Is Relativistic Mass Plausible?

[JANm]

Sure, you could do all of that, but why bother? All refeence frames are equivalent so you can use any that is convenient.

Hello DaleSpam
Here you relativate (=psychological indifference to stimuli smaller then threshold) the very fundament of relativity-theory.
In between Michelson with his experiment and Penzias and Wilson with their backgroundradiation: the relativity theory can survive with: benifit of the doubt. In common terms one does not know the absolute velocity of the Earth through space.
Since Penzias and Wilson we and for that matter all creatures in all of space can find an unique inertial system. For everybody the same, everywhere the same, radiation of 3 Kelvin has to be isotropic in all space. Astronomers have found that our particular velocity is 370 km/sec to the sign Leo. That is now a physical fact.
Why bother, while we have a theory which states it doesn't matter?
greetings Janm

 

[Dale]

In common terms one does not know the absolute velocity of the Earth through space.

In more precise terms there is no such thing as "absolute velocity" in any experimental sense.

Astronomers have found that our particular velocity is 370 km/sec to the sign Leo. That is now a physical fact.

Certainly it is a fact. Geographers have similarly found that the velocity of the Nile is 4 knots to the north. That is also a physical fact, with equal physical significance. If you wish you may also construct a reference frame where the Nile is at rest and compute all of your physics (including relativistic mass) wrt that reference frame.

Again, why bother? What prediction do we get correct by doing it your way that we get wrong by doing it the easy way?

 

[vitruvianman]

I've read contrasting sources concerning the concept of an object increasing in mass at relativistic velocities. Some of my older calculus texts mention this as being accepted by physicists, while a newer(by comparison) textbook called Principles of Physics: A Calculus-Based Text by Serway and Jewett claims that relativistic mass is outdated. Normally, the newer book would be correct, but I don't think there is any other reasonable explanation of the momentum of photons. Can the good people at PF please shed some light on this matter?

I think it's never been outdated and will never be. Relativistic mass is calculated so:

for example, kinetic enerji can be calculated with KE=mc^2-m_0c^2
it's also KE=mc^2-\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}c^2=mc^2-\frac{m_0c}{\sqrt{c^2-v^2}} and if we add momentum to this equ., it would be E^2=m_0^2c^4+p^2c^2=E=\sqrt{m_0^2c^4+p^2c^2} where momentum isn't relativistic. Remeber that relativistic p is p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}

after we make momentum relativistic in KE equ., we'll find out the relativistic value of energy, which includes p. It's so: E=\frac{m_0^2c^3}{\sqrt{c^2v^2}}+\frac{p^2cm_0}{\sqrt{c^2-v^2}}=\frac{m_0c(m_0c^2+p^2}{\sqrt{c^2-v^2}}

------------

we can also say for pc when E=\sqrt{m_0^2c^4+p^2c^2} pc is pc={\sqrt{p_0^2-m_0^2c^4}} and so, we can say p^2c^2=\frac{m_0^2v^2c^2}{1-\frac{v^2}{c^2}}=\frac{m_0^2\frac{v^2}{c^2}c^4}{1-\frac{v^2}{c^2}}

after a quick calculation you'll find p^2c^2=\frac{m_0^2c^4[\frac{v^2}{c^2}-1]}{1-\frac{v^2}{c^2}}+\frac{m_0^2c^4}{1-\frac{v^2}{c^2}}=-m_0^2c^4+m^2c^4=(mc^2)^2\Rightarrow E=pc

You can find the basic ideas of this calculations (made by me :) ) at the notes part of the book "Relativity: The Special and General Theory" by Albert Einstein.

If you see the equations as text, you can see them as pic.s by copy-pasting them onto http://www.sitmo.com/latex/

 

[Dale]

if we add momentum to this equ., it would be E^2=m_0^2c^4+p^2c^2=E=\sqrt{m_0^2c^4+p^2c^2} where momentum isn't relativistic. Remeber that relativistic p is p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}

Careful here. Your p is relativistic here. Remember m_0 and c are constant and E is unbounded so p must also be unbounded.

 

[vitruvianman]

Careful here. Your p is relativistic here. Remember m_0 and c are constant and E is unbounded so p must also be unbounded.

Well you're right I think I've gotten confused here. I made everything relativistic even it's already relativistic :)

But the relativistic value of energy calculated in my first part of calculations (before ---------), I mean E=\sqrt{m_0^2c^4+p^2c^2} is right I think. (am I right?)

After all, my primary aim was to show that E=pc is also right as the relativistic energy of photon.

And as you say p is unbounded because of the same feature of E.

 

[Dale]

Yes, that is correct, and it is equal to the expression I wrote above for the rest mass as the norm of the four-momentum.

 

[JANm]

Yes, that is correct, and it is equal to the expression I wrote above for the rest mass as the norm of the four-momentum.

Hello DaleSpam
The term norm is defined in linear analysis. So we have to call that mathematical norm.
The norm you mention is really different, so let us call that physical norm and that means the diagonal trace of a Minkowski matrix. As I pointed out a mathematical norm cannot have these minus signs because the thing could become zero in a case of a nonzero situation.
Norm is a partial ordening, so things can have the same norm while being different, but there is only one thing with norm zero and that is the zero element.

So one uses if you have || a-b||= 0 one may conclude a=b. With your physical norm this essential norm thesis is not true.
greetings Janm

 

[Dale]

What you say is all correct, the Minkowski norm is not really a norm at all. It is technically "a nondegenerate, symmetric, bilinear form with the signature (+,-,-,-) defined on a four-dimensional real space". However, that is rather cumbersome to write so "Minkowski norm" is the conventional shorthand.

In any case, whether you call it "Minkowski norm" or "nondegenerate, symmetric, bilinear form with signature (+,-,-,-) defined on a four-dimensional real space" does not substantially alter anything I wrote above.

 

[JANm]

Hello DaleSpam
The things I wrote will not be altered either. I found an answer To shakespeare and Einstein at the same time:

Of the many things relative: they are; and all the others they aren't.

As I explained within relativity theory restmass cannot be defined. It involves mass and velocity and especially the special case that velocity is zero. A mass of an individual object needs velocity to calculate m = gamma * m_0.

The orientation to the CMB would bring joy to Michelson and actual re honouring to Galileo Galilei. I might even say that if Lorentz and Einstein knew of the possibility to orientate to cosmic background radiation and know the velocity of the Earth through space, they would not have launched this cumbersome theory in the first place!

So hurray for Michelson and Hurray for Galileo Galilei and a little minus point for Lorentz and a little minus point for Einstein; let us say they did not know.

If relativity principle were true then the 3 kelvin radiation would be isotropic for every inertial system and by using the term of A.A. Robb; what can perfect translation of everything add to reality. Suppose you have everything in an encyclopedia in Greek language and you translate that perfectly to Latin, then the things sayd are exactly the same; so what use is there in translating and what possible effect can this translating have on the described physics in the firstplace: none.

Greetings Janm

 

[edpell]

... within relativity theory restmass cannot be defined.
... If relativity principle were true then the 3 kelvin radiation would be isotropic for every inertial system ...

Jan I do not understand either of these statements.

1) rest mass is the mass measured by one moving with the object.

2) why?

 

[Dale]

As I explained within relativity theory restmass cannot be defined.

Since I defined it just a couple of posts ago this comment seems disingenuous at best.

If relativity principle were true then the 3 kelvin radiation would be isotropic for every inertial system

That doesn't make any sense at all. That is like saying that Mt. Everest should have the same velocity in all reference frames. The CMBR is not a law of nature.

 

[vitruvianman]

The CMBR is not a law of nature.

Right!

 

[vitruvianman]

A mass of an individual object needs velocity to calculate m = gamma * m_0.

But not everytime. Let v be 0, then m=m_0 so m\geq m_0

 

[vitruvianman]

JANm how about we use \gamma = \frac{1}{\sqrt[]{1-\frac{v^2}{c^2}}}}
and say m = \gamma m_0 = m_0 + (\gamma -1)m_0 some people like to break it into these two parts the rest mass and the kinetic energy I think it is simpler to leave it as \gamma m_0.

Yes it's much simpler but not that effective. I mean you can't calculate what you want without using the original forms of equations.
For example the E^2=m_0^2c^4+p^2c^2 not only the famous E=mc^2 . Most of the time we "gotta" calculate momentum too... Simplifying doesn't work all the time. A quote from Einstein:

"Keep it simple, stupid — but never oversimplify."