Understanding Lorentz Transformations in Special Relativity

[Kaguro]

TL;DR Summary

x' = gamma*(x-vt). It infers that distance of event point is farther the faster you are.

The Lorentz tranformations are:
$x' = \gamma (x-vt)$
$t' = \gamma(t - \frac{vx}{c^2})$

Consider an event (x,t) happening in S frame. Let S' frame be moving w.r.t. S frame along x direction with speed v whose origins coincide at t=0.
We find that the new coordinates of this event are (x',t').

Classically, x'=x-vt. It makes sense. Time is absolute and so the more you wait (in S'), the closer it appears to the origin of the moving frame.In Relativity, this amount is increased by a factor $\gamma$.
Say the event happens at (x,0)
So, $x' = \gamma x$

What's the purpose of this? What happens physically here?

Then considering time,

t' = $\gamma(0 - vx/c^2)$
It's negative. Why negative?

Please help me guide through this. I already studied STR many times and still have not been able to visualize what actually happens..

 

[PeroK]

In Relativity, this amount is increased by a factor $\gamma$.
Say the event happens at (x,0)
So, $x' = \gamma x$

What's the purpose of this? What happens physically here?

Then considering time,

t' = $\gamma(0 - vx/c^2)$
It's negative. Why negative?

Please help me guide through this. I already studied STR many times and still have not been able to visualize what actually happens..

The event $(x, 0)$ represents a clock at position $x$ at time $t = 0$.

If we have similar synchronised clocks in the $S'$ frame, then these clocks are not (and cannot be) synchronised in the $S$ frame. We have no absolute simultaneity and, indeed, a "leading clock lags" rule, which is encapsulated by the Lorentz Transformation.

In particular, as we proceed along the positive $x/x'$ axis, the clocks in the two frames are progressively further out of synchronisation. When an x-clock shows $t =0$, the x'-clock will show a time before $t' = 0$. In the $S'$ frame, therefore, the x-clocks are progressively further ahead - and it's the x-clocks on the negative x-axis that are leading and lagging.

In terms of the $x'$ coordinate, again you have to look at the simulaneity issue. The event $(x, 0)$ for positive $x$ occurs at negative time in the $S'$ frame: before the origins have coincided, while the origin of $S'$ is still to the left of the origin of $S$. That's why, when you do the calculation, you end up with $x' = \gamma x > x$.

 

[Sagittarius A-Star]

Summary:: x' = gamma*(x-vt). It infers that distance of event point is farther the faster you are.

Please help me guide through this.

Maybe, it is easier to understand, if you note down the differences between the coordinates of 2 events:

The Lorentz tranformations are:
(1) $\ \ \ \ \Delta x' = x_2' - x_1' = \gamma (x_2-vt_2) - \gamma (x_1-vt_1) = \gamma (\Delta x-v \Delta t)$
(2) $\ \ \ \ \Delta t' = t_2' - t_1' =\gamma(t_2 - \frac{v}{c^2}x_2) - \gamma(t_1 - \frac{v}{c^2}x_1) = \gamma(\Delta t - \frac{v}{c^2}\Delta x)$

Then you get:

• Length contraction for $\Delta t = 0$, in equation (1):

$\Delta x'/ \gamma= \Delta x$​

• Time dilation for $\Delta x = 0$, in equation (2):

$\Delta t' / \gamma= \Delta t$​

• Relativity of "simultaneity" for $\Delta t = 0$, in equation (2):

$\Delta t' = \gamma(- \frac{v}{c^2}\Delta x)$​

• Relativity of "same location" for $\Delta x = 0$, in equation (1):

$\Delta x' = \gamma (-v \Delta t)$​

​

Explanation for length contraction: Assume a rod in frame $S'$. You have a ruler at rest in frame $S$. Then you must check the x-coordinates of both ends of the moving rod at the same time, therefore $\Delta t = 0$.

Explanation for time dilation: Assume for example the transversal Doppler effect of a moving optical clock. Both measurement events are in frame $S$ at the same x-coordinate, therefore $\Delta x = 0$.

 

[PeroK]

Let's do the calculation. Let's consider a portion of the x-axis between $x = 0$ and $x = L$ and describe its motion in the $S'$ frame. Let's call these points $P$ (x-origin) and $Q$ (point $x = L$). The motion of each in the $S'$ frame is (remember that $PQ$ is moving to the left in $S'$ and is length contracted):
$$P(t') = (-vt', t'), \ \ Q(t') = (\frac L \gamma - vt', t')$$
Now, let's add the time shown by a clock at $P$ and $Q$, where these clocks show the time in the $S$ frame. The one at the origin is simple dilated: $t_P(t') = \frac{ t'}{ \gamma}$. And the one at $Q$ is dilated and not synchronised with the clock at $P$: it's ahead by $+\frac{vL}{c^2}$. We can add those to show the time recorded at each point in its motion in the $S'$ frame:
$$P(t') = (-vt', t');[\frac{t'}{\gamma}], \ \ Q(t') = (\frac L \gamma - vt', t');[\frac{t'}{\gamma} + \frac{vL}{c^2}]$$
Now we can ask where is the clock at $Q$ when it reads zero? We have $t' = -\frac{\gamma vL}{c^2}$ and its position is: $$x' = \frac L \gamma - vt' = \frac L \gamma + \frac{\gamma v^2L}{c^2} = \gamma L(\frac{1}{\gamma^2} + \frac{v^2}{c^2}) = \gamma L$$ And we have hand-cranked what came out easily from the Lorentz Transformation.

 

[Sagittarius A-Star]

The Lorentz tranformations are:
(1) $\ \ \ \ \Delta x' = x_2' - x_1' = \gamma (x_2-vt_2) - \gamma (x_1-vt_1) = \gamma (\Delta x-v \Delta t)$
(2) $\ \ \ \ \Delta t' = t_2' - t_1' =\gamma(t_2 - \frac{v}{c^2}x_2) - \gamma(t_1 - \frac{v}{c^2}x_1) = \gamma(\Delta t - \frac{v}{c^2}\Delta x)$

@Kaguro : The spatial distance $\Delta x$ and the temporal distance $\Delta t$ between events are not invariant under Lorentz transformation. If you limit yourself for simplicity to events with $y=0$ and $z=0$, then you can define the "spacetime interval squared" as:
${\Delta s}^2 = c^2 {\Delta t}^2 - {\Delta x}^2$
Using the above Lorentz transformation, you can show, that this is equal to:
${\Delta s'}^2 = c^2 {\Delta t'}^2 - {\Delta x'}^2 (={\Delta s}^2)$

To show, that also the 4-dimensional spacetime interval squared
$c^2 {\Delta t}^2 - {\Delta x}^2 - {\Delta y}^2 - {\Delta z}^2$ is invariant, you can use the additional Lorentz transformations $\Delta y' = \Delta y$ and $\Delta z' = \Delta z$.

 

[Sagittarius A-Star]

• Time dilation for $\Delta x = 0$, in equation (2):

$\Delta t' / \gamma= \Delta t$​

...
Explanation for time dilation: Assume for example the transversal Doppler effect of a moving optical clock. Both measurement events are in frame $S$ at the same x-coordinate, therefore $\Delta x = 0$.

Correction to the explanation of time dilation in my posting #3 :

Explanation for time dilation: The clock is at rest in frame $S$ ($\Delta x = 0$). Ticks of the clock are events at the same x-coordinate in this frame. Descibed in frame $S'$, it ticks slower than the coordinate time $t'$.

 

[Guillermo Navas]

Summary:: x' = gamma*(x-vt). It infers that distance of event point is farther the faster you are.

The Lorentz tranformations are:
$x' = \gamma (x-vt)$
$t' = \gamma(t - \frac{vx}{c^2})$

Consider an event (x,t) happening in S frame. Let S' frame be moving w.r.t. S frame along x direction with speed v whose origins coincide at t=0.
We find that the new coordinates of this event are (x',t').

Classically, x'=x-vt. It makes sense. Time is absolute and so the more you wait (in S'), the closer it appears to the origin of the moving frame.In Relativity, this amount is increased by a factor $\gamma$.
Say the event happens at (x,0)
So, $x' = \gamma x$

What's the purpose of this? What happens physically here?

Then considering time,

t' = $\gamma(0 - vx/c^2)$
It's negative. Why negative?

Please help me guide through this. I already studied STR many times and still have not been able to visualize what actually happens..

Summary:: x' = gamma*(x-vt). It infers that distance of event point is farther the faster you are.

The Lorentz tranformations are:
$x' = \gamma (x-vt)$
$t' = \gamma(t - \frac{vx}{c^2})$

Consider an event (x,t) happening in S frame. Let S' frame be moving w.r.t. S frame along x direction with speed v whose origins coincide at t=0.
We find that the new coordinates of this event are (x',t').

Classically, x'=x-vt. It makes sense. Time is absolute and so the more you wait (in S'), the closer it appears to the origin of the moving frame.In Relativity, this amount is increased by a factor $\gamma$.
Say the event happens at (x,0)
So, $x' = \gamma x$

What's the purpose of this? What happens physically here?

Then considering time,

t' = $\gamma(0 - vx/c^2)$
It's negative. Why negative?

Please help me guide through this. I already studied STR many times and still have not been able to visualize what actually happens..

In the figure, the event A is simultaneous to B in the reference frame S' while A is simultaneous to O in the frame S. (The line of simultaneity is obtained by drawing a line parallel to the X axis of the reference system).
Event $A\equiv (x,0)_S=(\gamma x,-\gamma \frac v {c^2}x)_{S'}$
Event $B\equiv (-{\gamma}^2 {(\frac v c)}^2 x,-{\gamma}^2 \frac v {c^2}x)_S=(0,-\gamma \frac v {c^2}x)_{S'}$

 

[PeroK]

I am wondering why all these replies only discuss Lorentz transformations in 1+1 spacetime dimensions. That is the easy bit. The problems in understanding arise in 2+1 dimensions, and even more so in 3+1 dimensions. I understand Lorentz transformations in 1+1 dimensions, where we are talking about 2 independent observers. I struggle in 2+1 dimensions, where we have three independent observers. In 3+1 dimensions, with four independent observers, I cannot make any sense of the Lorentz group. And I am a group theorist.

In general, a Lorentz Transformation $\Lambda$ satisfies: $$\Lambda^T \eta \Lambda = \eta$$ where $\eta$ is the Minkowski metric.

The Lorentz group is generated by boosts and rotations: the product of two boosts is, in general, a boost plus a rotation. Thinking about it in terms of "observers" is generally not a good idea (even in 1 spatial dimension). It's really a relationship between coordinate systems.

 

[robwilson]

It's simple pragmatism on the part of the physicists, who often are interested in the math only to the extent that it is needed to effectively model the universe around them. As you say, the 1+1 case is easy - and that plus an assumption of isotropy, a coordinate rotation, and a wave of the hands is enough to work with problems in the locally flat space of special relativity.

Yes, indeed. It is the wave of the hands that worries me! It may work, it may not. My impression is that it doesn't work. I may be wrong.

 

[robphy]

I don't understand why you talk about 3 observers in 2+1 spacetime or 4 observers in 3+1 spacetime.

I also find this puzzling.
(By the way, I think this recent conversation has been forked off to another thread: https://www.physicsforums.com/threads/lorentz-transformations-1-1-spacetime-only.1000831/.)

 

[robwilson]

The most general Lorentz transformation can be represented as
$$\left(
\begin{array}{c|c}
\gamma & -\gamma \textbf{v}^T / c\\
\hline
-\gamma \textbf{v} / c & \textbf{I} + (\gamma - 1) \textbf{v} \textbf{v}^T / (\textbf{v}^T\textbf{v}) \\
\end{array}
\right)
\left(
\begin{array}{c|c}
1 & \textbf{0}^T \\
\hline
\textbf{0} & \textbf{U} \\
\end{array}
\right)
$$ where

• $\textbf{v}$ is the velocity of one observer relative to another, as a 3×1 column vector
• $\gamma$ is the usual Lorentz factor $1 / \sqrt{1 - \textbf{v}^T\textbf{v} / c^2}$
• $\textbf{U}$ is any orthogonal 3×3 matrix, representing a rotation of one observer's spatial axes relative to the other

(You can verify it by considering the special case $\textbf{v} = (v, 0, 0)^T$.)

That transformation represents a transformation between a pair of observers' coordinates. The whole group represents all possible transformations between pairs of observers, an infinite number of observers in total. I don't understand why you talk about 3 observers in 2+1 spacetime or 4 observers in 3+1 spacetime.

The essential point is that two observers cannot explore 2+1 spacetime. Or, they can only do so by making assumptions about what happens in the third dimension, that they cannot visit. Hence whatever theory they come up with can be tested only indirectly, not directly. They need to be able to communicate with a third observer in the third dimension in order to verify that their theoretical assumptions about what this observer observes are actually true in reality.

 

[DrGreg]

The essential point is that two observers cannot explore 2+1 spacetime. Or, they can only do so by making assumptions about what happens in the third dimension, that they cannot visit. Hence whatever theory they come up with can be tested only indirectly, not directly. They need to be able to communicate with a third observer in the third dimension in order to verify that their theoretical assumptions about what this observer observes are actually true in reality.

I've now moved my post to the other thread https://www.physicsforums.com/threads/lorentz-transformations-1-1-spacetime-only.1000831/ . Please continue the conversation there.

 

[Kaguro]

I'm sorry I am 1 week late. I had some exams. The first thing I see after coming back to the thread is complicated discussions about tensors and groups.

I'll try to visualize what happens through spacetime diagrams.

I imagine the event happening at some (x,0) in S frame.
This means that the light coming from the event reached origin of S at t=0 and the observer there noted the position as x.

The diagram says that:
In S' frame, the light reached the origin faster and since it was far behind, the position noted, x' was also greater than x.

But why did the light from the event reach S' faster? The origin of S' has not even crossed the origin of S yet, and the light should have passed the observer closer to it first.

Please explain me in terms of light.

 

[Kaguro]

Thinking about it in terms of "observers" is generally not a good idea (even in 1 spatial dimension).

Oh no...

I must try to understand basic STR. Today is Dr. Einstein's birthday! I must work hard.

 

[aperakh]

Summary:: x' = gamma*(x-vt). It infers that distance of event point is farther the faster you are.

What's the purpose of this? What happens physically here?

A lot of people already left [what I am sure are] good mathematical answers. But it sounds like what you were looking for is an intuitive explanation.
More, specifically let me try to answer your two actual question:

> What's the purpose of this?

There is no "purpose". Lorentz transformation and STR are description, nothing more. They describe, or offer one possible self-consistent model of, the experiment-derived realities.
The universe, on the other hand, simply is. It's properties don't have a purpose. So there is no way to really meaningfully answer that.
I'm guessing you actually already know this. So perhaps it was just an imprecise choice of words. The second question it, perhaps, more relevant.

> What happens physically here?

One way to answer is: "nothing". Nothing physically actually happens either to the moving object nor to the space around it. There is no physical change of some sort as such.
What changes are perceptions.
When object A accelerates relative to object B, B's frame of reference, that is its perception of A and anything in A's frame, changes.
Likewise A's perception of B and anything in B's frame changes.

Among those changes is the fact that B now perceived A as being squished shorter along the line of it's motion.
But either party perceives NO CHANGE it it's own size.

Now, saying that A appears to get shorter relative to the space it's traveling through is equivalent to saying that for A (who perceives no change in its own size) space appears to get longer. Which is what you essentially pointed out in your original post. It is just a another way to state length contraction.

But nothing actually "physically happens" to either of them or the space around them. Only their perception each other and A's perception of space changes.
Thus it is meaningless to ask what actually physically happens to them.

Hope it's what you were asking.

 

[Kaguro]

So perhaps it was just an imprecise choice of words

Yes, it was..
I have read everyone's answers. Now I need to take a step back and ask a few more questions.
I am having a hard time placing my foot on something concrete. I hope I don't annoy anyone by asking possibly redundant questions. Please bear with me.

 

[Kaguro]

The problem is with synchronization of clocks. An example from Kleppner's book says:

A person takes two clocks synchronized with each other and places one at the moon. Then comes back to Earth and tries to see the moon clock from a telescope.
The person sees the clock on moon delayed by 1 second since it takes light that much time to reach the earth.

Solution: Advance the moon clock by 1s. Now the person can see that the moon clock and the Earth clock are agreeing with each other. This is called synchronization of clocks.

-------------------------

Now back to my original question. After synchronization, the observer S at t=0 notes that a certain event E is happening at moon.

Now this means that in his frame, the light from moon event must have left moon earlier and has reached him only later. To notify that he has seen the event, the observer also emits another signal O at (x=0,t=0).There's another frame S' coming from behind moving with speed v relative to S. We tell him to set his clock at t'=0 when he reaches the origin of S(and sees his event O). So in S' frame O is at (0,0).

Now according to the Lorentz transformations, the coordinates of E are (γx, -γxv/c^2).
This means that the light from E reached S' when the clock of S' read t'= - γxv/c^2. And the position noted was also farther.

How is this possible? The origin of S' only reaches the origin of S at t'=0. But he supposedly has noted the event E earlier. So light would have reached him before he has reached S. This is strange.

 

[PeroK]

The problem is with synchronization of clocks. An example from Kleppner's book says:

A person takes two clocks synchronized with each other and places one at the moon. Then comes back to Earth and tries to see the moon clock from a telescope.
The person sees the clock on moon delayed by 1 second since it takes light that much time to reach the earth.

Solution: Advance the moon clock by 1s. Now the person can see that the moon clock and the Earth clock are agreeing with each other. This is called synchronization of clocks.

-------------------------

Now back to my original question. After synchronization, the observer S at t=0 notes that a certain event E is happening at moon.

Now this means that in his frame, the light from moon event must have left moon earlier and has reached him only later. To notify that he has seen the event, the observer also emits another signal O at (x=0,t=0).There's another frame S' coming from behind moving with speed v relative to S. We tell him to set his clock at t'=0 when he reaches the origin of S(and sees his event O). So in S' frame O is at (0,0).

Now according to the Lorentz transformations, the coordinates of E are (γx, -γxv/c^2).
This means that the light from E reached S' when the clock of S' read t'= - γxv/c^2. And the position noted was also farther.

How is this possible? The origin of S' only reaches the origin of S at t'=0. But he supposedly has noted the event E earlier. So light would have reached him before he has reached S. This is strange.

First, I don't think K&K is a good source to learn SR.

Second, SR has nothing to do with the delay in light signals reaching an observer.

Third, the "synchronisation" of clocks described above is not compatible with the Lorentz Transformation (LT). With the LT it is assumed that an observer on Earth would "see" the moon clock $1s$ behind. The LT uses the Einstein clock synchronisation. If a clock is at rest $1$ light second away and synchronised with a local clock (in your reference frame), then you must "see" it $1s$ behind.

You've got some of the basic ideas mixed up here.

 

[vanhees71]

I recommend to read the "kinematical part" of Einstein's original paper of 1905. There it's clearly explained how to construct the special-relativistic spacetime via gedanken experiments which are (in principle) really feasible in the lab.

https://www.fourmilab.ch/etexts/einstein/specrel/www/

 

[Kaguro]

I recommend to read the "kinematical part" of Einstein's original paper of 1905. There it's clearly explained how to construct the special-relativistic spacetime via gedanken experiments which are (in principle) really feasible in the lab.

https://www.fourmilab.ch/etexts/einstein/specrel/www/

I'll read this.

Thank you.

 

[Sagittarius A-Star]

View attachment 279714

I imagine the event happening at some (x,0) in S frame.
This means that the light coming from the event reached origin of S at t=0 and the observer there noted the position as x.

No. The light cannot be emitted at t=0 and received at a different location at t=0. Then it would have an infinite speed. Because the light speed is c, the world lines of light have an orientation of (+/-) 45° in the Minkowski diagram.

 

[Sagittarius A-Star]

The problem is with synchronization of clocks. An example from Kleppner's book says:

A person takes two clocks synchronized with each other and places one at the moon. Then comes back to Earth and tries to see the moon clock from a telescope.
The person sees the clock on moon delayed by 1 second since it takes light that much time to reach the earth.

Solution: Advance the moon clock by 1s. Now the person can see that the moon clock and the Earth clock are agreeing with each other. This is called synchronization of clocks.

Kleppner explicitely wrote, not to use such a synchronization in SR, see chapter "12.4.1 Synchronizing Clocks" in his book "An Introduction to Mechanics":
Source (Google books via Firefox)

You should read the chapter completely, not only the beginning.

 

[PeroK]

@Kaguro I would recommend talking at look at Morin's book. The first chapter is here:

https://scholar.harvard.edu/files/david-morin/files/relativity_chap_1.pdf

 

[Kaguro]

Kleppner explicitely wrote, not to use such a synchronization in SR, see chapter "12.4.1 Synchronizing Clocks" in his book "An Introduction to Mechanics":
Source (Google books via Firefox)

You should read the chapter completely, not only the beginning.

Oh... Sorry.

 

[Kaguro]

I'll read both David Morin's book and Einstein's paper and will ask questions in a few days.