configuration."

Am I wrong to assume that r = xi + yj + zk

Any help would be beautiful!

Thanx much
 
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"Show that the L2 transformation reduces to the L1 transformation when the two reference frames are in standard configuration."

I assume that the source of this question defines what the L2 and L1 transformations are, as well as what two reference frames are being compared. Please elaborate.
 
The L2 transformations are as follows:

r' = r + γv^[(1 - 1/γ)(r.v^) - βct];

ct' = γ(ct - r.β);

where β = v/c & v^ is the unit vector in the direction of v.

The L1 transformations are:

x' = γ(x - βct);
y' = y;
z' = z;

ct' = γ(ct - βx);

where β = v/c.

All are viewed in the S' frame.
 
We should also require that you define "standard configuration" but I'm going to assume that is with both reference frames moving in the direction of the x-axis :smile:.

Yes, you are correct to right r= xi+ yj+ zk. Notice that the difference is that "xi+ yj+ zk" assumes some particular coordinate system ("standard configuration") while "r" does not.

You may also assume ("standard configuration") that v= vi+ 0j+ 0k and that v^= i+ 0j+ 0k.
 
Sorry. Yes standard configuration is when both reference frames move in the direction of the x-axis.

Thanks I think I can solve it now.
 

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