Math of relativistic mass different from that of rest mass ?

[koti]

The planetoid is moving at 0,5c.

 

Case 1.

I'm the observer on earth, I weigh 200 lbs and I see a planetoid moving away at 0,5 c

 

Case 2.

I'm the observer on the planetoid, I weigh 200 pounds (?) and I see earth moving away at 0,5 c.

[Delta1212]

In Case 2, the planetoid isn't moving at 0.5c then. The Earth is.

[Strange]

I know now (thanks to Mordred) that I would weigh 200 pounds on the accelerated planetoid.

 

 

Note that when MigL says "weigh" he is automatically referring to weight rather than mass. Which is entirely reasonable. But may be confusing.

 

But it might have been a bit clearer if the example had been the Moon, so you weigh 1/6th what you do on Earth (while it is in orbit). When the space-aliens steal the moon and accelerate it so the relativistic mass is the same as the Earth, you would still weigh 1/6th of your Earth weight. (For the reasons given by others.)

Is the planetoid traveling at 0.5c or is the Earth?

 

 

Yes.

[Delta1212]

 

 

Note that when MigL says "weigh" he is automatically referring to weight rather than mass. Which is entirely reasonable. But may be confusing.

 

But it might have been a bit clearer if the example had been the Moon, so you weigh 1/6th what you do on Earth (while it is in orbit). When the space-aliens steal the moon and accelerate it so the relativistic mass is the same as the Earth, you would still weigh 1/6th of your Earth weight. (For the reasons given by others.)

 

 

 

Yes.

Yeah, I was hoping merely asking the question would help.

[koti]

Okay folks (Delta), youre not trapping me into the what is moving in reference to what, I get that. What bothers me is that both celestial bodies, now its the earth and the moon abducted by aliens and accelerated to 1G inertial mass like Strange proposed which is probably a more "down to earth" example (lol) I will weight 200 lbs on earth and 33,3 lbs on the moon at 0,5c. The question is will both frames have the same effect on spacetime curvature?

[Strange]

The question is will both frames have the same effect on spacetime curvature?

 

 

They won't. But, apparently, working out exactly what the gravitational effect of a moving object would be is non-trivial. Whenever this has come up on forums in the past the consensus view of those experts who could do the relevant math was always, "its complicated".

[Delta1212]

 

 

They won't. But, apparently, working out exactly what the gravitational effect of a moving object would be is non-trivial. Whenever this has come up on forums in the past the consensus view of those experts who could do the relevant math was always, "its complicated".

Noting also that which object is moving will depend on which frame you are in. If you are sitting on one of the objects traveling with it, you always calculate as if it is at rest.

[koti]

 

 

They won't. But, apparently, working out exactly what the gravitational effect of a moving object would be is non-trivial. Whenever this has come up on forums in the past the consensus view of those experts who could do the relevant math was always, "its complicated".

Phew...thats a relief. That stands in contradiction with Mordred's post #11 though doesn't it?

[Delta1212]

Phew...thats a relief. That stands in contradiction with Mordred's post #11 though doesn't it?

I assumed Mordred was talking about relativistic mass, which is not the same thing as rest mass in terms of how it interacts with gravity.

[koti]

Noting also that which object is moving will depend on which frame you are in. If you are sitting on one of the objects traveling with it, you always calculate as if it is at rest.

Lets forget about the observers and relative velocities. My original intention for this thread was to find out how spacetime curves "around" rest mass and relativistic mass.

 

Just to be sure...Extremely high relativistic mass of an extremely fastly moving body of a small rest mass could not lead to the formation of a black hole or could it?

[Delta1212]

Lets forget about the observers and relative velocities. My original intention for this thread was to find out how spacetime curves "around" rest mass and relativistic mass.

 

Just to be sure...Extremely high relativistic mass of an extremely fastly moving body of a small rest mass could not lead to the formation of a black hole or could it?

It could not. A black hole must be a black hole in all frames. If it isn't a black hole at rest, it cannot be a black hole in a frame moving very quickly with respect to it.

[Mordred]

I assumed Mordred was talking about relativistic mass, which is not the same thing as rest mass in terms of how it interacts with gravity.

Ok lets simplify this. Your on Earth planet observing a planetoid at high c.

 

That planetoid has two types of mass. Its invariant mass (lets get into the more accepted terminilogy/rest mass. It also has a variant mass we now call inertial mass. (replaces relativistic mass).

 

Now lets take a 1 kg test weight measured on Earth. ( note same reference frame as Observer). inertial mass is zero as it in the same frame as the Earth.

 

Now observer on Earth already noted the planetoid has its (at rest mass) as well as inertial mass. So the planetoid already has greater mass due to relativistic corrections. So when he places that test weight on the Planetoid he can still apply Newtons laws. (after all the planetoid now has a higher effective mass. aka greater gravity.

 

Now if the observer also switches location he removes the inertial mass as he is now back in the same reference frame. ( rule of thumb same reference frame as emitter =no inertial mass.)

Edited May 1, 2017 by Mordred

[koti]

correct

Mordred, we were talking about a 1kg fast moving steel ball with a relativistic mass resulting in 1G and an earth with a rest mass resulting in 1G. Was your post #11 a slip up or am I being dense again?

Edited May 1, 2017 by koti

[Mordred]

Not if you look at observer influence. An outside of reference frame can measure a 1 kg steel ball gain enough inertial mass to equal the gravitational potential of the Earths gravity.

 

Principle of equivalence still applies but you have to watch your observers and which mass applies to which observers.

 

Unfortunately you can't ignore this as its key to understanding how weight is influenced to different observers. Yes I do mean weight as well as mass.

 

There is no inertia mass if your in the same reference frame as the object being measured.

Edited May 1, 2017 by Mordred

[koti]

Okay, let me make sure I get this;

 

We have two frames of reference: The earth and the moon, both same diameter and our real world masses. Earth is at 1G at its rest mass, the moon is at 1/6G at its rest mass and is being accelerated to inertial mass resulting in 1G. Will spacetime regions occupied by those two frames be identical or not?

[Mordred]

How one measures a spacetime region depends on the observer. Different observers will have different spacetime measurements. Remember the geometry changes depending upon observer effects.

 

Observers in the same reference frame will always have identical spacetime conditions and dimensions. (lets include length contraction).

 

You contract length you increase energy density and therefore mass density. Same with giving an object inertia as inertia will also increase the energy density.

 

The (at rest frame) for a planet is accurately described by the Newton approximation under GR. Though once you hit relativistic effects you want the Schwartzchild metric. Though I should note, under GR all frames of reference are inertial.

Edited May 1, 2017 by Mordred

[koti]

How one measures a spacetime region depends on the observer. Different observers will have different spacetime measurements. Remember the geometry changes depending upon observer effects.

 

Observers in the same reference frame will always have identical spacetime conditions and dimensions. (lets include length contraction).

 

You contract length you increase energy density and therefore mass density. Same with giving an object inertia as inertia will also increase the energy density.

 

The (at rest frame) for a planet is accurately described by the Newton approximation under GR. Though once you hit relativistic effects you want the Schwartzchild metric.

I'm particularly interested in what happens with gravity in our two established frames of reference. We've established that on earth which is at rest mass and 1G I would weigh 200 pounds and that on the accelerated moon which is also at 1G but due to inertial mass I would weigh 33,3 lbs.

I'm curious what the spacetime regions occupied by those two frames would look like gravity wise. (Note that both the earth and the moon have the same diameter like you proposed in our earlier experiment)

[Delta1212]

If you are on the moon, it is not 1G. It is still 1/6G. It is only 1G when taking into account relativistic mass as observed from Earth.

 

I know you said earlier to forget the different observers and relative velocities but you literally cannot do that.

[MigL]

Sorry for the absence; I'm at work now...

 

Measured in their respective frames, the space-time geometries around the Earth or the relativistic moon will be exactly the same as if they were at rest.

This is because the geometry 'moves' at c along with the relativistic object.

 

From other frames the geometry will be different and much more complex as your original link stated.

This is due to the 'added' energy to the relativistic moon, and of course, is to be expected because energy is also frame dependent

[koti]

Asking the right questions is equally as important as the answer itself especially when dealing with complex subjects. It would have been so much easier for me to ask if spacetime curvature is frame dependant or not. Dang.

 

Delta, youre right, my evident error, ofcourse its 1/6 G and 1G dependant on frames. I didnt want to deviate into lenght contraction and time dilation and it ended up that I decided in my mind that curvature is independant of frames. Thanks everyone.

Edited May 2, 2017 by koti

[Mordred]

Knowing the right questions to ask is often the trick. +1 response though.

[Sensei]

If we start with a steel ball with the same diameter as the Earth the two spacetime regions will effectively be identical.

 

LOL.

If you start with steel ball with the same diameter as the Earth, they will have the same volume, but it'll be 43% more massive.

Density of Iron is ~ 7874 kg/m^3, and density of Earth is ~ 5514 kg/m^3

( 7874 kg / m^3 * volume of Earth ) / ( 5514 kg/m^3 * volume of Earth ) = 1.428

[Mordred]

well ok ya got me the steel ball is more dense Good catch completely slipped my mind lol +1 always love an accurate correction.

Edited May 6, 2017 by Mordred

[koti]

I assumed in my mind (and others I presume did the same) that the two bodies have same diameter and same density, otherwise the thought experiment wouldn't make sense. Good catch though Sensei. I see you like things to be coherent and thats a good thing so you will need to reconsider your calculations because steel is a bit less dense than iron

To be on the right side of right, make it a cast mixture of copper and aluminum to exactly match earths density and diameter

Edited May 6, 2017 by koti

[Sensei]

I see you like things to be coherent and thats a good thing so you will need to reconsider your calculations because steel is a bit less dense than iron

To do that we would have to get into details like what are other constituents of alloy..

"The density of steel varies based on the alloying constituents but usually ranges between 7750 and 8050 kg/m3"

https://en.wikipedia.org/wiki/Steel

(not to mention it varies with temperature/pressure)