Help with Lorentz Transformation

Theseus
Messages
19
Reaction score
0
In Einstein's book Relativity he provides a derivation of the LT.
link here
http://www.bartleby.com/173/a1.html"

In step 3 he brings in constants λ and μ and now I am lost.

In the equation (x'-ct') = λ(x-ct) - isn't this the same as "zero = anything X zero"?

How did λ and μ get in there?
Do they represent anything in particular?
 
Last edited by a moderator:
Physics news on Phys.org
The form (3-4) is equivalent to the form (5), so any justification for one form is also an acceptable justification for the other. To me, (5) is easier to justify; it's the simplest relationship between the (x,t) and (x',t') coordinates that has the following two properties:
(i) The origins of the two coordinate systems coincide at t=t'=0.
(ii) It respects the homogeneity of space.
Property i is just a trivial matter of convenience. Property ii has straightforward physical significance. Try to think up any other form for the relationship between (x,t) and (x',t'). For example, suppose you had something of the form x'=a/(x-b). The reason this form isn't physically acceptable is that something special happens at x=b (x' blows up). This would tell us that space had some special, violent property at x=b. But this is special relativity, where space is supposed to be homogeneous, i.e., have the same properties everywhere, so that's not admissible.
 
Theseus said:
In Einstein's book Relativity he provides a derivation of the LT.
link here
http://www.bartleby.com/173/a1.html"

In step 3 he brings in constants λ and μ and now I am lost.

In the equation (x'-ct') = λ(x-ct) - isn't this the same as "zero = anything X zero"?
The confusing part may be that he wants a general transformation which can translate any event with coordinates x and t to corresponding coordinates x' and t', but he starts by considering the special case of a light beam emitted from the origin at t=0, so x=ct for any point on the light beam's path (this would obviously not be true for arbitrary events that don't lie on this path). He's pointing out that as long as the general transformation has the property (x'-ct') = λ(x-ct) (regardless of the value of x and t), that will guarantee that the light beam has the same speed of c in both coordinate systems , because if x=ct, that equation implies x'=ct' too. I don't know if it'd be possible to come up with a coordinate transformation where it was true that any (x,t) satisfying x=ct would also satisfy x'=ct', but it wasn't true that any arbitrary x,t would satisfy (x'-ct') = λ(x-ct). I guess you could come up with a coordinate transformation that did satisfy that equation but where λ was a function of x and t rather than being a constant, but then it wouldn't be a linear coordinate transformation...
 
Last edited by a moderator:
Theseus said:
In step 3 he brings in constants λ and μ and now I am lost.

In the equation (x'-ct') = λ(x-ct) - isn't this the same as "zero = anything X zero"?

How did λ and μ get in there?
Do they represent anything in particular?

I stumbled on this same section when I studied that book, many years ago.

He is first taking a light pulse moving toward the right (increasing x for increasing t) and which is at x1 = 0 when t1 = 0, and gets

x1 - c t1 = 0.

And for another inertial frame (with coordinates x2 and t2), that SAME light pulse must satisfy

x2 - c t2 = 0,

where we have arbitrarily imposed the requirement that the coordinates for that second frame be chosen so that the point x2 = t2 = 0 corresponds to the point x1 = t1 = 0.

But both frames must also describe ANY spacetime point, and in general, spacetime points don't satisfy the two equations above...i.e., for an arbitrary point, x and t DON'T satisfy x - c t = 0. That equation is satisfied only by a light pulse moving toward increasing x.

He requires that the two sets of coordinates be linearly related. So he writes

( x2 - c t2 ) = lambda ( x1 - c t1 ),

which is linear, and which meets the requirement that FOR THE GIVEN LIGHT PULSE, if

x1 - c t1 = 0,

then the transformation between the two sets of coordinates must give

x2 - c t2 = 0.

But that will be true for ANY constant lambda. So the transformation hasn't been fully determined yet.

THEN, he does the same thing for a light pulse going in the other direction (toward decreasing x) and which is at x = 0 when t = 0 (for both frames), and he introduces another constant mu (because the two constants aren't necessarily the same).

That allows him to get the linear transformation in terms of the two unknowns a and b (which are linear functions of the lambda and mu).

Finally, he is able to determine the constants a and b from the arguments on the next couple of pages.

Hope that helps.

Mike Fontenot
 
Last edited:
I wanted to let you all know that each one of your responses was very helpful. I made it all the way to (7) and I'm currently slogging through that.
Next time I get fully stuck I'll know who to ask.
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. The Relativator was sold by (as printed) Atomic Laboratories, Inc. 3086 Claremont Ave, Berkeley 5, California , which seems to be a division of Cenco Instruments (Central Scientific Company)... Source: https://www.physicsforums.com/insights/relativator-circular-slide-rule-simulated-with-desmos/ by @robphy
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...

Similar threads

Replies
54
Views
3K
Replies
11
Views
9K
Replies
19
Views
4K
Replies
16
Views
2K
Replies
5
Views
2K
Replies
5
Views
3K
Replies
12
Views
2K
Replies
20
Views
2K
Back
Top