Lorentz Transformation Equations, time transformation

[Grimble]

t' = γ(t - vx/c²)
where t is the time in the stationary frame
t' is the time in the moving frame
v is the relative velocity between the two frames
and x is the distance traveled in the time t in the stationary frame.
∴ x = vt
substituting this into the Lorentz time equation gives us:
t' = γt(1 - v²/c²) = t/γ

So why is the formula not written this, t' = t/γ, much simpler way?

 

[Orodruin]

t' = γ(t - vx/c²)
where t is the time in the stationary frame
t' is the time in the moving frame
v is the relative velocity between the two frames
and x is the distance traveled in the time t in the stationary frame.
∴ x = vt
substituting this into the Lorentz time equation gives us:
t' = γt(1 - v²/c²) = t/γ

So why is the formula not written this, t' = t/γ, much simpler way?

Because it is only true when x = vt, i.e., for the world line of an observer at rest in S'.

 

[Grimble]

So can you define what those terms refer to in the equation?

 

[Orodruin]

The x, x', t, and t' are simply coordinates that describe any event in space-time, not necessarily an event along a given world line. The Lorentz transformations tell you how coordinates in one frame relate to coordinates in another frame, much like x' = x cosθ - y sinθ, y' = x sinθ + y cosθ would tell you how coordinates transform under a spatial rotation.

 

[Grimble]

Thank you, but that still leaves a question about the transformation equations; are the terms coordinates that are transformed (in which case there would surely be some factor that represnts the differences in the frames origins) or are they measurements between events?
I.e. is x really x1 - x2 and x', x'1 - x'2?

What is it that the LTE is transforming?
(I always thought it was how the relative velocity affected measurements made in one frame when converting them to be relative to another frame)

 

[Nugatory]

Thank you, but that still leaves a question about the transformation equations; are the terms coordinates that are transformed (in which case there would surely be some factor that represnts the differences in the frames origins) or are they measurements between events?
I.e. is x really x1 - x2 and x', x'1 - x'2?

What is it that the LTE is transforming?
(I always thought it was how the relative velocity affected measurements made in one frame when converting them to be relative to another frame)

The things being transformed are coordinates, the coordinates of a single point in spacetime. And there is indeed a term in them that represents the difference in the origins of the two coordinate systems: it's the $vx/c^2$ term that you were asking about when you started this thread (and the $vt$ in the corresponding transformation for the $x$ coordinate).

The length contraction and time dilation formulas tell you how to convert measurements between events, but these are both derived from the coordinate transforms by looking at the coordinates of two events. The length contraction formula comes from looking at the difference between the x coordinate of the event "one end of the object is here at time T" and the event "the other end of the object is there at the same time", for example.

It is a really good exercise to derive the time dilation and length contraction formulas directly from the Lorentz transforms. It's just simple algebra, but I doubt that it is possible to make sense of SR without going through this exercise at some point.

 

[Orodruin]

The things being transformed are coordinates, the coordinates of a single point.

Just to clarify what Nugatory means here. A "single point" (or "event") in space-time is a point in space and time. It is not enough to specify a location, you need to specify a time as well. Given a location and time for one event in one system, they tell you which time and location the event occurs at in the other system.

 

[Grimble]

The things being transformed are coordinates, the coordinates of a single point in spacetime. And there is indeed a term in them that represents the difference in the origins of the two coordinate systems: it's the vx/c^2 term that you were asking about when you started this thread (and the vt in the corresponding transformation for the xx coordinate).

So, if vt represents the spatial difference between the origins in the xx' transformation and x represents the same in the tt' transformation, then as x & vt represent the same value it still seems a reasonable substitution to simplify the formula as in my OP.

All I am doing is trying to reason from what you are telling me...

 

[harrylin]

t' = γ(t - vx/c²)
where t is the time in the stationary frame
t' is the time in the moving frame
v is the relative velocity between the two frames
and x is the distance traveled in the time t in the stationary frame.
∴ x = vt
substituting this into the Lorentz time equation gives us:
t' = γt(1 - v²/c²) = t/γ

So why is the formula not written this, t' = t/γ, much simpler way?

The much simpler way is like this, for small velocities (so that γ≈1): t' = t - vx/c²
That is called relativity of simultaneity and it dates from before SR.

It's very easy to work out if you think how someone who is in motion according to you will determine distant time by means of light or radio signals, if that person assumes to be in rest.

 

[PeroK]

So, if vt represents the spatial difference between the origins in the xx' transformation and x represents the same in the tt' transformation, then as x & vt represent the same value it still seems a reasonable substitution to simplify the formula as in my OP.

All I am doing is trying to reason from what you are telling me...

One of the good things about a formula is that you can plug some numbers in and see what you get! So, try the following. Take $v=\frac{3}{5}c$ and $\gamma = \frac{5}{4}$:

Find the coordinates in S' of:

1) The origin in S at times t = 0, 1, 2. I.e. find x' and t' for these "events".

2) The point at x = 1 in S at times t = 0, 1, 2. Again find x' and t' for these.

 

[Grimble]

When t = 0 , 1, 2
x' = 1.25(0 - 0.6t), 1.25(1 - 0.6t), 1.25(2 - 0.6t) where t is the time elapsed since the separation of the origins. Or are you specifying that O and O' are coincident?
(t in this case is not 0 as vt is the separation of the Origins as is x in the time transformations
t' = 1.25(0 - 0.6x), 1.25(1 - 0.6t), 1.25(2 - 0.6t)

Please note that I am not trying to be difficult but to pull together all the different things I am told, because what seems right in one context seems problematical elsewhere.

 

[Grimble]

It's very easy to work out if you think how someone who is in motion according to you will determine distant time by means of light or radio signals, if that person assumes to be in rest.

Why? Are we not dealing with Events - they are surely fixed, not moving?
Light or radio signals from an event will travel at c and the time taken will be added to the time of the event...

 

[Orodruin]

Why? Are we not dealing with Events - they are surely fixed, not moving?

What do you mean by fixed? Events are just a given place and a given time. There is no way of assigning a velocity to an event.

 

[PeroK]

When t = 0 , 1, 2
x' = 1.25(0 - 0.6t), 1.25(1 - 0.6t), 1.25(2 - 0.6t) where t is the time elapsed since the separation of the origins. Or are you specifying that O and O' are coincident?
(t in this case is not 0 as vt is the separation of the Origins as is x in the time transformations
t' = 1.25(0 - 0.6x), 1.25(1 - 0.6t), 1.25(2 - 0.6t)

Please note that I am not trying to be difficult but to pull together all the different things I am told, because what seems right in one context seems problematical elsewhere.

You're not alone in struggling with a formula and what it means and not realising that you can gain an insight just by plugging in some numbers. However, I've no idea what you've done above. What I meant was:

2) x = 1, t = 0:

$t' = \frac{5}{4}(0 - \frac{3}{5c}) = -\frac{3}{4c}$

So, the time of this event in S' is at t' slightly less than 0. Your simplified formula would not show this at all.

Perhaps a better thing to calculate next would be:

3) $x= 3 \times 10^8$, $t = 0$:

$t' = \frac{5}{4}(0 - \frac{3}{5}) = -\frac{3}{4}$

So, the further from the origin we go, the earlier the event (at t = 0) takes place in S'.

Finally:

4) $x= -3 \times 10^8$, $t = 0$:

$t' = \frac{5}{4}(0 + \frac{3}{5}) = +\frac{3}{4}$

So, times to left of the S origin (at t = 0) are actually ahead in S' (relative to S).

The point is that plugging in these numbers should give you an insight on what the LT means and why it's not a simple $t' = \gamma t$.

 

[Grimble]

What do you mean by fixed? Events are just a given place and a given time. There is no way of assigning a velocity to an event.

What do I mean by fixed? - I mean at a specific location in space at a specific point in time! i.e. Not moving! I.E having no velocity!

What can be confusing about the term fixed?

 

[Grimble]

You're not alone in struggling with a formula and what it means and not realising that you can gain an insight just by plugging in some numbers. However, I've no idea what you've done above. What I meant was:

2) x = 1, t = 0:

t′=54(0−35c)=−34ct' = \frac{5}{4}(0 - \frac{3}{5c}) = -\frac{3}{4c}

So, the time of this event in S' is at t' slightly less than 0. Your simplified formula would not show this at all.

What I did above was to substitute values into the Lorentz Equations.
BUT, in the x transformation the term vt refers to the separation of the Origins so the t in vt refers to the time it would take for the moving frame to move from O to where O' was at time t = 0, t' = 0!
t in this case is NOT the time coordinate!

x' is the point on the X axis when t = t' = 0, when the two origins are not collocated.
therefore one has to subtract vt - the distance that would be traveled by O' from O before the start of the transformation. So obviously x' is x - vt or the separation of the origins at the start.
If the two origins were collocated then the term vt would disappear.Come on now, that is not difficult to understand...

 

[PeroK]

What I did above was to substitute values into the Lorentz Equations.
BUT, in the x transformation the term vt refers to the separation of the Origins so the t in vt refers to the time it would take for the moving frame to move from O to where O' was at time t = 0, t' = 0!
t in this case is NOT the time coordinate!

x' is the point on the X axis when t = t' = 0, when the two origins are not collocated.
therefore one has to subtract vt - the distance that would be traveled by O' from O before the start of the transformation. So obviously x' is x - vt or the separation of the origins at the start.
If the two origins were collocated then the term vt would disappear.Come on now, that is not difficult to understand...

$t$ and $x$ in the Lorentz transformation are independent variables/coordinates. You can plug in any $t$ and $x$ and get $t'$ and $x'$.

You are restricting yourself to the case where $x = vt$. That is, in fact, the spatial origin of the S' frame over time: $x' = 0$ and $t' = \gamma t$.

The LT applies to any $x$ and any $t$; not just those that satisfy $x=vt$.

 

[harrylin]

Why? Are we not dealing with Events - they are surely fixed, not moving?
Light or radio signals from an event will travel at c and the time taken will be added to the time of the event...

The term (t - vx/c²) was introduced by Lorentz based on the propagation of EM radiation, and he called it "local time".
Here's Einstein's clarification with the use of events: http://www.bartleby.com/173/9.html

 

[Dale]

in the x transformation the term vt refers to the separation of the Origins so the t in vt refers to the time it would take for the moving frame to move from O to where O' was at time t = 0, t' = 0!
t in this case is NOT the time coordinate!

No. In the Lorentz transform t is the time coordinate of the unprimed frame. Always.

The Lorentz transform tells you how coordinates (t,x,y,z) transform to coordinates (t',x',y',z') in a frame moving at v relative to the first frame.

 

[PeterDonis]

What do I mean by fixed? - I mean at a specific location in space at a specific point in time! i.e. Not moving! I.E having no velocity!

What can be confusing about the term fixed?

You're confusing yourself, by using the terms "location in space" and "point in time" as if they were absolute. They're not. They're frame-dependent.

Take a simple example: you and I are both way out in empty space somewhere in our spaceships. You look at your clock and it reads exactly 12 noon. At that instant, I fly past you at relativistic speed, and you get an image of my clock, also reading exactly 12 noon. So our worldlines share an event: the event of me flying past you, at an instant where both our clocks read 12 noon. Call this event O.

In your frame, I am moving, and you are fixed: so the "location in space" at which event O takes place is your spaceship. But in my frame, you are moving, and I am fixed; so the "location in space" at which event O takes place is my spaceship. These two "locations in space" are not the same; they are moving relative to each other. For event O itself, this does not matter; the spatial coordinates of event O in both of our frames are (0, 0, 0), because O is the event at which our two "locations in space" coincide. But to transform coordinates of any event other than O from your frame to my frame, it's not enough just to adjust the time; you have to also adjust the spatial position. And because time and space are interlinked, you have to make the adjustment in both time and space. That is why the equation for $t'$ includes the $vx / c^2$ term, and why the equation for $x'$ includes the $vt$ term.

 

[Grimble]

No. In the Lorentz transform t is the time coordinate of the unprimed frame. Always.

So for an event at t = 15, x = 9, in system K, where system K' is traveling along the x-axis at velocity 0.6c, γ = 1.25. (Origin of system K, (t,x,y,z), (0,0,0,0); Origin of system K', (10, 6,0,0))
Then x' = γ(x' - vt) = 1.25( 9 - (0.6)15) = 0
where if the term vt were to represent the difference in the x coordinates for the origins
Then x' = γ(x' - vt) would give: x' = 1.25(9 - (0.6)10) = (1.25)3 = 4.25 or γ3
Explain, please...

What do you mean by fixed? Events are just a given place and a given time. There is no way of assigning a velocity to an event.

Yes, an Event cannot have a velocity as it cannot move as it is a location at a SINGLE SPECIFIC TIME. IT IS A FIXED POINT in SPACETIME.
Fixed in each and every frame whether inertial or not.
The COORDINATES of that event, as with any event will be particular to the individual frame. They will be FIXED relative to each Frame's origin.
What other meaning can fixed have other than having coordinates that are unchanging, that have particular defined values?
I wouls have said that the semantics were quite clear...

 

[Orodruin]

Origin of system K', (10, 6,0,0))

No, this is not the same event as that you just computed the LT for. The Lorentz transformations in standard form relate the coordinates for two systems with coinciding origins. The origin of K' is at (0,0,0,0).

 

[Grimble]

No, this is not the same event as that you just computed the LT for. The Lorentz transformations in standard form relate the coordinates for two systems with coinciding origins. The origin of K' is at (0,0,0,0).

But if they have the same origin then the '-vt' term is meaningless as it will always be 0.
With collocated origins, x' = γx, surely? What could vt represent here? The relative velocity of K' multiplied by the time coordinate? Which is surely the distance that K' has traveled in time t from the point of separation, their common origin!
One has to set vt to 0 with common origins or substituting x for vt (because x does equal vt) one gets x' = γ(x - x)

Please, please, please...
If I am wrong then shew me what other value vt has...

 

[Orodruin]

But if they have the same origin then the '-vt' term is meaningless as it will always be 0.

No, this is the location of the spatial origin of K', which is a world line and not an event. It may have a different t' coordinate and is described by (t',x',y',z') = (t',0,0,0).

With collocated origins, x' = γx, surely?

No, definitely not. The time coordinate is not equal to zero for all events. This is only true for events that occur at time t = 0. It is a very particular event (in the given coordinate system), namely an event which is simultaneous with the origin in K. The Lorentz transformation tell you:
x' = γ(x-vt/c)
and there is no a priori reason to think that a given event would have time coordinate t = 0 unless otherwise stated. You definitely cannot describe all of Minkowski space with t = 0. In the transformation, x is the position of any event in space-time and t is the time of said event, both in the system K, it may occur before or after the origin. The only thing that is particular for the given Lorentz transformation is the velocity v which is the velocity between two objects when one of them is at rest in K and the other in K'. You can check this by setting x' = 0, which is the spatial origin of K'. This gives you x = vt and thus an observer at rest in K' at x'=0 is traveling with velocity v in K.

If I am wrong then shew me what other value vt has...

vt can have any value, this depends only on when the event you are transforming occurs in K.

 

[Orodruin]

Question: Do you understand what the meaning of an inertial frame is? It is not equivalent to an event. It is a set of coordinates on Minkowski space and any event may be described using four coordinates in a given frame. The Lorentz transformations describe how the coordinates from different frames relate to each other and every event has coordinates in every frame.

 

[PeroK]

So for an event at t = 15, x = 9, in system K, where system K' is traveling along the x-axis at velocity 0.6c, γ = 1.25. (Origin of system K, (t,x,y,z), (0,0,0,0); Origin of system K', (10, 6,0,0))
Then x' = γ(x' - vt) = 1.25( 9 - (0.6)15) = 0
where if the term vt were to represent the difference in the x coordinates for the origins
Then x' = γ(x' - vt) would give: x' = 1.25(9 - (0.6)10) = (1.25)3 = 4.25 or γ3
Explain, please...Yes, an Event cannot have a velocity as it cannot move as it is a location at a SINGLE SPECIFIC TIME. IT IS A FIXED POINT in SPACETIME.
Fixed in each and every frame whether inertial or not.
The COORDINATES of that event, as with any event will be particular to the individual frame. They will be FIXED relative to each Frame's origin.
What other meaning can fixed have other than having coordinates that are unchanging, that have particular defined values?
I wouls have said that the semantics were quite clear...

I suspect you may have difficulty understanding the non-relativistic Galilean tranformation of classical physics. This transformation is simpler, in that t = t' for any two inertial frames, but you have the same issue of coincident origins at a shared time t = t' = 0, but thereafter the spatial coordinates in each frame will differ depending on the relative velocity between the frames.

It may be a good idea, therefore, to take at look at the Galilean tranformation. Make sure you understand that, then come back to Lorentz. It's not the Lorentz Transformation that you do not understnd, it's the concept of having two different reference frames in the first place.

 

[ghwellsjr]

So for an event at t = 15, x = 9, in system K, where system K' is traveling along the x-axis at velocity 0.6c, γ = 1.25.

This is a simple problem that we can work out.

But first you need to be corrected in your following statement:

(Origin of system K, (t,x,y,z), (0,0,0,0); Origin of system K', (10, 6,0,0))

No, the origin of every system is the event where all four coordinates equal zero. If you transform the four coordinates of the origin event of system K to see what the coordinates are for that same event in system K', you will see that they are all zero. I hate to ask why you thought they would be (10, 6,0,0) but you obviously didn't use the Lorentz Transformation process, did you?

Then x' = γ(x' - vt) = 1.25( 9 - (0.6)15) = 0

Excellent. You used the LT to calculate the x' coordinate in system K'.

where if the term vt were to represent the difference in the x coordinates for the origins
Then x' = γ(x' - vt) would give: x' = 1.25(9 - (0.6)10) = (1.25)3 = 4.25 or γ3
Explain, please...

Explain what? You already correctly calculated x', why are you dissecting the formula? Instead of doing whatever it is you are doing, you should instead calculate t', like this:

t' = γ(t' - vx) = 1.25( 15 - (0.6)9) = 1.25( 15 - 5.4) = 1.25( 9.6) = 12

Problem solved. Is that so difficult either in concept or in enactment?

Yes, an Event cannot have a velocity as it cannot move as it is a location at a SINGLE SPECIFIC TIME. IT IS A FIXED POINT in SPACETIME.
Fixed in each and every frame whether inertial or not.
The COORDINATES of that event, as with any event will be particular to the individual frame. They will be FIXED relative to each Frame's origin.
What other meaning can fixed have other than having coordinates that are unchanging, that have particular defined values?
I wouls have said that the semantics were quite clear...

You often describe things so well but then you don't follow through. Using the Lorentz Transformation is so simple if you would just do it.

 

[Dale]

So for an event at t = 15, x = 9, in system K, where system K' is traveling along the x-axis at velocity 0.6c, γ = 1.25.

For this event t'=12, x'=0

(Origin of system K, (t,x,y,z), (0,0,0,0); Origin of system K', (10, 6,0,0))

Uhh, no. The origin of any system is (0,0,0,0), by definition.

You may be thinking of combining the Lorentz transform (which preserves the origin) and a spacetime translation (which moves the origin). The full symmetry group of special relativity is the Poincare group which includes the Lorentz transform, spacetime translations, and spatial rotations. That is perfectly fine, but would be a different formula than what you posted (which was just the Lorentz transform in standard configuration). The formula that you have posted and correctly referred to as the Lorentz transform preserves the origin.

Then x' = γ(x' - vt) = 1.25( 9 - (0.6)15) = 0

Yes.

 

[Dale]

But if they have the same origin then the '-vt' term is meaningless as it will always be 0...
If I am wrong then shew me what other value vt has...

In the example you gave above $vt=0.6*15=9$. You even calculated it yourself.

 

[Grimble]

I will have to think about this, thank you one and all...

 

[jartsa]

t' = γ(t - vx/c²)
where t is the time in the stationary frame
t' is the time in the moving frame
v is the relative velocity between the two frames
and x is the distance traveled in the time t in the stationary frame.
∴ x = vt
substituting this into the Lorentz time equation gives us:
t' = γt(1 - v²/c²) = t/γ

So why is the formula not written this, t' = t/γ, much simpler way?

t' = γ(t - vx/c²)What does that formula mean concretely? I would guess it means something like this:

t is the time in the stationary frame, measured by a clock that is stationary in the stationary frame
t' is the time in the moving frame
v is the relative velocity between the two frames
x is the x-coordinate of the clock in the stationary frame

 

[Ibix]

t is the time in the stationary frame, measured by a clock that is stationary in the stationary frame

That is stationary and has always been stationary since it was last synchronised with other clocks stationary in that frame. And a similar statement about t' and clocks at rest in the "moving" frame. Otherwise, spot on.

 

[Grimble]

OK. I did not make it clear, but when I wrote:

So for an event at t = 15, x = 9, in system K, where system K' is traveling along the x-axis at velocity 0.6c, γ = 1.25. (Origin of system K, (t,x,y,z), (0,0,0,0); Origin of system K', (10, 6,0,0))

that I was referring to the coordinates of the Origin of 0' in the reference frame of system K.
So that starting from the origin of 0, that is (0,0,0,0), system K' (it's origin specifically) has traveled along the common x-axis for 10 seconds at 0.6c or 6 light seconds, arriving at (10,6,0,0). Which are the coordinates for the event at (10,6,0,0) in frame K, which is the same event that has coordinates (10,6,0,0) in frame K'.

The coordinates of the origin of system K' in system K are x' = γ(x - vt) = 1.25(6 - 10(0.6) = 0, and t' = 1.25(10 - 6(0.6)) = 8
while the coordinates after another 5 seconds when t = 15, x = 9 would be, x' = 1.25(9 - 15(0.6)) = 0 and t' = 1.25(15 - 9(0.6)) = 12

So the value calculated for the time coordinate, t' is fine = t/γ, whereas the x' coordinate always comes out at 0.
But it still seems reasonable to me that the x' coordinate is a function of the x coordinate γx, LESS the displacement of the O' frame when x' = 0.

 

[Orodruin]

system K' (it's origin specifically) has travelled

(My emphasis.) Wrong. The origin of K' is a single event, the same as that of K and does not travel. Unless you are talking about the spatial origin of K', i.e., its time axis.

Which are the coordinates for the event at (10,6,0,0) in frame K, which is the same event that has coordinates (10,6,0,0) in frame K'.

This is wrong even if you are talking about the spatial origin. The spatial origin of K' always has all spatial coordinates equal to zero in K'. The time coordinate is also not the same as the time coordinate in K.

Your errors after that seem to be results of misconceptions like this.

 

[Dale]

that I was referring to the coordinates of the Origin of 0' in the reference frame of system K.

K and K' share the same origin. I.e. the event (0,0,0,0) in K has the coordinates (0,0,0,0) in K'. Just look at the Lorentz transform and set t=x=y=z=0 and calculate the primed coordinates.

You yourself have pointed out several times, including using all caps, that events are fixed and do not travel. So why are you talking about an event, the origin, as traveling when you seem to clearly understand that they do not travel.

 

[Dale]

@Grimble

It may help you to draw the lines t=0, t=1, t=2, and x=0, x=1, x=2 on a spacetime diagram for K'. That was one of the key exercises that I did learning SR. The easiest way to do it is to write them as parametric equations and do substitution.

 

[ghwellsjr]

OK. I did not make it clear, but when I wrote:

"So for an event at t = 15, x = 9, in system K, where system K' is traveling along the x-axis at velocity 0.6c, γ = 1.25. (Origin of system K, (t,x,y,z), (0,0,0,0); Origin of system K', (10, 6,0,0))"

that I was referring to the coordinates of the Origin of 0' in the reference frame of system K.
So that starting from the origin of 0, that is (0,0,0,0), system K' (it's origin specifically) has traveled along the common x-axis for 10 seconds at 0.6c or 6 light seconds, arriving at (10,6,0,0). Which are the coordinates for the event at (10,6,0,0) in frame K, which is the same event that has coordinates (10,6,0,0) in frame K'.

The coordinates of the origin of system K' in system K are x' = γ(x - vt) = 1.25(6 - 10(0.6) = 0, and t' = 1.25(10 - 6(0.6)) = 8
while the coordinates after another 5 seconds when t = 15, x = 9 would be, x' = 1.25(9 - 15(0.6)) = 0 and t' = 1.25(15 - 9(0.6)) = 12

So the value calculated for the time coordinate, t' is fine = t/γ, whereas the x' coordinate always comes out at 0.
But it still seems reasonable to me that the x' coordinate is a function of the x coordinate γx, LESS the displacement of the O' frame when x' = 0.

As near as I can tell, you are describing an object "O" that is traveling at 0.6c in system K and passes through its origin. Here is a spacetime diagram depicting this scenario:

The origin of K is the black dot. The two events you specified are shown in green and red.

Now we transform all the dots to system K':

Notice how the origin of K' is still at (0,0,0,0,) and the other events transform as you calculated.

 

[Grimble]

Which are the coordinates for the event at (10,6,0,0) in frame K, which is the same event that has coordinates (10,6,0,0) in frame K'.

This should of course have read: "... which is the same event that has coordinates (0,0,0,0) in frame K'. "
Apologies for poor reviewing before posting.

 

[Grimble]

AAAAHHH! Now I have it!

I was as mentioned above still stuck in the rut of Galileian relativity!
It does in fact work just as you say it works. I was failing to grasp that while,the Origin of K' is an event at a fixed location in system K - (10,6,0,0) - the location of it, its spatial coordinates are constantly changing as that frame moves in system K. It moves as a function of the time t. So the x coordinate increases as a function of the time t. That function being -vt.
So I can see how it all works now.
If we take a spaceship 0.01ly in length where the rear is at (10,6,0,0) then the front would be at (10,6.01,0,0). And in system K' the front would have the coordinates
x' = 1.25(6.01 - 10(0.6)) = 0.0125 while t' would be t' = 1.25(10 - (6.01)(0.6)) = 7.9925 so the front would be 0.0075seconds in time reaching that point before the rear of the train reached (10,6,0,0)?

 

[ghwellsjr]

AAAAHHH! Now I have it!

I was as mentioned above still stuck in the rut of Galileian relativity!
It does in fact work just as you say it works. I was failing to grasp that while,the Origin of K' is an event at a fixed location in system K - (10,6,0,0) - the location of it, its spatial coordinates are constantly changing as that frame moves in system K. It moves as a function of the time t. So the x coordinate increases as a function of the time t. That function being -vt.
So I can see how it all works now.

How many times do you have to be told that the Origin of every system has the coordinates (0,0,0,0)? The origin, like all other events, doesn't move. Its spatial coordinates don't change. They remain all zeroes. If you are talking about an object or observer which passes through the origin of a system, its spatial coordinates can change with time in any arbitrary manner that you want. You can describe its motion as any function of time that you desire and then you can transform the coordinates of any events along its worldline to see how the coordinates appear in another system moving with respect to the original system. That's what I did for you in post #37. The motion of a frame is independent of the motion of any object or observer.

If we take a spaceship 0.01ly in length where the rear is at (10,6,0,0) then the front would be at (10,6.01,0,0). And in system K' the front would have the coordinates
x' = 1.25(6.01 - 10(0.6)) = 0.0125 while t' would be t' = 1.25(10 - (6.01)(0.6)) = 7.9925 so the front would be 0.0075seconds in time reaching that point before the rear of the train reached (10,6,0,0)?

Are you trying to understand Length Contraction with this example? If so, you are taking the wrong approach. If not, what are you trying to do?

Also, please spend some time proofing your posts. You started out with a spaceship and ended up with a train!

 

[Grimble]

How many times do you have to be told that the Origin of every system has the coordinates (0,0,0,0)? The origin, like all other events, doesn't move. Its spatial coordinates don't change. They remain all zeroes.

And that is EXACTLY what I said! The location of frame K', its origin will be moving in system K'. BECAUSE IT IS A MOVING FRAME in system K, the stationary frame.
The EVENT that is the origin, in space AND TIME, is fixed and cannot be moving BECAUSE it is a point in time, and movement is change of location over time!
But the ORIGIN of system K' the moving frame has to be changing location over time, BECAUSE it is moving!

My apologies for sending the train through space!

 

[Dale]

And that is EXACTLY what I said! The location of frame K', its origin will be moving in system K'. BECAUSE IT IS A MOVING FRAME in system K, the stationary frame.
The EVENT that is the origin, in space AND TIME, is fixed and cannot be moving BECAUSE it is a point in time, and movement is change of location over time!
But the ORIGIN of system K' the moving frame has to be changing location over time, BECAUSE it is moving!

My apologies for sending the train through space!

It is important to learn and use correct terminology. The origin of any frame is defined as the event (0,0,0,0). All frames agree on the coordinates of that event in the standard configuration (by design). The origin of any frame never moves.

The thing that you are occasionally referring to as the origin and claiming that it moves is not the origin. The correct word is the "time axis". In the unprimed frame the unprimed frame's time axis is $(t,x,y,z)=(\tau,0,0,0)$. You can easily use this to calculate the line of the unprimed time axis in the primed frame as a function of $\tau$.

Please use the correct terminology now that it has been explained to you. If you find yourself resorting to all caps it is probably a sign that someone is using incorrect terminology.

 

[Ibix]

In other words, you are mixing three and four dimensional thinking again. Or at least three andafour dimensional terminology.

 

[Ibix]

Here is a sequence of still images of a car passing by a tree, accompanied by close-ups of a clock that is stationary with respect to the tree. Note that time is increasing up the page, and that the car passes the tree (call this x=0) at time t=0.

Here is the same sequence of images viewed from the frame in which the car is at rest, with close-ups of the car's dashboard clock. Again, time is increasing up the page. In this frame it is the tree passing the car - again we call this x'=0 and note that t'=0.

I think that you are still confusing points and events, and that is making you confused about what the Lorentz transforms are doing. The tree and the car are both points (or, at least, their centers of mass are points). In the top picture, the car is moving but the tree is not. A point that is not moving appears in the same place in every picture. A point that is moving appears at a different place in each picture.

In contrast, an event (for example, the origin (t,x,y,z)=(0,0,0,0), where and when the car and tree pass each other) only appears in one picture. It is a point at a specific time. You keep insisting that events do not move; I would say that events neither move nor are stationary. Motion is change of position with time, and since events don't exist for any length of time, they cannot be said to be moving or stationary. They only exist at one instant, so velocity is not defined for them.

The origin of spatial coordinates - (x,y,z)=(0,0,0) - is a point, and it could be stationary or moving. For example, the origin of spatial coordinates in my second diagram is the location of the car, and you can clearly see that it is a moving point in the first diagram. However, this is not the origin for the frame in a 4d sense - that is the event when the car and tree pass (t,x,y,z)=(0,0,0,0). You are confusing the origin of spatial coordinates with the origin of space-time coordinates. The first can be moving, but is not what we mean by "origin" when we're talking about Lorentz transforms. The origin, in this context, is the origin of space-time coordinates, which is the event (0,0,0,0), which appears only in the middle image in each diagram. It is neither moving nor stationary since it is a position at an instant, and velocity cannot be defined in an instant.

The Lorentz transforms simple relate the positions and times of events measured in one frame to the positions and times measured in another. You've done some examples. The diagram below may help - it shows a grid of (x,t) coordinates in black and a grid of (x',t') coordinates in red, for the case where the primed frame is moving at 0.6c. If you know the coordinates of an event on the black grid, the Lorentz transforms simply tell you the coordinates on the red grid without you having to go out and measure them, too.

Hope all that was helpful.

 

[Dale]

I would say that events neither move nor are stationary. Motion is change of position with time, and since events don't exist for any length of time, they cannot be said to be moving or stationary. They only exist at one instant, so velocity is not defined for them.

Well said. Terms like "fixed" seem to imply that it has a well defined velocity and that velocity is equal to 0. But that isn't correct.

 

[Ibix]

Just to clarify something about the last diagram in my last post. In both the red and black frames, the origin of coordinates is the point where the axes cross - i.e. (x,t)=(0,0) or, equivalently, (x',t')=(0,0). The origin of spatial coordinates, however, is the t or t' axis - the vertical black line an the steeper red line. Anything at rest in the red frame is moving in the black one.

Hopefully you can see the relationship between the Minkowski diagram and the first picture. If not, draw a black line through the trees and a red one through the cars.

 

[Grimble]

Thank you all once again; I stand corrected, I must admit I find it a bit of a trial getting the terms correct as my working life has been devoted to English rather than science and I think of an event as being fixed in time and space, i.e only existing at that one point.
I will try and be more precise in my terminology from now on.
Thank you once again for your patience.