New Relativity FAQ

pmb_phy

Hi folks

Since we often speak of relativistic mass here, and many people here post the URL to the relativity FAQ on the subject, I wanted to inform the forum that some major changes have been made to it. The new FAQ is at

http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

Pete

Related Special and General Relativity News on Phys.org

robphy

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I wish it were written by a relativist.

MeJennifer

From the article:

"mass does increase with speed, because the faster an object moves, the more resistant it becomes to being accelerated."

Depending on the interpretation of the words I find this statement ranging from confusing to straight out incorrect.

"So rest mass, rest length, and proper time find their way into the tensor language of relativity because all observers agree on them"

I would be interested to know how the author thinks that all observers agree on rest length in general relativity.

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Garth

Gold Member
Whenever you use a term such as 'Mass', 'Length' or 'Time', you have to define how you measure them.

Defining c to be constant 'Length' L is simply cT, so the fundamental measurement is that of time.

One way is to measure 'Time' is in the observer's frame of reference, in which case clocks in other frames vary with relative velocity, time is frame dependent and therefore so is mass. So relativistic mass "increases with speed".

Another way is to use 'Proper Time', which is not frame dependent.

$$d{\tau}^2 = - g_{\mu \nu}dx^{\mu}dx^{\nu}$$

[using the signature (-,+++)],

in which case mass does not vary - the 'gamma' factor is absorbed into the time dilation.

It is simply a matter of convention which method to use and the standard convention is to use 'Proper' frame independent units as these are considered 'purer' and lead to less confusion.

Garth

pmb_phy

I would be interested to know how the author thinks that all observers agree on rest length in general relativity.
Do you think the term is "rest length" is ambiguous? I guess a better term would be proper length.

Pete

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robphy

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One thing that is missing from this answer to the FAQ is a clear mathematical definition of the rest-mass and the relativistic-mass of a point particle.

The web.archive.org/web/20060830132929/http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html version had at least the passage from Taylor-Wheeler's Spacetime Physics, which is now nowhere to be found in the current version.

20060830132929 version said:
The preference for invariant mass is stressed and justified in the classic relativity textbook Spacetime Physics by Taylor and Wheeler who write

"Ouch! The concept of `relativistic mass' is subject to misunderstanding. That's why we don't use it. First, it applies the name mass--belonging to the magnitude of a four-vector--to a very different concept, the time component of a four-vector. Second, it makes increase of energy of an object with velocity or momentum appear to be connected with some change in internal structure of the object. In reality, the increase of energy with velocity originates not in the object but in the geometric properties of space-time itself."
Allow me to propose a working definition:

Given the 4-momentum $$\widetilde P$$ of a point particle,

its invariant rest-mass $$m_{rest}$$ is given by $$\sqrt{\widetilde P \cdot \widetilde P}$$ (using the $$+,-,-,-$$ signature);

its relativistic-mass-according-to-an-observer-with-4-velocity-$$\widetilde u$$ is given by $$\widetilde P \cdot \widetilde u$$
(when that observer is the particle itself, then $$\widetilde P = m_{rest} \widetilde u$$ so that
$$\widetilde P \cdot \widetilde u \stackrel{\widetilde P = m_{rest} \widetilde u}{=} (m_{rest} \widetilde u)\cdot \widetilde u =m_{rest}$$).

The italicized phrase is the often implicit [and likely forgotten about] baggage that accompanies the use relativistic-mass. Relativistic-mass is not a property of the particle alone... it's a property of both the particle and the observer measuring that component. So, I am unhappy with the second paragraph that says
"when an object moves with speed v, three of its properties...".
Those aren't properties of the object alone... just like the x-component of a vector is not a property of the vector alone.

I'm not saying that "relativistic mass" isn't useful... that depends on the user.
But I think some clarity is needed is needed in presenting something that is fair and balanced.

In my opinion, the answer to the FAQ is essentially handled in the first paragraph... it depends on your definitions of the terms (which are distinct in Special Relativity and beyond... but degenerates into one term in the Galilean case). I think the rest of it should be split off into another FAQ "What is the controversy surrounding the use the term 'relativistic mass'?".

pmb_phy

One thing that is missing from this answer to the FAQ is a clear mathematical definition of the rest-mass and the relativistic-mass of a point particle.

The web.archive.org/web/20060830132929/http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html version had at least the passage from Taylor-Wheeler's Spacetime Physics, which is now nowhere to be found in the current version.

Allow me to propose a working definition:

Given the 4-momentum $$\widetilde P$$ of a point particle,

its invariant rest-mass $$m_{rest}$$ is given by $$\sqrt{\widetilde P \cdot \widetilde P}$$ (using the $$+,-,-,-$$ signature);

its relativistic-mass-according-to-an-observer-with-4-velocity-$$\widetilde u$$ is given by $$\widetilde P \cdot \widetilde u$$
(when that observer is the particle itself, then $$\widetilde P = m_{rest} \widetilde u$$ so that
$$\widetilde P \cdot \widetilde u \stackrel{\widetilde P = m_{rest} \widetilde u}{=} (m_{rest} \widetilde u)\cdot \widetilde u =m_{rest}$$).

The italicized phrase is the often implicit [and likely forgotten about] baggage that accompanies the use relativistic-mass. Relativistic-mass is not a property of the particle alone... it's a property of both the particle and the observer measuring that component. So, I am unhappy with the second paragraph that says
"when an object moves with speed v, three of its properties...".
Those aren't properties of the object alone... just like the x-component of a vector is not a property of the vector alone.

I'm not saying that "relativistic mass" isn't useful... that depends on the user.
But I think some clarity is needed is needed in presenting something that is fair and balanced.

In my opinion, the answer to the FAQ is essentially handled in the first paragraph... it depends on your definitions of the terms (which are distinct in Special Relativity and beyond... but degenerates into one term in the Galilean case). I think the rest of it should be split off into another FAQ "What is the controversy surrounding the use the term 'relativistic mass'?".
I disagree with the proposed definition of proper mass (aka rest mass). I think its circular since 4-momentum is defined in terms of proper mass. I think a good definition of proper mass is to define it implicitly by requiring that the product of proper mass and 4-velocity be a conserved quantity.

Pete

robphy

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I disagree with the proposed definition of proper mass (aka rest mass). I think its circular since 4-momentum is defined in terms of proper mass. I think a good definition of proper mass is to define it implicitly by requiring that the product of proper mass and 4-velocity be a conserved quantity.

Pete
From a geometric viewpoint, the 4-momentum vector is more primitive.
You need a metric or its equivalent to produce a scalar from it.
(You can certainly still have conservation of 4-momenta [an equation of 4-vectors] without making reference to mass [or a metric].)

Two photons may have the same rest-mass of zero but distinct 4-momenta.

maverick_starstrider

Relativity was never really my forte but is there actually any valid reason in terms of actual reality to be such a stickler for the concept of 'relativistic mass'? Ultimately a relativistic equation needs the relativistic correction gamma but is there some reason that I just never learned for saying that this gamma is a result of a dilation in the mass and not a dilation in one of the other factors? Is it really anything more than convention to couple the gamma with the idea of 'mass'

pmb_phy

From a geometric viewpoint, the 4-momentum vector is more primitive.
You need a metric or its equivalent to produce a scalar from it.
(You can certainly still have conservation of 4-momenta [an equation of 4-vectors] without making reference to mass [or a metric].)
But you need to define the 4-momentum first.

In your definition you required a metric too. In mine there was no requirement for one ... unless its required to define proper time!??

In any case both of our definitions require a geometric object. In the definition I propose all the quantities are well defined.

Did you get my PM? I'm anxioius to hear your thoughts.

Pete

Garth

Gold Member
Ultimately a relativistic equation needs the relativistic correction gamma but is there some reason that I just never learned for saying that this gamma is a result of a dilation in the mass and not a dilation in one of the other factors? Is it really anything more than convention to couple the gamma with the idea of 'mass'
In a fully relativistic, i.e. frame independent, representation, or 'convention', the 'gamma correction' is contained in the dilation of time.

Garth

Hurkyl

Staff Emeritus
Gold Member
But you need to define the 4-momentum first.
Isn't the usual modus operandi in physics these days to take momentum as a primitive notion?

Hurkyl

Staff Emeritus
Gold Member
Do you think the term is "rest length" is ambiguous? I guess a better term would be proper length.

Pete
"Rest length of a spacelike curve" is not ambiguous. "Rest length of an object" is ambiguous, because there is no canonical way to choose which spacelike curve expresses the rest length of the object.

You get the same problem in special relativity too -- there are only a few special cases which determine 'obvious' sorts of spacelike curve to measure. e.g. an inertially travelling object, or one undergoing uniform constant acceleration.

pmb_phy

Isn't the usual modus operandi in physics these days to take momentum as a primitive notion?
Not all quantities. 4-momentum is not a primative notion

pmb_phy

"Rest length of a spacelike curve" is not ambiguous. "Rest length of an object" is ambiguous, because there is no canonical way to choose which spacelike curve expresses the rest length of the object.

You get the same problem in special relativity too -- there are only a few special cases which determine 'obvious' sorts of spacelike curve to measure. e.g. an inertially travelling object, or one undergoing uniform constant acceleration.
That's true. The body could have different orientations in the field.

robphy

Homework Helper
Gold Member
But you need to define the 4-momentum first.

In your definition you required a metric too. In mine there was no requirement for one ... unless its required to define proper time!??

In any case both of our definitions require a geometric object. In the definition I propose all the quantities are well defined.

Did you get my PM? I'm anxioius to hear your thoughts.

Pete

It makes more sense to take the 4-momentum as more primitive than mass.

To describe a force, it is more sensible [geometrically] to treat it as a vector first,
rather than something with magnitude |F| [which needs a metric] and a direction.

In the case of relativity, it makes more sense to describe a particle by its worldline.
The particle's 4-momentum is a future-directed tangent vector to that worldline.

Using the metric, if it is timelike, then the particle has a nonzero rest-mass... and you can find a unit-tangent vector to call its 4-velocity... The invariant rest-mass is the proportionality factor between the 4-momentum and the 4-velocity.
If that 4-momentum is null, then the particle has zero rest-mass [or better zero invariant-mass] (like a photon)... and has no unit-tangent vector.
Thus, you can't define the 4-momentum of a photon in terms of its rest-mass (or invariant-mass) ... you'll need another way.... which is probably more roundabout than this procedure.

If I am describing a collision of particles,
then I might observe that a certain tangent-vector can be defined for each particle such that there is a conservation law [expressed as a vector sum, without reference to or need of a metric]. (I admit I don't have a nice way of defining the 4-momentum at this point... in fact, in my mind, most derivations I have seen are not very natural to me... other than "it works" or "it reduces to the ordinary situation". I'm sure it can be done... and hopefully it can be made digestible to an introductory student.)

pmb_phy

It makes more sense to take the 4-momentum as more primitive than mass.
That's a matter of opinion my friend.
In the case of relativity, it makes more sense to describe a particle by its worldline.
The particle's 4-momentum is a future-directed tangent vector to that worldline.
That gives a property of the 4-momentum but does not define it.
Using the metric, if it is timelike, then the particle has a nonzero rest-mass... and you can find a unit-tangent vector to call its 4-velocity... The invariant rest-mass is the proportionality factor between the 4-momentum and the 4-velocity.
If that 4-momentum is null, then the particle has zero rest-mass [or better zero invariant-mass] (like a photon)... and has no unit-tangent vector.
Thus, you can't define the 4-momentum of a photon in terms of its rest-mass (or invariant-mass) ... you'll need another way.... which is probably more roundabout than this procedure.
Actually I agree with that. I had a hard time convincing someone else of that as a matter of fact. In my opinion one defines 4-momentum as P = (m, p) and then, for a tardyon, define proper mass as m0 = m(0) (when measured in an inertial frame). p can then be defined by requiring that mv be a conserved quantity. I was describing the definition of proper mass as its found in some textbooks that I've read as well as in AJP.

I take it you have no comment on my PM. Okay. Thanks anyway.

Pete

robphy

Homework Helper
Gold Member
robphy said:
It makes more sense to take the 4-momentum as more primitive than mass.
That's a matter of opinion my friend.
That it make more sense... yes, an opinion.
That 4-momentum can [probably] be defined independent of the metric, whereas mass requires the metric for its definition.... suggests that 4-momentum can exist with fewer geometrical structure than rest-mass (as defined using the norm of the 4-momentum).

robphy said:
In the case of relativity, it makes more sense to describe a particle by its worldline.
The particle's 4-momentum is a future-directed tangent vector to that worldline.
That gives a property of the 4-momentum but does not define it.
From that part alone, yes... it's incomplete, as I said in the rest of that post.

I take it you have no comment on my PM. Okay. Thanks anyway.

Pete
...but my summer is very busy.

pmb_phy

That 4-momentum can [probably] be defined independent of the metric, whereas mass requires the metric for its definition.
I disagree. The metric is only required to establish an equality between the norm of P and the proper mass. This would make a poor definition though... at least in my opinion.
...but my summer is very busy.
It would take you less time to read that page than to read this thread. The idea is very simple. However I understand your time constrainst and welcome your thoughts whenever you get around to it. Thanks Rob!

Pete

Hurkyl

Staff Emeritus
Gold Member
I disagree. The metric is only required to establish an equality between the norm of P and the proper mass. This would make a poor definition though... at least in my opinion.
Having (enough of) that knowledge is equivalent to knowing a metric: there is a unique metric with the property that those 4-momenta have those rest masses as their norm.

pmb_phy

Having (enough of) that knowledge is equivalent to knowing a metric: there is a unique metric with the property that those 4-momenta have those rest masses as their norm.
I was addressing the differecnce between a definition and an equality.

Pete

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