Is Mass Relative? - Exploring the Question

[derek.basler]

Ok this may or may not sound like a very stupid idea. But having read the Black Hole War by Leonard Susskind, he makes a claim that as the Earth gets closer and closer to the center, squeezed by gravity, it loses potential energy. The lost potential energy is then radiated as heat, and therefore the Earth loses energy. And since energy and mass are inter-related, the Earth thereby loses mass. However, if my understanding is correct, potential energy is a relative measure [ P.E = mgh]. So would that also make mass a relative aspect as well, since energy is really mass? My understanding of the General Theory of Relativity is somewhat weak, so if anyone could just clear this up I would greatly appreciate it.

 

[HallsofIvy]

Of course it is. That's true in the special theory as well: if the mass of an object is m₀ in a coordinate system in which it is at rest, its mass as measured in a coordinate system moving at speed v with respect to it is
$$\frac{m_0}{\sqrt{1- \frac{v^2}{c^2}}}$$

 

[derek.basler]

So mass is not a set number at all then. If you were traveling next to an object at the same velocity, and you measured its mass, and then some how measured it while you were at rest and the object was still moving, their masses would differ?

 

[alxm]

Sounds weird. I was always told I'd _lose_ weight by jogging. :)

 

[Naty1]

You got it!
Mass, length, time, kinetic energy are all relative.
Electric charge and potential energy seem fixed. And I still wonder about potential energy.

 

[Dmitry67]

I wonder what physics mean by MASS by default?
Relativistic (like in this thread) or Invariant mass?
I was told in this forum that in modern physics mass=invariant mass as relativistic mass is just a synonym for E/c^2 and is useless.

 

[v2kkim]

The mass needs to be useful. One useful application should be for gravitational force calculation.

 

[derek.basler]

well it seems that you change in mass would be equal to your change in energy, potential or kinetic, then that divided by the speed of light squared plus your rest mass? Is there a mass that stays invariant no matter what reference frame you observe it from?

 

[jtbell]

Is there a mass that stays invariant no matter what reference frame you observe it from?

Yes, the invariant mass $m_0$ which can be calculated from

$$m_0 c^2 = \sqrt {E^2 - (pc)^2}$$

has the same value in all inertial reference frames.

 

[Dale]

The relativistic mass (aka total energy) is relative. The invariant mass (aka rest mass) is invariant. Typically the unqualified term "mass" refers to "invariant mass".

 

[Naty1]

I was told in this forum that in modern physics mass=invariant mass as relativistic mass is just a synonym for E/c^2 and is useless.

That does seem to be a convention favored by some here, yet in books I read by Smolin, Hawking, Penrose, Greene, Randall and others, they seem to routinely use relativistic mass...I don't see it as a big deal either way...

 

[Ich]

The lost potential energy is then radiated as heat, and therefore the Earth loses energy. And since energy and mass are inter-related, the Earth thereby loses mass.

True. The Earth loses invariant mass during this process. It's a pity that pervect is no longer around to give a short lecture on the meaning of "mass" in GR.

 

[Dmitry67]

That does seem to be a convention favored by some here, yet in books I read by Smolin, Hawking, Penrose, Greene, Randall and others, they seem to routinely use relativistic mass...I don't see it as a big deal either way...

The big deal is that people get replies without specifying the meaning of the "mass".
The same person can get inconsistent replies in parralel threads, like "is mass conserved? yes", "is mass invariant? yes"

 

[lightarrow]

That does seem to be a convention favored by some here, yet in books I read by Smolin, Hawking, Penrose, Greene, Randall and others, they seem to routinely use relativistic mass...I don't see it as a big deal either way...

Are they textbooks of physics currently used at university? Because if they were, it would be serious; if they are not, then I suggest you to look up in physics textbooks.