# B Understanding the Final Form of Lorentz Transformations

#### NoahsArk

Gold Member
In my last post I asked about the general form of the Lorentz Transformation for time. Now I am trying to understand the final form of it, and how it makes sense based on what's happening physically. The final form for t is:

t = γt1 + (γv/c2/)x1

It's the second part of this equation, the (γv/c2/)x1, which is throwing me way off. The first part of the equation represents time dilation. The second part of the equation, if I'm not mistaken, represents the fact that the distance in the direction of motion in which an event occurred in one frame of reference will also play a role in what time is recorded for that same event in another frame of reference. In other words, this demonstrates the relativity of simultaneity. It seems like the wrong formula though based on what's actually happening.

Say t is the earth frame and t1 is the rocket frame. Say t1 is moving at a speed of 3/4 c relative to t. If event 1 is a beam of light shooting from the tale of the rocket (which is the origin of the rocket frame), and event two is the beam of light striking the nose end of the rocket, and the rocket has a length of one light second, then the rocket frame will measure one light second of time between the events. In the earth frame, because the nose of the rocket has been moving according to an earth observer, the beam of light is catching up with the nose at a speed of only 1/4 c every second, and it will take 4 seconds for the light to strike the nose according to the earth observer. This is 3 seconds more than it took according to the rocket's measurement. I should also take into account who is measuring the distance from the tail to the nose to be 1 light second. If it is someone in the rocket making this measurement, then to get an earth observer's measurement I'd need to multiply 1 by γ. So, here is what the correct formula seems to be for the right part of the Lorentz Transformation for time:

(x1γ / c - v) - t1. Basically I divided the distance, from earth's measurement, by the difference in speed between the beam of light and the rocket to get earth's time for event two and then subtracted that by rocket's measured time to see how much more time elapsed according to an earth observer. The actual formula though is very different, and I don't see the logic of it based on what's actually happening. The actually formula says you take the speed of the rocket then multiply that by γ, then divide by C squared then multiply by x1?? I understand that when you do the derivation of the Lorentz Transformation from it's general form, that's what comes out, but still it doesn't make sense to me physically. My questions are:

1) How to make sense of the actual form of the second part of the LT for time, and where did I go wrong in thinking it should be in the other form?

2) Do we also need to take into account whether the event in question is happening inside the rocket (which would make it part of the rocket frame and therefore moving with respect to an earth observer), or outside the rocket which would might make it stationary with respect to the earth frame and moving with respect to the rocket? If so, shouldn't the formula for the LT change based on where the event is happening?

Thank you.

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#### Nugatory

Mentor
Do we also need to take into account whether the event in question is happening inside the rocket (which would make it part of the rocket frame and therefore moving with respect to an earth observer), or outside the rocket which would might make it stationary with respect to the earth frame and moving with respect to the rocket? If so, shouldn't the formula for the LT change based on where the event is happening?
Events don’t move. An event is a single point in spacetime, which is to say it is a single dot on the piece of paper that you’re drawing a Minkowski diagram on.

All events always happen in all frames.

So here’s how to do your example with the light signal leaving the tail of the rocket and reaching the nose. We have two events:
Event A: flash of light is emitted from light source at rocket’s tail.
Event B: flash of light reaches detector at nose of rocket.

Let’s also say for the sake of the example that event A has coordinates (x=7.5,t=10) in the earth frame. That is, it happened after ten seconds had passed and the rocket, moving at .75c, had travelled a distance of 7.5 light-seconds using the unprimed earth frame.

Use the Lorentz transformation to calculate the corresponding x’ and t’ coordinates for that event A. Post your work (and if you don’t get x’=0 you did something wrong, because the x’ coordinate of an event at the tail of the rocket will always be zero). If the t’ coordinate you calculate surprises you... remember that the events “earth clock reads ten seconds” and event A are simultaneous in the unprimed frame but not the primed frame. Call that t’ value you calculated (I’m not going to calculate it for you) T’A so using the rocket frame event A has the coordinates (x’=0,t’=T’A).

Now what are the coordinates of event B using the primed rocket frame? That’s easily calculated: the rocket is one light-second long so it takes one second for the light to reach the nose. The coordinates of event B using the rocket frame are (x’=1,t’=T’A+1). Don’t proceed beyond this point until you see how I did that and agree that it’s easy.

Next use the inverse Lorentz transformations to calculate the coordinates of event A in the unprimed frame from (x’=0,t’=T’A), the coordinates of event A in the primed rocket frame. Of course you already know the answer - it’s (x=7.5,t=10) - but the point of this step is to check that you have the inverse transformations right.

Finally, use the inverse transformation to calculate the coordinates of event B in the unprimed earth frame from its coordinates (x’=1,t’=T’A+1) in the primed rocket frame. Again, post your work. If you’ve done everything right, you will find that light travelled a longer distance in the unprimed frame, took more time to do it, and therefore moved at speed c in that frame as well as the primed frame.

#### pervect

Staff Emeritus
In my last post I asked about the general form of the Lorentz Transformation for time. Now I am trying to understand the final form of it, and how it makes sense based on what's happening physically. The final form for t is:

t = γt1 + (γv/c2/)x1

It's the second part of this equation, the (γv/c2/)x1, which is throwing me way off.
There is a name for the concept that this term of the Lorentz transform represents. It is "the relativity of simultaneity".

Many, many people have a hard time coming to grasp with this. I'd recommend at least looking at the paper by Scherr, "The challenge of changing deeply held student beliefs about the relativity of simultaneity". <<link>>

A brief quote from the abstract.

Previous research indicates that after standard instruction, students at all levels often construct aconceptual framework in which the ideas of absolute simultaneity and the relativity of simultaneityco-exist. We describe the development and assessment of instructional materials intended toimprove student understanding of the concept of time in special relativity, the relativity ofsimultaneity, and the role of observers in inertial reference frames.
Basically, that term in the equation means that two events, at different positions, are simultaneous in one frame, which we will call S, but are not simultaneous in a different frame S' that is moving with respect to S. Hence, simultaneity depends on the frame of reference one chooses. Two frames that are moving at different velocities will have different notions of what events are simultaneous, and which events are not.

Maybe it will help you, maybe it won't. Scherr is a good writer with experience teaching students about the issue that is confusing you, though.

The paper is oriented towards the instructor rather than the student, but it might still be helpful. I've found my own posts on the topic don't seem to get through that well - it's probably not entirely the fault of my posts, but it couldn't hurt to give the professional literature a look.

By the way, regardless of whether it does or doesn't help, if you read the paper I'd be interested in your feedback on how helpful (or not helpful) it was.

#### PeroK

Homework Helper
Gold Member
2018 Award
In my last post I asked about the general form of the Lorentz Transformation for time. Now I am trying to understand the final form of it, and how it makes sense based on what's happening physically. The final form for t is:

t = γt1 + (γv/c2/)x1
Here's how I would understand that formula. Imagine two clocks at rest in the primed frame: one at the origin and one at the coordinate $x'$.

An event, E, happens at coordinate $x'$ when the local clock at $x'$ reads $t'$.

Note: that is physically what it means for an event to have those coordinates in that reference frame.

What do we observe in the unprimed frame?

First, we have SoR in the form of the leading clock lags rule. The clock at $x'$ is permanently $\frac{v x_1}{c^2}$ behind the clock at the origin.

When event E takes place, the clock at the origin reads $t' + \frac{v x_1}{c^2}$ and the clock at $x'$ reads $t'$.

Now, the time shown on the clock at the origin is related to the time in the unprimed frame by simple time dilation:

$t = \gamma (t' + \frac{v x_1}{c^2})$

And that's it. That's the end of the argument. It can't be wrong. It's been proved.

Note that you have to learn to trust the mathematics. The Lorentz Transformation is mathematics, which eventually must be preferred to ad hoc analysis in every case.

#### Ibix

t = γt1 + (γv/c2/)x1
In the case of your rocket and the flash arriving at the nose, you have that $L=x^1=ct^1$ (where $L$ is the rest length of the rocket) because $x^1$ and $t^1$ are measured in the rocket frame. So the inverse Lorentz transform you quoted gives you $$\begin{eqnarray*}t&=&\gamma L\left(\frac 1c+\frac v{c^2}\right)\\ &=&\frac{\gamma L}{c^2}(c+v)\end{eqnarray*}$$

In the Earth frame the rocket is of length $L/\gamma$. It is moving at speed $v$, so at time $t$ its nose is at $x=vt+L/\gamma$. The light pulse coming from the tail is at $x=ct$. The light pulse meets the nose at the time when these positions are equal - i.e. when $vt+L/\gamma=ct$ or $$t=\frac L\gamma\frac 1{c-v}$$This is the equation you should have got when you got
(x1γ / c - v) - t1
I'm not clear how you arrived at that, and it doesn't seem to reduce to my intercept calculation. I don't really understand what you think you were doing to get this.

Anyway, if you note that $\gamma=1/\sqrt{1-v^2/c^2}=c/\sqrt{(c+v)(c-v)}$ you can show that my expressions are the same. Algebraic mucking around with $\gamma$ is, unfortunately, often a major part of relativistic calculations.

#### sweet springs

$$x'=\gamma(x-\beta ct)$$
Position x' depends on time t. Depending on x, past farther that way, now here and, in future farther this way.
$$ct'=\gamma(ct-\beta x)$$
Time t' depends on position x. Depending on t, farther past, here now and farther opposite way future.
We may think of the two on the same footing.

#### PeroK

Homework Helper
Gold Member
2018 Award
$$x'=\gamma(x-\beta ct)$$
Position x' depends on time t. Depending on x, past farther that way, now here and, in future farther this way.
$$ct'=\gamma(ct-\beta x)$$
Time t' depends on position x. Depending on t, farther past, here now and farther opposite way future.
We may think of the two on the same footing.
It looks like something has been lost in transalation there.

#### sweet springs

My bad English. People have been familiar in the first equation by Galilean transformation. Running train was far away in past, it is here passing by me now, and it will be far another way in future. Measures in original FR are put on wrong Origin.

The second equation suggests in far points clocks are wrong short setting, here correct and opposite far side wrong advanced setting. Clocks in original FR are badly set or under bad synchronization.

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#### SiennaTheGr8

Conceptually, it might help to regard the Lorentz boost as a kind of "rotation" of the $ct$ and $x$ axes.

It's actually quite analogous to a rotation of the Cartesian axes in the Euclidean plane, which you may already be familiar with. If you were to rotate your $x$ and $y$ axes counterclockwise through an angle $\theta$, your new "primed" coordinates would be related to the old ones by this matrix equation:

$\begin{bmatrix} x^\prime \\ y^\prime \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} .$

Similarly, a Lorentz boost is a hyperbolic rotation of your $ct$ and $x$ axes through the hyperbolic angle $\phi$. The new primed coordinates are related to the old unprimed coordinates by this matrix equation:

$\begin{bmatrix} ct^\prime \\ x^\prime \end{bmatrix} = \begin{bmatrix} \cosh \phi & -\sinh \phi \\ -\sinh \phi & \cosh \phi \end{bmatrix} \begin{bmatrix} ct \\ x \end{bmatrix} .$

The hyperbolic angle $\phi$ here represents the relative rapidity of the primed and unprimed frames. It's related to the relative speed $v$ of the frames by $v/c = \tanh \phi$, which also means that $\gamma = \cosh \phi$ and $\gamma v / c = \sinh \phi$ (do carry out the matrix multiplication and make these substitutions!).

#### NoahsArk

Gold Member
Thank you for the responses. I'm working through the math based on some of the responses and will post my results to see if I'm understanding it better.

#### NoahsArk

Gold Member
So here’s how to do your example with the light signal leaving the tail of the rocket and reaching the nose. We have two events:
Event A: flash of light is emitted from light source at rocket’s tail.
Event B: flash of light reaches detector at nose of rocket.

Let’s also say for the sake of the example that event A has coordinates (x=7.5,t=10) in the earth frame. That is, it happened after ten seconds had passed and the rocket, moving at .75c, had travelled a distance of 7.5 light-seconds using the unprimed earth frame.

Use the Lorentz transformation to calculate the corresponding x’ and t’ coordinates for that event A. Post your work (and if you don’t get x’=0 you did something wrong, because the x’ coordinate of an event at the tail of the rocket will always be zero).
This is a helpful example. I did get x1 = 0 for event A. First I found that γ for .75c = 1.5119. x1 = γx - γvt. So 1.5119(7.5) - .75(10)1.5119 = 0.
I got t1 for event A (t1A) as 6.6145 using the inverse LT for time: t1 = γt - (γv/c2/)x.
1.5119(10) - 1.5119(.75)/c2 (7.5) =
15.119 - 8.5044375 =
6.6145

I agree that for event B in the rocket frame it's easy to get the x1B and tB because you just add 1 to the t1 coordinates of event A which were (0, 6.6145). So the event B coordinates for t1 are (1, 7.7145).

When I do the inverse LT on the rocket's coordinates of event A to get the earth coordinate's of event A, I get (7.5, 10) again.

For the coordinates of event B in the earth frame, I get (13.1754, 12.80). For the x coordinate: 1(1.5119) + 7.7145(1.5119) = 1.5119 + 11.6635 = 13.1754.
For the t coordinate: t = 7.7145(1.5119) + (.75(1.5119)/c2)1 =
12.80.

I'm hoping I got it all correct!?

Many, many people have a hard time coming to grasp with this. I'd recommend at least looking at the paper by Scherr, "The challenge of changing deeply held student beliefs about the relativity of simultaneity". <<link>>
Thank you I am reading this now. Things are explained pretty well although I had trouble understanding the first diagram with the circles of light from Alan and Beth's perspective and what that is trying to illustrate.

First, we have SoR in the form of the leading clock lags rule. The clock at x′x′x' is permanently vx1c2vx1c2\frac{v x_1}{c^2} behind the clock at the origin.
That is helpful to see. I looked up the leading clocks lag rule and see that there is a lot of algebra that goes into deriving it. Once the lag rule is derived though, it makes sense on how you go from that to the right side of the LT equation for time.

Ibix: the two equations you wrote with the (c+v) and (c-v) coefficients are the times it takes for the light to reach the tail and nose respectively right? Then to get the lag in time we have to subtract time on the tail end from time on the nose end to get the form: t′+vx1/c2?

The second equation suggests in far points clocks are wrong short setting, here correct and opposite far side wrong advanced setting. Clocks in original FR are badly set or under bad synchronization.
If I am not mistaken, are you saying that, from earth's perspective, it takes light more time (than it does from rocket's perspective) to catch up with the nose, and it takes less time to reach the tail? If so, that makes sense to me.

Conceptually, it might help to regard the Lorentz boost as a kind of "rotation" of the ctctct and xxx axes.
What you described is an interesting way to think about it. I will need to polish up on trig, though, to better understand it. I'll do that and then go back to your example and see if I can follow.

Thank you.

#### sweet springs

If I am not mistaken, are you saying that, from earth's perspective, it takes light more time (than it does from rocket's perspective) to catch up with the nose, and it takes less time to reach the tail? If so, that makes sense to me.
I try with my poor English to explain in the case of train and Earth.

Earth is covered with clocks. The passengers of front, middle and end part of the train compare their watch and Earth clock they see in front of the window at 10:00.
Before observation all of them have confirmed that their watches are adjusted and synchronized properly.
Front passengers say " Earth clock I see is 10:03, 3 minutes ahead of mine."
Middle passengers say " Earth clock I see is 10:00, just time with mine."
End passengers say "Earth clock I see is 09:57, 3 minutes behind of mine."
Earth clocks are in poor synchronization for train passengers.

Train watches are in poor synchronization for Earth people as well.
Front Earth people say " At 10:03 train watch I see is 10:00, 3 minutes behind of mine."
Middle Earth people say " At 10:00 train watch I see is 10:00, just time with mine."
End Earth people say "At 09:57 train watch I see is 10:00, 3 minutes ahead of mine."

To be symmetric Earth people can plan to observe train watch in their 10:00 timing and observe similar law "front go too late, rear go too fast" ,contrary to the law for train passengers, "front go too fast, rear go too late".

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#### Ibix

Ibix: the two equations you wrote with the (c+v) and (c-v) coefficients are the times it takes for the light to reach the tail and nose respectively right? Then to get the lag in time we have to subtract time on the tail end from time
No - the light reaches the tail at time zero in noth frames, I think. They are both the time (in the Earth frame) that the light reaches the nose. By rewriting $\gamma$ as the longer expression I wrote you should be able to show that they are equal.

The first one is obtained by writing the time the light arrives at the nose in the rocket frame and transforming. The second is obtained by direct calculation in the Earth frame. You had done the former correctly in post #1, but your calculation of the latter did not seem to be correct.

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