Has relativistic mass gone out of fashion?

[7777777]

For a long time I have studied and read about whether the photon has mass.
On this forum, I could find a link to a document by Gary Oas: On the abuse and use of
the relativistic mass.
http://arxiv.org/pdf/physics/0504110v2.pdf

Oas' paper tells about an investigation done to 164 students:
"Einstein also showed that the mass of an object moving at close to the speed of light, as seen by
an outside observer, increases. [...] It explains, among other things, why the speed of light serves
as the ultimate speed limit in the universe. Suppose you’re in a spaceship, approaching the speed
of light. You think ‘I’ll just step on the accelerator a little harder and I’ll pass that pesky speed
limit, no problem.’ But it won’t work: to make your craft move faster, you have to use energy;
the more massive the spaceship, the more energy you need. And, thanks to Einstein’s special rel-
ativity, the mass keeps increasing, so you need more and more energy to further boost the speed.
And you’ll never quite make it. If you reached the speed of light, your spaceship, as seen by an outside observer, would have an infinite mass and it would have taken you an infinite amount of
energy to get there."

I remember that exactly this way I was also introduced into relativity while in school a very long time ago. Oas' paper says that in this way misconceptions develop among first time learners of relativity.Now I began to really wonder, can the concept of relativistic mass really be wrong or out of fashion. I am not an expert on relativity, but how is the ultimate speed limit, the speed of
light, now explained if not with velocity dependent mass?
If the velocity dependent mass does not prevent a massive object of achieving the speed of light,
what is going to do it?
The speed of a massive object cannot exceed the speed of light.
If the relativistic mass is abandoned and we use instead the energy, the kinetic energy, and if
there is no upper limit to how large the kinetic energy can be, doesn't this also mean that the speed of
a massive object can exceed the speed of light?

 

[phinds]

No, it means you would have to input infinite energy to get an object with rest mass to go at c and this remains impossible.

There are members on this forum who would prefer to hunt you down and hurt you if you use the term "relativistic mass" but they are much to polite to do so

 

[m4r35n357]

Now I began to really wonder, can the concept of relativistic mass really be wrong or out of fashion. I am not an expert on relativity, but how is the ultimate speed limit, the speed of
light, now explained if not with velocity dependent mass?
If the velocity dependent mass does not prevent a massive object of achieving the speed of light,
what is going to do it?

The speed limit is imposed by spacetime itself, via the Minkowski metric. Mass is not the issue here; there is simply no mathematical way to accelerate to c.

 

[ShayanJ]

Hi, I'm one of those that phinds mentions. But its OK, you're safe!
I don't know whether it is out of fashion or not, but if its not, I hope it becomes.
In some introductory SR books the author talks about relativistic mass and its increase with speed but it never uses it. I don't know why they define it when they don't need it!
Anyway, I have two arguments for you which makes relativistic mass unnecessary even here.

1- Imagine two persons at rest w.r.t. each other. One of them shines a laser at the other one. We can be sure that the guy receives it. But if another one passes by them with a speed greater than the speed of light, according to him, the laser pulse never reaches the other guy. But the fact that the laser pulse reaches the other guy shouldn't depend on the frame of reference! So its impossible for an object to move faster than light.

2- Imagine it is possible to send signals faster than the speed of light. That means I can send signals back in time. Now I have a device that is scheduled to send signals back in time only if it doesn't receive a signal from itself from future. But if it receives a signal, it means sent it in the future which means it didn't receive a signal. That's a paradox. Other paradoxes of this kind can be found too. So its impossible for energy to move faster than light.

 

[PeroK]

The speed limit is imposed by spacetime itself, via the Minkowski metric. Mass is not the issue here; there is simply no mathematical way to accelerate to c.

I'm not sure you can accelerate anything physically with mathematics! In any case, it's not mathematics that limits the relative velocity, it's physics.

 

[m4r35n357]

I'm not sure you can accelerate anything physically with mathematics! In any case, it's not mathematics that limits the relative velocity, it's physics.

You can accelerate mathematically by successive boosting of your velocity. You can do this as many times as you like and you will never get to c. That's maths.

 

[HallsofIvy]

But the reason the maths works that way is physics!

 

[ChrisVer]

Relativistic mass is not wrong... however it's not a useful quantity for physicists, who mostly prefer talking about frame invariant quantities... when you use relativistic mass you have to specify the ref frame from which you are using it [that is in the [itex]\gamma[/itex] factor that multiplies the rest mass, and [itex]\gamma[/itex] depends on the relative velocity].
In my opinion, I don't see how Minkowksi metric rules out velocities v>c...they exist in it but they are not causally connected. The minkowski metric gives the whole space of [itex](vec{x},t)[/itex] and from that you get timelike,spacelike and null vectors...
It's the energy-momentum relation which does that, if for example you try to find the speed of a particle which has [itex]E^{2}=p^{2}+m^{2}, ~(c=1)[/itex] then one can find that the velocity is given by:
[itex] |v|^2 = \frac{p^2}{E^2} [/itex] (frame independent)
or [itex]|v|^2= \frac{p^{2}}{p^{2}+m^{2}} \le 1[/itex]
In any case is the energy-momentum relation a result of the minkowski spacetime?
Maybe one could try to figure that out in GR frame where he'd have:
[itex] g_{\mu \nu} p^{\mu} p^{\nu}= m^{2} [/itex] and look at [itex]m \rightarrow 0[/itex]. However if I recall well, the results don't differ much? Plus I believe the velocities won't be the same as in SR (they'll be local)...

 

[PeroK]

You can accelerate mathematically by successive boosting of your velocity. You can do this as many times as you like and you will never get to c. That's maths.

##c - 1 + 2 = c + 1##

There it is. Nothing stopped me doing that! Not mathematically, anyway. What mathematics are you talking about?

 

[ShayanJ]

Relativistic mass is not wrong... however it's not a useful quantity for physicists, who mostly prefer talking about frame invariant quantities... when you use relativistic mass you have to specify the ref frame from which you are using it [that is in the [itex]\gamma[/itex] factor that multiplies the rest mass, and [itex]\gamma[/itex] depends on the relative velocity].
In my opinion, I don't see how Minkowksi metric rules out velocities v>c...they exist in it but they are not causally connected. The minkowski metric gives the whole space of [itex](vec{x},t)[/itex] and from that you get timelike,spacelike and null vectors...
It's the energy-momentum relation which does that, if for example you try to find the speed of a particle which has [itex]E^{2}=p^{2}+m^{2}, ~(c=1)[/itex] then one can find that the velocity is given by:
[itex] |v|^2 = \frac{p^2}{E^2} [/itex] (frame independent)
or [itex]|v|^2= \frac{p^{2}}{p^{2}+m^{2}} \le 1[/itex]
In any case is the energy-momentum relation a result of the minkowski spacetime?
Maybe one could try to figure that out in GR frame where he'd have:
[itex] g_{\mu \nu} p^{\mu} p^{\nu}= m^{2} [/itex] and look at [itex]m \rightarrow 0[/itex]. However if I recall well, the results don't differ much? Plus I believe the velocities won't be the same as in SR (they'll be local)...

How did you get those equations?

Anyway, I gave arguments that forbid faster than light motion and communication without using relativistic mass.

But what I mean, isn't that relativistic mass is wrong. It has no meaning to say that its right or wrong. Its a physicist's choice to let it in the theory or not. I only mean, why should we let something in that is of no use and only causes troubles?!

 

[ChrisVer]

How exactly did you reach that?
[itex] v^2 = \frac{p^2}{E^2} = \frac{p^2}{p^2 + m^2} [/itex]
if you go the other way:
[itex] v^2 = \frac{p^2}{E^2} = \frac{E^2-m^2}{E^2}= 1 - \frac{m^2}{E^{2}} \le 1 [/itex] again
except for if you are asking me why [itex]v^{2} = \frac{p^2}{E^{2}}[/itex]?
That's SR basics...
[itex] \frac{p^{i}}{E} = \frac{\gamma m v^{i}}{\gamma m} = v^{i}[/itex]
correction which doesn't depend on any relative velocity

Because I didn't want to have the vector indices, I just took the inner product...

 

[ShayanJ]

How exactly did you reach that?
[itex] v^2 = \frac{p^2}{E^2} = \frac{p^2}{p^2 + m^2} [/itex]
if you go the other way:
[itex] v^2 = \frac{p^2}{E^2} = \frac{E^2-m^2}{E^2}= 1 - \frac{m^2}{E^{2}} \le 1 [/itex] again
except for if you are asking me why [itex]v^{2} = \frac{p^2}{E^{2}}[/itex]?
That's SR basics...
[itex] \frac{p^{i}}{E} = \frac{\gamma m v^{i}}{\gamma m} = v^{i}[/itex]
The square then is a Lorentz invariant quantity.
Because I didn't want to have the vector indices, I just took the inner product...

Yeah, I got it.
But [itex] v^i [/itex] isn't frame independent.
The formulas [itex] \vec p=\gamma m \vec v [/itex] and [itex] E=\gamma m [/itex] contain a [itex] \gamma [/itex] and that is a function of a velocity, But what velocity? That's frame-dependent and so those formulas and fractions are frame-dependent.

EDIT:

correction which doesn't depend on any relative velocity

Yeah, that's better!

 

[vanhees71]

It's very important to keep in mind that the object
[tex]\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t}[/tex]
is not the spatial part of a three vector. It has no simple transformation properties in Minkowski space (it's of course a three vector under rotations).

That's why one introduces the idea of proper time, which is the time in the restframe of the particle (supposed it's not massless, since massless particles have no restframe). Its increment is given as
[tex]\mathrm{d} \tau= \mathrm{d} t \sqrt{1-\vec{v}^2/c^2}=\sqrt{c^2 \mathrm{d} t^2-\mathrm{d} \vec{x}^2}/c,[/tex]
which shows that it is a scalar. You can write it, using an arbitrary parameter of the worldline of the particle
[tex]\mathrm{d} \tau = \mathrm{d} \lambda \sqrt{\dot{x}_{\mu} \dot{x}^{\nu}},[/tex]
where the dot indicates the derivative wrt. [itex]\lambda[/itex].

Then the quantity
[tex]u^{\mu}=\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau},[/tex]
the four-velocity of the particle, is a four-vector.

The four-momentum is
[tex]p^{\mu}=m u^{\mu},[/tex]
where [itex]m[/itex] is the invariant mass of the particle, which is a scalar. The 0-component is [itex]E/c[/itex], where [itex]E[/itex] is the kinetic energy of the particle plus the rest energy [itex]E_{0}=m c^2[/itex]:
[tex]E=p^0 c=m c^2 \frac{\mathrm{d} t}{\mathrm{d} \tau}=m c^2 \gamma,[/tex]
with
[tex]\gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.[/tex]
The momentum is
[tex]\vec{p}=m \frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} = m \frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \frac{\mathrm{d} t}{\mathrm{d} \tau} = m \gamma \vec{v}.[/tex]
The notion of a relativistic mass is unnecessary and confusing. It's not used in physicists' everyday life anymore (at least not in the high-energy particle and nuclear physics community).

In addition it's irresponsible to teach the idea of photons in high school. It can only be properly understood in terms of relativistic quantum field theory and gauge symmetry. This is way beyond reach of high school students! The old idea of photons, introduced by Einstein in 1905, is the most misleading way to look at electromagnetic radiation. It's much better to teach it in terms of classical field theory. Most properties of em. radiation, including quantum phenomena as the photo effect can be explained using classical em. waves and quantized massive particles, and non-relativistic quantum theory is well in the reach of high-school students. The damage done with introducing these wrong concepts of the photon are obvious even at university level, when students are really able to learn the true thing. You always have to tell them to forget what they have learned about photons before it's treated in the proper way, i.e., using QED!

 

[Nugatory]

##c - 1 + 2 = c + 1##

There it is. Nothing stopped me doing that! Not mathematically, anyway. What mathematics are you talking about?

Speeds don't add that way (google for "relativistic velocity addition" or look for many threads here).

If you are moving at speed ##u<c## relative to me, and you boost your speed by ##v<c##, your speed relative to me will not be ##u+v##, it will be ##(u+v)/(1+uv)## (measuring time in seconds and distances in seconds so that ##c=1## - otherwise I'd need a ##c^2## in there too).

Try it with that math and you won't get above ##c##.

 

[ChrisVer]

Yeah, I got it.
But [itex] v^i [/itex] isn't frame independent.
The formulas [itex] \vec p=\gamma m \vec v [/itex] and [itex] E=\gamma m [/itex] contain a [itex] \gamma [/itex] and that is a function of a velocity, But what velocity? That's frame-dependent and so those formulas and fractions are frame-dependent.

EDIT:

Yeah, that's better!

Well that' almost right about the gamma, however the velocity in gamma refers not to the particle's velocity, but to the velocity between 2 frames which measure the same thing... (like the example with the 2 spaceships with the one launching a rocket).. the [itex]v^{i}[/itex] corresponds to the rocket's velocity as measured by one of them, and the velocity in [itex]\gamma[/itex] refers to the velocity between those two... that's why I initially called it frame independent- two frames will measure the velocity in the same way without having to correlate their measurements. However I think it sounded weird and I am somewhere wrong.
By getting rid of it, you can have the particle's velocity by knowing its energy and momentum at your ref frame, and without having to care about the different frames.
That's also why the quantity [itex]\frac{p^2}{E^2}
[/itex] appears in phase integrals which are Lorentz Invariant...

 

[7777777]

Anyway, I gave arguments that forbid faster than light motion and communication without using relativistic mass.

But what I mean, isn't that relativistic mass is wrong. It has no meaning to say that its right or wrong. Its a physicist's choice to let it in the theory or not. I only mean, why should we let something in that is of no use and only causes troubles?!

Ok, you did not need relativistic mass to give arguments that forbid faster than light motion and communication. You told that it is impossible for energy to move faster than light.

Now I am thinking in layman's terms: how does energy have speed? I think this means the speed of light, and light has energy, and the speed of this energy cannot exceed the speed limit.
I am looking for an explanation whether or not a massive object, not just energy or a photon with a zero rest mass, can travel with a speed exceeding the speed of light if the concept of relativistic mass is abandoned.

I think you don't abandon the concept of mass altogether. I mean that there is still
the concept called the velocity independent mass, or invariant mass, or "proper mass" or rest mass m[itex]_{0}[/itex] which can be nonzero for particles other than photons.
If the concept of relativistic mass is abandoned, can an object or a particle with rest mass m[itex]_{0}[/itex]≠0 if it has unlimited amount of kinetic energy travel with a speed v exceeding c, in other words v>c?

 

[jtbell]

It’s by now a well-established experimental fact that as an object moves faster, its “resistance” to going still faster, increases. Coming from a Newtonian background with F=ma, etc., it seems intuitively appealing to explain this by saying that m increases with velocity. The problem is that there’s no single way to do this mathematically that works in a wide variety of Newtonian formulas. In effect, there’s not one “relativistic mass”, but several.

Even in the early days of relativity (early 1900s), physicists recognized this. In the early literature about relativity, you find references to “transverse mass” for F=ma when the force is perpendicular to the velocity, and “longitudinal mass” for F=ma when the force is parallel to the velocity. If the force is at some other angle, you get something in between, and the acceleration isn’t parallel to the force any more!

It turns out that “transverse mass” also works for momentum, p=mv, but neither “mass” works for kinetic energy, K=(1/2)mv^2.

I suspect that “transverse mass” became “the relativistic mass” because the earliest experiments dealing with fast-moving particles were mass-spectrometer type experiments, in which particles travel in a circular path in a magnetic field, under the influence of a centripetal (perpendicular) magnetic force. Also, one of the first practical effects of “relativistic mass” was in developing high-energy circular particle accelerators (synchrotrons), which similarly use perpendicular magnetic forces, and in which relativistic effects have to be taken into account, unlike the earlier low-energy cyclotrons.

I remember that exactly this way I was also introduced into relativity while in school a very long time ago.

Same with me, in my second-year undergraduate “intro modern physics” course in the early 1970s. I don't remember if my high-school physics class did anything with relativity. However, when I was in graduate school in high-energy particle physics in the late 1970s and early 1980s, the only place I saw “relativistic mass” was in a textbook about accelerator design that had been written in the 1950s by someone who had worked with E. O. Lawrence on the first cyclotrons in the 1920s or 1930s, and later helped develop synchrotrons. Otherwise, when the people I worked with talked about “mass”, they always meant “rest mass” a.k.a. “invariant mass”.

Einstein himself said in his later years that he thought it was not a good idea to use a velocity-dependent mass:

It is not good to introduce the concept of the mass ##M = m / \sqrt{1-v^2/c^2}## of a moving body for which no clear definition can be given. It is better to introduce no other mass concept than the ’rest mass’ m. Instead of introducing M it is better to mention the expression for the momentum and energy of a body in motion.

(http://en.wikipedia.org/wiki/Mass_in_special_relativity)

I think introductory textbooks were slow to drop “relativistic mass” because they’re usually not written by specialists, and so they tend to be conservative with changes like this. But many of them have changed. I used to teach an intro modern physics course using Beiser’s textbook. I started with (I think) the 4th edition in the late 1980s, which had a section about “relativistic mass”, and some homework problems. In the 6th edition (late 1990s or early 2000s), “relativistic mass” appeared only in a short sidebar which acknowledged that many popular-level books and some introductory textbooks still use it, but indicated that most physicists don’t.

 

[ChrisVer]

yes, it can theoretically (mathematically) if you allow for [itex]m^{2}_0 <0 [/itex]
But even with infinite energy, a massive particle with [itex]m^{2}_0>0[/itex] won't reach c... it will always approach c...

 

[PeroK]

Speeds don't add that way (google for "relativistic velocity addition" or look for many threads here).

If you are moving at speed ##u<c## relative to me, and you boost your speed by ##v<c##, your speed relative to me will not be ##u+v##, it will be ##(u+v)/(1+uv)## (measuring time in seconds and distances in seconds so that ##c=1## - otherwise I'd need a ##c^2## in there too).

Try it with that math and you won't get above ##c##.

In a single reference frame, speeds do add like that, don't they? If I'm moving at u (in your reference frame) and I increase my speed by v (in your reference frame) then I'm moving at u + v (in your reference frame).

Also, suppose (in my reference frame) one object is moving at c/2 in one direction and another object is moving at c/2 in the opposite direction. Then (in my reference frame) the relative velocity between the two objects is c/2 + c/2 = c.

In any case, my point was that there's nothing mathematically stopping me adding velocities. That may not be the way the physics works in certain contexts (such as implicitly adding velocities from different reference frames!).

 

[ChrisVer]

In a single reference frame, speeds do add like that, don't they? If I'm moving at u (in your reference frame) and I increase my speed by v (in your reference frame) then I'm moving at u + v (in your reference frame).

Also, suppose (in my reference frame) one object is moving at c/2 in one direction and another object is moving at c/2 in the opposite direction. Then (in my reference frame) the relative velocity between the two objects is c/2 + c/2 = c.

In any case, my point was that there's nothing mathematically stopping me adding velocities. That may not be the way the physics works in certain contexts (such as implicitly adding velocities from different reference frames!).

this is not how the relativistic velocity addition is done though... by your idea, if you have something moving with +c and something moving with -c, then their relative velocity will be c+c=2c?!?
You have to understand that the way you add velocities in the Gallilean way (by simply adding them up) is just a low-velocity approximation of how it really works.

 

[ShayanJ]

Ok, you did not need relativistic mass to give arguments that forbid faster than light motion and communication. You told that it is impossible for energy to move faster than light.

Now I am thinking in layman's terms: how does energy have speed? I think this means the speed of light, and light has energy, and the speed of this energy cannot exceed the speed limit.
I am looking for an explanation whether or not a massive object, not just energy or a photon with a zero rest mass, can travel with a speed exceeding the speed of light if the concept of relativistic mass is abandoned.

1- Imagine two persons at rest w.r.t. each other. One of them shines a laser at the other one. We can be sure that the guy receives it. But if another one passes by them with a speed greater than the speed of light, according to him, the laser pulse never reaches the other guy. But the fact that the laser pulse reaches the other guy shouldn't depend on the frame of reference! So its impossible for an object to move faster than light.

!
This means if it is possible for any object to move faster than light, then, theoretically, one can be on it and see that paradox to happen and so no object can move faster than light.

2- Imagine it is possible to send signals faster than the speed of light. That means I can send signals back in time. Now I have a device that is scheduled to send signals back in time only if it doesn't receive a signal from itself from future. But if it receives a signal, it means sent it in the future which means it didn't receive a signal. That's a paradox. Other paradoxes of this kind can be found too. So its impossible for energy to move faster than light.

Also here, I didn't mean only light. What I proved in my second argument, is that you can't have faster than light communication but that's not just via EM waves. The most common way of communication is via EM waves but its not the only one. You can write a message on an object that can move faster than light and send a mail via that. You can encode it on gravitational waves. You can encode it on sound waves in gases, liquids or solids. You can use anything that is delivering energy or mass(which is a kind of energy BTW). And all of them are disproved by the second argument. Of course there are still faster than light motions possible. Like when you point a laser to the moon and move it. The point on the moon may move faster than light but it involves no delivery of mass or energy. Or Quantum Entanglement which is again unable to deliver information.

I think you don't abandon the concept of mass altogether. I mean that there is still
the concept called the velocity independent mass, or invariant mass, or "proper mass" or rest mass m[itex]_{0}[/itex] which can be nonzero for particles other than photons.
If the concept of relativistic mass is abandoned, can an object or a particle with rest mass m[itex]_{0}[/itex]≠0 if it has unlimited amount of kinetic energy travel with a speed v exceeding c, in other words v>c?

I proved that, assuming SR, there will be paradoxes if you assume faster than light motion is possible and I didn't use relativistic mass. I don't know what else you're telling!

 

[jtbell]

Also, suppose (in my reference frame) one object is moving at c/2 in one direction and another object is moving at c/2 in the opposite direction. Then (in my reference frame) the relative velocity between the two objects is c/2 + c/2 = c.

In relativity theory, when we say "an object's velocity can never be greater than c", we specifically mean "the [single] object's velocity in some inertial reference frame." The difference between two such velocities (taking +/- signs into account), in the same inertial reference frame, can be greater than c, up to 2c in fact.

In your example, people often distinguish between the "separation velocity" (or "closing velocity"), which is the difference between the velocities of the two moving objects in your reference frame, and the "relative velocity" of one object with respect to the other, which is the velocity of the first object in the reference frame in which the second object is at rest. In your example, the "separation velocity" is c, but the "relative velocity" is < c.

 

[PeroK]

this is not how the relativistic velocity addition is done though... by your idea, if you have something moving with +c and something moving with -c, then their relative velocity will be c+c=2c?!?

Yes, absolutely. In my reference frame.

You have to understand that the way you add velocities in the Gallilean way (by simply adding them up) is just a low-velocity approximation of how it really works.

Velocities add normally. What don't add normally are velocities from different reference frames (reference frames moving with respect to one another). Context is everything.

 

[PeroK]

In relativity theory, when we say "an object's velocity can never be greater than c", we specifically mean "the [single] object's velocity in some inertial reference frame." The difference between two such velocities (taking +/- signs into account), in the same inertial reference frame, can be greater than c, up to 2c in fact.

In your example, people often distinguish between the "separation velocity" (or "closing velocity"), which is the difference between the velocities of the two moving objects in your reference frame, and the "relative velocity" of one object with respect to the other, which is the velocity of the first object in the reference frame in which the second object is at rest. In your example, the "separation velocity" is c, but the "relative velocity" is < c.

Semantics to the rescue!

 

[ShayanJ]

Take a look at here.
You can find the formula [itex] \vec a=\frac{1}{m \gamma}(\vec F-\frac{\vec v \cdot \vec F}{c^2}\vec v) [/itex].
We can say that's the right formula instead of [itex] \vec a=\frac{\vec F}{m} [/itex]. That simple! no need for relativistic mass!

 

[jartsa]

If the velocity dependent mass does not prevent a massive object of achieving the speed of light,
what is going to do it?

Fast moving objects resist velocity changes. In other words fast moving objects have a large resistance to velocity changes.

Objects with large rest mass have a large resistance to velocity changes too.

The resistance to velocity changes of a fast moving object depends on the direction of the velocity change.
The resistance to velocity changes of an object at rest does not depend on the direction of the velocity change.

How do objects at rest resist velocity change? 1: They interact with Higgs bosons. 2: They have fast moving parts.

Oh yes, the resistance to velocity changes caused by those fast moving parts does not depend on direction of the velocity change.

Conclusion: Non-directional resistance to velocity changes is always called mass. Directional resistance to velocity changes is not called mass.

And finally the answer to the question: Velocity dependent resistance to velocity changes prevents a massive object of achieving the speed of light.

 

[ShayanJ]

I think OP thinks that in physics, proofs should always be "by construction" i.e. laws of physics should always be formulated starting from other laws. But that's not the case. Sometimes in physics, you set a law by observations. Sometimes by reason. And finding the relation of these laws with others, is the next step.
In this case, observations+reason tell us that the speed of light is the maximum speed limit. And for now, as far as I know, no other law can give us this. We have reasons for it, but no other physical law giving it!
You may say relativistic mass does that for us, but its a little advantage compared to its disadvantages and also its not needed any where else and we can do without it.
Another point is that, relativistic mass only provides us with a reason to prohibit faster than light motion for mass. Not for other kinds of energy.
So I prefer not to accept relativistic mass as part of SR.

 

[ChrisVer]

To be completely correct Shyan it's better to say "nothing can move at speeds greater than c".
However I think that theories don't really need this boundary...eg:
http://dottorato.dsf.unica.it/pdf/tesi/tesi_271-SERRA_MATTEO.pdf
Introduction here, which allows negative mass squared for AdS, without problems of the vacuum, only that it puts some boundary (B-F)

 

[ShayanJ]

To be completely correct Shyan it's better to say "nothing can move at speeds greater than c".
However I think that theories don't really need this boundary...eg:
http://dottorato.dsf.unica.it/pdf/tesi/tesi_271-SERRA_MATTEO.pdf
Introduction here, which allows negative mass squared for AdS, without problems of the vacuum, only that it puts some boundary (B-F)

Yeah, I don't refuse the possibility of theories allowing for faster than light motion. But they should cure the paradoxes and so should differ from SR in those regions. But as long as we have no generalization of SR beyond the speed of light, we should say its impossible to move faster than light. I should say I remember reading a paper which proposed such a generalization in which the speed of light itself was still a singularity in between but motions faster than light were allowed. Unfortunately I don't remember enough to find it now.

 

[CKH]

jarsta: You've convinced me that "relativistic mass" is not like rest mass, due to the lack of symmetry in the former. Is there a modern physical definition of mass then? E.g. the inertia of a body in its rest frame?

This also begs the question: how is force defined?

 

[Jonathan Scott]

jarsta: You've convinced me that "relativistic mass" is not like rest mass, due to the lack of symmetry in the former. Is there a modern physical definition of mass then? E.g. the inertia of a body in its rest frame?

This also begs the question: how is force defined?

This is all mostly a matter of terminology.

Instead of "relativistic mass", scientists now refer to the total "energy", of a particle, which is essentially the same thing expressed in energy units. They then use the word "mass" to mean what was previously called "rest mass". However, one could equally well refer to "rest energy" and this term is used as well.

Force is still the rate of change of momentum. In relativity, momentum is effectively energy times velocity (divided by c^2 if you want it in the usual units). If the particle has rest mass, this is also equal to the mass times the proper velocity. Note that a force with a component in the direction of motion will increase or decrease the kinetic energy as well as changing the velocity.

 

[Dadface]

Is't it so that the relativistic mass equation is still useful in that it gives a quick method of calculating relativistic KE?

 

[ghwellsjr]

Now I began to really wonder, can the concept of relativistic mass really be wrong or out of fashion. I am not an expert on relativity, but how is the ultimate speed limit, the speed of light, now explained if not with velocity dependent mass?
If the velocity dependent mass does not prevent a massive object of achieving the speed of light, what is going to do it?

I don't think you need mass or energy to explain why a massive object cannot achieve the speed of light. I think all you need is to understand how we establish the speed of an object. It is done by determining the distance the object has moved away from (or towards) an observer and applying that distance at a certain time. Distance is now defined as how far light travels in a particular period of time. For convenience, we use units where the speed of light is 1 such as minutes and light minutes or just the arbitrary ticking of a stable clock, ticks for time and light ticks for distance. The ticks are shown as dots. We also believe in the Principle of Relativity, Einstein's first postulate, that two inertial observers measure the speed of each other identically.

We can illustrate this on a spacetime diagram where the two observers are depicted as thick lines as they move away from each other at a constant speed. In the first case, the blue observer sends a radar or light signal to the red observer which reflects back to him. The blue observer sends the signal 1 tick after they separate and receives the echo at the fourth tick. He then takes the time between sending and receiving the signal, in this case, 4-1=3 and divides that by two which is 1.5 light ticks and applies that distance to the average or midpoint of his two measurements, in this case (4+1)/2 = 2.5 ticks. Now he can determine the speed as the distance divided by the time or 1.5/2.5 = 0.6c:

Applying the measurement of distance to the average of the sending and receiving signals is nothing more than Einstein's second postulate.

In the next spacetime diagram, we see how the red observer makes the same measurement of the blue observer as blue did of red, even though the red observer may have a different kind of clock that may tick at a different rate:

Now let's throw in a third black observer who is traveling in the opposite direction away from the blue observer but at the same speed:

It should be no surprise that the blue observer gets the same answer for the black observer as he did for the red observer because the only difference is the direction of travel.

And it should be no surprise that the black observer gets the same answer in measuring blue's speed as red did when he measured blue's speed, even though black's clock is ticking much faster than red's clock:

But where the surprise may come in is when black measures the speed of red as shown here:

Black sends his radar signal out at his clock time of 1 tick and waits until 16 ticks to get the echo. His calculation of speed is (16-1)/(16+1) = 15/17 = 0.882c. (You may have noticed that I left off 2 divisions by 2 as they just cancel each other out.) You may have thought that the answer should be 1.2c but it turns out it is less than c. As a matter of fact, we can add together any number of observers at any speed short of c and the final speed will remain under c.

Now what if an observer tried to measure the speed of an object departing from him at the speed of light (if such a thing were possible). Clearly, he could not use the same method as before since his outgoing radar signal could never catch up to the object as shown in this diagram:

 

[Jonathan Scott]

Is't it so that the relativistic mass equation is still useful in that it gives a quick method of calculating relativistic KE?

The equation is still important, but we now say that it gives the total energy for an item with a given (rest) mass and speed. You can therefore also calculate the kinetic energy by subtracting the rest energy.

Personally, I'm happy with using "relativistic" or "rest" in combination with "mass", and similarly using "total", "rest" and "kinetic" in combination with "energy". However, the important thing to note is that if we say "mass" on its own, the modern convention is that this means what used to be called the "rest mass", and if we say "energy" on its own in a relativity context, this means the total energy.

I'm not entirely happy with the way this modern terminology ignores factors of ##c##, as although this simplifies SR I think this can add confusion to GR later.

In special relativity, energy and mass only differ by factors of ##c##, which can be set to 1, so they are essentially equivalent. However, there is a complication in general relativity when describing events within a coordinate system, in that the coordinate speed of light can vary with direction. This means that coordinate values with dimensions of mass and energy actually vary in different ways with potential (and mass can be different in different directions).

When factors of ##c## are included explicitly and are treated as referring to the variable coordinate-dependent speed of light (which can of course be set to 1 in a local SR frame), various Newtonian equations of motion (relating for example to conservation of momentum, angular momentum and rate of change of momentum in free fall) have natural very similar and easily understood forms in GR.

For example, the rate of change of momentum of something with energy ##E## falling in a weak Newtonian gravitational field ##g## as represented in isotropic coordinates (where the scale factor between local space and coordinate space is the same in all directions) is given in GR by the following equation where all values are expressed as coordinate values including the coordinate speed of light ##c##:
$$\frac{d \mathbf{p}}{dt} = \frac{E}{c^2} \mathbf{g} \left ( 1 + \frac{v^2}{c^2} \right ) $$
This equation holds regardless of the direction of ##v## (radial, tangential or somewhere in between), just like Newtonian gravity, with the only difference being the additional term due to the curvature of space. Even that term only depends on the speed, not the direction. (For tangential motion, it leads to a curved path, and for radial motion it causes a change in momentum due to the change of the coordinate scale factor of space).

This notation is of course very unconventional, as ##c## normally represents the standard speed of light, so any equations using it for the coordinate speed of light either need very clear explanation of this unusual use or need some alternative notation, such as primed variables.

Of course this doesn't change the physics, but to write the above equation in the GR way without using the coordinate speed of light one would have to insert various scale factors from the metric which would obscure the relationship with the Newtonian view.

As I like to remain aware of these relationships between Newtonian and GR quantities, I personally prefer to continue to use "mass" for quantities with dimensions of mass and "energy" for quantities with dimension of energy, with appropriate qualifiers like "rest" when necessary to avoid ambiguity and explicit factors of ##c## when necessary to convert between them.

 

[7777777]

Relativistic mass appears in the formula:

[itex]E^2 = p^2 c^2 + m_{rest}^2 c^4 = m_{rel}^2 c^4 [/itex]
[itex]E = m_{rel}c^2[/itex]


these formulas are taken from link


One way to avoid writing relativistic mass is to write instead:

[itex]E = m_{rel}c^2 = \gamma m_{rest}c^2[/itex]

where
[itex]\gamma = \frac {1} {\sqrt{1 - v^2/c^2}}[/itex]

The problem arises if [itex]v = c[/itex] is inserted into the above equation. The value of
[itex]\gamma[/itex] becomes infinity because of division by zero.
The problem of dividing by zero can be avoided if [itex]v > c[/itex]. It will lead to imaginary
value of [itex]\gamma[/itex], imaginary value of energy, whatever that means. It appears
that speeds exceeding c are possible at least theoretically (mathematically) as also ChrisVer said.
I found useful info on link. Jesse Berezovsky writes about the problem
of how E becomes 0/0 if v=c and m[itex]_{0}[/itex]=0. He suggests looking at the limit of E
as mass goes to zero and v goes to c.
I am suggesting looking at v>c, and avoiding the relativistic mass [itex]m_{rel}[/itex] ,
using instead the invariant mass, the nonzero rest mass [itex]m_{rest}[/itex] = m[itex]_{0}[/itex].