Lorentz transformations, time intervals and lengths

[bernhard.rothenstein]

Prtesent the Lorentz transformations as
dx=g(dx'+Vdt')
dt=g(dx-Vdt)
In my oppinion dx and dx' represent proper lengths measured in I and in I', dt and dt' representing coordinate time intervals. Do you aggree.
Happy new year to all participamts on the Forum

 

[bernhard.rothenstein]

it should be
dx=g(dx'+Vdt')
dt=g(dt'+Vdx'/cc)

 

[Fredrik]

I don't agree. I don't see a reason to call one of variables "proper something", and I think that calling one of them "proper something" and the other "coordinate something" is the worst option. You might as well choose the opposite convention: dt is the proper time measured by an observer whose world line coincides with the t axis, and dx is a spatial coordinate in the coordinate system that we'd normally associate with this guy.

I'd just describe them both as coordinates.

 

[tiny-tim]

In my oppinion dx and dx' represent proper lengths measured in I and in I', dt and dt' representing coordinate time intervals. Do you aggree.
Happy new year to all participamts on the Forum

Happy new year to you too, bernhard

I agree with Fredrik …

there's no reason whatever to distinguish between x and t.

"proper" means independent of the observer …

the "proper length" of a rod (if anyone used the phrase) would mean its length as measured in a frame in which it is stationary.

 

[bernhard.rothenstein]

I don't agree. I don't see a reason to call one of variables "proper something", and I think that calling one of them "proper something" and the other "coordinate something" is the worst option. You might as well choose the opposite convention: dt is the proper time measured by an observer whose world line coincides with the t axis, and dx is a spatial coordinate in the coordinate system that we'd normally associate with this guy.

I'd just describe them both as coordinates.

Thanks for your answer. As I see we do not read the same textbooks. I like the definitions given in :A Travelers Guide to Sapcetime Mc.Graw-Hill,Inc. 1995
Proper time: The time between two events as measured by a single clock that is present at both events (its value depends on the world-line that the clock follows in getting from one event to the other.
Coordinate time interval: The time between two events in an inertial reference frame by a pair of synchronized clocks one present at one event, the other present at the other event.
The definitions of proper length and distorted length are well known.
So my question is how to qualify dx (dx') and dt (dt') in accordance with the definitions given above. As long as we do not use standard definitons...

 

[bernhard.rothenstein]

I'd just describe them both as coordinates.
dt and dt' are time intervals or coordinates?

 

[tiny-tim]

dt and dt' are time intervals or coordinates?

t and t' are coordinates,

dt and dt' are coordinate intervals.

 

[bernhard.rothenstein]

t and t' are coordinates,

dt and dt' are coordinate intervals.

Thanks again. But that does not answer my question?

 

[tiny-tim]

Thanks again. But that does not answer my question?

"dt and dt' are time intervals or coordinates"? …

ok … they can't be coordinates, they are intervals, so they're time coordinate intervals.

 

[JesseM]

it should be
dx=g(dx'+Vdt')
dt=g(dt'+Vdx'/cc)

What does it mean to put intervals like dx and dt in the Lorentz transformation rather than coordinates like x and t? Intervals imply you're talking about a pair of events, but normally the Lorentz transformation just deals with the coordinates of a single event E. Your notation only seems to make sense if dx is the distance along the x-axis between the event E you're looking at and an event at the origin at x=0, t=0 (or equivalently x'=0, t'=0). But if that's the case, dx cannot be a "proper length" unless E has a time-coordinate of t=0 (since proper length always refers to the spatial distance between two events in the frame where they are simultaneous).

 

[Fredrik]

A global coordinate system in SR is just a smooth function $x:M\rightarrow\mathbb R^4$, where M is Minkowski space. (Yes, M and $\mathbb R^4$ is the same set. Minkowski space is $\mathbb R^4$ with a manifold structure and a metric). A Lorentz transformation is just a transition function $x\circ y^{-1}:\mathbb R^4\rightarrow\mathbb R^4$, such that both x and y are inertial frames. It clearly doesn't matter if you express the 4-tuple that a Lorentz transformation acts on as (t,x,y,z) or (dt,dx,dy,dz).

 

[bernhard.rothenstein]

A global coordinate system in SR is just a smooth function $x:M\rightarrow\mathbb R^4$, where M is Minkowski space. (Yes, M and $\mathbb R^4$ is the same set. Minkowski space is $\mathbb R^4$ with a manifold structure and a metric). A Lorentz transformation is just a transition function $x\circ y^{-1}:\mathbb R^4\rightarrow\mathbb R^4$, such that both x and y are inertial frames. It clearly doesn't matter if you express the 4-tuple that a Lorentz transformation acts on as (t,x,y,z) or (dt,dx,dy,dz).

Thanks. Would you present all that to freshmans?

 

[bernhard.rothenstein]

What does it mean to put intervals like dx and dt in the Lorentz transformation rather than coordinates like x and t? Intervals imply you're talking about a pair of events, but normally the Lorentz transformation just deals with the coordinates of a single event E. Your notation only seems to make sense if dx is the distance along the x-axis between the event E you're looking at and an event at the origin at x=0, t=0 (or equivalently x'=0, t'=0). But if that's the case, dx cannot be a "proper length" unless E has a time-coordinate of t=0 (since proper length always refers to the spatial distance between two events in the frame where they are simultaneous).

Because the problem presents importance in my activity please read the following lines with the definitions given above in mind.
Consider a clock K"(0,0) located at the origin O" of the I" inertial reference frame. I" moves with constant speed u relative to a I inertial reference frame and with speed u' relative to I', I' moving with speed V relative to I. At the origin of time in the three mentioned frames their origins are located at the same point in space. After a given time of motion K"(0,0) arrives in front of a clock K(x,0) of I and in front of a clock K'(x',0) of I'. The clocks of I and I' are Einstein synchronized. When the three clocks are located at the same point in space they read t, t' and t" respectively. Detected from I" the trip of the moving clock lasts a proper time interval (t"-0). Detected from I it lasts a coordinate time interval (t-0) whereas detected from I' it lasts the coordinate time interval (t'-0). In accordance with the formula which accounts for the time dilation effect we have
(t-0)=g(u)(t"-0) (1)
(t'-0)=g(u')(t"-0) (2)
Combining (1) and (2) we obtain that the coordinate time intervals are related by
(t-0)=(t'-0)g(u)/g(u') (3)
Expressing the right side of (3) as a function of physical quantities measured from I' via the addition law of parallel speeds the result is
(t-0)=g(V)(t'-0)[1+Vu'/cc] (4)
During the time interval (t'-0) the origin O" generates in I' a "rod" of proper length
(x'-0)=u'(t'-0) (5)
with which (4) becomes
(t-0)=g(V)[1+V(x'-0)/cc]. (6)
Do you consider that the way in which the Lorentz-Einstein transformation (6) was derived proofs that (t-0) is a coordinate time interval whereas (x'-0) is a proper length.
All comments are highly appreciated.

 

[JesseM]

A global coordinate system in SR is just a smooth function $x:M\rightarrow\mathbb R^4$, where M is Minkowski space. (Yes, M and $\mathbb R^4$ is the same set. Minkowski space is $\mathbb R^4$ with a manifold structure and a metric). A Lorentz transformation is just a transition function $x\circ y^{-1}:\mathbb R^4\rightarrow\mathbb R^4$, such that both x and y are inertial frames. It clearly doesn't matter if you express the 4-tuple that a Lorentz transformation acts on as (t,x,y,z) or (dt,dx,dy,dz).

Your last sentence seems like a non sequitur to me--can you explain why what you said prior to that sentence implies "it clearly doesn't matter" if you transform between coordinates or intervals? Let's say v=0.6 (so gamma = 1.25), and we have two events A: (x=3 l.s., t=2 s.) and B: (x=5 l.s., t=7 s). In this case dx=2 l.s., dt=5 s. Now we can find the coordinates of A in the x',t' system:

x' = 1.25*(3 - 0.6*2) = 2.25
t' = 1.25*(2 - 0.6*3) = 0.25

and of B:

x' = 1.25*(5 - 0.6*7) = 1
t' = 1.25*(7 - 0.6*5) = 5

If we subtract the first from the second we get dx' = -1.25 and dt' = 4.75. This is the same as what we get if we take the original dx=2 and dt=5 and plug them into the equations:

dx' = 1.25(dx - 0.6*dt)
dt' = 1.25(dt - 0.6*dx)

...but I still don't understand why it works out this way. Suppose we instead used the following coordinate transformation:

x' = x^2 + 3t
t' = 4x + 2t^3

In this case A would have coordinates (x'=15, t'=28) and B would have coordinates (x'=46, t'=706), so dx'=31 and dt'=678. But this is different than what we get if we plug in dx=2, dt=5 into the equation:

dx' = dx^2 + 3dt
dt' = 4dx + 2dt^3

So why does it work for the Lorentz transformation but not the transformation above? Everything you said about the Lorentz transformation would seem to be true about it as well.

Thinking about it some more, I think the answer probably has to do with the fact that the x' and t' coordinates of an event in the second frame are each linear functions of the x,t coordinates of the same event in the first frame (a function being linear if f(A + B) = f(A) + f(B)), so if you have two events A: (x1, t1) and B: (x2, t2) and subtract their coordinates to get a new event C: (x3 = [x2 - x1], t3 = [t2 - t1]) then for C, x'(C) = gamma*x3 - gamma*v*t3 = gamma*(x2 - x1) - gamma*v*(t2 - t1) = [gamma*x2 - gamma*v*t2] - [gamma*x1 - gamma*v*t1] = x'(B) - x'(A), and similarly for t' of C.

 

[Fredrik]

Your last sentence seems like a non sequitur to me--can you explain why what you said prior to that sentence implies "it clearly doesn't matter" if you transform between coordinates or intervals?

I'm just saying that it doesn't matter what symbol you use for a variable. f(x) with x=3 is the same as f(y) with y=3.

 

[JesseM]

I'm just saying that it doesn't matter what symbol you use for a variable. f(x) with x=3 is the same as f(y) with y=3.

But if dx and dt are understood to mean intervals between a pair of events rather than coordinates of a single event, then you can't assume that you can replace x and t in a coordinate transformation with dx and dt and get correct answers for that coordinate transformation, although as it turns out this does work for the Lorentz transformation because it's a linear transformation (but see the example of a nonlinear coordinate transformation I gave, where it wouldn't work).

 

[Chrisc]

Feel free to ignore this comment as everyone knows my maths suck, but it seems to me Fredrik is describing a gauge transformation of local symmetry which happens in this case to be synonymous with a Lorentz transformation, or vise aversa.

Happy New Year

 

[JesseM]

Consider a clock K"(0,0) located at the origin O" of the I" inertial reference frame. I" moves with constant speed u relative to a I inertial reference frame and with speed u' relative to I', I' moving with speed V relative to I. At the origin of time in the three mentioned frames their origins are located at the same point in space. After a given time of motion K"(0,0) arrives in front of a clock K(x,0) of I and in front of a clock K'(x',0) of I'.

Does "in front of" mean they all meet at a negligible distance from one another, so we can treat the meeting as a single point in spacetime? Also, I'm confused by what the coordinates next to each K mean...I would have thought they were the coordinates in each frame of the point where the three clocks meet, but you said the clock K"(0,0) started at the origin of I'', so if the meeting occurred "after a given time of motion", wouldn't the K'' clock have a different t'' coordinate at the time of the meeting?

 

[bernhard.rothenstein]

What does it mean to put intervals like dx and dt in the Lorentz transformation rather than coordinates like x and t? Intervals imply you're talking about a pair of events, but normally the Lorentz transformation just deals with the coordinates of a single event E. Your notation only seems to make sense if dx is the distance along the x-axis between the event E you're looking at and an event at the origin at x=0, t=0 (or equivalently x'=0, t'=0). But if that's the case, dx cannot be a "proper length" unless E has a time-coordinate of t=0 (since proper length always refers to the spatial distance between two events in the frame where they are simultaneous).

I still think that dx is a "proper length". Consider the relative positions of the reference frames I and I' as detected from I when the standard synchronized clocks of I read t. Let E(x,t) and E'(x',t') be two events located on the overlapped OX(O'X') axes. The distance between the origins O and O' , V(t-0) , and (x-0) are proper lengths. The distance (x'-0) detected from I' is a proper length as well, (t-0) being a coordinate time interval.. Measured from I it is the contracted length
(x'-0)/g. Adding distances measured from I we have
(x'-0)/g=(x-0)-V(t-0) (1)
where from
(x'-0)=[(x-0)-V(t-0)]g (2)
in which we recognise the Lorentz transformation relating proper lengths and coordinate time intervals.
Let
In what concerns the speed, the distance traveled by a particle and the time interval during which the distance is covered we have
speed=proper distance covered/coordinate time interval
resulting that
proper distance covered=speedxcoordinate time interval.
As a starter of the thread I consider that the colateral discussions do not elucidate my problem.
Thanks for your help

 

[Fredrik]

Proper time: The time between two events as measured by a single clock that is present at both events (its value depends on the world-line that the clock follows in getting from one event to the other.
Coordinate time interval: The time between two events in an inertial reference frame by a pair of synchronized clocks one present at one event, the other present at the other event.

These are fine operational definitions of "proper time" and "coordinate time". I don't disagree with them, but I don't think of them as definitions, but rather as postulates about correspondences between things we measure and things in the mathematical model. For example, we define the "proper time" of a curve as the integral of $\sqrt{-g_{\mu\nu}dx^\mu dx^\nu}$ along that curve. The first of the "definitions" you quoted is the postulate that tells us how to measure proper time.

Thanks. Would you present all that to freshmans?

Yes I would, and you're not a freshman. You have 25 articles about SR at arxiv.org.

I still think that dx is a "proper length".

"Proper length" isn't a property of a number like dx. It's not a property of a pair of events. It's a property of a curve. That's why I wouldn't call dx a proper length even if it represents a number that agrees with some proper length measurement. (The definition of dx doesn't mention any curves. You could argue that it does implicitly, and that the relevant curve is the shortest possible curve that lies entirely in the plane of constant t that we're considering, but you could make a similar argument about dt).

 

[bernhard.rothenstein]

A global coordinate system in SR is just a smooth function $x:M\rightarrow\mathbb R^4$, where M is Minkowski space. (Yes, M and $\mathbb R^4$ is the same set. Minkowski space is $\mathbb R^4$ with a manifold structure and a metric). A Lorentz transformation is just a transition function $x\circ y^{-1}:\mathbb R^4\rightarrow\mathbb R^4$, such that both x and y are inertial frames. It clearly doesn't matter if you express the 4-tuple that a Lorentz transformation acts on as (t,x,y,z) or (dt,dx,dy,dz).

I started the thread in order to be helped to present the problem to beginner and tohave a confirmation for the fact that the way in which I approach the problem is correct and adequate for a transparent way of teaching. As a teacher of physics I think that starting with what you say the success is not ensured. But that is my opinion...

 

[bernhard.rothenstein]

Does "in front of" mean they all meet at a negligible distance from one another, so we can treat the meeting as a single point in spacetime? Also, I'm confused by what the coordinates next to each K mean...I would have thought they were the coordinates in each frame of the point where the three clocks meet, but you said the clock K"(0,0) started at the origin of I'', so if the meeting occurred "after a given time of motion", wouldn't the K'' clock have a different t'' coordinate at the time of the meeting?

Does "in front of" mean they all meet at a negligible distance from one another, so we can treat the meeting as a single point in spacetime?
YES


Also, I'm confused by what the coordinates next to each K mean...I would have thought they were the coordinates in each frame of the point where the three clocks meet.
YES

Clock K"(0,0) measures a proper time interval. Because it moves with speed u relative to I and with u' relative to I' it is differently related to to the coordinate time intervals measured from I and I' respectively.

Excuse the fact that my presentation of the problem is not very clear and thanks for helping me to understand where are the poor parts of my statements. Please continue to follow my argumentation.

 

[Meir Achuz]

I started the thread in order to be helped to present the problem to beginner and tohave a confirmation for the fact that the way in which I approach the problem is correct and adequate for a transparent way of teaching. As a teacher of physics I think that starting with what you say the success is not ensured. But that is my opinion...

I have avoided this thread up to now, but I think you can see that a "transparent" statement that leads to a thread of this length will not be transparent to a beginner.

 

[bernhard.rothenstein]

I have avoided this thread up to now, but I think you can see that a "transparent" statement that leads to a thread of this length will not be transparent to a beginner.

Thanks for your comment. Please read the following lines:

I still think that dx is a "proper length". Consider the relative positions of the reference frames I and I' as detected from I when the standard synchronized clocks of I read t. Let E(x,t) and E'(x',t') be two events located on the overlapped OX(O'X') axes. The distance between the origins O and O' , V(t-0) , and (x-0) are proper lengths. The distance (x'-0) detected from I' is a proper length as well, (t-0) being a coordinate time interval.. Measured from I it is the contracted length
(x'-0)/g. Adding distances measured from I we have
(x'-0)/g=(x-0)-V(t-0) (1)
where from
(x'-0)=[(x-0)-V(t-0)]g (2)
in which we recognise the Lorentz transformation relating proper lengths and coordinate time intervals.
Let
In what concerns the speed, the distance traveled by a particle and the time interval during which the distance is covered we have
speed=proper distance covered/coordinate time interval
resulting that
proper distance covered=speedxcoordinate time interval.
The problem is if you find it transparent enough and let me know your opinion about it. I think that the length arrises from the fact that no standard definitions are accepted by all the participants.

 

[Fredrik]

Please read the following lines:

No need to repost. You can just suggest that he read your #19.

Consider the relative positions of the reference frames I and I' as detected from I when the standard synchronized clocks of I read t. Let E(x,t) and E'(x',t') be two events located on the overlapped OX(O'X') axes.

You have a rather strange way of describing these things. I always find it difficult to read your descriptions, so I'm going to give you a few specific comments. (The questions I'm asking below are rhetorical).

1. The notation "OX(O'X')" is definitely not standard. I think you should just say that I and I' are two inertial frames with a common origin, and that the x and x' axes coincide (if that's what you mean).
2. The first sentence mentions "the relative positions" of I and I'. I can only assume that this is a reference to the spatial coordinates in I of the origin of I' and vice versa, but you're not using this information later, so why talk about it at all?
3. What does "as detected from I" mean? I think those words should be dropped, as they add nothing.
4. Events have nothing to do with coordinates, so why not just call the events e.g. E₁ and E₂?
5. What does your statement about those two events even mean? Did you mean to specify the coordinates of two events E and E' as E=(x,t) and E'=(x',t'). Why not specify both in the same frame? Why specify them at all if you're not going to refer to them later?

The distance between the origins O and O' , V(t-0) , and (x-0) are proper lengths.

6. Why would you call them that? A curve can have a "proper length", but you're not talking about curves (at least not explicitly).

The distance (x'-0) detected from I' is a proper length as well, (t-0) being a coordinate time interval..

7. This sentence is very strange, especially the part before the comma. x' is just the spatial coordinate in I' of an event, so why talk about it being "detected"? Also, t is the proper time measured by a clock at the origin, so why are you calling that a "coordinate time interval"? (I'm not saying that you should, but you should at least be consistent).

Measured from I it is the contracted length
(x'-0)/g.

8. What exactly does "it" refer to? Is it the distance between the two events that were specified in a strange way earlier? It doesn't make sense to talk about the distance between two events that aren't simultaneous. (If they are simultaneous in I, they aren't simultaneous in I'). Instead of talking about two events, you should have talked about two objects, and their world lines.

Adding distances measured from I we have
(x'-0)/g=(x-0)-V(t-0) (1)
where from
(x'-0)=[(x-0)-V(t-0)]g (2)
in which we recognise the Lorentz transformation relating proper lengths and coordinate time intervals.

9. In this case it's obvious that your "g" is $\gamma$, but the first time I saw you use this notation I had no idea what you meant. Why don't you use $\LaTeX$? You should at the very least use https://www.physicsforums.com/blog.php?b=347 .

It still doesn't make any sense to me to call the spatial coordinates "proper lengths". "Proper length" is a coordinate independent quantity, and a Lorentz transformation is just a transition ($x\circ y^{-1}$) from one inertial frame to another.

...no standard definitions are accepted by all the participants.

Doesn't everyone agree that a coordinate system is a function that maps a region of spacetime onto a region of $\mathbb R^4$? Doesn't everyone agree that proper time is the integral of $\sqrt{-g_{\mu\nu}dx^\mu dx^\nu}$ along a curve? Doesn't everyone agree that the definitions in #5 are excellent operational definitions of proper time and coordinate time? Doesn't everyone agree that we need both the mathematical and the operational definitions in order to be able to interpret mathematical calculations as predictions about the results of experiments? What exactly do you feel that there's a disagreement about?

As a teacher of physics I think that starting with what you say the success is not ensured. But that is my opinion...

I don't understand your objection. Do you think students shouldn't be told that a coordinate system is a function that assigns four numbers to each event?

 

[bernhard.rothenstein]

"4. Events have nothing to do with coordinhates?"
I think that it is hard to continue to discuss with somebody who makes such a statement.
I quote from Moore:
"An event is any physical occurence that can be considered to happen at a definite place in space at a definite instant of time, Because an event occurs at a specific point in space and at a specific time, we can quantify when and where the event occurs by four numbers three that specify the location of the event in some three dimensional space coordinate system and one that that speifies what time the event occured. These four numbers are called the spacetime coordinates of the event.
Please have a look at
J.M. Levy "A simple derivation of the Lorentz transformation and the accompanying velocity and acceleration changes," Am. J. Phys. 75, 615 (2007) chapter III. Lorentz transformation along the x axis. Would you apply the same criticism to it?

 

[stevebd1]

Prtesent the Lorentz transformations as
dx=g(dx'+Vdt')
dt=g(dt'+Vdx'/cc)
In my opinion dx and dx' represent proper lengths measured in I and in I', dt and dt' representing coordinate time intervals. Do you aggree. Happy new year to all participamts on the Forum

Maybe I'm stating the obvious here but what you have there are Gullstrand-Painlevé rain coordinates- http://en.wikipedia.org/wiki/Gullstrand-Painlevé_coordinates#Rain_coordinates

 

[tiny-tim]

"proper length"

The distance between the origins O and O' , V(t-0) , and (x-0) are proper lengths. The distance (x'-0) detected from I' is a proper length as well, (t-0) being a coordinate time interval.

what does that mean?

 

[Fredrik]

"4. Events have nothing to do with coordinhates?"
I think that it is hard to continue to discuss with somebody who makes such a statement.
I quote from Moore:

Don't quote mine. That's a really lame debating tactic that's mostly used by people who don't know the subject well enough to come up with an actual argument (e.g. creationists "arguing" against evolution). You know too much about SR to resort to that.

I could have chosen my words more carefully, but the point is still that it makes no sense to talk about an event as something that happens in a coordinate system. Events are points in spacetime. Coordinate systems are functions that assign four numbers (coordinates) to events. What does the statement "Let E(x,t) and E'(x',t') be two events" even mean? Are you calling the first event E or E(x,t)? Why mention coordinates there? Did you mean E=(x,t) or not?

 

[bernhard.rothenstein]

what does that mean?

My initial question, giving rise to so much noise, was:
Present the Lorentz transformations as
(1-bb)^1/2(x-0)=(x'-0)+V(t'-0)
Being new in English terminology, I asked how would you call
(x-0), (x'-0) and V(t'-0) taking into account that all have the physical dimension of length.
Teaching special relativity is it necessary to make a distinction between them?

 

[tiny-tim]

My initial question, giving rise to so much noise, was:
Present the Lorentz transformations as
(1-bb)^1/2(x-0)=(x'-0)+V(t'-0)
Being new in English terminology, I asked how would you call
(x-0), (x'-0) and V(t'-0) taking into account that all have the physical dimension of length.
Teaching special relativity is it necessary to make a distinction between them?

(have a square-root: √ and a gamma: γ )

oh i see …

you're saying that in the equations

dx = γ(dx' + vdt')
dt = γ(dt' + vdx'/c²)​

x and vt both have dimensions of length, so as a matter of English is it proper to call them both lengths?

in other words, just as x is naturally a "proper" length, is vt also a "proper" length?

My answer would be that, to familiarise students with "space-time" and the interchangeability of space and time, and particularly the rotational nature of a Lorentz boost (which obviously requires like to be rotated onto like),

it's best to use ct and v/c …

dx = γ(dx' + (v/c)d(ct'))
d(ct) = γ((d(ct') + (v/c)dx')​

… in other words, to present ct as a length (rather than vt), and v/c as an ordinary number …

and indeed to avoid using a "naked" vt at all.

 

[bernhard.rothenstein]

(have a square-root: √ and a gamma: γ )

oh i see …

you're saying that in the equations

dx = γ(dx' + vdt')
dt = γ(dt' + vdx'/c²)​

x and vt both have dimensions of length, so as a matter of English is it proper to call them both lengths?

in other words, just as x is naturally a "proper" length, is vt also a "proper" length?

My answer would be that, to familiarise students with "space-time" and the interchangeability of space and time, and particularly the rotational nature of a Lorentz boost (which obviously requires like to be rotated onto like),

it's best to use ct and v/c …

dx = γ(dx' + (v/c)d(ct'))
d(ct) = γ((d(ct') + (v/c)dx')​

… in other words, to present ct as a length (rather than vt), and v/c as an ordinary number …

and indeed to avoid using a "naked" vt at all.

Nice to meet you on the forum. I think that is the elegant way to share knowledge when somebody solicitates it.
From the way in which you present the Lorentz transformations (I will use them in my work)
I think that we could conclude that dx, dx' and d(ct') are proper lengths or they are not?
If we make a distinction between the different types of time interval it is not clear fro me what kind of time intervals are dt and dt'?
With kind regards

 

[tiny-tim]

hi bernhard!

… I think that we could conclude that dx, dx' and d(ct') are proper lengths or they are not?

I really don't think you should use the word "proper" …

a length is a length, and you don't have to praise it as being a "genuine" length or a "proper" length …

and (ct) is also a length … just a length!

Not only is it unnecessary, but it also risks confusion with the "proper length", tau, along a curve.

I think students should be told "spacetime has four dimensions, x y z and ct, and they're all lengths"

If we make a distinction between the different types of time interval it is not clear fro me what kind of time intervals are dt and dt'?

I honestly don't think one should talk about t (or dt) in terms of intervals … after all, they don't have lengths, do they?

And what good is an interval without a length? ​

 

[bernhard.rothenstein]

Thanks. I quote from a textbook from which I have learned a lot:
"ds=dt/g
this very useful equation links the spacetime interval ds measured by an inertial clock present at two events with the coordinate time separation dt between those events in some inertial frame and the speed v of the clock as measured in the same inertial frame".
When applying this equation, it is important to remember two things. First of all, coordinate time dt and spacetime interval ds represent the time interval between two events measured in two fundamental different ways. The coordinate time between events is measured with a pair of synchronized clocks in an inertial frame, while the spacetime interval is measured by an inertial clock present at both events."
Is all that an out of fashion way to teach special relativity? Is there some danger of missinterpretation?
With thanks and respect

 

[tiny-tim]

Thanks. I quote from a textbook from which I have learned a lot:
"ds=dt/g

(what happened to that γ i gave you? )

Of course, that's with units in which c = 1 …

so speed is a dimensionless number, and length and time are dimensionally the same.

… When applying this equation, it is important to remember two things. First of all, coordinate time dt and spacetime interval ds represent the time interval between two events measured in two fundamental different ways. The coordinate time between events is measured with a pair of synchronized clocks in an inertial frame, while the spacetime interval is measured by an inertial clock present at both events."
Is all that an out of fashion way to teach special relativity? Is there some danger of missinterpretation?

Personally, I dislike the word "interval".

"Interval" in English isn't a measurement, it's a one-dimensional region.

I'd use the standard term "separation" … "the separation is measured by an inertial clock present at both events."

Apart from that, the explanation seems fine.

(Though it doesn't work where the separation is negative … )​