Is Relativistic Mass Plausible?

  • Thread starter planck42
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  • #51
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Huh? Rest mass is easy to explain. It is simply the norm of the four-momentum. All reference frames agree on its value.
Hello DaleSpam
Thanks for this answer. Have to look into it. Do you also have an answer to the dt/dtau?
greetings Janm
 
  • #52
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I don't know what Taylor and Wheeler do with dt/dtau. Was that the question?
 
  • #53
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Rest mass is easy to explain. It is simply the norm of the four-momentum.
Hello DaleSpam
Momentum has to do with mass and velocity, while I expext that restmass has to do with mass and no velocity at all. So simple it is maybe for you; I have thought about what you wrote but I don't understand it at all...
greetings Janm
 
  • #54
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Sorry about the confusion, that is my fault. I was being sloppy with my factors of c (conceptually working in units where c=1). The four-momentum is:
[tex]\left(\frac{E}{c}, \, p_x, \, p_y, \, p_z\right)[/tex]
So it has units of momentum, but the timelike component is refered to as the total energy (even though it is actually total energy divided by c). The norm of this four-vector is:
[tex]m_0^2 c^2 = \frac{E^2}{c^2} - p_x^2 - p_y^2 - p_z^2[/tex]
So it also has units of momentum, but is often refered to as the rest mass or invariant mass (even though it is actually invariant mass times c).
 
  • #55
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Hello Rasalhague
The dt/dtau I find very interesting. Do Taylor & Wheeler put the time dilation factor in the formulae at this moment?
By [itex]\frac{\mathrm{d} t}{\mathrm{d} \tau}[/itex], Taylor & Wheeler denote the derivative of coordinate time with respect to proper time. This is the time dilation factor!

[tex]m_0c^2 \frac{\mathrm{d} t}{\mathrm{d} \tau} = \frac{m_0c^2}{1 - \left ( \frac{v}{c} \right )^2} = m_0c^2 \gamma = m_0c^2 \cosh \left ( \phi \right ) = E = T + m_0c^2,[/tex]

where [itex]E[/itex] is the relativistic total energy (which some people in the past have called "mass"), the time component of the energy-momentum 4-vector (sometimes called the momentum 4-vector), and [itex]m_0[/itex] is rest mass (which I think is usually considered nowadays more deserving of the name "mass" and often written simply [itex]m[/itex]).

Compare this with the space components of the energy-momentum 4-vector (the momentum part of it), which take the form:

[tex]m_0c \frac{\mathrm{d} x}{\mathrm{d} \tau} = m_0c \frac{\mathrm{d} x}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau} = m_0cv \gamma = m_0c \ \sinh \left ( \phi \right ),[/tex]

where [itex]x[/itex] stands for a representative direction in space.
 
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  • #56
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Hello planck42
So you are easily satisfied. I think that to say that mass-velocity relation is outdated is not a scientific statement. Science should be timeless. Theories are prooved true or false.
Theories are proven to be false only.
 
  • #57
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The norm of this four-vector is:
[tex]m_0^2 c^2 = \frac{E^2}{c^2} - p_x^2 - p_y^2 - p_z^2[/tex]
So it also has units of momentum, but is often refered to as the rest mass or invariant mass (even though it is actually invariant mass times c).
Hello DaleSpam
Where do these three minus signs come from? Shouldn't the norm be the thing with the plus signs? ||a+bi||=a^2+b^2 isn't it?
I found a formula for dynamic mass m=Sqrt(m_0+p^2/c^2)... I think this is the right relativistic formula. So the invariant mass you speak of is the mass with velocity and momentum; if the velocity is zero then p=0 and so dynamic mass m=m_0.
My proposal is use the inertial system orientated to the background, so the earth has velocity 370 km/sec. Things standing still in comparison to the background have velocity zero and dt/dtau is one. It is already known for years that dt/dtau in most common astronomical cases does not differ much from one.
Instead of using beta=sqrt(1-v^2/c^2) I am very fond of using the first order approximation: beta=1-v^2/(2*c^2).

So my mass velocity relation is m=m_0/(1-v^2/(2*c^2) and this mass velocity relation is realy very plausible...

greetings Janm
 
  • #58
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JANm how about we use [tex] \gamma = \frac{1}{\sqrt[]{1-\frac{v^2}{c^2}}}} [/tex]
and say [tex] m = \gamma m_0 = m_0 + (\gamma -1)m_0 [/tex] 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 [tex] \gamma m_0 [/tex].
 
  • #59
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Theories are proven to be false only.
I think there are three separate things experimental physics, theoretical physics and mathematical physics. I would agree that with the first two a theory can only be "not false" or "false" given our current available data and calculation techniques.

But I sure get the impress that mathematical physics would like to say they are correct by construction in the same way a math proof is correct by construction. Not sure if they are physics or math or some new in between ground.
 
  • #60
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Where do these three minus signs come from? Shouldn't the norm be the thing with the plus signs? ||a+bi||=a^2+b^2 isn't it?
The minus signs are from the usual Minkowski norm in relativity.

I found a formula for dynamic mass m=Sqrt(m_0+p^2/c^2)... I think this is the right relativistic formula. So the invariant mass you speak of is the mass with velocity and momentum; if the velocity is zero then p=0 and so dynamic mass m=m_0.
Your "dynamic mass" is more commonly known as "relativistic mass". It is equivalent to the total energy E (of course, divided by c^2 in order to make the units correct). There are many discussions here contrasting "relativistic mass" and "invariant mass".

My proposal is use the inertial system orientated to the background, so the earth has velocity 370 km/sec. Things standing still in comparison to the background have velocity zero and dt/dtau is one.
Sure, you could do all of that, but why bother? All refeence frames are equivalent so you can use any that is convenient.
 
  • #61
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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
 
  • #62
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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?
 
  • #63
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 [tex] KE=mc^2-m_0c^2 [/tex]
it's also [tex] 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}} [/tex] and if we add momentum to this equ., it would be [tex] E^2=m_0^2c^4+p^2c^2=E=\sqrt{m_0^2c^4+p^2c^2} [/tex] where momentum isn't relativistic. Remeber that relativistic p is [tex] p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]

after we make momentum relativistic in KE equ., we'll find out the relativistic value of energy, which includes p. It's so: [tex] 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}} [/tex]

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

we can also say for pc when [tex] E=\sqrt{m_0^2c^4+p^2c^2} [/tex] pc is [tex] pc={\sqrt{p_0^2-m_0^2c^4}} [/tex] and so, we can say [tex] 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}} [/tex]

after a quick calculation you'll find [tex] 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 [/tex]

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/
 
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  • #64
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if we add momentum to this equ., it would be [tex] E^2=m_0^2c^4+p^2c^2=E=\sqrt{m_0^2c^4+p^2c^2} [/tex] where momentum isn't relativistic. Remeber that relativistic p is [tex] p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}} [/tex]
Careful here. Your p is relativistic here. Remember m_0 and c are constant and E is unbounded so p must also be unbounded.
 
  • #65
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 [tex] E=\sqrt{m_0^2c^4+p^2c^2} [/tex] is right I think. (am I right?)

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

And as you say p is unbounded because of the same feature of E.
 
  • #66
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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.
 
  • #67
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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 realy 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
 
  • #68
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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.
 
  • #69
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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
 
  • #70
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... 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?
 
  • #71
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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.
 
  • #73
A mass of an individual object needs velocity to calculate m = gamma * m_0.
But not everytime. Let v be 0, then [itex]m=m_0[/itex] so [itex]m\geq m_0[/itex]
 
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  • #74
JANm how about we use [tex] \gamma = \frac{1}{\sqrt[]{1-\frac{v^2}{c^2}}}} [/tex]
and say [tex] m = \gamma m_0 = m_0 + (\gamma -1)m_0 [/tex] 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 [tex] \gamma m_0 [/tex].
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 [itex]E^2=m_0^2c^4+p^2c^2[/itex] not only the famous [itex]E=mc^2[/itex] . 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."
 

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