B Robert Wald's General Relativity: Energy-Momentum Relation

[carpinus]

Hello,
this is my first thread.
Robert Wald, in General Relativity, equation (4.2.8) says :
E = – p_a v^a
where E is the energy of a particle, p_a the energy-momentum 4-vector and v^a the 4-velocity of the particle. How can I see this is compatible with the common energy-momentum-relation E² – p² = m² ?
many thanks for help.

 

[PeroK]

Hello,
this is my first thread.
Robert Wald, in General Relativity, equation (4.2.8) says :
E = – p_a v^a
where E is the energy of a particle, p_a the energy-momentum 4-vector and v^a the 4-velocity of the particle. How can I see this is compatible with the common energy-momentum-relation E² – p² = m² ?
many thanks for help.

In this case, $v^a$ is the four-velocity of the observer who is measuring the energy of the particle. Wald's equation is a generalisation of the one you know, which is that equation in the rest frame of the observer, where: $p_a = (E, \vec p)$ and $v^a = (-1, 0, 0 ,0)$. In that reference frame $p_av^a = -E$.

The more general equation follows from the invariance of the inner product between coordinate systems.

The equation you quote is actually the invariance of the inner product $p_ap^a$. In the rest frame of the particle we have $p_a = (m, 0, 0, 0)$, hence $p_ap^a = -m^2$. Meanwhile, in a different frame we have $p_a = (E, \vec p)$ and $p_ap^a = -E^2 + p^2$, where $p = |\vec p|$.

 

[Ibix]

Wald's equation is a generalisation of the one you know, which is that equation in the rest frame of the observer, where: $p_a = (E, \vec p)$ and $v^a = (-1, 0, 0 ,0)$.

To be pedantic, you've written down the components of $p^a$ and $v_a$ there. And further pedantry, I think Wald would say that because you are using a specific coordinate system you should use Greek indices.

@carpinus: Geometrically, $p_av^a$ is the projection of $p^a$ on to the vector $v^a$. In a coordinate system where that four velocity is the timelike basis vector, this is equal to the zeroth comonent of $p^\mu$ in that basis (times -1 because of Wald's choice of -+++ metric signature). On the other hand, $E^2-p^2=m^2$ is more generally written $p_ap^a=-m^2$, which in geometrical terms says that the modulus-squared of the four momentum is minus the mass squared. So the equations are compatible, but are saying different things.

 

[PeroK]

And further pedantry, I think Wald would say that because you are using a specific coordinate system you should use Greek indices.

That's not pedantry. That's specificity to Wald's conventions. Mathematical conventions are not dictated to that extent.

 

[carpinus]

thank you for your answers.
As far as I understand, p_a and v^a both relate to the same laboratory frame, whereas E is the energy of the particle in the rest frame of the observer. However, the 3-momentum only enters equation (4.2.8) if the laboratory frame is not the rest frame of the observer. Therefore, equation (4.2.8) in fact does not relate energy and momentum (as measured in the same frame).
When reading I was confused by the line "… as measured by an observer – present at the site of the particle - … " .

 

[PeroK]

When reading I was confused by the line "… as measured by an observer – present at the site of the particle - … " .

Yes, in GR quantities are measued locally. So, we have an observer with four-velocity $v^a$ measuring the energy of a particle with four-momentum $p^a$, when it is local to the observer.

No overall "lab" frame is required or implied, as the local measurement itself is an invariant quantity: i.e. "everyone agrees" on the local measurement.

In whatever frame we analyse the measurement, where the $v^a$ and $p^a$ have specific components appropriate for that frame, the quantity $p_av^a$ is the same invariant quantity, owing to the invariance of the inner product.

 

[Ibix]

As far as I understand, pa and va both relate to the same laboratory frame, whereas E is the energy of the particle in the rest frame of the observer.

Not quite. $p_av^a$ is an invariant that can be calculated with any choice of coordinates.

However, $v^a$ defines a local frame (or at least the timelike direction of one), and the invariant is equal to the energy of the particle measured in this frame. You could calculate the particle momentum in this frame if you wanted (it is moving, in general) using $p_ap^a=-m^2$ or $E^2-p^2c^2=m^2$, but Wald is not interested in doing so at this point.

 

[PeterDonis]

How can I see this is compatible with the common energy-momentum-relation E² – p² = m² ?

By itself, you can't, because you also need a formula for the momentum $p$ of the particle as observed by the same observer.

Therefore, equation (4.2.8) in fact does not relate energy and momentum (as measured in the same frame).

That is correct. Equation (4.2.8) says that the energy of the particle as measured by a particular observer (the one whose 4-velocity is $v^a$) is (the negative of) the inner product of the particle's 4-momentum and the observer's 4-velocity. It says nothing at all about the momentum of the particle as measured by that observer.

 

[Ibix]

Incidentally, if you do want the three momentum of the particle in the frame defined by $v^a$, you can take the four momentum $p^a$ and subtract off $Ev^a$ (i.e., the component of the four momentum parallel to $v^a$ multiplied by a unit vector in that direction) to get $p^a-Ev^a$. That's a four vector whose modulus is the three-momentum in that frame (if you can't see why, think about it in the frame where $v^a=(1,0,0,0)$). If you calculate the modulus-squared of that four vector, remembering that $p_ap^a=-m^2$ and $p_av^a=p^av_a=-E$ and keeping careful track of the minus signs, you'll get a familiar looking result.