Is Mass Relative to the Observer?

[Mr.Illusion]

I know that time and velocity are relative to the observer, but what about mass? For example, if my friend were to go 0.999c, we know that his mass would increase to me. But would he measure the same increase, or would he feel the same?

 

[yuiop]

I know that time and velocity are relative to the observer, but what about mass? For example, if my friend were to go 0.999c, we know that his mass would increase to me. But would he measure the same increase, or would he feel the same?

You would measure your friend to be length contracted, his clock to be running slower than your clock and his "relativistic mass" to have increased. Your friend will think exactly the same of you. You are length contracted, time dilated and mass increased from his point of view and yet you would not feel any different. It is the same for your friend. He does not feel any different. The concept of "relativistic mass" is a delicate matter and its use is discouraged these days. It is something that has to handled with care.

 

[DaveC426913]

I know that time and velocity are relative to the observer, but what about mass? For example, if my friend were to go 0.999c, we know that his mass would increase to me. But would he measure the same increase, or would he feel the same?

He would not experience any increase, no.

In fact, as far as he is concerned, he is stationary and you are moving at .999c.

In fact, there is no test either of you could do (inside your closed spaceship) to prove that you are moving at all.

 

[Shackleford]

In my university physics textbook, it says that most physicists rather not say the mass increases, but rather there is an increase in momentum, when dealing with the relativistic equations and their consequences.

 

[starthaus]

but rather there is an increase in momentum,

...or energy

 

[Shackleford]

...or energy

Yeah. The increase in total energy stems from the increase in momentum, since the rest energy is constant.

 

[starthaus]

Yeah. The increase in total energy stems from the increase in momentum, since the rest energy is constant.

$$E^2=(\vec{P}c)^2+(mc^2)^2$$
$$\vec{P}=\gamma(u)m\vec{u}$$

 

[Shackleford]

$$E^2=(\vec{P}c)^2+(mc^2)^2$$

That's what I was basing my comment on.

 

[skycastlefish]

The concept of "relativistic mass" is a delicate matter and its use is discouraged these days. It is something that has to handled with care.

Why? Who do they think they're going to offend?

 

[Shackleford]

Why? Who do they think they're going to offend?

Maybe because it's pedagogically undesirable or misleading.

 

[Mr.Illusion]

Thanks guys, I have a better grasp of this now.

 

[hawkingfan]

The conservation of mass still holds in special relativity despite the Lorentz transformation for it. OK, this sounds incorrect but I assure you that it isn't.

Assume that the rest mass of the particle stays constant even while the object is moving. That means that the amount of matter in a point-like particle remains constant. When at rest, this pointlike particle has a spherical geometry with differential volume dV = 4/3(pi)(dr)^3. It also has it's own specific spherically symmetrical gravitational vector field in 3 dimensions.

When this point-like particle begins moving, the length of it in it's direction of motion decreases by the Lorentz factor. (only in that direction) The geometry of the point-like particle becomes like an oblate ellipsoid due to the length contraction in one of it's directions. Correspondingly, the geometry of the gravitational vector field is distorted as the field lines become closer together. (if you know multivariable calculus, it means that the divergence of the field at any specified point becomes more negative with increasing velocity and increasing "oblateness") This increase of the particle's gravitational field at any specified point causes it to appear to have more "mass". (because according to Newton's inverse square law, if the distance between the CMs are fixed, increasing gravitational fields are due to the increased "mass" of the object creating the field)

OK, I will now give an example to clarify this. Assume that there are two particles; particle A and particle B. Particle A now begins moving at a nonzero constant speed relative to B. In B's reference frame, an observer would see A's length contract by the Lorentz factor. Correspondingly, an observer in frame B would also see the gravitational field lines of particle A lump closer and closer together. (the closeness depends on the value of the Lorentz factor) Also since particle B is at rest in it's own reference frame, it sees it's pointlike diameter as unchanged. Therefore more field lines pass through particle B since it's diameter is constant while field lines lump together. This causes particle A to appear as having more "mass" in B's frame.

Similarly, viewed from the reference frame of particle A, particle B is moving at with respect to A at that same speed (but in opposite direction) In A's frame, it sees itself as having the same diameter while the field lines of B lump together increasing the "mass" of B in A's frame.

Let's also say that we have three particles, C, D and E. Assume that both D and E move at the same velocity with respect to C. An observer in frame C would see that the "mass" of both D and E become larger at the same rate.

Now let's focus only only particles D and E which are at rest with respect to each other. The length/semi-major diameter of D is decreased and it's gravitational flux at any point is correspondingly increased. Why doesn't it's mass appear to increase with respect to E? This is because particle E also decreases it's length/ semi-major diameter and therefore less field lines will be able to pass through it. The increase of D's gravitational flux is compensated by the reduced semimajor diameter of E such that the "mass" of D appears the same.

I've deduced this explanation of mass increases based on Einstein's argument for mass-energy equivalence in special relativity found at the website below:

http://aether.lbl.gov/www/classes/p139/exp/gedanken.html

Experiment 9 concerns the apparent increase in mass of objects but it does not give the explanation of gravitational fields. It simply states that the mass of an object must change at the same rate as the time dilation in that frame so that the momentum remains the same. The field explanation seems plausible based upon my own reasoning and the fact that this is how electromagnetic waves are propagated. The geometry of an electric vector field of a charged point particle depends on it's oblateness which in turn is dependent on it's velocity. If a point charge oscillates back and forth, the variation of it's electric field is periodic (like a sine wave) and it's magnetic field oscillates correspondingly with it's electric field. Changes in this information propagates outward from the point particle at the speed of light and this is what constitutes an electromagnetic wave. This is the relativistic explanation of Maxwell's electromagnetism which states that the speed of light must always the same in any reference frame.

 

[GRDixon]

The concept of "relativistic mass" is a delicate matter and its use is discouraged these days. It is something that has to handled with care.

Yes, things have changed since Feynman gave his lectures. In Sect 15-1 (Volume 1) he writes "m=gamma*m0") and then adds, "...that is all there is to the theory of relativity --- it just changes Newton's laws by introducing a correction factor to the mass."

 

[qiuhd]

If we accelerate a nuclear bomb to 0.999c, then detonate it, will the exploration release more energy? because the mass difference caused by the exploration maybe increased.

 

[Shackleford]

If we accelerate a nuclear bomb to 0.999c, then detonate it, will the exploration release more energy? because the mass difference caused by the exploration maybe increased.

Hm. We may relativistically observe more energy in the system due to its velocity/momentum, but the nuclear reaction in its frame will not produce more energy.