Straight line through spacetime.

[stevmg]

Do you really not understand the difference between the distance between A and B:
\sqrt{c^2(t_B-t_A)^2 - (x_B-x_A)^2 - (y_B-y_A)^2 - (z_B-z_A)^2}

and the length of a path from A to B:
\int_A^B \sqrt{c^2 \left(\frac{dt}{d\lambda}\right)^2 - \left(\frac{dx}{d\lambda}\right)^2 - \left(\frac{dy}{d\lambda}\right)^2 - \left(\frac{dz}{d\lambda}\right)^2 } d\lambda

Dale -

The top expression is the "shortest" distance between the two points in space-time

The bottom exprerssion is the actual length of the path that one travels from A to B which may not be the shortest distance

Am I right? Did I get it?

 

[Dale]

Dale -

The top expression is the [STRIKE]"shortest"[/STRIKE] distance between the two points in space-time

The bottom exprerssion is the actual length of the path that one travels from A to B which may not be the shortest distance

Am I right? Did I get it?

Yes, with one small correction. There is only one distance between two points, but there are an infinite number of paths, each with a different length. The distance is the extremum of the lengths of all possible paths.

 

[stevmg]

Yes, with one small correction. There is only one distance between two points, but there are an infinite number of paths, each with a different length. The distance is the extremum of the lengths of all possible paths.

We are on the same sheet of music, DaleSpam. An analogy to cartesian coordinates would be a quarter circle path with point A at (2, 0) and point B at (0, 2) and the center at the origin and the path is the quarter circle.

The distance from A to B is 2\sqrt{2} = 2.2828 while the path from A to B is \pi = 3.1416 which are clearly different. There are clearly other paths from A to B.

In the time-space coordinates there is a "subtraction" if you will when you deviate from the "straight" path which can actually make the combined interval shorter

The twin paradox is a great example of that.

 

[DrGreg]

In the Euclidean geometry of space, the distance between two points is the length of shortest path between them.

Also, in (flat) spacetime, for two events with spacelike separation, i.e. where one event is outside the light cone of the other, the interval between two events is the length of shortest spacelike path between them.

But for two events with timelike separation, i.e. where one event is inside the light cone of the other, the interval between two events is the length of longest timelike path between them.

That's why the zero-length zig-zag path mentioned earlier is an irrelevance.

 

[stevmg]

In the Euclidean geometry of space, the distance between two points is the length of shortest path between them.

Also, in (flat) spacetime, for two events with spacelike separation, i.e. where one event is outside the light cone of the other, the interval between two events is the length of shortest spacelike path between them.

But for two events with timelike separation, i.e. where one event is inside the light cone of the other, the interval between two events is the length of longest timelike path between them.

That's why the zero-length zig-zag path mentioned earlier is an irrelevance.

Guilty as charged!

 

[stevmg]

So I get this right...

If one follows a zig-zag path (each "zig" and "zag" along the local light cone and going forward in time,) the sum total of all the lengths of these "zigs" and "zags" is 0 (not taking into account GR.) Is that right? Is that called the "interval?"

The distance between the origin of the zigzag path and its destination would be the equation as eumerated above by DaleSpam and that distance would not be zero. Is that right?

 

[stevmg]

More terminology questions:

in the expression m(v) = m₀/SQRT[1 - v²/c²]
m(v) is the inertial mass
m₀ is the rest mass

Am I right?

Finally, in a system of particles which do not interact other than banging together, the sum of all momenta for each particle is the total momentum of the system.

Likewise the total energy of the system is the sum of all the resting energy for each particle + the kinetic energy for each particle added up. Right?

Since momenta are linear operators can one prove that the above is true by induction:

Assume true for n particles and from that show it is true for n+1 particles.

I don't get how they do the energy, which the kinetic energy is not linear because it goes with the square of the velocity. It is non linear. Can't use induction then as it is non linear.

 

[Dale]

So I get this right...

If one follows a zig-zag path (each "zig" and "zag" along the local light cone and going forward in time,) the sum total of all the lengths of these "zigs" and "zags" is 0 (not taking into account GR.) Is that right? Is that called the "interval?"

The distance between the origin of the zigzag path and its destination would be the equation as eumerated above by DaleSpam and that distance would not be zero. Is that right?

That is correct, the length of two different paths between A and B in general may be different from each other and different from the distance between A and B (which is the length of the shortest path). That applies both for Euclidean geometry (spatial distance) and Minkowski geometry (spacetime interval).

 

[stevmg]

That is correct, the length of two different paths between A and B in general may be different from each other and different from the distance between A and B (which is the length of the shortest path). That applies both for Euclidean geometry (spatial distance) and Minkowski geometry (spacetime interval).

Any comment about the energy-momentum. You have already pointed me in the direction of the vectors as you did with twin paradox.

My real concern is, how do you work this - at least an elementary way and is the proof by induction valid?

The total energy of a system is invariant but is not conserved when looking from different frames of reference (I think).

Starthaus has a proof that he worked out for this and I would like to see something like that but if he has copyrighted it, then I can understand.

 

[starthaus]

Any comment about the energy-momentum. You have already pointed me in the direction of the vectors as you did with twin paradox.

My real concern is, how do you work this - at least an elementary way and is the proof by induction valid?

No. I already told you that several times.

The total energy of a system is invariant

No, it is not invariant since it is speed dependent:

E=\gamma(v)m_0c^2
E'=\gamma(v')m_0c^2

Nor is the momentum frame-invariant, since it depends on speed and velocity:

\vec{p}=\gamma(v)m_0\vec{v}

It is the norm of the energy-momentum (E , c\vec{p}) that is invariant. Indeed:

E^2-(c\vec{p})^2=(m_0c^2)^2=E'^2-(c\vec{p'})^2

Wait until you receive Spacetime Physics and you study it.

but is not conserved when looking from different frames of reference (I think).

In a closed system, the total momentum \vec{p} and the total energy E are conserved, each one of them. You managed to get everything backwards.

Starthaus has a proof that he worked out for this and I would like to see something like that but if he has copyrighted it, then I can understand.

I haven't copyrighted anything, this is all easily available in textbooks.I wrote a little piece for Battleimage to help him out understanding the energy-momentum conservation, see here. Why don't you wait for your own copy of Spacetime Physics, it is all there.

 

[DrGreg]

More terminology questions:

in the expression m(v) = m₀/SQRT[1 - v²/c²]
m(v) is the inertial mass
m₀ is the rest mass

Am I right?

Yes. m(v) is also called relativistic mass, although its use is deprecated by most modern physicists who prefer to work with rest mass only.

Finally, in a system of particles which do not interact other than banging together, the sum of all momenta for each particle is the total momentum of the system.

Likewise the total energy of the system is the sum of all the resting energy for each particle + the kinetic energy for each particle added up. Right?

Yes. Usually the energy isn't explicitly separated into "rest energy" and "kinetic energy", we just talk about energy.

Since momenta are linear operators can one prove that the above is true by induction:

Assume true for n particles and from that show it is true for n+1 particles.

I don't get how they do the energy, which the kinetic energy is not linear because it goes with the square of the velocity. It is non linear. Can't use induction then as it is non linear.

It's not the "linearity of momentum" that is used here. In fact the function \textbf{p}(m_0,\textbf{v})=\gamma(v) m_0 \textbf{v} is only linear in m₀ and not linear in v. Induction works simply because addition (adding two momenta to get the total momentum) is linear, and therefore similarly for energy. That is true as long as you are looking at this within anyone frame only.

There is a technical difficulty if you want to go further than this and show that the total "energy-momentum" is a well-behaved 4-vector that transforms from one frame to another the way that 4-vectors are supposed to transform (i.e. via the Lorentz transform). The proof isn't entirely trivial due to the relativity of simultaneity.

The total energy of a system is invariant but is not conserved when looking from different frames of reference (I think).

You got "invariant" and "conserved" the wrong way round there!

 

[stevmg]

As stated my book on Spacetime Physics will get here. It is taking me a while to get the terminology straight (look at all the corrections made to my earlier statements inthe immediately preceding posts) and I still need to bounce things off you folks as I learn them. Had to do that all the time in medicine. I never relied totally on my interpretation when it came to new concepts. Never killed anybody and I wish I can say that about others.

I understand that you (starthaus) have stated that the proof is not an induction proof - just getting a second opinion.

When I saw my pulmonolgist earlier this year and he told me that I had sleep apnea, I said "Hey, I want a second opinion!"

He gave it to me...

"OK, Steve, here it is... You're ugly, too.!"

I never win.

 

[Dale]

Any comment about the energy-momentum. You have already pointed me in the direction of the vectors as you did with twin paradox.

I am not sure what you are asking for here. My comment is that I like them

Oh, one thing to remember is that photons are massless so you always know that the energy is equal to the momentum times c.

My real concern is, how do you work this - at least an elementary way and is the proof by induction valid?

Personally, I wouldn't worry too much about the proof. It will be good to read through since it will really make you think about the relativity of simultaneity and how a system of particles is even defined in different frames. But all proofs require some assumptions which then must be validated experimentally.

 

[starthaus]

There is a technical difficulty if you want to go further than this and show that the total "energy-momentum" is a well-behaved 4-vector that transforms from one frame to another the way that 4-vectors are supposed to transform (i.e. via the Lorentz transform). The proof isn't entirely trivial due to the relativity of simultaneity.

Yes, the two page essay by Rindler on this subject is very interesting. A mechanical application of the rule "the sum of four-vectors is a four-vector" does not apply in the case of the energy-momentum four-vector due to the very definition of relativistic total energy and relativistic three-momentum. Relativity of simultaneity plays a big role in defining the values that enter the conservation equations. Though Rindler's discussion ends up with the conclusion that the resultant energy-momentum of an isolated system of particles is indeed a four-vector, it is the proof that is worth following.

 

[stevmg]

Let's get this correct (for me)

The sum of all energy in a system of particles in which the total momentum is zero is invariant

If the momentum of the system is not zero then the relation:

E² -(cp)² = E₀²

in which E is the energy of the system, cp is the energy due to the momentum of the system, and E₀ is the invariant energy of the system if the momentum were to be brought down to zero.
This invariant quantity E₀ would be the same form all frames of reference.
The conservation applies to all the frames of reference each frame at a time. Different frames of reference will change the total momentum of the system and the total energy of the system but not the E₀ which is the invariant part.

Is this right?

 

[DrGreg]

Let's get this correct (for me)

The sum of all energy in a system of particles in which the total momentum is zero is invariant

If the momentum of the system is not zero then the relation:

E² -(cp)² = E₀²

in which E is the energy of the system, cp is the energy due to the momentum of the system, and E₀ is the invariant energy of the system if the momentum were to be brought down to zero.
This invariant quantity E₀ would be the same form all frames of reference.
The conservation applies to all the frames of reference each frame at a time. Different frames of reference will change the total momentum of the system and the total energy of the system but not the E₀ which is the invariant part.

Is this right?

I'd agree with all that except for one small detail. I wouldn't describe cp as "the energy due to the momentum" (although I suppose in a sense it is). The equation in which it appears has energy-squared in it. But you could describe (E-E₀) as "kinetic energy" of the system, and E₀/c² is the "invariant mass" of the system (which isn't the sum of the individual particles' rest masses except in the special case when they're all at rest relative to each other).

 

[stevmg]

I'd agree with all that except for one small detail. I wouldn't describe cp as "the energy due to the momentum" (although I suppose in a sense it is). The equation in which it appears has energy-squared in it. But you could describe (E-E₀) as "kinetic energy" of the system, and E₀/c² is the "invariant mass" of the system (which isn't the sum of the individual particles' rest masses except in the special case when they're all at rest relative to each other).

If cp isn't the energy of the particle associated with the momentum as seen from a given frame of reference, then what is it?

Of course the point of what I did write was to make sure my concept of invariant energy etc. was correct and that momentum and energy is conserved for each frame of reference while the E₀ is invariant - across all frames of reference.

 

[Dale]

I also would not describe cp as the energy associated with the momentum. "Energy associated with momentum" sounds like another way to say "kinetic energy", and for v<<c cp reduces to cmv which is not the same as 1/2 mv². While cp has units of energy, I would think of it as momentum, not energy. Or, preferably, you could think of it as what it is: the spacelike component of the energy-momentum four-vector expressed in units of energy. Mentioning that it is the spacelike component of a four-vector ensures that its relationship to energy is properly characterized.

 

[stevmg]

I also would not describe cp as the energy associated with the momentum. "Energy associated with momentum" sounds like another way to say "kinetic energy", and for v<<c cp reduces to cmv which is not the same as 1/2 mv². While cp has units of energy, I would think of it as momentum, not energy. Or, preferably, you could think of it as what it is: the spacelike component of the energy-momentum four-vector expressed in units of energy. Mentioning that it is the spacelike component of a four-vector ensures that its relationship to energy is properly characterized.

Thank you. This is new to me. I knew that total energy was mc² + 1/2mv²

I will treat this as a new entity for me - cmv which I will have to wrap my brain around.

I do understand what "spacelike" means.

This is going to take time.

 

[starthaus]

Thank you. This is new to me. I knew that total energy was mc² + 1/2mv²

Not quite, this is the resultant of doing the Taylor expansion of the correct formula:
E=\frac{m_0c^2}{\sqrt{1-(v/c)^2}}.
Like any Taylor expansion, it is valid only for \frac{v}{c}&lt;&lt;1.
It is not valid at large speeds.

 

[stevmg]

Not quite, this is the resultant of doing the Taylor expansion of the correct formula:
E=\frac{m_0c^2}{\sqrt{1-(v/c)^2}}.
Like any Taylor expansion, it is valid only for \frac{v}{c}&lt;&lt;1.
It is not valid at large speeds.

So we are clear in future reference:

m₀ is "rest mass"

m(v) = m₀/SQRT[1 - (v/c)²] You have stated before andI have read that the term "relativistic mass" is in disfavor. It is not "inertial mass" is it?

My only references are AP French and you follks. I have no other sources here.

 

[starthaus]

So we are clear in future reference:

m₀ is "rest mass"

m(v) = \frac{m_0c^2}{\sqrt{1-(v/c)^2}} is called what?

relativistic mass.

You have stated before and I have read that the term "relativistic mass" is in disfavor.

correct

It is not "inertial mass" is it?

nope.

 

[stevmg]

So, what is it? What's it called? Does it have a new name? I haven't found one in the few books that I have.

 

[starthaus]

So, what is it? What's it called? Does it have a new name?

It still has the old name but we prefer not to call it by it.

 

[DrGreg]

Just to get this clear

E = \frac{m_0 c^2}{\sqrt{1-v^2/c^2}}​

is the energy of a particle. This is the same E that appears in

E^2 = (m_0 c^2)^2 + (pc)^2​

and it's also equal to rest energy (m₀c²) plus kinetic energy. If v/c is very small, the kinetic energy approximates to ½m₀v².

If you want to give the quantity

M = \frac{m_0}{\sqrt{1-v^2/c^2}}​

a name, you can call it "relativistic mass", but the modern preference is not to use M at all but to use E instead. (The two are related by E = Mc² and so are "almost the same thing".)

 

[DrGreg]

It is also worth mentioning that the modern physicists who don't use relativistic mass will refer to rest mass (=invariant mass) as just "mass".

Old timers who do use relativistic mass will refer to relativistic mass as just "mass".

Hardly anyone uses the term "relativistic mass" except in discussions like this comparing alternative definitions. So when an author refers to "mass" (without qualifying it as "rest ..." or "relativistic ..." etc) you will need to work out which of the two camps they fall into. It's unfortunate, but there are no Physics Police enforcing the use of one term or the other.

 

[Passionflower]

Old timers who do use relativistic mass will refer to relativistic mass as just "mass".

Indeed, together with their endlessly annoying claim that if one uses an infinite amount of energy to accelerate an object with mass it will actually travels at c.

 

[starthaus]

Indeed, together with their endlessly annoying claim that if one uses an infinite amount of energy to accelerate an object with mass it will actually travels at c.

Riight :-)

 

[stevmg]

Just to get this clear

E = \frac{m_0 c^2}{\sqrt{1-v^2/c^2}}​

is the energy of a particle. This is the same E that appears in

E^2 = (m_0 c^2)^2 + (pc)^2​

and it's also equal to rest energy (m₀c²) plus kinetic energy. If v/c is very small, the kinetic energy approximates to ½m₀v².

If you want to give the quantity

M = \frac{m_0}{\sqrt{1-v^2/c^2}}​

a name, you can call it "relativistic mass", but the modern preference is not to use M at all but to use E instead. (The two are related by E = Mc² and so are "almost the same thing".)

It is also worth mentioning that the modern physicists who don't use relativistic mass will refer to rest mass (=invariant mass) as just "mass".

Old timers who do use relativistic mass will refer to relativistic mass as just "mass".

Hardly anyone uses the term "relativistic mass" except in discussions like this comparing alternative definitions. So when an author refers to "mass" (without qualifying it as "rest ..." or "relativistic ..." etc) you will need to work out which of the two camps they fall into. It's unfortunate, but there are no Physics Police enforcing the use of one term or the other.

Indeed, together with their endlessly annoying claim that if one uses an infinite amount of energy to accelerate an object with mass it will actually travels at c.

Riight :-)

Hardly anyone uses the term "relativistic mass" except in discussions like this comparing alternative definitions. So when an author refers to "mass" (without qualifying it as "rest ..." or "relativistic ..." etc) you will need to work out which of the two camps they fall into. It's unfortunate, but there are no Physics Police enforcing the use of one term or the other.

Ha! Ha! Now that is clear... Seems like my original question as to the definition of "mass" or "relativistic mass" was a legitimate one as there is no consistency and no "Physics Police" to enforce one.

It is also clear as to why one cannot use the induction method of proving the momentum-energy equations for a system of particles. To wit,

E² - (cp)² = E₀² would be true for one particle. If you were to have two particles of the same mass but going in opposite directions at v treating the combination of the two particles as a "system" would yield a total momentum of 0 and thus (cp)² would be zero in that equation as if the velocity were zero. But, in actuality, the velocity of these particles is not zero and an erroneous answer would result.

I will have to go through Space-Time Physics as suggested by starthaus to see if I can wrap my brain aound this. No more comments from me until I have done so and that may take a very long time.

Until later.