# Topic 4: Special Relativity - Lorentz transformations

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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.

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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 by Ghideon
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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.

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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?

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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.

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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.

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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?

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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

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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.

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22 minutes ago, Bufofrog said:

just use x and x'.  So x = x' + vt.

In a Galilean transformation...

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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?

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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.)

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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 by Ghideon
some minor format issues
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Is there no one who wants to go through this thread and either confirm or argue against Fig. 4-05?

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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.

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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 by PervPhysProf
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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.

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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

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Who is Darron﻿ Arr﻿onfs﻿k﻿y﻿ and what all does this have to do with special relativity/Lorentz transforms?

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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.

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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.

On 6/17/2019 at 6:45 PM, Jan Slowak said:

Is not it﻿ like tha﻿t?﻿

no, it is not like that.

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13 minutes ago, Ghideon said:

Ok.

no, it is not like that.﻿﻿

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27 minutes ago, Jan Slowak 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.

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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!

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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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