Simple Physic, Celerity traveling 101

[Splitsecond]

But doesnt the E=MC^2 still apply? then whatever non zero E is, according to m=E/c^2, the mass would still exist and be non zero, even if its increadibly small, is that not correct?

No. The mass of the photon is zero but the momentum isn't. The magnitude of the momentum of a photon is P=h/lamda. The energy of a mass-less particle is E=pc.

[thekillman]

And about the innertial dampeners, do they realy reduce mass? all i remember about them is "take innertia out of the equation"

no, they reduce the EFFECT of mass, not the mass itself as far as i know. essentially, instead of F=ma, Force applied=Mass times Acceleration, the formula changes to reduce the effect of mass.

so it becomes like F=(x.m).a. 1=x>0 x=1 is zero reduction, x>0 is reduction.

so the effect of mass becomes less, eventually to the point where it approaches F=a.

this means the force applied is (nearly) equal to the accelleration. when you are propelling a million kilograms, then you dont need a million kilograms of thrust to create 1m/s/s thrust, you'd get a million kilograms of thrust.

inertia is mass' resistance to acceleration. therefore, inertial reduction means the effect of mass becomes less

[Quadhelix]

No. The mass of the photon is zero but the momentum isn't. The magnitude of the momentum of a photon is P=h/lamda. The energy of a mass-less particle is E=pc.

Well, there are two things that a person can mean when they say "mass": they can mean "rest mass" or they can mean "relativistic mass." IIRC, photons have relativistic mass, but no rest mass.

E=1/2mv^2 'v is for speed from french:Vitesse, m for mass
E=ma*distance 'a is for acceleration

Well, the first equation is the equation for Kinetic Energy, and is actually only the low velocity approximation. The relativistic equation for kinetic energy is K=(1-1/Sqrt(1-v^2/c^2))m*c^2. When v is much smaller than c, that approximates to K=(1/2)m*v^2

The second equation actually describes work, which is a transfer of energy, and is properly written as W={F*ds where { is my stand-in for the integral symbol, F is the force vector exerted on a object, and ds is the instantaneous change in that object's position (also a vector). The reason that I make note of these differences from what you have said is that your second equation gives the impression that an object's energy is 0 if the object isn't accelerating. Another import that distinction is that it is only movement in the same direction as the force "contributes" to the work.

so the effect of mass becomes less, eventually to the point where it approaches F=a.

If F=a were ever to become a true statement, then you would either have to measure acceleration in units of force or force in units of acceleration. That would be like saying "the speed limit is 60 miles."

this means the force applied is (nearly) equal to the accelleration. when you are propelling a million kilograms, then you dont need a million kilograms of thrust to create 1m/s/s thrust, you'd get a million kilograms of thrust.

A kilogram is not a units of force, it is a unit of mass. The appropriate unit of force is newtons. One newton is equivalent to a single kilogram being accelerated at 1 m/(s^2).

[Exthalion]

Or to sum up everything that has been said, even with a zpm, some Ancient magic toy that works like a supped up inertial damper, and a puddle jumper you could not send a jumper through the receiving gate of a wormhole.

And as a corollary of the above, the writers will never make an episode where this happens.

[[email protected]]

No. The mass of the photon is zero but the momentum isn't. The magnitude of the momentum of a photon is P=h/lamda. The energy of a mass-less particle is E=pc.

Ok E = pc witch mean that energy is the product of the momentum(p) and the celerity(c) witch is the speed of light

But the momentum (p) equals the product of the mass(m) and speed(v)

so

p = mv ' v = c

p = mc

by the end of this factorisation you end up with:

E = mcc = pc = mc^2

so in case of photon this mean heat transfer(so acceleration) of atomic particule

In any way's when you factorize your Energy formula, you end up with a mass witch is very small, negligeable, but still present...

It's high scholl physic, go argue with my teacher...

[Exthalion]

Why are we arguing the viability of an impossible situation in a fictional universe?

Considering the fact that a hyperdrive exists the laws of physics have already been shot to hell. How will minutia about particle physics change the fact that the basic premise is flawed? You cannot get anything not already traveling at c to c without infinite energy.

Nothing is every going to trick a stargate into letting matter go through the receiving gate by moving fast. Plot device or hacking the receiving gate maybe. But this is petty.

Even if we all agreed on every point about definitions, masses, and particle speeds it still doesn't change the fact that the original story will never happen and unless the writers think the fans are idiots and change the principles of gate mechanics it couldn't even happen.

Can't we all find something more fruitful to expend talent, energy, and Google searches on?

[thekillman]

A kilogram is not a units of force, it is a unit of mass. The appropriate unit of force is newtons. One newton is equivalent to a single kilogram being accelerated at 1 m/(s^2).

my fault, i memorized POUNDS of thrust and thought the same applied to kilograms.

[Splitsecond]

Ok E = pc witch mean that energy is the product of the momentum(p) and the celerity(c) witch is the speed of light

But the momentum (p) equals the product of the mass(m) and speed(v)

so

p = mv ' v = c

p = mc

by the end of this factorisation you end up with:

E = mcc = pc = mc^2

so in case of photon this mean heat transfer(so acceleration) of atomic particule

In any way's when you factorize your Energy formula, you end up with a mass witch is very small, negligeable, but still present...

It's high scholl physic, go argue with my teacher...

No. You missed the other equation I gave for the magnitude of the momentum |p|=h/lamda.
For particles with zero rest mass you don't use mass to calculate the energy. You don't strictly speaking do this for any other particles either. It's always the momentum not the rest mass. This all comes down to the fact that special relativity means that the momentum continues to change at a set rate where as the velocity will slowly stop increasing. The result is that the energy continues to increase with the momentum.

[[email protected]]

Ok

p = h / Lamda AND p = mv

if p = p then

h / Lamda = mv

so

m = h / vLamda

v = c

m = h / cLamda

it's algerba

[Splitsecond]

No it doesn't work like that. p=h/lamda is for zero mass particles. p=gamma.mv is for positive mass particles.

[K^2]

Ok E = pc witch mean that energy is the product of the momentum(p) and the celerity(c) witch is the speed of light

But the momentum (p) equals the product of the mass(m) and speed(v)

so

p = mv ' v = c

p = mc

by the end of this factorisation you end up with:

E = mcc = pc = mc^2

so in case of photon this mean heat transfer(so acceleration) of atomic particule

In any way's when you factorize your Energy formula, you end up with a mass witch is very small, negligeable, but still present...

It's high scholl physic, go argue with my teacher...

Photon has relativistic mass. It has no rest mass.

Matter has rest mass. It's relativistic mass at c diverges to infinity.

No, your method of traveling backwards does not work. Stargate prevents anything with rest mass from traveling backwards. But that's not even your biggest problem. Your biggest problem is that the SG operation is to take something going in, store energy signature in buffer, then send contents of buffer. Receiving gate stores incoming energy signature in the receiving buffer, and when the whole object is received, reconstructs it.

If you go backwards, you are added to the send buffer of the destination gate, but destination gate never sends contents of its send buffer. It is only used for situations where a person might stick their hand through event horizon to hold the gate open, or something along these lines. The send buffer is cleared when the connection is dropped. So if you go backwards through the gate, you get trapped in the send buffer, and then effectively destroyed when the connection closes.

P.S. In your case, E=mc² is incorrect. Formula you should use is E²=p²c²+(mc²)²

[thekillman]

oh and Splitsecond, because F=a can not be, the inertial dampening can only get infinitely close to max. ergo, the mass or the effect of mass can not reach zero

[Splitsecond]

oh and Splitsecond, because F=a can not be, the inertial dampening can only get infinitely close to max. ergo, the mass or the effect of mass can not reach zero

Firstly, thanks for the name check, not sure why though. Secondly, if the effective mass reached zero the equation would be F=0 not F=a. It will never be F=a because the mass is never canceled from F=ma. You can make the mass equal to 1kg to get the equation to F=1a. If you want to do something that appears to be scientifically impossible and use an inertial dampener you could have the equation F=nma, where n is the damping factor and 0<n<1, but what is being damped, the mass or the acceleration? You can always view it as multiplying the effective thrust but that isn't a dampener. Either way it doesn't seem to have a real meaning in science.

[thekillman]

the mass is dampened.

its why it has to be

f= (n.m).a

inertia is mass' resistance to acceleration. therefore, reducing the inertia is reducing the (effect of ) mass

[Quadhelix]

the mass is dampened.

its why it has to be

f= (n.m).a

inertia is mass' resistance to acceleration. therefore, reducing the inertia is reducing the (effect of ) mass

Except that, as Splitsecond pointed out, (n*m)*a=m*(n*a). In other words, physically speaking, there is no difference between reducing the mass and increasing the thrust, at least not in that equation.