# Why matter can't reach the speed of light ?

1. Mar 21, 2013

### big_bounce

Hello all .

We know when matter like electron reach near speed of light its mass increase and Limits to Infinity .

Why we can not reach electron to c ?

Because for changing speed at near speed of light :
1 - We need infinite momentum ?
2 - We need infinite energy ?
3 - We need infinity force to give it acceleration ?
4 - If electron wants to reach c , it must change form to a mass-less particle like photon and conversation of charge violate .
or
All of them maybe correct / None / other idea ?

Thanks

2. Mar 21, 2013

### phinds

what do you think and why?

3. Mar 21, 2013

### Mentz114

I think it is more subtle than this. It is the geometry of Minkowski spacetime that prevents v -> c.

Every observer can 'coordinatetize' other observers. This means we can convert their local measurements into numbers corresponding to our local rulers and clocks. This process is governed by the Lorentz tranformation, and the LT can never convert a velocity to c.

4. Mar 21, 2013

### ghwellsjr

Once you understand that light comes from accelerated charges then you would see that your question is like, "Why can't I pull myself up by my own bootstraps?"

5. Mar 21, 2013

### 1977ub

External agent finds that increasing relativistic mass/energy brings diminishing returns and acceleration won't bring the particle to c relative to them.

6. Mar 21, 2013

### big_bounce

I think 4 is more acceptable .
Even we have infinite "energy , force, momentum" we can not annihilate a electron and release it's rest mass since this is a blatant violation of charge conservation .
Unless we'll find a mass-less particle with charge or a massive particle travels at c .

Thank you .

Last edited: Mar 21, 2013
7. Mar 21, 2013

### Mentz114

It would be a paradox in any case, if some observers could see the body reach c and others could not.

8. Mar 23, 2013

### RealityQuest

Has anyone run an experiment with two particle emitters aimed in opposite directions? Or would the result simply be obvious?

By measuring the time it takes for a particle to travel from the emitter to some target, determining velocity becomes a simple matter of d/t. If another emitter is fired at a similarly placed target in the opposite direction, it's particle's velocities would be calculated the same: d'/t'. If, by design, d' is set equal to d then t'=t assuming the same type of emitter is used.

If the emitters are fired at the same time, how would you explain the velocity of separation between particles moving in opposite directions? The math would be (d+d')/t. If the velocity of the particles in relation to the observer is anything greater than 0.5c, than their velocity of separation would appear to the observer to be >c.

9. Mar 23, 2013

### Staff: Mentor

The "velocity of separation" of two particles, as calculated in the rest frame of the emitters, is not the same sort of thing as the "velocity of one particle with respect to the other", which is more precisely stated as "the velocity of one particle in the rest frame of the other."

The latter is subject to the rule that it cannot be > c; the first is not, i.e. it can be > c, as in the situation you describe.

There is no reference frame in which either particle has a velocity equal to the "velocity of separation" in the emitter's rest frame.

10. Mar 23, 2013

### RealityQuest

Then if two observers, each assigned to accompany opposite-moving particles were to take whatever time they need to later measure d+d' (which are equal and traversed in equal t), they would both agree on the distance along with a third observer assigned to the emitters.

But while the emitter observer would have perceived the two travelers reaching their targets simultaneously, each of the travelers would have perceived themselves as beating the other to their respective target.

Is that correct?

Suppose each traveler clocked their journeys with a stopwatch and the met later in the middle to discuss their situation. How would they reconcile the fact that d=d' and t=t', but both observed the other reaching their target in time >t ?

11. Mar 23, 2013

### Naty1

The simplest explanation I have seen about why you can't get to light speed is that in any local observational frame [ where you are] light always passes you by at c, no matter your speed. Another way to say this is that all inertial observers measure light locally at c. So how could you ever 'catch up'??

This perspective, like that of Mentz's Lorentz based answer, is unfortunately not based on any first principles....like many things, electron mass and charge for example, or the four fundamental forces, or the statistics of quantum mechanics, we have to use observations as we have no first principles. Sure we have lots of math, but why this math when there is a lot of other math that doesn't work in this universe?

12. Mar 23, 2013

### Staff: Mentor

That bit about meeting in the middle changes the problem into something completely different, as the only way that can happen is for at least one of the two to accelerate or otherwise change his speed. The simple time dilation equation doesn't apply in this case (you'll notice that it is written for v being constant); instead we have the well-known "twin paradox" (google it, or look for explanations from the mentors and science advisers here).

In all of the cases where the two observers eventually reunite, it is possible for the two observers to measure different amounts of time on their journeys so that their clocks (and the grayness of their hair) disagree at the reunion. Which one will be more aged and by how much will depends on the exact details of their journeys, but it can be calculated using either observer's reference frame, or that of some unrelated third observer and the answer will come out the same.

Last edited: Mar 23, 2013
13. Mar 23, 2013

### RealityQuest

A better way to frame the experiment that doesn't require hypothetical observers...

Place the two emitters 2d apart aimed directly at each other so that their particles are on a collision course at a central target. If the emitters accelerate the particles to 0.8c, their approaching speed would be 1.6c to an observer on the ground. How would one explain that the relative approach velocity of the particles to each other is <= c ? The measurable distance from which they started was closed in the same time they each closed the distance to the target. Their collision at the target would seem to prove that it was simultaneous for all observers.

14. Mar 23, 2013

### RealityQuest

Further, since the particles are traveling at high velocity, less time would have passed for them than for the observer on the ground, yielding a smaller denominator with an equal numerator. So it would seem, from the particles' perspectives, the velocity at which they are approaching each other should be significantly higher than 1.6c. Same distance in less time.

Obviously I'm wrong, but I don't know where I get off track logically.

15. Mar 23, 2013

### Staff: Mentor

Why can't matter reach c? Because the rules governing how the universe works prevents it from occurring. This has several consequences, such as the aforementioned laws that we know and use. None of them answer 'Why' it can't happen, they only tell us what to expect when we move objects around at high velocities.

The collision would not be simultaneous to all observers.

Length contraction. Each particle travels less distance, from their own perspective, since length contraction affects how they view space as well. From one particles perspective, the distance between the center target and the other particle, as well as the distance between both of them and itself, is less than the center target would measure.

16. Mar 23, 2013

### OmCheeto

It was explained to me, quite a while back, as such:

The original question actually was;

I've reworded my original question to match your question.

The final comment by one of the moderators was;

The way I interpret this now is:

"c" is defined as the "invariable speed" of something that is massless, that fits Einsteins equations. Whether or not such a beast really exists, remains to be measured. See also FTL Neutrino discussions. :tongue2:

(ref)

-----------------------
ok to delete and infract, as I have a bit of the flu going on right now, and need some rest.

17. Mar 23, 2013

### rbj

the logic might be okay but your axioms are wrong.

velocities and lengths do not add linearly along the line of motion. the velocity addition mathematics is different than just $u+v$. i think it's

$$\frac{u+v}{1 + \frac{uv}{c^2}}$$

18. Mar 23, 2013

### Staff: Mentor

That is correct. There are two things worth pointing out here:
1) If u and v are both small compared to the speed of light, this formula yields an answer which is very close to u+v. For example, if two cars traveling at 30 meters/sec relative to the ground pass each in opposite directions, their closing speed won't be 60 meters/sec, it will be 59.9999.... meters per second, out to about 12 decimal places or so. This is far smaller than we can measure, which is why we've never noticed that u+v is not exactly right.
2) If either u or v is equal to c, the result comes out to c. That's why people moving at different speeds relative to each other will both get c for the speed of a light signal relative to themselves.

19. Mar 23, 2013

### Popper

Simply put, it's the properties of spacetime which determines how the reltivistic mass of a particle, it's energy and it's momentum change with speed v. As v -> c, E -> infinity, mrel -> infinity and p -> infinity. However there is nothing that says that a particle can't be created traveling faster than the speed of light other than the problem of being able to use it to cause a paradox. Such a class of particles are called tachyons.

20. Mar 24, 2013

### RealityQuest

By "all" I was referring to the two colliding particles and the target. How are those three not experiencing the collision at precisely the same time? And how is such simultaneity possible given the c speed limit?

Except any measuring device would be contracted as well, so.... wouldn't the distance still measure out the same for all three?

How would the above formula apply to each observer--the one on the ground and the two in the passing cars? Their calculations for u and v would be slightly different. The cars' time would be measurably smaller (with a 13-decimal-point precision stopwatch). How would the shrinkage of d be determined since all the mile markers would have shrunk as well?

I need some help understanding how length contraction really works. I'm picturing the particles in a supercollider spinning around thousands of time a second at a foreword velocity near c. To them the universe is contracting and re-expanding along their ever-changing vector. It would seem clear that the universe isn't ACTUALLY contracting and re-expanding--that that feature of relativity is more a mathematical necessity than the physical reality.

It reminds me of i. We know the sqr(-1) is undefined so we assign it a designation and hope to cancel it out in the rest of the formula. With length contraction, it allows the formulas of relativity to work so we just accept it. I have no doubt the math works out perfectly. Has anyone devised a way to verify it empirically?

Thanks for indulging my curiosity.