Real physical effects of Relative Motion

1. Feb 17, 2010

mogra

I have always been confused by SR and GR in terms of what each observer observes of their
own frame of reference.

e.g. Does an observer moving at a high v/c measure length contraction in their onw F of R? I assume not since all apparatus would be similarly contracted.

Nor do they measure time dilation in their own F of R ---only observe it in other frames?

Can they confirm that their mass is increased by applying a force to an object in their F of R?

If mass is increased and length contracted then if density were to not increase the width
perpendicular to the direction of motion would need to increase?

In short, this begs the metaphysical question of whether these changes are real or simply
distorted observations on the part of an oberver in another F of R and the lorentz transforms provide the description.

iunderstand that time dilation has real, permanent and measurable effects but if
mass is increasing then what is the physical mechanism- a gain in no. of molecules?
If length is contracting what forces are compressing it and/or increasing it width wise?

2. Feb 17, 2010

Matterwave

No, he cannot observe anything different in his own frame. Such is the result of the first postulate of special relativity. All inertial observers are equal. If I could measure masses changing in my own frame of reference, it would imply that there is a preferred frame of absolute rest. As far as I'm concerned, if I'm in an inertial frame of reference, I am standing still while everything else moves around me.

3. Feb 17, 2010

Staff: Mentor

Consider the situation where there are three observers, A, B, and C, all moving relative to each other with different speeds. If observer C could in fact measure his own length contraction (or time dilation, etc.), which one would he measure? The length contraction relative to A or to B?

4. Feb 17, 2010

mogra

Thanks. But if i had a powerful enough instrument and watched the molecular structure
of the high V spacecraft as it flew past, how would i account for the increased mass?
Would I see more molecules? If i have a mole of carbon atoms before acceleration i must
have a mole at the end of that acceleration. How can a carbon atom become more massive?

5. Feb 17, 2010

ZapperZ

Staff Emeritus
Relativistic "mass" effect is an everyday consideration for particle accelerators. This is so well-studied and verified, that I can dig up an 1941 paper that already well into dealing with this:

http://physics.princeton.edu/~mcdonald/examples/accel/kerst_pr_60_53_41.pdf [Broken]

Zz.

Last edited by a moderator: May 4, 2017
6. Feb 17, 2010

mogra

I'm sure it's well studied- i just haven't seen an explanation on the molecular/atomic level.

Thanks for the paper. So an electron in an accelerator presumably "sees" my electrons as
similarly increased in mass?

If so, assuming i was made of U235 would there be an increase in the fissile energy
i could release if i detonated?

Last edited by a moderator: May 4, 2017
7. Feb 17, 2010

ZapperZ

Staff Emeritus
Er.. no. I think you don't understand basic SR. Do you see your "mass" increasing? No? Yet, you are moving very, very fast to another creature in another galaxy.

All of the apparent "mass increase" in SR are always from the observer in another inertial reference frame.

What kind of "explanation" are you actually looking for here? The way this is presented, it appears that you are looking for an explanation of an "apple".

Zz.

8. Feb 17, 2010

mogra

Thanks again. I admit i'm not being clear.

HOw's this:

At T1, 3 identical observers are at rest relative to each other.

All observers are made of fissile U235 but each also has a robust instrument that
can survive the nuclear blast.

At T2, Observer C detonates himself and observers A and B get an identical reading.

A fires his engine at T3 and is now accelerating relative to observer B.

At T4, A see's B's mass increasing. B see's A's mass increasing even though he has done nothing.

Both observers then detonate their fissile material (themselves).

Robust instruments record the energy yield of the other observer.

Both record that the yield is higher than that of observer C's original detonation at T1 since both have gained U235 mass as A accelerated.

Yet observer B has merely sat there watching the acceleration of A. Either his objective
mass is constant (in which case he should have released the same energy as C) and B's measurement is impossible or else he has gained objective mass just by sitting there being
observed.

IS this the twin paradox in another form. If so, how?

9. Feb 17, 2010

Ich

But you're aware that there's not much objective about "Energy", and that "relativistic mass" is just another word for energy? So that B gained an enormous amount of kinetic energy (as measured by A) by doing nothing?
As to the fission energy: Depending on how you define it, it will generally change with speed, but not necessarily proportional to total energy.

"relativistic mass" is definitely not what you call "objective mass". That would be "rest mass" rather, aka "mass" nowadays. And fission energy would also be "rest fission energy".

10. Feb 17, 2010

Staff: Mentor

The "yield" or "energy released" is the same in any inertial reference frame, at least by my definition of "energy released." Perhaps your definition is different. Here's a numerical example for a simplified situation so you can judge for yourself.

Start with a "parent" particle with rest mass m_0 = 1000 MeV/c^2, at rest in reference frame S. In this frame, its momentum p and kinetic energy K are both zero, obviously, and its total energy E is 1000 MeV.

Let it decay or fission into two identical "daughter" particles, each with rest mass m_0 = 400 MeV/c^2. Because of conservation of momentum, in frame S they must travel in opposite directions with equal speed, kinetic energy and magnitude of momentum. Suppose for the sake of discussion that they're traveling along the x-axis. I get the following results.

In frame S:

Parent:
m_0 = 1000 MeV/c^2
v = 0
p = 0
K = 0
E = 1000 MeV (m_rel = 1000 MeV/c^2)

Daughter #1:
m_0 = 400 MeV/c^2
v = +0.600c
p = +300 MeV/c
K = 100 MeV
E = 500 MeV (m_rel = 500 MeV/c^2)

Daughter #2:
m_0 = 400 MeV/c^2
v = -0.600c
p = -300 MeV/c
K = 100 MeV
E = 500 MeV (m_rel = 500 MeV/c^2)

My definition of "energy released" is the change in total kinetic energy, which is 100 + 100 - 0 = 200 MeV. This is accounted for by the change in the total rest mass which is 400 + 400 - 1000 = -200 MeV.

Now let's look at this in reference frame S' which is moving relative to S in the +x direction with speed 0.800c. I calculated the following quantities for the three particles in frame S':

Parent:
m_0 = 1000 MeV/c^2
v = -0.800c
p = -1333 MeV/c
K = 667 MeV
E = 1667 MeV (m_rel = 1667 MeV/c^2)

Daughter #1:
m_0 = 400 MeV/c^2
v = -0.385c (using relativistic velocity addition)
p = -167 MeV/c
K = 33 MeV
E = 433 MeV (m_rel = 433 MeV/c^2)

Daughter #2:
m_0 = 400 MeV/c^2
v = -0.946c
p = -1167 MeV/c
K = 833 MeV
E = 1233 MeV (m_rel = 1233 MeV/c^2)

The change in total kinetic energy is 33 + 833 - 667 = 199 MeV which agrees with frame S except for a bit of roundoff error in the calculations.

Last edited: Feb 17, 2010
11. Feb 17, 2010

mogra

Thank you for your help on this. However, i thought the point of SR was that the observer from the S' frame of reference would observe a greater rest mass in the parent particle
compared to the observer from the FoR.

i.e. shouldn't m_0 for the S' observer be greater than 1000 MeV/c^2 as the rest
mass of the parent? i.e. the observer in the S frame sees 1000 but shouldn't the
S' observer moving at -0.8C see more than this rest mass.

Thanks again

12. Feb 18, 2010

Matterwave

Why would we call it "rest mass" if it could change just by moving (would we then call it a "moving mass")? How do we define a "rest mass"?

The rest mass of any particle is defined as the mass observed by an observer at rest with respect to that particle.

In this way, rest mass can never change due to motion.

What DOES change is the kinetic energy. This extra energy yields an "apparent" change in mass or what is called the "relativistic mass". We only have this notion because if you incorporate the gamma and rest mass together, you get equations that look like your everyday equations from classical mechanics. It's a confusing notion, and nowadays people use it less and less.

For example, the equation for momentum is: $$p=mv$$

But this is wrong in the relativistic case since in relativity $$p=\gamma mv$$

People thought "hey if I just put the gamma inside the m, I could call that a "relativistic mass, and all my equations look the same!".

So, you get $$p=m_{rel}v$$ where $$m_{rel} = \gamma m$$

Now your equation looks like the previous one, but you've just confused a hellova lot of people...

13. Feb 18, 2010

yuiop

Relativistic mass is old concept found in old text books and even in Einstein's original paper but the concept has since been rejected in more modern textbooks and by Einstien too. It is better to think of relativistic mass in terms of total energy which is a combination of rest energy and kinetic energy. A better term for rest mass is invariant mass, because that makes clear that the mass does not change with a change of reference frame. The only way to change the rest mass is in a nuclear reaction where the energy of the rest mass is converted to kinetic energy.

There are a couple of reasons that the concept of relativistic mass was rejected. One, is that despite an apparent increase in relativistic mass an object does not increase its gravitational field and collapse to black hole at very high relativistic speeds. In the universe somewhere there is a probably a particle going at 0.9999c relative to you and from the point of view of that particle you have the mass of a black hole, but if you are still reading this then you have not turned into a black hole. Nor does an object increase in its nuclear energy just because it has relative energy. As ZZ showed, the kinetic energy profit realised by the decaying parent particle is 200 in either frame of reference. (Nice demonstration by the way, ZZ.) Another unpleasant aspect of the concept of relativistic mass is that the effort to accelerate a particle parallel to its relative velocity is different from the effort to accelerate a particle transverse to its initial relative velocity. This meant having two versions of relativistic mass. One is its parallel relativistic mass and the other is its transverse relativistic mass. For all these reasons, the concept was rejected and we simply accept that the equations for momentum and acceleration of a mass in relativity is different from the equations in Newtonian physics.
What the observer in the S' frame sees that is different is the momentum, kinetic energy and total energy. In frame S the total energy (E) is 1000 and in frame S' the total energy is 1667. Note that the total energy is the same before and after the decay of the parent in either FoR. In both cases the total energy (which replaces the concept of relativistic mass) is the sum of the rest mass energy and the kinetic energy. In frame S the initial kinetic energy is zero and the final kinetic energy is 200 and in frame S' the initial kinetic energy is 667 and the final kinetic energy is 867 so the gain in kinetic energy is +200 in both cases.

So wherever you see relativistic mass mentioned, just think old fashioned or popularist beginner treatment. The mass simply does not change in a way that increases its gravitational influence or potential nuclear power output due to relative motion. There is also the positive aspect that in a universe where nearly everything just seems to be a point of view that there are some things like rest mass (invariant mass) that do not change. It certainly makes analysis a lot easier.

14. Feb 18, 2010

yuiop

You are right. An observer can not measure length contraction in their own FoR for the reason you give.
Right again.
As mentioned in the last post the mass does not increase, but I also mentioned that the equations of momentum and acceleration are different in relativity from Newtonian mechanics. Let us say that an observer in S applies a force of 1 Newton to a kilogram mass and it accelerates by 1 meter per second per second. Now he observes someone in frame S' apply a force of one Newton (as measured in S') to a mass of one kilogram (also in S') and the mass accelerates by less than one meter per second per second as measured in S. The observer in S' on the other hand, also measures the mass as accelerating at one meter per second per second because the time dilation of the clocks in S' makes that observer measure the acceleration to be faster than the acceleration measured by S. The end result is that changes in inertia can not be measured in your own FoR for similar reasons to why changes in length can not be measured in your own FoR. In fact you can not measure any relativistic changes in your own FoR and if anyone ever does, the Theory of Relativty would be invalidated.
The width perpendicular to the direction of motion does not increase (or decrease). The mass does not change. The parallel length does decrease as does the volume and in some sense the density does increase. What you have to think how this increase in density would manifest itself and how an observer would try to measure it. In the case of a contained volume of gas, there are other changes that ensure the pressure measured in the FoR at rest with gas is always what you would expect. You could measure density by floating objects in a liquid, but that gets complicated because you need a gravitational field or a centrifuge that that takes you beyond Special Relativity. When you do find a method, you will not detect any relativistic changes in the density, in the FoR at rest with the material.
Some people think of relativity that way, but as you acknowledge......
What you call "distorted observations" are brought about by real changes in the measuring devices such as clocks and rulers. It is not simply a problem of simultaneity or signal delay times. It is not possible to construct a theory that denies any change in ruler lengths or clock rates that agrees with actual experiments.

15. Feb 18, 2010

mogra

Thank you for this. It is very clear. You have a real gift for teaching. I hope you are
using it. See other points below:

16. Feb 18, 2010

mogra

Sorry Kev, i neglected to bold my own conclusion at the end of your quote. i.e. i said:

i.e. the number of molecules remains invariant in any FoR but time dilation effects changes to observations of energy and distance for an observer in an inertial FoR?

17. Feb 18, 2010

Staff: Mentor

But in fact, relativistic equations don't look the same as classical ones, just with "relativistic mass" substituted for "classical mass." You can do that trick only with the momentum equation. It doesn't work for kinetic energy, for example.

18. Feb 18, 2010

Staff: Mentor

I just realized that people might interpret the very name "rest mass" as indicating that you're only allowed to use it if the object is at rest. In fact, its importance comes from the fact that you can calculate it from the energy and momentum via $E^2 = (pc)^2 + (m_0 c^2)^2$, and always get the same result for a particular object, regardless of which inertial reference frame you measure E and p in.

In particle physics experiments at places like CERN or Fermilab, one technique for identifying particles is to send them through detectors that measure energy and momentum separately, and then calculate

$$m_0 c^2 = \sqrt {E^2 - (pc)^2}$$

This is why physicists usually call $m_0$ the "invariant mass" when they need to distinguish it from other "kinds" of mass. Actually, they usually call it just "mass" because it's the default kind of mass that they talk about. If you ask a particle physicist, "what's the mass of that electron?" he'll answer "0.511 MeV/c^2" regardless of whether it's at rest or in a 50-GeV storage ring.

I say "usually" because "relativistic mass" is in fact customarily used in at least one subfield of physics (accelerator-design physics). When I was in graduate school in experimental high-energy particle physics thirty years ago, the only time I saw anybody talk or write about "relativistic mass" (outside of pop-science books or old introductory textbooks) was in a course about accelerator design.

19. Feb 18, 2010

yuiop

Thanks for you kind words, but I still have a LOT to learn from the other members on this forum. My new comments in blue below:

Last edited: Feb 18, 2010
20. Feb 18, 2010

Matterwave

I suppose so. However, since Newton's second law in the form: $$F=\frac{dp}{dt}$$ is correct for relativistic physics, one may look at that equation as $$F=\frac{d}{dt}(m_{rel}v)$$ in which case, you could then tell the class "Newton's equation works, but the mass just changes as you go faster!".

Anyways, this is not a big deal. The main point is that the relativistic mass is a concept which may raise confusion in people thinking "where's the matter come from?" when in fact the object hasn't grown more molecules or anything, it's just gotten more energy.

21. Mar 2, 2010

stevmg

Somebody - without using the
so-called relativistic mass equation (m = m_0/SQRT[1 - v^2/c^2]
PROVE the relativistic mass equation without being tautological. That equation was not a gift from God.

It is this relativistic mass equation which is the basis for the derivation of E=mc^2 and therefore one cannot use E=mc^2 as a basis either.

You have to start from scratch.