# Are covariant components of 4-gradient of scalar product made of contravariant components of vector?

Trying to find normal to hypersurface $p^ip_i=m^2c^2$, I take gradient of corresponding function $F=p^ip_i-m^2c^2$. What I get is:

$$n_i=\frac{\partial F}{\partial p^i}=\frac{\partial}{\partial p^i}\left(p^ip_i-m^2c^2\right)=\\ =\frac{\partial}{\partial p^i}\left(\left(p^0\right)^2-\left(p^1\right)^2-\left(p^2\right)^2-\left(p^3\right)^2-m^2c^2\right)=2(p^0,-p^1,-p^2,-p^3)$$

This looks strangely to me: on the LHS we have $n_i$, which are covariant components of normal, and on the RHS I have $p^i$'s, which are contravariant components of 4-momentum. So it looks like covariant components of normal are made of contravariant components of 4-momentum. How could this be? Have I made a mistake somewhere?

• The vector on the RHS may involve contravariant components of the 4-momentum, but it is certainly not equal to $$(p^0,p^1,p^2,p^3) \,,$$ which are the contravariant components of the 4-momentum. Using the relation $$p_\mu = \eta_{\mu \nu} p^\nu$$ we see that the list of components on the right are indeed the covariant components of the 4-momentum. The covariant and contravariant components of an object are closely related, and so it is natural that we could write them in terms of each other --- we have $p_1 = -p^1$, for instance. Feb 1 '14 at 13:12
• @gj255 indeed, I didn't see the obvious possibility to lower the indices on the RHS and get the simple relation. Have to do more index gymnastics to get used to it :) Feb 2 '14 at 6:58

## 1 Answer

You should use only contravariant components when differentiating, but making explicit the presence of the metric, since you differentiate with respect to these components. So: $$n_i=\frac{\partial F}{\partial p^i}=\frac{\partial}{\partial p^i}\left(g_{hk}p^hp^k-m^2c^2\right)= g_{hk} \delta^h_i p^k + g_{hk} p^h\delta^k_i = g_{ik} p^k + g_{hi} p^h = g_{ik} p^k + g_{ki} p^k= g_{ik} p^k + g_{ik} p^k =2g_{ik}p^k = 2p_i\:.$$