Why is the scalar product of two four-vectors Lorentz-invariant?

For instance, given two four-vector $A^\mu$ and $B^\mu$, so their scalar product is $A\cdot B=A^\mu B_\mu=A^\mu g_{\mu\nu}B^{\nu}$.

Why is $A^\mu g_{\mu\nu}B^{\nu}$ Lorentz-invariant?

  • 3
    $\begingroup$ Have a look at eranreches' answer to Scalar invariance under Lorentz-transformation. That should answer your question or at least help you understand it. $\endgroup$ – John Rennie Nov 20 '18 at 11:25
  • $\begingroup$ That's the only way the speed of light can be the same in all frames. Any standard reference on special relativity should explain this. I personally like the book "Introduction to Special Relativity" by Rindler. $\endgroup$ – Eric David Kramer Nov 20 '18 at 12:14
  • $\begingroup$ A scalar product like $A^\mu B_\mu$ is invariant under any smooth coordinate transformation. The only thing that's special about Lorentz transformations is that they leave the components of the metric invariant. $\endgroup$ – user4552 Nov 20 '18 at 15:00
  • $\begingroup$ Possible duplicate of Lorentz invariance of the Minkowski metric $\endgroup$ – AccidentalFourierTransform Nov 20 '18 at 18:52

Frankly, you're looking at this backwards.

Why is $A^\mu g_{\mu\nu}B^{\nu}$ Lorentz-invariant?

That's the wrong way around: $A^\mu g_{\mu\nu}B^{\nu}$ is Lorentz invariant because Lorentz transformations are defined as the class of transformations that leaves $A^\mu g_{\mu\nu}B^{\nu}$ invariant.

Generally, if you transform $A^\mu$ and $B^\mu$ by some linear transformation with the transformation matrix $\Lambda^\mu_{\ \ \nu}$, then their transformed values will be $\tilde A^\mu = \Lambda^\mu_{\ \ \nu}A^\nu$ and $\tilde B^\mu = \Lambda^\mu_{\ \ \nu} B^\nu$ (using Einstein summations). This means that the transformed inner product will be \begin{align} \tilde A^\mu g_{\mu\nu} \tilde B^{\nu} & = (\Lambda^\mu_{\ \ \alpha}A^\alpha) g_{\mu\nu} (\Lambda^\nu_{\ \ \beta}B^\beta) \\ & = A^\alpha (\Lambda^\mu_{\ \ \alpha} g_{\mu\nu} \Lambda^\nu_{\ \ \beta} )B^\beta \\ & = A^\mu (\Lambda^\gamma_{\ \ \mu} g_{\gamma\delta} \Lambda^\delta_{\ \ \nu} )B^\nu \qquad\qquad\text{(by re-labelling)} \\ & \stackrel{\text{required}}= A^\mu g_{\mu\nu} B^\nu. \end{align} Thus, for $A^\mu g_{\mu\nu} B^\nu$ to be invariant we require that $$ A^\mu (\Lambda^\gamma_{\ \ \mu} g_{\gamma\delta} \Lambda^\delta_{\ \ \nu} )B^\nu = A^\mu g_{\mu\nu} B^\nu $$ for all $A^\mu$ and $B^\mu$, and by judicious choices of those vectors (basically running each independently over the basis in use) that can only be the case if $$ \Lambda^\gamma_{\ \ \mu} g_{\gamma\delta} \Lambda^\delta_{\ \ \nu} = g_{\mu\nu}, $$ which forms the core requirement on $\Lambda^\mu_{\ \ \nu}$ for it to be a Lorentz transformation.


Here's the way to think about this -- why is the standard Euclidean dot product, $\sum x_iy_i$ interesting? Well, it is interesting primarily from the perspective of rotations, due to the fact that rotations leave dot products invariant. The reason this is so is that this dot product can be written as $|x||y|\cos\Delta\theta$, and rotations leave magnitudes and relative angles invariant.

Is the standard Euclidean norm $|x|$ invariant under Lorentz transformations? Of course not -- for instance, $\Delta t^2+\Delta x^2$ is clearly not invariant, but $\Delta t^2-\Delta x^2$ is. Similarly, $E^2+p^2$ is not important, but $E^2-p^2$ is. The reason this is the case is that Lorentz boosts are fundamentally skew transformations, which means the invariant locus is a hyperbola, not a circle. So you have $\cosh^2 \xi - \sinh^2 \xi = 1$, and $x_0^2-x_1^2$ is the right way to think of the norm on Minkowski space.

Similarly, Lorentz boosts change the rapidity $\xi$ by a simple displacement, so $\Delta \xi$ is invariant. From this point, it's a simple exercise to show that


(as for the remaining dimensions -- remember that the standard Euclidean dot product is still relevant in space, so you just need to write $x_0y_0-x\cdot y=x_0y_0-x_1y_1-x_2y_2-x_3y_3$.)


A vector $\mathbf{v} = v^i \, \mathbf{e}_i = q^j \,\mathbf{u}_j$ has different vector components ($v^i$, $q^j$ in this case) in different bases ($\{\mathbf{e}\}$,$\{\mathbf{u}\}$, in our example) which we can interpret as different reference frames (different axes with different origins).

Physicists are lazy: they refer to the vector components $v^i$ as vectors, which is a misnomer! A true vector $\mathbf{v}$ exists out width whichever basis you choose to work in but to know its entries you must reference these with respect to a given basis: this is just elementary linear algebra.

Now, the magnitude of a vector is independent of whichever bases you choose for its description (that is, geometrically speaking its length is fixed):

$$ v^2 = v^i \; v_i \, (\mathbf{e}^i \cdot \mathbf{e}_i) = q^{j} \, q_j\,(\mathbf{u}_j\cdot \mathbf{u}^j) . \tag{assuming orthonormal bases} $$

Hence, scalars do not transform upon of change of basis. In fact it doesn't make much sense to talk about basis for scalars since intuitively these are just numbers.

However, one other way to look at this is to consider a scalar a special type of vector with only one entry and one orthonormal basis (the number 1): its "length" must also be fixed. Hence, this one-dimensional "vector" is the same independent of reference frame.

This is true for all vectors, including special relativistic four-vectors.

As a sanity check, one of the tenets of special relativity is that $c$, the speed of light and a scalar, is the same for all observers. This could not be so if it was somehow different in different frames.

  • $\begingroup$ @EmilioPisanty Yes, indeed! Fixed. $\endgroup$ – OldTomMorris Nov 20 '18 at 21:59

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