Mass of a body when Acceleration is 0 and Force is 0.

1. Jun 6, 2015

Ankit Mathur

We know that Force= Mass x Acceleration. When the force applied on a body is 0 the accelaration produced in the body is also 0. (F=m a). So, what will be the mass of a body on which no force is applied and no acceleration is produced?

[ mentor note: post moved from New Member Introductions forum ]

2. Jun 6, 2015

Staff: Mentor

Hi Ankit Mathur.

What are your thoughts on this conundrum?

3. Jun 7, 2015

Ankit Mathur

0 cannot be divided by 0. So I think from now val. mass shold be given in the numericlas of physics.

4. Jun 7, 2015

davenn

don't overthink it .... get back to basics
how about googling .... what is mass? and see what you come up with
any thing you don't quite understand, ask a question

5. Jun 7, 2015

Staff: Mentor

Why not try some values? Does m of 1kg work in your equation?

Try x kg, where x is any value > 0.

6. Jun 8, 2015

logico

A.M.

If you mean that mass can only be measured when there is an acceleration (F=ma) you are right, but there is no problem. Consider a massive particle at the centre of gravity of a massive system. The particle itself experiences no net force, no acceleration, but it exerts a net force on (let's say) every other element of the system. Of course, if you can imagine a universe containing only one point particle, the concept of mass would be meaningless.

7. Jun 8, 2015

Ok

8. Jun 8, 2015

sophiecentaur

If one is used to the ideas in Calculus then the 'Limit as δx approaches zero' being the gradient of a curve will be a familiar idea. The Limit for f=ma, as f approaches zero will give the value of mass. i.e. your force can be a small as you like. (And the curve is, of course, a straight line in tis case.)

9. Jun 8, 2015

logico

In words: When we look at any body accelerating in response to an external force, we find that F=ka, and we call that constant of proportionality "mass". But A.M. wanted to know what happens if there is no F, no a. I think that you are ignoring the problem (and what has calculus fot to do with it?) The problem is real, and there are two schools of thought in the philosophy of science. Some (count me in) hold that mass is just the measured quantity k - a number we find (or could find) by observation of acceleration in response to a force. Others hold that mass is a stuff that exists and persists between observations. I think such a view is inconsistent with QM.

10. Jun 8, 2015

sophiecentaur

It has everything to do with it. The Mass is the gradient of the f/a graph (at least that's one way of looking at it), which is where the Calculus comes into it and it also takes care of the 'zero force' situation by introducing the idea of a limit for that particular relationship. Whether or not you choose to believe that Mass is any more real than other physical quantities, it can still be regarded as a ratio and it just depends where you happen to be starting from.
My mathematical response to the OP has my excellent School (and further) Maths tuition to blame. I have never found Maths to do anything but to help to add to my personal feeling of understanding of the Physical world.

11. Jun 8, 2015

Drakkith

Staff Emeritus
That doesn't look right to me. Mass is a fundamental property, not something that can change between observations, like the position of a particle or the orientation of its spin.

12. Jun 8, 2015

logico

I'm afraid you got that wrong. Gradient is tan theta. It is only when there is no theta, no unique angle, that we use differential calculus - we extend a property of straight lines to curved lines by a clever mathematical argument. You are suggesting the opposite: that we can only discover the properties of straight lines by finding the limiting cases of more fundamental rules for curved lines. Don't blame your teachers - you learned tangents long before you learned calculus - necessarily so.

13. Jun 8, 2015

PWiz

You only need to explore the mathematical implications of the question you are describing to get the answer - no qualitative explanation is required.

There are two cases of division by 0 which are forbidden in math:
1)$\frac {x}{0}$ where x is non-zero
2) $\frac {0}{0}$
The first case is disallowed because there is no number which when multiplied by 0 gives you a non-zero number. (There is no answer)
The second case is disallowed because every number gives you 0 when it is multiplied by 0, so the set of all numbers is the solution. Since there is no unique solution to this problem, this kind of expression is usually considered meaningless.

So there is your answer - the mass can take any numerical value that you care to specify in your scenario. It is impossible to determine a unique value for the mass with the information that has been provided.

14. Jun 8, 2015

logico

The short answer is that we don't know any properties to be fundamental. Mass is more problematic than most. In GR mass depends on your frame of reference, and changes if you are in a changing frame of reference. But the big issue is not whether mass changes from one observation to another, but whether there is any meaning to the word for an unobserved particle. Don't get me wrong - in my everyday life I am perfectly happy that the kilo of flour on my shelf will persist until needed, as will the standard kilogram ingot in Paris (I think). But that property is is not so very different from the properties of solidity and continuousness which break down into something entirely different and more fundamental.

15. Jun 8, 2015

PWiz

This is not true.

16. Jun 8, 2015

logico

I hope you mean that the masses of objects sharing the observer's frame of reference behave in the everyday way. That would be true, but irrelevant to my point. Two identical bodies making a close approach at relativistic velocities will each measure the mass of the other as greater than their own. If you believe there is some god-like perspective from which absolutely true values for mass can be observed, you have missed the point of relativity. (I could have just as easily included Special Relativity - but I didn't want to get into the special case of the "invariant mass" of an isolated system.)

17. Jun 8, 2015

logico

Sorry, that was unfair. Do blame your teachers. They didn't sufficiently emphasise that dy/dx is not actuially a gradient. It's not actually a fraction.

18. Jun 8, 2015

PWiz

On the contrary, I would be forced to advise you to look at the definition of mass in relativity again.
Non-relativistically, mass is defined as a measure of inertia, which tells us how much matter is present in an object. This definition starts to break down is SR, as the coordinate acceleration of an object approaching relativistic velocities starts to differ from the proper acceleration experienced by the accelerating object (the newtonian acceleration $a$ is related to the proper acceleration of $\alpha$ by the relation $\alpha={ \gamma }^3 a$ , and you can clearly see that the coordinate acceleration decreases as the relative velocity between the frames increases even if the proper acceleration stays constant), so the familiar $F=ma$ cannot readily be used (which is the non-relativistic definition).

The relativistic mass was introduced to restore the "classical" version of many equations, primarily the linear momentum expression $p=\gamma m_0 v=mv$ where $m=\gamma m_0$. This is VERY misleading, so much so that the mentors have set up an entire FAQ thread on it. Whether the observer is moving or not wrt to the object will make 0 physical difference to the actual intrinisic structure of the object, and its mass in its rest frame will remain unchanged. You're using "relativistic mass" to describe "mass," which is very different from the usual interpretation of mass meaning "invariant mass."
I also don't get why you mentioned GR in your post when this is clearly an SR issue - we are not concerned with intrinsic spacetime curvature over here.

And by the way, I have not made any allusions to a "true" reference frame in my previous post.

Last edited: Jun 8, 2015
19. Jun 8, 2015

Staff: Mentor

OK, this thread seems to have gone far off into deep water with no navigational aids.....

It is closed.

Last edited: Jun 9, 2015