I Can we deal with relativistic mass once and for all?

[stevendaryl]

SR is not a matter of 3- or 4-vectors. They are just different formalisms to describe the same physics.

I guess I would agree that using 3-vectors or 4-vectors amounts to different formalisms for expressing the same physics. But that applies to both Newtonian and SR physics. You can describe Newtonian physics using 4-vectors (as I did above) and you can describe Special Relativity using 3-vectors.

 

[stevendaryl]

That's why Newton's definition of momentum is p=m·v and not P=m·V. The equations may look similar but they are actually different and have different physical meanings. Replacing one by the other has concequences. One of them is a different concept of mass.

If you are going to say that SR can be described in a 3-vector or 4-vector formalism, you should be able to describe Newtonian physics in a 3-vector or 4-vector formalism. The physics is not changed. What I think is nice about the 4-vector formalism is that

• conservation of mass is a consequence of conservation of momentum in Newtonian physics
• 4-velocity, 4-momenta and 4-force are true vectors (while 3-velocity, 3-momenta and 3-force are only vectors under changes of pure spatial coordinates, not Galilean or Lorentz transformations)
• mass is a true scalar in the 4-vector formulation, both in SR and Newtonian physics.

 

[DrStupid]

If you are going to say that SR can be described in a 3-vector or 4-vector formalism, you should be able to describe Newtonian physics in a 3-vector or 4-vector formalism.

How does that contribute to the topic?

 

[stevendaryl]

How does that contribute to the topic?

I think it shows the relationship between mass in SR and mass in Newtonian physics most clearly. In the 4-vector approach, it's not necessary to have a velocity-dependent mass in order for $F = m \frac{dV}{ds}$ to hold.

 

[DrStupid]

I think it shows the relationship between mass in SR and mass in Newtonian physics most clearly.

Can you please demonstrate that?

 

[PeroK]

One thing is clear:

Q. Can we deal with relativistic mass once and for all?

A. Apparently not!

 

[stevendaryl]

Can you please demonstrate that?

I thought I did. In the 4-vector formalism:

1. 4-velocity, 4-momentum, 4-force are true vectors under all coordinate changes (while 3-velocity, 3-momentum and 3-force are only vectors under a change of spatial coordinates)
2. Mass is a true scalar under coordinate changes.
3. Momentum is just the product of the mass and the 4-velocity.
4. Force and velocity are related by Newton's law: $F = m \frac{dV}{ds}$

In the 4-vector formalism, all 4 statements are true without change in both Newtonian physics and Special Relativity.

 

[stevendaryl]

I thought I did. In the 4-vector formalism:

1. 4-velocity, 4-momentum, 4-force are true vectors under all coordinate changes (while 3-velocity, 3-momentum and 3-force are only vectors under a change of spatial coordinates)
2. Mass is a true scalar under coordinate changes.
3. Momentum is just the product of the mass and the 4-velocity.
4. Force and velocity are related by Newton's law: $F = m \frac{dV}{ds}$

In the 4-vector formalism, all 4 statements are true without change in both Newtonian physics and Special Relativity.

It's sort of interesting that in the 4-vector formalism of Newtonian physics, there are two different places where mass appears:

1. It's a scale factor to convert a 4-velocity into a 4-momentum.
2. It's also the 4th component of the 4-momentum.

In SR, #1 corresponds to invariant mass, and #2 corresponds to "relativistic mass". So I guess the 4-vector formalism gives support to both claims:

• Invariant mass in SR corresponds to Newtonian mass (as the scalar multiple of 4-velocity to produce 4-momentum)
• Relativistic mass in SR corresponds to Newtonian mass (as the 4th component of the 4-momentum)

 

[DrStupid]

I thought I did.

I still don't see it.

Mass is a true scalar under coordinate changes.

Yes, that's true for mass but where is the relation to mass as used in Newtonian physics?

Momentum is just the product of the mass and the 4-velocity.

No, momentum is defined the product of mass (as used in Newtonian physics) and velocity. What you mean is 4-momentum.

Force and velocity are related by Newton's law: $F = m \frac{dV}{ds}$

That's true if m is mass. But i still do not see the the relationship between mass in SR and mass in Newtonian physics.

 

[stevendaryl]

I still don't see it.

Yes, that's true for mass but where is the relation to mass as used in Newtonian physics?

It's the same: In both Newtonian physics and SR, mass is what you multiply velocity by to get momentum. Mass is what you multiply acceleration by to get force.

No, momentum is defined the product of mass (as used in classical mecatnis) and velocity. What you mean is 4-momentum.

3-momentum is just the first 3 components of 4-momentum.

That's true if m is mass. But i still do not see the the relationship between mass in SR and mass in Newtonian physics.

They are the SAME! Mass is the scalar quantity that you multiply velocity by to get momentum.

 

[DrStupid]

It's the same: In both Newtonian physics and SR, mass is what you multiply velocity by to get momentum. Mass is what you multiply acceleration by to get force.

That's only true for closed systems in classical mechanics. In SR mass is what you multiply 4-velocity by to get 4-momentum and what you multiply with 4-acceleration to get 4-force.

3-momentum is just the first 3 components of 4-momentum.

That's just another way to say that they are different.

 

[stevendaryl]

That's only true for closed systems in classical mechanics. In SR mass is what you multiply 4-velocity by to get 4-momentum and what you multiply with 4-acceleration to get 4-force.

It's true for both Newtonian physics and SR, it's just that nobody bothered with the 4th component of 4-velocity or 4-momentum in Newtonian physics since it's so boring.

That's just another way to say that they are different.

That doesn't make any sense to me. If multiplying mass by 4-velocity produces 4-momentum, and 3-velocity is the spatial components of 4-velocity, then it follows that multiplying 3-velocity by mass produces 3-momentum. Why would you say that it's a different concept of mass?

 

[DrStupid]

It's true for both Newtonian physics and SR, it's just that nobody bothered with the 4th component of 4-velocity or 4-momentum in Newtonian physics since it's so boring.

In SR 4-momentum is

P = m \cdot V

but momentum is

p = \frac{{m \cdot v}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}

and not

p = m \cdot v

If multiplying mass by 4-velocity produces 4-momentum, and 3-velocity is the spatial components of 4-velocity, then it follows that multiplying 3-velocity by mass produces 3-momentum.

Obviously not (see above).

 

[stevendaryl]

In SR 4-momentum is

P = m \cdot V

but momentum is

p = \frac{{m \cdot v}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}

Yes, but the spatial components of the 4-momentum $P$ are equal to the components of the 3-momentum $p$. So I don't understand what point you are making.

Newtonian physics doesn't make the distinction between clock time $s$ and coordinate time $t$ because the two are trivially related: $ds = dt$. But formulating Newtonian physics in terms of $s$ produces a theory where the generalization to SR is a lot clearer. Almost everything is the same in the two theories---the only difference is a different notion of how $s$ is related to coordinate time $t$.

It is a fact of SR that clock time $s$ is not equal to coordinate time $t$, so you have to introduce that change, regardless. But once you're introduced that change, there is no additional change needed to go from $m$ to $\gamma m$. The $\gamma$ can be seen as just the factor $\frac{dt}{ds}$.

 

[PeterDonis]

and 3-velocity is the spatial components of 4-velocity

Which it isn't. The spatial components of the 4-velocity are the 3-velocity times $\gamma$, not the 3-velocity itself.

 

[PeterDonis]

The $\gamma$ can be seen as just the factor $\frac{dt}{ds}$.

Yes, but that doesn't change the definition of "3-velocity". It changes the relationship between 3-velocity and 4-velocity. When you measure the 3-velocity of an object in SR, you still measure $\mathbf{v}$, not $\gamma \mathbf{v}$.

 

[stevendaryl]

Which it isn't. The spatial components of the 4-velocity are the 3-velocity times $\gamma$, not the 3-velocity itself.

I think you missed my "retcon" of Newtonian kinematics in terms of clock time s, rather than coordinate time t. In retrospect, everything could have been formulated in a 4-vector formalism. The reason it wasn't was, first, because nobody thought of it, and second, because the 4-the component of most vectors is pretty boring in Newtonian physics: $V^t = \frac{dt}{ds} = 1$, $P^t = m V^t = m$.

 

[PeterDonis]

I think you missed my "retcon" of Newtonian kinematics in terms of clock time, rather than coordinate time

No, I didn't miss it. I'm just pointing out that your retcon doesn't change the meaning of the term "3-velocity". It formulates Newtonian mechanics in terms of "4-velocity", but that just means, as I said, that the relationship between 4-velocity and 3-velocity is different in Newtonian mechanics and SR. That shouldn't be surprising, since Newtonian mechanics and SR are different theories that make different predictions, so there will be no way to "retcon" one so that everything "matches" the other; there will always be a difference somewhere, it's just a matter of preference for where you want to put it.

 

[DrStupid]

In retrospect, everything could have been formulated in a 4-vector formalism.

It could but it doesn't. You can't simply replace Newton's p=m·v by P=m·V and expect that it is still the same definition.

 

[stevendaryl]

No, I didn't miss it. I'm just pointing out that your retcon doesn't change the meaning of the term "3-velocity". It formulates Newtonian mechanics in terms of "4-velocity", but that just means, as I said, that the relationship between 4-velocity and 3-velocity is different in Newtonian mechanics and SR.

But I would trace that difference to the fact that the relationship between clock time (or proper time) and coordinate time is more complicated in SR.

 

[stevendaryl]

It could but it doesn't.

I thought you said that it doesn't change the physics whether you use 4-vectors or 3-vectors. So Newton may not have used 4-vectors, but it's the same theory, whether you do or not.

 

[PeterDonis]

I would trace that difference to the fact that the relationship between clock time (or proper time) and coordinate time is more complicated in SR.

Sure, but it's still a difference, and other people might have different preferences for how to express it. Nobody appears to be disagreeing anywhere in this thread on the actual physics--the actual predictions for experimental results. It's all just about how to describe things in ordinary language, and such arguments will never end.

 

[PeterDonis]

You can't simply replace Newton's p=m·v by P=m·V and expect that it is still the same definition.

No, but that's not what "4-vector formalism" means. It just means you can derive all of the same experimental predictions: the two formulations of the same theory are mathematically equivalent. It doesn't mean every symbol has to have the same definition. As I've already noted, that's impossible, since the two theories (Newtonian mechanics and SR) are different theories and make different experimental predictions; it's just a matter of personal preference how you want to express the differences in words or symbols.

 

[PeterDonis]

Nobody appears to be disagreeing anywhere in this thread on the actual physics--the actual predictions for experimental results. It's all just about how to describe things in ordinary language, and such arguments will never end.

it's just a matter of personal preference how you want to express the differences in words or symbols

And with that, this discussion has run its course and this thread is closed.