Modern Special Relativity and Mass

[DW]

Some introductory texts and most "popular" literature has not kept up with modern conventions and mathematical techniques employed in the modern world of relativity. I see some of the confusions caused by this has influenced board members here as well as many other forums. So, I decided to write up a summery of how quantities in dynamics are formulated and defined in modern "special" relativistic physics. Hopefully this will serve to plant a nail in the coffin of obsolete conventions such as "relativistic mass" and "ict".
Consider the rest frame of something we wish to write the equations of physics for. According to this frame the spatial components of its momentum will be zero. The amount of energy that it may have according to this frame is how we define mass m and the energy according to its rest frame is written $$E_{0}$$. So
$$m \equiv \frac{E_0}{c^2}$$
We define a four element vector constructed according to this frame by its time element $$P^{0}$$ given by
$$P^{0} = \frac{E}{c}$$
and the three other components zero for this frame is what we will call the momentum. This results in the following.
$$\left[P'^\mu\right] = \left[\begin{array}{cc}\frac{E_0}{c}\\0\\0\\0\end{array}\right]$$
Equivalently:
$$\left[P'^\mu\right] = \left[\begin{array}{cc}mc\\0\\0\\0\end{array}\right]$$
The four vector momentum according to an arbitrary inertial frame is given by the Lorentz transform of this vector. So in general this results in
$$\left[P^\mu\right] = \left[\begin{array}{cc}\gamma \frac{E_0}{c}\\\gamma\frac{E_0}{c}\frac{u^x}{c}\\\gamma\frac{E_0}{c}\frac{u^y}{c}\\\gamma\frac{E_0}{c}\frac{u^z}{c}\end{array}\right]$$
Equivalently:
$$\left[P^\mu\right] = \left[\begin{array}{cc}\gamma mc\\\gamma mu^x\\\gamma mu^y\\\gamma mu^z\end{array}\right]$$
From this we can extrapolate a few things. The time component of the four vector momentum was what we called energy divided by c and the spatial components were what we call momentum. This then gives us the special relativistic expressions for momentum and energy according to an arbitrary inertial frame.
$$E = \gamma mc^2$$
$$P^i = \gamma mu^i$$
Kinetic energy it the amount of energy we associate with motion only therefor we also arrive at the expression for the kinetic energy according to an arbitrary inertial frame
$$KE = E - E_{0} = (\gamma - 1)mc^2$$
The four component coordinate velocity which is not a true vector is given by
$$\left[u^\mu\right] = \left[\begin{array}{cc}c\\u^x\\u^y\\u^z\end{array}\right]$$
The velocity four vector is defined by
$$U^\mu = \frac{dx^\mu}{d\tau}$$
where $$\tau$$ is called proper time and can be thought of as time according to a hypothetical watch that rides along with the mass. The coordinate and four vector velocities can then be related through special relativistic time dilation.
$$U^\mu = \gamma u^\mu$$
One can then refer back to the result for four vector momentum and arrive at
$$P^\mu = mU^\mu$$
The relativistic force law is that four vector force is the proper time derivative of four vector momentum.
$$F^\mu = \frac{dP^\mu}{d\tau}$$
Inserting the expression in term of four vector velocity results in
$$F^\mu = m\frac{dU^\mu}{d\tau}$$
Four vector acceleration is defined as
$$A^\mu = \frac{dU^\mu}{d\tau}$$
resulting in the following special relativistic force equation analog of Newton's second law.
$$F^\mu = mA^\mu$$
Given this form of the dynamics equation for special relativity it becomes evident that a mass "changing with speed" is not the correct explanation for why a massive object can not be accelerated up to the speed of light. The mass m here does not change with speed. The correct explanation is that given this law of motion an arbitrary amount of ordinary force produces a diminishing coordinate acceleration as the coordinate velocity approaches c due to the time dilations involved in relating coordinate to four-vector expressions. These time dilations are in turn due to the Lorentzian structure of spacetime(the Lorentzian structure being why we defined the momentum four vector in terms of a Lorentz transform in the first place). So the reason really that nothing with mass can be accelerated up to c speeds in special relativity is ultimately that spacetime has a Lorentzian structure.

 

[Integral]

DW,
Thank you. While I have not completely comprehended your math, (I get tenser around tensors!) I believe your post contains answers to questions I have been asking myself.



Edited to remove referrence to a deleted post

 

[cragwolf]

Originally posted by DW
Some introductory texts and most "popular" literature has not kept up with modern conventions...

But that's just it: it's a convention. There's no correct way to go about it. There has been a long debate about this in the American Journal of Physics and Physics Today, and nothing has been settled, except that you can do physics either way; invariant mass or relativistic mass, it doesn't really matter which formulation you use, as long as you are consistent. Here's some references (pro and con) on the subject:

Online refs.:

Relativistic Mass
What is mass?
Rest mass or inertial mass?

Printed refs.:

Spacetime Physics - 2nd edition, Taylor and Wheeler, W H Freeman & Co. (April 1992)

The Advantage of Teaching Relativity with Four-Vectors, Robert W. Brehme. Am. J. Phys. 36 (10), October 1968

Does mass depend on velocity dad?, Carl Adler, Am. J. Phys. 55 (8), August 1987

The concept of mass, Lev B. Okun, Physics Today, June 1989

Letter to the Editor in Physics Today, Wolfgang Rindler, Physics Today, May 1990

In defense of relativistic mass, T.R. Sandin, Am. J. Phys. 59 (11), November 1991

Relativistic Generalizations of Mass, R.P. Bickerstaff AND G. Patsakos, Eur. J. Phys. 16 (1995), pg 63-66

Basic Relativity, Richard A. Mould, Springer Verlag, (1996)

Concepts of Mass in Contemporary Physics and Philosophy, Max Jammer, Princeton Univ. Press, (2000)

Letters to the editors, Uspekhi: What is mass? R I Khrapko, Physics Uspekhi., 43 (12) (2000)

Reply to R I Khrapko, Lev Okun, Physics Uspekhi., 43 (12) (2000)

 

[Arcon]

Originally posted by cragwolf
But that's just it: it's a convention. There's no correct way to go about it. There has been a long debate about this in the American Journal of Physics and Physics Today, and nothing has been settled, except that you can do physics either way; invariant mass or relativistic mass, it doesn't really matter which formulation you use, as long as you are consistent. Here's some references (pro and con) on the subject:...

I quite agree. It never ceases to amaze me how often this topic comes up. I mean, who really cares? It’s all just a matter of semantics. You don’t see people arguing over any other word in physics as much as you do the term mass. It all seems very silly to me.

With regards to concept of mass, some of the same arguments apply to the concept of time as well. E.g. DW's post addresses what the term mass –should- mean. He/she is, of course, referring to proper mass (DW incorrectly implies that only introductory texts use the concept of relativistic mass. That is incorrect.). However it's just a word. It’s a word that different relativists use to mean different things in modern relativity. Only a student would be confused over such trivial things, not someone with a solid understanding of relativity. Such confusion can only be eliminated by study - not by renaming terms when a few students become confused. Perhaps DW is/was a student who is/was confused by relativistic mass.

However this kind of multiple meaning of widely used terms happens throughout all of physics, not just relativity. E.g. when physicists use the term momentum it could refer to either mechanical momentum or generalized momentum. Sometimes physicists use the term Lagrangian when they really mean Lagrangian density. There's really no big deal here. It's all semantics. Nobody who knows physics solid is ever confused by things like this. Most would say "Big deal. Just say what you mean by it and don't worry about it."

However if one chooses to work in 4-d spacetime geometry one uses geometric quantities like proper time and proper mass. The components of the 4-vectors might then be labeled time and mass. Jammer explains this in the text you listed as you may have seen. However unlike regular vectors Euclidean geometry, 4-vectors in spacetime geometry are of quite a different nature. In regular Euclidean geometry the components of vectors have the -same- physical meaning. However in spacetime geometry the components have a –different- physical meaning. One of my favorite sayings is that you can rotate a rod into a rod but you can't rotate a rod into a clock. Also, the geometric quantities don’t carry the same meaning in spacetime geometry as their names imply. For example; the magnitude of a massive particle’s four-velocity is the speed of light. The four-velocity of a massive particle can never be zero, even for a particle at rest. There is no four-velocity for the speed of light. However one can just as correctly work in the 3+1 view of relativity. In fact it’s sometimes very preferable and productive to do so.

DW – the expression you gave for force is not a definition of force but a relationship between mass and force under certain conditions. In Newtonian mechanics F = mA is Euler’s expression for force. It only holds for particles of constant mass. Newton’s definition, the correct definition, is f = dp/dt.

For the reason above it is best to distinguish between mass and proper mass just as its best to distinguish between time and proper time. Notice how DW uses a Greek symbol for proper time. To be consistent DW should have used a Greek symbol for proper mass. But nobody does that. I suppose its because it would be too confusing since the Greek letter for "m" is mu, which usually represents reduced mass.

However it is incorrect to refer to the time component as energy. It is more appropriate to call it –mass-. The time component of (ct, r) is time. Therefore the time component of (Mc, p) should be called mass where M = m/(1-v^c/c^2)^1/2 (See Jammer). This convention is readily extendable to general relativity since the time component, P^0, of (mechanical) four-momentum is not energy as reading DW’s comments would suggest. P_0 is energy. The quantity M = m dt/d(tau) represents the inertia of a body, just as Einstein held.

Notice how DW writes

Given this form of the dynamics equation for special relativity it becomes evident that a mass "changing with speed" is not the correct explanation for why a massive object can not be accelerated up to the speed of light.

In his formulation one might get the impression that the speed of a particle is the magnitude of the four-velocity. However that magnitude is always equal to the speed of light. DW is mixing up the 3+1 view with the geometric view.

DW also writes

The correct explanation is that given this law of motion an arbitrary amount of ordinary force produces a diminishing coordinate acceleration as the coordinate velocity approaches c due to the time dilations involved in relating coordinate to four-vector expressions.

One is not right while the other is wrong. They are both equivalent ways of describing the same thing. However DW claims that the diminishing coordinate acceleration is due to time dilation. That is partially true since part of the diminishing coordinate acceleration is due to length contraction.

Arcon

 

[Per T]

Dear Arcon

I would be very happy if you could help a novis with a simple ansver!

I struggle to get a picture of the specetime.

"Curvature tells matter how to move, and matter tells space how to curve".

I will challenge you - not with a question but with an invitation - tell me the story that explains why my key drops to the floor when released from my hand! You are allowed to use entities as wraped space, equvalence principle and so forth but not any equations or math.

I do want to understand gravity from the wraped spacetime perspective. No rubber sheets - I want the truth... why does it fall down. What is ment by following a streight line in wraped spacetime? What causes the movemet? How does time get into the picture? I know these questions are kind of naive but I believe you can get my point; I know there is no Newtonian force but wraped spacetime but how do I get the real feeling for what happens when the key drops?

Per

 

[Jimmy]

Originally posted by Per T
I do want to understand gravity from the wraped spacetime perspective. No rubber sheets - I want the truth... why does it fall down. What is ment by following a streight line in wraped spacetime? What causes the movemet? How does time get into the picture? I know these questions are kind of naive but I believe you can get my point; I know there is no Newtonian force but wraped spacetime but how do I get the real feeling for what happens when the key drops?

Per

Well, since the key is always in motion in space-time, the curvature tells the key where to move. I think the curvature doesn't make the key move, it just redirects it's motion. It follows the curvature.

 

[DW]

Originally posted by Arcon
He/she is, of course, referring to proper mass (DW incorrectly implies that only introductory texts use the concept of relativistic mass. That is incorrect.).

Any modern "relativity" text that does use it is either introductory or isn't a good text.

Only a student would be confused over such trivial things, not someone with a solid understanding of relativity. Such confusion can only be eliminated by study - not by renaming terms when a few students become confused.

The confusions would never arise if the improper terminology were never used in the first place. There would never have been a question posted here whether photon mass was infinite for example if the concept of mass changing with speed were never taught. And giving that poster the correct modern definitions seemed to satisfy him as to the answer.

Perhaps DW is/was a student who is/was confused by relativistic mass.

I was a student, but wasn't confused by it. You shouldn't speculate about my history.

There is no four-velocity for the speed of light.

Actually there is a path parameterized four velocity of light.

In Newtonian mechanics F = mA is Euler’s expression for force. It only holds for particles of constant mass. Newton’s definition, the correct definition, is f = dp/dt.

Any "particle" has constant mass. (outside of QM mass oscillation)

For the reason above it is best to distinguish between mass and proper mass...

No, because they are the same thing by definition.

Notice how DW uses a Greek symbol for proper time. To be consistent DW should have used a Greek symbol for proper mass.

No because argument by analogy is not a logical arguement. These are two different variables with two different definitions.

However it is incorrect to refer to the time component as energy. It is more appropriate to call it –mass-.

You can say its inappropriate to call it either and then call it a dog if you want, but that doesn't change what it is.

(See Jammer).

why? Jammer is not very good.

(snipped some repetition)

In his formulation one might get the impression that the speed of a particle is the magnitude of the four-velocity.

Its not "my" formulation. It is general relativity and if you are confused by it take your own advice and go study.

DW is mixing up the 3+1 view with the geometric view.

I am not mixing anything up.

That is partially true since part of the diminishing coordinate acceleration is due to length contraction.

No, length contraction has nothing to do with it. Describe for example the acceleration of a particle of no significant length. The relation between the coordinate acceleration and four-vector acceleration is determined only by the transformation of the time derivatives which are the only differences in their definitions.

 

[Arcon]

Originally posted by DW
No, length contraction has nothing to do with it. Describe for example the acceleration of a particle of no significant length.

That is incorrect. The length of an object is not relevant to the question of diminishing coordinate acceleration, assuming a constant force that is. The relationship between force and acceleration is based on distance traveled and the time it takes to travel that distance and is reflected in the Lorentz factor 1/(1-v^c/c^2)^1/2. Since distances contract in the direction of motion and not perpendicular to the direction of motion, the resistance is therefore different in the direction of motion given the same acceleration in the rest frame in each direction.

Consider two situations L and T. In situation L there is no force transverse to the direction of motion. In situation T there is no force in the direction of motion. Let the magnitude of the force, as measured in a frame in which the particle is moving, have the same magnitude in each case. Then the ratio of transverse acceleration to longitudinal acceleration equals the Lorentz factor. It follows from that the acceleration transverse to the direction of motion is greater that it is to the acceleration in the direction of motion by a factor of 1/(1-v^c/c^2)^1/2. The reason is due to Lorentz contraction of the distance traveled.

Arcon

 

[DW]

Originally posted by Arcon
That is incorrect. The length of an object is not relevant to the question of diminishing coordinate acceleration, assuming a constant force that is.

I was the one who said it wasn't relevant and it isn't for any angle. The only difference in SR dynamics from Newtonian dynamics is that the time derivatives in the SR definitions are proper time derivatives instead of coordinate time derivatives as the Lorentzian structure of spacetime yields time dilation. You might argue that Lorentz Vs Gallilean transformation is also indirectly relavent as in how $$dx^\mu$$ is related between frames, but as I show in the following equations the only explicit difference is in the time derivatives.

Newton - coordinate velocity
$$u^i = \frac{dx^i}{dt}$$
Modern Relativity - four vector velocity
$$U^\lambda = \frac{dx^\lambda}{d\tau}$$

Nowton - coordinate acceleration
$$a^i = \frac{du^i}{dt}$$
Modern Relativity - four vector acceleration
$$A^\lambda = \frac{DU^\lambda}{d\tau}$$

Newtonian momentum
$$p^i = mu^i$$
Modern Relativity - four vector momentum
$$P^\lambda = mU^\lambda$$

Newtonian force
$$f^i = \frac{dp^i}{dt}$$ - Your Newtonian expression
or
$$f^i = ma^i$$
Modern Relativity - four vector force
$$F^\lambda = \frac{DP^\lambda}{d\tau}$$
or
$$F^\lambda = mA^\lambda$$

Now the result I'm pointing out is that the mass m in the dynamics equation for relativity above does not change with speed. The reason the coordinate acceleration diminishes in SR and not in Newtonian mechanics obviously has nothing to do with a mass changing with speed. It has to do with the fact that the SR expressions differ from the Newtonian expressions by the time dilation in the proper time derivatives. Time dilates due to the Lorentzian structure of spacetime and so this structure, not a changing mass and certainly not length contraction is what is responsible for coordinate acceleration diminishing with speed.

 

[that_guy]

I think having the acceleration decrease with increasing speed makes more sense asthetically as well. Large relative speeds change lengths, time and the addition of velocities. So it shouldn't be too surpising when relativity changes our understanding of meters, meters/s and seconds that meters/s^2 changes.

Keeping the mass constant also avoids the "if mass travels fast enough can't it become a black hole?" mess.

 

[cragwolf]

Originally posted by that_guy
Keeping the mass constant also avoids the "if mass travels fast enough can't it become a black hole?" mess.

What mess? You mean the "mess" that students who cram the night before exams find themselves in? No physicist or conscientious physics student finds themselves in a "mess" because some books choose to present the "relativistic mass" formulation. I think it's valuable for the serious student to have two different ways of approaching this issue.

 

[Arcon]

Originally posted by cragwolf
What mess? You mean the "mess" that students who cram the night before exams find themselves in? No physicist or conscientious physics student finds themselves in a "mess" because some books choose to present the "relativistic mass" formulation. I think it's valuable for the serious student to have two different ways of approaching this issue.

I'm happy to see that someone realizes that fact. Due to the recent attempts to downplay the concept of relativistic mass Sandin wrote article in the American Journal of Physics (Rinlder wrote a letter in Physics today too) on the subject in hopes of warding off such attempt. The paper is

In defense of relativistic mass, T.R. Sandin, Am. J. Phys. 59 (11), November 1991

Rindler's article is found at
http://www.geocities.com/physics_world/rindler_article.htm
(It has been placed there with the permission of the author)

re - if mass travels fast enough can't it become a black hole?

If one tries to use this argument as an excuse to not teach relativistic mass then all one is doing is not uncovering a missunderstanding held by the person who asks such a question and gives the student the idea that the gravitational field of a moving particle is not a function of speed - which is incorrect. This too has been explained in the American Journal of Physics in the article

Measuring the active gravitational mass of a moving object, D. W. Olson and R. C. Guarino, Am. J. Phys., 53, 661 (1985)

If a student asks this 'black hole' question then all they are doing is revealing the misunderstanding that an object is a black hole becuase it has a large mass. That is a sufficient reason for an object to become a black hole since when an object reaches a certain mass it will collapse to a black hole. However it is not a neccesary condition for an object to become a black hole. Its quite possible for any object of any mass to become a black hole. All one has to do is crunch the matter to a small enough size. Hawking postulated the existence of mini black holes if I recall correctly.

 

[DW]

Originally posted by Arcon
re - if mass travels fast enough can't it become a black hole?

If one tries to use this argument as an excuse to not teach relativistic mass then all one is doing is not uncovering a missunderstanding held by the person who asks such a question and gives the student the idea that the gravitational field of a moving particle is not a function of speed - which is incorrect.

Gravitation in relativity is not at all the Newtonian $$\frac{GM}{r^2}$$ acceleration field of Newtonian gravitation where you are replacing the M with relativistic mass. This is bad physics and the fact that people propose this sort of thing is yet another example of what is wrong with teaching the very concept of "relativistic mass" outside of a mere historical footnote. Take a frame according to which a gravitational source is not in motion. Look at the motion of a test particle from the perspective of that frame. Now transform the results to a frame according to which the source is in motion. The source didn't somehow gain any matter making the test particle in any way heavier. The difference in the observed behavior of the test particle then is not in any way due to the motion of the source. The difference is mearly a matter of using a different frame to describe the motion of the test particle. The correct physics is (instead of writting an equation relating the Newtonian gravitational acceleration field to a mass source) is to write Einstein's field equations. These relate the correct source, the stress energy tensor (not relativistic mass) to second order derivatives of the metric. These differential equations are then solved for the metric at least in approximation and the results are used to arrive at the Riemann tensor. The Riemann tensor is the expression for Riemannian spacetime curvature and the modern relativistic concept of the gravitational field is the spacetime curvature field, not the Newtonian acceleration field. The reason modern relativity defines things this way is that it leaves the physics invariant to frame which is the very principle of relativity. The stress energy tensor, not relativistic mass, is the source. The Reimann tensor, not the Newtonian acceleration field, is the spacetime curvature field. The metric is the differential spacetime geometry. In the absence of real forces i.e. four vector forces the four vector law of motion for general relativity reduces to geodesic motion. Newton's law of gravitation only comes out of all this in the case of linearized weak field gravitation at low speeds. At higher speeds it isn't even a good approximation to replace mass with relativistic mass in Newtonian gravitation, because instead geodesic motion yields something that is even more analogous to electromagnetism than Newtonian gravitation.

 

[GijXiXj]

Good stuff here.

But ... tell me. Doesn't a given body moving with a certain speed relative to me not actually exert greater "gravitational" force on me as it passes by at a certain distance, than if it were stationary relative to me at that same distance?

I mean at all speeds, "relativistic" or not (hopefully to avoid possible simultaneity problems). If this is so, then can I not say the body exhibits larger gravitational mass when it is moving than when it is not? Furthermore, isn't gravitational mass equivalent to inertial mass?

Alternatively or addtitionally, anybody care to reiterate exactly what the decomposition of the stress-energy tensor is into things more human beings can understand?

Ta.





If you're not confused, you don't understand the problem.

 

[Arcon]

Originally posted by GijXiXj
Good stuff here.

But ... tell me. Doesn't a given body moving with a certain speed relative to me not actually exert greater "gravitational" force on me as it passes by at a certain distance, than if it were stationary relative to me at that same distance?

Yes. Without question it most certainly does. However increase with velocity is neglegible at relativistic speeds.

Mind you - the gravitational force is an inertial force. When I say that the gravitational force on a pariticle I'm speaking of the force as meaured by an observer who is in a frame in which the gravitational force has not be transformed away. For example: for simplicity of discussion consider the example of an infinite sheet of matter (i.e. the matter lies in a plane) with a uniform mass distribution. Suppose the matter is such that it generates what is appoximately a uniform gravitational field as observed by an observer who is in frame S where S is at rest with respect to the sheet in the frame in which the matter is not moving. Let g = the local acceleration due to gravity at z = 0 (sheet lies in z = 0 plane). g is proportional to the mass density rho.


Now move relative to that sheet in the x-direction while remaining on the sheet at z = 0 - i.e. drive your car on the sheet real fast. Then in your car the local acceleration due to gravity will be increased by a factor of gamma^2 where gamma = 1/sqrt[1-(v/c)^2]. This is due to the fact that the mass density increased by a factor of gamma while the volume of the mass decreased by a factor of gamma leaving the mass density to increase by a factor of gamma^2. If a particle is weighed then the weight will be gamma^2 what it was in the rest frame.

However due to the momentum of the matter there are other gravitational effects just as in EM where instead of only an increased electric field (only one factor of gamma there since charge does not increase with speed but density still does of course) but in the moving frame there is now a magnetic field. But if the particle is not moving in the new frame there will be no magnetic force on it.

I mean at all speeds, "relativistic" or not (hopefully to avoid possible simultaneity problems). If this is so, then can I not say the body exhibits larger gravitational mass when it is moving than when it is not?

You betcha!

Furthermore, isn't gravitational mass equivalent to inertial mass?

Also - You betcha!

Alternatively or addtitionally, anybody care to reiterate exactly what the decomposition of the stress-energy tensor is into things more human beings can understand?

To a certain extent you can think of it as you would charge and current. Think of the charge as relativistic mass and think of the current as momentum. (there is also "stress" in GR but you get the idea).

 

[cragwolf]

Originally posted by DW
This is bad physics and the fact that people propose this sort of thing is yet another example of what is wrong with teaching the very concept of "relativistic mass" outside of a mere historical footnote.

Students will do bad physics whether or not we use the concept of relativistic mass. It's called not studying hard enough or last-night cramming. I just don't care about those "students". A student who understands time dilation, length contraction and the "twin paradox" will have no problems understanding relativistic mass.

The stress energy tensor, not relativistic mass, is the source.

And not invariant or rest mass either. It seems to me that you have this idea that relativistic mass is an inherently confusing concept. I'm pretty dumb compared to the average person, and I understood it just fine. Indeed, I felt I understood things better when I was presented with more than one perspective (so even the pedagogical angle is covered).

 

[Arcon]

Originally posted by cragwolf
Students will do bad physics whether or not we use the concept of relativistic mass. It's called not studying hard enough or last-night cramming.

Bravo! Excellant obseration.

A student who understands time dilation, length contraction and the "twin paradox" will have no problems understanding relativistic mass.

And a student who does not understand time dilation will think that there are internal changes to the physical structure of the clock which causes the time dilation. A student who does not understand length contraction will think that there are internal changes to the physical structure of rods which causes the Lorentz-Fitzgerald contraction.

And not invariant or rest mass either.

The claim The stress energy tensor, not relativistic mass, is the source. is inaccurate.

In the first place a mathematical quantity is not a physical entity which acts as a source of anything. The mathematical quantity is simply that: The mathematical quantity which describes the physical quantity.

In the second place (relativistic) mass and energy are proportional and as such one can be replaced by the other by m = E/c². In both special and general relativity (relativistic) mass and thus energy are both correctly and fully described by the stress-energy-momentum tensor T^uv. However one can also describe the same physics with the tensor M^uv = T^uv/c² and call that the "mass-momentum tensor".

For purposes of illustration I've described such a tensor here
http://www.geocities.com/physics_world/sr/mass_tensor.htm

In the third place it was Einstein himself who said, in his famous GR review paper on 1916, that

The special theory of relativity has led to the conclusion that inert mass is nothing more or less than energy, which finds its complete mathematical expression in a symmetrical tensor of second rank, the energy-tensor.

The reason is similar to the EM case. For example: The mathematical entity which plays the role of source in EM is the 4-current J^u = (c*rho, j) where rho = charge density and j = current density. A charge at rest in one frame becomes charge and current in another frame. Same with relativistic mass. Rest mass in one frame becomes relativistic mass, momentum and stress in another frame.

Charge is to EM as relativistic mass is to GR.

It seems to me that you have this idea that relativistic mass is an inherently confusing concept.

I have a question for the entire group: Please explain why relativistic mass confuses you. If it doesn't confuse you then please tell us if it once did at one time and why you were confused at that time.

Thank you

 

[GijXiXj]

Arcon:

Thanks for your reply to my queries. I'm rusty as hell on most of this stuff, but thought I had a handle on it a few years ago. Your answers are pretty much as I expected them to be. ;-)

In relation to your query in a later post:

"I have a question for the entire group: Please explain why relativistic mass confuses you. If it doesn't confuse you then please tell us if it once did at one time and why you were confused at that time."

Just see my 'signature'. ;-)

 

[selfAdjoint]

It's annoying to have to convert back and forth between the two systems. I have old relativity books that use relativistic mass, and it's a drag to convert their statements to the relativistic energy format to communicate with people trained on the modern system.

The fact that the two systems are fully equivalent while saying different things just suggests to me that the math isn't very deeply coupled to the physics. Maybe we need to go to equivalence classes or something.

 

[Arcon]

Originally posted by selfAdjoint
It's annoying to have to convert back and forth between the two systems. ...

That's the nature of modern relativity. What you use should relfect the problem at hand. Sometimes the geometric view of relativity is better to solve a problem and sometimes its better to use the 3+1 view of relavitity.

This can also be explained rather easily by considering this from the stand point of the center of mass. See
http://www.geocities.com/physics_world/sr/center_of_mass.htm

If the bullet is spinning at a constant rate then center of mass is moving at with uniform velocity. The momentum of the system can then be readily computated since the all the particles which make up the bullet are moving at constant speed and thus the calculation in the above link are valid.

 

[DW]

Originally posted by GijXiXj
Good stuff here.

But ... tell me. Doesn't a given body moving with a certain speed relative to me not actually exert greater "gravitational" force on me as it passes by at a certain distance, than if it were stationary relative to me at that same distance?

I just answered that above.

Furthermore, isn't gravitational mass equivalent to inertial mass?

The law of motion for general relativity is

$$F^\lambda = mA^\lambda$$
where the mass m is invariant. The four-vector acceleration can be written as the sum of two parts, an ordinary derivative and an affine connection part so that this expression becomes
$$F^\lambda = m(\frac{dU^\lambda}{d\tau} + \Gamma ^{\lambda}_\mu _\nu U^\mu U^\nu)$$
for gravitation alone $$F^\lambda = 0$$ which results in
$$m\frac{dU^\lambda}{d\tau} = - m\Gamma ^{\lambda}_\mu _\nu U^\mu U^\nu$$
In a Newtonian limit this is written
$$ma^i = mg^i$$
The m on the left is what Newton would refer to as the inertial mass. The m on the right is what Newton would refer to as the gravitational mass. These are what are equivalent and of cource they have to be as they were both the exact same thing multiplied through, and they are both the mass which is of course invariant.

 

[DW]

Originally posted by Arcon
Yes. Without question it most certainly does.

As I have demonstrated that is bad physics.

Now move relative to that sheet in the x-direction while remaining on the sheet at z = 0 - i.e. drive your car on the sheet real fast. Then in your car the local acceleration due to gravity will be increased by a factor of gamma^2 where gamma = 1/sqrt[1-(v/c)^2].

Actually the spacetime curvature field will not be changed one $$i$$. What will be different is the Normal force required to keep something away from geodesic motion in order to keep it in the same motion state as the vehicle and the car frame coordinate acceleration on a test mass dropped from the passenger.

This is due to the fact that the mass density increased by a factor of gamma while the volume of the mass decreased by a factor of gamma leaving the mass density to increase by a factor of gamma^2.

It was the energy density that went up two factors of $$\gamma$$, not the mass.

 

[Arcon]

Originally posted by GijXiXj
Furthermore, isn't gravitational mass equivalent to inertial mass?

I was incomplete in answering this. For a complete explanation you need to consider both active gravitational mass and passive gravitational mass.

It can be said that the source of gravity is active gravitational mass which is completely described the mass tensor (which is proportional to the energy-momentum tensor). The mass tensor is defined here
http://www.geocities.com/physics_world/sr/mass_tensor.htm

As far as passive gravitational mass and inertial mass then each equals relativistic mass. See
http://www.geocities.com/physics_world/gr/grav_force.htm

As to how that can be applied see
http://www.geocities.com/physics_world/gr/weight_move.htm

 

[DW]

Originally posted by cragwolf
Students will do bad physics whether or not we use the concept of relativistic mass.

My point above was that using it in the way he was, WAS bad physics.

And not invariant or rest mass either.

I never said that mass was the source on the right hand side of the field equations. I said the stress energy tensor was. I am also saying that in the linearized weak field limit one does not actually get Newton's equation modified by relativistic mass replacing the mass either. One instead gets equations analogous to electromagnetism. It is only when one also does a slow speed limit as well that something like Newtons law of gravitation appears and in such a limit the question of whether the mass should be replaced with relativistic mass is mute.

 

[DW]

Originally posted by Arcon
Bravo! Excellant obseration.
The claim The stress energy tensor, not relativistic mass, is the source. is inaccurate.

Not only is it accurate, but irrefutable. The source term, the right hand side of Einstein's field equations, is nothing other than the stress energy tensor and a proportionality constant.

However one can also describe the same physics with the tensor M^uv = T^uv/c² and call that the "mass-momentum tensor".

You've done nothing here but divide the stress energy tensor by c^2 and call inappropriately call it mass.

For purposes of illustration I've described such a tensor here
http://www.geocities.com/physics_world/sr/mass_tensor.htm

It would be better to learn about the stress energy tensor at
http://www.geocities.com/zcphysicsms/
the online modern relativity text.

In the third place it was Einstein himself who said, in his famous GR review paper on 1916, that

The reason is similar to the EM case. For example: The mathematical entity which plays the role of source in EM is the 4-current J^u = (c*rho, j) where rho = charge density and j = current density. A charge at rest in one frame becomes charge and current in another frame. Same with relativistic mass. Rest mass in one frame becomes relativistic mass, momentum and stress in another frame.

Charge is to EM as relativistic mass is to GR.

Fortunately he eventually changed his mind about the use of Plank's relativistic mass concept and went back to mass as invariant as he originally defined it.

 

[DW]

Originally posted by Arcon

This can also be explained rather easily by considering this from the stand point of the center of mass. See
http://www.geocities.com/physics_world/sr/center_of_mass.htm

Except for that it isn't good to refer to it as "center of mass" any more. For example, the velocity four vector to boost to the center of momentum frame is just that, the "center of momentum" velocity etc.

 

[Stingray]

You guys shouldn't be mixing special and general relativity. The stress-energy tensor is the source for gravity in GR, but there is no quasilocal (and only a restricted global) definition of mass in that theory. The stress-energy tensor is it. The comparison to SR's rest mass doesn't make any sense.

 

[Arcon]

Originally posted by Stingray
You guys shouldn't be mixing special and general relativity. The stress-energy tensor is the source for gravity in GR, ..

Mass is the source of a gravitational field in the same way the charge is the source of of an EM field. In the later case charge in one frame is current in another frame so that a complete description requires the 4-current vector J = (rho*c, j) where rho = charge density and j = current density. Same kind of thing in GR. However since mass and energy are proportional one can use the energy-momentum tensor or anything which is proportional to it such as the mass-momentum tensor. Semantics at this point.

..but there is no quasilocal (and only a restricted global) definition of mass in that theory. The stress-energy tensor is it. The comparison to SR's rest mass doesn't make any sense.

We were talking about densities which are local quantities. The energy-momentum tensor is actually the energy-density/momentum-density tensor.

However from what you said one might get the impression that all problems in special relativity can be solved without the energy-momentum tensor. That is not true. For example: When you have a body, such as a rod, which is under stress then, in general, you must use the stress-energy tensor to find the (relativistic) mass of the body in the moving frame. That is why Einstein said

The special theory of relativity has led to the conclusion that inert mass is nothing more or less than energy, which finds its complete mathematical expression in a symmetrical tensor of second rank, the energy-tensor.

 

[Stingray]

Originally posted by Arcon
Mass is the source of a gravitational field in the same way the charge is the source of of an EM field. In the later case charge in one frame is current in another frame so that a complete description requires the 4-current vector J = (rho*c, j) where rho = charge density and j = current density. Same kind of thing in GR. However since mass and energy are proportional one can use the energy-momentum tensor or anything which is proportional to it such as the mass-momentum tensor. Semantics at this point.

The current 4-vector can be made one number in a certain frame (charge), so its reasonable to say that charge sources the EM field. The energy-momentum tensor cannot in general be reduced to a single number (mass density). It can at most be made diagonal, where it still has up to 4 components. Gravity depends on more than just one number.

And yes, this is semantics, but that's what this whole thread is about .

We were talking about densities which are local quantities. The energy-momentum tensor is actually the energy-density/momentum-density tensor.

True, but you can't integrate these densities in a very meaningful way in general cases. Nothing's gauge invariant, so there's little point in talking about it I think. To be fair though, it wouldn't be wrong. Just inelegant.

However from what you said one might get the impression that all problems in special relativity can be solved without the energy-momentum tensor. That is not true. For example: When you have a body, such as a rod, which is under stress then, in general, you must use the stress-energy tensor to find the (relativistic) mass of the body in the moving frame. That is why Einstein said

That's not what I was implying, but ok. I agree it sounded like that.

 

[Arcon]

Originally posted by Stingray
The current 4-vector can be made one number in a certain frame (charge), so its reasonable to say that charge sources the EM field.

That is not always true. Only in certain special cases. If you have a current carrying wire with a non-vanishing charge density then there is no frame in which both components vanish.

If you have a frame in which all charges are at rest (no current) then there is a frame such that only the time component does not vanish. However the 4-current vector can always be decomposed as the sum of other 4-current vectors each of which does reduce to one component in a certain frame (each 4-vector reducing to one component in different frames in general).

The energy-momentum tensor cannot in general be reduced to a single number (mass density).

Same here as with the 4-current. If you have a frame of referance in which all particles are at rest (such as is the case with the energy-momentum tensor of "dust") then only the $$T^{00}$$ does not vanish. As is in case in EM - When the energy-momentum tensor is describing particles (which have non-zero rest mass) which do not interact you can also decompose it into a sum of energy-momentum tensors which do have only one component in a given frame (as above with EM these frames may be different)

It can at most be made diagonal, where it still has up to 4 components.

Not at most. The case of dust is a counter example. It has only 1 non-vanishing component in the rest frame.

True, but you can't integrate these densities in a very meaningful way in general cases.

Why not? It's done all the time.

 

[DW]

Originally posted by Arcon
Mass is the source of a gravitational field in the same way the charge is the source of of an EM field.

Actually the analogy is closer to the four vector current being the source to the vector potential which is like the stress energy tensor being the source for the differential spacetime geometry. The comparison being compared are
A^\mu ;^\nu _\nu = \mu _{0}J^\mu
and
G^\mu ^\nu = kT^\mu ^\nu .

Originally posted by Stingray
You guys shouldn't be mixing special and general relativity. The stress-energy tensor is the source for gravity in GR ... The current 4-vector can be made one number in a certain frame (charge), so its reasonable to say that charge sources the EM field. The energy-momentum tensor cannot in general be reduced to a single number (mass density).

This point is in fact why I kept saying that the stress energy tensor, NOT either mass, was the source term for gravitation in relativity. So you shouldn't say "you guys" here.

Originally posted by Arcon
That is not always true.

So what you should be arguing is that like the stress energy tensor not mass is the source in gravitation, the four vector current not charge is the source in electromagnetism.