Jan Slowak Posted June 15, 2019 Posted June 15, 2019 I really want to understand how Lorentz Transformations work! I refer to the picture Fig. 4-01: There we have an inertial reference system, S, that I depict with the axes x, y, x. In the future, we will only process the points on the x-axis, so in the next pictures I will draw only the x-axis. We regard this reference system as stationary, ie the distance between the reference system origin and the point where the event occurs is the same all the time.Then I would like help with defining what the coordinates x, t mean for S.
Ghideon Posted June 15, 2019 Posted June 15, 2019 (edited) 21 minutes ago, Jan Slowak said: I really want to understand how Lorentz Transformations work! That sounds as a good approach to the topic of special relativity! If you have one (stationary) frame of reference and x and t are defined as coordinates in the S frame of reference then: t is the time a stationary clock would display in the S frame of reference x is the distance that would be measured by some measurement device stationary in the S frame of reference. An alternative way to formulate the above answer could be: Observers, stationary in the S frame of reference, will agree about the position x and the time t for the event E. Edited June 15, 2019 by Ghideon added last sentence
Strange Posted June 16, 2019 Posted June 16, 2019 14 hours ago, Jan Slowak said: We regard this reference system as stationary, ie the distance between the reference system origin and the point where the event occurs is the same all the time. This is always true within a single frame of reference (whether you consider it stationary or not). That is almost the definition of a frame of reference. 14 hours ago, Jan Slowak said: Then I would like help with defining what the coordinates x, t mean for S. I’m not sure what you mean by “mean” The coordinate system of the frame of reference is x,y,z,t (you haven’t drawn the t axis). These are fixed and independent axes which can be used to define the location of an event in space time. We can transform between these and the coordinates of another frame S’ (x’,y’,z’,t’). If you draw the x and t axes, you can see that this transformation is actually a rotation. 15 hours ago, Jan Slowak said: I really want to understand how Lorentz Transformations work! It is just a function that maps from one coordinate system to another. They predate SR. SR explains how they work based on the invariant speed of light.
Jan Slowak Posted June 16, 2019 Author Posted June 16, 2019 5 hours ago, Strange said: This is always true within a single frame of reference (whether you consider it stationary or not). That is almost the definition of a frame of reference. I’m not sure what you mean by “mean” The coordinate system of the frame of reference is x,y,z,t (you haven’t drawn the t axis). These are fixed and independent axes which can be used to define the location of an event in space time. We can transform between these and the coordinates of another frame S’ (x’,y’,z’,t’). If you draw the x and t axes, you can see that this transformation is actually a rotation. It is just a function that maps from one coordinate system to another. They predate SR. SR explains how they work based on the invariant speed of light. Maybe I was unclear. We look at Fig. 4-01. I am sure that x from E = (x, t) means the distance between the S-origo and the point on the x-axis where the event occurred, the point x. But what does t from E = (x, t) represent? What time do you mean here? We must also ask ourselves the question: How does S know that the event E has occurred?
Strange Posted June 16, 2019 Posted June 16, 2019 1 hour ago, Jan Slowak said: But what does t from E = (x, t) represent? What time do you mean here? It is your notation, so I suppose only you can answer that. I assume t represents the time the event occurred (ie the position on the t axis corresponding to the time of the event in the same way that x,y,z represent the spatial location). In other words, the coordinates of the event are x,y,z,t. 1 hour ago, Jan Slowak said: We must also ask ourselves the question: How does S know that the event E has occurred? S is a frame of reference and doesn’t know anything. We know it has occurred and assign the x,y,z,t coordinates to it.
Bufofrog Posted June 16, 2019 Posted June 16, 2019 6 hours ago, Jan Slowak said: Maybe I was unclear. We look at Fig. 4-01. I am sure that x from E = (x, t) means the distance between the S-origo and the point on the x-axis where the event occurred, the point x. But what does t from E = (x, t) represent? What time do you mean here? We must also ask ourselves the question: How does S know that the event E has occurred? An event occurs at some distance x. Let's assume it is a flash of light. The observer at the origin of S would detect the flash at an elapsed time of t = x/c.
Jan Slowak Posted June 17, 2019 Author Posted June 17, 2019 20 hours ago, Bufofrog said: An event occurs at some distance x. Let's assume it is a flash of light. The observer at the origin of S would detect the flash at an elapsed time of t = x/c. I thought the same way as Bufofrog. We look at the picture Fig. 4-03. If we believe that t = 0 when the event occurred at point x, then t = x/c, time when S gets to know that the event occurred. Then E = (x, x/c). I would also like to point out that we cannot say much about the event unless you know the distance to it. Now we move on and also add the second reference system S' which moves to the right at constant speed v > 0. When the time is t = 0, t' = 0, both reference systems are located in the same point. Then, an event at distance x from the S-origo (S'-origo) occurs. See picture Fig. 4-04. What is special about S' is that while the light signal fails the distance x', S' moves the distance vt'. Am I right here?
swansont Posted June 17, 2019 Posted June 17, 2019 19 minutes ago, Jan Slowak said: I thought the same way as Bufofrog. We look at the picture Fig. 4-03. If we believe that t = 0 when the event occurred at point x, then t = x/c, time when S gets to know that the event occurred. Then E = (x, x/c). No. Einstein defined a clock synchronization technique, so that everyone in a frame can agree that an event that happened at some value of t happened at that time. It is the same time everywhere in a given frame. So whatever time it happens for someone co-located and at rest with respect to the event, that's the time for everyone in that frame You don't get to redefine this. And misapplying it will lead to errors. So if an observer at the origin learns of the event at t = x/c (from light travel) they know the event happened at t=0
Bufofrog Posted June 17, 2019 Posted June 17, 2019 26 minutes ago, Jan Slowak said: Now we move on and also add the second reference system S' which moves to the right at constant speed v > 0. When the time is t = 0, t' = 0, both reference systems are located in the same point. Then, an event at distance x from the S-origo (S'-origo) occurs. See picture Fig. 4-04. I think the vt' should just be vt. No need to add an E, that complicated things, just use x and x'. So x = x' + vt.
swansont Posted June 17, 2019 Posted June 17, 2019 22 minutes ago, Bufofrog said: just use x and x'. So x = x' + vt. In a Galilean transformation...
Jan Slowak Posted June 17, 2019 Author Posted June 17, 2019 22 minutes ago, Bufofrog said: I think the vt' should just be vt. No need to add an E, that complicated things, just use x and x'. So x = x' + vt. This is not true! In the figure Fig. 4-03, we have processed and calculated the coordinates of the event in S. In the picture Fig. 4-04, we have processed and calculated the coordinates of the event in S'.I draw a new picture only with the components for S', see Fig. 4-05. This applies: during the same time t' as the light signal reaches S', the S' moves a distance vt'. S' knows that it moves at speed v > 0. S' registers the light signal at time t'. Then S' knows that it is from S at a distance equal to vt'. Is not it like that?
Ghideon Posted June 17, 2019 Posted June 17, 2019 6 minutes ago, Jan Slowak said: Is not it like that? I think you need to be more rigorous. When you add a second frame of reference and also introduces travel time for signals you must tell in which frame of reference you are doing calculations and if you move between frames of reference you need to use Lorentz transforms. May I suggest that you use the good advice from @Strange and draw diagrams including time t? (I'll try to post a diagram later today.)
Ghideon Posted June 17, 2019 Posted June 17, 2019 (edited) Here is an example of a diagram with coordinates x and t* for one frame of reference S. Diagram 1: One observer O1, stationary in origo, detects an event, in this case a light pulse. Since x=0 and t=0 the event is in point (0,0). We also have a second observer O2 located at some other point in S. Since O2 is in the same frame of reference and stationary there is no relativistic effects due to any relative movement. O2 will agree with O1 that the event takes place at (0,0). Due to the non-infinite speed of signal travel O2 will be able to register the event at a later time than t=0 but that has no effect on the fact that the event happened at t=0 according to O1 and O2 (or according to any other, stationary, observer in S). Of course both x and t must be known to be able to tell at what point in space and time the event happened. If O2 or O1 are unable to measure x or t to agree about the point x,t that is an engineering issue, not a relativistic issue. Here is a second example. This is the same situations as Diagram 1 but at some later time when the light signal reaches O2. O1 and O2 are still at the same place in space but at a later point in time. Since vertical axis is “ct” the light pulse draws a 45 degree line. Note again: No relative movement and no relativistic effects. 5 minutes ago, FreeWill said: Does this mean that there is a common moment of Now (the present) i.e time is ticking everywhere where space is. I think you could say there is a common moment of now for stationary observers in one frame of reference. As soon as there is relative movement there is no universal "now". *) vertical axis is "ct" to allow for easier drawing of light pulses; they will make a 45 degree slope Edited June 17, 2019 by Ghideon some minor format issues 1
Jan Slowak Posted June 19, 2019 Author Posted June 19, 2019 Is there no one who wants to go through this thread and either confirm or argue against Fig. 4-05?
Ghideon Posted June 19, 2019 Posted June 19, 2019 39 minutes ago, Jan Slowak said: Is there no one who wants to go through this thread and either confirm or argue against Fig. 4-05? Lack of feedback & responses on answers already given may cause members to spend their efforts elsewhere? (bold by me) On 6/17/2019 at 6:45 PM, Jan Slowak said: This applies: during the same time t' as the light signal reaches S', the S' moves a distance vt'. S' knows that it moves at speed v > 0. S' registers the light signal at time t'. Then S' knows that it is from S at a distance equal to vt'. Is not it like that? 1: Stating that the S' moves with velocity v looks as if the S' frame is not considered stationary in the calculations. Then it is not clear if the calculations are done in S' with S' stationary or in S with S' moving with a relative velocity v. Having posted in other threads regarding SR I must stress that details regarding frame of reference is important, misunderstandings leads to issues and incorrect responses. 2: When talking about times t and t' in frames of reference S and S' in SR that often means the actual time of and event, as registered by clocks in each of the frame of reference. You can include the time the signal takes to reach some point in the frame of reference but some clarification might be good.
PervPhysProf Posted June 19, 2019 Posted June 19, 2019 (edited) You have to account for length contraction by finding variable(sub(2-n)) and how their coordinates evolve based on t=n that's 4d calc Since t=n+1 always in 4d (as opposed to negative time in higher dimensional analyses) time dilates because length gets into a lower +/-(x)/n for the (x) value each time .. At least until it gets too small. Issue is standard model does that for larger volumes like cubes in flat space when Darron Arronfsky's Pi: Faith Chaos Novelization clearly shows nature as a sphere. Cone centered, the concentric curve for the in-between variables Edited June 19, 2019 by PervPhysProf -1
Ghideon Posted June 19, 2019 Posted June 19, 2019 2 hours ago, PervPhysProf said: You have to account for length contraction by finding variable(sub(2-n)) and how their coordinates evolve based on t=n that's 4d calc Since t=n+1 always in 4d (as opposed to negative time in higher dimensional analyses) time dilates because length gets into a lower +/-(x)/n for the (x) value each time .. At least until it gets too small. Issue is standard model does that for larger volumes like cubes in flat space when Darron Arronfsky's Pi: Faith Chaos Novelization clearly shows nature as a sphere. Cone centered, the concentric curve for the in-between variables Can you clarify and add a source? It is not obvious how this relates to the question in OP; "want to understand how Lorentz Transformations work". I do not posses the required knowledge regarding "negative time in higher dimensional analysis" and google dd not help within the limited amount of time I had available.
PervPhysProf Posted June 19, 2019 Posted June 19, 2019 9 minutes ago, Ghideon said: Can you clarify and add a source? It is not obvious how this relates to the question in OP; "want to understand how Lorentz Transformations work". I do not posses the required knowledge regarding "negative time in higher dimensional analysis" and google dd not help within the limited amount of time I had available. That's the real trick isn't it. Having all the hacks -2
swansont Posted June 19, 2019 Posted June 19, 2019 Who is Darron Arronfsky and what all does this have to do with special relativity/Lorentz transforms?
Jan Slowak Posted June 20, 2019 Author Posted June 20, 2019 On 6/17/2019 at 6:55 PM, Ghideon said: I think you need to be more rigorous. When you add a second frame of reference and also introduces travel time for signals you must tell in which frame of reference you are doing calculations and if you move between frames of reference you need to use Lorentz transforms. May I suggest that you use the good advice from @Strange and draw diagrams including time t? (I'll try to post a diagram later today.) I analyze and talk about the derivation of LT. We don't have them yet! You cannot use them as counter-arguments. On 6/17/2019 at 10:17 PM, Ghideon said: Here is an example of a diagram with coordinates x and t* for one frame of reference S. Diagram 1: One observer O1, stationary in origo, detects an event, in this case a light pulse. Since x=0 and t=0 the event is in point (0,0). We also have a second observer O2 located at some other point in S. Since O2 is in the same frame of reference and stationary there is no relativistic effects due to any relative movement. O2 will agree with O1 that the event takes place at (0,0). Due to the non-infinite speed of signal travel O2 will be able to register the event at a later time than t=0 but that has no effect on the fact that the event happened at t=0 according to O1 and O2 (or according to any other, stationary, observer in S). Of course both x and t must be known to be able to tell at what point in space and time the event happened. If O2 or O1 are unable to measure x or t to agree about the point x,t that is an engineering issue, not a relativistic issue. Here is a second example. This is the same situations as Diagram 1 but at some later time when the light signal reaches O2. O1 and O2 are still at the same place in space but at a later point in time. Since vertical axis is “ct” the light pulse draws a 45 degree line. Note again: No relative movement and no relativistic effects. I think you could say there is a common moment of now for stationary observers in one frame of reference. As soon as there is relative movement there is no universal "now". *) vertical axis is "ct" to allow for easier drawing of light pulses; they will make a 45 degree slope Please, do not send other pictures. We talk about my picture Fig. 4-05.
Ghideon Posted June 20, 2019 Posted June 20, 2019 43 minutes ago, Jan Slowak said: I analyze and talk about the derivation of LT. We don't have them yet! You cannot use them as counter-arguments. Ok. Then the answer is : On 6/17/2019 at 6:45 PM, Jan Slowak said: Is not it like that? no, it is not like that.
Jan Slowak Posted June 20, 2019 Author Posted June 20, 2019 13 minutes ago, Ghideon said: Ok. Then the answer is : no, it is not like that. This is no answer. If you answer NO, you must counter-argument.
swansont Posted June 20, 2019 Posted June 20, 2019 27 minutes ago, Jan Slowak said: This is no answer. If you answer NO, you must counter-argument. Oh, come on. You can't tell someone to not post something as a counterargument and then complain that they haven't posted counterarguments. You wouldn't want to give the impression that you aren't arguing in good faith.
Jan Slowak Posted June 20, 2019 Author Posted June 20, 2019 16 minutes ago, swansont said: Oh, come on. You can't tell someone to not post something as a counterargument and then complain that they haven't posted counterarguments. You wouldn't want to give the impression that you aren't arguing in good faith. When I describe a physical phenomenon, make a mathematical model of it, presents my ideas, arguments, then I do not want to deviate from the main track. It's not possible! I present something, ask if I am right or wrong, then I want either an approval, Yes, or a No, but then you have to counter-argument. It should not be more difficult than that! -1
swansont Posted June 20, 2019 Posted June 20, 2019 22 minutes ago, Jan Slowak said: When I describe a physical phenomenon, make a mathematical model of it, presents my ideas, arguments, then I do not want to deviate from the main track. It's not possible! I present something, ask if I am right or wrong, then I want either an approval, Yes, or a No, but then you have to counter-argument. It should not be more difficult than that! Ghideon gave you some feedback about shortcomings in your drawings, and you ignored it. Then you scolded him for trying to give better drawings. Others have given feedback as well, and much of it has been ignored. I'm leaning more and more toward "not arguing in good faith" here. You can show that this is not the case by actually addressing the comments that have been made.
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