Invariant Mass in Gravitational Fields: Special Relativity

[DuckAmuck]

TL;DR Summary

What happens to invariant mass of an object when it gets closer or further from a gravitational body?

In Special Relativity, you learn that invariant mass is computed by taking the difference between energy squared and momentum squared. (For simplicity, I'm saying c = 1).
$$m^2 = E^2 - \vec{p}^2$$
This can also be written with the Minkowski metric as:
$$m^2 = \eta_{\mu\nu} p^\mu p^\nu$$
More generally, if there is a different metric (for example Schwartzchild), you would write it as:
$$m^2 = g_{\mu\nu} p^\mu p^\nu$$

Now the question is, if invariant mass does not change from one metric to the other, you get the equation:
$$0 = (g_{\mu\nu} - \eta_{\mu\nu})p^\mu p^\nu$$

This seems to give unphysical results.

I solved for a photon in the Schwartzchild metric, and the only physical solution available is if the Schwartzchild radius is 0. So this seems to imply that invariant mass (or lack thereof) is not invariant under gravitational fields.

Any help here would be much appreciated. Thank you.

 

[Orodruin]

Now the question is, if invariant mass does not change from one metric to the other, you get the equation:

No you do not. The Minkowski metric has nothing to do with the manifold you are now considering. The invariant mass does not depend on the coordinate system you choose and you can, in a single point, always find a coordinate system so that the metric takes the form of the Minkowski metric. However, as you transform your expression to that coordinate system, the components of the 4-momentum will also change.

 

[PeterDonis]

What happens to invariant mass of an object when it gets closer or further from a gravitational body?

Short answer: nothing. An object's invariant mass is an invariant property of the body.

 

[PeroK]

Now the question is, if invariant mass does not change from one metric to the other, you get the equation:
$$0 = (g_{\mu\nu} - \eta_{\mu\nu})p^\mu p^\nu$$

Note that the inner product of a particle's four-velocity with itself is always $1$:

$g_{\mu\nu}u^\mu u^\nu = 1$

Therefore, the inner product of a particle's four-momentum with itself is always $m^2$.

If, as you move through spacetime, the components of the metric change then the components of your four-velocity and four-momentum change to preserve these invariant quantities.

 

[Orodruin]

OP is using a timelike metric convention. Let’s stick to that in order to reduce confusion.

 

[Dale]

Now the question is, if invariant mass does not change from one metric to the other, you get the equation:

0=(gμν−ημν)pμpν0=(gμν−ημν)pμpν​

This seems to give unphysical results.

That equation doesn't make any sense. If the components of p are given in local Minkowski coordinates then you use $\eta$, if the components of p are given in some other coordinates then you use g. You cannot contract vector components given in one coordinate system with the metric components given in another coordinate system. That is indeed unphysical!

 

[DuckAmuck]

That equation doesn't make any sense. If the components of p are given in local Minkowski coordinates then you use $\eta$, if the components of p are given in some other coordinates then you use g. You cannot contract vector components given in one coordinate system with the metric components given in another coordinate system. That is indeed unphysical!

I see now. So it should be written:
$$0 = g_{\mu\nu}p^\mu p^\nu - \eta_{\mu\nu} p'^\mu p'^\nu$$

 

[DuckAmuck]

Note that the inner product of a particle's four-velocity with itself is always $1$:

$g_{\mu\nu}u^\mu u^\nu = 1$

Therefore, the inner product of a particle's four-momentum with itself is always $m^2$.

If, as you move through spacetime, the components of the metric change then the components of your four-velocity and four-momentum change to preserve these invariant quantities.

Thanks for clarifying this. It was what I was thinking but did the math wrong.

 

[Dale]

I see now. So it should be written:
$$0 = g_{\mu\nu}p^\mu p^\nu - \eta_{\mu\nu} p'^\mu p'^\nu$$

Yes, and that equation will hold.

 

[DuckAmuck]

Some follow-up questions. If there's a photo traveling toward the center of a gravitational body, we have:
$$m^2 = 0 = g_{\mu\nu} p^\mu p^\nu$$
If we simplify by saying motion is along the x-axis:
$$g_{00} E^2 + g_{11} p^2 = 0$$
Plug in the Schwartzchild metric, and we get
$$\frac{\left(1 - \frac{r_s}{4R}\right)^2}{\left(1 +\frac{r_s}{4R}\right)^4} = \frac{p^2}{E^2}$$
This seems to imply light slowing down to be subluminal in a gravitational field?
$$v = \left|\frac{p}{E}\right| = \frac{1 - \frac{r_s}{4R}}{\left(1 + \frac{r_s}{4R}\right)^2} \leq 1$$

If you were to try to indirectly measure the mass of the photon here by measuring energy and momentum, you would get a non-zero mass?
$$m_{\text{fake}}^2 = E^2 \left( 1 - \frac{\left(1 - \frac{r_s}{4R}\right)^2}{\left(1 +\frac{r_s}{4R}\right)^4} \right)$$

 

[PeterDonis]

This seems to imply light slowing down to be subluminal in a gravitational field

No, it just illustrates that coordinate speed has no physical meaning.

If you were to try to indirectly measure the mass of the photon here by measuring energy and momentum

Evaluating coordinate components $E$ and $p$ of the 4-momentum is not "measuring energy and momentum". Try computing the measured energy and momentum of a photon according to an observer hovering at rest at the same value of $R$. You will find that $E = p$ (and if you compute the measured speed of that photon according to that observer, you will find that it is $1$).

 

[DuckAmuck]

No, it just illustrates that coordinate speed has no physical meaning.Evaluating coordinate components $E$ and $p$ of the 4-momentum is not "measuring energy and momentum". Try computing the measured energy and momentum of a photon according to an observer hovering at rest at the same value of $R$. You will find that $E = p$ (and if you compute the measured speed of that photon according to that observer, you will find that it is $1$).

Ok. What is it I am computing? How do I compute the velocity, momentum and energy of the photon in this scenario?

 

[Orodruin]

You need to express those quantities as invariants and then compute them. For example, the energy of a photon as observed by an observer is given by $E = V \cdot P$, where $V$ is the observer's 4-velocity (normalised to 1) and $P$ is the 4-momentum of the photon.

 

[PeterDonis]

What is it I am computing?

You started an "A" level thread on this topic; anyone with the appropriate background knowledge for an "A" level thread should already know the answer to this. Every GR textbook I've read assigns exercises like this as homework problems.

What background do you have in GR?

 

[DuckAmuck]

You started an "A" level thread on this topic; anyone with the appropriate background knowledge for an "A" level thread should already know the answer to this. Every GR textbook I've read assigns exercises like this as homework problems.

What background do you have in GR?

None. I took a class that covered SR. That's why I am asking here.

 

[PeterDonis]

None. I took a class that covered SR.

Then this thread should not be an "A" level thread, since you don't have the requisite background. Changing thread level to "I".

 

[PeterDonis]

None. I took a class that covered SR.

I would also recommend taking some time to work through the basics of GR. Sean Carroll's online lecture notes give a decent introduction:

https://arxiv.org/abs/gr-qc/9712019

 

[Ibix]

Ok. What is it I am computing?

Coordinate velocity. Or something like that. Note that $p^0\neq E$ (assuming $E$ is the energy measured by an observer at rest with respect to the coordinate system) in general, which may be confusing you.