According to Einstein, when a mass is accelerated and achieves at a new velocity, its mass increases. Then what happens when a mass is moving toward earth (asteroid); its mass must also be increasing, but its gain in kinetic energy is taken from its mass (as it loses potential energy and gains kinetic energy its mass decreases accordingly, thus providing the energy for its increased velocity. In the sum total of the event, does the object have a net gain or loss of mass?

The energy to increase its speed doesn't come from its mass. It comes from whatever external force is causing it to accelerate. Its (rest) mass stays the same.

By the way, when people say mass increases with velocity, they are talking about something called "relativistic mass", which most physicists don't really think is a good concept to use any more. These days, we prefer to just use rest mass, which never changes, and talk about kinetic energy separately.

Rest mass? I thought it was common knowledge that the mass of a body increases with its speed, whatever is causing it Yes this velocity increase is occurs by whatever force that is causing it (in this case gravity) but what is providing the energy? Isn't the mass of an object greater at a distance away from earth than at its surface; this mass differential is equal to the kinetic energy developed dropping from such a height-sounds neat to me, all the energy and mass are accounted for.

Rest mass? I thought it was common knowledge that the mass of a body increases with its speed, whatever is causing it Yes this velocity increase is occurs by whatever force that is causing it (in this case gravity) but what is providing the energy? Isn't the mass of an object greater at a distance away from earth than at its surface; this mass differential is equal to the kinetic energy developed dropping from such a height-sounds neat to me, all the energy and mass are accounted for.
Mass doesn't change with speed, no not according to modern relativity. You are using a dead end, useless, ill informed, unjustified, outdated, bad definition for mass. In order for me to best prove my point let me take you through this step by step, the first by posing a simple question. Given a spaceship that undergoes an ordinary force
fⁱ = dpⁱ/dt
imparting to it a coordinate acceleration
a<SUP>i</sup> = d²xⁱ/dt²
according to a Newtonian law
fⁱ = Ma<SUP>i</sup>
we can make this special relativistically descriptive by imbuing the proportionality M with a velocity dependence at least in the case that the force is in the direction of motion. Call the value of that proportionality at zero velocity the mass m. For that case of force in the direction of motion without calculation what do you assume the relation is between M and m and why?

According to Newton's equation of motion: force is equal to the the rate of change of mass multiplied by the rate of change of velocity, where is this in your equation. What about the second part of my statement. You can omit the personal disparagments.

According to Newton's equation of motion: force is equal to the the rate of change of mass multiplied by the rate of change of velocity, where is this in your equation. What about the second part of my statement. You can omit the personal disparagments.
No,according to Newton's second law, the force is not equal to the rate of change of mass multiplied by the rate of change of velocity. For a point particle (with constant mass) it is equal to the mass multiplied by the acceleration. If you have a changing mass object, the force is equal to the rate of change of mass multiplied by the velocity plus the mass multiplied by the acceleration.

According to Newton's equation of motion: force is equal to the the rate of change of mass multiplied by the rate of change of velocity,
No it isn't.

where is this in your equation.It isn't in anyones equation.

No,according to Newton's second law, the force is not equal to the rate of change of mass multiplied by the rate of change of velocity. For a point particle (with constant mass) it is equal to the mass multiplied by the acceleration. If you have a changing mass object, the force is equal to the rate of change of mass multiplied by the velocity plus the mass multiplied by the acceleration.
Mass doesn't change with speed, but lets pretend that it did. If so you could still wright for linear motion a proportionality between ordinary force and coordinate acceleration according to
fⁱ = Maⁱ
The question is, now without doing the calculus that you are refering to what do you *guess* would be the final relation after doing your product rule between THIS M, and m the value that M would be at rest?

According to Newton's equation of motion: force is equal to the the rate of change of mass multiplied by the rate of change of velocity
d(mv)/dt &ne; dm/dt * dv/dt
d(mv)/dt = m*dv/dt + v*dm/dt

steponit:


Rest mass? I thought it was common knowledge that the mass of a body increases with its speed, whatever is causing it

Relativistic mass increases, as I said, but that's not the best way to look at it.


Yes this velocity increase is occurs by whatever force that is causing it (in this case gravity) but what is providing the energy?

The force. If you push your car, what gives it its energy of motion? Answer: your pushing force.


Isn't the mass of an object greater at a distance away from earth than at its surface

No. Distance has nothing to do with mass.


this mass differential is equal to the kinetic energy developed dropping from such a height-sounds neat to me, all the energy and mass are accounted for.

When you drop an object from a height, gravitational potential energy is converted to kinetic energy as the object falls. The (rest) mass is not affected. The increase in speed is due to the gravitational force. Energy doesn't come from the mass.

According to Einstein, when a mass is accelerated and achieves at a new velocity, its mass increases.

Actually, Einstein did not like the idea of relativistic mass. In a 1948 letter to Lincoln Barnett, Einstein wrote

"It is not good to introduce the concept of the mass M = m/(1-v2/c2)1/2 of a body for which no clear definition can be given. It is better to introduce no other mass than `the rest mass' m. Instead of introducing M, it is better to mention the expression for the momentum and energy of a body in motion."

I could be mistaken, but I believe that in addition to the problems already mentioned you would have to make your relativistic mass into some sort of a tensor because I think the relativistic effects only occur in the direction of motion and do not cause any changes to forces and accelerations in orthogonal directions.

-Dale

I could be mistaken, but I believe that in addition to the problems already mentioned you would have to make your relativistic mass into some sort of a tensor because I think the relativistic effects only occur in the direction of motion and do not cause any changes to forces and accelerations in orthogonal directions.

-Dale
Not presicely, but you're close. Good intuition! His version of mass can be expressed as a matrix. It does have different effects in different directions.

Not presicely, but you're close. Good intuition! His version of mass can be expressed as a matrix. It does have different effects in different directions.
What would this matrix look like? For the generic case of velocity v in the direction x.

Due to image number restrictions I must break this answer into seperate posts. Consider the ordinary force as given by the column vector
<IMG SRC="http://www.geocities.com/zcphysicsms/ford.gif" WIDTH=90 HEIGHT=165>
and the coordinate acceleration given by the column vector
<IMG SRC="http://www.geocities.com/zcphysicsms/aord.gif" WIDTH=74 HEIGHT=141>.

These can be related by
<IMG SRC="http://www.geocities.com/zcphysicsms/fmaord.gif" WIDTH=63 HEIGHT=26>
if we define the matrix by
<IMG SRC="http://www.geocities.com/zcphysicsms/mord.gif" WIDTH=418 HEIGHT=176>
Unfortunately this might get used to get the answer to the question I posed.

So the gamma m terms are an isotropic "mass" increase that is purely due to time dilation effects. What do the gamma^3 terms mean and where do they come from? That is the term that only occurs in the direction of motion.

-Dale

So the gamma m terms are an isotropic "mass" increase that is purely due to time dilation effects.The mass which is really m does not change with speed, but your right that the <FONT FACE=SYMBOL>g</font> comes about purely due to time dilation.

What do the gamma^3 terms mean and where do they come from? Also from time dilation. The relativistic law of dynamics is
F<SUP><FONT FACE=SYMBOL>m</FONT></SUP> = mA<SUP><FONT FACE=SYMBOL>m</FONT></SUP>
(four-vector force equals mass times four-vector acceleration) The mass doesn’t change with speed. Four-vector force is related to four-vector momentum by
F<SUP><FONT FACE=SYMBOL>m</FONT></SUP> = Dp<SUP><FONT FACE=SYMBOL>m</FONT></SUP>/d<FONT FACE=SYMBOL>t</FONT>
which in special relativity due to time dilation can be written
F<SUP><FONT FACE=SYMBOL>m</FONT></SUP> = <FONT FACE=SYMBOL>g</FONT>dp<SUP><FONT FACE=SYMBOL>m</FONT></SUP>/dt
but the last bit is the ordinary force so
F<SUP><FONT FACE=SYMBOL>m</FONT></SUP> = <FONT FACE=SYMBOL>g</FONT>f<SUP><FONT FACE=SYMBOL>m</FONT></SUP>
So far we now have
<FONT FACE=SYMBOL>g</FONT>f<SUP><FONT FACE=SYMBOL>m</FONT></SUP> = mA<SUP><FONT FACE=SYMBOL>m</FONT></SUP>
In special relativity the four-vector acceleration is given by
A<SUP><FONT FACE=SYMBOL>m</FONT></SUP> = d<SUP>2</SUP>x<SUP><FONT FACE=SYMBOL>m</FONT></SUP>/d<FONT FACE=SYMBOL>t</FONT><SUP>2</SUP>
Due to time dilation this can be written
A<SUP><FONT FACE=SYMBOL>m</FONT></SUP> = <FONT FACE=SYMBOL>g</FONT>(d/dt)(<FONT FACE=SYMBOL>g</FONT>dx<SUP><FONT FACE=SYMBOL>m</FONT></SUP>/dt)
After use of the product and chain rules this can be written
<B>A</B><SUP> </SUP>= <FONT FACE="Symbol">g</FONT>[<FONT FACE=SYMBOL>g</FONT><B>a</B> + <FONT FACE="Symbol">g</FONT><SUP>3</SUP><B>u</B>(<B>u<FONT FACE="Symbol">×</FONT>a</B>)/c<SUP>2</SUP>]
Inserting this into
<FONT FACE=SYMBOL>g</FONT>f<SUP><FONT FACE=SYMBOL>m</FONT></SUP> = mA<SUP><FONT FACE=SYMBOL>m</FONT></SUP>
we then have
<B>f</B><SUP> </SUP>= <FONT FACE=SYMBOL>g</FONT>m<B>a</B> + <FONT FACE="Symbol">g</FONT><SUP>3</SUP>m<B>u</B>(<B>u<FONT FACE="Symbol">×</FONT>a</B>)/c<SUP>2</SUP>
This can then be written in the matrix equation form that I gave. You see all the gammas came from time dilation. None had anything to do with an increasing mass. The mass which is m is invariant.

This can then be written in the matrix equation form that I gave.
Thanks, that is one of the most clear and sensible responses that I have seen on this forum.

I am not 100% familiar with SR notation. Does the superscript mu simply indicate that the associated quantity is a 4-vector? Is tau the dilated time in the moving reference frame? Finally, is u the relative velocity of the two frames (bold indicating a simple 3-vector)?

If so then I think I follow everything.

-Thanks
Dale

Thanks, that is one of the most clear and sensible responses that I have seen on this forum.

I am not 100% familiar with SR notation. Does the superscript mu simply indicate that the associated quantity is a 4-vector?
It is an index. When using i,j,k etc they usually represent 1,2, or 3. When using greek like <FONT FACE=SYMBOL>m</FONT> they usually represent 0,1,2 or 3. Each number represents a coordinate and usually 0 is reserved for the timelike coordinate. For example, you may choose Cartesian coordinates with an independent time coordinate so that
x<SUP>0</SUP> = ct
x<SUP>1</SUP> = x
x<SUP>2</SUP> = y
x<SUP>3</SUP> = z
In that case the equation U<SUP><FONT FACE=SYMBOL>m</FONT></SUP> = dx<SUP><FONT FACE=SYMBOL>m</FONT></SUP>/d<FONT FACE=SYMBOL>t</FONT>
means
U<SUP>0</SUP> = dct/d<FONT FACE=SYMBOL>t</FONT>
U<SUP>1</SUP> = dx/d<FONT FACE=SYMBOL>t</FONT>
U<SUP>2</SUP> = dy/d<FONT FACE=SYMBOL>t</FONT>
U<SUP>3</SUP> = dz/d<FONT FACE=SYMBOL>t</FONT>
or you might choose to represent this set of equations as
U<SUP>ct</SUP> = dct/d<FONT FACE=SYMBOL>t</FONT>
U<SUP>x</SUP> = dx/d<FONT FACE=SYMBOL>t</FONT>
U<SUP>y</SUP> = dy/d<FONT FACE=SYMBOL>t</FONT>
U<SUP>z</SUP> = dz/d<FONT FACE=SYMBOL>t</FONT>

Is tau the dilated time in the moving reference frame?
Yes it would be the time according to the test particle for events that are at its location.

Finally, is u the relative velocity of the two frames (bold indicating a simple 3-vector)?

If so then I think I follow everything.

-Thanks
Dale
I usually use v to represent the coordinate velocity between coordinate frames and lower case u to represent the coordinate velocity of a test particle with respect to a lab frame, so for all practicle purposes yes they are the same thing for this discussion. I use upper case U to represent four-vector velocity. Sometimes I will allow what are ordinarily expressed as 3-vectors to take a fourth component which you can do for u, but it if you do that it doens't tell you much because u<SUP>0</SUP> exactle equal to c always. U<SUP>0</SUP> is more interesting because it represents time dilation.