Question about equivalence principle

  • #26
887
98
Wow, some really good posts from everyone. I think I get it now. Thank you. But I have one more question. Why has there been so many tests of the equivalence of inertial mass to passive gravitational mass and so very little tests of the equivalence of active gravitational mass to passive gravitational mass?
 
  • #27
Al68
They are not equal according to the equivalence principle - they are proportional. The equality depends on a choice of units -so the weakness is in large part hidden in G, Newton's universal gravitational constant.
There is no practical difference between "equal" and proportionally related by a constant function.

But your point is the right one, there is no equivalence between the force necessary to accelerate a test mass and the active gravitational effect ("force") that test mass has on other masses. Those "forces" are proportional to inertial mass and gravitational mass respectively, and to each other since those masses are equal, but the "forces" are related to mass by variable functions and therefore not "equal".
 
  • #28
DrGreg
Science Advisor
Gold Member
2,282
860
Why has there been so many tests of the equivalence of inertial mass to passive gravitational mass and so very little tests of the equivalence of active gravitational mass to passive gravitational mass?
So I would say the equivalence of active and passive mass isn't a testable hypothesis, it's a definition. And any "test of active mass" is really just another name for "measuring the value of G".
....
 
  • #29
887
98
Thanks Greg. Your post #20 was one of the more helpful for me.

DrGreg said:
Let's say active mass is redefined to be twice passive mass. That wouldn't make any difference as long as we also halve the value of G. We might as well just take the constant to be one to keep things simple.
What if instead of redefining the ratio and adjusting G, we actually make the ratio slightly different. Would that mean that our measured value of G would be slightly different?
 
  • #30
29,554
5,882
But the value of G may in principle vary in space-time, and this is indeed testable.
Only the values of dimensionless physical constants have any physical significance. The value of G (or any other dimensionful constant) depends entirely on our choice of units. There is no other physical significance.
 
  • #31
Only the values of dimensionless physical constants have any physical significance. The value of G (or any other dimensionful constant) depends entirely on our choice of units. There is no other physical significance.
This old myth raises its ugly head again. The reason that quantities with dimension are in general regarded as less fundamental than dimensionless quantities, is that there generally is no physical principle that separates out any specific dimensionful quantity. To take an example, the fine structure constant. If it is changing, is it the electron charge, the Planck constant or the speed of light? There is no way to tell experimentally since there is no physical principle which separates out one over any other. But for gravitation it is different. We have the EEP, tested to umpteenth decimal places. The EEP says that you may perform a gazillion local non-gravitational experiments without involving gravity at all.

Experimentally this means that if you detect nothing unusual in these experiments, but find a consistent
change in all local gravitational experiments measuring the value of G, it is clear as day that this is due to a changing value of G and not any other dimensionful quantity used to construct a dimensionless quantity (such as the gravitational counterpart to the fine structure constant, for example). In this case, the only formal ambiguity is whether to consider a variable active gravitational mass or a variable G which is essentially the same thing.
 
  • #32
29,554
5,882
This old myth raises its ugly head again. The reason that quantities with dimension are in general regarded as less fundamental than dimensionless quantities, is that there generally is no physical principle that separates out any specific dimensionful quantity. To take an example, the fine structure constant. If it is changing, is it the electron charge, the Planck constant or the speed of light? There is no way to tell experimentally since there is no physical principle which separates out one over any other. But for gravitation it is different. We have the EEP, tested to umpteenth decimal places. The EEP says that you may perform a gazillion local non-gravitational experiments without involving gravity at all.

Experimentally this means that if you detect nothing unusual in these experiments, but find a consistent
change in all local gravitational experiments measuring the value of G, it is clear as day that this is due to a changing value of G and not any other dimensionful quantity used to construct a dimensionless quantity (such as the gravitational counterpart to the fine structure constant, for example). In this case, the only formal ambiguity is whether to consider a variable active gravitational mass or a variable G which is essentially the same thing.
No, this is not a myth, it is correct. Your example of the fine structure constant is a good example. The corresponding dimensionless constants for gravity and inertia are the dimensionless ratios of the particle masses to the Planck mass. If that changes then is it the particle masses, the Planck constant, or G that has changed? The situation is no different in that respect for G than for other dimensionful constants.

What you are probably thinking of is a situation where the Planck mass is changed but the various particle masses have remained constant so that the dimensionless ratios of particle masses to the Planck mass (see http://math.ucr.edu/home/baez/constants.html) have changed. That is indeed measurable since dimensionless constants have changed. However, if the dimensionless constants do not change then the difference is not measurable (see https://www.physicsforums.com/showpost.php?p=2011753&postcount=55 and https://www.physicsforums.com/showpost.php?p=2015734&postcount=68).
 
Last edited by a moderator:
  • #33
No, this is not a myth, it is correct. Your example of the fine structure constant is a good example. The corresponding dimensionless constants for gravity and inertia are the dimensionless ratios of the particle masses to the Planck mass. If that changes then is it the particle masses, the Planck constant, or G that has changed? The situation is no different in that respect for G than for other dimensionful constants.
The EEP says that G has changed, since every local non-gravitational experiment, where the particle masses and Planck constant may enter in all possible combinations, show no change. To further illustrate this, the dimensionless quantity (used in the literature) [itex]{\alpha}_G=Gm_p^2/({\hbar }c)[/itex], where [itex]m_p[/itex] is the proton mass, is an appropriate
gravitational counterpart to the fine structure constant. If [itex]{\alpha}_G[/itex] changes, the EEP says that this is due to a change of G, and not to a change of [itex]{\hbar}[/itex] or c, since the local non-gravitational physics is unaffected. Thus a measurable change of G is equivalent to a measurable change of [itex]{\alpha}_G[/itex].
What you are probably thinking of is a situation where the Planck mass is changed but the various particle masses have remained constant so that the dimensionless ratios of particle masses to the Planck mass (see http://math.ucr.edu/home/baez/constants.html) have changed. That is indeed measurable since dimensionless constants have changed. However, if the dimensionless constants do not change then the difference is not measurable (see https://www.physicsforums.com/showpost.php?p=2011753&postcount=55 and https://www.physicsforums.com/showpost.php?p=2015734&postcount=68).
Of course a measurable change in G
implies a corresponding change in a relevant dimensionless quantity. But that is not the point. The point is that it is perfectly legitimate to speak of a measurable change in G since the EEP ensures that this is equivalent to a change in the corresponding relevant dimensionless quantity.

In short, all units may be operationally defined from local non-gravitational experiments. Using these units to measure gravitational quantities, the EEP says that these quantities (dimensionless or not) do not have any extra ambiguities associated with the choice of units.
 
Last edited by a moderator:
  • #34
atyy
Science Advisor
13,848
2,127
The EEP says that G has changed, since every local non-gravitational experiment, where the particle masses and Planck constant may enter in all possible combinations, show no change. To further illustrate this, the dimensionless quantity (used in the literature) [itex]{\alpha}_G=Gm_p^2/({\hbar }c)[/itex], where [itex]m_p[/itex] is the proton mass, is an appropriate
gravitational counterpart to the fine structure constant. If [itex]{\alpha}_G[/itex] changes, the EEP says that this is due to a change of G, and not to a change of [itex]{\hbar}[/itex] or c, since the local non-gravitational physics is unaffected. Thus a measurable change of G is equivalent to a measurable change of [itex]{\alpha}_G[/itex].
Of course a measurable change in G
implies a corresponding change in a relevant dimensionless quantity. But that is not the point. The point is that it is perfectly legitimate to speak of a measurable change in G since the EEP ensures that this is equivalent to a change in the corresponding relevant dimensionless quantity.

In short, all units may be operationally defined from local non-gravitational experiments. Using these units to measure gravitational quantities, the EEP says that these quantities (dimensionless or not) do not have any extra ambiguities associated with the choice of units.
From your old post EEP = WEP + LLI + LPI. Can we drop LPI?
 
  • #35
From your old post EEP = WEP + LLI + LPI. Can we drop LPI?
No we cannot. The LPI is crucial. Without it, the EEP would be unable to ensure that the local non-gravitational physics is independent of gravitation.
 
  • #36
29,554
5,882
Of course a measurable change in G implies a corresponding change in a relevant dimensionless quantity. But that is not the point.
Well, that was the only point that I was making, so it seems that we agree.
 
Last edited:
  • #37
DrGreg
Science Advisor
Gold Member
2,282
860
What if instead of redefining the ratio and adjusting G, we actually make the ratio slightly different. Would that mean that our measured value of G would be slightly different?
I'm not really sure what you think the difference is between "redefining the ratio" and "actually make the ratio slightly different". Seems like the same thing to me.

But, yes, I'm saying that if the ratio was different, we'd measure a different value of G.

All the equations of physics that have "active gravitational mass" in them always contain a GM term rather than M on its own. That's true in General Relativity as well as Newtonian theory. So any constant of proportionality is accounted for within the value of G.


Technicality: relativists often choose to measure mass in units such than G = 1 (instead of kg) so their equations seem not to have a G in them. But if you switch to other units, what I said above is true.
 
  • #38
887
98
DrGreg said:
I'm not really sure what you think the difference is between "redefining the ratio" and "actually make the ratio slightly different". Seems like the same thing to me.
I was just trying to make the point that the equivalence of active gravitational mass and passive gravitational mass is a testable hypothesis. (I admit I did not do a very good job.)

This is the of type of experiment I have been looking for:
http://www.iop.org/EJ/article/1742-...uest-id=69a8b2bc-bdaa-4b3a-939b-daae73156a2a"
On page 4 beginning with 3. Newton-I Experiment, is a test of the composition dependence of the gravitational constant. I believe this is in effect a test for the equivalence of active gravitational mass and passive gravitational mass for different compositions of matter.
 
Last edited by a moderator:
  • #39
887
98
DrGreg said:
So I would say the equivalence of active and passive mass isn't a testable hypothesis, it's a definition. And any "test of active mass" is really just another name for "measuring the value of G".
Yes, I think I understand what you are saying. And you are correct. But it is based on the assumption that gravitational mass and inertial mass are the same for all compositions of matter. And indeed, that is what EP postulates. So I am not disputing your claim. However, I do believe that EP should be tested.

So let me ask my second question again, being more precise.

There is a test of the equivalence principle which is sometimes referred to as a test for composition-dependent forces. This is a valid test, because EP states that all gravitational forces are independent of mass composition. Many of these tests involve the http://www.npl.washington.edu/eotwash/intro/intro.html" [Broken]. While this is a perfectly good test of the EP, it does not address the question of composition dependence based on active gravitational mass. The test masses in the composition dipole are essentially responding to a source gravitational field. In other words, they are the passive mass. There are, and have been, many other similar types of tests, such as STEP. But they are all the same in that the test masses are passive.

Now take a look at the experiment (for composition dependence of gravitational constant) linked in my previous post. Notice that the test masses are the attractors. In this experiment the test masses are the "source" of the gravitational field. So this is a test for composition dependence based on active gravitational mass.

So my question is: Why so many EP tests for composition dependent gravity based on Mgp, and so very few EP test for composition dependent gravity based on Mga?

Thank you to everyone who has responded. And thank you to those who have PM me with links and information.

Turtle
 
Last edited by a moderator:
  • #40
887
98
Here is the equation for measuring the value of G:

G = Fr2 / (Mtarget Mattractor)

My third question: Does it make any difference whether the test mass is the target or the attractor? Keep in mind that the attractor is the source (large mass), and that the experiment is to measure the value of G using different compositions of mass as the test mass. This is done by measuring G with a test mass of composition "a", and then repeating the experiment with a test mass of composition "b". The two values of G are then compared. A null experiment is Ga - Gb = 0.
 
Last edited:

Related Threads on Question about equivalence principle

  • Last Post
Replies
14
Views
2K
Replies
1
Views
2K
Replies
3
Views
863
Replies
100
Views
11K
Replies
14
Views
2K
Replies
14
Views
969
Replies
2
Views
895
  • Last Post
2
Replies
38
Views
2K
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
2
Views
2K
Top