Another interpretation of Lorentz transformations

  • I
  • Thread starter ilasus
  • Start date
  • #1
32
3
I consider three material points O, O', M, in uniform rectilinear motion in a common direction, so that in relation to the point O, the points O' and M move in the same direction with the constant velocities v and u (u>v>0). Assuming that at the initial moment (t0=0), the points O, O', M were in the same initial place (x0=0), the distances traveled by the points M and O' in relation to O, denoted by x and x1, respectively, are given by the laws of motion

x = ut, x1 = vt

where time t is measured from the initial moment. It follows that in relation to point O, point M travels the distance x during t expressed by the relations

(1) x = ut, t = (1/u)x

On the other hand, taking into account that the point M travels the distance

x1 = (v/u)x

during

t1 = x1/u = (v/u2)x

it follows that between points O and O', point M travels the distance x1 during t1 expressed by the relations

(2) x1 = vt, t1 = (v/u2)x

and from (1) and (2), it results that in relation to the point O’, the point M moves on the distance x2 during t2 given by the relations

(3) x2 = x - vt, t2 = t - (v/u2)x

The hypothesis that the points O' and M are moving in the same direction with respect to O, which led to relations (1), (2), (3), is in fact the point of view of an observer located at the origin O of a referential S. An observer located at the origin O' of a referential S' will have another point of view, namely that the points O and M are moving in opposite directions in relation to the point O'. Specifically, assuming that at the initial moment (t'0=0), the points O and M were in the same initial place (x'0=0), respectively at the origin O' of the referential S', then in relation to the point O', the point M travels the distance x' during t' given by the relations

(1') x' = ut', t' = (1/u)x'

between points O and O', point M travels the distance x'1 during t'1 given by the relations

(2') x'1 = vt', t'1 = (v/u2)x'

and in relation to the point O, the point M travels the distance x'2 during t'2 expressed by the relations

(3') x'2 = x' + vt', t'2 = t' + (v/u2)x '

However, even if x’0 = x0 and t’0 = t0, the homologous distances and time intervals traveled by the point M in the referentials S, S ’, expressed in the cases presented above, cannot be equal. Specifically, the distances and time intervals expressed in the relations (1'), (3) and (1), (3'), ie the distances x', x2 and x, x'2, respectively the time intervals t', t2 and t, t'2, can be at most proportional. In other words, the factor k in the equations

(4) x' = k(x - vt), t' = k(t - (v/u2)x)

(4') x = k(x' + vt'), t = k(t' + (v/u2)x')

it cannot be unitary. For example, if we try to solve the system of Cramer equations (4) in the unknowns x, t, or the system of Cramer equations (4') in the unknowns x', t', we find that they have the solutions (4') and respectively (4), only if we assign the value to the factor k

(5) k = 1/(1 - v2/u2)1/2

Note that if u = c, so M is a light signal, then the relations (4) and (4') are identified with the Lorentz transformations, and k given by (5) is the Lorentz factor.

What do you think about this?
 
Last edited by a moderator:
  • Skeptical
Likes Motore

Answers and Replies

  • #2
Ibix
Science Advisor
Insights Author
2020 Award
8,307
7,711
What do you think about this?
I think you need a good diagram, probably a displacement-time graph on O's frame.

You may also like to look up one-postulate derivations of the Lorentz transforms. "Nothing but relativity" by Palash Pal is easy to find and follow, although the history goes back a long way.
 
  • Like
Likes bhobba and vanhees71
  • #3
32
3
See Fig.1
 

Attachments

  • Fig.1.PNG
    Fig.1.PNG
    14.3 KB · Views: 42
  • #4
Ibix
Science Advisor
Insights Author
2020 Award
8,307
7,711
Your relations 1, 2 and 3 seem to be correct. Your relations 1', 2' and 3' are not correct because they assume that the speed of ##M## as measured by ##O'## is ##u##, the same as that measured by ##O##.

I would repeat my recommendation of looking up "one postulate" derivations of the Lorentz transforms. Essentially you use symmetry and an assumption of linearity to restrict the possible transforms to a small class of linear transforms, and then further restrict them by asserting that the composition of two transforms must also be a transform. The end results includes both Galilean and Lorentz transforms, and we discard the Galilean transforms by observation of the real world.
 
  • Like
Likes bhobba and vanhees71
  • #5
anorlunda
Staff Emeritus
Insights Author
9,759
6,853
we discard the Galilean transforms by observation of the real world.
That is an important advantage that physics has over mathematics. In physics, we can ignore potential solutions that do not conform to observations of nature. Mathemeticians can't do that.
 
  • #6
32
3
Your relations 1, 2 and 3 seem to be correct. Your relations 1', 2' and 3' are not correct because they assume that the speed of ##M## as measured by ##O'## is ##u##, the same as that measured by ##O##.
Relationships 1,2,3 express the point of view of an observer located at the origin O of a reference S. According to the observer in S, in relation to point O point M travels the distance x with velocity u during t, between O and O' point M travels the distance x1 with velocity u during t1, and in relation to the point O' the point M travels the distance x2 with velocity u during t2.

Relationships 1', 2', 3' express the point of view of an observer located at the origin O' of a reference S'. In this case we used other notations: x', t' instead of x2, t2, x'1, t'1 instead of x1, t1 and x'2, t'2 instead of x, t. According to the observer of S', in relation to point O' point M travels the distance x' with speed u during t', between O and O' point M travels the distance x'1 with speed u during t'1, and in relation to point O point M travels the distance x'2 with speed u during t'2.
 
Last edited:
  • #7
32
3
That is an important advantage that physics has over mathematics. In physics, we can ignore potential solutions that do not conform to observations of nature. Mathemeticians can't do that.
In other words, if we do not know the physical interpretation of relationships 1,2,3 and 1', 2', 3 ', then we can ignore these relationships and die happy.
 
  • #8
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
18,170
9,843
Relationships 1', 2', 3' express the point of view of an observer located at the origin O' of a reference S'. In this case we used other notations: x', t' instead of x2, t2, x'1, t'1 instead of x1, t1 and x'2, t'2 instead of x, t. According to the observer of S', in relation to point O' point M travels the distance x' with speed u during t', between O and O' point M travels the distance x'1 with speed u during t'1, and in relation to point A point M travels the distance x'2 with speed u during t'2.
In the ##O'## frame, you need to use ##u'## as the speed of ##M## in that frame, where ##u' \ne u##; and, in fact, ##u' \ne u-v##.
 
  • Like
Likes bhobba
  • #9
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
18,170
9,843
Note that if u = c, so M is a light signal, then the relations (4) and (4') are identified with the Lorentz transformations, and k given by (5) is the Lorentz factor.

What do you think about this?
When ##M## is a light signal, we have ##u = c## and ##u' = u = c##. That's the only case where ##u' = u## and your analysis holds up.
 
  • #10
32
3
In the ##O'## frame, you need to use ##u'## as the speed of ##M## in that frame, where ##u' \ne u##; and, in fact, ##u' \ne u-v##.
It is enough that I use another space (x') and another time (t'), it is not necessary to use another speed (u'), if I refer to another referential (S'). This is because I can go at the same speed u (for example u=5km/h) and on a road, and in a plane and in any other S’ reference. Try it and you will see that you can too.
 
  • #11
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
18,170
9,843
It is enough that I use another space (x') and another time (t'), it is not necessary to use another speed (u'), if I refer to another referential (S'). This is because I can go at the same speed u (for example u=5km/h) and on a road, and in a plane and in any other S’ reference. Try it and you will see that you can too.
That's simply wrong. The only speed that is frame invariant is ##c##.
 
  • #12
32
3
But I'm not referring to the t'=t case you're referring to ...
 
  • #13
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
18,170
9,843
But I'm not referring to the t'=t case you're referring to ...
I think you're seriously confused here.
 
  • #14
Ibix
Science Advisor
Insights Author
2020 Award
8,307
7,711
It is enough that I use another space (x') and another time (t'), it is not necessary to use another speed (u'), if I refer to another referential (S'). This is because I can go at the same speed u (for example u=5km/h) and on a road, and in a plane and in any other S’ reference. Try it and you will see that you can too.
Consider a spider capable of covering 1 mile every hour. It is walking in a car that is doing 60mph. So in your scenario O is a person standing on the street corner, O' is the car and M the spider. You say that from the perspective of O the spider is doing speed u, approxinately 61mph in this case. You also say that from the perspective of O', the spider is doing speed u - but the spider cannot be doing 61mph from the perspective of the car. Certainly not at the same time as doing 61mph as seen by the person on the street.
 
  • #15
32
3
The relations 1,2,3 and 1',2',3' describe an experiment (displacement of the points O, O', M in relation to each other) from the point of view of an observer located in the origin O of a referential S and respectively from the point of view of an observer located at the origin O' of a frame S', but the observer in S does not see the experiment in S ', and the observer in S' does not see the experiment in S. So it is not an observer standing at the corner of the street and see what's going on in the other observer's yard.
 
Last edited:
  • #16
32
3
I think you're seriously confused here.
You can say anything about me, but this topic is about relationships 1,2,3 and 1',2',3', not about me.
 
  • Sad
Likes weirdoguy
  • #17
Ibix
Science Advisor
Insights Author
2020 Award
8,307
7,711
I'm sorry but you don't seem to be making any sense.

In which frame is ##M## travelling at speed ##u##? ##S## or ##S'##? Who is at rest in that frame?

What is ##M##'s speed in the other frame? Who is at rest in that frame?
 
  • #18
32
3
In which frame is ##M## travelling at speed ##u##? ##S## or ##S'##? Who is at rest in that frame?
What is ##M##'s speed in the other frame? Who is at rest in that frame?
I think that in order to clarify these questions you will have to read answer #6 carefully.
 
  • Sad
Likes weirdoguy
  • #19
Ibix
Science Advisor
Insights Author
2020 Award
8,307
7,711
I think that in order to clarify these questions you will have to read answer #6 carefully.
...which says that ##M## has speed ##u## in both frames ##S## and ##S'##.

If that is a correct statement of your claims then you imply an assumption that there is a finite invariant speed that you are denoting this ##u##. You don't seem to have stated this explicitly. Is this an assumption you are making?
 
  • #20
32
3
...which says that ##M## has speed ##u## in both frames ##S## and ##S'##.

If that is a correct statement of your claims then you imply an assumption that there is a finite invariant speed that you are denoting this ##u##. You don't seem to have stated this explicitly. Is this an assumption you are making?
The speed marked with u is not special, it is a normal speed, but u>v - for example v=3km/h, u=5km/h - that is why your example with the spider is wrong. Here's a simple example, if you will. On a rectilinear road S on which a landmark O is fixed, a platform S' on which a landmark O' is fixed and a mobile M moves with speeds v and u, respectively. Then, on the road S, the mobile M and the landmark O' move in the same direction according to relations (1), (2), (3) (this is the point of view of the observer on the road), and on the platform S', the mobile M and the landmark O it moves in opposite directions according to relations (1’), (2’), (3’) (this is the point of view of the observer on the platform). Obviously, if the displacement according to (1), (2), (3) of the mobile M and the landmark O' in S is real, then the displacement according to (1'), (2'), (3') of the mobile M and the landmark O in S' is virtual, and vice versa.
 
  • Skeptical
Likes weirdoguy and Motore
  • #21
Doc Al
Mentor
45,180
1,502
On a rectilinear road S on which a landmark O is fixed, a platform S' on which a landmark O' is fixed and a mobile M moves with speeds v and u, respectively.
Those speeds are with respect to frame S. What are the velocities of O and M according to frame S'?
 
  • #22
Ibix
Science Advisor
Insights Author
2020 Award
8,307
7,711
Then, on the road S, the mobile M and the landmark O' move in the same direction according to relations (1), (2), (3) (this is the point of view of the observer on the road),
Ok.
on the platform S', the mobile M and the landmark O it moves in opposite directions according to relations (1’), (2’), (3’) (this is the point of view of the observer on the platform).
But the velocity of M in this frame cannot be ##u## - unless you are assuming that M is moving at the same speed in both frames.
Obviously, if the displacement according to (1), (2), (3) of the mobile M and the landmark O' in S is real, then the displacement according to (1'), (2'), (3') of the mobile M and the landmark O in S' is virtual, and vice versa.
Please define 'real' and 'virtual' displacement as these are not standard terminology.
 
  • Like
Likes Doc Al
  • #23
vanhees71
Science Advisor
Insights Author
Gold Member
18,551
9,425
Your relations 1, 2 and 3 seem to be correct. Your relations 1', 2' and 3' are not correct because they assume that the speed of ##M## as measured by ##O'## is ##u##, the same as that measured by ##O##.

I would repeat my recommendation of looking up "one postulate" derivations of the Lorentz transforms. Essentially you use symmetry and an assumption of linearity to restrict the possible transforms to a small class of linear transforms, and then further restrict them by asserting that the composition of two transforms must also be a transform. The end results includes both Galilean and Lorentz transforms, and we discard the Galilean transforms by observation of the real world.
I like this one:

https://doi.org/10.1063/1.1665000

It only assumes the special principle of relativity, homogeneity of space-time, and isotropy of space for any inertial observer to show that you have on two spacetime models: Galilean or Minkowski spacetime. As anything in physics which one to prefer is a question of observations, and it's clear that Minkowski is the more comprehensive description.
 
  • Like
Likes Ibix
  • #24
32
3
What are the velocities of O and M according to frame S'?
The mobile M and the marker O move in opposite directions on the platform (in the S’ reference with the origin O’) with the velocities –v and u respectively. In this case, in relation to the landmark O', the mobile M travels the distance x' during time t' according to (1'), between the landmarks O and O', the mobile M travels the distance x'1 during time t’1 according to (2’), and in relation to the landmark O, the mobile M travels the distance x'2 during t’2 according to (3').

Note. Regarding the role and necessity of relations (1), (2), (3) and (1'), (2'), (3'), there are no explanations in official physics. I also did not present a physical interpretation of these relations, but I only showed how one can go from relations (1), (2), (3) and (1'), (2'), (3') to relations (4) and (4').
 
  • Sad
Likes weirdoguy
  • #25
Ibix
Science Advisor
Insights Author
2020 Award
8,307
7,711
The mobile M and the marker O move in opposite directions on the platform (in the S’ reference with the origin O’) with the velocities –v and u respectively.
This cannot be correct unless you assume that ##u## is an invariant speed, since ##u## is the speed of ##M## in ##S##.
 
  • Like
Likes Doc Al

Related Threads on Another interpretation of Lorentz transformations

Replies
44
Views
5K
O
Replies
4
Views
803
Replies
2
Views
103
Replies
2
Views
2K
Replies
2
Views
1K
Replies
6
Views
2K
Replies
2
Views
1K
Top