Question about full Lorentz transformation

[Chenkel]

Hello everyone,

I recently have learned about space time intervals and how these intervals between two events are invariant across all inertial frames and this can be proven by using the full Lorentz transformation.

I wanted to learn more about the full Lorentz transformation and I read the following on Wikipedia "where (t, x, y, z) and (t′, x′, y′, z′) are the coordinates of an event in two frames with the origins coinciding at t=t′=0, where the primed frame is seen from the unprimed frame as moving with speed v along the x-axis," Wikipedia

It seems to me (and I might be misunderstanding) that there is a convention that the primed frame is seen as the moving frame and the unprimed frame is seen as the rest frame.

I see in the Lorentz transformation on Wikipedia that the following is true for x equal to zero $t' = \gamma{t}$

How can t primed be in a moving frame and measure to be more than t unprimed? Doesn't this go against the theory that moving clock's tick slower than rest clocks?

I thought gamma usually multiplies the clock that is moving but I think I need maybe to look at things differently for the full Lorentz transformation to make sense.

Thanks in advance for any help on this matter.

 

[robphy]

I suggest first looking at the Euclidean analogue: https://en.wikipedia.org/wiki/Rotation_matrix

 

[Chenkel]

I suggest first looking at the Euclidean analogue: https://en.wikipedia.org/wiki/Rotation_matrix

How will understanding rotation matrices help me understand Lorentz transformations?

 

[Dale]

It seems to me (and I might be misunderstanding) that there is a convention that the primed frame is seen as the moving frame and the unprimed frame is seen as the rest frame.

No. The convention is that “the primed frame is … moving with speed $v$ along the $x$-axis” while the unprimed frame is moving with speed $-v$ along the $x’$-axis. The only difference is which is $v$ and which is $-v$, and that isn’t even a strong convention.

The primed and unprimed designations are deliberately intended to not convey the distinction of “rest” and “moving”.

I see in the Lorentz transformation on Wikipedia that the following is true for x equal to zero t′=γt

How can t primed be in a moving frame and measure to be more than t unprimed?

What happens if, instead of $x=0$ we have $x’=0$?

 

[robphy]

How will understanding rotation matrices help me understand Lorentz transformations?

The Lorentz Transformation is to Minkowski spacetime
as
the rotation matrix is to Euclidean space.

 

[Chenkel]

No. The convention is that “the primed frame is … moving with speed $v$ along the $x$-axis” while the unprimed frame is moving with speed $-v$ along the $x’$-axis. The only difference is which is $v$ and which is $-v$, and that isn’t even a strong convention.

The primed and unprimed designations are deliberately intended to not convey the distinction of “rest” and “moving”.

What happens if, instead of $x=0$ we have $x’=0$?

If $x' = 0$ then does that mean $x = vt$?

 

[Chenkel]

If $x' = 0$ then does that mean $x = vt$?

I took $x' = \gamma(x - vt)$ and set x' to zero and solved for x

 

[Chenkel]

What happens if, instead of x=0 we have x′=0?

I think this implies that the moving frame is technically the unprimed frame since x=vt (for all events where x' equals zero) which means the time dilation expression makes sense to me now.

 

[PeroK]

I think this implies that the moving frame is technically the unprimed frame since x=vt (for all events where x' equals zero) which means the time dilation expression makes sense to me now.

The first thing to note about the Lorentz Transformation is the symmetry between the primed and unprimed frames. This symmetry is physical and mathematical.

If the primed frame is moving with velocity $v$ relative to the unprimed frame, then (by implication) the unprimed frame is moving with velocity $-v$ relative to the primed frame. Neither frame is moving and neither frame is stationary. They are moving relative to each other.

This symmetry is also seen in the equations:
$$t' = \gamma(t - \frac{vx}{c^2}), \ \ x' = \gamma(x - vt), \ \ y' = y, \ \ z' = z$$$$t = \gamma(t' - \frac{(-v)x'}{c^2}), \ \ x = \gamma(x' - (-v)t'), \ \ y = y', \ \ z = z'$$Note that these two sets of equations are identical, except the first set has $v$ and the second set $-v$. And which has $v$ is purely a matter of how we choose the orient the positive x-axis. Although the second set is called the inverse Lorentz Transformation, there is no conceptual difference between the two. The first set is also the inverse of the second set. Finally, note that usually we cancel the two minus signs in the "inverse" transformation to give:
$$t = \gamma(t' + \frac{vx'}{c^2}), \ \ x = \gamma(x' +vt'), \ \ y = y', \ \ z = z'$$

 

[Ibix]

I think this implies that the moving frame is technically the unprimed frame since x=vt (for all events where x' equals zero) which means the time dilation expression makes sense to me now.

Neither frame is "the moving frame" in any absolute sense. Both frames are moving if you use the other frame. Both frames are stationary if you use the frame itself. This is kind of a key point.

If you set $x=0$ then you are describing a single point that is at rest in the unprimed frame. If you plug this into the Lorentz transforms you will find that $x'=-\gamma vt$ and $t'=\gamma t$, which you can combine to get $x'=-vt'$. That is, a clock that is at rest in the unprimed frame is moving in the primed frame, so you find that $t'=\gamma t$ along this line - this clock ticks slowly when described from the primed frame.

Conversely, if you set $x'=0$ then you are describing a single point that is at rest in the primed frame. If you plug this into the inverse Lorentz transforms you will find that $x=\gamma vt'$ and $t=\gamma t'$, which you can combine to get $x=vt$. That is, a clock that is at rest in the primed frame is moving in the unprimed frame, so you find that $t=\gamma t'$ along this line - this clock ticks slowly when described from the unprimed frame.

Note that my last paragraph is a word-for-word copy of the one before, just with "primed" and "unprimed" switched, "inverse" added before the Lorentz transforms, and the opposite sign on the velocity. I wrote it that way to emphasise the point in my first paragraph, also made by @PeroK while I was typing, that there is no difference between the primed and unprimed frame except for the sign on $v$. Both frames are "the stationary frame" by their own measure and clocks moving with respect to them tick slow by that frame's measure. Both frames are "the moving frame" by the other frame's measures, and clock at rest in them tick slowly using the other frame's measures.

 

[vanhees71]

I think this implies that the moving frame is technically the unprimed frame since x=vt (for all events where x' equals zero) which means the time dilation expression makes sense to me now.

It doesn't make sense to say the one or the other frame were "moving" or "at rest". There is no absolute notion of motion or being at rest. You can only observer, whether one body is moving against another body, and physically in the lab a frame of reference must somehow be specified by material bodies, e.g., three rods of equal length fastened at one point in space and being perpendicular to each other defining a Cartesian basis to describe positions and a set of clocks at rest relative to these rods and distributed over space and synchronized according to the Einstein synchronization formalism defines the corresponding four-vector components of Minkowski spacetime (an affine space with the Minkowski product as "fundamental form") in terms of a Lorentzian basis, which is the analogue to a Cartesian basis in an Euclidean affine space.

 

[Dale]

I see in the Lorentz transformation on Wikipedia that the following is true for x equal to zero $t′=γt$

What happens if, instead of $x=0$ we have $x’=0$?

For $x=0$ you get $t’=\gamma t$.

For $x’=0$ you get $t=\gamma t’$.

 

[Sagittarius A-Star]

I wanted to learn more about the full Lorentz transformation

I like, that you now do calculations with the Lorentz transformation. I recommend, that you calculate also some problems in Morin's book.

I think it would help to train your intuition regarding the Lorentz transformation, if you would do the calculation that I proposed at the end of this posting.

 

[Chenkel]

I like, that you now do calculations with the Lorentz transformation. I recommend, that you calculate also some problems in Morin's book.

I think it would help to train your intuition regarding the Lorentz transformation, if you would do the calculation that I proposed at the end of this posting.

I'll see if I can do it, thanks for the suggestion.

 

[Chenkel]

This symmetry is also seen in the equations:
t′=γ(t−vxc2), x′=γ(x−vt), y′=y, z′=zt=γ(t′−(−v)x′c2), x=γ(x′−(−v)t′), y=y′, z=z′Note that these two sets of equations are identical, except the first set has v and the second set −v. And which has v is purely a matter of how we choose the orient the positive x-axis. Although the second set is called the inverse Lorentz Transformation, there is no conceptual difference between the two. The first set is also the inverse of the second set. Finally, note that usually we cancel the two minus signs in the "inverse" transformation to give:

So in the Lorentz transformation a clock at rest in the unprimed frame is seen as moving relative to the primed frame?

And in the inverse Lorentz transformation a clock at rest in the primed frame is seen as moving relative to the unprimed frame?

 

[Nugatory]

So in the Lorentz transformation a clock at rest in the unprimed frame is seen as moving relative to the primed frame?

A clock at rest in the one frame will be moving in another frame. This has nothing to do with the Lorentz transformations, it's a property of frames that are moving relative to one another. Because the primed and unprimed frames are moving relative to one another a clock that is at rest in the primed frame will be moving in the unprimed frame and a clock that is at rest in the unprimed frame will be moving in the primed frame.

The Lorentz transformations are used to relate the coordinates used in one frame to the coordinates used in another frame. By convention we label the coordinates used in the primed frame x' and t' and the coordinate used in the unprimed frame x and t. The Lorentz transformation is used to calculate t' and x' when when we know t and x; the inverse Lorentz transformation is used to calculate t and x when we know t' and x'

 

[PeterDonis]

a clock at rest in the unprimed frame is seen as moving relative to the primed frame?

a clock at rest in the primed frame is seen as moving relative to the unprimed frame?

These statements have nothing to do with the Lorentz transformation in particular. They are true statements based solely on the definition of what different frames means.

The Lorentz transformation is a transformation of coordinates of events (points in spacetime) from one frame to another. In other words, if we just restrict to one spatial dimension, if you have coordinates $(t, x)$ of some event in the unprimed frame, you use the Lorentz transformation equations to get the coordinates $(t', x')$ of that event in the primed frame; and if you have coordinates $(t', x')$ of some event in the primed frame, you use the inverse Lorentz transformation equations to get the coordinates $(t, x)$ of that event in the unprimed frame.

You seem to be confusing yourself by using notations like $t$ to refer to the elapsed time on some clock. But to even talk about elapsed time, you need to look at at least two events, and you need to carefully distinguish their coordinates. For example, suppose we have a clock at rest at the spatial origin in the unprimed frame. Then, if we pick two events along the clock's worldline, say event A, coordinates $(t, x) = (0, 0)$ and event B, coordinates $(t, x) = (1, 0)$, the elapsed time on the clock between those events is $1$; we obtain that by computing the spacetime interval $\sqrt{(t_B - t_A)^2 - (x_B - x_A)^2}$, which is of course trivial in this frame.

However, the result we get is true in any frame; the elapsed time on a particular clock between two particular events on its worldline is an invariant. We can use the Lorentz transformation to check this by transforming the coordinates into the primed frame: we get event A coordinates $(t', x') = (0, 0)$ and event B coordinates $(t', x') = (\gamma, - \gamma v)$. Note that the $x'$ coordinate of event B is negative, indicating that the clock is moving to the left (the minus $x'$ direction) in the primed frame. We then compute $\sqrt{(t'_B - t'_A)^2 - (x'_B - x'_A)^2} = \sqrt{\gamma^2 \left( 1 - v^2 \right)} = 1$.

Note also, though, that the coordinate time in the primed frame between events A and B is not $1$--it's $\gamma$. In other words, for this clock, we have $\Delta t' = \gamma \Delta t$ (I have put the $\Delta$ in front of each time to make it clear that we are talking about differences in time between two events--notice that I had to do two Lorentz transformations, one for event A and one for event B, in order to get the difference in coordinate time between them in the primed frame). This corresponds to what you read in the Wikipedia article, that for $x = 0$ (i.e., for a clock at rest at the spatial origin of the unprimed frame), we have $t' = \gamma t$ (but without the $\Delta$ in front of each time it's less clear exactly what is meant).

I would recommend now looking at the case of a clock at rest at the spatial origin in the primed frame, i.e., with $x' = 0$, picking the events A, coordinates $(t', x') = (0, 0)$ (note that this is the same event A as above) and C, coordinates $(t', x') = (1, 0)$ (note that this is not the same as event B above) on its worldline, and seeing what you get for an elapsed coordinate time between those events in the unprimed frame.

 

[Chenkel]

These statements have nothing to do with the Lorentz transformation in particular. They are true statements based solely on the definition of what different frames means.

The Lorentz transformation is a transformation of coordinates of events (points in spacetime) from one frame to another. In other words, if we just restrict to one spatial dimension, if you have coordinates $(t, x)$ of some event in the unprimed frame, you use the Lorentz transformation equations to get the coordinates $(t', x')$ of that event in the primed frame; and if you have coordinates $(t', x')$ of some event in the primed frame, you use the inverse Lorentz transformation equations to get the coordinates $(t, x)$ of that event in the unprimed frame.

You seem to be confusing yourself by using notations like $t$ to refer to the elapsed time on some clock. But to even talk about elapsed time, you need to look at at least two events, and you need to carefully distinguish their coordinates. For example, suppose we have a clock at rest at the spatial origin in the unprimed frame. Then, if we pick two events along the clock's worldline, say event A, coordinates $(t, x) = (0, 0)$ and event B, coordinates $(t, x) = (1, 0)$, the elapsed time on the clock between those events is $1$; we obtain that by computing the spacetime interval $\sqrt{(t_B - t_A)^2 - (x_B - x_A)^2}$, which is of course trivial in this frame.

However, the result we get is true in any frame; the elapsed time on a particular clock between two particular events on its worldline is an invariant. We can use the Lorentz transformation to check this by transforming the coordinates into the primed frame: we get event A coordinates $(t', x') = (0, 0)$ and event B coordinates $(t', x') = (\gamma, - \gamma v)$. Note that the $x'$ coordinate of event B is negative, indicating that the clock is moving to the left (the minus $x'$ direction) in the primed frame. We then compute $\sqrt{(t'_B - t'_A)^2 - (x'_B - x'_A)^2} = \sqrt{\gamma^2 \left( 1 - v^2 \right)} = 1$.

Note also, though, that the coordinate time in the primed frame between events A and B is not $1$--it's $\gamma$. In other words, for this clock, we have $\Delta t' = \gamma \Delta t$ (I have put the $\Delta$ in front of each time to make it clear that we are talking about differences in time between two events--notice that I had to do two Lorentz transformations, one for event A and one for event B, in order to get the difference in coordinate time between them in the primed frame). This corresponds to what you read in the Wikipedia article, that for $x = 0$ (i.e., for a clock at rest at the spatial origin of the unprimed frame), we have $t' = \gamma t$ (but without the $\Delta$ in front of each time it's less clear exactly what is meant).

I would recommend now looking at the case of a clock at rest at the spatial origin in the primed frame, i.e., with $x' = 0$, picking the events A, coordinates $(t', x') = (0, 0)$ (note that this is the same event A as above) and C, coordinates $(t', x') = (1, 0)$ (note that this is not the same as event B above) on its worldline, and seeing what you get for an elapsed coordinate time between those events in the unprimed frame.

I had a little difficulty following the math but I will read again a little later.

But your distinction made between what the Lorentz transformation does and what the inverse Lorentz transformation does I found useful.

 

[Chenkel]

In the Lorentz transformation (non inverse) is the observer at rest relative to the primed frame?

This seems to be the only way I can make sense of it.

If $x' = 0$ we find $x = vt$ which means that the primed frame is moving along the positive x axis at velocity v in the Lorentz transformation.

But we also find that $t' = \gamma{t}$ which must mean that the observer is in the primed frame observing events.

Furthermore for the inverse Lorentz transformation we find that for $x = 0$ we find that $x' = -vt$ that is in the inverse transformation the unprimed frame is moving in the negative direction of this x axis at velocity v, but we also have $t = \gamma{t'}$ which means the observer of events must be at rest in the unprimed frame for the inverse Lorentz transformation.

Let me know if my math makes sense, thank you!

 

[PeroK]

In the Lorentz transformation (non inverse) is the observer at rest relative to the primed frame?

This seems to be the only way I can make sense of it.

If $x' = 0$ we find $x = vt$ which means that the primed frame is moving along the positive x axis at velocity v in the Lorentz transformation.

But we also find that $t' = \gamma{t}$ which must mean that the observer is in the primed frame observing events.

Furthermore for the inverse Lorentz transformation we find that for $x = 0$ we find that $x' = -vt$ that is in the inverse transformation the unprimed frame is moving in the negative direction of this x axis at velocity v, but we also have $t = \gamma{t'}$ which means the observer of events must be at rest in the unprimed frame for the inverse Lorentz transformation.

Let me know if my math makes sense, thank you!

There is no observer involved in the Lorentz transformation, which is a transformation of coordinates between reference frames.

 

[Chenkel]

There is no observer involved in the Lorentz transformation, which is a transformation of coordinates between reference frames.

But don't we need to an observer to record events in the reference frame? Furthermore doesn't the observer have to be at rest with respect to a specific reference frame?

 

[Ibix]

But don't we need to an observer to record events in the reference frame?

Not really. Things are what they are and they can be described without anyone to record them. Suppose you take a shot on a pool table. You're the only one in the room, and you close your eyes. The balls won't suddenly fly off and do anything unexpected just because there's no observer. You can still write down the paths they took.

Furthermore doesn't the observer have to be at rest with respect to a specific reference frame?

It is trivially true that there is a reference frame in which any arbitrarily chosen object is at rest. You are not required to work with that frame and it is quite common not to. For example, we typically say that we were driving at 30mph, using the rest frame of the Earth's surface even though we are moving in that frame.

 

[PeroK]

But don't we need to an observer to record events in the reference frame? Furthermore doesn't the observer have to be at rest with respect to a specific reference frame?

No. There are no observers at rest relative to the Sun, but that didn't stop Copernicus or Kepler figuring out the heliocentric model.

Moreover, the Lorentz transformation can be seen as a mathematical object, used to describe the nature of spacetime in our universe.

 

[Dale]

But don't we need to an observer to record events in the reference frame? Furthermore doesn't the observer have to be at rest with respect to a specific reference frame?

A lot of times people setting up “thought experiments” will spend a lot of time talking about observers. But they are not part of the math of relativity.

The principle of relativity says that all inertial frames are equivalent. So you can use any inertial frame you like. You, as an observer, are allowed to use a reference frame where you are at rest. But you are equally allowed to use a frame where you are not at rest.

You can say “I went to the store” you are not required to say “the store came to me”.

The Lorentz transform relates a primed frame and an unprimed frame that move with respect to each other at $v$. There is no specification of one at rest and one as moving. They are each at rest with respect to themselves. They are each moving with respect to the other.

 

[Chenkel]

Not really. Things are what they are and they can be described without anyone to record them. Suppose you take a shot on a pool table. You're the only one in the room, and you close your eyes. The balls won't suddenly fly off and do anything unexpected just because there's no observer. You can still write down the paths they took.

It is trivially true that there is a reference frame in which any arbitrarily chosen object is at rest. You are not required to work with that frame and it is quite common not to. For example, we typically say that we were driving at 30mph, using the rest frame of the Earth's surface even though we are moving in that frame.

$t' = \gamma{t}$ and
$t = \gamma{t'}$ are not true simultaneously, the first equation works with the events being recorded relative to an event recorder at rest in the primed frame, and the 2nd equation works with events being recorded where the event recorder is at rest relative to the unprimed frame.

Notice the first equation is used in the Lorentz transformation and the 2nd equation is used in the inverse Lorentz transformation.

Instead of calling the thing an "observer" I'm calling it an "event recorder"

 

[PeroK]

$t' = \gamma{t}$ and
$t = \gamma{t'}$ are not true simultaneously, the first equation works with the events being recorded relative to an event recorder at rest in the primed frame, and the 2nd equation works with events being recorded where the event recorder is at rest relative to the unprimed frame.

Notice the first equation is used in the Lorentz transformation and the 2nd equation is used in the inverse Lorentz transformation.

Instead of calling the thing an "observer" I'm calling it an "event recorder"

Note that in the Lorentz transformation $t, x$ are general coordinates of an arbitrary event. If you are dealing with a specific event or series of events, you need to identify the coordinates in some way. E.g. $t_A, x_A$.

Something like $t = \gamma t'$ strictly speaking makes no sense. Whereas,$t_A = \gamma t'_A$ may hold for a specific event A.

 

[PeroK]

Or, you could write $t = \gamma t' \ (x'=0)$

 

[FactChecker]

But don't we need to an observer to record events in the reference frame? Furthermore doesn't the observer have to be at rest with respect to a specific reference frame?

You should think of an "observer" as an inertial reference frame that can record the position of events in its spacetime coordinate system, no matter where or when they occur.
This is so important for keeping SR simple that Bernard Schutz describes it immediately on page 3 of "A First Course in General Relativity":

"1.2 Definition of an inertial observer in SR
It is important to realize that an 'observer' is in fact a huge information-gathering system, not simply one person with binoculars. In fact, we shall remove the human element entirely from our definition and say that an inertial observer is simply a spacetime coordinate system, which makes an observation simply by recording the location (x, y, z) and the time (t) of any event. ...."

I recommend reading the first chapter of his book because, IMO, it addresses many misconceptions that people have about SR.

 

[Ibix]

$t' = \gamma{t}$ and
$t = \gamma{t'}$ are not true simultaneously

You are measuring different things in your two equations. That's why they are different, not anything to do with observers.

 

[Chenkel]

You are measuring different things in your two equations. That's why they are different, not anything to do with observers.

I agree that they're measuring different things, but maybe that's because observers in different reference frames will measure different things?

 

[Sagittarius A-Star]

$t' = \gamma{t}$ and
$t = \gamma{t'}$
...
Notice the first equation is used in the Lorentz transformation and the 2nd equation is used in the inverse Lorentz transformation.

You can i.e. derive the first equation also with the inverse Lorentz transformation. It is only a little bit more complicated.

Inverse LT:
$x=\gamma(x'+vt') \ \ \ \ \ (1)$
$t=\gamma(t'+vx'/c^2) \ \ \ \ \ (2)$

The clock is at rest in the unprimed frame at $x=0$.
With equation (1) follows:
$\require{color} 0=\gamma(x'+vt') \ \ \ \ \ (3)$

To eliminate $x'$ in equation (3), I solve equation (2) for $x'$...
$t=\gamma(t'+vx'/c^2)$
${t \over \gamma}-t' = vx'/c^2$
$x'=\color{blue}{c^2 \over v}({t\over \gamma}-t')\color{black} \ \ \ \ \ (4)$
... and put the right side of equation (4) for $x'$ into equation (3):

$0=\gamma(\color{blue}{c^2 \over v}({t\over \gamma}-t')\color{black}+vt')$
$0={c^2 \over v}({t\over \gamma}-t')+vt'$
$0={c^2t\over \gamma v}-t'{c^2 \over v}+vt'$
$t= {\gamma v\over c^2 }({c^2 \over v} -v)t'$
$t= \gamma(1-{v^2\over c^2})t'$
$$t' = \gamma{t} \ \ \ \ \ (\text{condition:}\ x=0)$$

 

[Dale]

$t' = \gamma{t}$ and
$t = \gamma{t'}$ are not true simultaneously

I agree with this.

the first equation works with the events being recorded relative to an event recorder at rest in the primed frame, and the 2nd equation works with events being recorded where the event recorder is at rest relative to the unprimed frame

I do not agree with this. The first equation works when $x=0$. The second equation works when $x'=0$.

There is no event recorder in the equations.

Instead of calling the thing an "observer" I'm calling it an "event recorder"

I don't think that this change in terminology is helpful. You may as well call it an observer like everyone else.

 

[PeroK]

You can i.e. derive the first equation also with the inverse Lorentz transformation. It is only a little bit more complicated.

Inverse LT:
$x=\gamma(x'+vt') \ \ \ \ \ (1)$
$t=\gamma(t'+vx'/c^2) \ \ \ \ \ (2)$

The clock is at rest in the unprimed frame at $x=0$.
With equation (1) follows:
$0=\gamma(x'+vt') \ \ \ \ \ (3)$

To eliminate $x'$ in equation (3), I solve equation (2) for $x'$...
$t=\gamma(t'+vx'/c^2)$
${t \over \gamma}-t' = vx'/c^2$
$x'={c^2 \over v}({t\over \gamma}-t') \ \ \ \ \ (4)$
... and put the right side of equation (4) for $x'$ into equation (3):

$0=\gamma({c^2 \over v}({t\over \gamma}-t')+vt')$
$0={c^2 \over v}({t\over \gamma}-t')+vt'$
$0={c^2t\over \gamma v}-t'{c^2 \over v}+vt'$
$t= {\gamma v\over c^2 }({c^2 \over v} -v)t'$
$t= \gamma(1-{v^2\over c^2})t'$
$$t' = \gamma{t}$$

That's fine, except that it doesn't formally distinguish between general coordinates and a specific coordinate. This is a shorthand that most of us use. But, if the student has yet to learn or understand this distinction, then it can't resolve that confusion.

 

[Ibix]

I agree that they're measuring different things, but maybe that's because observers in different reference frames will measure different things?

Can't be, since both are in both reference frames.

They're measuring different things because you chose that's what they should measure. You need to have set up two reference systems in order to make the measurements of $t$ and $t'$, but nothing obliges anybody to use the system in which they are at rest. Nothing stops me looking at clocks and rulers at rest with respect to you, and we've given you multiple examples of this already.

 

[Histspec]

$t' = \gamma{t}$ and
$t = \gamma{t'}$ are not true simultaneously, the first equation works with the events being recorded relative to an event recorder at rest in the primed frame, and the 2nd equation works with events being recorded where the event recorder is at rest relative to the unprimed frame.

In $t' = \gamma{t}$, variable t is the "proper" time between two events measured in S by clock C1 at rest in the same frame, whereas t' is the "coordinate" time measured in frame S' with respect to which clock C1 is moving.
In $t = \gamma{t'}$, variable t' is the "proper" time between two events measured in S' by clock C2 at rest in the same frame, whereas t is the "coordinate" time measured in frame S with respect to which clock C2 is moving.

As you can see, the equations describe two different experiments.