# Exactly Why C is impossible for Massive Objects?

1. Jul 10, 2013

### AndromedaRXJ

Everytime I look this up, the answer always is "because mass increases as you approach C, therefore infinite energy is required.", but from what I understand, Relativistic Mass is a misconception and there's no actual increase in mass.

So if infinite energy is required, then why? And why else is it impossible to reach C with mass?

2. Jul 10, 2013

### pervect

Staff Emeritus
You don't need dynamics (i.e masses, forces, etc) to determine that it's impossible to reach 'c'. All you need is kinematics. In case this term is unfamiliar, I'll copy the definition from wiki.

Suppose you have an series of numbered objects 1,2,3,4,5,6,7....etc, where the number of the last object is finite, but as large as you like.

Now suppose the speed of each object is 1 meter/second in the frame of the object before it.

I.e. the speed of object 2 is 1 m/s in the frame of object 1, the speed of object 3 is 1 m/2 in the frame of object 2, the speed of object 100 is 1 m/s in the frame of object 99.

When you use the relativistic velocity addition laws, you will find that the speed of object n relative to object 1 does not grow without limit, but instead approaches 'c' , as n increases without bound.

The relativistic velocity laws say that v1+v2 = (v1+v2)/(1+v2*v2/c^2)

I don't currently have a closed form expression for the series summation of this, but it's easy to show that if v1 and v2 are less than c, the sum v1+v2 is less than c.

It might be interesting and instructive to find (or attempt to find, at least) a closed form solution for the series.

You can look at dynamics and at energy if you like in that case you just use the relativistic expressions for energy, E = gamma*m, m being the invariant mass.

3. Jul 10, 2013

### Bill_K

It's the formula for tanh(a + b), right? tanh(a + b) = (tanh a + tanh b)/(1 + tanh a tanh b).

So let v be the relative velocity between object n and object n+1, and let a = tanh-1(v). Also let bn = tanh-1(vn).

Then the recursion relation becomes bn+1 = a + bn, so bn = na, and vn = tanh(n tanh-1(v)).

4. Jul 10, 2013

### Naty1

Well, some would say that, maybe, but it's a subject of controversy in these forums....

For a discussion of the pros and cons see here:

On the concept of Mass

Wikipedia points out Okun, and Taylor and Wheeler don't like the concept:

http://en.wikipedia.org/wiki/Relativistic_mass#Controversy

You can see why people say this from the first formula in the 'Controversy' section above....which happens to be for momentum, but the same idea holds... when you have the Lorentz
factor 1/ sq rt[1 -v2/c2] as v approaches c, the expression diverges.

The easiest way to think about this: If the speed of light is 'c' in all inertial frames, how could you ever catch up?? Even at .9999c a mass observer would see light whizzing by locally at good old 'c'.
Of course the fact that every inertial frame observer sees light at 'c' has no explanation other than it is observed. There is no set of first principles that demands that and excludes all other possibilities. It's the way stuff works in this universe.

Edit: I see Bill_k posted while I was composing...the hyperbolic functions he mention derive from the Lorentz relationship I posted.......two sides of the same coin....

You can fuss with the math from here if desired:

http://en.wikipedia.org/wiki/Lorentz_factor

See the first formula and then its relationship to hyperbolic functions under RAPIDITY.

Last edited: Jul 10, 2013
5. Jul 10, 2013

### DrGreg

In fact, you don't need to know any mathematical formulas at all, you just need to know Einstein's 2nd postulate. In pervect's example in post #2, object 1 measures the speed of light relative to itself and gets the answer c. Object 2 measures the speed of light relative to itself and, by the 2nd postulate, must also get the same answer c. And so must object 3. And so must object 4. And so on, for all the objects. None of the objects are "getting any closer" to the speed of light, according to their own measurements.

6. Jul 10, 2013

### DrGreg

Just to be pedantic, and for the benefit of readers not familiar with the convention of using units where $c = 1$, that would be$$v_n = c \, \tanh \left( n \, \tanh^{-1}\frac{v}{c}\right)$$

$v_n \rightarrow c$ as $n \rightarrow \infty$.

7. Jul 10, 2013

### nitsuj

How come it's not the geometry first and all else follows?

The "rules" of spacetime are the postulates, "drilling down" to the concept of c, specifically how length and time are defined limits the speed; to what ever magnitude. Further with that, it's causality.

So speed is limited, "invariantly", because one thing leads to another; and the whole universe has to "see" this physically "play out" the same. We measure these gaps across spacetime between "happenings" using time and length. now reading the underlined part above helps

It seems strange because we move so slow compared to things around us, but speed is really something physically fundamental to the causal system as a whole (Universe), as fundamental as the "every action there is an equal but opposite reaction" we are more familiar with. When we examine this very fundamental perspective, it's separated it into two measures. That makes sense in a Newtonian world, since we could all agree on the measured results. In SR the measurement across space time is never just time or length but a perfect balance of the two if you are inertial. consider measuring time/length where there is gravity/tidal gravity. With gravity potential you are not inertial and comparative measures of length / time from different potentials would not be exact.

I like that question "And why else is it impossible to reach C with mass? ".

lol because gravity can only go c, you can break the sound barrier but not the causality barrier. In fact if you ever leave your personal gravity bubble you'd be in trouble, so don't even think of it!

Wow what a neat perspective, just by "being/existing " matter is causally connected at c via gravity, hmmm as of my birth my gravity field would be nearly 66 light years across today...now that's quite a presence! :rofl: I wonder what that would be cubed for volume. And what physical property is attributed to that value. Oh right it's just fictitious lol gravity, pfft Who's the mathematician that ruined GR? if gravity is fictious so is speed.

Last edited: Jul 10, 2013
8. Jul 10, 2013

### pervect

Staff Emeritus
Last I looked most if not all of the Science Advisors and Mentors recommended against relativistic mass. Not much of a "controversy" in my opinion.

It's not IMPOSSIBLE to use relativistic mass correctly, but rare. "Energy" is a direct synonym for it, and when "relativistic mass" is called "energy" it seems to be misused much less frequently.

I'd preach a little more but I feel like it won't actually help stem the tide of elementary misunderstandings that seem to be highly correlated with use of the term, so I'll leave it at that.

9. Jul 10, 2013

### pervect

Staff Emeritus
Ah yes, if one remembers that rapidities add, the answer becomes clear.

10. Jul 10, 2013

Staff Emeritus
If you'll do a search, you'll also see that the "controversy" has one long-term member who supports it. And a bunch of crackpots, of course. Injecting this does absolutely nothing to help the OP understand.

AndromedaRXJ, Science Advisors have a long track record of correct and helpful posts. You probably want to keep that in mind.

The easiest way I think there is to show this is to calculate the velocity of an object with mass m and energy E:

$$v = c \frac{\sqrt{E^2-m^2c^4}}{E}$$

Note that the numerator is always less than E, so the velocity is always less than c.

11. Jul 10, 2013

### Raze

I know I really shouldn't ask this, but what if m was imaginary? Then wouldn't you always have a velocity greater than c? (sorry for the off topic but I read something on wikipedia about tachyons having this property some time ago)

12. Jul 10, 2013

### PAllen

That's the definition of a tachyon: imaginary mass, postive energy, speed > c. They were hypothesized, searched for and not found. A majority of physicists (but by no means a near universal consensus) believes they are unlikely to exist. They produce causality problems given certain reasonable assumptions.

13. Jul 10, 2013

### anuraj.b

velocity and quantum particles

In quantum physics the particles move with velocity c(=velocity of light) , each particle energy is quantised. E=nhc/λ. this particles called quantum particles.
Where n is an integer.
consider massive bodies there is infinite number of quantum particles inside the nucleus and they are bonded with each other. so massive body doesn't move with the velocity c
the reason is quantum particle are bonded due to nuclear forces,columb force etc. if these particles are separately each particle move with the velocity c.

14. Jul 11, 2013

### nikkkom

Formulas of special relativity's Lorenz transform show that however you increase the velocity, v+dv will never reach c.

Mathematical models don't answer the question "why?".

If you are asking "why do we use Lorenz transform?", the answer is "because it matches observations".

15. Jul 11, 2013

### harrylin

No matter how you name it, the increased kinetic energy brings with it an increase of resistance against acceleration - even to acceleration perpendicular to its motion. And for acceleration in the direction of motion additional work must be done for the increase in kinetic energy, which is disproportional to the increase in speed as shown by the equation in post #10 by Vanadium.

Last edited: Jul 11, 2013
16. Jul 11, 2013

Staff Emeritus
Let's not hijack this thread. The question was about massive particles, not hypothetical particles with imaginary masses.

That's not true.

17. Jul 11, 2013

### Staff: Mentor

Can you provide a source for that claim?

I doubt that you can draw that conclusion like that.
The proton mass, for example, is a fixed value as well, and if you fix its wavelength (like you did for light), you get a similar expression for the energy.

That is wrong, electrons and quarks have rest masses.

18. Jul 11, 2013

### AndromedaRXJ

Thank you everyone, for responding to my question.

So let me see if I got this down. Mass particles can't reach c simply because c is constant in all inertial frames. And, through experimental observation, we find that length contracts, time dilates and simultaneity is relative, which preserve this fact about c. And as a particle approaches c, these relativistic effects are extreme just enough to prevent a particle from reaching c.

Why the universe is this way is unknown, but we know that it simply just is(through experiments). Is that right?

Okay, so help me to understand this resistance to accelerate.

Is this more to do with what's going on with the space-time around the mass approaching c, rather than the internal structure of the mass? Like length contracts just enough to where each change in speed requires proportionally more energy?

That's what I got out of the discussion that Naty1 posted a link to.

Wait, really? So our change in velocity becomes harder?

So if we travel in a circle at constant speed of 0.99c, the rate of energy expense is needed to steadily go up to stay in the circle? Or am I misunderstanding?

Last edited: Jul 11, 2013
19. Jul 11, 2013

### Staff: Mentor

Right

Length is not a property of space. It is a measurement you do. In special relativity, you do not change spacetime at all - you just use it to make your measurements in.

The required force grows. You do not need energy to keep a particle in a circle.*

Needs more force (and gives more energy), too.

*unless you consider synchrotron radiation, but that is beyond the scope of this thread.

20. Jul 11, 2013

### nitsuj

I'd beg to differ, it's kinda clear why it would be this way, especially if some time is spent with Pythagoras theorem, understanding 4D & causality.

21. Jul 11, 2013

### AndromedaRXJ

So going at a constant speed of 0.99c in a circle, the force needed to maintain the circular path constantly GROWS?

Like, for instance, a tether keeping us in the circle would have to constantly grow in tensile strength, AS IF the mass of the object is increasing, classically speaking?

22. Jul 11, 2013

### PAllen

Yes. A better way to look at it, though, is momentum increases with minimal increase in speed. Force is really dp/dt. So momentum keeps increasing as force is applied, but the relation between momentum and speed in SR is very different from Newtonian physics.

Last edited: Jul 11, 2013
23. Jul 11, 2013

### pervect

Staff Emeritus
I was thinking it might be helpful to simplify this a bit more. Setting v=1 m/s, as in my original example, we can write

$v_n \approx c \, (\tanh 3.335641 \,10^{-9} (n-1))$

(to avoid any concerns about the approximation, we could also write)
$c \, \tanh (3.33564\,10^{-9} (n-1)) < v_n < c \, (\tanh 3.335641 \,10^{-9} (n-1))$

where c = 299792458 m/s

and I"ve chosen my indexing such that the velocity of spaceship 1 is zero and the velocity of spaceship 2 is 1, hence the n-1.

This makes it really clear, I hope, that in the limit as n->$\infty$, v_n goes to c * tanh($\infty$) = c and that v_n is always lower than c for any finite n.

We didn't need any consideration of forces , masses, or any other dynamical elements to compute this - all we needed to know was the relativistic velocity addition formula.

24. Jul 12, 2013

### Samshorn

That's self-contradictory. The very definition of a "frame" (by which you must mean a standard inertial frame, since otherwise the "speeds" defined in terms of it need not satisfy that composition formula) is based on the dynamics and inertia of physical entities. The relativistic velocity composition formula is a consequence of how inertial coordinate systems are related to each other, which in turn is a consequence of the fact that all forms of energy - including kinetic energy - have inertia (something which Newton didn't realize).

But the physical meaning of the relativistic velocity composition formula (including the identification of the symbol "v" with a physical "speed") is based entirely on the dynamical inertia of physical objects and processes, which is the basis for the definitions of frames and speeds. The answer to the OP's question is that all forms of energy, including kinetic energy, have inertia. It isn't possible to give a meaningful account of special relativity without reference to inertia.

25. Jul 12, 2013

### harrylin

Space-time is hardly affected - according to SR, not at all (it's a GR correction). Instead, the energy is carried along with the object, and at impact this energy is given off as heat (this was shown in a famous demonstration experiment by Bartocci).
Yes, in a particle accelerator* more force is needed to laterally accelerate a particle when it moves at high speed: its momentum p = γ m0 v. Thus it is sometimes said that the particle becomes more "massive" at higher speed, as it has increased inertia: it is harder to deflect.

Neglecting radiation, no energy is needed ("no work is done") to keep a particle in circular motion; it is just harder to pull it inwards than one would expect from Newton's laws.