# Effects of relativistic mass on astronaut

1. Jan 1, 2015

### SpiderET

Lets assume we have a starship which is flying from Earth to star XY which is in distance for example 100 lightyears. The computer of that ship is programmed that way, that it maintains acceleration 1 g. After some time the speed of ship reaches some significant part of speed of light and to maintain the acceleration, the engins must burn more and more fuel to maintain 1 g. But there is also the effect of time dilation, contraction and increase of mass.
Lets say the ship reaches 87%of speed of light, which would mean that there are some significant relativistic effects for example the relativistic mass would double compared to original rest mass of the ship.
Now Im getting to the question: Would the astronaut inside the ship feel that he weights 160 kilograms when his original weight was 80 kilograms?
What I know, the accepted consensus is that he would not feel the increase of weight, because he is in his own reference frame inside the ship. But is this view hardwired in quotations of SR and GR or is this just a philosophical point of view which was never confirmed by any experiment?

2. Jan 1, 2015

### Bandersnatch

You can perform the experiment without leaving your chair by changing the reference frame in which you describe your own motion to any one moving at 87%c relative to you (or even 99.99999999%c for that matter). Can you feel yourself being crushed by your suddenly-increased weight?

3. Jan 1, 2015

### SpiderET

You are talking about something completely different and to avoid such misundestanding I have described the situation of the astronaut in detail.

4. Jan 1, 2015

### ShayanJ

OK, to account for the difference: Consider a transformation where as time passes, you go to a frame moving a bit faster than the last one w.r.t. you. Do you feel something now?

5. Jan 1, 2015

### Staff: Mentor

Not at all: the point is that choices of reference frames are largely arbitrary. But what matters to you is that you are stationary with respect to your ship and scale.

And yes, this is built into physics and long predates Einstein, though it is one of the two postulates of SR. It is called the principle of relativity.

6. Jan 1, 2015

### SpiderET

I dont feel anything special, because Im not near to speed of light compared to any reference frame.

7. Jan 1, 2015

### SpiderET

OK, but lets say Im saying that the astronaut feels like he is having 160 kilograms and Im using the SR quotation to calculate his weight (relativistic mass increase). Is there any experimental evidence against it or is this just based on philosophical explanation based on principle of relativity? And please keep in mind that standard experiments like Michaelson Morley null result would not disprove possibility of this astronaut having 160 kilograms weight.

8. Jan 1, 2015

### Bandersnatch

That's most emphatically not true. There's an infinite number of reference frames in which you're travelling at arbitrary speeds. Particlular examples include the frames of muons created by solar wind in the upper atmosphere of Earth, or of the particles accelerated to near c in any of the many particle accelerators.

The built-in bit in the equations is the gamma (Lorentz) factor $$\gamma=\frac{1}{\sqrt{1-\frac{V^2}{c^2}}}$$ present in all relativistic equations. In a stationary frame it's alwas 1 (as V=0). Since the astronaut is measuring his own weight in a frame in which he's stationary there can be no relativistic effects he could ever notice with regards to himself.

9. Jan 1, 2015

### Staff: Mentor

Sure you are. You are traveling near to the speed of light compared to a reference frame that is traveling near the speed of light relative to you.

10. Jan 1, 2015

### Staff: Mentor

Yes, you are. There are an infinite number of perfectly legitimate frames where your current speed is within one part per billion of the speed of light.

11. Jan 1, 2015

### Staff: Mentor

That isn't true. You are near the speed of light in an infinite number of reference frames.

12. Jan 1, 2015

### Staff: Mentor

Yes, tons. SR and GR are exquisitely well tested.

13. Jan 1, 2015

### SpiderET

That means that quotations of SR and GR are giving good predictions compared to experiments and astronomical reality. But the question was different. I could use SR quotation to calculate that this increase of relativistic mass was leading to 160 kilograms of weight of astronaut. This would be not in line with relativity principle, but is there any specific experiment which would prove this calculation wrong?

14. Jan 1, 2015

### ShayanJ

You're not paying attention. Let me give you an example:
Imagine spaceship A and B moving w.r.t. each other with velocity v(which is comparable to that of light). According to SR, all inertial frames are equivalent and there is no preferred inertial frame and so each of A and B has equal right to consider itself as stationary and the other one as moving. So A says that he's at rest and B is moving and so A sees relativistic effects happening for B. But B can also consider consider himself as being at rest and A as moving and so B sees relativistic effects happening for A. For A, everything is as before for things in its own reference frame and the same for B.

15. Jan 1, 2015

### Staff: Mentor

Yes, decay rates of highly relativistic muons and electrons is an example. A more massive muon or electron would be less stable than an actual muon or electron. In the case of electrons, it would lead to spontaneous decay of relativistic electrons, and in the case of muons it would lead to faster decay than the experimentally verified decay rates.

16. Jan 1, 2015

### SpiderET

Thank you very much, this is exactly what I needed. I will check it.

17. Jan 1, 2015

### Staff: Mentor

ALL predictions of Relativity, including relativistic mass change(apologies for using the out of date term), are exquisitely well tested.

18. Jan 1, 2015

### Staff: Mentor

19. Jan 1, 2015

### Orodruin

Staff Emeritus
This seems to me a typical misunderstanding based on the relativistic mass concept.

Last edited by a moderator: May 7, 2017
20. Jan 1, 2015

### pervect

Staff Emeritus
With the right analysis, you can get rid of a lot of these factors. Interestingly enough, though, people seem to resist doing this, even when they are told how it can be accomplished.

How do you do this? Well, the first key observation is this. The value of the coordinate acceleration of the ship depends on the observer, or the observers frame of reference. This is probably the most important step to working the problem. The notion of "proper acceleration" would also be useful to understand the problem, but it appears to be unfamiliar to many readers.

Now I'll start some analysis using a 'question and answer' format.

1) We noted that the acceleration of the ship depends on the observer. For w hat observer, in what frame of reference, is the coordinate acceleration of the ship equal to specified value of 1g.

answer: this frame of reference is the reference frame of the ship. Note that this is the only frame where the proper acceleration is equal to the more familiar coordinate acceleration.

2) What is the "relativistic mass" of the ship in this frame.

answer: it is equal to the rest mass of the ship.

3) Doesn't this make the relativistic mass pretty much irrelevant to solving the problem

4) Why do people go on and on about the relativistic mass, then, if it's really not that useful in solving the problem

answer beats me :-). If you find out, let me know. I do suspect that the usual reason for this focus on relativistic mass is related to a belief that saying "it requires infinite energy to accelerate close to the speed of light" (true) is the best way to describe why objects can't reach the speed of light (probably false).

5) Wait - this is getting a bit off topic.

6) OK, lets get back on track. We've determined that we find the acceleration of the ship in its own frame, where we don't need to worry about relativistic mass, and we only need its non-relativistic mass and it's thrust. How do we go about finding the acceleration (and velocity, and position) of the ship versus time in a wholly inertial reference frame - i.e the initial inertial reference frame the ship was in before it started accelerating?

answer: This starts to get a bit mathematical, but the math required doesn't need any dynamics at all, only kinematics. The kinematics required are the Lorentz transform and perhaps the velocity addition formula. See http://en.wikipedia.org/wiki/Velocity-addition_formula

In some amount of ship time, (also called proper time), $\tau$, we know that the ships velocity increases by a multipled by $\tau$ in the ship frame. The velocity equation in general says that we add two velocities v1 and v2 using the relativistic formula

v_tot = (v1 + v2) / (1 + v1 v2 / c^2)

In this case we have v1 = $v_{ship}$ and v2 = $a \tau$ so we get

$$\frac {v_{ship} + a \, \tau}{1 + v_{ship} a \tau / c^2}$$

where$v_{ship}$ is the current velocity of the ship
a is the proper acceleration of the ship (measured in the ship frame)
$\tau$ is the proper time of the ship (measured in the ship frame)
c is the speed of light

If you then consider $v_{ship}$ to be a function of $\tau$, you can write a differential equation and solve it as a function of $\tau$. You would still need to do some work to express $\tau$ in terms of what you probably want, which would be "t", the amount of coordinate time in the initial inertial reference frame of the ship before it started accelerating.

You can use the well known time dilation equation $d\tau = {dt}{\sqrt{1-v^2/c^2}}$ to relate t and $\tau$

This gets detailed enough that I'll give a link to the answer instead of trying to wade through it - see "The Relativistic Rocket" http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

The end result for the distance, velocity, and acceleration of the ship in an inertial where the ships initial velocity at t=0 is 0 is:

distance = ( (c2/a) (sqrt[1 + (at/c)2] - 1)
velocity = at / sqrt[1 + (at/c)2]

The reference doesn't give acceleration, we can derive it easily enough by differentiating velocity with respect to t
acceleration = a / [1 + (at/c)^2)] ^ (3/2)

Note that while the proper acceleration a of the ship in its own frame remains constant, the coordinate acceleration drops off as time increases, becoming lower and lower.

Last edited by a moderator: May 7, 2017