Deciding if a transformation is a Lorentz transformation

[spaghetti3451]

Homework Statement

Is the transformation $Y:(t,x,y,z)\rightarrow (t,x,-y,z)$ a Lorentz transformation? If so, why is it not considered with P and T as a discrete Lorentz transformation? If not, why not?

Homework Equations

The Attempt at a Solution

A Lorentz transformation $\Lambda$ satisfies the relation $\Lambda^{T}g\Lambda = \Lambda$, where $g$ is the Minkowski metric.

In our case, the transformation $Y = \text{diag}(1,1,-1,1)$.

Therefore, $Y^{T}gY = \text{diag}(1,1,-1,1) \cdot{\text{diag}(1,-1,-1,-1)}\cdot{\text{diag}(1,1,-1,1)} = \text{diag}(1,-1,-1,-1) =$ the parity operator.

Therefore, the relation $Y^{T}gY = Y$ is not satisfied.

Therefore, $Y$ is not a Lorentz transformation.

Is my answer correct?

 

[strangerep]

A Lorentz transformation $\Lambda$ satisfies the relation $\Lambda^{T}g\Lambda = \Lambda$, [...]

Are you sure about that? Are you sure it's not $\Lambda^{T}g\Lambda = g$ ?

Alternatively, you could just check whether the Minkowski line element $ds^2 = t^2 - x^2 - y^2 - z^2$ is preserved.

 

[spaghetti3451]

Oh, right! In that case, the transformation $Y$ satisfies $Y^{T}gY=g$, so the transformation $Y$ is a Lorentz transformation.

Second way:

An inner product of Lorentz four-vectors is Lorentz invariant.

(A Lorentz four-vector $V^{\mu}$ is a four-vector which transforms under the Lorentz transformation ${\Lambda^{\mu}}_{\nu}$ with the transformation law $V'^{\mu} = {\Lambda^{\mu}}_{\nu}V^{\nu}$.)

(A mathematical object (i.e. Lorentz scalar, Lorentz four-vector, Lorentz tensor and combinations thereof) is Lorentz invariant if it does not change under a Lorentz transformation. Not all mathematical objects are necessarily Lorentz invariant.)

The only four-vectors mentioned in the problem are the position four-vectors $x^{\mu}$ before and $x'^{\mu}$ after the transformation $Y$.

We know that the position four-vector is a Lorentz four-vector, so the inner products $x^{\mu}x_{\mu}$ before and $x'^{\mu}x'_{\mu}$ after a Lorentz transformation must be equal to each other.

So, let's take the inner products $x^{\mu}x_{\mu}$ before and $x'^{\mu}x'_{\mu}$ after the transformation $Y$ and check if the product is invariant:

Before the transformation $Y$, $x^{\mu}x_{\mu}=g_{\mu\nu}x^{\mu}x^{\nu}=g_{00}x^{0}x^{0}+g_{11}x^{1}x^{1}+g_{22}x^{2}x^{2}+g_{33}x^{3}x^{3} = (x^{0})^{2}-(x^{1})^{2}-(x^{2})^{2}-(x^{3})^{2}=(t)^{2}-(x)^{2}-(y)^{2}-(z)^{2}$.

After the transformation $Y$, $x'^{\mu}x'_{\mu}=g_{\mu\nu}x'^{\mu}x'^{\nu}=g_{00}x'^{0}x'^{0}+g_{11}x'^{1}x'^{1}+g_{22}x'^{2}x'^{2}+g_{33}x'^{3}x'^{3} = (x'^{0})^{2}-(x'^{1})^{2}-(x'^{2})^{2}-(x'^{3})^{2}=(t)^{2}-(x)^{2}-(-y)^{2}-(z)^{2}=(t)^{2}-(x)^{2}-(y)^{2}-(z)^{2}$.

The inner products are the same under the transformation $Y$, so the transformation $Y$ is a Lorentz transformation.Is my solution accurate?

 

[strangerep]

Second way:

An inner product of Lorentz four-vectors is Lorentz invariant.

Yes, but you don't need to do all that work if you realize that $Y^T = Y^{-1}$ for a Lorentz transformation (since this is true for all "SO(...)" transformations). Just write out the inner product in matrix notation...

 

[spaghetti3451]

So, do you mean that,

$x'^{\mu}x'_{\mu} = x'^{T}x' = (Yx)^{T}(Yx) = x^{T}Y^{T}Yx = x^{T}x$,

so that the inner product is invariant?

 

[strangerep]

That's the general idea, but your details are not quite right. For arbitrary 4-vectors $x,y$ we have
$$y'^T g' x' ~=~ y^T Y^T \; g' \; Y y ~,$$ so if $Y^T g' Y = g$, then the rhs is $y^T g x$, showing that the inner product between 2 arbitrary 4-vectors is preserved.