Relativity rotations rockets Doppler shift

[SuperFinGuy]

Also as it is well known objects traveling at the same speed have zero kinetic energy to each other, so for the engine the rocket is relatively stationary and weight the same as at rest. I don't see how the relativistic KE would not allow light speed.

[WaiteDavidMSPhysics]

@SuperFinGuy The *weight* you refer to is called the proper acceleration. For example if you burn the rocket at constant proper acceleration so that the crew always feels like their weight never changes then the velocity is

v = ctanh(gt'/c)

It doesn't matter how long you run it, that tanh function asymtotes to 1 limiting the speed to less than c. Thats just the way it is.

[SuperFinGuy]

I agree that mass is invariant (contrary to popular interpretation), what actually changes with speed is the momentum of the mass. So it's the energy that is increasing exponentially to infinity.

[WaiteDavidMSPhysics]

@SuperFinGuy

If math is the language of physics than your notation is your dialect. The invariant mass definition in a choice of notation, the best choice, but not one that actually changes the physics. It only changes the way you describe the physics. Either way, the relativistic physics yields the rocket equation that I derived given my notation, and that equation does not let the rocket exceed the speed of light.

[WaiteDavidMSPhysics]

@WaiteDavidMSPhysics

typo. should read

is a choice of notation

not

in a choice of notation

[SuperFinGuy]

@WaiteDavidMSPhysics Yeah different notations are different ways of expressing a theory, but by invariant mass I mean the density of the mass doesn't change, you don't have a proportional increase in mass and momentum, as in the gravitational case. Why your equation didn't let the rocket exceed c? I think c is an observable limit from our dimensions, not an actual speed limit.

[WaiteDavidMSPhysics]

@SuperFinGuy Its not just an observational limit. As long as the physics is special relativistic, c is an actual speed limit. The rocket's physics is special relativistic. The relativistic equation for the speed of the rocket was

v = ctanh(theta) and gave what theta was in terms of how much mass you burned. Regardless of the invariance of the mass

v = ctanh(theta) is the final calculation to get the velocity. The tanh function has an asymtote to a maximal value of 1.

This yields v < c.

[SuperFinGuy]

@WaiteDavidMSPhysics You mean v < c in what frame of reference? Because for an observer on the rocket's frame it is the outside stationary frame that is changing dimensionally so that the speed of light remains constant and always c to the moving observer. I have to ask what about for an observer at a stationary frame?

[WaiteDavidMSPhysics]

@SuperFinGuy I mean v < c for every frame of reference actually.

The rockets velocity is

v = ctanh(theta) for any frame. The value of theta is different for different frames and that yields a different value of v for different frames, but no matter what theta is for your chosen frame tanh(theta) is always less than 1 so for every frame always

v < c.

[SuperFinGuy]

@WaiteDavidMSPhysics But space/time and velocity are changed by reciprocal factors. S/t is multiplied if v is c's fraction. c*tanh(theta) = v so tanh(theta) = v/c yeah, but that is a factor of the s/t change, theta changes according to v/c but it is not a measure of v but of s/t rotation. If we have a velocity equal to c, v/c = 1 means that theta is infinite or unitary. The limit for the asymptote is infinity. We have a limit for s/t dimensions not speed.

[WaiteDavidMSPhysics]

@SuperFinGuy No absolutely not. v is not the proper velocity, it is the coordinate velocity dx/dt.

[SuperFinGuy]

Relativistic geometry shows this, objects separated by a light ray have zero separation in s/t. Take for example that the time dilation factor blows up to infinity when v is the fraction c/1. If v = c/1 than the factor is

1/sqrt (1-(299792458/299792458)^2) or infinity (division by zero).

v can be equal to c and it seems that our s/t dimensions that are limited not v. How can v be limited?

[WaiteDavidMSPhysics]

@SuperFinGuy Its limited *because* that factor blows up. That factor is a factor in how much energy something with mass would need, according to a frame for which its speed is v, in order to travel at that speed v. In special relativistic physics because that factor diverges as v goes to c it would take an infinite amount of energy in order to get a cockpit with mass up to the speed of light, even if you consider the factor on the relativistic energy of the fuel itself. Look at the derivation.

[WaiteDavidMSPhysics]

@WaiteDavidMSPhysics The derivation is at the link in the description bar. The factor is on the fuel.

[SuperFinGuy]

@WaiteDavidMSPhysics In Einstein's special relativity as in his 1905 paper, the energy calculated when v = c is the energy of motion at that speed. It is not the energy that the mass needs to reach c, it is the energy it has at c. It is never shown it's impossible for a mass to reach c. You have included g, but in special relativity the mass remains constant, what changes with speed is the momentum/energy.

[WaiteDavidMSPhysics]

@SuperFinGuy NO. The energy calculated for a speed v IS the energy required to reach that speed. And actually it has been shown to be impossible for a mass to reach c in special relativistic physics, in fact that IS what my derivation shows. You're just plain wrong. Deal with it.

[SuperFinGuy]

@WaiteDavidMSPhysics I have to disagree, here's the proof that the energy is the energy of motion of the mass or its kinetic energy. Where x = sqrt (1-(v/c)^2) is the relativistic fraction, kinetic E = m*v^2/x is equal to the energy of a mass in special relativity or m*c^2/sqrt (1-(v/c)^2). As expected when v = c the energy reaches infinity.

I think the twist here is because the Lorentz factor is a reciprocal of x, 1/x = c/sqrt(c^2-v^2). The relativistic fraction that is the real number.

[WaiteDavidMSPhysics]

@SuperFinGuy That is not a proof. That is a statement which is wrong in terms of your equation for the kinetic energy. The correct kinetic energy for something with nonzero mass is

KE = (gamma - 1)mc^2

You can't just assert an equation and assume that its representative of special relativistic physics, and even if the equation were correct YOUR own wrong equation for kinetic energy would still also not allow it to reach the speed of light.