Couldn't we always redefine units so that inertial mass and gravitational mass are equal?
It is a known fact that inertial and gravitational masses are the same thing, and therefore are numerically equal. This is not an obvious thing, since there are even experiments trying to find a difference between the two kinds of masses. What I don't understand is: why is this not obvious? Usually when something that isn't considered obvious seems obvious to me there's something deep that I'm not getting.
Here's my line of thought:
The inertial mass is defined by
$$
{bf{F}} = m_i {bf{a}} tag{1}
$$
The gravitational mass is derived from the fact that the gravitational force between two objects is proportional to the product of the masses of the objects:
$$
{bf{F_g}} = -G frac{m_{G1} m_{G2}}{|{bf{r}}_{12}|^2} hat{{bf{r}}} tag{2}
$$
Now if the only force acting on the object $1$ is the gravitational force, I can equate equations $(1)$ and $(2)$, and I can always fix the constant $G$ in such a way that the gravitational mass and the inertial mass are numerically equal.
What's wrong with this line of thought and why is the equivalence not really so obvious?
newtonian-mechanics newtonian-gravity mass dimensional-analysis inertia
edited Dec 15 at 16:31
Ruslan
9,11742969
asked Dec 14 at 12:39
Run like hell
1,048724
|
show 2 more comments
It is a known fact that inertial and gravitational masses are the same thing, and therefore are numerically equal. This is not an obvious thing, since there are even experiments trying to find a difference between the two kinds of masses. What I don't understand is: why is this not obvious? Usually when something that isn't considered obvious seems obvious to me there's something deep that I'm not getting.
Here's my line of thought:
The inertial mass is defined by
$$
{bf{F}} = m_i {bf{a}} tag{1}
$$
The gravitational mass is derived from the fact that the gravitational force between two objects is proportional to the product of the masses of the objects:
$$
{bf{F_g}} = -G frac{m_{G1} m_{G2}}{|{bf{r}}_{12}|^2} hat{{bf{r}}} tag{2}
$$
Now if the only force acting on the object $1$ is the gravitational force, I can equate equations $(1)$ and $(2)$, and I can always fix the constant $G$ in such a way that the gravitational mass and the inertial mass are numerically equal.
What's wrong with this line of thought and why is the equivalence not really so obvious?
newtonian-mechanics newtonian-gravity mass dimensional-analysis inertia
edited Dec 15 at 16:31
Ruslan
9,11742969
asked Dec 14 at 12:39
Run like hell
1,048724
3
Your procedure only works if the two masses are proportional. In that case yes, we can redefine the units to make them equal. Otherwise we can't.
– Javier
Dec 14 at 13:46
8
Technically inertial and gravitational mass are merely proportional, not necessarily equal. One could say that the "true" gravitational mass is really $m_g = sqrt{G} m_i$...
– Aetol
Dec 14 at 15:15
How are they equal? They have different units in the end because G isn't dimensionless.
– Joshua
Dec 14 at 20:36
the equivalence principle can be restated in this way "gravitation is an acceleration field, not a force field"
– lurscher
Dec 14 at 21:46
3
Run like hell, note that units are arbitrary. In the U.S., I measure mass with pounds. Everywhere else, mass is measured with kg. Somewhat related to units are dimensions, such as length, time, mass, etc. Physics is based on the dimensions in the problem, not the units. This means that physics is actually independent of the units used. Thus, you can't redefine the physics of the problem by choosing different units.
– David White
Dec 15 at 5:22
|
show 2 more comments
It is a known fact that inertial and gravitational masses are the same thing, and therefore are numerically equal. This is not an obvious thing, since there are even experiments trying to find a difference between the two kinds of masses. What I don't understand is: why is this not obvious? Usually when something that isn't considered obvious seems obvious to me there's something deep that I'm not getting.
Here's my line of thought:
The inertial mass is defined by
$$
{bf{F}} = m_i {bf{a}} tag{1}
$$
The gravitational mass is derived from the fact that the gravitational force between two objects is proportional to the product of the masses of the objects:
$$
{bf{F_g}} = -G frac{m_{G1} m_{G2}}{|{bf{r}}_{12}|^2} hat{{bf{r}}} tag{2}
$$
Now if the only force acting on the object $1$ is the gravitational force, I can equate equations $(1)$ and $(2)$, and I can always fix the constant $G$ in such a way that the gravitational mass and the inertial mass are numerically equal.
What's wrong with this line of thought and why is the equivalence not really so obvious?
newtonian-mechanics newtonian-gravity mass dimensional-analysis inertia
edited Dec 15 at 16:31
Ruslan
9,11742969
asked Dec 14 at 12:39
Run like hell
1,048724
It is a known fact that inertial and gravitational masses are the same thing, and therefore are numerically equal. This is not an obvious thing, since there are even experiments trying to find a difference between the two kinds of masses. What I don't understand is: why is this not obvious? Usually when something that isn't considered obvious seems obvious to me there's something deep that I'm not getting.
Here's my line of thought:
The inertial mass is defined by
$$
{bf{F}} = m_i {bf{a}} tag{1}
$$
The gravitational mass is derived from the fact that the gravitational force between two objects is proportional to the product of the masses of the objects:
$$
{bf{F_g}} = -G frac{m_{G1} m_{G2}}{|{bf{r}}_{12}|^2} hat{{bf{r}}} tag{2}
$$
Now if the only force acting on the object $1$ is the gravitational force, I can equate equations $(1)$ and $(2)$, and I can always fix the constant $G$ in such a way that the gravitational mass and the inertial mass are numerically equal.
What's wrong with this line of thought and why is the equivalence not really so obvious?
newtonian-mechanics newtonian-gravity mass dimensional-analysis inertia
newtonian-mechanics newtonian-gravity mass dimensional-analysis inertia
edited Dec 15 at 16:31
Ruslan
9,11742969
asked Dec 14 at 12:39
Run like hell
1,048724
edited Dec 15 at 16:31
Ruslan
9,11742969
asked Dec 14 at 12:39
Run like hell
1,048724
edited Dec 15 at 16:31
Ruslan
9,11742969
edited Dec 15 at 16:31
Ruslan
9,11742969
edited Dec 15 at 16:31
Ruslan
9,11742969
9,11742969
asked Dec 14 at 12:39
Run like hell
1,048724
asked Dec 14 at 12:39
Run like hell
1,048724
asked Dec 14 at 12:39
Run like hell
1,048724
1,048724
3
Your procedure only works if the two masses are proportional. In that case yes, we can redefine the units to make them equal. Otherwise we can't.
– Javier
Dec 14 at 13:46
8
Technically inertial and gravitational mass are merely proportional, not necessarily equal. One could say that the "true" gravitational mass is really $m_g = sqrt{G} m_i$...
– Aetol
Dec 14 at 15:15
How are they equal? They have different units in the end because G isn't dimensionless.
– Joshua
Dec 14 at 20:36
the equivalence principle can be restated in this way "gravitation is an acceleration field, not a force field"
– lurscher
Dec 14 at 21:46
3
Run like hell, note that units are arbitrary. In the U.S., I measure mass with pounds. Everywhere else, mass is measured with kg. Somewhat related to units are dimensions, such as length, time, mass, etc. Physics is based on the dimensions in the problem, not the units. This means that physics is actually independent of the units used. Thus, you can't redefine the physics of the problem by choosing different units.
– David White
Dec 15 at 5:22
|
show 2 more comments
3
Your procedure only works if the two masses are proportional. In that case yes, we can redefine the units to make them equal. Otherwise we can't.
– Javier
Dec 14 at 13:46
8
Technically inertial and gravitational mass are merely proportional, not necessarily equal. One could say that the "true" gravitational mass is really $m_g = sqrt{G} m_i$...
– Aetol
Dec 14 at 15:15
How are they equal? They have different units in the end because G isn't dimensionless.
– Joshua
Dec 14 at 20:36
the equivalence principle can be restated in this way "gravitation is an acceleration field, not a force field"
– lurscher
Dec 14 at 21:46
3
Run like hell, note that units are arbitrary. In the U.S., I measure mass with pounds. Everywhere else, mass is measured with kg. Somewhat related to units are dimensions, such as length, time, mass, etc. Physics is based on the dimensions in the problem, not the units. This means that physics is actually independent of the units used. Thus, you can't redefine the physics of the problem by choosing different units.
– David White
Dec 15 at 5:22
3
3
Your procedure only works if the two masses are proportional. In that case yes, we can redefine the units to make them equal. Otherwise we can't.
– Javier
Dec 14 at 13:46
Your procedure only works if the two masses are proportional. In that case yes, we can redefine the units to make them equal. Otherwise we can't.
– Javier
Dec 14 at 13:46
8
8
Technically inertial and gravitational mass are merely proportional, not necessarily equal. One could say that the "true" gravitational mass is really $m_g = sqrt{G} m_i$...
– Aetol
Dec 14 at 15:15
Technically inertial and gravitational mass are merely proportional, not necessarily equal. One could say that the "true" gravitational mass is really $m_g = sqrt{G} m_i$...
– Aetol
Dec 14 at 15:15
How are they equal? They have different units in the end because G isn't dimensionless.
– Joshua
Dec 14 at 20:36
How are they equal? They have different units in the end because G isn't dimensionless.
– Joshua
Dec 14 at 20:36
the equivalence principle can be restated in this way "gravitation is an acceleration field, not a force field"
– lurscher
Dec 14 at 21:46
the equivalence principle can be restated in this way "gravitation is an acceleration field, not a force field"
– lurscher
Dec 14 at 21:46
3
3
Run like hell, note that units are arbitrary. In the U.S., I measure mass with pounds. Everywhere else, mass is measured with kg. Somewhat related to units are dimensions, such as length, time, mass, etc. Physics is based on the dimensions in the problem, not the units. This means that physics is actually independent of the units used. Thus, you can't redefine the physics of the problem by choosing different units.
– David White
Dec 15 at 5:22
Run like hell, note that units are arbitrary. In the U.S., I measure mass with pounds. Everywhere else, mass is measured with kg. Somewhat related to units are dimensions, such as length, time, mass, etc. Physics is based on the dimensions in the problem, not the units. This means that physics is actually independent of the units used. Thus, you can't redefine the physics of the problem by choosing different units.
– David White
Dec 15 at 5:22
|
show 2 more comments
5 Answers
5
active
oldest
votes
To make it clear that it is not obvious it is better to stop using the word "mass" in both cases. So it is better to say that it is not obvious that the inertial resistance, meaning the property that scales how different objects accelerate under the same given force, is the same as the "gravitational charge", meaning the property that scales the gravitational field that different objects produce.
Just to make it clear, the problem with the equivalence of masses is not "does $m_i=1$ imply $m_{G}=1$?" in whatever units. The real question is, "does doubling the inertial mass actually double the gravitational mass?". So your procedure of redefinition of $G$, while keeping it as a constant, is only valid if the ratio ${m_i}/{m_G}$ is constant, meaning if there is a constant scale factor between the inertial mass and the gravitational mass. By the way, this scale factor would actually never be noticed since from the begginning it would already be accounted for in the constant $G$.
You could do the same thing with electrical forces, using Coulombs law. You can check if electrical charge is the same as inertial mass, since you have:
$${bf{F}} = m_i {bf{a}} tag{1}$$
$${bf{F_e}} = -K frac{q_1 q_2}{|{bf{r}}_{12}|^2} hat{{bf{r}}} tag{2}$$
You can ask, is $q_1$ the same as $m_i$? And it is true that for one specific case you could redefine $K$ such that it would come out that $q_1$ and $m_i$ are the same, but it would not pass the scaling test, since doubling the inertial mass does not double the charge.
answered Dec 14 at 13:36
Hugo V
1,486616
8
As far as I can see, the fundamental issue is not with doubling, but already with singling. That is: do any two objects with the same inertial mass also have the same gravitational mass?
– Emil Jeřábek
Dec 14 at 15:40
6
Sure, I agree, but I used the "doubling" as an example. I meant that the question is if the ratio of inertial mass to gravitational mass is the same for all objects. So that covers both the case of doubling $m_i$ and getting double $m_g$, as well as your case, which is if different objects with $m_i$ will have the same $m_g$. If the ratio is the same for all objects of all masses, then the masses are equivalent no matter what.
– Hugo V
Dec 14 at 15:52
add a comment |
Scenario I: I have a white ball and a black ball. In the system of units I've adopted, I discover that:
Gravitational mass of white ball = 2
Inertial mass of white ball = 3
Gravitational mass of black ball = 10
Inertial mass of black ball = 15
But I want each ball's gravitational mass to equal its inertial mass. So I fix the constant $G$, multiplying it by $2/3$ so that all my gravitational masses will be multiplied by $3/2$. Problem solved, just like you said!
Scenario II: I have a white ball and a black ball. In the system of units I've adopted, I discover that:
Gravitational mass of white ball = 2
Inertial mass of white ball = 3
Gravitational mass of black ball = 10
Inertial mass of black ball = 20
But I want each ball's gravitational mass to equal its inertial mass. So I fix the constant $G$, multiplying it by $2/3$ so that the white ball's gravitational mass will be multiplied by $3/2$. Or maybe I should fix the constant $G$, multiplying it by $1/2$ so the black ball's gravitational mass will be multiplied by $2$. Uh oh. Neither solution manages to cover both cases.
The full content of the statement that "gravitational mass equals inertial mass" is that "gravitational mass equals inertial mass provided you choose the right units". This doesn't rule out Scenario I, but it rules out Scenario II. The fact that Scenario II can't occur is a substantive statement.
edited Dec 15 at 5:06
Volker Siegel
2,50811642
answered Dec 14 at 18:20
WillO
6,61622132
This is the most intuitive and straightforward explanation.
– Greg Schmit
Dec 14 at 20:34
add a comment |
Try it like this; "All objects follow the same trajectory in a gravitational field".
That means that the force of gravity on an object, which alters its momentum, scales exactly with its inertial mass, which resists that alteration of momentum.
answered Dec 14 at 21:29
Jerome
511
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It's kind of difficult since we have so much prior intuition that gravitational and inertial mass are the same thing because we know that something heavy to lift also takes more force to accelerate, but there's not really any inherent reason why this should be the case. I think it helps if you call the quantity that goes into the gravitational force law "gravitational charge", to differentiate it from the one in $F = ma$. The question then is "why is the gravitational charge the quantity that determines how much inertia an object has?" (up to a constant scaling factor that you can, indeed, incorporate into $G$).
You could imagine a universe where the magnitude of the electric charge is what determines an object's inertia, and now Newton's second law becomes $F = Sigma |q_i|a$ for an object made up of a number of charged particles. Now an electron and a proton would accelerate to the same speed if you subjected them to the same potential difference, but an atom of hydrogen (mass ~1 GeV) and an atom of positronium (a bound electron and positron, mass ~1 MeV) would fall at different speeds - the force of gravity would still be the same as it is in our universe for each particle, so it would be larger for the hydrogen, but would no longer be balanced by the larger inertial mass of hydrogen proportionally decreasing its acceleration.
Interestingly, if the physical laws of our universe suddenly switched to this behaviour it actually wouldn't be immediately obvious. Since regular matter is made up of equal numbers of protons and electrons, doubling the amount of stuff in something would double both this "electrical inertia" and the regular inertia due to mass, I think the most obvious effect would be plasmas behaving very strangely. Note that I'm ignoring neutrons in this because I didn't think of them and quark charges until now.
answered Dec 14 at 19:05
llama
22116
add a comment |
It is a known fact that Inertial and gravitational mass are the same
thing
Actually not.
(Edit: I have rewritten the answer totally).
To be clear, let us differentiate between facts and theories.
The only facts we have in science are things we can measure. Assume that you measure the ratio between a circles diameter and its circumference. You measure the ratio to 3 plus minus 0.5. This is a fact. Get a better measurement and you might reach that the ratio is 3.1 plus minus 0.1. And so on.
We have tried to measure the difference between inertial and gravitational mass and have not been able to show any differences, so far. This is a fact.
But does it imply that there is no and can never be any difference and that they are identical? In science this is a theory. And the bad thing about theories is that they can never be shown to be true, only falsified.
Let us be inside a space station circling earth. No windows and we feel absolutely no gravity and no acceleration. So both a and F are zero. What happened to the gravity? Well, Einstein solved that by saying that there is no force, but instead it is space that is curved. The space station moves around the earth in a straight line (no force) because a straight line is a circle. Go figure.
If we measure very tiny forces things start to change. Quantuum mechanics did show that at very small scales energy comes in small packades, called quanta. Classical mechanics, example F=ma, has an implied belief that any value of F,m or a is possible. Quantuum mechanics shows that not all values of F or m are possible, there seems to be some minimum "unit" to them. So not even Einstein had it all in place (he did spend a lot of time trying to disprove some of the parts of quantuum mechanics that are now experimentally shown to be more true than false).
So, just maybe, in the future we will be able to measure a difference between inertial and gravitational mass. That measurement would be a fact forcing us to create a new theory. Or we would not be able to measure any difference and then the theory holds so far.
answered Dec 15 at 9:47
ghellquist
1553
add a comment |
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5 Answers
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To make it clear that it is not obvious it is better to stop using the word "mass" in both cases. So it is better to say that it is not obvious that the inertial resistance, meaning the property that scales how different objects accelerate under the same given force, is the same as the "gravitational charge", meaning the property that scales the gravitational field that different objects produce.
Just to make it clear, the problem with the equivalence of masses is not "does $m_i=1$ imply $m_{G}=1$?" in whatever units. The real question is, "does doubling the inertial mass actually double the gravitational mass?". So your procedure of redefinition of $G$, while keeping it as a constant, is only valid if the ratio ${m_i}/{m_G}$ is constant, meaning if there is a constant scale factor between the inertial mass and the gravitational mass. By the way, this scale factor would actually never be noticed since from the begginning it would already be accounted for in the constant $G$.
You could do the same thing with electrical forces, using Coulombs law. You can check if electrical charge is the same as inertial mass, since you have:
$${bf{F}} = m_i {bf{a}} tag{1}$$
$${bf{F_e}} = -K frac{q_1 q_2}{|{bf{r}}_{12}|^2} hat{{bf{r}}} tag{2}$$
You can ask, is $q_1$ the same as $m_i$? And it is true that for one specific case you could redefine $K$ such that it would come out that $q_1$ and $m_i$ are the same, but it would not pass the scaling test, since doubling the inertial mass does not double the charge.
answered Dec 14 at 13:36
Hugo V
1,486616
8
As far as I can see, the fundamental issue is not with doubling, but already with singling. That is: do any two objects with the same inertial mass also have the same gravitational mass?
– Emil Jeřábek
Dec 14 at 15:40
6
Sure, I agree, but I used the "doubling" as an example. I meant that the question is if the ratio of inertial mass to gravitational mass is the same for all objects. So that covers both the case of doubling $m_i$ and getting double $m_g$, as well as your case, which is if different objects with $m_i$ will have the same $m_g$. If the ratio is the same for all objects of all masses, then the masses are equivalent no matter what.
– Hugo V
Dec 14 at 15:52
add a comment |
To make it clear that it is not obvious it is better to stop using the word "mass" in both cases. So it is better to say that it is not obvious that the inertial resistance, meaning the property that scales how different objects accelerate under the same given force, is the same as the "gravitational charge", meaning the property that scales the gravitational field that different objects produce.
Just to make it clear, the problem with the equivalence of masses is not "does $m_i=1$ imply $m_{G}=1$?" in whatever units. The real question is, "does doubling the inertial mass actually double the gravitational mass?". So your procedure of redefinition of $G$, while keeping it as a constant, is only valid if the ratio ${m_i}/{m_G}$ is constant, meaning if there is a constant scale factor between the inertial mass and the gravitational mass. By the way, this scale factor would actually never be noticed since from the begginning it would already be accounted for in the constant $G$.
You could do the same thing with electrical forces, using Coulombs law. You can check if electrical charge is the same as inertial mass, since you have:
$${bf{F}} = m_i {bf{a}} tag{1}$$
$${bf{F_e}} = -K frac{q_1 q_2}{|{bf{r}}_{12}|^2} hat{{bf{r}}} tag{2}$$
You can ask, is $q_1$ the same as $m_i$? And it is true that for one specific case you could redefine $K$ such that it would come out that $q_1$ and $m_i$ are the same, but it would not pass the scaling test, since doubling the inertial mass does not double the charge.
answered Dec 14 at 13:36
Hugo V
1,486616
8
As far as I can see, the fundamental issue is not with doubling, but already with singling. That is: do any two objects with the same inertial mass also have the same gravitational mass?
– Emil Jeřábek
Dec 14 at 15:40
6
Sure, I agree, but I used the "doubling" as an example. I meant that the question is if the ratio of inertial mass to gravitational mass is the same for all objects. So that covers both the case of doubling $m_i$ and getting double $m_g$, as well as your case, which is if different objects with $m_i$ will have the same $m_g$. If the ratio is the same for all objects of all masses, then the masses are equivalent no matter what.
– Hugo V
Dec 14 at 15:52
add a comment |
To make it clear that it is not obvious it is better to stop using the word "mass" in both cases. So it is better to say that it is not obvious that the inertial resistance, meaning the property that scales how different objects accelerate under the same given force, is the same as the "gravitational charge", meaning the property that scales the gravitational field that different objects produce.
Just to make it clear, the problem with the equivalence of masses is not "does $m_i=1$ imply $m_{G}=1$?" in whatever units. The real question is, "does doubling the inertial mass actually double the gravitational mass?". So your procedure of redefinition of $G$, while keeping it as a constant, is only valid if the ratio ${m_i}/{m_G}$ is constant, meaning if there is a constant scale factor between the inertial mass and the gravitational mass. By the way, this scale factor would actually never be noticed since from the begginning it would already be accounted for in the constant $G$.
You could do the same thing with electrical forces, using Coulombs law. You can check if electrical charge is the same as inertial mass, since you have:
$${bf{F}} = m_i {bf{a}} tag{1}$$
$${bf{F_e}} = -K frac{q_1 q_2}{|{bf{r}}_{12}|^2} hat{{bf{r}}} tag{2}$$
You can ask, is $q_1$ the same as $m_i$? And it is true that for one specific case you could redefine $K$ such that it would come out that $q_1$ and $m_i$ are the same, but it would not pass the scaling test, since doubling the inertial mass does not double the charge.
answered Dec 14 at 13:36
Hugo V
1,486616
To make it clear that it is not obvious it is better to stop using the word "mass" in both cases. So it is better to say that it is not obvious that the inertial resistance, meaning the property that scales how different objects accelerate under the same given force, is the same as the "gravitational charge", meaning the property that scales the gravitational field that different objects produce.
Just to make it clear, the problem with the equivalence of masses is not "does $m_i=1$ imply $m_{G}=1$?" in whatever units. The real question is, "does doubling the inertial mass actually double the gravitational mass?". So your procedure of redefinition of $G$, while keeping it as a constant, is only valid if the ratio ${m_i}/{m_G}$ is constant, meaning if there is a constant scale factor between the inertial mass and the gravitational mass. By the way, this scale factor would actually never be noticed since from the begginning it would already be accounted for in the constant $G$.
You could do the same thing with electrical forces, using Coulombs law. You can check if electrical charge is the same as inertial mass, since you have:
$${bf{F}} = m_i {bf{a}} tag{1}$$
$${bf{F_e}} = -K frac{q_1 q_2}{|{bf{r}}_{12}|^2} hat{{bf{r}}} tag{2}$$
You can ask, is $q_1$ the same as $m_i$? And it is true that for one specific case you could redefine $K$ such that it would come out that $q_1$ and $m_i$ are the same, but it would not pass the scaling test, since doubling the inertial mass does not double the charge.
answered Dec 14 at 13:36
Hugo V
1,486616
edited Dec 16 at 10:09
answered Dec 14 at 13:36
Hugo V
1,486616
answered Dec 14 at 13:36
Hugo V
1,486616
answered Dec 14 at 13:36
Hugo V
1,486616
1,486616
8
As far as I can see, the fundamental issue is not with doubling, but already with singling. That is: do any two objects with the same inertial mass also have the same gravitational mass?
– Emil Jeřábek
Dec 14 at 15:40
6
Sure, I agree, but I used the "doubling" as an example. I meant that the question is if the ratio of inertial mass to gravitational mass is the same for all objects. So that covers both the case of doubling $m_i$ and getting double $m_g$, as well as your case, which is if different objects with $m_i$ will have the same $m_g$. If the ratio is the same for all objects of all masses, then the masses are equivalent no matter what.
– Hugo V
Dec 14 at 15:52
add a comment |
8
As far as I can see, the fundamental issue is not with doubling, but already with singling. That is: do any two objects with the same inertial mass also have the same gravitational mass?
– Emil Jeřábek
Dec 14 at 15:40
6
Sure, I agree, but I used the "doubling" as an example. I meant that the question is if the ratio of inertial mass to gravitational mass is the same for all objects. So that covers both the case of doubling $m_i$ and getting double $m_g$, as well as your case, which is if different objects with $m_i$ will have the same $m_g$. If the ratio is the same for all objects of all masses, then the masses are equivalent no matter what.
– Hugo V
Dec 14 at 15:52
8
8
As far as I can see, the fundamental issue is not with doubling, but already with singling. That is: do any two objects with the same inertial mass also have the same gravitational mass?
– Emil Jeřábek
Dec 14 at 15:40
As far as I can see, the fundamental issue is not with doubling, but already with singling. That is: do any two objects with the same inertial mass also have the same gravitational mass?
– Emil Jeřábek
Dec 14 at 15:40
6
6
Sure, I agree, but I used the "doubling" as an example. I meant that the question is if the ratio of inertial mass to gravitational mass is the same for all objects. So that covers both the case of doubling $m_i$ and getting double $m_g$, as well as your case, which is if different objects with $m_i$ will have the same $m_g$. If the ratio is the same for all objects of all masses, then the masses are equivalent no matter what.
– Hugo V
Dec 14 at 15:52
Sure, I agree, but I used the "doubling" as an example. I meant that the question is if the ratio of inertial mass to gravitational mass is the same for all objects. So that covers both the case of doubling $m_i$ and getting double $m_g$, as well as your case, which is if different objects with $m_i$ will have the same $m_g$. If the ratio is the same for all objects of all masses, then the masses are equivalent no matter what.
– Hugo V
Dec 14 at 15:52
add a comment |
Scenario I: I have a white ball and a black ball. In the system of units I've adopted, I discover that:
Gravitational mass of white ball = 2
Inertial mass of white ball = 3
Gravitational mass of black ball = 10
Inertial mass of black ball = 15
But I want each ball's gravitational mass to equal its inertial mass. So I fix the constant $G$, multiplying it by $2/3$ so that all my gravitational masses will be multiplied by $3/2$. Problem solved, just like you said!
Scenario II: I have a white ball and a black ball. In the system of units I've adopted, I discover that:
Gravitational mass of white ball = 2
Inertial mass of white ball = 3
Gravitational mass of black ball = 10
Inertial mass of black ball = 20
But I want each ball's gravitational mass to equal its inertial mass. So I fix the constant $G$, multiplying it by $2/3$ so that the white ball's gravitational mass will be multiplied by $3/2$. Or maybe I should fix the constant $G$, multiplying it by $1/2$ so the black ball's gravitational mass will be multiplied by $2$. Uh oh. Neither solution manages to cover both cases.
The full content of the statement that "gravitational mass equals inertial mass" is that "gravitational mass equals inertial mass provided you choose the right units". This doesn't rule out Scenario I, but it rules out Scenario II. The fact that Scenario II can't occur is a substantive statement.
edited Dec 15 at 5:06
Volker Siegel
2,50811642
answered Dec 14 at 18:20
WillO
6,61622132
This is the most intuitive and straightforward explanation.
– Greg Schmit
Dec 14 at 20:34
add a comment |
Scenario I: I have a white ball and a black ball. In the system of units I've adopted, I discover that:
Gravitational mass of white ball = 2
Inertial mass of white ball = 3
Gravitational mass of black ball = 10
Inertial mass of black ball = 15
But I want each ball's gravitational mass to equal its inertial mass. So I fix the constant $G$, multiplying it by $2/3$ so that all my gravitational masses will be multiplied by $3/2$. Problem solved, just like you said!
Scenario II: I have a white ball and a black ball. In the system of units I've adopted, I discover that:
Gravitational mass of white ball = 2
Inertial mass of white ball = 3
Gravitational mass of black ball = 10
Inertial mass of black ball = 20
But I want each ball's gravitational mass to equal its inertial mass. So I fix the constant $G$, multiplying it by $2/3$ so that the white ball's gravitational mass will be multiplied by $3/2$. Or maybe I should fix the constant $G$, multiplying it by $1/2$ so the black ball's gravitational mass will be multiplied by $2$. Uh oh. Neither solution manages to cover both cases.
The full content of the statement that "gravitational mass equals inertial mass" is that "gravitational mass equals inertial mass provided you choose the right units". This doesn't rule out Scenario I, but it rules out Scenario II. The fact that Scenario II can't occur is a substantive statement.
edited Dec 15 at 5:06
Volker Siegel
2,50811642
answered Dec 14 at 18:20
WillO
6,61622132
This is the most intuitive and straightforward explanation.
– Greg Schmit
Dec 14 at 20:34
add a comment |
Scenario I: I have a white ball and a black ball. In the system of units I've adopted, I discover that:
Gravitational mass of white ball = 2
Inertial mass of white ball = 3
Gravitational mass of black ball = 10
Inertial mass of black ball = 15
But I want each ball's gravitational mass to equal its inertial mass. So I fix the constant $G$, multiplying it by $2/3$ so that all my gravitational masses will be multiplied by $3/2$. Problem solved, just like you said!
Scenario II: I have a white ball and a black ball. In the system of units I've adopted, I discover that:
Gravitational mass of white ball = 2
Inertial mass of white ball = 3
Gravitational mass of black ball = 10
Inertial mass of black ball = 20
But I want each ball's gravitational mass to equal its inertial mass. So I fix the constant $G$, multiplying it by $2/3$ so that the white ball's gravitational mass will be multiplied by $3/2$. Or maybe I should fix the constant $G$, multiplying it by $1/2$ so the black ball's gravitational mass will be multiplied by $2$. Uh oh. Neither solution manages to cover both cases.
The full content of the statement that "gravitational mass equals inertial mass" is that "gravitational mass equals inertial mass provided you choose the right units". This doesn't rule out Scenario I, but it rules out Scenario II. The fact that Scenario II can't occur is a substantive statement.
edited Dec 15 at 5:06
Volker Siegel
2,50811642
answered Dec 14 at 18:20
WillO
6,61622132
Scenario I: I have a white ball and a black ball. In the system of units I've adopted, I discover that:
Gravitational mass of white ball = 2
Inertial mass of white ball = 3
Gravitational mass of black ball = 10
Inertial mass of black ball = 15
But I want each ball's gravitational mass to equal its inertial mass. So I fix the constant $G$, multiplying it by $2/3$ so that all my gravitational masses will be multiplied by $3/2$. Problem solved, just like you said!
Scenario II: I have a white ball and a black ball. In the system of units I've adopted, I discover that:
Gravitational mass of white ball = 2
Inertial mass of white ball = 3
Gravitational mass of black ball = 10
Inertial mass of black ball = 20
But I want each ball's gravitational mass to equal its inertial mass. So I fix the constant $G$, multiplying it by $2/3$ so that the white ball's gravitational mass will be multiplied by $3/2$. Or maybe I should fix the constant $G$, multiplying it by $1/2$ so the black ball's gravitational mass will be multiplied by $2$. Uh oh. Neither solution manages to cover both cases.
The full content of the statement that "gravitational mass equals inertial mass" is that "gravitational mass equals inertial mass provided you choose the right units". This doesn't rule out Scenario I, but it rules out Scenario II. The fact that Scenario II can't occur is a substantive statement.
edited Dec 15 at 5:06
Volker Siegel
2,50811642
answered Dec 14 at 18:20
WillO
6,61622132
edited Dec 15 at 5:06
Volker Siegel
2,50811642
edited Dec 15 at 5:06
Volker Siegel
2,50811642
edited Dec 15 at 5:06
Volker Siegel
2,50811642
2,50811642
answered Dec 14 at 18:20
WillO
6,61622132
answered Dec 14 at 18:20
WillO
6,61622132
answered Dec 14 at 18:20
WillO
6,61622132
6,61622132
This is the most intuitive and straightforward explanation.
– Greg Schmit
Dec 14 at 20:34
add a comment |
This is the most intuitive and straightforward explanation.
– Greg Schmit
Dec 14 at 20:34
This is the most intuitive and straightforward explanation.
– Greg Schmit
Dec 14 at 20:34
This is the most intuitive and straightforward explanation.
– Greg Schmit
Dec 14 at 20:34
add a comment |
Try it like this; "All objects follow the same trajectory in a gravitational field".
That means that the force of gravity on an object, which alters its momentum, scales exactly with its inertial mass, which resists that alteration of momentum.
answered Dec 14 at 21:29
Jerome
511
New contributor
Jerome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Try it like this; "All objects follow the same trajectory in a gravitational field".
That means that the force of gravity on an object, which alters its momentum, scales exactly with its inertial mass, which resists that alteration of momentum.
answered Dec 14 at 21:29
Jerome
511
New contributor
Jerome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Try it like this; "All objects follow the same trajectory in a gravitational field".
That means that the force of gravity on an object, which alters its momentum, scales exactly with its inertial mass, which resists that alteration of momentum.
answered Dec 14 at 21:29
Jerome
511
New contributor
Jerome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Try it like this; "All objects follow the same trajectory in a gravitational field".
That means that the force of gravity on an object, which alters its momentum, scales exactly with its inertial mass, which resists that alteration of momentum.
answered Dec 14 at 21:29
Jerome
511
New contributor
Jerome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 14 at 21:29
Jerome
511
New contributor
Jerome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 14 at 21:29
Jerome
511
answered Dec 14 at 21:29
Jerome
511
511
New contributor
Jerome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jerome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jerome is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
It's kind of difficult since we have so much prior intuition that gravitational and inertial mass are the same thing because we know that something heavy to lift also takes more force to accelerate, but there's not really any inherent reason why this should be the case. I think it helps if you call the quantity that goes into the gravitational force law "gravitational charge", to differentiate it from the one in $F = ma$. The question then is "why is the gravitational charge the quantity that determines how much inertia an object has?" (up to a constant scaling factor that you can, indeed, incorporate into $G$).
You could imagine a universe where the magnitude of the electric charge is what determines an object's inertia, and now Newton's second law becomes $F = Sigma |q_i|a$ for an object made up of a number of charged particles. Now an electron and a proton would accelerate to the same speed if you subjected them to the same potential difference, but an atom of hydrogen (mass ~1 GeV) and an atom of positronium (a bound electron and positron, mass ~1 MeV) would fall at different speeds - the force of gravity would still be the same as it is in our universe for each particle, so it would be larger for the hydrogen, but would no longer be balanced by the larger inertial mass of hydrogen proportionally decreasing its acceleration.
Interestingly, if the physical laws of our universe suddenly switched to this behaviour it actually wouldn't be immediately obvious. Since regular matter is made up of equal numbers of protons and electrons, doubling the amount of stuff in something would double both this "electrical inertia" and the regular inertia due to mass, I think the most obvious effect would be plasmas behaving very strangely. Note that I'm ignoring neutrons in this because I didn't think of them and quark charges until now.
answered Dec 14 at 19:05
llama
22116
add a comment |
It's kind of difficult since we have so much prior intuition that gravitational and inertial mass are the same thing because we know that something heavy to lift also takes more force to accelerate, but there's not really any inherent reason why this should be the case. I think it helps if you call the quantity that goes into the gravitational force law "gravitational charge", to differentiate it from the one in $F = ma$. The question then is "why is the gravitational charge the quantity that determines how much inertia an object has?" (up to a constant scaling factor that you can, indeed, incorporate into $G$).
You could imagine a universe where the magnitude of the electric charge is what determines an object's inertia, and now Newton's second law becomes $F = Sigma |q_i|a$ for an object made up of a number of charged particles. Now an electron and a proton would accelerate to the same speed if you subjected them to the same potential difference, but an atom of hydrogen (mass ~1 GeV) and an atom of positronium (a bound electron and positron, mass ~1 MeV) would fall at different speeds - the force of gravity would still be the same as it is in our universe for each particle, so it would be larger for the hydrogen, but would no longer be balanced by the larger inertial mass of hydrogen proportionally decreasing its acceleration.
Interestingly, if the physical laws of our universe suddenly switched to this behaviour it actually wouldn't be immediately obvious. Since regular matter is made up of equal numbers of protons and electrons, doubling the amount of stuff in something would double both this "electrical inertia" and the regular inertia due to mass, I think the most obvious effect would be plasmas behaving very strangely. Note that I'm ignoring neutrons in this because I didn't think of them and quark charges until now.
answered Dec 14 at 19:05
llama
22116
add a comment |
It's kind of difficult since we have so much prior intuition that gravitational and inertial mass are the same thing because we know that something heavy to lift also takes more force to accelerate, but there's not really any inherent reason why this should be the case. I think it helps if you call the quantity that goes into the gravitational force law "gravitational charge", to differentiate it from the one in $F = ma$. The question then is "why is the gravitational charge the quantity that determines how much inertia an object has?" (up to a constant scaling factor that you can, indeed, incorporate into $G$).
You could imagine a universe where the magnitude of the electric charge is what determines an object's inertia, and now Newton's second law becomes $F = Sigma |q_i|a$ for an object made up of a number of charged particles. Now an electron and a proton would accelerate to the same speed if you subjected them to the same potential difference, but an atom of hydrogen (mass ~1 GeV) and an atom of positronium (a bound electron and positron, mass ~1 MeV) would fall at different speeds - the force of gravity would still be the same as it is in our universe for each particle, so it would be larger for the hydrogen, but would no longer be balanced by the larger inertial mass of hydrogen proportionally decreasing its acceleration.
Interestingly, if the physical laws of our universe suddenly switched to this behaviour it actually wouldn't be immediately obvious. Since regular matter is made up of equal numbers of protons and electrons, doubling the amount of stuff in something would double both this "electrical inertia" and the regular inertia due to mass, I think the most obvious effect would be plasmas behaving very strangely. Note that I'm ignoring neutrons in this because I didn't think of them and quark charges until now.
answered Dec 14 at 19:05
llama
22116
It's kind of difficult since we have so much prior intuition that gravitational and inertial mass are the same thing because we know that something heavy to lift also takes more force to accelerate, but there's not really any inherent reason why this should be the case. I think it helps if you call the quantity that goes into the gravitational force law "gravitational charge", to differentiate it from the one in $F = ma$. The question then is "why is the gravitational charge the quantity that determines how much inertia an object has?" (up to a constant scaling factor that you can, indeed, incorporate into $G$).
You could imagine a universe where the magnitude of the electric charge is what determines an object's inertia, and now Newton's second law becomes $F = Sigma |q_i|a$ for an object made up of a number of charged particles. Now an electron and a proton would accelerate to the same speed if you subjected them to the same potential difference, but an atom of hydrogen (mass ~1 GeV) and an atom of positronium (a bound electron and positron, mass ~1 MeV) would fall at different speeds - the force of gravity would still be the same as it is in our universe for each particle, so it would be larger for the hydrogen, but would no longer be balanced by the larger inertial mass of hydrogen proportionally decreasing its acceleration.
Interestingly, if the physical laws of our universe suddenly switched to this behaviour it actually wouldn't be immediately obvious. Since regular matter is made up of equal numbers of protons and electrons, doubling the amount of stuff in something would double both this "electrical inertia" and the regular inertia due to mass, I think the most obvious effect would be plasmas behaving very strangely. Note that I'm ignoring neutrons in this because I didn't think of them and quark charges until now.
answered Dec 14 at 19:05
llama
22116
answered Dec 14 at 19:05
llama
22116
answered Dec 14 at 19:05
llama
22116
answered Dec 14 at 19:05
llama
22116
22116
add a comment |
add a comment |
It is a known fact that Inertial and gravitational mass are the same
thing
Actually not.
(Edit: I have rewritten the answer totally).
To be clear, let us differentiate between facts and theories.
The only facts we have in science are things we can measure. Assume that you measure the ratio between a circles diameter and its circumference. You measure the ratio to 3 plus minus 0.5. This is a fact. Get a better measurement and you might reach that the ratio is 3.1 plus minus 0.1. And so on.
We have tried to measure the difference between inertial and gravitational mass and have not been able to show any differences, so far. This is a fact.
But does it imply that there is no and can never be any difference and that they are identical? In science this is a theory. And the bad thing about theories is that they can never be shown to be true, only falsified.
Let us be inside a space station circling earth. No windows and we feel absolutely no gravity and no acceleration. So both a and F are zero. What happened to the gravity? Well, Einstein solved that by saying that there is no force, but instead it is space that is curved. The space station moves around the earth in a straight line (no force) because a straight line is a circle. Go figure.
If we measure very tiny forces things start to change. Quantuum mechanics did show that at very small scales energy comes in small packades, called quanta. Classical mechanics, example F=ma, has an implied belief that any value of F,m or a is possible. Quantuum mechanics shows that not all values of F or m are possible, there seems to be some minimum "unit" to them. So not even Einstein had it all in place (he did spend a lot of time trying to disprove some of the parts of quantuum mechanics that are now experimentally shown to be more true than false).
So, just maybe, in the future we will be able to measure a difference between inertial and gravitational mass. That measurement would be a fact forcing us to create a new theory. Or we would not be able to measure any difference and then the theory holds so far.
answered Dec 15 at 9:47
ghellquist
1553
add a comment |
It is a known fact that Inertial and gravitational mass are the same
thing
Actually not.
(Edit: I have rewritten the answer totally).
To be clear, let us differentiate between facts and theories.
The only facts we have in science are things we can measure. Assume that you measure the ratio between a circles diameter and its circumference. You measure the ratio to 3 plus minus 0.5. This is a fact. Get a better measurement and you might reach that the ratio is 3.1 plus minus 0.1. And so on.
We have tried to measure the difference between inertial and gravitational mass and have not been able to show any differences, so far. This is a fact.
But does it imply that there is no and can never be any difference and that they are identical? In science this is a theory. And the bad thing about theories is that they can never be shown to be true, only falsified.
Let us be inside a space station circling earth. No windows and we feel absolutely no gravity and no acceleration. So both a and F are zero. What happened to the gravity? Well, Einstein solved that by saying that there is no force, but instead it is space that is curved. The space station moves around the earth in a straight line (no force) because a straight line is a circle. Go figure.
If we measure very tiny forces things start to change. Quantuum mechanics did show that at very small scales energy comes in small packades, called quanta. Classical mechanics, example F=ma, has an implied belief that any value of F,m or a is possible. Quantuum mechanics shows that not all values of F or m are possible, there seems to be some minimum "unit" to them. So not even Einstein had it all in place (he did spend a lot of time trying to disprove some of the parts of quantuum mechanics that are now experimentally shown to be more true than false).
So, just maybe, in the future we will be able to measure a difference between inertial and gravitational mass. That measurement would be a fact forcing us to create a new theory. Or we would not be able to measure any difference and then the theory holds so far.
answered Dec 15 at 9:47
ghellquist
1553
add a comment |
It is a known fact that Inertial and gravitational mass are the same
thing
Actually not.
(Edit: I have rewritten the answer totally).
To be clear, let us differentiate between facts and theories.
The only facts we have in science are things we can measure. Assume that you measure the ratio between a circles diameter and its circumference. You measure the ratio to 3 plus minus 0.5. This is a fact. Get a better measurement and you might reach that the ratio is 3.1 plus minus 0.1. And so on.
We have tried to measure the difference between inertial and gravitational mass and have not been able to show any differences, so far. This is a fact.
But does it imply that there is no and can never be any difference and that they are identical? In science this is a theory. And the bad thing about theories is that they can never be shown to be true, only falsified.
Let us be inside a space station circling earth. No windows and we feel absolutely no gravity and no acceleration. So both a and F are zero. What happened to the gravity? Well, Einstein solved that by saying that there is no force, but instead it is space that is curved. The space station moves around the earth in a straight line (no force) because a straight line is a circle. Go figure.
If we measure very tiny forces things start to change. Quantuum mechanics did show that at very small scales energy comes in small packades, called quanta. Classical mechanics, example F=ma, has an implied belief that any value of F,m or a is possible. Quantuum mechanics shows that not all values of F or m are possible, there seems to be some minimum "unit" to them. So not even Einstein had it all in place (he did spend a lot of time trying to disprove some of the parts of quantuum mechanics that are now experimentally shown to be more true than false).
So, just maybe, in the future we will be able to measure a difference between inertial and gravitational mass. That measurement would be a fact forcing us to create a new theory. Or we would not be able to measure any difference and then the theory holds so far.
answered Dec 15 at 9:47
ghellquist
1553
It is a known fact that Inertial and gravitational mass are the same
thing
Actually not.
(Edit: I have rewritten the answer totally).
To be clear, let us differentiate between facts and theories.
The only facts we have in science are things we can measure. Assume that you measure the ratio between a circles diameter and its circumference. You measure the ratio to 3 plus minus 0.5. This is a fact. Get a better measurement and you might reach that the ratio is 3.1 plus minus 0.1. And so on.
We have tried to measure the difference between inertial and gravitational mass and have not been able to show any differences, so far. This is a fact.
But does it imply that there is no and can never be any difference and that they are identical? In science this is a theory. And the bad thing about theories is that they can never be shown to be true, only falsified.
Let us be inside a space station circling earth. No windows and we feel absolutely no gravity and no acceleration. So both a and F are zero. What happened to the gravity? Well, Einstein solved that by saying that there is no force, but instead it is space that is curved. The space station moves around the earth in a straight line (no force) because a straight line is a circle. Go figure.
If we measure very tiny forces things start to change. Quantuum mechanics did show that at very small scales energy comes in small packades, called quanta. Classical mechanics, example F=ma, has an implied belief that any value of F,m or a is possible. Quantuum mechanics shows that not all values of F or m are possible, there seems to be some minimum "unit" to them. So not even Einstein had it all in place (he did spend a lot of time trying to disprove some of the parts of quantuum mechanics that are now experimentally shown to be more true than false).
So, just maybe, in the future we will be able to measure a difference between inertial and gravitational mass. That measurement would be a fact forcing us to create a new theory. Or we would not be able to measure any difference and then the theory holds so far.
answered Dec 15 at 9:47
ghellquist
1553
edited Dec 15 at 11:26
answered Dec 15 at 9:47
ghellquist
1553
answered Dec 15 at 9:47
ghellquist
1553
answered Dec 15 at 9:47
ghellquist
1553
1553
add a comment |
add a comment |
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Your procedure only works if the two masses are proportional. In that case yes, we can redefine the units to make them equal. Otherwise we can't.
– Javier
Dec 14 at 13:46
8
Technically inertial and gravitational mass are merely proportional, not necessarily equal. One could say that the "true" gravitational mass is really $m_g = sqrt{G} m_i$...
– Aetol
Dec 14 at 15:15
How are they equal? They have different units in the end because G isn't dimensionless.
– Joshua
Dec 14 at 20:36
the equivalence principle can be restated in this way "gravitation is an acceleration field, not a force field"
– lurscher
Dec 14 at 21:46
3
Run like hell, note that units are arbitrary. In the U.S., I measure mass with pounds. Everywhere else, mass is measured with kg. Somewhat related to units are dimensions, such as length, time, mass, etc. Physics is based on the dimensions in the problem, not the units. This means that physics is actually independent of the units used. Thus, you can't redefine the physics of the problem by choosing different units.
– David White
Dec 15 at 5:22