
#1
Dec1603, 12:44 PM

P: 328

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 [tex]E_{0}[/tex]. So [tex]m \equiv \frac{E_0}{c^2}[/tex] We define a four element vector constructed according to this frame by its time element [tex]P^{0}[/tex] given by [tex]P^{0} = \frac{E}{c}[/tex] and the three other components zero for this frame is what we will call the momentum. This results in the following. [tex]\left[P'^\mu\right] = \left[\begin{array}{cc}\frac{E_0}{c}\\0\\0\\0\end{array}\right][/tex] Equivalently: [tex]\left[P'^\mu\right] = \left[\begin{array}{cc}mc\\0\\0\\0\end{array}\right][/tex] 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 [tex]\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}\fr ac{u^y}{c}\\\gamma\frac{E_0}{c}\frac{u^z}{c}\end{array}\right][/tex] Equivalently: [tex]\left[P^\mu\right] = \left[\begin{array}{cc}\gamma mc\\\gamma mu^x\\\gamma mu^y\\\gamma mu^z\end{array}\right][/tex] 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. [tex]E = \gamma mc^2[/tex] [tex]P^i = \gamma mu^i[/tex] 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 [tex]KE = E  E_{0} = (\gamma  1)mc^2[/tex] The four component coordinate velocity which is not a true vector is given by [tex]\left[u^\mu\right] = \left[\begin{array}{cc}c\\u^x\\u^y\\u^z\end{array}\right][/tex] The velocity four vector is defined by [tex]U^\mu = \frac{dx^\mu}{d\tau}[/tex] where [tex]\tau[/tex] 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. [tex]U^\mu = \gamma u^\mu[/tex] One can then refer back to the result for four vector momentum and arrive at [tex]P^\mu = mU^\mu[/tex] The relativistic force law is that four vector force is the proper time derivative of four vector momentum. [tex]F^\mu = \frac{dP^\mu}{d\tau}[/tex] Inserting the expression in term of four vector velocity results in [tex]F^\mu = m\frac{dU^\mu}{d\tau}[/tex] Four vector acceleration is defined as [tex]A^\mu = \frac{dU^\mu}{d\tau}[/tex] resulting in the following special relativistic force equation analog of Newton's second law. [tex]F^\mu = mA^\mu[/tex] 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 fourvector 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. 



#2
Dec1603, 02:55 PM

Mentor
P: 7,292

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 



#3
Dec1603, 07:55 PM

P: 210

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 FourVectors, 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 6366 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) 


#4
Dec2103, 11:00 AM

P: n/a

Modern Special Relativity and MassWith 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 4d spacetime geometry one uses geometric quantities like proper time and proper mass. The components of the 4vectors 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, 4vectors 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 fourvelocity is the speed of light. The fourvelocity of a massive particle can never be zero, even for a particle at rest. There is no fourvelocity 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/(1v^c/c^2)^1/2 (See Jammer). This convention is readily extendable to general relativity since the time component, P^0, of (mechanical) fourmomentum 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 DW also writes Arcon 



#5
Dec2103, 12:44 PM

P: 3

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 



#6
Dec2103, 04:58 PM

P: 633





#7
Dec2103, 10:36 PM

P: 328

Any modern "relativity" text that does use it is either introductory or isn't a good text. (snipped some repetition) 


#8
Dec2203, 05:44 AM

P: n/a

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/(1v^c/c^2)^1/2. The reason is due to Lorentz contraction of the distance traveled. Arcon 



#9
Dec2203, 10:14 AM

P: 328

Newton  coordinate velocity [tex]u^i = \frac{dx^i}{dt}[/tex] Modern Relativity  four vector velocity [tex]U^\lambda = \frac{dx^\lambda}{d\tau}[/tex] Nowton  coordinate acceleration [tex]a^i = \frac{du^i}{dt}[/tex] Modern Relativity  four vector acceleration [tex]A^\lambda = \frac{DU^\lambda}{d\tau}[/tex] Newtonian momentum [tex]p^i = mu^i[/tex] Modern Relativity  four vector momentum [tex]P^\lambda = mU^\lambda[/tex] Newtonian force [tex]f^i = \frac{dp^i}{dt}[/tex]  Your Newtonian expression or [tex]f^i = ma^i[/tex] Modern Relativity  four vector force [tex]F^\lambda = \frac{DP^\lambda}{d\tau}[/tex] or [tex]F^\lambda = mA^\lambda[/tex] 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. 



#10
Jan804, 09:56 PM

P: 11

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. 



#11
Jan904, 01:20 AM

P: 210




#12
Jan904, 05:47 AM

P: n/a

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_wor...er_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 arguement 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 existance of mini black holes if I recall correctly. 



#13
Jan904, 12:25 PM

P: 328





#14
Jan904, 01:38 PM

P: 21

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 stressenergy tensor is into things more human beings can understand? Ta. If you're not confused, you don't understand the problem. 


#15
Jan904, 02:14 PM

P: n/a

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 xdirection 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. 



#16
Jan904, 11:37 PM

P: 210




#17
Jan1004, 03:21 AM

P: n/a

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/c^{2}. In both special and general relativity (relativistic) mass and thus energy are both correctly and fully described by the stressenergymomentum tensor T^{uv}. However one can also describe the same physics with the tensor M^{uv} = T^{uv}/c^{2} and call that the "massmomentum tensor". For purposes of illustration I've described such a tensor here http://www.geocities.com/physics_wor...***_tensor.htm In the third place it was Einstein himself who said, in his famous GR review paper on 1916, that Charge is to EM as relativistic mass is to GR. Thank you 



#18
Jan1004, 03:40 AM

P: 21

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'. ;) 


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