Relativity and Cosmology

Jorrie, as far as I know, mass is a constant thing, while weight depends on the gravity in your vicinity. Surely, my ideal weight will be different on Mars than on Earth. What puzzles me is that it may alter my calculated mass - I'm confused!

SL

Hi SL, go and calculate your ideal mass in the two scenarios (assume that you did not grow in length) and compare! You should find that your ideal mass remains constant, but what about your rest energy?

Regards, Jorrie

Jorrie, yes I've done that (calculating mass on earth and on Mars) and I agree that rest mass is invariant. Turning to energy, your equation 6.7 on page 94 of Relativity 4 Engineers indicates that total energy depends on the mass, the gravitational field (g_tt, g_rr) and the velocity components v_r and v_t.

Now, since we are talking rest mass here, we can ignore the velocity components, but one cannot ignore the gravity components, right? If so, my mass-energy (mc²) must be more on Mars than here on earth, since I'll be farther out of the potential well of the sun.

Before I attempt these calculations, I would like to know if I'm on the right track.

SL

Hi SL, you asked: "Now, since we are talking rest mass here, we can ignore the velocity components, but one cannot ignore the gravity components, right?"

I you want to include the gravitational (g_tt, g_rr) components, you have to choose a reference frame, say Schwarzschild coordinates with the Sun permanently at rest at the origin. Then you have to include the effects of the Sun, Earth, Mars and at least Jupiter in your calculations - all very messy!

Remember also that the equations given in Relativity 4 Engineers are for energy as measured by a distant static observer (this is what Schwarzschild coordinates are all about). Think a bit on what "rest mass-energy" means (look at this Wikipedia article).

Regards, Jorrie

PS: in the end, the answer is surprisingly simple!

???? e=mc^2 :. m=E/c^2 :. about 3*10^10 J of atomic bonds everywhere???? not familiar with GR

1. Case 1 :

Force or weight = Mass x Acceleration or Constant of Gravity

Work or Energy = (Force) x Distance = (Mass x Acceleration) x Distance

For the same Mass and same Distance, the Work will depends on Acceleration.

Due to the Constant of Gravity on Earth is differs from Constant of Gravity on Mars, so the rest of energy will be differ.

2. Case 2 (dynamic)

Energy = Mass x (Speed)^2

= Kg (mass) x (meter/second)^2

= Kg (mass) x (meter/second^2) x meter

= Kg (mass) x Acceleration x meter ------- (1)

From equation (1), it is obvious that the Energy depends on Acceleration, and we know that acceleration on Earth differs from acceleration on Mars due to gravity.

So, the Energy is different.

Hi Abdel, you said

"Energy = Kg (mass) x Acceleration x meter ------- (1)

From equation (1), it is obvious that the Energy depends on Acceleration, and we know that acceleration on Earth differs from acceleration on Mars due to gravity."

This question is about "rest energy", meaning "energy when you are at rest". So where does velocity and acceleration come in?

SL

the acceleration here would means the gravitational acceleration or more commonly known as "g". Your weight is what pulls you to the center of earth. so "g", the acceleration here is what makes you fall fassster. Uncle "g" differs with the length between you and the center of the heavenly mass. So not just on mars, the "rest" energy at the poles will be different than when you are at the equator

Hi, the other Guest, after carefully considering what Jorrie said before, I think you are wrong. Your "ideal rest mass" does not change, wherever you may go. Therefore I think that your "ideal rest energy" will also be the same everywhere.

SL

My wife sez my jules is big maybe 1/4 lb....

Seriously though, I believe your rest energy would remain the same, as would your mass....but I'm just guessing...

Hi Guest, you wrote: "Seriously though, I believe your rest energy would remain the same, as would your mass....but I'm just guessing..."

You and guest SL (if you are in fact different people?) are on the right track, so who will give the first correct answer?

Regards, Jorrie

PS: Appreciate your wife...

guest with jules... 6.8488e+18 joules

My jules in joules...

1.0192e+16

Hi Guest and Guest, your answers differ by a factor ~600! This humanly impossible!

Do you care to show us your calculation so that we can decide on the reason for this discrepancy?

Regards, Jorrie

I believe the difference could be possible, but not at the stated quantities...

If you will read carefully, the first post is referencing a mass of 168lbs. The second is referencing the .25lb weight of the family jewels...These figures were obtained by merely multipling the mass quantity, times the speed of light squared...

c² = 89,875,517,800,000,000 mtr²/sec² which rounds to
9 x 10¹⁶ mtr²/sec².
So, the energy equals
168lbs times 0.4536 to covert to kg= 76.2048 (mass) x 9 x 10¹⁶ mtr²/sec² (c²),
which equals 685.8432 x 10¹⁶kg • mtr²/sec²
which equals 6.8488e+18 joules or 1.902444444e+15 watt hours

or 6,491,409,897,416,502 Btu..... or the equivalent of 52 bil gal of gasoline, about 1/3 the amount used in the US annually....or 884 billion lbs of switchgrass, about 30 mil acres....

Of course the gasoline and switchgrass equivalents, are that energy which is available chemically...Nuclear energy is limited to uranium at present, and even that has to be processed...That is why nuclear energy is so productive compared to other man made sources....Nuclear power plants are said to be some 60 times over unity, and even their efficiency, theoretically speaking, is quite low...Hydroelectric is the leading method for efficient production of electricity because of gravity...but they are disruptive to the environment...

Hi Guest, very correctly and very elaborately answered - congrats!

The official answer, elaborating on the reasons for the simplicity of the result will be posted over this weekend as an edit of the question post.

Regards, Jorrie

PS: I will post a real "over-unity", whatever that may mean, energy converter design in about a week from now - still working on the design documentation...

I'm humbled by my greatness....(applause)

SolarEagle