B Inertial, gravitational masses

jeremyfiennes

Summary
Under what circumstances are they different?
Under what circumstances are they different?

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PeterDonis

Mentor
Under no circumstances at all. They're the same.

pervect

Staff Emeritus
In Newtonian theory, the two are completely separate concepts that just happen to be identical for no clear reason. This suggests a common origin between inertial forces and gravitational forces, but Newtonian theory can't account for why they are the same.

In General relativity, the principle of equivalence basically assumes that they are the same, so the identity is part of the theory. This assumption is based on a fair number of experimental observations (look up 'tests of the equivalence principle'). As such, there is no need to distinguish between inertial and gravitatioanl mass in General relativity.

This is slightly oversimplified, in particular I haven't adressed the subtleities in the Principle of Equivalence, such as the weak, strong, and Einstein variants. But it should be a good start.

jeremyfiennes

My question came from a comment I read in Hatch, R.R. (2007) "A New Theory of Gravity", Physics Essays 20:1 "Inertial and gravitational mass diverge in value as a function of velocity." So he is wrong?
Thanks.

PeterDonis

Mentor
Hatch, R.R. (2007) "A New Theory of Gravity", Physics Essays 20:1
This, from the abstract, looks like a speculative paper from a questionable journal. It appears to be behind a paywall so I can't read the full paper.

Also, if you are asking a question based on something you read in a reference, you should give that reference in the OP.

"Inertial and gravitational mass diverge in value as a function of velocity." So he is wrong?
No such effect has ever been observed experimentally. The current limits are pretty tight:

Also, since in relativity "velocity" has no invariant meaning, it's hard to see what "as a function of velocity" would mean physically.

metastable

Also, since in relativity "velocity" has no invariant meaning, it's hard to see what "as a function of velocity" would mean physically.
I’m confused by this as I believe it was explained in a recent time dilation thread that the time dilation experienced by a particle has dependence on the length of its world line compared to the length of another particle’s world line, when the particles are separated and then brought back together.

PeterDonis

Mentor
the time dilation experienced by a particle has dependence on the length of its world line compared to the length of another particle’s world line, when the particles are separated and then brought back together
This is true. Why does it make you confused about what I said about velocity?

metastable

If 2 particles are at rest, one then moves around wildly and is then brought back to rest with respect to the other particle, isn’t it now younger than the other particle that had a shorter world line or in plain language younger than the particle that had “less velocity?” Isn’t the “age” of the particle a function of its velocity?

PeterDonis

Mentor
Isn’t the “age” of the particle a function of its velocity?
No. Velocity is not an invariant, just as "at rest" is not an invariant.

Orodruin

Staff Emeritus
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If 2 particles are at rest, one then moves around wildly and is then brought back to rest with respect to the other particle, isn’t it now younger than the other particle that had a shorter world line or in plain language younger than the particle that had “less velocity?” Isn’t the “age” of the particle a function of its velocity?
Although it can be expressed as a functional of the velocity, it is an invariant quantity. However, you have misunderstood what it means to have a short world-line. The length of your world-line is the amount that you have aged. Hence, the shorter the world-line, the younger the object. Now, geometry in spacetime does not work the same way as geometry works in a "normal" space so the object moving around actually has a shorter world-line.

Ibix

If 2 particles are at rest, one then moves around wildly and is then brought back to rest with respect to the other particle, isn’t it now younger than the other particle
If, by "at rest", you actually mean "at rest in an inertial frame" then yes. Otherwise, maybe. It depends on what you mean by "at rest".
isn’t it now younger than the other particle that had a shorter world line or in plain language younger than the particle that had “less velocity?”
The length (more precisely, the interval) of the worldline is the key thing here, the invariant on which everyone will agree. Whether or not an object had "more" or "less" velocity is dependent on your choice of frame.
Isn’t the “age” of the particle a function of its velocity?
If you pick a frame then you can express the length of the worldline as a function solely of the velocity of the particle with respect to the frame. The form of the expression depends on your choice of frame, however, which is how the result can be the same even if you choose to use a frame where the velocity of the "wildly moving" one is zero.

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stevendaryl

Staff Emeritus
If 2 particles are at rest, one then moves around wildly and is then brought back to rest with respect to the other particle, isn’t it now younger than the other particle that had a shorter world line or in plain language younger than the particle that had “less velocity?” Isn’t the “age” of the particle a function of its velocity?
The following comment is not relevant to the issue of inertial versus gravitational mass, but I would like to suggest that interpreting time dilation as "speed slowing aging" is a bad way to think about it.

There are two different notions of "time" in Special Relativity: (1) coordinate time is one dimension of 4-dimensional spacetime, and as such is much like the other three spatial dimensions, (2) proper time is a measure of a traveler's progress through spacetime. If you travel for one hour, according to your watch, you will end up in a different point in spacetime than when you started. Exactly where you end up in spacetime depends on your proper velocity. How much your x-coordinate changes depends on the x-component of your velocity, $\dfrac{\Delta x}{\Delta \tau}$ (where $\tau$ is the time on your watch). How much your t-coordinate changes depends on the t-component of your velocity, $\dfrac{\Delta t}{\Delta \tau}$.

metastable

How much your t-coordinate changes depends on the t-component of your velocity
If I understand this correctly,"not accelerating" an unstable particle can "hasten" its loss of inertial mass.

stevendaryl

Staff Emeritus
If I understand this correctly,"not accelerating" an unstable particle can "hasten" its loss of inertial mass.
No, I wasn't saying that. What I was saying was that if you're traveling to a particular point in spacetime--say, the point "Seattle, July 1, 2019", you can get there quicker (according to your watch) if you have a larger 4-velocity.

The appropriate notion of "time" to use when measuring such things as decay rates is proper time, not coordinate time.

Ibix

larger 4-velocity.
Should that say "3-velocity"? 4-velocity is normalised to $c$. Or am I missing your point?

PeterDonis

Mentor
you can get there quicker (according to your watch) if you have a larger 4-velocity.
I assume that, as @Ibix suggested, you mean a larger 3-velocity. 4-velocity is a unit vector; all 4-velocities are "the same size".

stevendaryl

Staff Emeritus
Should that say "3-velocity"? 4-velocity is normalised to $c$. Or am I missing your point?
Yes, that's what I meant.

Nugatory

Mentor
say, the point "Seattle, July 1, 2019", you can get there quicker (according to your watch) if you have a larger 43-velocity.
Although you’ll also have to start later and closer.

stevendaryl

Staff Emeritus
Although you’ll also have to start later and closer.
If you want to keep a constant 4-velocity, yes. But you can zig-zag to Seattle, July 1, 2019 and get there in arbitrarily small amount of proper time.

"Inertial, gravitational masses"

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