# A problem concerning the Lorentz transformations

#### bernhard.rothenstein

I state my problem in the following way.
Consider the Lorentz transformations
x=g(x’+Vt’) (1)
t=g(t’+Vx’/c2). (2)
They relate the space-time coordinates of the same events E(x,t) and E’(x’,t’) i.e. the space coordinates of the points M(x,0) and M’(x’,0) where the events take place and the readings of the clocks C(x,0) and C’(x’,0) located at that points when they read t and t’ respectively. In each of the involved inertial reference frames I and I’, the corresponding clocks are synchronized a la Einstein. We mention the clocks C0(0,0) and C’0(0,0) of the two reference frames located at theirs origins display the same running time as C(x,0) and C’(x’0) do respectively, as a result of the synchronization procedure. It is obvious that
t=x/c (3)
t’=x’/c. (4)
Multiplying both sides of (2) by c and taking into account (3) and (4) we obtain
x=g(x’+Vx’/c)=g(x’+Vt’) (5)
and we recover (1)!
Dividing (1) with c and taking into account (3) and (4) we obtain
t=g(t’+Vt’/c)=g(t’+Vx’/c2)
and we recover 2)! Is it correct to say that equation (1) is self contained in (2) and vice-versa?

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#### Galileo

Homework Helper
I can't say if I followed your explanation exactly. But given (1) and the fact that lightspeed is the same in both frames, then (2) follows directly.
Mathematically the proof is the same as yours, but the interpretation is different. (What does (5) say? Is it the time that is SEEN on a clock at position x? Since it takes a time t=x/c to reach the origin? not sure I understand.)

A correct interpretation would be: Suppose a light signal is shot in the positive x-direction at t=t'=0. Then the coordinates of the beam in I and I' should be:
x=ct
x'=ct'
Now use (1) to find the relation of t' as a function of x and t and out pops (2).

#### bernhard.rothenstein

I can't say if I followed your explanation exactly. But given (1) and the fact that lightspeed is the same in both frames, then (2) follows directly.
Mathematically the proof is the same as yours, but the interpretation is different.which kind of proof and interpretation do you mention? (What does (5) say? Is it the time that is SEEN on a clock at position x? Since it takes a time t=x/c to reach the origin?i think to reach the point where clock C(x,0) is located? not sure I understand.)

A correct interpretation would be: Suppose a light signal is shot in the positive x-direction at t=t'=0.that statement is equivalent with saying that the clocks in the two involved inertial reference frame are synchronized a la Einstein? Then the coordinates of the beam in I and I' should be:
x=ct
x'=ct'
that is a direct consequence of the synchronization in the invoved inertial reference frame a fact not always mentioned in the literature of the subject
Now use (1) to find the relation of t' as a function of x and t and out pops (2).[/QUOTE
Thank you for your contribution to my problem. My questions are inserted in bold in your message

#### nakurusil

I can't say if I followed your explanation exactly. But given (1) and the fact that lightspeed is the same in both frames, then (2) follows directly.
Mathematically the proof is the same as yours, but the interpretation is different. (What does (5) say? Is it the time that is SEEN on a clock at position x? Since it takes a time t=x/c to reach the origin? not sure I understand.)

A correct interpretation would be: Suppose a light signal is shot in the positive x-direction at t=t'=0. Then the coordinates of the beam in I and I' should be:
x=ct
x'=ct'
Now use (1) to find the relation of t' as a function of x and t and out pops (2).
Correct. Still , one cannot and should not draw the conclusion that (2) is the consequence of (1).
(1) and (2) are independent coordinate transformations by virtue of the way that they have been inferred originally.
Making
x=ct (3)
x'=ct' (4)
into (1) and reobtaining (2) only shows that (2) is consistent with (1) for the particular case  , something that we already knew!.
Bernhard, we discussed this type of issue, these exercises result into the irrelevant papers that on sees in Am.Jour.Phys. They tend to exhibit a type of hidden flaw in the form of circular thinking. (1) and (2) are independent transformations that show how the event (x,t) in S is viewed as (x',t') in S'
Creating a linkage of the form x=f(t) , will produce a linkage x'=f'(x') and this should not be misconstrued as having (1) and (2) dependent of each other. Because they are not.

#### lalbatros

I would say it rather in this way:

x' = c t' <==> x = c t

since, by hypothesis, a spherical wave in one frame remains a spherical wave in the other frame. This is the fundamental property of the Lorentz transformation: leaving spherical waves invariant in any (inertial) frame.

Therefore it should be no surprise if that pops up from the Lorentz transformation in one way or the other.

Michel

#### bernhard.rothenstein

Correct. Still , one cannot and should not draw the conclusion that (2) is the consequence of (1).
(1) and (2) are independent coordinate transformations by virtue of the way that they have been inferred originally.
Making
x=ct (3)
x'=ct' (4)
into (1) and reobtaining (2) only shows that (2) is consistent with (1) for the particular case  , something that we already knew!

Bernhard, we discussed this type of issue, these exercises result into the irrelevant papers that on sees in Am.Jour.Phys. They tend to exhibit a type of hidden flaw in the form of circular thinking. (1) and (2) are independent transformations that show how the event (x,t) in S is viewed as (x',t') in S'
Creating a linkage of the form x=f(t) , will produce a linkage x'=f'(x') and this should not be misconstrued as having (1) and (2) dependent of each other. Because they are not.

Thank you for your answer even if I do not understand all you say. It is not up to me to discuss the quality of the papers published by Am.J.Phys.
Do you know a better journal?

#### nakurusil

Thank you for your answer even if I do not understand all you say. It is not up to me to discuss the quality of the papers published by Am.J.Phys.
Do you know a better journal?
Yes, try Physical Reviews

#### bernhard.rothenstein

why the plural (Reviews?).