# How to understand gravitational potential

1. Oct 31, 2013

### nenyan

See the attached

#### Attached Files:

• ###### 2.jpg
File size:
35.8 KB
Views:
92
2. Oct 31, 2013

### Staff: Mentor

Where are you getting these definitions for potential and "field strength"? They look like electromagnetism to me, not gravity.

"Gravitational potential" is not even defined in general; it's only defined in stationary spacetimes, and in those cases it's a scalar, there is no "vector potential". The gradient of the scalar potential, in cases where it's defined, is the "acceleration due to gravity", but this is not usually referred to as the "gravitational field strength".

3. Oct 31, 2013

### WannabeNewton

None of that makes any sense I'm afraid. In electromagnetism the gauge freedom is $A_{a} \rightarrow A_{a} + \nabla_{a}\varphi$; in general relativity the gauge freedom is the pullback of $g_{ab}$ under any diffeomorphism of the space-time manifold.

As Peter pointed out, in general relativity there is no way to define a gravitational potential of any kind for arbitrary space-times solving the Einstein field equation; if a space-time is stationary then one can define a gravitational potential using its stationary killing field. In electromagnetism, the homogeneous Maxwell equation $\nabla_{[a}F_{bc]} = 0$ guarantees that there exists a 4-potential $A_{a}$ locally such that $F_{ab} = \nabla_{[a}A_{b]}$ as a consequence of the Poincare lemma. In general relativity the closest thing you have is $g_{ab}$ but it cannot be interpreted in the general case as a potential (it can be interpreted as such in the weak field limit; see for example section 4.4 of Wald "General Relativity"-here we can draw a very beautiful correspondence between electric/magnetic fields and gravitoelectric/gravitomagnetic fields).

Last edited: Oct 31, 2013
4. Oct 31, 2013

### Bill_K

In the weak field approximation, g00 is a scalar gravitational potential and g0i is a vector potential. The field is not required to be static or stationary.

5. Oct 31, 2013

### Staff: Mentor

Is it $g_{00}$ or $h_{00}$ (where $g_{ab} = \eta_{ab} + h_{ab}$)? (And similarly for the 0i case.)

6. Oct 31, 2013

### Bill_K

An even better quantity to use for the potential is $\bar{h}_{\mu \nu} \equiv h_{\mu \nu} - \frac{1}{2} \eta_{\mu \nu} h$, because it obeys the flat space wave equation.

EDIT: Oops, the actual reason to prefer $\bar{h}_{\mu \nu}$ over $h_{\mu \nu}$ is that the Hilbert gauge condition in terms of $\bar{h}_{\mu \nu}$ is simpler: ${\bar{h}_{\mu \nu ;}}^{\nu} = 0$

Last edited: Oct 31, 2013
7. Oct 31, 2013

### Staff: Mentor

Ah, ok. But then a gauge transformation looks a bit different than what the OP wrote, correct? For any 4-vector field $\varepsilon_{\mu}$ we can write:

$$\bar{h}'_{\mu \nu} = \bar{h}_{\mu \nu} + \partial_{\mu} \varepsilon_{\nu} + \partial_{\nu} \varepsilon_{\mu}$$

Then we just pick $\varepsilon_{\mu}$ as necessary to satisfy whatever constraint we want to impose, such as the Hilbert gauge condition.

8. Oct 31, 2013

### WannabeNewton

Well relative to a background global inertial frame $(t,\vec{x})$ the gravitational analogue of the EM 4-potential is defined as $A_{\mu} = -\frac{1}{4}\bar{h}_{\mu t}$ and the gravitoelectric/gravitomagnetic fields are defined in the same way as electric/magnetic fields in EM i.e. $\vec{E}_g = -\vec{\nabla }A_t - \partial_{t}\vec{A}$ and $\vec{B}_g = \vec{\nabla}\times \vec{A}$; the coordinate acceleration of a freely falling particle is then decomposed as $\vec{a} = -\vec{E}_g - 4\vec{v}\times \vec{B}_g$, where $\vec{v}$ is the 3-velocity of the particle in this background global inertial frame. So perhaps the OP was referring to the usual gauge transformations leaving $E_g$ and $B_g$ invariant?

9. Oct 31, 2013

### Staff: Mentor

Possibly, but there are still a couple of issues:

(1) The gauge transformation is more complicated than in EM because there is an extra index: $\bar{h}_{\mu \nu}$ is a tensor whereas the EM $A_{\mu}$ is a 4-vector. What the OP wrote down might work out to a special case of this for the particular coordinates he adopted (I haven't checked); but to generalize properly, the transformation still needs to be written in covariant form.

(2) $\vec{a}$ can't be a "field strength" in the same sense as $F_{\mu \nu}$ is in EM, because the latter is a tensor, i.e., a properly covariant geometric object, and the former is not; it's a 3-vector in a particular coordinate chart. Furthermore, there is no way to make it covariant, because it basically corresponds to the connection, and the connection is not a tensor. In fact, from what I can see, the OP's "gauge transformation" that makes $\vec{a} = 0$ is basically just an alternate way of showing that you can make the connection vanish at a point by choosing the chart appropriately.

10. Oct 31, 2013

### nenyan

Yes. That is what I said.

compare to electromagnetic field: E=dF/dQ
we can define gravitational field strength: a=dF/dm
dm is the mass of test mass point (herer dx^i/dt=0)
Then we can define the potential and find a specific transformation to vanish the potential but keep the strength (a).

If the potential is 0 then the strength will be 0. How to keep the strength? I misunderstood sth?

11. Oct 31, 2013

### Staff: Mentor

But there's a key difference: dF/dQ can vary for different objects. dF/dm is the same for all objects, because of the equivalence of inertial and gravitational mass: all objects "fall" with the same acceleration in a gravitational field. So dF/dm can't be a field strength in the sense that dF/dQ is for electromagnetism. That's why you're running into problems.

12. Nov 1, 2013

### nenyan

Could you please explain it in detail? Why I run in to problems?

What is field strength? In my opinion, it is a parameter to describe a property of the field. The property is belong to the field, it is the nature feature of the field.
dF/dQ is determined by the EM filed itself and has nothing to do with the electric quantity of test charge.
dF/dm_g is determined by the G field itself and has nothing to do with the gravitational quantity of test mass point (m_g). From the equivalence of inertial and gravitational mass, we obtain dF/dm=a. So we use "a" to describe the strength of G field. "a" is determined by G field itself.

13. Nov 1, 2013

### Staff: Mentor

Because you're looking at something that is not invariant but thinking of it as though it *were* invariant (see below).

Also, you're ignoring key differences between electromagnetism and gravity. As I said in my last post, all objects fall with the same acceleration in a gravitational field; but all objects do not "fall" with the same acceleration in an electromagnetic field. Furthermore, when objects are acted on by an electromagnetic field, they feel a force; but when objects are acted on by a gravitational field, they do not feel a force: they are in free fall. So your "a" does not work the same as the "field strength" in electromagnetism; yet you are trying to treat it the same. That doesn't work.

This doesn't really say very much; lots of different things can meet this description. For example, the potential itself has the same property. What picks out "field strength" from among all the different things that can be properties of the field?

But "a" is also frame-dependent; that's what you showed in your OP. You can always find a frame that makes "a" zero at a chosen event. So "a" cannot be an intrinsic property of the field itself; your choice of frame also affects it. (And since changing frames corresponds to a "gauge transformation" in GR, your "a" is not gauge invariant either, so it can't be a "field strength" in the sense that E and B are in electromagnetism, since E and B *are* gauge invariant.)

If you want to find something that is purely a property of the field, you need to look at invariants: things that don't change when you change frames. For example, $h_{\mu \nu} h^{\mu \nu}$.

(Note, by the way, that the EM counterpart, dF/dQ, is also frame-dependent; you may not be able to find a frame that makes it zero at a given event, but you can certainly *change* it by changing frames. So your EM definition of "field strength" isn't really correct either; you should be looking at the EM field tensor $F_{\mu \nu}$, or more precisely the invariants associated with it, such as $F_{\mu \nu} F^{\mu \nu}$.)

14. Nov 1, 2013

### nenyan

Thank you Peter. I kind of understand...

Last edited: Nov 1, 2013
15. Nov 1, 2013

### WannabeNewton

In other words, for a test particle interacting with an electromagnetic field we have $\xi^b \nabla_b \xi^a = \frac{q}{m}F^{a}{}{}_{b}\xi^b$ whereas for a test particle interacting with the "gravitational field" (I put it in quotes because it isn't a well defined object for arbitrary space-times in general relativity) we have $\xi^b \nabla_b \xi^a =0$; there is a very clear difference between these two equations of motion. The first asserts that the acceleration $a^a = \xi^b \nabla_b \xi^a$ depends on the ratio $\frac{q}{m}$ of the particle as well as the 4-velocity $\xi^a$ and the electromagnetic field $F_{ab}$ (and in fact also on the "gravitational field", which is codified in $\nabla_a$) whereas the second asserts that all freely falling test particles have identically zero acceleration in a "gravitational field" regardless of their mass.