Does Relativistic Mass Affect Scale Readings Inside a Moving Rocket?

[Platformance]

From relativity, if a rocket moves close to c, then its mass increases along with everything inside the rocket.

Now what if there was a scale inside the rocket? Would the scale read a higher value?

I assume the scale should stay the same because if it didn't, then you can do an experiment to determine whether you are moving or not, and being able to detect a moving reference frame is not allowed.

So the true question should be, why would the scale read the same number even though mass is being increased?

 

[pervect]

IT seems to me that you've simply made a reasonably good argument for NOT using relativistic mass, but rather using invariant mass.

ANd in fact pretty much all physicists do use invariant mass nowadays , even though the concept of "relativistic mass" is popular in popularizations, for reasons that aren't clear.

Invariant mass generally makes things much simpler, though.

 

[HallsofIvy]

You see to be missing the piont of "relative". The scale is moving with the ship so "relative to the scale" everything is motionless. The "mass" would be increased only relative to a person outside the ship such that the ship is moving at high speed relative to him.

 

[Platformance]

So the scale doesn't change to a person inside the rocket but it changes to a person outside the rocket.

In this case, how can a scale read 2 different numbers at once? Shouldn't both observers read the same value on the scale?

 

[Passionflower]

So the scale doesn't change to a person inside the rocket but it changes to a person outside the rocket.

In this case, how can a scale read 2 different numbers at once? Shouldn't both observers read the same value on the scale?

No, if the scale reads 100 in the spaceship all observers would agree the reading is 100.

 

[1977ub]

No, if the scale reads 100 in the spaceship all observers would agree the reading is 100.

An observer in relative motion would find that that spring used by the scale has a different coefficient I reckon than the observer at rest wrt the scale/spring, to equalize forces at the '100' reading.

 

[Platformance]

So if both observers see the same mass on the scale, then how could an observer outside the rocket know that the rocket is heavier?

 

[1977ub]

So if both observers see the same mass on the scale, then how could an observer outside the rocket know that the rocket is heavier?

Weight is a measure of gravitational force related to a mass. You stand on a spring and it compresses. You read your weight off of the device related to the spring's compression.

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

... if an electron in a cyclotron is moving in circles with a relativistic velocity, the weight of the cyclotron+electron system is increased by the relativistic mass of the electron, not by the electron's rest mass.

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

 

[Nugatory]

So if both observers see the same mass on the scale, then how could an observer outside the rocket know that the rocket is heavier?

He could apply a transverse force, and see how much the rocket is accelerated sideways (which will appear as a deflection from its initial course). He knows the force he applies, he measures the acceleration, and F=ma gets the mass from there.

This is an old-fashioned and mathematically inconvenient way of thinking about it; as other posters have already pointed out, it's generally better to think in terms of the invariant rest mass and the total energy.

 

[jtbell]

He could apply a transverse force, and see how much the rocket is accelerated sideways (which will appear as a deflection from its initial course). He knows the force he applies, he measures the acceleration, and F=ma gets the mass from there.

Or he could apply a longitudinal force (along the direction of motion), measure the change in the rocket's speed, and get the mass from F=ma. The problem is that this produces a different value for m than Nugatory's procedure does. In the very early days of relativity, some physicists used the terms "transverse mass" and "longitudinal mass."

 

[pervect]

The scale reading is a force. The 4-force transforms via the Lorentz transform. So in the 4-vector formalism one can say:

F = ma

where F is the 4-force, m is the invariant mass (not the relativistic mass), and a is the 4-acceleration.

Relating the 4-force to the three force is only slightly trickier. The 4-force is dP/dtau, P being the 4-momentum (also know as the energy-momentum 4-vector) and tau being proper time. The 3-force is dP/dt (you need to throw away the energy part of the energy-momentum 4-vector and just keep the part that appears to be momentum in the frame of interest).

You can find some info on the 4-vector formalism in wiki, http://en.wikipedia.org/w/index.php?title=Four-vector&oldid=539690212,

from which you'll find that the 4 force is gamma times the 3 force plus an extra component (the time component).

To learn this stuff from scratch, you'd be much better off getting a relativity textbook that treats relativity using 4-vectors , the wiki treatment is to abrupt and abstract to learn from. I'd suggest "Space time physics" by Taylor and Wheeler.

The botom line is pretty simple:

The 4-acceleration and the 4-force can be read directly from the scale in the space-ship, using the properties of the scale (which are known in it's rest frame) and the invariant mass (also known in the rest frame).

The 4-acceleration has the property that it's the same for all observers, so you compute it where it's easiest (in a frame in which the ship is at rest), and then you also know it in any other frame (including the frame where the spaceship is moving at a velocity v).

Finding the coordinate acceleration from the 4-acceleration will take some math, unless there's some specific interest in it I'm not going to bother working that out.