Relativistic Momentum: Mass, Time & Distance

[gulsen]

1. Why do we have to assume mass doesn't change? And always use m_0?
2. OK, let's assume we always use m_0. Then why is momentum is defined as \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}. If measuring in object's frame of reference, shouldn't we use his distance and time, where they are \frac{x_0}{\gamma(v)} and t_0 \gamma(v), and v would be \frac{x_0}{t_0 \gamma(v)^2}. It seems that we're using x_0 either t_0 and not both, nor none. Isn't this inconsistent?

 

[jtbell]

It's basically a matter of taste whether one defines relativistic momentum with one formula using rest mass, or a different formula using relativistic mass. They give the same numeric value for momentum in either case. We use one of these definitions because that is the quantity whose sum turns out to be conserved in collisions and other processes involving two or more particles. The way to derive it is to analyze a collision carefully and look for a quantity whose sum is conserved.

See http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000054000009000804000001&idtype=cvips&gifs=yes for one derivation. This is not the one that is usually given in textbooks, but the paper lists some textbooks that do contain the usual derivation. It might be on the net somewhere, but I'm not going to spend a half hour looking for it. You're welcome to do that if you want.

 

[bernhard.rothenstein]

please have a look at
http://arxiv.org./abs/physics/0505025 where a derivation of the transformation equations for mass, momentum and energy does not use conservation laws. A post use review is appreciated.

 

[gulsen]

jtbell, that link you gave seems not to be available freely (username+password, or $$$). Having not paid to AIP because I think physics is a science and not a subject of commerce, I couldn't see that paper. Thanks anyway.

Bernhard, thanks for that paper. However, according to that paper, mass change seems to be necessary. So, that would quite disagree with usenet??

 

[pmb_phy]

1. Why do we have to assume mass doesn't change?

You don't have to assume that. It follows from the definition of mass(speed) = |momentum|/speed. Those who chose rest mass over relativistic mass do something similar but defined "mass" m to be mass(0) = m. However that is a limited definition which does not hold in general. m defined in this way is not always the intrinsic property of an object as some might believe. The mass(0) = m definition holds only for isolated (aka 'closed') systems or for point particles. When it comes to fields or extended objects under stress
(or for mass density) then the definition where "m" is an invariant makes no sense.

2. OK, let's assume we always use m_0. Then why is momentum is defined as \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}.

Because those who use the term "mass" to mean "proper mass" and label it "m" express it as "m" rather than m₀.

If measuring in object's frame of reference, shouldn't we use his distance and time, where they are \frac{x_0}{\gamma(v)} and t_0 \gamma(v), and v would be \frac{x_0}{t_0 \gamma(v)^2}. It seems that we're using x_0 either t_0 and not both, nor none. Isn't this inconsistent?

If you're measuring in the objects frame of reference then \gamma = 1. I don't understand what you mean by the quantities \frac{x_0}{\gamma(v)} and t_0 \gamma(v). What is x₀? Do you mean the x-coordinate as measured in the rest frame of the object? If so then since \gamma = 1 you have not really done anything. Please clarify for me.

Note: (ct, x, y, z) is the coordinate of an event while distances and time intervals are the components of the a spacetime displacement.

Pete

 

[jtbell]

jtbell, that link you gave seems not to be available freely (username+password, or $$$).

Sorry about that. We must have an institutional account here which gives access to everyone who connects from my college's network.

Anyway, I'm sure you can find the "standard" derivation that uses a glancing collision, in many second-year level "introdution to modern physics" textbooks. I know it's in Beiser and in Taylor/Zafiriatos/Dubson. It might even be on a Web page somewhere if you plow through the hits that Google gives you for "derivation of relativistic momentum".

 

[gulsen]

pmb_phy, I'm pretty certain I'm wrong at somewhere, I just don't know where.
I assume a coordinate system, moving with -v according to the mass. So, according to mass, there should be a length contraction in coordinate system right? My question is, what if mass thinks "oh, I'm moving at v, and coordinate system stands still, from this POV, what is my momentum? my speed is: in a time interval I measure the distance I've taken (which was contracted according to and outside observer, standing still to the coordinate system). and my m is m_0. so my momentum should be m_0 (dr/gamma(dr/dt))/dt"?

What I mean by t_0 and l_0 are proper time and proper length. Okay, it seems his time is proper time, but there should be some contraction in length, right?

jtbell, in Taylor/Zafaritos, it first assumes that momentum is mdr/dt_0, then tests it in a thought experiment (well, actually it doesn't do it even, full solution it's given as an exercise!). Good assumption I say, but why to assume that.

Or is this the whole story: "find an expression that is consistent with conservation of momentum, oh, and it should include gamma somewhere".

 

[pervect]

2. OK, let's assume we always use m_0. Then why is momentum is defined as \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}.

Momentum is defined in this manner so that it remains a conserved quantity when frames of reference are changed via the Lorentz transform.

The Newtonian definition of momentum, p=mv, is compatible with the Gallilean transform - the relativistic deveition of momentum, p = mv/sqrt(1-(v/c)^2) is compatible with the Lorentz transform.

There is a detailed proof in many physics textbooks, Goldstein's "Classical mechanics" comes to mind.

 

[bernhard.rothenstein]

Momentum is defined in this manner so that it remains a conserved quantity when frames of reference are changed via the Lorentz transform.

The Newtonian definition of momentum, p=mv, is compatible with the Gallilean transform - the relativistic deveition of momentum, p = mv/sqrt(1-(v/c)^2) is compatible with the Lorentz transform.

There is a detailed proof in many physics textbooks, Goldstein's "Classical mechanics" comes to mind.

Have a look please at a paper
why p=gamamv? because of the principle pf relativity
http:arxiv.org/abs/physics/0404059

 

[DrGreg]

For what it's worth, you can find Einstein's own derivation for relativistic momentum in his paper Elementary derivation of the equivalence of mass and energy, Bull. Amer. Math. Soc. 37 (2000), 39 - 44.

(In this paper, Einstein uses the word "impulse" instead of "momentum", and adopts the unstated convention that c = 1.)

But more modern texts are probably easier to read than this.

 

[DrGreg]

Have a look please at a paper
why p=gamamv? because of the principle pf relativity
http:arxiv.org/abs/physics/0404059

I am having some problems understanding this paper. I don't see how equation (11) was obtained, and equation (14) doesn't even make sense, as it equates a velocity to a momentum!

 

[bernhard.rothenstein]

I am having some problems understanding this paper. I don't see how equation (11) was obtained, and equation (14) doesn't even make sense, as it equates a velocity to a momentum!

Please move (m) from the right side to the left one in (14).
With a little perseverence you can derive (11) using the addition law of relativistic velocities.

 

[pmb_phy]

pmb_phy, I'm pretty certain I'm wrong at somewhere, I just don't know where.

Then perhaps what you're looking for is in this paper -

http://www.geocities.com/physics_world/mass_paper.pdf

I wrote it specifically for people who have questions regarding mass in relativity, just like yourself. I also shined some light on well-established but little known facts of mass in relativity. I used one in a paradox regarding the mass density of a magnetic field.

Let me know what you think of it. I've been having a hard time getting feedback on it. What little I got was very good though.

Pete

 

[duordi]

I enjoyed your paper.

Let me describe what I think I understand and you can point out the errors.

Mass A is near several masses B, C, and D which have random velocities.

Each of the masses B, C, and D see a distortion of mass A depending on their velocity with respect to A.

Each of the masses B, C, and D have a distortion component and the “rest energy” component.

The energy required to accelerate A in a particular direction as perceived by B will depend on the “rest energy” of A and the velocity A has with respect to B in the vector direction A is to be accelerated.

Mass C and D do not experience the velocity mass B has attributed to A but only the “rest energy” of A and then of course C and D each have a perceived mass distortion depending on their differential velocity with respect to A.

I also have a question.

If A is accelerating with respect to B will B perceive A to have an apparent mass distortion from the “rest energy” mass of A?

Duane

 

[pmb_phy]

I enjoyed your paper.

I'm glad to hear that Duane. Thanks for letting me know.

Let me describe what I think I understand and you can point out the errors.

I'll do my best.

Mass A is near several masses B, C, and D which have random velocities.

So particles A, B, C and D for a system of particles with random velocities..

Each of the masses B, C, and D see a distortion of mass A depending on their velocity with respect to A.

What kind of distortion? Gravitational?

Each of the masses B, C, and D have a distortion component and the “rest energy” component.

I don't understand what this means. Particles B, C and D have a "rest energy" component? Do you mean that there is a rest mass associated with the system of the B,C.D particles?

The energy required to accelerate A in a particular direction as perceived by B will depend on the “rest energy” of A and the velocity A has with respect to B in the vector direction A is to be accelerated.

If you're asking whether the gravitational force is a velocity dependent one then yes, it is.

Mass C and D do not experience the velocity mass B has attributed to A but only the “rest energy” of A and then of course C and D each have a perceived mass distortion depending on their differential velocity with respect to A.

Sorry. You lost me. There is nothing about which indicates that C and D do not interact with A and B.

I also have a question.

If A is accelerating with respect to B will B perceive A to have an apparent mass distortion from the “rest energy” mass of A?

Duane

What kind of distortion?

Pete

 

[duordi]

Sorry for the terminology.

Mass A has a rest mass.

Mass B has a velocity with respect to mass A.

Mass B has a gravitational force toward A and the mass which is used in the gravitational equation to determine the gravitational force is the relativistic mass of A as perceived by B.

So mass A would "appear" to have more mass then its rest mass.

Is this correct?

 

[pervect]

Sorry for the terminology.

Mass A has a rest mass.

Mass B has a velocity with respect to mass A.

Mass B has a gravitational force toward A and the mass which is used in the gravitational equation to determine the gravitational force is the relativistic mass of A as perceived by B.

So mass A would "appear" to have more mass then its rest mass.

Is this correct?

If you intend to use the Newtonian force equation GmM/r^2 with m being the relativistic mass of the moving mass, this is most definitely not correct.

To get a really correct answer, one needs to think about the gravitational field as being "tidal forces", because this is what one can actually measure at a point. One can also describe these tidal forces as "curved space-time" - these two descriptions are equivalent.

To get an approximately correct answer, one can think of the gravitational field as being not spherically symmetrical, i.e. one can think of the field as being "squished" so that the transverse field is stronger, much like the electric field of a moving charge.

However, if one does not take into account the issues raised with regards to how one defines the gravitational field, the above description will still not be exactly correct, though it does give one a rough idea of what happens.

Using the spherically symmetrical Newtonian formula is "right out".

 

[duordi]

I understand that curved space has complicated equations.

Does it simplify in the case I just described if masses A and B are moving directly away from each other and what would the true equation be for the force of gravity on B which is considered the stationary objcect toward A with is considered to be moving?

If it is to difficult to explain don’t bother posting it.

I don’t know if I can handle 10 pages of equations.

Duane

 

[duordi]

I have a second question.

Mass A and B are as before.

A is seen by B as being flattened along the line of site between A and B.

If A consisted of two objects, A1 and A2.
A1 AND A2 are on a line of site from B to A.
A2 is farther away from B then A1.
Would the distance between the two objects A1 and A2 also seem shorter to B due to the contraction?

Duane

 

[MeJennifer]

I have a second question.

Mass A and B are as before.

A is seen by B as being flattened along the line of site between A and B.

If A consisted of two objects, A1 and A2.
A1 AND A2 are on a line of site from B to A.
A2 is farther away from B then A1.
Would the distance between the two objects A1 and A2 also seem shorter to B due to the contraction?

Duane

Yes, in the special theory of relativity it is not only objects that contract, the whole space in the direction of the relative motion contracts.


Look at the beauty of relativity:

When a smaller box s is situated, relatively at rest, inside the hollow space of a larger box S, then the hollow space of s is a part of the hollow space of S, and the same "space", which contains both of them, belongs to each of the boxes. When s is in motion with respect to S, however, the concept is less simple. One is then inclined to think that s encloses always the same space, but a variable part of the space S. It then becomes necessary to apportion to each box its particular space, not thought of as bounded, and assume that these two spaces are in motion with respect to each other.

Before one has become aware of this complication, space appears as an unbounded medium or container in which material objects swim around. But it must be remembered that there is an infinite number of spaces, which are in motion with respect to each other.

The concept of space as something existing objectively and independent of things belongs to pre-scientific thought, but not so the idea of the existence of an infinite number of spaces in motion relatively to each other. This latter idea is indeed unavoidable, but is far from having played a considerable role even in scientific thought.

Fascinating isn't it?

 

[duordi]

Reality just kind of sucks you into it doesn't it.

I had two posts.

I was kind of hoping there would be an easy transformation from Newton’s gravity equation to space time if certain conditions were met without eliminating the velocity difference, like that A and B are moving directly away from each other or something to make the system symmetrical.

Did you miss it or was the relationship just to complicated to discuss?

Also, I have always imagined that relativistic mass was just a distortion of the information received about mass A as observed by mass B due to the motions of mass A.

Is this a dangerous view point?

would the masses A1 and A2 appear from B to interact ( fall toward each other) as if they had relativistic mass ( or whatever space time predicted ) or would they interact as if they have their rest mass.

I know that to an observer on A1 or A2 which are moving together they would experience rest mass attractions.

Duane

 

[pmb_phy]

Sorry for the terminology.

Mass A has a rest mass.

Mass B has a velocity with respect to mass A.

Mass B has a gravitational force toward A and the mass which is used in the gravitational equation to determine the gravitational force is the relativistic mass of A as perceived by B.

This isn't neccesarily true Duane. If body B is moving towards body A then the hte gravitational force may not neccesarily be parallel to the line connecting the two. I'd have to calculate it to see that. Also the force on body A is due to more than just the relativistic mass, the force is also due to the fact that the gravitational force is velocity dependent. There are two velocity dependancies here (1) one due to relativistic mass and (2) one due to the velocity dependent gravitational force. I worked out an example and placed it on my website here

http://www.geocities.com/physics_world/gr/force_falling_particle.htm

Notice Eq. (12). There is an extra factor for the velocity dependent force.

So mass A would "appear" to have more mass then its rest mass.

Is this correct?

Actually it would be the moving (i.e. relativistic) mass which "appears" to have more mass. However (relativistic) mass gets altered just by its presence of being in the field.

Pete

 

[pervect]

I have some serious issues with Pete's calculations, as I've mentioned before. I think Chris Hillman does too, though he generally doesn't think it's worth his time to do "debunking".

 

[pervect]

If you have a small test mass falling into a black hole on a radial path, one can say a couple of things:

1) The tidal stretching force experienced by the infalling observer will be 2GM/r^3, where r is the schwarzschild coordinate. This is the same formula as the formula for the Newtonian tidal force (except for replacing the distance r by the Schwarzschild coordinate r) and is independent of the radial velocity of the falling object.

(Interestingly enough, the non-radial velocity of the falling object does matter, it increases the tidal force, though this result is not in the textbooks but must be calculated.)

See for instance "Gravitation" by Misner, Thorne, Wheeler for a detailed derivation. (I can look up the page # on request if anyone is really interested).

The tidal stretching force on an observer falling through the event horizon of a black hole is thus finite and the same as the tidal stretching force due to Newtonian gravity. A naive interpretation using "relativistic mass" would give the incorrect result of an infinite tidal force for an object falling into a black hole (as the velocity of the infalling mass approaches the speed of light at the event horizon relative to a stationary observer).

Also, there is an interesting paper which describes the total imparted velocity from a relativistic flyby. The result is also not equal to the "relativistic mass", there is an additional factor of (1+\beta^2)

http://dx.doi.org/10.1119/1.14280

 

[duordi]

Beggars can't be choosers.
It’s my responsibility to find out if this is correct is it not?
Besides if it is too easy I will not learn anything.

I will see how much of it I can understand.

I may go over you web page as it seems to have some sort of tutorials.

Duane

 

[pmb_phy]

Beggars can't be choosers.

Huh? Where did that come from?

It’s my responsibility to find out if this is correct is it not?

I personally have taken a great deal of responsibility when I chose to undertake helping people understand physics so I'm hear as long as you need me. It may just be that I need to come to understand the language that you're lossing better or you mine.

Besides if it is too easy I will not learn anything.

That's the spirit. You have a fine way of looking at learning physics. Bravo good sir, bravo!

I may go over you web page as it seems to have some sort of tutorials.

Duane

Duane. I neglected to mention that just because something is more gravitationally attracted than something else it doesn't mean that its motion will otherwise be differrent. But in the case I showed you it is the gravitational force whose magnitude I was referring to and that tells you how hard it is to puch such an object off of its natural course (aka "worldline").

Pete

 

[pervect]

And I have serious issuse with your poor attitude. Posts like this with an attitude such as yours clearly violates forum posting guidelines. It appears to me that you are unable to restrain yourself from these childish responses which serve no purpose other than to inflate your own ego. Please do not drag another poster in and attempt to speak for him. If he wanted to he to could post an insulting post which will also be submitted for deletion.

Its rude attitudes from people like you that make posting on the internet a lot less enjoyable and you need to take a good look at yourself before you continue to act in this purile manner.

I basically am trying to be quite polite, and I really don't see how informing people that in my opinion your pet theories are "personal pet theories" rather than part of the mainstream is violating forum guidelines.

Hopefully, the moderators will inform me if they feel like I am stepping out of bounds, or even if they just have some suggestions as to how I could handle the situation better. I am interested in how to handle this issue as peacefully and with as little drama as possible, while still meeting my goals of providing accurate information about relativity theory to those who are interested. (On a personal note, it seems to me that we've had more than our share of drama here at PF recently, we really don't need any more.)

If you could show us where your pet theories were published in a peer reviewed journal, things might be different.

While I'm sorry we haven't been able to resolve this issue, I really do have some serious issues with your approach.

At the moment I wouldn't have all that much time to argue the details anyway, so I'll stick with providing a clue to people who are not well-read in the field that your views may not be widely accepted unless some better idea presents itself.

The example of finite tidal forces for someone approaching a black hole at arbitrarily large relativistic velocities is, I think, a good caution to give anyone in any event, even if turns out that there is some way of plugging up what I perceive as holes in your argument.

Someone who takes the idea of "relativistic mass" seriously in a naive manner would probably predict an infinite tidal force for an object approaching a black hole at a relativistic velocity, and this turns out not to be the case. In fact, the tidal forces for two objects at the same event are independent of radial velocity (though they are not independent of other "transverse" components of velocity).

 

[Gokul43201]

pete, you must admit that at the very least, your article opposes the mainstream from the points of view of aesthetics and more importantly, pedagogy. That last one alone is sufficient reason to disallow it here. There's a place on these forums, where we specifically permit such submissions, and you know it is not here.

pervect, I understand your desire to state your objections, but I think you should perhaps let Chris speak for himself (even if there's little doubt as to what he will likely say).

 

[MeJennifer]

pete, you must admit that at the very least, your article opposes the mainstream from the points of view of aesthetics and more importantly, pedagogy. That last one alone is sufficient reason to disallow it here.

I think that is a rather restricted and dogmatic approach to teaching.
As long as what is shown is not wrong and instead is just another perspective on the theory of relativity it would be in the interest of pedagogy IMHO rather than counter to it. To see things from different perspectives enriches one's understanding.

And if we want to be consistent it seems to me that for instance frame fields are not mainstream either. But that does not mean that they are not useful. Would you argue to treat those the same way?

One personal note though, I could live without endless identical arguments in each topic that contains the word mass, that certainly does not serve any good.

 

[pmb_phy]

pete, you must admit that at the very least, your article opposes the mainstream from the points of view of aesthetics and more importantly, pedagogy.

What I posted came from "Gravitation," by Misner, Thorne and Wheeler as well as "A first course in general relativity," by Bernard F. Schutz. These two text are about as mainstream as you can get. Then there are text such as thaty Mould, Rindler and D'Inverno. All rather new and all contain the idea of relativistic mass. Because I hold relativistic mass (aka "inertial mass") one shouldn't attack me for holding this perfectly valid opinion. When I think that a p[roblem could exist in interpreation I make sure to state that this is a subject matter of my own personal opinion.

That last one alone is sufficient reason to disallow it here. There's a place on these forums, where we specifically permit such submissions, and you know it is not here.

I see no valid reason to prohiit these things from what a person posts. If 35% of physicsts use inertia mass (aka "relativistic mass") then count me in that 35%. Should we ban that 35% of the relativistic community because their field makes things simplier when using relativistic mass?

Mind you I normally use rel-mass here when people are asking about it or when he answere is easily answred by using rel-mass. This does not make me a bad person. Frankly I really don't care what a person uses so long as they're clear about it.

Pete