# Relativistic rocket - where is the relativistic mass?

• I

## Main Question or Discussion Point

A simple problem with a constant acceleration, ignoring mass of the fuel.

The velocity of a rocket, which moves withe a constant acceleration g, is equal:
v(t) = gt

but I want to keep the const acceleration inside the rocket, not in the absolute sense.

An acceleration has a dimension: L/T^2 [m/s^2];
and in a moving rocket the meter is contracted gamma times,
and a time is dilated gamma times too,
so finally the absolute acceleration must be: ##g / \gamma^3##

therefore:
##dv=\frac{g}{\gamma^3}dt##

which can be easili integrated:
##\gamma^3dv = gdt\to\, v\gamma = gt\to\, \frac{v}{\sqrt{1-v^2}}=gt##

finally:
##v(t) = \frac{gt}{\sqrt{1+(gt)^2}}##

Is this a correct speed of the relativistic rocket?

Related Special and General Relativity News on Phys.org
PeterDonis
Mentor
2019 Award
The velocity of a rocket, which moves withe a constant acceleration g, is equal:
v(t) = gt
Not in relativity.

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

Is this a correct speed of the relativistic rocket?
I think the expression you end up with for ##v(t)## is consistent with what is in the article I linked to above; but your derivation is hand-waving, whereas that article does it using the correct relativistic starting point.

What a hand-waving?

It's rather obvious, that an absolute acceleration of the rocket should be equal to: g / gamma^3, to keep a constant 'gravity' inside the rocket.

[m/s^2]
the length contraction: +1, and dilation of time: +2 = 3.

But it's strange somewhat for me, because there is nothing about the relativistic mass increase.
Probably the relativistic mass is irrelevant.

I can use a reversed version of the relativistic rocket, using the g = const in the absolute sense.

In general: for an absolute acceleration a, the acceleration measured in the rocket is: a' = a x gamma^3, thus the standard work-energy equation is now:

##dE = Fdx = m a\gamma^3dx##
but: ##a = dv/dt## thus:

##dE=m\gamma^3\frac{dv}{dt}dx=\gamma^3 vdv=m\gamma == mc^2 \gamma##

The result is still correct without any relativistic mass usage!

PeterDonis
Mentor
2019 Award
What a hand-waving?
You started with

The velocity of a rocket, which moves withe a constant acceleration g, is equal:
v(t) = gt
and then used the term "constant acceleration in an absolute sense" to describe this. That's not correct. The only "absolute" (in the sense of invariant) acceleration in relativity is proper acceleration, which is what you called "constant acceleration inside the rocket". There is no such thing as "acceleration in an absolute sense" other than that. So the concept you started from is not valid. It just happened to lead you to the correct answer (because ##v = g t## happens to be the non-relativistic limit of the correct formula).

It's rather obvious, that an absolute acceleration of the rocket should be equal to: g / gamma^3, to keep a constant 'gravity' inside the rocket.
This is another example of the same conceptual issue: g / gamma^3 is not the relativistic version of "absolute acceleration inside the rocket". It's just the coordinate acceleration in a particular inertial frame when the proper acceleration is g.

Probably the relativistic mass is irrelevant.
Yes. Relativistic mass is really an outdated concept; you don't need to use it at all.

I know from the school and other, more advanced literature, the acceleration is the real quantity, probably because it's a second derivative.

And if the relativistic mass is outdatet, then why it's still used in the relativity?

For example: p = gamma mv, what is this?

further:
F-relativistic = dp/dt = ... what is this?
a transversal vs longitudinal force - what is a purpose still to use these fantastic ideas?

Ibix
##p=\gamma mv## is the relativistic momentum. What we don't do is write ##p=mv## where ##m=\gamma m_0## is the relativistic mass. This is because all sorts of different "relativistic masses" are needed in different circumstances.

PeterDonis
Mentor
2019 Award
I know from the school and other, more advanced literature, the acceleration is the real quantity, probably because it's a second derivative.
You may think you "know" that, but it isn't true. The only invariant acceleration in relativity is proper acceleration. Coordinate acceleration, the second derivative of position with respect to time, is not invariant; it can change depending on your choice of coordinates.

If you have any references (textbooks or peer-reviewed papers) that you think support your claim, please post them.

if the relativistic mass is outdatet, then why it's still used in the relativity?
It almost never is in modern sources. If you are seeing it in sources you are using, they are probably older ones.

The sense of an absolute quantity doesn't mean it's invariant automaticaly for any observer.
The absolute means the thing is a real only, but the real things are not invariant, in the relativistic sense at least!

What is invariant in the reality?
Nothing is invariant, maybe a pure numbers only:
for example: there are seven flies in my garden.
The number 'seven' is preserved everywhere.

##p=\gamma mv## is the relativistic momentum. What we don't do is write ##p=mv## where ##m=\gamma m_0## is the relativistic mass.
We use term "mass" for ##m_0## only. That's all. It makes no difference if we write ##p=\gamma m_0 v## or ##p=mv## whith ##m=\gamma m_0##. Both variants are identical. And it doesnt matter how we name m or if we name it at all.

Nugatory
Mentor
The absolute means the thing is a real only, but the real things are not invariant, in the relativistic sense at least!
....
What is invariant in the reality?
"Invariant" has a precise meaning in relativity: A quantity is invariant if its value is the same no matter which frame you use to calculate it.
F-relativistic = dp/dt = ... what is this?
That's the standard definition of force since Newton's day. ##F=ma## is the simplified version that we teach in high school physics when the students are not yet familiar with calculus, so not ready for the more complete and generally useful ##F=\frac{dp}{dt}##.
A transversal vs longitudinal force - what is a purpose still to use these fantastic ideas?[/QUOTE]
None. The advantage of the modern treatment of relativity based on invariants such as proper acceleration is that it eliminates the need to use these clumsy ideas.

PeterDonis
Mentor
2019 Award
The absolute means the thing is a real only, but the real things are not invariant, in the relativistic sense at least!

What is invariant in the reality?
Nothing is invariant, maybe a pure numbers only:
for example: there are seven flies in my garden.
The number 'seven' is preserved everywhere.