# B Trying to understand Lorentz Transformations

#### NoahsArk

Gold Member
I am trying to understand the general form of the Lorentz Transformations before I even get into the long process of deriving that into the specific equations. In Taylor and Wheeler's, Spacetime Physics book they give this as the general form:

t= Bx1 + Dt1
x= Gx1 + Ht1

In the equation for t, why is x being added into the equation? I don't understand, if we are trying to convert time in one reference frame to time in another, why we'd be adding x when x is distance.

Similarly, in the equation for x, why are we adding t?

Also, in the equation for t, I know that the coefficient, D, in front of t1 is γ which is derived from the invariance of the interval. In the equation for, x, though, the authors give vγ (where v is the velocity of the second (rocket) frame from the perspective of the first (earth) frame), as the coefficient of t1. What is the explanation for that?

Thank you.

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#### Ibix

In the equation for t, why is x being added into the equation? I don't understand, if we are trying to convert time in one reference frame to time in another, why we'd be adding x when x is distance.

Similarly, in the equation for x, why are we adding t?
It's just the most general linear transformation. They're not really adding x or t into the equation - they're just not assuming they can leave either out. So this is the form of transformation with the fewest possible assumptions. If the transformations had turned out not to depend on x or t (like the Galilean t transform) the relevant coefficient would be zero.

Also, in the equation for t, I know that the coefficient, D, in front of t1 is γ which is derived from the invariance of the interval. In the equation for, x, though, the authors give vγ (where v is the velocity of the second (rocket) frame from the perspective of the first (earth) frame), as the coefficient of t1. What is the explanation for that?
The derivation should justify the values somewhere. It depends what derivation you are following. Where are you looking in Taylor and Wheeler?

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#### PeroK

Homework Helper
Gold Member
2018 Award
Similarly, in the equation for x, why are we adding t?
If you go back to basic Newtonian physics and have one reference frame moving at velocity $v$ relative to another, then:

$x' = x - vt$

That's because the origin of the second reference frame is moving at velocity $v$.

This is called the Galilean transformation, the other part of which is the obvious $t' =t$.

The basic idea of the Lorentz transformation is that you want to generalise these two equations:

$x' = x - vt$ becomes in general $x' = Ax + Bt$

And $t' = t$ becomes $t' = Ct + Dx$

And, to see whether anything else makes sense, apart from the Galilean solution: $A = 1, B = -v, C = 1, D = 0$. Are there any other possible solutions that make physical sense?

#### NoahsArk

Gold Member
It's just the most general linear transformation. They're not really adding x or t into the equation
I'm not sure why, when trying to figure out t based on t1[/S
UP], we need to know what x is at all.

The derivation should justify the values somewhere. It depends what derivation you are following. Where are you looking in Taylor and Wheeler?
In my book it's the "special topic" chapter which is between chapters 3 and 4. It's on page 100.

PeroK: "The basic idea of the Lorentz transformation is that you want to generalize these two equations:"

For the case where x1 = 0, then the equation to convert time would be t = γt1. I am still not sure why when x1 is greater than zero, we have to add x1 into the equation.

#### pervect

Staff Emeritus
t' = t is very familiar, but it's just not compatible with special relativity. Basically, you'll have to come to terms with the fact that t' depends on t and x, and not just t, in order to understand special relativity. Truly understaning all of the consequences of this will take a while, some are not obvious. Unfortunately, just rejecting the idea that t' depends on t and x out of hand isn't going to get you anywhere useful :(.

#### Dale

Mentor
I'm not sure why, when trying to figure out t based on t1, we need to know what x is at all.
Why wouldn’t you? You are just assuming that you can neglect x, but there is no physical basis for that assumption.

The idea here is to make as few assumptions as possible and then derive as much as you can with as few assumptions as possible. You are assuming that B=0, which is a very strong and unjustified assumption. The derivation is not assuming any value for B, it could turn out to be 0, but they are not assuming that. It could turn out to be non-zero, but they are not assuming that either.

So leaving B in is not an assumption, so it needn’t be justified. Setting B=0 a priori is an assumption and requires justification

#### NoahsArk

Gold Member
Why wouldn’t you? You are just assuming that you can neglect x, but there is no physical basis for that assumption.
My understanding of the Lorentz transformations is that we use them for situations where some event in the rocket frame (x1) occurred some distance in front of the rocket. To know that distance from earth's frame, we need to know how far away that event occurred from the rocket (using the rocket measurement) and multiply that distance by γ, and add that number to the distance that the rocket traveled from us (v times γt1). So, it would make sense to me why we use t1 to try and figure out x. The reverse, though, I can't explain- i.e. why we need x1 to figure out t.

So leaving B in is not an assumption, so it needn’t be justified.
.

To me though, putting X into the equation has no more reason than putting any other variable in it. The only reason I can think of is my vague notion that it has something to do with relativity of simultaneity, and if an event occurred some distance away from the rocket, the mere distance alone from the rocket will affect x1's time measurement of that event. Seems like there should be a simpler explanation than that though.

#### Pencilvester

To me though, putting X into the equation has no more reason than putting any other variable in it.
What other variable would you put in? Are there any other coordinates you would use to label an event in an inertial frame other than three for position in space and one for time?

#### Dale

Mentor
putting X into the equation
X isn’t being put into the equation. If B=0 then x is not in, which is not rejected a priori.

Again, what is your physical justification for asserting B=0, a priori?

Seems like there should be a simpler explanation than that though.
You have already been given the simpler explanation multiple times. The goal is to make as few a priori assumptions as possible. If you don’t make assumptions then you have to allow the possibility that x might be in the equation or it might not.

#### Nugatory

Mentor
My understanding of the Lorentz transformations is that we use them for situations where some event in the rocket frame (x1) occurred some distance in front of the rocket.
They are much more general than that. They answer the question:

I say that something happened at point x and time t. You are moving relative to me. Where and when did it happen according to you?

#### sweet springs

t= Bx1 + Dt1
x= Gx1 + Ht1

In the equation for t, why is x being added into the equation? I don't understand, if we are trying to convert time in one reference frame to time in another, why we'd be adding x when x is distance.
From this postulate
$$c^2t^2-x^2=(c^2D^2-H^2)(t^1)^2-(G^2-c^2B^2)(x^1)^2+2(c^2BD-GH)x^1 t^1$$

The SR relation, square of invariant distance, $$c^2t^2-x^2=c^2(t^1)^2-(x^1)^2$$ requires non zero B. Thus
$$cD=cosh\ \alpha,\ H=sinh\ \alpha$$
$$G=cosh\ \gamma,\ cB=sinh\ \gamma$$
$$cosh\ \alpha \ sinh\ \gamma=cosh\ \gamma \ sinh\ \alpha$$ so $$\alpha=\gamma=arctanh\ \beta$$
Introducing beta=v/c above, we get familiar Lorentz transformation formula, assured by uniqueness of series expansion by beta.

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#### sweet springs

In the equation for t, why is x being added into the equation? I don't understand, if we are trying to convert time in one reference frame to time in another, why we'd be adding x when x is distance.

Similarly, in the equation for x, why are we adding t?
Say all the lattice points of the two frames of reference are occupied with synchronized clocks.
The transformation formula tell correspondence of the superposed two FR in space coordinates and clock readings or time coordinates. In this correspondence
$$x \approx x'+vt'$$ has been obvious at latest since time of Galilei, but
$$t \approx t'+v/c^2 x'$$ has not been noticed until 20th century because of smallness of vx'/c^2 in most cases.

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#### NoahsArk

Gold Member
Thank you for the responses.

To take an example, let's say a rocket is moving ahead of me at v = .8 c. For the time between any two events happening in the rocket, I multiply by 1.67 to get my measurement of the time between the events. No need for Lorentz transformations.

Now say two events occurred at a distance of one light year in front of the rocket. It wouldn't cross my mind to say that I now have to take whatever the rocket time was, multiple by 1.67, and then add one light year to that amount? I'm pretty confused.

Also, if the rocket were very long, do we consider any events happening inside the rocket, whether at the nose or tail of the rocket, as being at X1 = zero in the rocket frame? I am assuming that Lorenz transformations come into play in situations where there is a third frame of reference, like another rocket traveling in front of the first one, but even then I can't see why we'd be thinking of adding X1 to get t.

#### NoahsArk

Gold Member
Following up on this, I was just watching a coursera video on SR (second video on the Lorentz transformations) where he says that the X1 in the equation for t is basically the leading clocks lag principle built into the Lorentz transformations. This makes sense, but I don't see how we could assuming from the beginning that the equation for t must take that general form without prior knowledge that leading clocks lag.

#### Nugatory

Mentor
To take an example, let's say a rocket is moving ahead of me at v = .8 c. For the time between any two events happening in the rocket, I multiply by 1.67 to get my measurement of the time between the events.
Be precise. There are four events here: the two events you're talking about which we will call E1 and E2, but also E3 ("you look at your wristwatch at the same time as E1") and E4 ("you look at your wristwatch at the same time as E2"). The rocket observer is present at both E1 and E2 so he can directly measure the time between them (call that $\Delta{T}'$). You multiply $\Delta{T}'$ by 1.67, and that gives us..... the time you'll measure between E3 and E4. We then assert (because E1 and E3 happen at the same time, as do E2 and E4) that this is also our measurement of the time between E1 and E2.
No need for Lorentz transformations.
How so? The "multiply by 1.67" formula you used is the result of applying the Lorentz transformations in the specific case in which event E1 has coordinates $(x'=X_0,t'=T_0)$ in the spaceship frame and event E2 has coordinates $(x'=X_0,t'=T_0+\Delta{T}')$.

Please, please, please.... don't just take my word for this. First, satisfy yourself that the coordinates I just wrote down for E1 and E2 are right for two events that happen at the same place and different times according to the spaceship observer. Then plug them into the Lorentz transformations to see that according to you they happened at different places (separated by $1.67v\Delta{T}'$) and that if events E3 and E4 happened at the same time as E1 and E2 according to you, then E4 happened $1.67v\Delta{T}'$ after E3.
Now say two events occurred at a distance of one light year in front of the rocket. It wouldn't cross my mind to say that I now have to take whatever the rocket time was, multiple by 1.67, and then add one light year to that amount?
In this case, the coordinates for E1 in the spaceship frame would be $(x'=1ly,t'=T_0)$ and the coordinates of E2 would be $(x'=1ly,t'=T_0+\Delta{T}')$. Plug this into the Lorentz transformations and you'll find that the corresponding E3 and E4 are still separated by $1.67\Delta{T}'$.
Also, if the rocket were very long, do we consider any events happening inside the rocket, whether at the nose or tail of the rocket, as being at X1 = zero in the rocket frame?
No. suppose the length of the rocket is $L$. We could choose the tail of the rocket to be $x'=0$, and then an event happening at time T (using the rocket frame) at the nose of the rocket would have coordinates $(x'=L,t'=T)$ while an event happening at the same time at the tail of the rocket would have coordinates $(x'=0,t'=T)$. (This is exactly your previous question, if the rocket were one lightyear long).

Also, note that although these events happen at the same time using the rocket frame (they both have the same $t'$ coordinate) they do not happen at the same time using the frame in which you are at rest.

I am assuming that Lorenz transformations come into play in situations where there is a third frame of reference
No. The Lorentz transformations come into play any time that we know an event happened at location X and time T according to one frame, and we want to calculate the location and time at which that event occurred using another frame.

#### Nugatory

Mentor
This makes sense, but I don't see how we could assuming from the beginning that the equation for t must take that general form without prior knowledge that leading clocks lag.
We start with the most general possible transformation. We're making no assumption about whether leading clocks might lag, trailing clocks might lag, or whether they all stay synchronized... we just don't know, so we assume nothing.

This most general assume-nothing transformation will be of the form$$t'=Ax+Bt\\x'=Cx+Dt$$We use what we know about the behavior of light (the same speed in all frames tells us that $x\pm{ct}=0$ implies $x'\pm{ct'}=0$) to solve for the values of A, B, C, and D and we end up with the Lorentz transformations...

And then we look at these transformations and see that they're telling us, among other things, that leading clocks lag. So we didn't have any prior knowledge of this, it's the conclusion that we came to at the end.

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#### PeroK

Homework Helper
Gold Member
2018 Award
Following up on this, I was just watching a coursera video on SR (second video on the Lorentz transformations) where he says that the X1 in the equation for t is basically the leading clocks lag principle built into the Lorentz transformations. This makes sense, but I don't see how we could assuming from the beginning that the equation for t must take that general form without prior knowledge that leading clocks lag.
It's not a question of prior knowldege, it's a question of exploring the alternatives to $t' =t$.

Let's suppose you were a physics student in 1850 and you and a friend were discussing time and thought of this question: is there any alternative to $t' = t$? You might say "no, nothing else could possibly make any physical sense". Your friend might say "but, what if, $t'$ depended in some way on $x$ and $t$? And, you'd say: "it's impossible". Your friend says: "but what if we try?" Just imagine that it's linear (because that's the simplest alternative). What if $t' = At + Bx$ for some $A, B$?"

You'd say: "you must have prior knowldege of the leading clocks lag rule"! And your friend would say "what's that. then?".

So, your friend simply wants to try something - to see where it leads. But, you are resolute, there is no point in considering an alternative to $t'= t$ unless you already know what the answer will be! You're not allowed even to try.

#### Dale

Mentor
let's say a rocket is moving ahead of me at v = .8 c. For the time between any two events happening in the rocket, I multiply by 1.67 to get my measurement of the time between the events. No need for Lorentz transformations.
Not really.

The multiplication only works if the events happen at the same location in the rocket’s frame. If one happens in the front of the rocket and the other happens in the rear of the rocket then the simplified formula fails and you have to use the full Lorentz transform.

Furthermore, if they are in the same location in the rocket’s frame then the simplified formula is the Lorentz transform. Since $\Delta x=0$ that term drops out and you are left with the multiplication. So saying “no need for the Lorentz transform” fails to recognize that the simple multiplication is the Lorentz transform in this specific case.

Now say two events occurred at a distance of one light year in front of the rocket. It wouldn't cross my mind to say that I now have to take whatever the rocket time was, multiple by 1.67, and then add one light year to that amount? I'm pretty confused.
No matter what, you will apply the Lorentz transform. If you are in a special scenario then the Lorentz transform will automatically simplify to the time dilation formula or some other relevant simple formula. Always using the Lorentz transform will prevent mistakes that arise from incorrectly using a simplified formula and confusion about when to apply the simplified formulas.

#### Dale

Mentor
This makes sense, but I don't see how we could assuming from the beginning
For like the fourth time now, no such assumption is made.

Do you understand the concept of a general formula? A general formula is one that captures all of the possibilities. It makes no assumptions about which specific formula is correct, but allows you to pick between the various possible specific formulas by setting one or more parameters. In this case the parameter is B. We make no assumptions about the value of B.

This is getting very frustrating. It seems that you are not even making an effort to understand this point. You keep on saying that we are making an assumption when in fact it is you that are the only one making an assumption, and you have not once even attempted to justify your assumption! You have this totally backwards and are repeating it over and over with no attempt to figure it out. Stop! We are not assuming anything! Stop saying that we are. The only assumption here is your unjustified assumption that B=0. The next time you use the word “assumption” please make sure that it refers specifically to your assumption that B=0.

#### Ibix

@NoahsArk - are you asking your question backwards? Is the underlying question "why $x$ and not $z$ or temperature or the price of gold or something"?

#### NoahsArk

Gold Member
are you asking your question backwards? Is the underlying question "why xxx and not zzz or temperature or the price of gold or something"?
Yes, that is exactly what I am asking.

Dale: It's a difficult concept for me. I understand what the general form of an equation is. I just don't see how people originally came up with the idea that the general equation took that form.

#### PeroK

Homework Helper
Gold Member
2018 Award
Yes, that is exactly what I am asking.

Dale: It's a difficult concept for me. I understand what the general form of an equation is. I just don't see how people originally came up with the idea that the general equation took that form.
Maybe they took a guess!

How much of physics started with an educated or inspired guess? Where did Newton get his law of gravitation? Maybe he guessed:

$F = \frac{GMm}{r}$

And that didn't work.

Then he guessed:

$F = \frac{GMm}{r^2}$

And that worked.

If that hadn't worked he might have tried:

$F = \frac{GMm}{r^3}$

In fact, it doesn't matter at the end of the day where the equation came from. As long as you can demonstrate that it has the properties you want. The law of gravitation being a case in point.

#### NoahsArk

Gold Member
In fact, it doesn't matter at the end of the day where the equation came from. As long as you can demonstrate that it has the properties you want. The law of gravitation being a case in point.
Hmmm. Are you saying that as long as the equation is giving the right results, I should just learn it and use it and not get too bogged down in how it originated?

#### Dale

Mentor
Dale: It's a difficult concept for me. I understand what the general form of an equation is. I just don't see how people originally came up with the idea that the general equation took that form.
OK, that is fine, but please stop asserting that the equation with B is an assumption, it is the opposite of an assumption. So the question is what reasoning led them to believe that the linear expression was the correct generalization.

The first thing that we need to do is to understand what it is that a transformation is doing and what kind of transformation we are looking for. A transformation is intended to change one set of coordinates into another set of coordinates. So if we have Cartesian coordinates we might change them into polar coordinates. Coordinates are just ways to numerically label different times and places.

Now, we are particularly interested in inertial coordinate systems, that is we like coordinates where free particles go in a straight line at constant velocity, following Newton's first law. Any coordinates where all free particles have straight lines as their worldlines are inertial coordinates, so if we want to study the transformations from one inertial frame to another inertial frame then we want to study transformations that map straight lines to other straight lines.

The simplest such transformation is a linear transformation, which is the form chosen in the derivation you cited. So the reason for choosing that generalization is that it is the simplest generalization that has the necessary property of mapping straight lines to straight lines.

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#### NoahsArk

Gold Member
No. suppose the length of the rocket is LLL. We could choose the tail of the rocket to be x′=0x′=0x'=0, and then an event happening at time T (using the rocket frame) at the nose of the rocket would have coordinates (x′=L,t′=T)(x′=L,t′=T)(x'=L,t'=T) while an event happening at the same time at the tail of the rocket would have coordinates (x′=0,t′=T)
Thank you for clarifying that.

The simplest such transformation is a linear transformation, which is the form chosen in the derivation you cited.
In the equation y = mx + b, the first time I read it and the description for it it made sense to me right away. It makes sense because I visualize that what it's describing is a straight light which isn't curved, and I understand why we use each of the terms. The Lorentz equation for time, though, doesn't jump out at me in the same way as being intuitive.

"Trying to understand Lorentz Transformations"

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