# Using the Minkowski metric show that $u_au^a = -c^2$

Using the Minkowski metric show that $$u_au^a = -c^2$$.

I am trying to solve the above problem.

Firstly $$4$$-velocity can be defined as $$u_a = \frac{dx^a}{dt'}$$ where $$dt' = dt\sqrt{1-\frac{v^2}{c^2}}$$ and we use the usual summation convention.

My question is what exactly is $$u^a$$ and what is it from a strictly mathematical perspective? Moreover how exactly can we multiply $$u_au^a$$ since they are both vectors (and you can't multiply elements of a vector space).

The only way this would make sense is if both $$u_a$$ and $$u^a$$ were tangent vectors in the tangent space (at a point) to Minkowski space and then $$u_au^a := g(u_a, u^a)$$ where $$g = -(dx^0)^2 + (dx^1)^2 + (dx^2)^2 + (dx^3)^2$$ is the Minkowski metric and this is the only way I see that this question could make any sense.

So my question is what exactly is $$u^a$$ and how can I show that $$u_au^a = -c^2$$?

• If $u^a$ is an element of a vector space, then $u_a$ is an element of the dual vector space. We refer to $u_a$ and $u^a$ as covariant and contravariant vectors, respectively — which are related by raising and lowering the indices, which is essentially the musical isomorphism. If you want to really understand what's going on, you should probably read those links. – Mike Oct 28 '18 at 16:56
• I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. – Ben Crowell Oct 28 '18 at 16:57
• @Mike In that case $u^a$ is a one-form, and again multiplication $u_au^a$ doesn't make sense, the only way that would make sense is if $u^a$ took $u_a$ as an input i.e. $u^a(u_a)$ or if we identified $u_a$ with the element in the bi-dual space and then evaluated $u_a(u^a)$, then only would we get an element of $\mathbb{R}$ as a result – Perturbative Oct 28 '18 at 17:46
• Actually, $u_a$ is the one-form according to the usual conventions. (It might be a little confusing because the basis vectors for one-forms are denoted $\mathbf{e}^i$. Another link that might help clear up that point is abstract index notation.) And you're not multiplying them; you're contracting them... – Mike Oct 28 '18 at 17:56
• And you can certainly think of $u^a$ as taking $u_a$ as input to arrive at $u^a(u_a)$ because the dual of the dual space is the original space. But it is more common to think of $u_a u^a$ as being precisely $u_a(u^a)$. – Mike Oct 28 '18 at 17:57

First of all the correct definition of the 4-velocity is $$u^i = \frac{dx^i}{dt'}$$, where $$i=0,\ldots,3$$. Secondly in physics in 90% of all cases we work with components of vectors or tensors or forms, so $$u^i$$ is meant to be the ensemble of the components of the 4-velocity, so these are just coefficients(numbers) which might follow under coordinate change (here only Lorentz transformations are permitted) some transformation laws which actually define which type of object they belong to.

$$u^i$$ are called contravariant components, and $$u_i$$ are called covariant components of the 4-velocity and are just defined like $$u_i = g_{ik}u^k$$ with Einstein's summation convention applied, i.e. over double appearing indices is summed. $$g_{ik}$$ are the components of the metric tensor here of the Minkowski space, again it is be understood as ensemble of coefficients which transform under change of coordinates specifically which shows that they belong to a tensor. In Minkowski space $$g_{ik} =diag(-1,1,1,1)$$.

A more mathematical approach is to define a 1-form $$g(u,\cdot)$$ with the definition given in the post $$g = - (dx^0)^2 + (dx^1)^2+(dx^2)^2+(dx^3)^2$$. This form would have the following representation in coordinates ($$u= u^i \partial_i)$$:

$$g(u,\cdot)=-(dx^0)^2(u) + (dx^1)^2(u)+(dx^2)^2(u)+(dx^3)^2(u)=-u^0dx^0+u^1dx^1+u^2dx^2+u^3dx^3$$

We remember that the covariant components of $$u$$ are defined like $$u_i = g_{ik}u^k$$ Applied on $$g(u,\cdot)$$ it yields ($$u_0= -u^0, u_{1,2,3} = u^{1,2,3}$$ applying the metric tensor components)

$$g(u,\cdot)=u_0dx^0+u_1dx^1+u_2dx^2+u_3dx^3\equiv u_i dx^i$$

We got the 1-form in coordinate representation.

We actually want to compute $$g(u,u)$$ which is in components exactly what we are looking for: $$g(u,u)= g_{ik}u^i u^k = u^i u_i$$:

$$g(u,u)=u_i dx^i(u) = u_i u^i$$

In the last step finally physics comes in. According to special theory of relativity the invariant line element does not change under coordinate change (here exclusively Lorentz transformations are permitted) $$x\longrightarrow x'$$ $$- (dx^0)^2 + (dx^1)^2+(dx^2)^2+(dx^3)^2 =- (dx'^0)^2 + (dx'^1)^2+(dx'^2)^2+(dx'^3)^2$$

We choose the $$x'$$ coordinate system to be the comoving eigen system of the particle whose 4-velocity we are looking at. In this system the particle has the coordinates $$x'=(x'^0,0,0,0)=(ct',0,0,0)$$. Applied on the last expression we get:

$$- (dx^0)^2 + (dx^1)^2+(dx^2)^2+(dx^3)^2 =- (dx'^0)^2\equiv - c^2 dt'^2$$

Now we apply the definition of the components of the 4-velocity:

$$u_i u^i = \frac{dx_i}{dt'} \frac{dx^i}{dt'}= g_{ik} \frac{dx^k}{dt'} \frac{dx^i}{dt'} = \frac{- (dx^0)^2 + (dx^1)^2+(dx^2)^2+(dx^3)^2 }{(dt')^2} = -c^2$$

QED.