Conservation of Mass

[curiousphoton]

Simple question:

Rest Frame: In their respective rest frames, the following objects have the following rest masses: Object A Mass = 1 kg ; Object B Mass = 1 kg. Total Rest Mass of system: 2 kg.

A & B accelerate and reach some constant Velocity. At this velocity, their relative masses are determined to be 1.5 kg each. They head straight at each other and collide. Their collision is perfectly inelestic.

Let's call the resulting object, Object C. It is my understanding that the rest mass of this object is 3 kg. Total rest mass of system = 3 kg.

So rest mass of A & B seperate = 2 kg. Rest mass of A+B = 3 kg. Is this correct?

A simple Duh, Yes of Course will suffice.

[Bob_for_short]

Yes, it is correct because you added 1 kg of kinetic energy.

P.S. It is not a conservation of mass but conversation of mass.

[A.T.]

Simple question:

Rest Frame: In their respective rest frames, the following objects have the following rest masses: Object A Mass = 1 kg ; Object B Mass = 1 kg. Total Rest Mass of system: 2 kg.

A & B accelerate and reach some constant Velocity.

How do they accelerate? Do you put energy from outside into the system?

At this velocity, their relative masses

You mean the relativistic mass?

are determined to be 1.5 kg each. They head straight at each other and collide. Their collision is perfectly inelestic.

So the energy you put in for acceleration is converted to thermic energy.


Let's call the resulting object, Object C. It is my understanding that the rest mass of this object is 3 kg.

Total rest mass of system = 3 kg.

So rest mass of A & B seperate = 2 kg. Rest mass of A+B = 3 kg. Is this correct?


Rest mass of A+B+thermic energy = 3 kg

[lightarrow]

Simple question:

Rest Frame: In their respective rest frames, the following objects have the following rest masses: Object A Mass = 1 kg ; Object B Mass = 1 kg. Total Rest Mass of system: 2 kg.Said in this way, it's already wrong, because the objects can move with respect the system's centre of mass. If that is the case, Total Rest Mass of system > 2 kg.

A & B accelerate and reach some constant Velocity. At this velocity, their relative masses are determined to be 1.5 kg each. Better if you don't speak of relativistic mass, but of total energy.


They head straight at each other and collide. Their collision is perfectly inelestic.

Let's call the resulting object, Object C. It is my understanding that the rest mass of this object is 3 kg. Total rest mass of system = 3 kg.

So rest mass of A & B seperate = 2 kg. Rest mass of A+B = 3 kg. Is this correct?Yes, *if* the system doesn't radiate away some energy because of the collision, in other terms, it's correct if you measure the system's rest mass before it radiates away any form of energy.

[curiousphoton]

Said in this way, it's already wrong, because the objects can move with respect the system's centre of mass. If that is the case, Total Rest Mass of system > 2 kg.

Since we are looking at the objects at rest in this 'rest frame', and each object has a rest mass of 1 kg, why would the system's mass be greater than 2 kg?

[pervect]

Simple question:


A simple Duh, Yes of Course will suffice.

It sounds to me like you've got the idea basically right, though your wording is rather awkward in a few spots.

Since rest masses don't add, people don't generally talk about "total rest mass", for instance. And "relative mass" doesn't make much sense, I assume this is a typo for "relativistic mass".

Let me write what I think you're trying to say in a totally different manner. If this isn't equivalent, but more detailed, than what you meant to write, then it's possible that there is still some communication problem ....

In computing the mass (i.e the rest mass, or invariant mass) of a system of many particles, we do not simply add together the masses of the component parts. Rather, we go to the definition of mass, which is m = sqrt(E^2/c^2 - p^2/c^4), where E is the total energy of the system, and p is its momentum. Thus one way of computing the mass of a system is to compute its total energy E in a frame in which the total momentum is zero - and then divide by c^2.

It is important that said multi-particle system be a "closed system" if we wish the mass that we compute in this manner to be independent of the observer.

[DaleSpam]

Since we are looking at the objects at rest in this 'rest frame', and each object has a rest mass of 1 kg, why would the system's mass be greater than 2 kg?This is simply due to basic Minkowski geometry, the Euclidean analog is the triangle inequality.

Euclid: the length of the sum of two vectors is less than or equal to the sum of the lengths of the two vectors.

Minkowski: the mass (the invariant length of an energy-momentum four-vector) of the sum of two particles is greater than or equal to the sum of the masses of the two particles.

Each statement is a basic geometric fact that follows directly from the respective definitions of length.

[curiousphoton]

This is simply due to basic Minkowski geometry, the Euclidean analog is the triangle inequality.

Euclid: the length of the sum of two vectors is less than or equal to the sum of the lengths of the two vectors.

Minkowski: the mass (the invariant length of an energy-momentum four-vector) of the sum of two particles is greater than or equal to the sum of the masses of the two particles.

Each statement is a basic geometric fact that follows directly from the respective definitions of length.

Yes, I get that. Notice you said for Minowski: greater than OR equal to.

In this case, why would it not be EQUAL TO the mass of Object A and Object B. They are at rest in their rest frame.

[DaleSpam]

Yes, I get that. Notice you said for Minowski: greater than OR equal to.

In this case, why would it not be EQUAL TO the mass of Object A and Object B. They are at rest in their rest frame.There is no inertial reference frame where both object A and B are at rest.

In Euclidean geometry the length of the sum of two vectors is less than or equal to the sum of the lengths of the two vectors, with equality only in the special case that the vectors are parallel.

In Minkowski geometry the "length" of the sum of two four-vectors is greater than or equal to the sum of the "lengths" of the two four-vectors, with equality only in the special case that the four-vectors are parallel. This special case implies that the objects are at rest wrt each other, i.e. that there exists some intertial frame where they are both at rest.

[curiousphoton]

There is no inertial reference frame where both object A and B are at rest.

I'm sitting at my desk with Object A (Pen) and Object B (Pen #2) in front of me. My two pens and are at rest. Since you are saying there is no RF where both A & B are at rest, which one is moving, A or B?

[DaleSpam]

I'm sitting at my desk with Object A (Pen) and Object B (Pen #2) in front of me. My two pens and are at rest. Since you are saying there is no RF where both A & B are at rest, which one is moving, A or B?Huh? This conflicts with your previous description:A & B accelerate and reach some constant Velocity. At this velocity, their relative masses are determined to be 1.5 kg each. They head straight at each other and collide.For the original description there is no reference frame where they are both at rest.

As I said in my previous post in the special case that the objects are at rest wrt each other then their energy-momentum four-vectors are parallel and the mass of the sum is equal to the sum of the masses. That special case obviously applies to your altered description.