# Special Relativity - Lorentz Transformation

If I assume only 1-dimensional space i.e. every event is defined by just 2 space-time co-ordinates $$(x,t)$$

Can I derive the Lorentz Transformation in this setup?

Only condition I can use in 1-D setup is that light beam travelling in $$+ve$$ $$x$$-axis and $$-ve$$ $$x$$-axis will both have same speed '$$c$$' in both the inertial reference frames $$S$$ and $$S$$'. And then I'm stuck there.

In 3-D I think we have a rather "stronger" condition that the radius of the light sphere grows at the rate of '$$c$$' for both $$S$$ and $$S'$$ and this can then be solved mathematically to get Lorentz equations.

I'm unable to do same in 1-D

So is it correct that Lorentz equations can't be derived considering only the 1-D setup?

• What stops you from doing the same procedure without writing $y$ and $z$ parts?
– ACat
May 16, 2020 at 14:40
• How do you do the 3d case? I don't see how an expanding "light sphere" helps, or even has any relevance. May 16, 2020 at 14:50
• @DvijD.C. - Good point, I guess I'm just confused let me try and get back, Thanks May 16, 2020 at 14:51
• @garyp I guess it is just some convoluted way of saying that the speed of light remains the same in all frames. I don't exactly remember the details but I remember reading it in Resnick's book on relativity (it's filled with other such confusing "explanations" for simple things) ;)
– ACat
May 16, 2020 at 14:57
• @DvijD.C.Indeed Resnick book is what I'm reading. Guess I then picked a wrong one. :( May 16, 2020 at 15:04

Yes you can always derive the Lorentz transformations in one dimension of space, and then generalise them. The point is that the Lorentz Transformations have rotational symmetry, so you could first work the equations out in the radial direction (parallel to the velocity) and then just apply coordinate rotation. Here's an excerpt from my book on SR

## Relativity of Simultaneity

As the name suggests, Relativity of simultaneity means that simultaneity depends on the reference frame. Two events which are simultaneous in one frame need not be so in another frame. In Galilean Relativity, time was absolute, and it flowed the same for everyone, but with Special Relativity the time had come to abolish the old ideas. Time was no longer absolute, and the rate at which it flows could now be changed.

## Synchronization of Clocks

Now we come to an experiment, whose results we will use to derive the Lorentz transformations. The experiment is synchronization of clocks. It's a simple experiment which uses the second postulate. This time however we would need more than two people. Say you are standing on the ground while Vis and Sandy are on a bus. They wish to synchronize their clocks using the second postulate, so what should thy do? To help them, they have an equipment which will detect if the light shone onto it reaches it simultaneously. There are many possible ways in which they could synchronize their clocks, but we will be taking the simplest case. So they both sit at the same distance from the equipment and shine their lasers at what they think at the same instant, and the equipment buzzes, but what do you, the observer standing outside, looking at the bus move, what do you see? Obviously the light from both Vis and Sandy reaching the equipment should be simultaneous, but what you do not see to be simultaneous is at the time at which they shoot their lasers. Let's consider the following Space—Time diagram to aid our analysis of the situation.

Check this for the Space Time Diagram

Now Let's Solve for the point $$A$$ first, the light ray released by Sandy has the equation $$x=t$$ And the equipment's timeline/worldline equation is $$x=vt+1$$ Their intersection corresponds to point $$A$$ whose coordinates are $$x=t=\frac{1}{1-v}$$ We know that the light ray from Vis must pass through A and in our Space-Time diagrams they must have either a slope of 1 or -1 Thus the equation of the light released from Vis's laser is $$x+t=\frac{2}{1-v}$$ Now this must have come out of Vis's laser at the same time as Sandy's laser in the bus frame, thus wherever the light ray has intersected Vis's worldline, that is where she had turned her lights on, and everywhere on the line connecting the two points which they two had turned their lasers on, would correspond to $$t'=0$$. Now let's find out the coordinates of the point B, Where Vis turns her laser on. The point B is the point of intersection of the two lines $$x+t=\frac{2}{1-v}$$ which is the equation of the light released by Vis, and $$x=vt+2$$ which is Vis's worldline.

Solving for the point B we get $$t=\frac{2v}{1-v^2}$$ and $$x=\frac{2}{1-v^2}$$ Thus the equation of $$t'=0$$ is the same as $$t=vx$$ We shall be using this to derive the Lorentz transformation, which are the equivalent of Galilean Transformations in Special Relativity.

## Lorentz Transformations

We shall now talk about Lorentz transformations, which we said were the equivalent of Galilean transformations in Special Relativity. But how exactly do we get to the Lorentz transformations? We shall use the facts about $$x$$, $$x'$$, $$t$$ and $$t'$$ which we derived in the previous section. First of all we know that when $$t'=0$$, $$t=vx$$, hence we may write the following $$t'=(t-vx)f$$

Similarly we can say $$x'=(x-vt)g$$

I will leave it as an exercise for you to prove that $$f$$ and $$g$$ are equal. I'll give you a hint as well, write down the transformation equations for light.

Now let us write down the reverse transformations as well which give me the transformation equations for $$x'$$ and $$t'$$ in terms of $$x$$ and $$t$$. Also since $$f=g$$, I will write them as $$f=g=\gamma$$ Thus we can write $$t=(t'+vx')\gamma$$ and $$x=(x'+vt')\gamma$$

This is how you derive the Lorentz Transformations for (1+1)dimensional motion

Yes, you can derive the Lorentz transformation in just 2D (1D space, and 1D time). It is done here. But right now, if you have an object moving only in the $$x$$ direction, when you look at it, the Lorentz transforms are defined as: $$t^{'} = \gamma (t - \frac{vx}{c^2})$$ $$x^{'} = \gamma (x - vt)$$ $$y^{'} = y$$ $$z^{'} = z$$

Here the only axes that are transformed are the are the $$x$$ and the $$t$$. That's because the particle moves only in the $$x$$ direction and through time. So you can entirely ignore the $$y$$ and $$z$$ components, as the object does not have a velocity in the $$y$$ and $$z$$ directions.

What you get is analogous to what you would get if you had only 1 dimension of space and 1 dimension of time.

Thanks PNS and SK Das - That you have to use the symmetry argument of the transformation equations with respect to S and S' is something I was not using when I was trying it.

(While I think perfectly commonsense - I just feel bit unconformable about this symmetry argument - not saying it is wrong though)