Does inertia vary from place to place?

• November 16th, 2013, 06:15 AM
vinoo
Does inertia vary from place to place?
Does the force needed to produce unit acceleration in say a 1kg in free space object vary depending on the location?
Basically is inertia stronger or weaker in other parts of our universe or is it uniform throughout?
• November 16th, 2013, 07:15 AM
Markus Hanke
Quote:

Originally Posted by vinoo
Does the force needed to produce unit acceleration in say a 1kg in free space object vary depending on the location?

No. It depends only on mass.

Quote:

Basically is inertia stronger or weaker in other parts of our universe or is it uniform throughout?
All currently available evidence is consistent with the laws of gravity being the same everywhere in the universe.
• November 16th, 2013, 05:33 PM
Hill Billy Holmes
If inertia is dependant upon mass and mass increases due to relative speed?
• November 16th, 2013, 06:33 PM
Dywyddyr
For someone who claimed to have "spent a few years studing Einstien exclusivly" you're remarkably uninformed.
It must have taken a great effort to remain so ignorant.
• November 16th, 2013, 07:38 PM
Hill Billy Holmes
Sir Daffystein duck, actually it takes appsolutley very little effort. Sorry for my above stupid question.
• November 16th, 2013, 08:30 PM
Daecon
Quote:

Originally Posted by Hill Billy Holmes
If inertia is dependant upon mass and mass increases due to relative speed?

I've only recently learned myself that there's a difference between rest mass (invariant) and relativistic mass.

I'm guessing that inertia is dependent on invariant rest mass and not of relativistic mass, which increases at near-light speeds?
• November 16th, 2013, 08:33 PM
Hill Billy Holmes
I am corrected. Thank you Deacon.
• November 17th, 2013, 03:03 AM
Markus Hanke
Quote:

Originally Posted by Daecon
I'm guessing that inertia is dependent on invariant rest mass and not of relativistic mass, which increases at near-light speeds?

Inertia - being the resistance to change in the state of motion - is a property of all forms of energy; it thus applies to relativistic mass, and not just to rest mass. This simply means that as an object approaches the speed of light, it becomes harder and harder to accelerate it more, or, in other words, that a constant acceleration has less and less effect as you get closer to c.
• November 17th, 2013, 03:41 AM
KJW
Quote:

Originally Posted by Daecon
I'm guessing that inertia is dependent on invariant rest mass and not of relativistic mass, which increases at near-light speeds

It depends on the way you view it. For example, Newton's second law can be relativistically expressed in terms of four-dimensional vectors:

( is the absolute differential operator, the covariant form of the ordinary differential operator)

Here, is the invariant (rest) mass.

Alternatively, a force density can be defined in terms of the covariant divergence of the energy-momentum density tensor :

Here, has to be only a part of the total energy-momentum density because for the total energy-momentum density:

which may be considered as a form of Newton's third law. Nevertheless, the individual components of the force density is the result of the individual components of the energy-momentum density tensor (i.e. relativistic mass).
• November 17th, 2013, 04:58 AM
Markus Hanke
Quote:

Originally Posted by KJW
( is the absolute differential operator, the covariant form of the ordinary differential operator)

Interesting, I haven't come across the notation before; I would have written it as

with capital letters. Not that it really makes much difference :)
• November 17th, 2013, 05:44 AM
KJW
Quote:

Originally Posted by Markus Hanke
Quote:

Originally Posted by KJW
( is the absolute differential operator, the covariant form of the ordinary differential operator)

Interesting, I haven't come across the notation before; I would have written it as

with capital letters. Not that it really makes much difference :)

Actually, it's the notation used in my first textbook on Tensor Calculus. It also uses the semicolon notation for the covariant derivative (i.e. ), but I dislike that notation, so I use the nabla operator (i.e. ) which I got from another textbook. On forums where symbol fonts are unavailable, I would use for the covariant differential operator and for both the partial and ordinary differential operators.
• November 17th, 2013, 06:25 AM
Markus Hanke
Quote:

Originally Posted by KJW
Actually, it's the notation used in my first textbook on Tensor Calculus. It also uses the semicolon notation for the covariant derivative (i.e. ), but I dislike that notation, so I use the nabla operator (i.e. ) which I got from another textbook. On forums where symbol fonts are unavailable, I would use for the covariant differential operator and for both the partial and ordinary differential operators.

I share your sentiments in that I deeply dislike the "," and ";" notations; from my first textbook I learned the following notations for the partial and covariant derivatives respectively :

I use them myself to this day, but most people don't seem to be familiar with them.
• November 17th, 2013, 06:31 AM
KJW
An alternative way of expressing the covariant form of the ordinary differential operator is:

which is an application of the chain rule for the total derivative ()

I tend to prefer this way as it doesn't introduce any new notation to a set of expressions that already uses the covariant differential operator ().
• November 17th, 2013, 06:37 AM
KJW
Quote:

Originally Posted by Markus Hanke
I use them myself to this day, but most people don't seem to be familiar with them.

I'm not familiar with the specific notation, though I know you are referring to covariant and/or partial derivatives. But I do need to have a closer look to see whether you are referring to covariant or partial derivatives.