PDA

View Full Version : Seen any equation similar to this?

Singularitarian
15th August 2009, 05:02 AM
In the standard equation which was used to derive the many Planckian relationships given as M^2=\hbar(c/G), i wondered if anyone had ever seen a similar equation to the following:

v^2 (\hbar c/G)= E^2/c^2 - p [1]

Where the right hand side can be interpreted to be quite small, a quantization effort to gain the mass squared. In the appearance of velocity squared, we have the same dimensionality as what is found from derivations of the 4-momentum relativistically-invariant relationship describing the length of a vector on the left hand side.

Normally, one replaces the energy and the momentum on the left hand side for i \hbar(\partial / \partial t) and the momentum operator -i \hbar \nabla. But without inducing this, has this equation [1] in this form ever been used to anyone knowledge here?

sol invictus
15th August 2009, 06:53 AM
As written, it's not even dimensionally consistent. Perhaps you meant p^2. But even if so, the equation is wrong: the right side is a Lorentz invariant, and the left side is not.

I Ratant
15th August 2009, 07:01 AM
Another note from the fringe wondering why 1+1=Rabbit.

Monketey Ghost
15th August 2009, 07:04 AM
As written, it's not even dimensionally consistent. Perhaps you meant p^2. But even if so, the equation is wrong: the right side is a Lorentz invariant, and the left side is not.

So, no? :D

Perpetual Student
15th August 2009, 09:34 AM
:rolleyes:

"...not dimensionally consistent."

"...the right side is a Lorentz invariant, and the left side is not."

:rolleyes:

Singularitarian
15th August 2009, 09:34 AM
As written, it's not even dimensionally consistent. Perhaps you meant p^2. But even if so, the equation is wrong: the right side is a Lorentz invariant, and the left side is not.

For M^2, we have \hbar c/G. v^2 on to this expression yields the value M^2v^2. According to a wiki source, (E/c)^2 - p^2 = M^2v^2. So i thought they where dimensionally consistent?

Source:

In the chapter ''Comparison with the Schrödinger equation''

http://en.wikipedia.org/wiki/Dirac_equation

edit; yes it was supposed to be p^2

Singularitarian
15th August 2009, 09:39 AM
And to keep invariance true, its easy to apply gamma into the left hand side.

sol invictus
15th August 2009, 09:41 AM
P^2

Multiplying by gamma doesn't fix the problem.

Singularitarian
15th August 2009, 09:51 AM
Yes, i had already edited my post just above admitting to accidently leaving out the sqaured factor.

Though i am curious, as to why the equation doesn't work?

It about as abitrary as Einsteins equations for light which must be lorentz invariant Mc - 2\Delta P= (M- E/c^2)v?

ben m
15th August 2009, 10:03 AM
According to a wiki source, (E/c)^2 - p^2 = M^2v^2.

That's nonsense; what Wiki did you get this from?

ben m
15th August 2009, 10:05 AM
Yes, i had already edited my post just above admitting to accidently leaving out the sqaured factor.

You've "forgotten a squared factor" (or two) in basically every math-containing post you've made. Please proofread before you post. You want to know a good way to find these mistakes? Check your units.

Singularitarian
15th August 2009, 10:27 AM
That's nonsense; what Wiki did you get this from?

I thought that too, believe me or not.

http://en.wikipedia.org/wiki/Dirac_equation

ben m
15th August 2009, 10:53 AM
I thought that too, believe me or not.

http://en.wikipedia.org/wiki/Dirac_equation

That says "m^2c^2", not "m^2v^2". If it ever said the latter, it was wrong.

Please note that in this case "c" is not standing in for the velocity of some particle, it's standing for a universal constant whose value is 2.99e8 m/s.

~enigma~
15th August 2009, 10:58 AM
That says "m^2c^2", not "m^2v^2". If it ever said the latter, it was wrong.

Please note that in this case "c" is not standing in for the velocity of some particle, it's standing for a universal constant whose value is 2.99e8 m/s.
Your saying once again our resident troll is wrong?

Monketey Ghost
15th August 2009, 11:01 AM
Your saying once again our resident troll is wrong?

He's never wrong, go read some books. I assure you, he knows what he's talking about.

~enigma~
15th August 2009, 11:04 AM
He's never wrong, go read some books. I assure you, he knows what he's talking about.
Read some books? You mean Singularitarian is correct although he once again omitted a relevant term? Maybe you need to read a few books...

Monketey Ghost
15th August 2009, 11:06 AM
Read some books? You mean Singularitarian is correct although he once again omitted a relevant term? Maybe you need to read a few books...

;)

Singularitarian
15th August 2009, 11:08 AM
That says "m^2c^2", not "m^2v^2". If it ever said the latter, it was wrong.

Please note that in this case "c" is not standing in for the velocity of some particle, it's standing for a universal constant whose value is 2.99e8 m/s.

I know. I was setting it for the special case of v=c, i thought that would have been obvious. For instance, as is found:

pc^2=Ev

reference:

http://en.wikipedia.org/wiki/Special_relativity

ben m
15th August 2009, 11:19 AM
pc^2=Ev

That equation is always true, not just in the special case v=c. Note that the true equation contains "v" on the right and a "c" on the left. They are different variables and they mean different things; "v" is the velocity of the particle whose energy is E and momentum is p. "c" is a constant.

E^2 = m^2c^4 + p^2c^2 is also true. It does not contain v. "c" is a constant. You cannot substitute "v" in and get a true equation. If you have a particle with v=c ... well, picking this case and doing the substitution anyway is stupid; you didn't specify, and it was "not obvious"; your inclusion of M suggests that you didn't mean to specify that at all, in fact.

Singularitarian
15th August 2009, 11:36 AM
That equation is always true, not just in the special case v=c. Note that the true equation contains "v" on the right and a "c" on the left. They are different variables and they mean different things; "v" is the velocity of the particle whose energy is E and momentum is p. "c" is a constant.

E^2 = m^2c^4 + p^2c^2 is also true. It does not contain v. "c" is a constant. You cannot substitute "v" in and get a true equation. If you have a particle with v=c ... well, picking this case and doing the substitution anyway is stupid; you didn't specify, and it was "not obvious"; your inclusion of M suggests that you didn't mean to specify that at all, in fact.

Well, that is what is implied.

Also, i usually use M_0 to represent rest mass.

ben m
15th August 2009, 11:41 AM
Well, that is what is implied.

That is what you claim you meant to imply. Fine, whatever. Let's rewrite your original post, then:

i wondered if anyone had ever seen a similar equation to the following:

v^2 (\hbar v/G)= E^2/v^2 - p^2 = E^2/c^2 - p^2 = 0

You see, Sing, if v=c, then E^2/c^2 = p^2 and the difference is zero.

The "m" in the famous E^2 = m^2 + p^2 equation is, in fact, the rest mass.

Raze
15th August 2009, 04:20 PM
Don't know whether to laugh or to sigh.

Wowbagger
15th August 2009, 04:31 PM
Has anyone seen any equations similar to this?

a = b = 1
a2 = ab
a2 - b2 = ab - b2
(a - b)(a + b) = b(a - b)
a + b = b
a = 0
1 = 0

Raze
15th August 2009, 04:55 PM
Ah, that's a classic, lol.

Has anyone seen any equations similar to this?

a = b = 1
a2 = ab
a2 - b2 = ab - b2

At this step you have 0 = 0

So it's done. You can't go further.

(a - b)(a + b) = b(a - b)
a + b = b
a = 0
1 = 0

ben m
15th August 2009, 06:41 PM
Though i am curious, as to why the equation doesn't work?

It about as abitrary as Einsteins equations for light which must be lorentz invariant Mc - 2\Delta P= (M- E/c^2)v?

Why doesn't the equation work? I have a guess.

You are making specific type of mistake that is very, very common in intro physics classes; we call it "equation mining". You're looking up random equations on Web pages; you notice that you can find two equations, both containing (for example) an "m" term, and you figure you solve either equation for "m" and substitute it into the other. You do this with very little understanding of what the equation represents and what its terms are.

For example, M=sqrt(hc/G) isn't some general property of masses. The terms on the right are constants, so this is simply calculating some particular mass; it's the definition of the Planck mass.

What does, for example, E^2=m^2c^4+p^c^2 mean? It means that if you are considering a particle with rest mass m and momentum p, E is is its total energy. So when you "substitute" sqrt(hc/G) for m, you are just stating "if you have a particle whose rest mass is the Planck mass and whose momentum is p, its energy is E". Nothing universal, nothing generally interesting. Want to keep plugging? Go ahead, there are other true equations you can apply to "a particle whose rest mass is the Planck mass", if you want to ... but not all equations so apply. v=c, an interesting special case in other circumstances, does not apply here for example. E=mc^2 does not apply, since that's a different E than the one in the E^2 = m^2 + p^2 equation. And so on.

Singularitarian
16th August 2009, 04:16 PM
Well yeh, but i never surfed the net intentionally as you are making out. I was already aware of the relationship GM^2=\hbar c.

It wasn't until investigating the length of a vector did i come across a dimesional relationship: The equation in the OP was just an accident of other equations i was writing, so it had nothing to do with me generally surfing the net.

Also, you must remember ben, the form you have given above is indeed a planck mass value, however, i never intend to create this value when in the original equation in the post. Under the special condition of v=c, the equations dimensions are perfectly fine. As i said, when i define rest mass, i write M_0 usually.

(And please, ben, don't talk to me about the energy-momentum equation as if i am some idiot. I know fine well what it is all about, probably more than you).

Wangler
16th August 2009, 04:25 PM
You want to know a good way to find these mistakes? Check your units.

I had a great undergrad physics professor, who taught me that valuable problem solving skill.

He would always have us work with the variables, never numbers ("they get in the way")...solve the problem, check units....units o.k.? answer seems reasonable? you are likely right, then. plug in numbers to your hearts content.

ben m
16th August 2009, 04:31 PM
Also, you must remember ben, the form you have given above is indeed a planck mass value, however, i never intend to create this value when in the original equation in the post. Under the special condition of v=c, the equations dimensions are perfectly fine. As i said, when i define rest mass, i write M_0 usually.

Two questions, then: since v=c only for a massless particle, why did you write the energy-momentum equation for a particle of mass M_Planck? Second: why did you use the variable v at all?

(And please, ben, don't talk to me about the energy-momentum equation as if i am some idiot. I know fine well what it is all about, probably more than you).

I know! Why don't you post a relativity quiz (http://forums.randi.org/showthread.php?t=148751)? Oh, wait ...

ben m
16th August 2009, 04:39 PM
Under the special condition of v=c, the equations dimensions are perfectly fine.

No, after you fixed the p^2 the dimensions were fine. v has the same units/dimensions as c.

v != c doesn't break the unit consistency; as Sol pointed out, it breaks the Lorentz invariance of the left-hand side of your equation.

As i said, when i define rest mass, i write M_0 usually.

First: bad idea; no one else in physics does this. The rest mass is m. If you want to write the "relativistic mass" (which is rarely used anyway) just use E/c^2.

Second: well, you took the thing that you called "M" and you plugged it into the energy-momentum equation in a place that called for a rest mass. If you did not intend M to be the rest mass, you should not have plugged it into the "M" of M^2 = E^2 - p^2.

Singularitarian
16th August 2009, 04:58 PM
Well, i missed the sqaured momentum by accident. But it was really what i meant originally.

And i don't know about the last. I've had lecturers telling me not to use \gamma M=m.

Singularitarian
16th August 2009, 05:00 PM
I had a great undergrad physics professor, who taught me that valuable problem solving skill.

He would always have us work with the variables, never numbers ("they get in the way")...solve the problem, check units....units o.k.? answer seems reasonable? you are likely right, then. plug in numbers to your hearts content.

I have to admit, yes, i never checked it over when i wrote it out for any errors, but to be honest, i know my units.

ben m
16th August 2009, 05:42 PM
And i don't know about the last. I've had lecturers telling me not to use \gamma M=m.

Yep---nobody calls (gamma x rest mass) the "mass" anymore. You'd never guess from reading pop-science accounts of Special Relativity, which frequently report the long-dead early-20th-century usage.

Anyway, I am not asking you to use "gamma M = m" (where M=rest mass and m="relativistic mass"). I'm asking: you plugged some sort of mass into the rest mass slot in M^2 = E^2 + p^2, then specified that v=c which implies zero rest mass. Why?

ben m
16th August 2009, 09:28 PM
M^2 = E^2 + p^2

(Typo alert: I meant M^2 = E^2 - p^2)

Singularitarian
20th August 2009, 07:48 PM
I refer you to the Springer Link journal paper, from the Rutherford Observatory, Columbia University, New York, N. Y by L. Motz, published under the Italian
Physical Society, and the summery was thus described;

''Summary In a previous paper (referred to as I in the text) it was shown that the Weyl principle of gauge invariance leads to the relationship
Gm^2 =\hbar c for a particle of inertial mass M obeying the Dirac equation, where G is the Newtonian gravitational constant. Instead of interpreting
this equation to mean that G takes on the extremely large value \hbar c/m^2 inside a particle like an electron (as we did in I), we now write it
in the form Gm^2/c = \hbar and treat it as a quantization condition on the square of the gravitational charge \sqrt{Gm}. We show that
this same quantization condition can be obtained from an angular-momentum component in the general-relativistic two-body problem as well as from the
Machian definition of inertial mass in a rotating universe by using the Dirac-Schwinger procedure for quantizing charge. From this quantization condition
we now deduce that the fundamental particle in Nature (the uniton) has an inertial mass equal to about 10^{-5}g. The possibility of using the uniton to shed
light on the mystery of the « missing mass » in the Universe is discussed. Other cosmological implications of the uniton are also discussed and it is
suggested that unitons can clear up the solar-neutrino discrepancy.''

Now, i know for a fact that Dirac's equation is relativistically invariant, so it seems now that the union of the two can be expressed relativistically, since the above is referring to the equation that obeys the Dirac Equation itself.

Reality Check
20th August 2009, 08:52 PM
I refer you to the Springer Link journal paper, from the Rutherford Observatory, Columbia University, New York, N. Y by L. Motz, published under the Italian
Physical Society, and the summery was thus described;

''Summary In a previous paper (referred to as I in the text) it was shown that the Weyl principle of gauge invariance leads to the relationship
http://www.randi.org/latexrender/latex.php?Gm^2 =\hbar c for a particle of inertial mass http://www.randi.org/latexrender/latex.php?M obeying the Dirac equation, where G is the Newtonian gravitational constant. Instead of interpreting
this equation to mean that http://www.randi.org/latexrender/latex.php?G takes on the extremely large value http://www.randi.org/latexrender/latex.php?\hbar c/m^2 inside a particle like an electron (as we did in I), we now write it
in the form http://www.randi.org/latexrender/latex.php?Gm^2/c = \hbar and treat it as a quantization condition on the square of the gravitational charge http://www.randi.org/latexrender/latex.php?\sqrt{Gm}. We show that
this same quantization condition can be obtained from an angular-momentum component in the general-relativistic two-body problem as well as from the
Machian definition of inertial mass in a rotating universe by using the Dirac-Schwinger procedure for quantizing charge. From this quantization condition
we now deduce that the fundamental particle in Nature (the uniton) has an inertial mass equal to about http://www.randi.org/latexrender/latex.php?10^{-5}g. The possibility of using the uniton to shed
light on the mystery of the « missing mass » in the Universe is discussed. Other cosmological implications of the uniton are also discussed and it is
suggested that unitons can clear up the solar-neutrino discrepancy.''

Now, i know for a fact that Dirac's equation is relativistically invariant, so it seems now that the union of the two can be expressed relativistically, since the above is referring to the equation that obeys the Dirac Equation itself.
It is the particle that is obeying Dirac's equation (emphasis added).

ben m
20th August 2009, 09:17 PM
Yes, sqrt(hc/G) is an allowable rest mass for a particle. (So is any other value)

Yes, a particle of any rest mass will obey the relativistic energy-momentum equation. (The relativistic energy-momentum equation, E^2 = m^2 + p^2, is not the same thing as the Dirac equation; do you have them mixed up? They do both occur in the same Wikipedia article. In any case, if your particle is a fermion it may also obey the Dirac equation.)

No, such a particle cannot travel at v=c as you (eventually) specified.

What in Motz's paper do you think contradicts that statement?

Singularitarian
21st August 2009, 09:43 AM
I had a few vodkas last night, and when i read it, it said something else.

Sorry.

Singularitarian
21st August 2009, 09:44 AM
If it where a fermion though, i better start doing some math to try and prove that.

Singularitarian
21st August 2009, 09:46 AM
For instance, you can derive not only the Dirac equation from above if it where a fermion, but you could also derive the schrodinger equation.

Olowkow
21st August 2009, 07:52 PM
If we were where we were wired, where we're wary of wearing wiry war ware we wore at the whirring weir, why were warring wair wielding weary werewolves where we're weird?