Conservation of momentum a universal truth

[stevmg]

Wherever I have read, the conservation of momentum is a universal truth. This is the basis, as I undertsand it, to the mass equation:

m = m₀/SQRT(1 - v²/c²) in which m₀ = the mass in the rest frame and m is the mass in the moving frame relative to the rest frame moving at v.

The idea behind it, as I understand it, that in order to conserve momentum, one must increase the mass to keep the mv product the same, as the "moving frame" is time dilated and v is functionally less than the v in the rest frame.

Does this automatically happen?. The interesting part about this is that the v in the moving frame is "slowed down" by time dilation yet the v is perpendicular to the path of the moving frame. I guess time dilation must occur in all three axes (x, y, z) as it would be impossible for an object to exist at one time in x and "simultaneously" other times in y and z.

 

[map19]

At speeds we are used to everyday, momentum is a reduction of that formula to p = mv.
Normally we only consider v to be in the x direction.
v in several axes at once gets very confusing.
while momentum is a handy measure for doing calcs of energy distribution it is not in itself very useful.
The kinetic energy of a moving body, relative to the rest frame is, as I'm sure you know, is Ek = 1/2mv^2
v is not 'slowed down' the time dilation refers to any processes on the moving body as seen from the rest frame. The observers on the moving body see the rest frame as being slower. this is the basis of relativity.

 

[stevmg]

At speeds we are used to everyday, momentum is a reduction of that formula to p = mv.
Normally we only consider v to be in the x direction.
v in several axes at once gets very confusing.
while momentum is a handy measure for doing calcs of energy distribution it is not in itself very useful.
The kinetic energy of a moving body, relative to the rest frame is, as I'm sure you know, is Ek = 1/2mv^2
v is not 'slowed down' the time dilation refers to any processes on the moving body as seen from the rest frame. The observers on the moving body see the rest frame as being slower. this is the basis of relativity.

I know, v is not slowed down but momentum is "slowed down" so to speak which requires the gamma kick on the mass to up it back to where it should be.

Ain't life grand?

 

[starthaus]

Wherever I have read, the conservation of momentum is a universal truth. This is the basis, as I undertsand it, to the mass equation:

m = m₀/SQRT(1 - v²/c²) in which m₀ = the mass in the rest frame and m is the mass in the moving frame relative to the rest frame moving at v.

The idea behind it, as I understand it, that in order to conserve momentum, one must increase the mass to keep the mv product the same, as the "moving frame" is time dilated and v is functionally less than the v in the rest frame.

Does this automatically happen?. The interesting part about this is that the v in the moving frame is "slowed down" by time dilation yet the v is perpendicular to the path of the moving frame. I guess time dilation must occur in all three axes (x, y, z) as it would be impossible for an object to exist at one time in x and "simultaneously" other times in y and z.

Energy-Momentum conservation has nothing to do with "relativistic mass", nor does it have anything to do with the definition of relativistic momentum as $$\vec{p}=\gamma m_0\vec{v}$$.

 

[Mentz114]

Stevmg, energy and momentum are not invariant under Lorentz transformations.

 

[stevmg]

Energy-Momentum conservation has nothing to do with "relativistic mass", nor does it have anything to do with the definition of relativistic momentum as $$\vec{p}=\gamma m_0\vec{v}$$.

Stevmg, energy and momentum are not invariant under Lorentz transformations.

Well, I guess you are both saying the same thing so it would appear that momentum DOES change when looking at it from different frames of reference.

Back to AP French and to try to decipher what the ---- this is all about. The preservation of momentum was presented to me in a math blog - Karl's Calculus Tutor just to let you know that I really wasn't on Pluto when I came up with this.

I've attached the Karl's Calculus Tutor reply to my question.

 

[Phrak]

It may be useful to consider the prerelativity, Galilean transformation. The momentum of a particle is not the same for two observers in relative motion. It continues to hold true in relativity. The conserved quantity in special relativity is E²- c²p².

 

[map19]

Let's try a practical example.
two meteorites collide in space.
This collision is observed from 3 different planets. Thus the meteorites have 3 different sets of relative velocities.
each planetary observer sees total momentum before the collision equal to total momentum after the collision.
Thus the 3 frames of reference all see momentum conserved, regardless of the relative speeds of the meteorites.
But these are three Different Values of momentum. because they are measured from 3 different frames of reference.
See: Modern Physics by Krane.

 

[starthaus]

Well, I guess you are both saying the same thing so it would appear that momentum DOES change when looking at it from different frames of reference.

You are comingling two DIFFERENT terms:

-frame invariance
-conservation

we are telling you that:

-momentum is NOT frame-invariant
-momentum IS conserved

 

[Dale]

Starthaus is correct. And the same applies to energy. It is not frame invariant but it is conserved.

The quantity Phrak mentioned is both frame invariant and conserved.

 

[stevmg]

"Momentum is not frame invariant..." [Double negative here...]

Does that mean it varies according to frame?

"Momentum is conserved." Does that mean within each frame it is "preserved?" Or does it mean that momentum is preserved across all frames.

Got to get on the same page.

The article from Karl's Calculus Tutor is trying to show that by conserving momentum across frames, the mass must go up (by gamma) to preserve the momentum.

When I used the term relativistic mass above I meant $$\gamma$$m₀. Maybe that is the wrong term.

 

[DrGreg]

"Momentum is not frame invariant..." [Double negative here...]

Does that mean it varies according to frame?

"Momentum is conserved." Does that mean within each frame it is "preserved?" Or does it mean that momentum is preserved across all frames.

I will quote myself from another thread (click on the little blue arrow to see it in context):

It is perhaps worth mentioning two concepts, "conservation" and "invariance", which can sometimes be confused with each other.

A conserved quantity is something measured by a single observer that doesn't change over time; for example it has the same value before and after a collision, and typically it is the sum of several measurements, e.g. of multiple particles. Examples are energy (a 1D number), momentum (a 3D vector), four-momentum (a 4D vector), all when there are no external forces, of course. In Newtonian physics, mass is also conserved. In relativity, relativistic mass may be conserved (in the absence of any other form of energy) but rest mass isn't.

An invariant quantity is a single measurement whose value all observers agree upon, i.e. a frame-independent value. Examples are proper time, (scalar) proper acceleration, and rest mass. Or anything that can be expressed in the form $g_{ab}U^aV^b$ (where U and V are genuine 4-vectors).

So, energy and momentum are both conserved but neither is invariant. In relativity, rest mass is invariant but not necessarily conserved across multi-particle interactions. (In Newtonian physics, mass is both conserved and invariant.)

 

[starthaus]

"Momentum is not frame invariant..." [Double negative here...]

Does that mean it varies according to frame?

I would recommend that you read carefully DrGreg's excellent post above mine. I will put DrGreg's post in a mathematical format.

Yes, momentum, being dependent on velocity, varies from frame to frame. If you have two frames , F and F' moving wrt each other , the velocity of a partcle is $$\vec{u}$$ in F and $$\vec{u'}$$ in F'. Thus , the momentum is $$\gamma(u)m_0 \vec{u}$$ in F and $$\gamma(u')m_0 \vec{u'}$$ in F'. So, momentum is not frame invariant. Neither is total energy $$\gamma(u)m_0c^2$$.
Yet, the energy-momentum $$E^2-(\vec{p}c)^2$$ is frame invariant for a system of particles.

"Momentum is conserved." Does that mean within each frame it is "preserved?" Or does it mean that momentum is preserved across all frames.

Neither. It means exactly what it means in Newtopnian mechanics, the variation of momentum of a system is equal to the sum of the external forces applied to the system:

$$\frac{d \vec{P}}{dt}=\Sigma \vec{F}$$

If $$\Sigma \vec{F}=0$$ then $$\vec{P}=constant$$.
If total momentum is conserved in one frame, it is conserved in all frames.

 

[stevmg]

I will quote myself from another thread (click on the little blue arrow to see it in context):

Maybe my use of "relativistic mass" was incorrect to describe what I was trying to say. the equation of m = m₀$\gamma$ I used m₀$\gamma$ as "relativistic mass."

Karl, from Karl's Calculus Tutor tries to explain this that one must use whatever you want to call this above equation to preserve momemntum over different frames.

The question was, how did we derive m = m₀$\gamma$? What is the basis of it? Why does mass increase in the moving frame in relation to the orignal "rest frame?"

 

[stevmg]

I would recommend that you read carefully DrGreg's excellent post above mine. I will put DrGreg's post in a mathematical format.

Yes, momentum, being dependent on velocity, varies from frame to frame. If you have two frames , F and F' moving wrt each other , the velocity of a partcle is $$\vec{u}$$ in F and $$\vec{u'}$$ in F'. Thus , the momentum is $$\gamma(u)m_0 \vec{u}$$ in F and $$\gamma(u')m_0 \vec{u'}$$ in F'. So, momentum is not frame invariant. Neither is total energy $$\gamma(u)m_0c^2$$.
Yet, the energy-momentum $$E^2-(\vec{p}c)^2$$ is frame invariant for a system of particles.



Neither. It means exactly what it means in Newtopnian mechanics, the variation of momentum of a system is equal to the sum of the external forces applied to the system:

$$\frac{d \vec{P}}{dt}=\Sigma \vec{F}$$

If $$\Sigma \vec{F}=0$$ then $$\vec{P}=constant$$.
If total momentum is conserved in one frame, it is conserved in all frames.

Exellent post. Makes AP French more "palatable."

But, the question is - what makes this rest mass in different frames change according to the frame? As you said, if momentum is preserved in one frame it is preserved in all. Is that what "forces" the rest mass to vary according to what frame you are in? According to Karl, it does, but he is a mathematician and not a physicist.

 

[starthaus]

.

The question was, how did we derive m = m₀$\gamma$? What is the basis of it? Why does mass increase in the moving frame in relation to the orignal "rest frame?"

"We" didn't. Einstein did, long ago. See here

The usage of "relativistic mass" is in disfavor these days. We use instead the relativistic energy $$\gamma(u) m_0c^2$$ and the relativistic momentum$$\gamma(u)m_0\vec{u}$$.
If you want to learn how we arrived to these definitions, you need to get a good book, like "Spacetime Physics" by Taylor and Wheeler and start studying. Cobbling info from websites like "Kurt Calculus" will not do, sorry.

 

[stevmg]

"We" didn't. Einstein did, long ago. See here

The usage of "relativistic mass" is in disfavor these days. We use instead the relativistic energy $$\gamma(u) m_0c^2$$ and the relativistic momentum$$\gamma(u)m_0\vec{u}$$.
If you want to learn how we arrived to these definitions, you need to get a good book, like "Spacetime Physics" by Taylor and Wheeler and start studying. Cobbling info from websites like "Kurt Calculus" will not do, sorry.

That's what I have read ("relativistic mass" is in disfavor.)

I have seen that book Spactime Physics somewhere recently. Not a cheap one so I will get it from the library, even if it takes three months to get it. Will take a lot of energy to get into.

Don't knock Karl - he does OK but he is not a physicist.

Will get back later on it as this is NOT number one on my priority list right now. This is mental exercising. There is a real world out there.

 

[starthaus]

That's what I have read ("relativistic mass" is in disfavor.)

I have seen that book Spactime Physics somewhere recently. Not a cheap one so I will get it from the library, even if it takes three months to get it. Will take a lot of energy to get into.

Don't knock Karl - he does OK but he is not a physicist.

Will get back later on it as this is NOT number one on my priority list right now. This is mental exercising. There is a real world out there.

I am not knocking Karl, I am criticizing your approach to learning. Spacetime Physics can be bought (used) for about 15$ from Amazon.com. I highly recommend it.

 

[Altabeh]

A conserved quantity is something measured by a single observer that doesn't change over time...

Greg, you have to modify this in such a way that the conservation of momentum can also be included in there in general. That is just correct for the conservation of energy; as we all know the momentum has to remain invariant over the spatial translations not time translations in order for it to be conserved.

AB

 

[stevmg]

I am not knocking Karl, I am criticizing your approach to learning. Spacetime Physics can be bought (used) for about 15$ from Amazon.com. I highly recommend it.

I am 67 years old, hold a college degree, a MA, a MD degree, am board certified in 2 specialties and have been practicing medicine for longer than you are alive. I have read countless medical books, journals, monographs and have outstanding board scores in both my specialties. I think my approach to learning has spoken for itself. This venture into spacetime physics, etc. is to satisfy a curiosity I have had for years but never pursued because I was reading medical articles (I didn't want to kill anybody through ignorance) rather than "free-lancing." Fredrik and others have told me to continue to write "simplistic" questions on this forum (I did ask them for a another site so as to not clog up the forum with "trivial" questions, but they said "no, continue to write." ) I have learned more by this method in the past two months than ever before on this subject and I have accompanied the questions by reading textbooks suggested by DaleSpam and JesseM. Spacetime Physics was one of them but it was too advanced for me but the AP French Special Relativity book was more in line with what I knew and my level of comprehension of physics.

I wish I could find Taylor's Spacetime Physics book for $15 but it isn't there. I have looked and looked. It is not on Amazon, that's for sure. If you hear of a site please let me know at stevmg@yahoo.com as I will be most grateful.

Verstehst du? Caspisci? Entiendes?

 

[matheinste]

I wish I could find Taylor's Spacetime Physics book for $15 but it isn't there. I have looked and looked. It is not on Amazon, that's for sure. If you hear of a site please let me know at stevmg@yahoo.com as I will be most grateful.

Verstehst du? Caspisci? Entiendes?

If you are referring to Taylor and Wheeler, Spacetime Physics, you can pick it up on line, secondhand, from AbeBooks, at around that price.

Matheinste.

 

[starthaus]

Spacetime Physics was one of them but it was too advanced for me but the AP French Special Relativity book was more in line with what I knew and my level of comprehension of physics.

AP French is also good.

I wish I could find Taylor's Spacetime Physics book for $15 but it isn't there. I have looked and looked. It is not on Amazon, that's for sure.

http://www.abebooks.com/servlet/SearchResults?an=Taylor&sts=t&tn=Spacetime+Physics&x=44&y=5

 

[stevmg]

If you are referring to Taylor and Wheeler, Spacetime Physics, you can pick it up on line, secondhand, from AbeBooks, at around that price.

Matheinste.

AP French is also good.




http://www.abebooks.com/servlet/SearchResults?an=Taylor&sts=t&tn=Spacetime+Physics&x=44&y=5

Gentlemen,

I have ordered the Taylor book from AbeBooks.com for the $13 price. When I have time I will peruse it along with the AP French book I already have.

Thank you both for the ordering info.

Steve Garramone, MD
Melbourne, FL

 

[starthaus]

Gentlemen,

I have ordered the Taylor book from AbeBooks.com for the $13 price. When I have time I will peruse it along with the AP French book I already have.

Thank you both for the ordering info.

Steve Garramone, MD
Melbourne, FL

Excellent, once you get it, check chapter 2 , Energy and Momentum.

 

[Dale]

But, the question is - what makes this rest mass in different frames change according to the frame? As you said, if momentum is preserved in one frame it is preserved in all. Is that what "forces" the rest mass to vary according to what frame you are in? According to Karl, it does, but he is a mathematician and not a physicist.

Hi stevmg,

In https://www.physicsforums.com/showthread.php?t=382591&highlight=stevmg&page=9" I briefly introduced you to the concept of four-vectors which you at first were hesitant about but then seemed to really like. So, I would like to extend that approach just a little bit for you.

You have already seen the position-time four-vector, (ct,x,y,z), and seen how it can be used with a matrix representation of the Lorentz transform, and you have also seen how the spacetime interval s²=c²dt²-dx²-dy²-dz² is the same in any reference frame (invariant) and is proportional to the time measured by a clock.

Similarly, there is another four-vector called the energy-momentum four-vector or the http://en.wikipedia.org/wiki/Four-momentum" . It is defined as (E/c,px,py,pz) where E is the total energy and px, py, and pz are the x, y, and z components of the momentum respectively. The nice thing about this four-vector is that it also transforms the same way, so if you want to know what the energy and momentum are like in another frame all you have to do is to multiply it by that same matrix that you used to transform the position and time coordinates. Also, just like the interval is the same in any reference frame, so also is the quantity m²c²=E²/c²-px²-py²-pz² where m is known as the invariant mass since all frames will agree on its value (aka "rest mass", although I prefer the term "invariant mass").

So, when you look at the four-momentum you see that energy and momentum are frame variant, or relative, in the same sense that time and space are. Furthermore, you see that energy has the same relationship to momentum as time has to space. Also, although energy and momentum are frame variant, there is a combination of the two which is frame invariant and is proportional to the invariant mass.

 

[DrGreg]

Greg, you have to modify this in such a way that the conservation of momentum can also be included in there in general. That is just correct for the conservation of energy; as we all know the momentum has to remain invariant over the spatial translations not time translations in order for it to be conserved.

My explanation was aimed at people just beginning to get to grips with relativity, not intended to be a rigorous definition.

 

[starthaus]

Also, just like the interval is the same in any reference frame, so also is the quantity m²c²=E²/c²-px²-py²-pz² where m is known as the invariant mass since all frames will agree on its value (aka "rest mass", although I prefer the term "invariant mass").

This is a very nice post. Have you found a reference to the fact that m²c²=E²/c²-px²-py²-pz² holds for a system of particles also, not only for individual particles? I managed to prove it (the proof isn't trivial) but I haven't seen the proof in any book. (maybe in Griffiths' "Introduction to Particle Physics"? I don't have the book).

ETA: I found a proof in Rindler , pages 117-118. Takes him 1.5 pages of calculations to prove it. Interesting discussion on the fact that the total energy-momentum of the systm is not trivially a four-vector, one needs to sweat it out.

 

[stevmg]

This is a very nice post. Have you found a reference to the fact that m²c²=E²/c²-px²-py²-pz² holds for a system of particles also, not only for individual particles? I managed to prove it (the proof isn't trivial) but I haven't seen the proof in any book. (maybe in Griffiths' "Introduction to Particle Physics"? I don't have the book).

The concept that this energy-momentum "invariance" fpr a system of particles when looked at in toto is stated in AP French but not proven.

The only way I could think of proving it using what I remember from mathematics would be by induction In other words if one can show that it is true for n particles, it is true for n+1 (or if true for n, was true for n-1) then one can claim the principle is true for n = 1 to any number of particles.

I presume that is what you did.

I await the book Spacetime Physics to see this more clearly. But, again, there are other pressing matters on me as this is an intellectual past time and not something I use in the "real world."

 

[starthaus]

The concept that this energy-momentum "invariance" fpr a system of particles when looked at in toto is stated in AP French but not proven.

Because the proof is quite tricky.

The only way I could think of proving it using what I remember from mathematics would be by induction In other words if one can show that it is true for n particles, it is true for n+1 (or if true for n, was true for n-1) then one can claim the principle is true for n = 1 to any number of particles.

I presume that is what you did.

No, I didn't. It is much more complicated than that.

 

[stevmg]

So much for my guessing. Yes, I am sure it was much more complicated than that.

Enough of speculation on my part. I have to get into that Taylor book to even get a miniscule concept of what this is all about.

I cannot believe that I was the one who even started this thread.

 

[stevmg]

This is a very nice post. Have you found a reference to the fact that m²c²=E²/c²-px²-py²-pz² holds for a system of particles also, not only for individual particles? I managed to prove it (the proof isn't trivial) but I haven't seen the proof in any book. (maybe in Griffiths' "Introduction to Particle Physics"? I don't have the book).

ETA: I found a proof in Rindler , pages 117-118. Takes him 1.5 pages of calculations to prove it. Interesting discussion on the fact that the total energy-momentum of the systm is not trivially a four-vector, one needs to sweat it out.

I like that m²c²=E²/c²-px²-py²-pz² - looks neat.

I looked this book up on Google Books and it got me to page 117 - just for grins - and that's just the start of the proof - The Zero Momentum Frame. The book is $97 somewhere. AbeBooks doesn't have it.

You're right... It isn't simple. We'll forgo that pleasure. I still like my old induction approach - if it's good for one, it's good for two, etc. all the way on up. I know that isn't right but at least I can conceptualize it. In AB French he sort of does an "induction" approach to justifying the concept verbally (no formal proof.) He uses the fact that the momenta are linearly related which allows for this "addition" or $$\sum$$ concept. I have used that before when we studied vectors in ejection bailout systems for jet aircraft. If the three-space vectors are not linear, you cannot add them up and get a good answer. Had to use the numerical solutions of calculating each minute step along the way.

My son lives a few blocks from UT Texas Dallas where Wolfgang Rindler teaches and if I really needed it I could get him to go the llibrary there and Xerox those pages.

 

[stevmg]

Is this the proof you saw in 2006 Rindler on page 117-18?

No way would I ever understand that. I would have to spend the last 50 years since high school getting it straight and even then wouldn't get it.

I'll accept the premise that one can find energy-momentum equations for systems of particles without understanding the proof.

The equation that DaleSpam gave (m²c²=E²/c²-px²-py²-pz²) is really another way of stating the AP French equation on page 213 of Special Relativity

-E₀²/c² = p_x² + p_y² + p_z² + (iE/c)²

That's good enough for me.

 

[starthaus]

Is this the proof you saw in 2006 Rindler on page 117-18?

Yes.

No way would I ever understand that. I would have to spend the last 50 years since high school getting it straight and even then wouldn't get it.

No one said that physics is easy. I have a more direct proof than Rindler's but it isn't trivial by any stretch of imagination.

I'll accept the premise that one can find energy-momentum equations for systems of particles without understanding the proof.

This is a bad idea, we need to strive to understand the proofs. Otherwise, we become robots that plug in numbers into equations that we don't understand. Or worse, we use equations that do not apply.

 

[stevmg]

OK, but if it takes me 50 more years to understand it, I will give you the derivation in the old soldiers' home.

I will first read space-time physics and then maybe invest in Rindler.

The price we pay for our intellectual curiosity.