Can we choose different functional forms for Lorentz transformation?

[zenterix]

TL;DR

When deriving the Lorentz transformation, it seems we can pick specific functions to be used in the equations. If we choose functions that lead to a slightly different transformation, what do we make of this transformation?

Susskind's book "Special Relativity and Classical Field Theory" derived the Lorentz transformations

$$x'=(x-vt)\frac{1}{\sqrt{1-v^2}}$$

$$t'=(t-vx)\frac{1}{\sqrt{1-v^2}}$$

$$x=(x'+vt')\frac{1}{\sqrt{1-v^2}}$$

$$t=(t+vx')\frac{1}{\sqrt{1-v^2}}$$

While redoing the calculations, I reached a point where it seems to me I could choose the transformation to be

$$x'=x-vt$$

$$t'=t-vx$$

$$x=(x'+vt')\frac{1}{1-v^2}$$

$$(t'+vx')\frac{1}{1-v^2}$$

Are these latter four equations also Lorentz transformations?

I'm going to post here my exact calculations to reach these equations.

I started with a stationary frame A (coordinates x and t) and a moving frame B (coordinates x' and t', with three railcars)

From this scenario we conjecture a transformation of the form

But then what if we think about frame B being stationary and the three railcars being in frame A (which is now moving relative to frame B). We have

Here are the calculations to find points a and b

So we conjecture that the transformation from coordinates in frame A to frame B are

$$x=(x'+vt)f_2(v^2)$$

$$t=(t'+vx')g_2(v^2)$$

At this point, taking into consideration the four equations that we have conjectured, we have

But we can show that $f_1=g_1$ and $f_2=g_2$ by considering the fact that a light ray has equations

$$x=t$$

$$x'=t'$$

Thus,

Hence our equations become

And now if we solve for $f_1$ or $f_2$ we obtain

One option is to say $f_1=f_2$ and then we obtain the usual Lorentz transformation.

But can we not choose other functions?

For example suppose $f_1=1$ and $f_2=\frac{1}{1-v^2}$. Then we get

 

[Dale]

TL;DR Summary: When deriving the Lorentz transformation, it seems we can pick specific functions to be used in the equations. If we choose functions that lead to a slightly different transformation, what do we make of this transformation?

But can we not choose other functions?

Only the usual function respects the principle of relativity. The others respect the 2nd postulate, but not the 1st.

 

[PeroK]

Your equations are inconsistent. If you let $v = \pm \frac c 2$, or any other speed, you get different transformations for the same relative velocity.

You can't just stop at $f_1f_2 = \frac 1 {1-v^2}$. As my first comment shows, there are further necessary conditions.

 

[Orodruin]

Put slightly differently: You can make whatever coordinate transformations you want, but you cannot do so expecting the metric to stay the same. Much like you cannot expect the metric to stay the same using polar coordinates on $\mathbb R^2$.

Only the Lorentz transformations keep the metric invariant.

 

[Dale]

Put slightly differently: You can make whatever coordinate transformations you want, but you cannot do so expecting the metric to stay the same. Much like you cannot expect the metric to stay the same using polar coordinates on $\mathbb R^2$.

Only the Lorentz transformations keep the metric invariant.

That is a good way to put it

So for this particular case, if we start in an inertial frame with the metric $$ds^2=-dt^2 + dx^2$$ then the transformations $$x'=x-vt$$$$t'=t-vx$$ give $$ds^2 = -\frac{1}{1-v^2} dt'^2 + \frac{1}{1-v^2} dx'^2$$ which is a perfectly valid metric, but just not the inertial frame metric. A clock at rest in these coordinates must be scaled to read coordinate time.

 

[Orodruin]

That is a good way to put it

So for this particular case, if we start in an inertial frame with the metric $$ds^2=-dt^2 + dx^2$$ then the transformations $$x'=x-vt$$$$t'=t-vx$$ give $$ds^2 = -\frac{1}{1-v^2} dt'^2 + \frac{1}{1-v^2} dx'^2$$ which is a perfectly valid metric, but just not the inertial frame metric. A clock at rest in these coordinates must be scaled to read coordinate time.

One can also note that, since the metric is simply scaled by a factor $\gamma^2$:

• The coordinate speed of light is still 1 because $dx’ = \pm dt’$ leads to $ds^2 = 0$
• The coordinate system is still affine and orthogonal. Just not orthonormal.

 

[PeroK]

That is a good way to put it

So for this particular case, if we start in an inertial frame with the metric $$ds^2=-dt^2 + dx^2$$ then the transformations $$x'=x-vt$$$$t'=t-vx$$ give $$ds^2 = -\frac{1}{1-v^2} dt'^2 + \frac{1}{1-v^2} dx'^2$$ which is a perfectly valid metric, but just not the inertial frame metric. A clock at rest in these coordinates must be scaled to read coordinate time.

The Lorentz Transformation is not a single, specific coordinate transformation. That could be anything. It's supposed to be a transformation that has the same form for any $v$. That transformation fails as a generic transformation, as the inverse transformation (which is still a boost) explicitly has a different form.

 

[Orodruin]

The Lorentz Transformation is not a single, specific coordinate transformation. That could be anything. It's supposed to be a transformation that has the same form for any $v$. That transformation fails as a generic transformation, as the inverse transformation (which is still a boost) explicitly has a different form.

This is a bit ambiguous without specifying what you mean by ”having the same form”. Surely you are not saying that the inverse is the same as the transformation itself.

I can also easily construct a parametrization such that chamging the sign of the parameter does not give the inverse.

The bottom line is that the Lorentz transformations are defined as the group of linear coordinate transformations that preserve the Minkowski metric. As they form a group, the inverse of any Lorentz transformation must also be a Lorentz transformation.

 

[PeroK]

The bottom line is that the Lorentz transformations are defined as the group of linear coordinate transformations that preserve the Minkowski metric.

That's not what the OP is trying to do. He has:
$$x' = x - vt, \ x =\frac 1 {1 -v^2}(x +vt)$$That gives two different transformations for $v = \frac 1 2$, depending on whether we treat it as the transformation, or the inverse transformation of $v = -\frac 1 2$.

In the derivation from first principles, the inverse is a key condition of determining the functions $f_1, f_2$. See the original post.

The point about the metric, although valid, is probably beyond the scope of the Susskknd material he's covered so far.

 

[Orodruin]

That's not what the OP is trying to do. He has:
$$x' = x - vt, \ x =\frac 1 {1 -v^2}(x +vt)$$That gives two different transformations for $v = \frac 1 2$, depending on whether we treat it as the transformation, or the inverse transformation of $v = -\frac 1 2$.

Again, there is nothing stating that you cannot find a parametrization where changing the parameter sign does not give the inverse. It just so happens to be the case for the standard form of the Lorentz transformations because of how v is defined as a relative velocity between frames.

In the derivation from first principles, the inverse is a key condition of determining the functions $f_1, f_2$. See the original post.

It is not a necessary condition. You can easily derive the form of the Lorentz transformations (defined as the set of transformations that preserve the metric) without ever mentioning v or -v.

In fact, this is the route I take in my classes.

The point about the metric, although valid, is probably beyond the scope of the Susskknd material he's covered so far.

Probably so, but that part was more about the general discussion and placing the context rather than necessarily for the OP.

 

[Ibix]

While redoing the calculations, I reached a point where it seems to me I could choose the transformation to be

$$x'=x-vt$$

$$t'=t-vx$$

$$x=(x'+vt')\frac{1}{1-v^2}$$

$$(t'+vx')\frac{1}{1-v^2}$$

Are these latter four equations also Lorentz transformations?

What you've got here is a boost and a scaling; that's why your inverse transforms are also a boost and a scaling with the inverse scale factor. As others have already noted it's not wrong, but it adds additional unnecessary complexity (you need different units in every frame) and conceals the principle of relativity under it.