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And at that time, what speed is M doing in S and what speed is it doing in S'?ilasus said:only the initial place and the initial moment can be equal, ie x0=x'0 and t0=t'0.
This discussion centers on the interpretation of Lorentz transformations through the analysis of two reference frames, S and S', involving points O, O', and M moving with constant velocities u and v. The relationships derived (1, 2, 3 for frame S and 1', 2', 3' for frame S') illustrate the discrepancies in observed distances and time intervals between the two frames. The factor k in the equations (4) and (4') is established as k = 1/(1 - v²/u²)^(1/2), which becomes the Lorentz factor when u equals the speed of light, c. The discussion emphasizes the necessity of understanding the relative motion of M as perceived from different frames and the implications for the validity of the derived equations.
PREREQUISITESStudents of physics, particularly those studying special relativity, theoretical physicists, and anyone interested in the mathematical foundations of motion in different reference frames.
And at that time, what speed is M doing in S and what speed is it doing in S'?ilasus said:only the initial place and the initial moment can be equal, ie x0=x'0 and t0=t'0.
Because you've assumed ##u = c##, the invariant speed.ilasus said:My idea is that the relations (1), (2), (3) and (1'), (2'), (3') exist. The question is why?
@Ibix, @PeroK, and others have been trying to get you to see the hidden assumption underlying your statement, without much luck.ilasus said: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.
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?PeroK said:Because you've assumed ##u = c##, the invariant speed.
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?Doc Al said:@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'?
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 said: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?
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.Ibix said:And at that time, what speed is M doing in S and what speed is it doing in S'?
Therefore, ##u = c##, the invariant speed.ilasus said: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 said: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.
And flies fly at the same speed u in any inertial reference system, but that doesn't mean u=c!PeroK said:Therefore, ##u = c##, the 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.Ibix said:Then it seems that you agree that you are assuming that ##u## is an invariant speed.
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##.ilasus said:And flies fly at the same speed u in any inertial reference system, but that doesn't mean u=c!
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?ilasus said:For example, to prove to anyone that I can travel at the same speed in any inertial reference system, I use a speedometer.
ilasus said:For example, to prove to anyone that I can travel at the same speed in any inertial reference system, I use a speedometer
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 said: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.
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 said:What does the speedometer have to do with the movement of the Earth around the Sun?
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 said: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.
I don't think it is, but you said it was:ilasus said: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'm just exploring the implications of that.ilasus said:For example, to prove to anyone that I can travel at the same speed in any inertial reference system, I use a speedometer.
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 said:I don't think it is, but you said it was:
I'm just exploring the implications of that.
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?ilasus said:Let's change the subject.
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.ilasus said: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?
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.Ibix said: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?
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##.ilasus said: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.