Questions about inertial mass

[Johnny5]

I know this is going to sound like a stupid question, but is inertial mass a scalar or a vector?

 

Another question I have is, what is the definition of inertial mass? Is there an operational definition for it?

 

Thank you

[5614]

inertial mass is the mass of a body as determined by the second law of motion from the acceleration of the body when it is subjected to a force that is not due to gravity aka f=ma

[Johnny5]

inertial mass is the mass of a body as determined by the second law of motion from the acceleration of the body when it is subjected to a force that is not due to gravity aka f=ma

 

That sounds like an excellent answer. Let me ask you something, why did you exclude gravity?

 

Thanks

[5614]

because that's the definition of inertial mass!!!!

[Johnny5]

because that's the definition of inertial mass!!!!

 

Will the answer depend upon whether or not the object I push in a vacuum is spinning?

 

Regards

[Martin]

I know this is going to sound like a stupid question' date=' but is inertial mass a scalar or a vector?

 

Another question I have is, what is the definition of inertial mass? Is there an operational definition for it?

 

Thank you[/quote']

 

there is a trend in physics to get rid of other kinds of mass as

unnecessary (even confusing) verbal baggage and say that

the only kind of mass is rest mass

 

which is the object's inertia in its rest frame

 

so for many physicists nowadays "mass" just means rest mass,

there is no reason to say "rest" because there is no other kind

 

you asked about OPERATIONAL definition.

 

people typically use gravity. they measure g the acceleration of gravity at a particular time and place and they measure the force F of the object's weight.

basically they know that if they dropped the object it would accelerate

with acceleration a = g.

So the mass m is defined as m=F/a

 

for a microscopic object like a proton or electron one can apply to it a known force by an electric field (a known voltage gradient)

and observe the acceleration

so, again, one is operationally defining the mass by m = F/a

 

when high speeds are involved one needs to make relativistic corrections because F = ma is not exactly true and ideally one should have the object at rest, to measure the rest mass. so it is more complicated.

 

but the ideal situation where the object is at rest or nearly at rest is not a bad picture to have.

 

(when an object is moving it does not have a single scalar value for inertia because then its resistance to being accelerated depends on direction, but people insist that mass is inertia and that it be independent of direction, so this pretty much forces them to focus on inertia at rest---rest mass. then all the other stuff can be calculated from that depending on how the thing moves)

[Johnny5]

there is a trend in physics to get rid of other kinds of mass as

unnecessary (even confusing) verbal baggage and say that

the only kind of mass is rest mass

 

which is the object's inertia in its rest frame

 

so for many physicists nowadays "mass" just means rest mass' date='

there is no reason to say "rest" because there is no other kind [/quote']

 

I like this response, seems consistent with my way of doing physics. Rest mass is the objects 'inertia' (resistance to acceleration) as measured in a frame of reference in which the object is at rest, and remaining so, hence it's measured in the vacuum, where the gravitational potential is approximately zero.

 

 

you asked about OPERATIONAL definition.

 

people typically use gravity. they measure g the acceleration of gravity at a particular time and place and they measure the force F of the object's weight.

basically they know that if they dropped the object it would accelerate

with acceleration a = g.

 

Isn't this incorrect, as you just said? And I have a few questions about weight, now that you bring it up. Is my weight on earth, the same kind of thing as when I am in a spaceship being accelerated? I mean on the earth, there is a force F upon me, and on the ship there is a force F, upon me. The origins of the force upon me are different. In the one case, my body is in a nonzero gravitational field, and in the other case, my body is in a zero gravitational field. So if weight is strictly caused by gravity, then is it correct to write W=mA, where A is the absolute acceleration of the spaceship?

 

Or to ask the question succinctly, are gravitational and inertial mass equivalent?

 

Regards

[Johnny5]

(when an object is moving it does not have a single scalar value for inertia because then its resistance to being accelerated depends on direction' date=' but people insist that mass is inertia and that it be independent of direction, so this pretty much forces them to focus on inertia at rest---rest mass. then all the other stuff can be calculated from that depending on how the thing moves)[/quote']

 

This confuses me.

 

Why does an object's resistance to being accelerated depend on direction???

 

Regards

[swansont]

This confuses me.

 

Why does an object's resistance to being accelerated depend on direction???

 

Regards

 

If you don't use rest mass, you can then cause a larger acceleration perpendicular to the motion, (or opposed the motion I would think) than by applying the same force in the direction of motion. This is very apparent when v approcaches c.

[Johnny5]

If you don't use rest mass, you can then cause a larger acceleration perpendicular to the motion, (or opposed the motion I would think) than by applying the same force in the direction of motion. This is very apparent when v approcaches c.

 

How in the world do you reach the conclusion that this becomes very apparent when v approaches c?

 

Where are you getting the idea that if something is moving in your rest frame with speed v, and you push it with force F in the direction of motion and give it acceleration a1, that the same force applied perpendicular to v would give it an acceleration a2, which is different from a1.

 

Are you using relativistic mass?

 

Regards

 

PS Did you draw the inference from experiments using particle accelerators? Could there be a drag force at work? I have many more questions.

[J.C.MacSwell]

If you don't use rest mass, you can then cause a larger acceleration perpendicular to the motion, (or opposed the motion I would think[/b']) than by applying the same force in the direction of motion. This is very apparent when v approcaches c.

 

Wouldn't that be especially true?

[swansont]

How in the world do you reach the conclusion that this becomes very apparent when v approaches c?

 

Where are you getting the idea that if something is moving in your rest frame with speed v' date=' and you push it with force F in the direction of motion and give it acceleration a1, that the same force applied perpendicular to v would give it an acceleration a2, which is different from a1.

 

Are you using relativistic mass?

[/quote']

 

Yes - if you use relativistic or inertial mass, instead of rest mass.

[Johnny5]

Yes - if you use relativistic or inertial mass, instead of rest mass.

 

Well exactly how have you reached the conclusion that the formula for relativistic mass is true?

 

Regards

[swansont]

Well exactly how have you reached the conclusion that the formula for relativistic mass is true?

 

I haven't. I'm saying it's not, because of the contradiction I gave. You get the right answer, mathematically, for things like energy and momentum. But you run into the problem I described.

[Johnny5]

I haven't. I'm saying it's not, because of the contradiction I gave. You get the right answer, mathematically, for things like energy and momentum. But you run into the problem I described.

 

Ok, help me out here.

You are saying the relativistic mass formula isn't true?

 

What is the problem you described?

 

Regards

[swansont]

Ok' date=' help me out here.

You are saying the relativistic mass formula [b']isn't[/b] true?

 

What is the problem you described?

 

Regards

 

I'm saying that if you use that as your definition of mass, then you get a given force causing different accelerations in different directions. You have no velocity in a direction perpendicular to your motion, thus you get a different answer. Your term for mass that is direction dependent, i.e. it's no longer a scalar. That's the problem. Using rest mass and relativistic formulae alleviates this problem.

[Johnny5]

I'm saying that if you use that as your definition of mass, then you get a given force causing different accelerations in different directions. You have no velocity in a direction perpendicular to your motion, thus you get a different answer. Your term for mass that is direction dependent, i.e. it's no longer a scalar. That's the problem. Using rest mass and relativistic formulae alleviates this problem.

 

Exactly why do you get a given force causing different accelerations in different directions? I understand what you are saying, I just want to know why you are saying it.

 

Suppose that:

 

[math] M = \frac{m_0}{\sqrt{1-v^2/c^2}} [/math]

 

I don't see any vector symbols there, so could you explain yourself more?

 

Regards

[J.C.MacSwell]

Exactly why do you get a given force causing different accelerations in different directions? I understand what you are saying' date=' I just want to know why you are saying it.

 

Suppose that:

 

[math'] M = \frac{m_0}{\sqrt{1-v^2/c^2}} [/math]

 

I don't see any vector symbols there, so could you explain yourself more?

 

Regards

 

It is assumed to be in the direction of the velocity, and even then only marginally as it will change again as any acceleration takes place.

[swansont]

Exactly why do you get a given force causing different accelerations in different directions? I understand what you are saying' date=' I just want to know why you are saying it.

 

Suppose that:

 

[math'] M = \frac{m_0}{\sqrt{1-v^2/c^2}} [/math]

 

I don't see any vector symbols there, so could you explain yourself more?

 

Regards

 

There are no vector symbols - that's the problem.

 

F=dP/dt = mdv/dt + vdm/dt

 

if the force is in the direction of motion, the mass must change: dm/dt is not zero

 

but an identical magnitude force perpendicular to the motion does no work, and so does not change the speed, so dm/dt = 0

 

The only terms that can be different is dv/dt, which is the acceleration, or m

 

If you want to use the inertial idea that m = F/a, the only way to do that is if the mass is no longer a scalar

[Johnny5]

There are no vector symbols - that's the problem.

 

F=dP/dt = mdv/dt + vdm/dt

 

if the force is in the direction of motion' date=' the mass must change: dm/dt is not zero [/quote']

 

Lets do an actual problem please.

 

Proton being linearly accelerated

 

Initially, a proton is at rest inside a linear accelerator.

Let the rest mass m0 of the proton be exactly 1.67 x 10^-27 kilograms.

 

So the proton is just floating there, it is at rest and remaining at rest. There are currently no external forces acting upon this proton. This frame is an inertial reference frame.

 

Then at some moment in time, an external force acts upon this proton, to accelerate it linearly.

 

Assume the relativistic mass formula is correct. Therefore, the mass of the proton in this frame at all moments in time is given by:

 

[math] M = \frac{m_0}{\sqrt{1-v^2/c^2}} [/math]

 

Since this is an inertial reference frame the following statement is true in this frame:

 

[math] \vec F = \frac{d\vec P}{dt} [/math]

 

The momentum P of this proton in this reference frame is given by

 

[math] \vec P = M \vec v [/math]

 

If you carry out the differentiation you get:

 

[math] \vec F = \frac{m_0}{(1-v^2/c^2)^{3/2}} \vec a [/math]

 

The formula above is valid for a proton starting at rest, and having the velocity and acceleration be in the same direction, which is the case with linear acceleration, so the formula is valid for this exact problem.

 

Now you say:

 

but an identical magnitude force perpendicular to the motion does no work' date=' and so does not change the speed, so dm/dt = 0

 

[/quote']

 

Would you be kind enough to prove this for me? (Thanks in advance)

 

Next you say:

 

The only terms that can be different is dv/dt' date=' which is the acceleration, or m

 

[/quote']

 

I don't understand this, can you explain it more please?

 

Lastly you say:

 

If you want to use the inertial idea that m = F/a' date=' the only way to do that is if the mass is no longer a scalar[/quote']

 

After you answer my other questions, lets focus on this question, which I think is the most important one of all. I have had discussions with several others on this exact question. Some say that division of a vector by a vector is undefined, but I myself realize that if F and a have the same direction, then you can just cancel out the directions, and are left with a scalar.

 

Others (electrical engineers) have mentioned something about this as being similar to how to define an electrical impedance (R=V/i), and insist that this is the way to define m to represent inertia, in the Galilean sense. Then they talk about drag force formulas, and lose me very fast. One of them mentioned something about a phenomenon that happens when you reach a certain speed, (sonic boom), and talked about some kind of mathematical miracle that would happen if you ever got something to reach the speed of light. (The effect they were talking about was something like what happens when you break the sound barrier, but they weren't very clear). And lastly, I recall one of the EE's saying something about the mathematics needed for this kind of analysis, but I cannot remember the name. I think it was vector geometry, but I'm not sure.

 

Anyway, I have many questions about m=F/a, but I will wait for you to answer my first few questions.

 

Thank you

 

J5

[swansont]

an identical magnitude force perpendicular to the motion does no work, and so does not change the speed, so dm/dt = 0

 

This is by definition. W = F.dr, and if they are perpendicular, the work is zero. No work, no change in kinetic energy, no change in speed

 

F=dP/dt = mdv/dt + vdm/dt

 

If I label the two cases 1 and 2, then m₁dv₁/dt + v₁dm₁/dt = m₂dv₂/dt

 

and a = dv/dt

 

so a₁ and a₂ are different or m₁ and m₂ are different

 

 

 

There are objects that exceed the speed of light in a medium, and there is an effect that is analogous to a "sonic boom" known as Cerenkov radiation. But the speed of sound is an engineering barrier, and the speed of light is not.

[Johnny5]

an identical magnitude force perpendicular to the motion does no work' date=' and so does not change the speed, so dm/dt = 0

[/quote']

 

This is what I don't understand. If whilst the object was being accelerated linearly, say along an x axis, in the direction of increasing x coordinates, you instantaneously applied any force whatsover f, in the j^ direction (as the particle passed by you), then you would alter the trajectory of the particle. It would no longer have been purely linearly accelerated.

 

At the moment the force in the j^ direction was applied, the particle would move in that direction, and so dR in that direction would be nonzero, and therefore an identical magnitude force F, perpendicular to the original velocity vector V, would have done work, since F*dR would be nonzero.

 

So I don't understand what you are trying to say I guess, but I'm trying to.

 

Thank you though

[swansont]

The forces can always be broken down into their components. The perpendicular component does no work.

[Johnny5]

The forces can always be broken down into their components. The perpendicular component does no work.

 

Why do you say that the perpendicular component does no work?

 

A proton is flying through a linear accelerator, something is applying a force to it, to cause it to linearly accelerate.

 

Now, someone way down the tube is waiting for it to pass them. As the proton flies by, they "punch" the proton, in a direction perpendicular to its original motion.

 

In that scenario, work is done by the "punch" force. So I am not following you. I do better with equations.

[swansont]

I already gave the equation. F.d (dot product). By definition, the force can do no work. The direction will change, but not the energy.