Does the density or the volume change when you have an object moving near c?

[Endercreeper01]

When you have an object moving near the speed of light, would the density or the volume of an object change from mass increase?

You either need to have the density or the volume to change in order for the mass to change

[mathematic]

The length of the object in the direction it is going gets shorter, other dimensions stay the same. Net result - the observer sees an increase in density.

In the frame of the object, the density and its dimensions are unchanged.

[swansont]

There is no mass increase. The effect is purely from length contraction.

[decraig]

When you have an object moving near the speed of light, would the density or the volume of an object change from mass increase?

You either need to have the density or the volume to change in order for the mass to change

 

Are you asking whether or not the increase in relativistic mass is due to a change in volume?

 

If so, the answer to this within the context of special relativity is 'no'.

 

Within the general theory, I'm no so sure. In special relativity we can pretend that relativistic mass is an element of a vector where the other vector elements are three momentum components. The invariant quantity is the rest mass. With general relativity we have to begin with something a little more fundamental--the action with units of energy-time, or even the action 4-density.

[Endercreeper01]

The mass actually does increase. If it didn't, then it wouldn't take infinite energy to accelerate something to c.

[md65536]

There are a ton (estimated) of threads that mention why relativistic mass should not be used as a definition of mass. The latest one I've seen is a link posted by swansont to this:

 

The mass actually does increase. If it didn't, then it wouldn't take infinite energy to accelerate something to c.

Your argument is problematic because for example if an observer accelerates itself relative to something else, then it is "adding mass" to that something else. How does it do that without transferring energy to it? It's just a different type of energy... kinetic energy, which is relative, but not mass energy. So it is gaining "relativistic mass" which isn't the same as "mass". Edited November 16, 2013 by md65536

[Endercreeper01]

when you get close to light speed, it takes more then four times the energy to double the velocity of something, even though the equation for kinetic energy is 1/2mv² . This is because of the mass increasing by a factor of gamma, making it take more energy.

And also, when you have a constant force acting upon something, the acceleration slows, even though you have the same force acting upon it. This is because of mass increase.

Edited November 17, 2013 by Endercreeper01

[xyzt]

When you have an object moving near the speed of light, would the density or the volume of an object change from mass increase?

You either need to have the density or the volume to change in order for the mass to change

This is an example of an ill posed question resulting into a lot of confusing/meaningless answers. The issue is that indeed, "relativistic" mass increases with speed by [math]\gamma[/math] while volume decreases by [math]1/\gamma[/math], so it would appear that COORDINATE-DEPENDENT density would increase by [math]\gamma^2[/math]. But, this is a meaningless exercise, since COORDINATE-DEPENDENT density is not a meaningful physics quantity.

Edited November 17, 2013 by xyzt

[Janus]

when you get close to light speed, it takes more then four times the energy to double the velocity of something, even though the equation for kinetic energy is 1/2mv² . This is because of the mass increasing by a factor of gamma, making it take more energy.

And also, when you have a constant force acting upon something, the acceleration slows, even though you have the same force acting upon it. This is because of mass increase.

 

[math] \frac{mv^2}{2}[/math], is the Newtonian equation for kinetic energy. The relativistic kinectic energy(for something with a non-zero rest mass) is found by:

 

[math] mc^2 \left ( \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \right )[/math]

 

If you convert this to the total energy, including rest mass energy equivalence, you get

 

[math]\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}[/math]

 

If you take this equation and expand it into a series, you get:

 

[math]mc^2 + \frac{mv^2}{2}+ \frac{3mv^4}{c^2}+...[/math]

 

Note the first two terms. One is the energy equivalence for the rest mass and the the other is Newtonian kinetic energy. Essentially, if you subtract out the rest mass part, you get a formula that ends up being close to the newtonian value when v is very small, but diverges as v approaches c.

 

IOW, [math] \frac{mv^2}{2}[/math] is not the equation for kinetic energy, but a only a close approximation when v is small.

Edited November 17, 2013 by Janus

[Toffo]

This is an example of an ill posed question resulting into a lot of confusing/meaningless answers. The issue is that indeed, "relativistic" mass increases with speed by [math]\gamma[/math] while volume decreases by [math]1/\gamma[/math], so it would appear that COORDINATE-DEPENDENT density would increase by [math]\gamma^2[/math]. But, this is a meaningless exercise, since COORDINATE-DEPENDENT density is not a meaningful physics quantity. Case and point, charge density is frame invariant.

 

I checked wikipedia, it says: "charge density is a relative concept"

 

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

 

Wikipedia also says that Anthony French has described how magnetism is a result of relativity of charge density.

 

So therefore I say that gravitomagnetism is a result of the relativity of mass density.

When you have an object moving near the speed of light, would the density or the volume of an object change from mass increase?

You either need to have the density or the volume to change in order for the mass to change

 

Oh yes. And also from volume decrease.

Edited November 17, 2013 by Toffo

[MigL]

You stand on a scale watching a spaceship pass by at a uniform 0.95c and you apply the SR equations to find that the ship's relativistic mass has increased by a significant amount.

The pilot of the ship is in an inertial frame and measures your speed to be 0.95c relative to his ship ( which one is actually at 'rest' has no meaning ). He calculates your relativistic mass to have increased substantially.

 

You look down at the scale, What does it say ??? Is it substantially increased from normal ??? Or does it still say your rest mass ???

[xyzt]

 

I checked wikipedia, it says: "charge density is a relative concept"

 

This is what I get for quoting from memory, I opened my Landau and Lifschitz and, sure enough, charge denisty transforms under the Lorentz transform as :

[math]\rho=\gamma(\rho'+j' \frac{v}{c^2})[/math]

[decraig]

There are a ton (estimated) of threads that mention why relativistic mass should not be used as a definition of mass. The latest one I've seen is a link posted by swansont to this:[video removed]

 

 

As much as I appreciate and admire Sean Carroll, I find this ongoing argument of invariant mass vs. relative mass empty. There is no capital "The mass", in my opinion. It's perfectly fine that two physical quantities may have the same units.

Edited November 18, 2013 by decraig

[Endercreeper01]

When things are accelerating, the velocity is:

 

v=at/(1-a²t²/c²)

 

You can use this for the relativistic mass.

But, I agree with decraig

And also, why would energy be mc²y if the actual equation is E²=m²c⁴+p²c² ?

But now, let's assume it was mc²y. It would not be mc² and would be mc²y because of the increase in mass.

Edited November 18, 2013 by Endercreeper01

[Sensei]

There are a ton (estimated) of threads that mention why relativistic mass should not be used as a definition of mass. The latest one I've seen is a link posted by swansont to this:

 

 

Quote from

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

 

"The mass of composite systems"

"The rest mass of a composite system is not the sum of the rest masses of the parts, unless all the parts are at rest. The total mass of a composite system includes the kinetic energy and field energy in the system."

 

When we say f.e. proton has mass 938.272 MeV/c^2 we are talking about sum of relativistic masses of quarks, not rest masses of quarks..

 

Your argument is problematic because for example if an observer accelerates itself relative to something else, then it is "adding mass" to that something else. How does it do that without transferring energy to it?

 

 

It doesn't have to, if there is absolute frame of reference (which we cannot find).

 

With extreme acceleration to relativistic velocities, new particles are created.

p+ + p+ -> p+ + p+ + p+ + p-

(proton colliding with proton is creating three protons and one antiproton)

If relativistic mass is not real, then it has completely no sense.

But it's real experimentally confirmed result.

 

[swansont]

And also, why would energy be mc²y if the actual equation is E²=m²c⁴+p²c² ?

 

 

Why can't both be true?

 

With extreme acceleration to relativistic velocities, new particles are created.

p+ + p+ -> p+ + p+ + p+ + p-

(proton colliding with proton is creating three protons and one antiproton)

If relativistic mass is not real, then it has completely no sense.

 

No, it's perfectly consistent with physics and explainable without relativistic mass. Relativistic mass is simply a proxy for total energy; all you have is a lazy shorthand that conveys no new information. You also have a concept that doesn't fit in with other physics when you get into a little more depth. Using rest mass/invariant mass doesn't introduce these problems.

[Endercreeper01]

You know how mass comes from the higgs field? If you have an object moving through the higgs field, it is interacting more with the field, and so, it has more mass.

[xyzt]

When things are accelerating, the velocity is:

 

v=at/(1-a²t²/c²)

 

 

 

 

Nope, it is [math]v=\frac{at}{\sqrt{1+(at/c)^2}}[/math]. You rolled multiple errors in one formula.

 

 

 

And also, why would energy be mc²y if the actual equation is E²=m²c⁴+p²c² ?

 

Because the formula

E²=m²c⁴+p²c²

 

is a trivial consequence of [math]E=\gamma mc^2[/math] and [math]p=\gamma mv[/math].

 

 

But now, let's assume it was mc²y. It would not be mc² and would be mc²y because of the increase in mass.

 

When v=0 [math]\gamma=1[/math] and [math]E=mc^2[/math].

[LaurieAG]

 

There is no capital "The mass", in my opinion. It's perfectly fine that two physical quantities may have the same units.

 

I suppose you think it's perfectly fine to be out by 2 * Pi as well? The difference between Planck's constant h as used in rest mass calculations, and the reduced constant h_bar as used in relativistic mass calculations, is 2 * Pi. It's the same difference between the standard Compton wavelength and the reduced Compton wavelength.

 

Another character could be used to avoid confusion.

 

₥

[Endercreeper01]

I have lag and computer issues, so It won't always be what I meant to write. Correct me when that happens.

And for the energy equation, mass is a measure of the amount of energy an object has, according to Sean Carroll. If an object has more energy from moving, it therefore has more mass.

Edited November 19, 2013 by Endercreeper01

[swansont]

I have lag and computer issues, so It won't always be what I meant to write. Correct me when that happens.

And for the energy equation, mass is a measure of the amount of energy an object has, according to Sean Carroll. If an object has more energy from moving, it therefore has more mass.

 

Mass is related to the energy it has at rest. When it's moving, we call the additional energy kinetic energy. There are advantages to doing it this way.

 

Sean Carroll is on record as being against the concept of relativistic mass, so citing him means something is being taken out of context.

[Endercreeper01]

One thing no one has answered is my argument with the higgs field. Mass is a measure of how much something interacts with the higgs field, so therefore, if something is moving, it interacts more with the higgs field, and so, more mass.

[decraig]

 

I suppose you think it's perfectly fine to be out by 2 * Pi as well? The difference between Planck's constant h as used in rest mass calculations, and the reduced constant h_bar as used in relativistic mass calculations, is 2 * Pi. It's the same difference between the standard Compton wavelength and the reduced Compton wavelength.

 

Another character could be used to avoid confusion.

 

₥

 

How is this being "out"?

 

Relativistic mass is Newtonian inertial mass. It is that which resists a change in velocity,

m = F/a. It can also be measured with a scale, not the invariant mass.

 

You are free to use E= Fc^2/a.

 

Should I have to need to use both masses, there already exists standard notation, without using the funny 'm'.

 

Would you also object to defining torque because it has the same units as energy?

 

There is also the action 4-density and energy density. They share the same units. Each are useful and measure different things.

 

 

 

This was a thread on density, and has gone far afield, nit-picking over definitions that are already well defined. We could simply recognize that Endcreeper could be answered in terms of energy density.

Edited November 20, 2013 by decraig

[Endercreeper01]

This thread should have been on density and not on relativistic mass. That should be another thread.

[decraig]

Going back to the original post:

 

When you have an object moving near the speed of light, would the density or the volume of an object change from mass increase?

You either need to have the density or the volume to change in order for the mass to change

 

You propose a cause and effect problem. Since there is no order of events in how these quantities relate, I don't really think it is a so easily answered as posed. However...

 

The volume of an object contracts just as the length.

 

[math]\gamma = \frac{1} { \sqrt{1-v^2/c^2} }[/math]

 

[math]L=L_0/\gamma[/math]

 

[math]V=V_0/\gamma[/math]

 

Density of a uniform extended object:

 

[math]\rho_0 = m/V_0[/math]

 

r₀ is the density, m is the invariant mass and V₀ is the volume in the rest frame of the object.

 

In the inertial frame where the object is in motion with velocity v.

 

[math]\rho =m/V = \gamma \rho_0[/math]

 

Gamma is greater than or equal to 1, so the density of the invariant mass increases with velocity.

Edited November 24, 2013 by decraig