Covariant vs. contravariant time component

1. Oct 4, 2008

snoopies622

...of the four-momentum vector.

Why is the energy of a particle identified with p0 instead of p0? Is there a theoretical basis for this, or was it simply observed that p0 is conserved in a larger set of circumstances?

2. Oct 4, 2008

atyy

How is the metric written in that book (-+++) or (+---)?

Covariant and contravariant components are related by the metric.

Just make sure the energy is positive.

3. Oct 4, 2008

clem

In the more usual metric (+---) of SR, p^0 and p_0 are equal, so it doesn't matter which you use. In any event, although they are equal, it should be p^0 that is the physical energy since the four-momentum is a contravariant vector.
If you are reading a book that uses the metric (-+++), then everything could be different.

4. Oct 4, 2008

snoopies622

I did not have the Minkowski metric specifically in mind. If one uses the Schwarzschild metric -- or any other diagonal metric with $$|g_{00}|\neq1$$ -- p0 and p0 differ by more than just the sign; they have different magnitudes, so the energy of a particle cannot have both values. It's been my impression that in such circumstances one uses the covariant form instead of the contravariant form, but I don't know why.

5. Oct 4, 2008

George Jones

Staff Emeritus
If one uses standard Schwarzschild coordinates, then

$$k = \frac{\partial}{\partial t}$$

is a timelike Killing vector. If $u$ is the 4-velocity of a freely falling particle, then

$$E = g \left( k , u \right)$$

is constant along the particle's worldline.

What is the coordinate expression of the above coordinate-free expression?

6. Oct 4, 2008

snoopies622

What I had in mind was

$$p^0=m_0 c \frac{dt}{d\tau}$$ while $$p_0= g_{00}m_0 c \frac{dt}{d\tau}=(1-\frac{r_s}{r})m_0 c \frac{dt}{d\tau}$$

where $$d\tau=ds/c$$ and $$ds$$ is defined using the Schwarzschild metric. Since $$(1-\frac{r_s}{r})$$ won't equal 1 while r is finite, these two terms (p0 and p0) have different values, and I don't know which one (if either) represents the energy of the particle.

Regarding the coordinate-free expression, when you say that $$k = \frac{\partial}{\partial t}$$ is a vector, do you mean the partial derivative of the displacement vector with respect to coordinate time? or of a different vector?

Last edited: Oct 4, 2008