Deriving Lorentz Transformation

[Arman777]

How can we derive Lorentz Transformation ?

I used one approach using the length contraction and time dilation and simultaneity but my prof wasnt much happy about it. Is there any other way to derive it ?

 

[Orodruin]

Can you show us your derivation and we can discuss why your professor was not happy about it? Typically, the inference is made the other way, but it is possible both ways.

 

[Arman777]

I guess I don't need to proof time dilation and length contraction.

For position lorentz transformation now I can realize I made a mistake.. Any idea how can I derive it ?

For the time part, from simultaneity I derived that for an outside observer the clock difference will be $ΔT = vx/c^2$. So I thought I can say t'=t-ΔT with some constant in front which it is $γ$. I didnt do any math do put it there. Well maybe we can say " we know that $t' = tγ$ so that we can say $t'=γ(t-ΔT)$".. Here its not like mathematical approach but more like a thought appraoch ?

I guess my proofs are wrong. Is there a good way to approach it somehow ?

 

[Orodruin]

I guess I don't need to proof time dilation and length contraction.

If you are going to use them to derive the Lorentz transformations, yes you do. You need to show that they follow from the basic postulates of SR if you want to use them. The typical way of deriving them is to first derive the Lorentz transformations and then use the Lorentz transformations to derive time dilation and length contraction.

For the time part, from simultaneity I derived that for an outside observer the clock difference will be $ΔT = vx/c^2$. So I thought I can say t'=t-ΔT with some constant in front which it is $γ$. I didnt do any math do put it there. Well maybe we can say " we know that $t' = tγ$ so that we can say $t'=γ(t-ΔT)$".. Here its not like mathematical approach but more like a thought appraoch ?

It is difficult to understand what you are saying here. You need to properly define all of the events you are using to make your derivation, not just throw coordinates around.

 

[Arman777]

If you are going to use them to derive the Lorentz transformations, yes you do. You need to show that they follow from the basic postulates of SR if you want to use them.

yes you are right. I meant I know how to derive them so it would be waste of time to shoe it here. So let's assume we derived those quantities. Then what can we do ?

It is difficult to understand what you are saying here. You need to properly define all of the events you are using to make your derivation, not just throw coordinates around.

My approach is wrong actually ... I need some ideas I can try the math

The typical way of deriving them is to first derive the Lorentz transformations and then use the Lorentz transformations to derive time dilation and length contraction.

Maybe we should focus on this part. How can I approach to the problem ?

 

[Nugatory]

Maybe we should focus on this part. How can I approach to the problem

Googling for "Lorentz transformation derivation" will find some good approaches (and this being the Internet, also some bad ones of course). There's a classic one in the appendices of Einstein's popularization "Relativity: The special and general theory".

We have some good threads as well.

 

[Kely]

One of the most intuitive ways of deriving the transformations is the following. Consider an infinitesimal homogeneous Lorentz transformation $\lambda = \delta + \epsilon K$. Put this into the Lorentz condition $$\Lambda_\mu{}^\sigma \eta_{\sigma \rho} \Lambda^\rho{}_\nu = \eta_{\mu \nu} \leftrightarrow \Lambda^T \eta \Lambda = \eta \\ (\delta_\rho{}^\mu + \epsilon K_\rho{}^\mu) \eta_{\mu \nu}(\delta^\nu{}{\sigma} + \epsilon K^\nu{}_\sigma) = \eta_{\rho \sigma} \leftrightarrow K^T \eta = -\eta K$$ Use the convention $(t, x, y, z)$ for the coordinates. Then a boost in the x direction has to be of the form $K_x = \text{diag}(a, 0)$ where $a$ is a 2x2 matrix to be found. By inserting into the above condition leads to $a_{12} = a_{21} = 1, a_{11} = a{22} = 0$. (The non vanishing elements of $a$ can be set equal to 1 because $K_x$ gets multiplied by an infinitesimal $\epsilon$ anyway.)

Now you exponentiate to find the actual boost $$\exp(\varphi a) = \Lambda_x = \begin{pmatrix}\cosh\varphi&\sinh\varphi&0&0\\\sinh\varphi&\cosh\varphi&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$ where now $\varphi$ is a 'finite' parameter. Then $$\begin{pmatrix}t'\\x'\\y'\\z'\end{pmatrix} = \Lambda_x \begin{pmatrix}t\\x\\y\\z\end{pmatrix} \\ t' = t\cosh\varphi - x\sinh\varphi \\ x' = x\sinh\varphi - t\cosh\varphi \\ y' = y \\ z' = z$$ where $\varphi$ is taken negative in order to fulfill the conventions.

 

[Sorcerer]

How can we derive Lorentz Transformation ?

I used one approach using the length contraction and time dilation and simultaneity but my prof wasnt much happy about it. Is there any other way to derive it ?

EDIT- this post is for the student who has NOT mastered tensors or linear algebra, etc. This entire derivation is pure elementary algebra
In my opinion, the best way to get started is to use a light clock and use the triangle that forms from the motion on light according to the at rest observer to dervive the Lorentz factor. That will at least get you time dilation.

But if you want to really dig in, I’d start like this:

Generic, easy, non-rigorous Lorentz transformation derivation method

(1) Start with two reference frames S and S’, with x-axes coinciding, moving at a relative speed of v.
(2) Assume the laws of physics are the same in both frames, so that things like distance = speed multiplied by time hold, and that the speed of light is c in both frames.
(3) Shoot a beam of light in S. Its distance is given by x = ct. Rearrange to x - ct = 0. Shoot it in the other direction and its distance is given by x = -ct, which is x + ct = 0. With S’, those beams’ distances are x’ = ct’ and x’ = -ct’, which can be written, respectfully, as x’ - ct’ = 0 and x’ + ct’ = 0.
(4) Assume those frames are connected by two functions like this:

x - ct = δ(x’ - ct’)

and

x + ct = ζ(x’ + ct’)

Add those two together to get:

2x = (δ + ζ)x’ + (ζ - δ)ct’

Divide by 2

x = (δ + ζ)x’/2 + (ζ - δ)ct’/2

Then, to clean things up, let γ = (δ + ζ)/2 and η = (ζ - δ)/2

Leaving:

x = γx’ + ηct’

At this point you should begin to see where this is going.

(5) Do this again, this time taking x = ct and making it ct - x = 0, and so on. You’ll get an equation similar to the bold, but it will be in terms of ct = such and such instead of x = such and such. Do the same things you just did in step 4, and out should pop a similar equation connecting ct with ct’ and x’.

(6) Now invert your primed and unprimed coordinates and swap the signs to get your inverse transformation equarions (e.g., x = γx’ + ηct’ would belcome x’ = γx - ηct).

(7) Let x’ = 0 in the inverse transformation equation (meaning the object moving according to S is at rest in S’ for at least a moment), and solve for η. Note that v = x/t.

(8) At this point the only remaining unknown should be γ. To solve for this, first plug what you found for η into the x transformation equation. Then substitute in your previously found equations for ct’ and x’. Simplify, and solve for γ.

(9) Go back and replace all γ‘s with what you solved for (although more often than not, no one bothers to do this).

(10) Realize that since the motion of coordinates is along both x axes, y = y’ and z = z’.Sometimes instructors or tutorials just start at “this is a generic transformation: x = ax + bct,” but I think it’s important to show why, with the whole “shoot a beam along the x-axis” thing.
Anyway, that’s my two cents on it. First time I did it I used a light clock and the Pythagorean theorem to derive the Lorentz factor. But I am much more satisfied doing it this longer way.

 

[robphy]

In my opinion, Bondi's method is the best approach since it is more closely tied to physical radar measurements that one might do.
The primary source is
Bondi's https://archive.org/details/RelativityCommonSense

I used his approach in my Insight ... about 70% into the page
https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/

 

[Arman777]

Thanks for your help. But all these things are seems complex. WE didnt see matrix derivation but thanks.

I find a great source though. Look pages after 35.

 

[PeroK]

Thanks for your help. But all these things are seems complex. WE didnt see matrix derivation but thanks.

I find a great source though. Look pages after 35.

That text looks pretty good, but it does have a questionable statement on page 9:

Einstein wrote two theories of relativity; the 1905 work is
known as “special relativity” because it deals only with the
special case of uniform (i.e. non-accelerating) motion.

It may be just a slip of the word processor, but it's fascinating that a professor of physics could ever assert such a thing! Are you, for example, unable to study kinematics and forces or circular motion in SR?

Surely, SR deals with flat spacetime and GR with the curved spacetime (associated with gravity) and that is the difference.

 

[vela]

It may be just a slip of the word processor, but it's fascinating that a professor of physics could ever assert such a thing! Are you, for example, unable to study kinematics and forces or circular motion in SR?

If you continue reading that paragraph, it's pretty clear he was referring to the motion of one reference frame relative to another. He could have been a bit clearer about that, though.

 

[Orodruin]

but it's fascinating that a professor of physics could ever assert such a thing!

Oh, you would be surprised what I have heard physics professors get wrong about relativity... you do not need to specialize in anything that has to do with relativity to become a physics professor.

 

[PeroK]

If you continue reading that paragraph, it's pretty clear he was referring to the motion of one reference frame relative to another. He could have been a bit clearer about that, though.

You can study an accelerating reference frame in SR. Either linear acceleration or circular motion.

Also, on page 31-32, re the twin paradox, he says:

In fact, the two reference frames are not equivalent. Paul’s
reference frame has been accelerating, and he knows this; he
is pushed towards the rear of the spaceship as it speeds up,
and towards its front as it slows down. There is an absolute
difference between the frames, and it is clear that Paul is the
one who has been moving; he really does end up younger than
Peter.

This gives the strong impression that if you accelerate, then you are absolutely moving.

 

[Sorcerer]

Oh, you would be surprised what I have heard physics professors get wrong about relativity... you do not need to specialize in anything that has to do with relativity to become a physics professor.

I can attest to that. I have one physics professor at my school who doesn't know much about tensors, and confessed it to the class. He said in order to truly learn general relativity (which he never did in his education) he'd first have to take an entire class on tensor calculus. His point was the degree of difficulty in jumping from SR to GR, in terms of minimal mathematical knowledge required, was very high, and in so doing confessed that he really only knew the bare minimum for tensors, and as such only was familiar with the minimum requirement for relativity.

I understand that now days people prefer to use the same tensor formulation for special relativity to ease the transition, but in his case, he point blank admitted to the class he was not very familiar with GR or the math needed for it.

 

[Orodruin]

I can attest to that. I have one physics professor at my school who doesn't know much about tensors, and confessed it to the class. He said in order to truly learn general relativity (which he never did in his education) he'd first have to take an entire class on tensor calculus. His point was the degree of difficulty in jumping from SR to GR, in terms of minimal mathematical knowledge required, was very high, and in so doing confessed that he really only knew the bare minimum for tensors, and as such only was familiar with the minimum requirement for relativity.

I understand that now days people prefer to use the same tensor formulation for special relativity to ease the transition, but in his case, he point blank admitted to the class he was not very familiar with GR or the math needed for it.

I was talking about SR ... There are many research fields where you do not need SR and so professors from those fields are not required to have any knowledge beyond a vague memory of their own undergrad course. This leads to confusions that, sadly, are passed on to a new generation of students.

 

[Sorcerer]

I was talking about SR ... There are many research fields where you do not need SR and so professors from those fields are not required to have any knowledge beyond a vague memory of their own undergrad course. This leads to confusions that, sadly, are passed on to a new generation of students.

Truthfully, I'm not even sure this particular professor is knowledgeable about SR either. He specialized in QM in his grad work, and the only class he teaches that even touch it are the first two year classes (your first two semesters of intro physics and then your first two semesters of intro to modern physics).

Still, every physics graduate has to pass electricity and magnetism, and I would hope SR is covered to at least a reasonable degree in the graduate level versions of those classes, but probably not enough to cover all the subtleties.
I do feel every university or college should have an entire class on SR at minimum. My school does not, but we do have a senior class where the student chooses their area of emphasis, which includes SR as an option (and really any topic that is approved).

 

[Arman777]

Hey all again, I now defenitly need to derive the lorentz transformation using the length contraction and time dilation. I ll write down the equations for time part of the lorentz transformation but I couldn't derive the length part.Lets think a train and in the middle there's a light beam sender ,which sends 2 beams of light into the opposite directions. The observer on the train will see that both beams arrive there at the same time. But for the observer outside there will be a time delay. Let's call it ΔT. to the delay.

$$ΔT = t_1-t_2=L/2(c+v)-L/2(c-v)$$

Now we have $ΔT=Lvγ^2/c^2$ and then we are applying the length contraction so we have $ΔT=Lvγ/c^2$ So then Lorentz Time transformation will be,

$t' = tγ-ΔTγ$ and that's equal to our case.

Now what about length..How can I find it..

 

[Sorcerer]

Hey all again, I now defenitly need to derive the lorentz transformation using the length contraction and time dilation. I ll write down the equations for time part of the lorentz transformation but I couldn't derive the length part.Lets think a train and in the middle there's a light beam sender ,which sends 2 beams of light into the opposite directions. The observer on the train will see that both beams arrive there at the same time. But for the observer outside there will be a time delay. Let's call it ΔT. to the delay.

$$ΔT = t_1-t_2=L/2(c+v)-L/2(c-v)$$

Now we have $ΔT=Lvγ^2/c^2$ and then we are applying the length contraction so we have $ΔT=Lvγ/c^2$ So then Lorentz Time transformation will be,

$t' = tγ-ΔTγ$ and that's equal to our case.

Now what about length..How can I find it..

Why don't you try assuming the inverse time transformation, plugging it in and the solving for whatever you're calling x?

I was playing around with doing this and got something vaguely similar to what it's supposed to look like, but with your notation using $ΔT$ to equal two different things it's throwing me off some.

 

[vela]

Hey all again, I now defenitly need to derive the lorentz transformation using the length contraction and time dilation. I ll write down the equations for time part of the lorentz transformation but I couldn't derive the length part.Lets think a train and in the middle there's a light beam sender ,which sends 2 beams of light into the opposite directions. The observer on the train will see that both beams arrive there at the same time. But for the observer outside there will be a time delay. Let's call it ΔT. to the delay.

$$ΔT = t_1-t_2=L/2(c+v)-L/2(c-v)$$

Now we have $ΔT=Lvγ^2/c^2$ and then we are applying the length contraction so we have $ΔT=Lvγ/c^2$ So then Lorentz Time transformation will be,

$t' = tγ-ΔTγ$ and that's equal to our case.

Now what about length..How can I find it..

I don't get what you're doing here. I've attached a spacetime diagram which may help. No labels, though, so you have to figure out what's going on.

 

[Arman777]

I don't get what you're doing here
View attachment 232520

I just used this.

 

[Sorcerer]

Hey all again, I now defenitly need to derive the lorentz transformation using the length contraction and time dilation. I ll write down the equations for time part of the lorentz transformation but I couldn't derive the length part.Lets think a train and in the middle there's a light beam sender ,which sends 2 beams of light into the opposite directions. The observer on the train will see that both beams arrive there at the same time. But for the observer outside there will be a time delay. Let's call it ΔT. to the delay.

$$ΔT = t_1-t_2=L/2(c+v)-L/2(c-v)$$

Now we have $ΔT=Lvγ^2/c^2$ and then we are applying the length contraction so we have $ΔT=Lvγ/c^2$ So then Lorentz Time transformation will be,

$t' = tγ-ΔTγ$ and that's equal to our case.

Now what about length..How can I find it..

I was thinking about this for no particular reason and I'm suddenly having an algebra crisis. If I take

$$ΔT = \frac{L}{2(c+v)} - \frac{L}{2(c-v)}$$
I do not get $ΔT = \frac{Lvγ^2}{c^2}$ out of it. I get this:

$$ΔT = \frac{L}{2(c+v)} - \frac{L}{2(c-v)}$$
$$ΔT = \frac{L(c-v)}{2(c+v)(c-v)} - \frac{L(c+v)}{2(c-v)(c+v)}$$
$$ΔT = \frac{L(c-v)-L(c+v)}{2(c+v)(c-v)}$$
$$ΔT = \frac{L(c-v)-L(c+v)}{2(c^2 - v^2)}$$
$$ΔT = \frac{L(c-v)-L(c+v)}{2c^2(1 - \frac{v^2}{c^2})}$$
$$ΔT = \frac{L(c-v)-L(c+v)}{2c^2\frac{1}{γ^2}}$$
$$ΔT = \frac{L}{2} \frac{(c-v)-(c+v)}{c^2\frac{1}{γ^2}}$$
$$ΔT = \frac{L}{2} \frac{(c-v-c-v)}{c^2\frac{1}{γ^2}}$$
$$ΔT = \frac{L}{2} \frac{-v-v}{c^2\frac{1}{γ^2}}$$
$$ΔT = \frac{L}{2} \frac{-2v}{c^2\frac{1}{γ^2}}$$
$$ΔT = \frac{L}{2} \frac{-2vγ^2}{c^2}$$
$$ΔT = \frac{-Lvγ^2}{c^2}$$

Why do I have a negative sign on mine and you do not? Did I screw up some algebra somewhere?

If, on the the other hand, I swap the two original terms, I get:

$$ΔT = \frac{L}{2(c-v)} - \frac{L}{2(c+v)}$$
$$ΔT = \frac{L(c+v)}{2(c-v)(c+v)} - \frac{L(c-v)}{2(c+v)(c-v)}$$
$$ΔT = \frac{L(c+v)-L(c-v)}{2(c-v)(c+v)}$$
$$ΔT = \frac{L}{2}\frac{(c+v)-(c-v)}{(c-v)(c+v)}$$
$$ΔT = \frac{L}{2}\frac{c+v-c+v}{(c^2-v^2)}$$
$$ΔT = \frac{L}{2}\frac{2v}{(c^2-v^2)}$$
$$ΔT = L\frac{v}{(c^2-v^2)}$$
$$ΔT = L\frac{v}{c^2(1-\frac{v^2}{c^2})}$$
$$ΔT = \frac{Lvγ^2}{c^2}$$
So, did you write the first one backwards or did I make some sort of error here?

 

[Arman777]

Well, you are right. I probably misplaced the two terms