# Problem understanding Lorentz invariance [duplicate]

So they usually started with "...This is obviously Lorentz invariant, because of the 4-vector character of the quantity,..., (and after a two page long derivation) another quantity is also obviously Lorentz invariant". I have encountered this problem a lot of times in several physics textbooks.

I feel they are implying that as long as a quantity is a tensor, it is automatically invariant. But is there a formal way of knowing/proving this?

For example:

$$\int d^4 p \: \delta \left( p^2 - m^2 \right) \, \theta \left( p^0 \right) = \int \frac{d^3p}{2E_{\bf p}}.$$ "The relation shows that the expression $d^3p/2E_{\bf p}$ is Lorentz invariant."

How do they see this?

• Tensor (components) are not invariant. Scalars are. – Danu May 27 '15 at 10:02
• – Heterotic May 27 '15 at 14:06
• I think you should dig up some information on tensors and Einstein's summation rule. – Prof. Legolasov May 27 '15 at 16:23
• Possible duplicate: physics.stackexchange.com/q/167813/2451 – Qmechanic May 27 '15 at 17:31

Under a Lorentz transformation $$d^4 p \to | \det \Lambda | d^4 p = d^4 p$$ $$p^2 \to (\Lambda p)^2 = p^2 \implies \delta(p^2 - m^2) \to \delta(p^2 - m^2)$$ and $$dp^0 \text{sign}~p^0 \to dp^0 \text{sign}~p^0 \implies dp^0 \theta(p^0) \to dp^0 \theta(p^0)$$ Thus, the LHS is Lorentz invariant. Therefore the RHS is Lorentz invariant.
BTW, a general tensor, e.g. $T_{\mu\nu}$ is Lorentz covariant, NOT Lorentz invariant. Only a scalar is Lorentz invariant. The only exception is the metric $g_{\mu\nu}$ and the Levi-Civita tensor $\varepsilon_{\mu\nu\rho\sigma}$. The latter is a only invariant under proper Lorentz transformations.