# Active vs Passive Gravitational mass

## Recommended Posts

I was wondering if any of you folks have heard the terms Active Gravitational Mass and Passive Gravitational Mass? Theoretically they are proportional to each other. You can choose constants of proportionality such that they're differerent, However, regardlesso how you chose those constants the densities of these quantities have the same value.

A longtime ago I wrote a web page addressing this matter. I placed it here

http://home.comcast....e_grav_mass.htm

The density is seen in Eq. (3). Let rho = density of active gravitational density. As you can see from

Eq. (3) rhoactive g-mass = u0 + 3p. I don't derive it on that page (but will create a new page soon) but the passive gravitational mass density equals rhopassive g-mass = u0 + p. It is the case tht passive g-mass = active g-mass. I think that's equal by definition. Not sure. I forgot.

The point I'm making is that even though you can choose constants so that they're equal, there is a very good reason to define two equantites since they have different densities in general.

Consider now a particle moving in a static gravitational field. The 4-momentum of the particle is P = (mc, p) where m is the passive gravitational mass as well as the inertial mass. Then for a particls at rest in the field m2 = P2. The energy if the particle is P0 where the mass of the particle is The energy if the particle is P0.

Note: m(0) is the rest mass butit is not equal to the particles proper mass. I.e. m(0) != |P|/c. In this thread I'm hoping to make it clear that one has to be careful with what they call mass in SR/GR. It's not as easy as it appears.

Pete

ps - When I create a new page on the subject I'll post a URL to it if there is a desire for somonme to read it.

Edited by pmb
##### Share on other sites

I was wondering if any of you folks have heard the terms Active Gravitational Mass and Passive Gravitational Mass? Theoretically they are proportional to each other. You can choose constants of proportionality such that they're differerent, However, regardlesso how you chose those constants the densities of these quantities have the same value.

A longtime ago I wrote a web page addressing this matter. I placed it here

http://home.comcast....e_grav_mass.htm

The density is seen in Eq. (3). Let rho = density of active gravitational density. As you can see from

Eq. (3) rhoactive g-mass = u0 + 3p. I don't derive it on that page (but will create a new page soon) but the passive gravitational mass density equals rhopassive g-mass = u0 + p. It is the case tht passive g-mass = active g-mass. I think that's equal by definition. Not sure. I forgot.

The point I'm making is that even though you can choose constants so that they're equal, there is a very good reason to define two equantites since they have different densities in general

Pete

ps - When I create a new page on the subject I'll post a URL to it if there is a desire for somonme to read it.

Eq. (9) is identical to Eq. (2), which was obtained through Einstein’s equations.

You must have ment "Eq. (9) is identical to Eq. (3),

##### Share on other sites

You must have ment "Eq. (9) is identical to Eq. (3),

I stated Eq. (3) whereupon I decide to prove/derive Eq. (9) which I did. I just didn't say "this equals 3" since it was 3 which I stated I was proving. I assume that the reader doesn't forget what I stated 6 equations ago.

Thanks for pointing it out though. I think that just might make it clearer.

##### Share on other sites

Just some quick comments on your page:

-Poisson's equation is that the Laplacian of the potential is proportional to the mass density, not the gradient (i.e. 2ϕ=4πGρ). The latter doesn't make any sense because a gradient is a vector whereas ρ is a scalar.

-The proportionality constant in the EFE's you give us are incorrect: Gμν= (8πG/c4) Tμν

As for what you write above, you haven't given a definition of active/passive gravitational mass. It seems like, from what you say about four-momentum, you consider PG-mass to be equal to rest/invariant mass. But then you go on to say that invariant mass doesn't equal the magnitude of the four-momentum, so I'm not sure what to make of it.

Edited by elfmotat
##### Share on other sites

Just some quick comments on your page: ...

Thanks for pointing out the oversites on my part. So many equaions can make me go blind sometimes. Please don't hesitate to point them out in the future. And I thank you vedy vedy much!

As for what you write above, you haven't given a definition of active/passive gravitational mass.

The definition of active and passive gravitational mass are as follows

1) inertial mass mi is the constant of proportionality between velocity and momentum.

2) passive gravitational mass is the mass that gravity acts on

3) active gravitational mass is the mass that is the source of gravity

It seems like, from what you say about four-momentum, you consider PG-mass to be equal to rest/invariant mass.

Yes. That is correct.

But then you go on to say that invariant mass doesn't equal the magnitude of the four-momentum, so I'm not sure what to make of it.

I don't know where you got that from. Sometime people confuse proper mass with m(0) where m = m(v) = inertial mass = relativistic mass. If a particleis at rest in a gravitational field then then, as always,

|P|/c is the invariant mass. However when v = 0 and the particle is in a g-field (or non-inertial frame of reference) then

|P|/c != m(0) != proper mass

I recommend that you try it out for the most simplest case you can think of. E.g. let the particle be at rest in a uniform g-field or something and see how it works out.

##### Share on other sites

Something occured to me this morning. I've read/heard people argue about the distinction between active and passive gravitational mass. If these quantities were different then there'd be a violation in conservation of momentum. The conservtion of momentum is based on observation. The equality is therefore a law of physics, and not something which is true by definition.

Let us compare this with the equality of a particle's inertial mass and passive gravitational mass. If this equality were not true then the Equivalence Principle would be wrong. If the equivalence principle was wrong then the rate of a particles rate of fall would depend on the particle's rest mass. Therefore this is also something which is true based on observed. The equality is therefore a law of physics and not something which is true based on definition.

To take not take note of passive g-mass and active g-mass is therefore inconsistent with the taking note of the equality of passive g-mass and inertial g-mass.

Edited by pmb
##### Share on other sites

Conservation of momentum is a consequence of the symmetry of of the Lagrangian under a translation in space. Under that condition, it is true by definition.

##### Share on other sites

Conservation of momentum is a consequence of the symmetry of of the Lagrangian under a translation in space. Under that condition, it is true by definition.

Establishing a relation by using the Lagrangian, Lagranges equations, and a conservation relationhip is a matter of derivation. It's not true because it was defined that way.

There is something else which bothers me about your arguement which I can't put my finger on. If I figure it out I'll get back to this. I have a feeling that there is trouble forming a Lagrangian. In any case I'm having trouble trying to write down a Lagrangian for a simple ayarwn for, say, a system of three particles moving in the xy-plain. E.g. how does on express the potential terms in tems of acitve and gravitational mass?

Edited by pmb
##### Share on other sites

Establishing a relation by using the Lagrangian, Lagranges equations, and a conservation relationhip is a matter of derivation. It's not true because it was defined that way.

Conservation of momentum is a result of translational symmetry in the Lagrangian. It's easy to see: consider a Lagrangian which satisfies the following property:

$L(x^i,\dot{x}^i,t)=L(x^i+\delta x^i,\dot{x}^i,t)$

This means that $\partial L / \partial x^i=0$, so the Euler-Lagrange equation reduces to:

$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{x}^i} \right )=0$

Now, $\partial L / \partial \dot{x}^i =p^i$ by definition, therefore momentum doesn't change over time.

##### Share on other sites

Conservation of momentum is a result of translational symmetry in the Lagrangian.

...

Yep. I'm quite aware of that, thanks. I.e. I've spent a great deal of time working on Lagrangian dynamics, as an under graduate student, in graduate school, and in personal independant studies, e.g.

http://home.comcast....stic_energy.htm as well as Noether's theorem.

It's easy to see: consider a Lagrangian which satisfies the following property:

$L(x^i,\dot{x}^i,t)=L(x^i+\delta x^i,\dot{x}^i,t)$

This means that $\partial L / \partial x^i=0$, so the Euler-Lagrange equation reduces to:

$\frac{d}{dt}\left ( \frac{\partial L}{\partial \dot{x}^i} \right )=0$

Now, $\partial L / \partial \dot{x}^i =p^i$ by definition, therefore momentum doesn't change over time.

Yep. I'm also aware of that as well as Noether's theorem which is what you're discussing. That does't tell us anything more than what swansont was trying to say .... unless you thought I didn't understand Lagrangian dynamics and Noether's theorem. In either case, I understand them both. Yay!

Yep. I'm quite aware of that, thanks. I.e. I've sent a great deal of time working on Lagrangian dynamics, as an under graduate student, in graduate school, and in personal independant studies, as in e.g.

http://home.comcast....stic_energy.htm

Yep. I'm also aware of that as well as Noether's theorem which is what you're discussing. That does't tell us anything more than what swansont was trying to say .... unless you thought I didn't understand Lagrangian dynamics and Noether's theorem. In either case, I understand them both. Yay!

Does anybody want to take a crack at figuring out the Lagrangian of a system of three particles, and expressed in terms of each particle's passive gravitational mass and active gravitational mass? Thanks.

Pete

Edited by pmb
##### Share on other sites

Yep. I'm quite aware of that, thanks. I.e. I've spent a great deal of time working on Lagrangian dynamics, as an under graduate student, in graduate school, and in personal independant studies, e.g.

http://home.comcast....stic_energy.htm as well as Noether's theorem.

Yep. I'm also aware of that as well as Noether's theorem which is what you're discussing. That does't tell us anything more than what swansont was trying to say .... unless you thought I didn't understand Lagrangian dynamics and Noether's theorem. In either case, I understand them both. Yay!

Yep. I'm quite aware of that, thanks. I.e. I've sent a great deal of time working on Lagrangian dynamics, as an under graduate student, in graduate school, and in personal independant studies, as in e.g.

http://home.comcast....stic_energy.htm

Yep. I'm also aware of that as well as Noether's theorem which is what you're discussing. That does't tell us anything more than what swansont was trying to say .... unless you thought I didn't understand Lagrangian dynamics and Noether's theorem. In either case, I understand them both. Yay!

I wasn't sure whether or not you were confused about swansont's post, but clearly you weren't.

Does anybody want to take a crack at figuring out the Lagrangian of a system of three particles, and expressed in terms of each particle's passive gravitational mass and active gravitational mass? Thanks.

Relativistically, no, I'd rather not even try.

Doing it with good old classical mechanics isn't too hard though. The Lagrangrian of the ith particle with inertial mass mi, passive g-mass μi and active g-mass Mi would be:

$L = \frac{1}{2} m_i |\bold{\dot{x}}_i|^2 + G \mu_i \sum_{j} \frac{M_j}{|\bold{x}_i-\bold{x}_j|}$

The sum is obviously over j where j≠i.

Edited by elfmotat
##### Share on other sites

I wasn't sure whether or not you were confused about swansont's post, but clearly you weren't.

I had a feeling that was the case. No harm done.

Relativistically, no, I'd rather not even try.

I'm going to try it in Newtonian mechanics first and then relativistically later. I'll share the intermediate results with you if you'd like? In all likeyhood I'll have some help from experts on the subject along the way.

Doing it with good old classical mechanics isn't too hard though. The Lagrangrian of the ith particle with inertial mass mi, passive g-mass μi and active g-mass Mi would be:

$L = \frac{1}{2} m_i |\bold{\dot{x}}_i|^2 + G \mu_i \sum_{j} \frac{M_j}{|\bold{x}_i-\bold{x}_j|}$

The sum is obviously over j where j≠i.

In first term it seems as if you'er using Einstein's summation convention but in the second term it doesn't appear that way. Confusing!

Ya have to be sure when it comes to things like this. For example; Let's consider the system at hand. We have a three particle system moving in the z = 0 plane and therefore V_3 = Vz = 0. The kinetic energy of the system T is given as found to be

$T = \frac{1}{2} m_1 |\bold{\dot{x}}_1|^2 + \frac{1}{2} m_2 |\bold{\dot{x}}_2|^2 + \frac{1}{2} m_3 |\bold{\dot{x}}_3|^2$

The general potential for a multi-particle system is derived in Classical Mechanics - 3rd Ed., Goldstein, Poole and Safko on page 11 and is found to be

$V = \sum_{i} V_{i} + \frac{1}{2}\sum_{ij} V_{ij}$

We can now state the full Lagrangian (without active/passive mass terms)

$L = T - V = \frac{1}{2} m_1 |\bold{\dot{x}}_1|^2 + \frac{1}{2} m_2 |\bold{\dot{x}}_2|^2 - \sum_{i} V_{i} - \frac{1}{2}\sum_{ij} V_{ij}$

where i does not equal j in last summation. I haven't got this Latext thing down. I was fluent a while back but their Latex thingy shut down and since then my Latex became atrophied.

Now! The trick is to formulate this in terms of active and passive g-mass. That I don't know how to do. I may figure it out. That takes time. Sometimes the answer just pops into my head. I love it when hat happens!

Thanks,

Pete

Edited by pmb
##### Share on other sites

pmb,

In case you hadn't already seen them, I thought the following might be useful for you:

And you are welcome to use this next one to make a thread for tesing things out in:

http://www.sciencefo...99-the-sandbox/

Hope that helps

##### Share on other sites

pmb,

In case you hadn't already seen them, I thought the following might be useful for you:

And you are welcome to use this next one to make a thread for tesing things out in:

http://www.sciencefo...99-the-sandbox/

Hope that helps

Yes! Excellant! Thank you!

##### Share on other sites

In first term it seems as if you'er using Einstein's summation convention but in the second term it doesn't appear that way. Confusing!

Nope, I wasn't summing the first term. I was actually considering the Lagrangian of a single particle acted on by others.

##### Share on other sites

Nope, I wasn't summing the first term. I was actually considering the Lagrangian of a single particle acted on by others.

To what end? The system I was considering is a system of three interacting particles. What would you use your Lagrangian for?

##### Share on other sites

To what end? The system I was considering is a system of three interacting particles. What would you use your Lagrangian for?

True. My initial Lagrangian would only be valid if the other particles in the system were held static.

Here's what it would be for the system:

$L=\frac{1}{2} \sum_k m_k|\dot{\bold{x}}_k|^2+ \frac{1}{2} G \sum _{i,j} \frac{\mu_i M_j}{|\bold{x}_i-\bold{x}_j|}$

The reason this works is because $\mu_1 M_2= \mu_2 M_1$, or else Newton's third law fails. I think this is reason enough to conclude that passive and active gravitational mass are equivalent, or at least the constant of proportionality between them is a universal constant which is absorbed into G. In that sense I don't see how you could perform any experiment to determine whether or not they differed, so I don't know why it would be useful to think of them as not being equivalent.

The general potential for a multi-particle system is derived in Classical Mechanics - 3rd Ed., Goldstein, Poole and Safko on page 11 and is found to be

$V = \sum_{i} V_{i} + \frac{1}{2}\sum_{ij} V_{ij}$

The first term is actually an external potential, so it's zero in the case we're considering.

Edited by elfmotat
##### Share on other sites

The first term is actually an external potential, so it's zero in the case we're considering.

Oy! Of course. I had this sinking feeling in the pit of my stomach as to the V_i and I couldn't resolve it. Thanks elmotat! Problem solved!

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account