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...