I Another interpretation of Lorentz transformations

[ilasus]

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?

 

[Ibix]

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.

 

[ilasus]

See Fig.1

 

[Ibix]

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.

 

[anorlunda]

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.

 

[ilasus]

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.

 

[ilasus]

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.

 

[PeroK]

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, $u' \ne u-v$.

 

[PeroK]

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.

 

[ilasus]

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.

 

[PeroK]

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 $c$.

 

[ilasus]

But I'm not referring to the t'=t case you're referring to ...

 

[PeroK]

But I'm not referring to the t'=t case you're referring to ...

I think you're seriously confused here.

 

[Ibix]

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.

 

[ilasus]

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.

 

[ilasus]

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.

 

[Ibix]

I'm sorry but you don't seem to be making any sense.

In which frame is $M$ traveling 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?

 

[ilasus]

In which frame is $M$ traveling 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.

 

[Ibix]

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?

 

[ilasus]

...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.

 

[Doc Al]

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'?

 

[Ibix]

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.

 

[vanhees71]

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.

 

[ilasus]

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').

 

[Ibix]

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 $S$.

 

[ilasus]

This cannot be correct unless you assume that $u$ is an invariant speed, since $u$ is the speed of $M$ in $S$.

If mobile M travels at speed u on the road, then mobile M does not travel at speed u on the platform. That is correct. And this is especially true that moving the mobile with speed u on the road is effective, it happens in reality, ie it is 'real', and in this case moving the mobile with speed u on the platform is just an idea, a possibility, ie it is 'virtual'. In this case, the relations (1), (2), (3) are real, and the relations (1'), (2'), (3') are virtual. But if, starting from a certain moment, the referential in which the mobile M moves changes, ie the mobile M passes from the referential S associated with the road to the referential S' associated with the platform and therefore the mobile M continues its movement with speed u on the platform, then the displacement of the mobile M with speed u on the platform becomes real, and the movement of the mobile on the road becomes virtual. In this case, the relations (1'), (2'), (3') become real, and the relations (1), (2), (3) become virtual.

 

[Ibix]

That is a long-winded way of saying that only one of the primed and unprimed relationships can be true at a given time, unless we assert that $u$ is an invariant speed. I agree.

The problem is that you then try to draw conclusions from the assumption that both sets are true at once without making that assertion.

 

[ilasus]

That is a long-winded way of saying that only one of the primed and unprimed relationships can be true at a given time, unless we assert that $u$ is an invariant speed. I agree.

The problem is that you then try to draw conclusions from the assumption that both sets are true at once without making that assertion.

I don't understand what your idea is. The relations (1), (2), (3) and (1'), (2'), (3') are both true, but we cannot speak of something 'simultaneously', because we refer to different distances and to different time intervals - only the initial place and the initial moment can be equal, ie x₀=x'₀ and t₀=t'₀.

 

[PeroK]

I don't understand what your idea is.

And none of us understands your idea.

 

[ilasus]

And none of us understands your idea.

My idea is that the relations (1), (2), (3) and (1'), (2'), (3') exist. The question is why?

 

[Ibix]

only the initial place and the initial moment can be equal, ie x₀=x'₀ and t₀=t'₀.

And at that time, what speed is M doing in S and what speed is it doing in S'?

 

[PeroK]

My idea is that the relations (1), (2), (3) and (1'), (2'), (3') exist. The question is why?

Because you've assumed $u = c$, the invariant speed.

 

[Doc Al]

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.

@Ibix, @PeroK, and others have been trying to get you to see the hidden assumption underlying your statement, without much luck.

How about we forget special relativity (and the Lorentz transformations) for the moment and have you simply answer the question using Galilean relativity: What are the velocities of O and M according to frame S'?

 

[ilasus]

Because you've assumed $u = c$, the invariant speed.

That is, do you consider that formulas (1)-(3 ') have no explanations for the case v<u<c and have explanations only for the case v<u=c? That is, do you consider, for example, that the relations (2) x₁=vt, t₁=(v/c²)x have an obvious and clear physical interpretation?

 

[ilasus]

@Ibix, @PeroK, and others have been trying to get you to see the hidden assumption underlying your statement, without much luck.

How about we forget special relativity (and the Lorentz transformations) for the moment and have you simply answer the question using Galilean relativity: What are the velocities of O and M according to frame S'?

If I remember the Lorentz transformations, it does not mean that I am thinking of special relativity. Regarding the transformation of Galilei, it is included in relations (3), ie it is identified with the first equality in (3) - or with the first equality in (3'). So according to Galilei, t₂=t. On the other hand, Ibix considers that the relations (1), (2), (3) are correct, and this means that the equality t₂=t is wrong, because t₂=t- (v/u²)x according to (3). Under these conditions, before asking questions about speeds, we will have to think about the Galilean composition of speeds. We can start with the question: do you consider the relations (1), (2), (3) correct?

 

[PeroK]

That is, do you consider that formulas (1)-(3 ') have no explanations for the case v<u<c and have explanations only for the case v<u=c? That is, do you consider, for example, that the relations (2) x₁=vt, t₁=(v/c²)x have an obvious and clear physical interpretation?

I think relations (1), (2), (3) are wrong. Instead, you should have:$$x = ut, \ x_1 = vt, \ x_2 = (u-v)t$$You can then substitiute $t = \frac x u$ into (2) and (3) if you wish.

 

[ilasus]

And at that time, what speed is M doing in S and what speed is it doing in S'?

The motion of M is uniform rectilinear with constant velocity u both at S, at time t (including at time t₀), and at S', at time t' (including at time t₀). In other words, M ‘does not know’ to move in an inertial reference system with a speed other than u.

 

[PeroK]

The motion of M is uniform rectilinear with constant velocity u both at S, at time t (including at time t₀), and at S', at time t' (including at time t₀). In other words, M ‘does not know’ to move in an inertial reference system with a speed other than u.

Therefore, $u = c$, the invariant speed.

 

[Ibix]

The motion of M is uniform rectilinear with constant velocity u both at S, at time t (including at time t₀), and at S', at time t' (including at time t₀). In other words, M ‘does not know’ to move in an inertial reference system with a speed other than u.

Then it seems that you agree that you are assuming that $u$ is an invariant speed.

 

[ilasus]

Therefore, $u = c$, the invariant speed.

And flies fly at the same speed u in any inertial reference system, but that doesn't mean u=c!

 

[ilasus]

Then it seems that you agree that you are assuming that $u$ is an invariant speed.

It is not about a certain speed u, but about any speed. The fact that ‘speed has the same value in any inertial reference system’ does not seem to be an assumption, but an experiment-based conclusion. For example, to prove to anyone that I can travel at the same speed in any inertial reference system, I use a speedometer.

 

[PeroK]

And flies fly at the same speed u in any inertial reference system, but that doesn't mean u=c!

Yes, it does. The speed of a particle is certainly not invariant across inertial reference frames. There is only one invariant speed, denoted by $c$.

It's taken 40 posts, but we have got to the root of the problem. You cannot assume speeds are invariant. That's why your analysis is only valid (if at all) where $M$ is a light pulse, with invariant speed $c$.

 

[Ibix]

For example, to prove to anyone that I can travel at the same speed in any inertial reference system, I use a speedometer.

The Earth is doing nearly 30km/s in an inertial system relative to an observer hovering stationary with respect to the Sun. If you are only doing 30km/hour (because that's what your speedometer says), how do you keep up?

In other words, you seem to have some fundamental misunderstandings about how speed is defined in different reference frames.

 

[weirdoguy]

For example, to prove to anyone that I can travel at the same speed in any inertial reference system, I use a speedometer

You are traveling at the moment with nearly light speed in reference frame of particle accelerated in LHC. Does your speedometer show that?

You clearly do not understand what a reference frame is and how speed/velocity is defined. And it seems that you are not here to gain that knowledge...

 

[ilasus]

The Earth is doing nearly 30km/s in an inertial system relative to an observer hovering stationary with respect to the Sun. If you are only doing 30km/hour (because that's what your speedometer says), how do you keep up?

In other words, you seem to have some fundamental misunderstandings about how speed is defined in different reference frames.

What does the speedometer have to do with the movement of the Earth around the Sun? Have you seen a speedometer that shows the speed of the Earth relative to the Sun? We seem to be referring to different things. I propose to return to Earth, that is, to relations (1), (2), (3). Do you still think that these relationships are correct?

 

[Ibix]

What does the speedometer have to do with the movement of the Earth around the Sun?

You said that you would use your speedometer to show that you were doing 30km/hour in all reference frames. I picked a reference frame and applied your reasoning: by your argument you are doing 30km/hour with respect to the Sun because your speedometer says so. If the result is bizarre, that should tell you something about your reasoning.

 

[ilasus]

You said that you would use your speedometer to show that you were doing 30km/hour in all reference frames. I picked a reference frame and applied your reasoning: by your argument you are doing 30km/hour with respect to the Sun because your speedometer says so. If the result is bizarre, that should tell you something about your reasoning.

But how did you find out (if it's not a secret of yours) that the speedometer is a measuring instrument that can be used to indicate the speed relative to the frame of reference you imagine - relative to the Sun, for example?

 

[Ibix]

But how did you find out (if it's not a secret of yours) that the speedometer is a measuring instrument that can be used to indicate the speed relative to the frame of reference you imagine - relative to the Sun, for example?

I don't think it is, but you said it was:

For example, to prove to anyone that I can travel at the same speed in any inertial reference system, I use a speedometer.

I'm just exploring the implications of that.

 

[ilasus]

I don't think it is, but you said it was:

I'm just exploring the implications of that.

Let's change the subject. PeroK decided to consider the relations (1), (2), (3) wrong, but does not say what mistakes he sees. Do you still consider relationships (1), (2), (3) correct?

 

[Ibix]

Let's change the subject.

Let's not, because it's key to where you are going wrong. Do you believe us now when we say that, in general, an object doing speed $u$ in one frame is not doing $u$ in another?