Another interpretation of Lorentz transformations

[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?

 

[PeroK]

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?

In general, if we are looking at a set of particles with uniform velocities $v_1, v_2 \dots$, then the position of each particle is given by $x_1 = v_1t, x_2 = v_2t, \dots$. It's not necessary to introduce a different time parameter, $t_1, t_2, \dots$ for each particle. Doing so suggests that you are not really understanding the basic concept of unform motion in a given reference frame.

That's a separate issue from the invariant speed misconception.

 

[ilasus]

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?

You refer to the consequence of the transformation of Galilee, ie to the mechanical composition of the velocities from different referentials, if the time is the same in the respective referentials. In this case, it is obvious that a mobile is moving at different speeds in different references. But I did not intend to talk about this case in this topic.

 

[PeroK]

You refer to the consequence of the transformation of Galilee, ie to the mechanical composition of the velocities from different referentials, if the time is the same in the respective referentials. In this case, it is obvious that a mobile is moving at different speeds in different references. But I did not intend to talk about this case in this topic.

This is absurd. You can't make every speed an invariant just by tinkering with time parameters. As before, it only works for a single invariant speed, $c$.

 

[bhobba]

Ilasus is very confused, as many have pointed out. Here is the correct math that does not even assume a constant speed of light:
http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf

This thread is mercifully shut. Ilasus please study the deviation and start a new thread if you can fault it.

Thanks
Bill